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1800	https://www.quora.com/What-is-the-speed-of-light-in-meters-per-second	What is the speed of light in meters per second? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Physics EM Radiation Measurement Units Velocity of Light Science Light (science) The Electromagnetic Spect... Science Physics Speed of Light 5 What is the speed of light in meters per second? All related (44) Sort Recommended Amit Pant Studied Computer Science&Physics (Graduated 2023) · Author has 77 answers and 4M answer views ·4y The speed of light is calculated to be 299,792,458m/s. Often equated to 310^8 m/s as it makes our calculations very easy. But, the one-way speed of light(from source to object only) has never been practically measured. All those are theoretical calculations. Whereas the two-way speed ( from source to object and then back to the source) of light has been practically measured and is then divided by two to calculate the one-way speed of light. The problem with practically calculating the one-way speed of light is that we cannot synchronize clocks accurately. Continue Reading The speed of light is calculated to be 299,792,458m/s. Often equated to 310^8 m/s as it makes our calculations very easy. But, the one-way speed of light(from source to object only) has never been practically measured. All those are theoretical calculations. Whereas the two-way speed ( from source to object and then back to the source) of light has been practically measured and is then divided by two to calculate the one-way speed of light. The problem with practically calculating the one-way speed of light is that we cannot synchronize clocks accurately. Upvote · 9 6 9 6 Promoted by The Hartford The Hartford We help protect over 1 million small businesses ·Updated Sep 19 What is small business insurance? Small business insurance is a comprehensive type of coverage designed to help protect small businesses from various risks and liabilities. It encompasses a range of policies based on the different aspects of a business’s operations, allowing owners to focus on growth and success. The primary purpose of small business insurance is to help safeguard a business’s financial health. It acts as a safety net, helping to mitigate financial losses that could arise from the unexpected, such as property damage, lawsuits, or employee injuries. For small business owners, it’s important for recovering quickl Continue Reading Small business insurance is a comprehensive type of coverage designed to help protect small businesses from various risks and liabilities. It encompasses a range of policies based on the different aspects of a business’s operations, allowing owners to focus on growth and success. The primary purpose of small business insurance is to help safeguard a business’s financial health. It acts as a safety net, helping to mitigate financial losses that could arise from the unexpected, such as property damage, lawsuits, or employee injuries. For small business owners, it’s important for recovering quickly and maintaining operations. Choosing the right insurance for your small business involves assessing your unique needs and consulting with an advisor to pick from comprehensive policy options. With over 200 years of experience and more than 1 million small business owners served, The Hartford is dedicated to providing personalized solutions that help you focus on growth and success. Learn about our coverage options! Upvote · 999 555 9 1 9 3 Viktor T. Toth IT pro, part-time physicist -- patreon.com/vttoth · Upvoted by Kim Aaron , Has PhD in fluid dynamics from Caltech and Kasper Emil Feld , PhD Physics & Nanotechnology, University of Copenhagen (2012) · Author has 10.1K answers and 172.7M answer views ·10y Related How can the meter be defined as the distance light travels in 1/299 792 458 of a second, if the speed of light itself is defined as meter per second? Think of it this way: You are a space alien with the ability to count and in possession of exquisite instruments, but with no knowledge of terrestrial units of measurements. We meet and find a way to communicate. So I tell you to take a look at the chemical element which has 55 electrons and protons inside its neutral atoms. Good. Now I tell you to observe these atoms in their ground state and a hyperfine transition that you can detect with your instruments. Cool. Now I say count 9,192,631,770 of these transitions. I call this my unit of time. I have a name for it: "second". So now we agreed on Continue Reading Think of it this way: You are a space alien with the ability to count and in possession of exquisite instruments, but with no knowledge of terrestrial units of measurements. We meet and find a way to communicate. So I tell you to take a look at the chemical element which has 55 electrons and protons inside its neutral atoms. Good. Now I tell you to observe these atoms in their ground state and a hyperfine transition that you can detect with your instruments. Cool. Now I say count 9,192,631,770 of these transitions. I call this my unit of time. I have a name for it: "second". So now we agreed on a unit of time. Next I tell you to take a look at a ray of light in space. Measure the distance that it covers during those 9,192,631,770 transitions. Divide that distance into 299,792,458 equal segments. One of those segments is what I call my unit of distance. My name for this unit distance is: "meter". Note that I am able to communicate these measurements to you without any a priori definition of a "speed of light". Once you have acquired my definitions for the "second" and the "meter", it is kind of self-evident that the velocity of that ray of light will be 299,792,458 meters every second, but you did not need this fact in order to learn from me how to measure seconds and how to measure meters. That is because in the definition that I provided to you, I was referencing an actual beam of light (that you could measure in an experiment), not some abstract description of its velocity in a textbook. Upvote · 99 60 Tom Hartley Thrown out of studies in Physics, The University of York · Author has 503 answers and 807.7K answer views ·Updated 5y The speed of light is actually DEFINED to be 299,792,458m/s. It is so close to 3 3 x 10 8 m s−1 10 8 m s−1 that that is the figure most people use. Originally the meter was defined in terms of the circumference of the Earth (in fact, one ten-millionth of the distance from the Equator to the North Pole) over two hundred years ago by the French. This was hard to measure exactly so a standard meter was made and kept in Paris. Using this standard unit the speed of light was very close to 299,782,458 m/s but was a little bit over, and the size of that little bit over depended on the error inherent in knowing the s Continue Reading The speed of light is actually DEFINED to be 299,792,458m/s. It is so close to 3 3 x 10 8 m s−1 10 8 m s−1 that that is the figure most people use. Originally the meter was defined in terms of the circumference of the Earth (in fact, one ten-millionth of the distance from the Equator to the North Pole) over two hundred years ago by the French. This was hard to measure exactly so a standard meter was made and kept in Paris. Using this standard unit the speed of light was very close to 299,782,458 m/s but was a little bit over, and the size of that little bit over depended on the error inherent in knowing the size of this standard meter. Later the speed of light was found by Einstein to be a fixed constant everywhere, so it seemed to make more sense to define the meter in terms of the speed of light than the other way round. The meter is actually defined as one 299,792,458′th of the distance light travels in one second in a perfect vacuum. This is what then makes the speed of light such a perfectly known number; it is defined to be that number. Upvote · 9 1 9 1 9 2 Related questions More answers below What is the speed of light in meters/seconds? What is the speed of light in 1 second? What is the speed of light in inches per second? Why is a metre defined in terms of the speed of light? What is the speed of light in miles per second and meters per second? Alan Feldman Ph.D. in Physics, University of Maryland, College Park (Graduated 1991) · Author has 17.3K answers and 4.9M answer views ·1y Related What is the speed of light measured in? What is the speed of light measured in? The numerical value of the speed of light is defined to be 299792458 when expressed in meters per second. By measuring the speed of light, you are actually defining what a meter is, and that is where the error shows up. Regardless, you are still measuring the speed, though indirectly. If we found the meter to be twice as long, light would be going twice as fast. But it’s not good to overthink this. Ignoring that advice (!) : c = 299792458 meters/second OK, the numerical value is fixed, but what is a meter? By determining what a meter is, you are in effect m Continue Reading What is the speed of light measured in? The numerical value of the speed of light is defined to be 299792458 when expressed in meters per second. By measuring the speed of light, you are actually defining what a meter is, and that is where the error shows up. Regardless, you are still measuring the speed, though indirectly. If we found the meter to be twice as long, light would be going twice as fast. But it’s not good to overthink this. Ignoring that advice (!) : c = 299792458 meters/second OK, the numerical value is fixed, but what is a meter? By determining what a meter is, you are in effect measuring the speed. It is only the number or numerical value that is fixed. And that number will vary with the system of units you use. The speed also includes the meter and the second. So by determining what a meter is, you are still measuring its speed. Say you have a room that was, say, 20% bigger than another room. If the light gets from one end to the other of one room in 100 ns, then you get a certain length for the meter. If the light would instead get from one end of the other room to the other in 100 ns, the speed of light would be 20% bigger. Keep in mind that only the numerical value is fixed, assuming a given system of units. The speed still includes the meter and the second. So if you use this to determine what the meter is, you’re still measuring the speed because the speed includes the meter. Without the meter, all you have is a number. Now, you can measure the speed in any units you like, as long as it’s dimensionally a distance over a duration, like meters/second, etc. You could do meters/second, feet/minute, whatever suits your heart’s desire. Upvote · 9 4 Assistant Bot · 1y The speed of light in a vacuum is approximately 299,792,458 299,792,458 meters per second. This is often rounded to 3.00×10 8 3.00×10 8 meters per second for simplicity in calculations. Upvote · 9 1 Related questions More answers below What is the approximate speed of light in miles per second? Since speed of light is defined as 299,792,458 meters per second (about 186,000 miles per second), what is the speed of dark? What are some alternative units for measuring the speed of light besides meters per second or kilometers per hour? What is the speed of light in feet per second? What is the value with equation if the speed is within 100 meters per second of the speed of light? Jameson Garnett Bachelor's in Business&Applied Mathematics, University of Kansas (Graduated 2011) · Author has 430 answers and 143K answer views ·9mo The speed of light. It never changes. It takes a year to go a light year. Actually that’s not true. It takes light a year, to reach a spot that is a light year away. It takes you zero because you don’t live nearly long enough for even our fastest shuttle to make the journey. The actual number doesn’t really matter, the fact that is never changes - does. That being said it’s 300 million meters per second but we typically say kilometers, so 300 grando. 30 “racks” if you will. Racks of meters… Per second. Lol. Upvote · Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Work faster with AI, while ensuring your writing always makes the right impression. Download 999 207 Arun VK Science communicator, What The Science (YouTube channel) · Author has 526 answers and 735.4K answer views ·7y Light travels at approximately 300,000 kilometers per second in a vacuum, which has a refractive index of 1.0, but it slows down to 225,000 kilometers per second in water (refractive index = 1.3) and 200,000 kilometers per second in glass (refractive index of 1.5). Light has constant speed, only its frequency and wavelength varies Speed of light does not change, it has to travel more in a medium than in vacuum, When light is passing through a medium, the electrons in the medium absorbs the energy from the light and gets excited and releases them back. This absorption and re emission of light g Continue Reading Light travels at approximately 300,000 kilometers per second in a vacuum, which has a refractive index of 1.0, but it slows down to 225,000 kilometers per second in water (refractive index = 1.3) and 200,000 kilometers per second in glass (refractive index of 1.5). Light has constant speed, only its frequency and wavelength varies Speed of light does not change, it has to travel more in a medium than in vacuum, When light is passing through a medium, the electrons in the medium absorbs the energy from the light and gets excited and releases them back. This absorption and re emission of light gives objects colour. Thus light interacts with the particle in the medium, which causes delay. But it’s speed remains same, only it has to travel more distance in the given time, so it seems that it’s speed is varying but it is not. It is impossible for any physical object to travel at or more than speed of light. The only reason why a photon can travel at light speeds is because it is mass less. Any object having mass when tires to reach light speeds their mass increases, called relativistic mass. So as it accelerates to reach light speeds it becomes more massive and a infinite energy is required to reach light speeds. Photons have no mass so it can travel at ‘c’ Now even light can’t travel faster than ‘c’, because if it does, it will also gain relativistic mass which acts as a deterrent for further increment in speed. So max value anything can reach is “c” Upvote · 9 5 9 3 9 1 Scott Brickner Programming almost every day for the last 40 years. · Upvoted by Andy Buckley , MA MSci PhD Physics, University of Cambridge (2006) and Ahmad Mbark , PhD Particle Physics, University of London (2025) · Author has 19.1K answers and 6.5M answer views ·Updated 6y Related Why does the speed of light equal 299,792,458 meters per second? You should understand that while we call it the “speed of light”, that’s not really what it is. What relativity tells us is that there’s a specific velocity that’s measured the same by all observers. It turns out that something that’s massless will travel at that velocity when it’s not interacting with other energy, and photons in a vacuum are a pretty good approximation to those conditions. Because of this, and for some other historical reasons, we call this speed “the speed of light”. So, there’s a speed that’s a constant for everyone. That’s simply a fundamental fact of the universe. Now, when Continue Reading You should understand that while we call it the “speed of light”, that’s not really what it is. What relativity tells us is that there’s a specific velocity that’s measured the same by all observers. It turns out that something that’s massless will travel at that velocity when it’s not interacting with other energy, and photons in a vacuum are a pretty good approximation to those conditions. Because of this, and for some other historical reasons, we call this speed “the speed of light”. So, there’s a speed that’s a constant for everyone. That’s simply a fundamental fact of the universe. Now, when you measure the speed, you have to choose some units. In meters per second, the speed is 299,792,458 m/s, but in miles per second it’s about 186,282 miles per second. The actual number you get depends on the units you choose. But the meter, mile, and second—along with every other unit of length or time—is mostly a historical accident. The meter is what it is because it’s a good approximation to one ten-millionth of the distance from the north pole to the equator. The second is just a sixtieth of a minute, which is a sixtieth of an hour, which is a twenty-fourth of a day, which is the mean time it takes for the Earth to rotate on its axis. Had we evolved on a different planet, these values would likely be slightly different from the ones we chose. From a theoretical perspective, the speed of light is more fundamental. It makes more sense to simply say that the speed of light is 1, and then pick units for distance and time that make it come out that way. For example, if you made the base unit of length equal to about a foot and the base unit of time equal to a nanosecond, the speed of light would be 1. Of course, it’s not really all that convenient to measure time in nanoseconds. It’s only relatively recently that we even could measure time that precisely. Because of that, we stick with the more traditional units. About 35 years ago, the international standards bodies decided to change the definition of the meter. Back in 1967, they had already agreed to define the second to be "the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium-133 atom at a temperature of 0 K”. The new definition of the meter was “the distance that light travels in a vacuum in 1/299,792,458 of a second”. What this is really doing is acknowledging the existence of that universal constant—c c, the speed of light—and defining the meter in terms of it. So there’s your simple answer—the speed of light is that number because that’s how we define the meter. We chose that number because it agreed closely with the historical definition of the meter. Upvote · 999 382 99 56 9 4 Sponsored by MEDvi Prescription Weight Loss Why does Tirzepatide shed fat 2x as fast as Ozempic? Ozempic is made for diabetes. Tirzepatide is specifically for weight loss. It starts at $179/month. Start Now 999 109 Paul Grimshaw IT Architect in the Computer Industry (1983–present) · Author has 3.5K answers and 14.9M answer views ·1y Related What is the speed of light measured in? Curiously it’s the other way around - since 1983 we have defined our distance measurement units using the speed of light. Specifically the metre length is defined to be the distance travelled by light in a vacuum in precisely 1/299792458 of a second. And all other length units nowadays are defined in terms of a metre, so for example a mile is defined to be precisely 1609344 millimetres (or 5280 feet, where a foot is precisely 0.3048 metres). And so fundamentally a mile is defined to be the distance travelled by light in a vacuum in 1609344/299792458000 of a second. So we no longer measure the sp Continue Reading Curiously it’s the other way around - since 1983 we have defined our distance measurement units using the speed of light. Specifically the metre length is defined to be the distance travelled by light in a vacuum in precisely 1/299792458 of a second. And all other length units nowadays are defined in terms of a metre, so for example a mile is defined to be precisely 1609344 millimetres (or 5280 feet, where a foot is precisely 0.3048 metres). And so fundamentally a mile is defined to be the distance travelled by light in a vacuum in 1609344/299792458000 of a second. So we no longer measure the speed of light. We just know it to be 299792458 metres per second because that’s what a metre itself means. And the same would apply in whatever speed units you decide to convert this into. Upvote · 9 1 Jack Fraser-Govil Doctor of Physics, Writer of Code, Player of Games · Featured on Forbes and Mental Floss · Upvoted by Eurico Chagas , MS Civil Engineering & Physics, University of Illinois (1978) and Don van der Drift , In PhD Physics program for 2.5 years at Technische Universiteit Eindhoven, former Physics researcher at LBNL · Author has 2.6K answers and 51.5M answer views ·Updated 8y Related How do scientists measure the speed of light? We don't. No, seriously, we don't measure the speed of light (which always refers to the speed in a vacuum). We know exactly what the speed of light is. It is: c=c=299792458 299792458 m s−1 m s−1 And that is absolutely 100% accurate, with no measurement errors. But Jack, I hear you say, what the bloody hell are you talking about? The reason we know that that's exactly the speed of light, is that we defined it to be that number. We then take our definition of a second (the length of time for a certain number of periods of the radiation emitted in hyperfine transitions in caesium-133), and from that we define a metre Continue Reading We don't. No, seriously, we don't measure the speed of light (which always refers to the speed in a vacuum). We know exactly what the speed of light is. It is: c=c=299792458 299792458 m s−1 m s−1 And that is absolutely 100% accurate, with no measurement errors. But Jack, I hear you say, what the bloody hell are you talking about? The reason we know that that's exactly the speed of light, is that we defined it to be that number. We then take our definition of a second (the length of time for a certain number of periods of the radiation emitted in hyperfine transitions in caesium-133), and from that we define a metre. So the thing we would be measuring is what a metre is! We use the speed of light as a fixed velocity, from which all observers can define their own length scale. To measure the speed of light would require an external definition of what a metre is - and since about the 1970s, we don't have one! And if you did want to measure the speed of light using this external distance reference, it's easy to test - you just release a light pulse at t=0, towards a mirror - and then time how long it takes to get back to you. This is the exact principle that Radar/Sonar work on (although again, they measure the distance knowing the speed - but it works either way round). Some background: The metre was originally defined after the French Revolution, in about 1799. It was defined as 1 10,000,000 1 10,000,000 the distance between the equator and the pole. The “metre” was formally defined from 1889 as the length of a platinum rod, held in a vault in Paris. From this definition of a metre (and an old definition of a second - I forget what that was), we measured (using the mirror-timing method, or based on astronomical observations) the speed of light to be about 299792458 299792458, plus a non-integer bit, and error bars from the measurement errors. Eventually, we realised that having a metre defined by something there was only one of was a bit annoying. So, we attempted to define it in a way that anyone could replicate - without having to refer to a “standard object”. Therefore, we redefined the metre - using the speed of light. The official definition of a metre today is: 1 299792458 1 299792458 of the distance travelled by light in a vacuum, in 1 second.. Using the caesium definition of a second. Therefore, this was exactly equivalent to defining the speed of light to be the number given above. We chose that number (and not a more convenient number like 300,000,000), because that number changed the definition of a metre by only a fraction of a fraction of a percent - but made everything all nice and integer-y. A consequence of using this definition is that any attempt to measure the speed of light is cyclical - you must use a “metre” to measure it at some point - which relies on the speed of light. Therefore what you actually do now, when you “measure” the speed of light (in a vacuum), is actually “measure how accurate your measuring instruments are”! Upvote · 2.9K 2.9K 999 174 9 6 Sponsored by Stake Stake: Online Casino games - Play & Win Online. Play the best online casino games, slots & live casino games! Unlock VIP bonuses, bet with crypto & win. Play Now 999 286 Jnalis Jmson Studied at Mbeya University of Science and Technology ·7y 299,792,458 meter per second (approximately 3.00×108 m/s) Upvote · 9 2 Steve Baker Senior Software Engineer (2013–present) · Upvoted by Jesse Raffield , Master's degree in physics · Author has 40.1K answers and 295.1M answer views ·2y Related How fast is the speed of light 299,792,458 m/s or 186,000 miles per second? The speed of light is exactly equal to 299,792,458 metres per second and only approximately 300,000 kilometres per second; 186,000 miles per second; 671 million miles per hour. The 299,792,458 metres per second number is 100% perfectly exact - and can never be wrong to even the billionth decimal place! The reason for that is that since 1983, the metre is defined to be the distance travelled by a bea Continue Reading The speed of light is exactly equal to 299,792,458 metres per second and only approximately 300,000 kilometres per second; 186,000 miles per second; 671 million miles per hour. The 299,792,458 metres per second number is 100% perfectly exact - and can never be wrong to even the billionth decimal place! The reason for that is that since 1983, the metre is defined to be the distance travelled by a beam of light in a vacuum in 1/299792458th of a second…so 299,792,458 metres per second will ALWAYS be correct - even if scientists measure the speed of light more precisely…a fact which hurts my head! Since 1959, the “mile” has been officially defined in both the UK and the USA as exactly 1,609.344 meters, the exact speed of l... Upvote · 999 115 9 8 Thomas Yee Senior software engineer · Author has 377 answers and 717.1K answer views ·7y Related How can the meter be defined as the distance light travels in 1/299 792 458 of a second, if the speed of light itself is defined as meter per second? The modern history of weights and measures has been one of concerted efforts to improve the accuracy and reproducibility of measurement standards. Historically, for instance, time standards were based on Earth’s rotational period. In the 19th century, it was realized that the Earth’s rotation not only was gradually slowing, it showed measurable small-scale irregularities, and time standards were switched to being based on the motions of the Moon. The invention in 1955 of the cesium atomic clock led to the replacement of astronomical time standards with atomic time standards, with some concessio Continue Reading The modern history of weights and measures has been one of concerted efforts to improve the accuracy and reproducibility of measurement standards. Historically, for instance, time standards were based on Earth’s rotational period. In the 19th century, it was realized that the Earth’s rotation not only was gradually slowing, it showed measurable small-scale irregularities, and time standards were switched to being based on the motions of the Moon. The invention in 1955 of the cesium atomic clock led to the replacement of astronomical time standards with atomic time standards, with some concessions, in the form of leap second adjustments, being made to to keep atomic time standards in line with the practical need to correlate with the Earth’s irregular motions. In 1793, the meter was defined as one ten-millionth of the distance from the equator to the North Pole. Practical difficulties in reproducing this measurement to better than 10−4 10−4 led to a 1799 redefinition as the distance between two lines scratched on a prototype meter bar, this length being reproducible to about 10−5 10−5 in 1799, to 10−7 10−7 in 1889 (using a new, improved meter bar). Closeup of National Prototype Meter Bar No. 27, made in 1889 by the International Bureau of Weights and Measures (BIPM) and given to the United States, which served as the standard for defining all units of length in the US from 1893 to 1960. Source: File:US National Length Meter.JPG - Wikipedia In 1960, the meter was redefined as 1,650,763.73 wavelengths of the orange spectral line of krypton-86, reproducible to about 10−8 10−8. An international standards body, the General Conference on Weights and Measures (CGPM), is the senior of three international organizations that evaluates the primary standards of measurement, establishes best practice methods for realizing secondary standards of measurement, and otherwise sees to the interests of its member states. In 1979, a group at the US National Bureau of Standards measured the speed of light to be 299792456.2±1.1 m/s. By 1983, the 17th CGPM, carefully evaluating sources of error in speed-of-light determinations, found that the dominant source of error came from the inability to reproduce the meter to much better than 10−8 10−8. By way of contrast, time measurement error in 1983 had reached levels of 10−13 10−13 and was steadily getting better. Given that strong theoretical arguments existed for the speed of light to be a defined constant of nature, the CGPM redefined the meter to be “the length of the path traveled by light in vacuum during a time interval of 1/299792458 of a second,” with the speed of light, therefore, having an exact defined value of 299792458 m/s. Henceforth, instead of improving the precision of speed of light measurements in terms of a difficult-to-reproduce meter, the focus in experimental metrology has been to improve the precision and reproducibility of meter measurements in terms of the exactly defined speed of light. Upvote · 9 3 9 1 Related questions What is the speed of light in meters/seconds? What is the speed of light in 1 second? What is the speed of light in inches per second? Why is a metre defined in terms of the speed of light? What is the speed of light in miles per second and meters per second? What is the approximate speed of light in miles per second? Since speed of light is defined as 299,792,458 meters per second (about 186,000 miles per second), what is the speed of dark? What are some alternative units for measuring the speed of light besides meters per second or kilometers per hour? What is the speed of light in feet per second? What is the value with equation if the speed is within 100 meters per second of the speed of light? Why is the speed of light 10 to the power 8 metres per second? What is the speed of light in a vacuum (in meters per second)? What does the meter have to do with the value of the speed of light? What does it mean if the speed is within 100 meters per second of the speed of light? Why do we say that light travels at 300 million meters per second and not 30 billion meters per second? Related questions What is the speed of light in meters/seconds? What is the speed of light in 1 second? What is the speed of light in inches per second? Why is a metre defined in terms of the speed of light? What is the speed of light in miles per second and meters per second? What is the approximate speed of light in miles per second? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1801	https://simple.wikipedia.org/wiki/Travelling_salesman_problem	Jump to content Travelling salesman problem العربية Asturianu Azərbaycanca Català Čeština Dansk Deutsch Ελληνικά English Español Euskara فارسی Français Galego 한국어 Hrvatski Bahasa Indonesia Italiano עברית Lietuvių Magyar മലയാളം Монгол Nederlands 日本語 Norsk bokmål Polski Português Romnă Русский Slovenčina Slovenščina Српски / srpski Srpskohrvatski / српскохрватски Suomi Svenska Türkçe Українська Tiếng Việt 粵語中文 Change links From Simple English Wikipedia, the free encyclopedia The Traveling Salesman Problem (often called TSP) is a classic algorithmic problem in the field of computer science and operations research. It is focused on optimization. In this context, a better solution often means a solution that is cheaper, shorter, or faster. TSP is a mathematical problem. It is most easily expressed as a graph describing the locations of a set of nodes. The traveling salesman problem was defined in the 1800s by the Irish mathematician W. R. Hamilton and by the British mathematician Thomas Kirkman. Hamilton’s Icosian Game was a recreational puzzle based on finding a Hamiltonian cycle. The general form of the TSP appears to have been first studied by mathematicians during the 1930s in Vienna and at Harvard, notably by Karl Menger. Menger defines the problem, considers the obvious brute-force algorithm, and observes the non-optimality of the nearest neighbour heuristic: We denote by messenger problem (since in practice this question should be solved by each postman, anyway also by many travelers) the task to find, for ﬁnitely many points whose pairwise distances are known, the shortest route connecting the points. Of course, this problem is solvable by finitely many trials. Rules which would push the number of trials below the number of permutations of the given points, are not known. The rule that one first should go from the starting point to the closest point, then to the point closest to this, etc., in general does not yield the shortest route. Hassler Whitney at Princeton University introduced the name traveling salesman problem soon after. Stating the problem [change \| change source] The Travelling Salesman Problem describes a salesman who must travel between N cities. The order in which he does so is something he does not care about, as long as he visits each once during his trip, and finishes where he was at first. Each city is connected to other close by cities, or nodes, by airplanes, or by road or railway. Each of those links between the cities has one or more weights (or the cost) attached. The cost describes how "difficult" it is to traverse this edge on the graph, and may be given, for example, by the cost of an airplane ticket or train ticket, or perhaps by the length of the edge, or time required to complete the traversal. The salesman wants to keep both the travel costs and the distance he travels as low as possible. The Traveling Salesman Problem is typical of a large class of "hard" optimization problems that have intrigued mathematicians and computer scientists for years. Most important, it has applications in science and engineering. For example, in the manufacture of a circuit board, it is important to determine the best order in which a laser will drill thousands of holes. An efficient solution to this problem reduces production costs for the manufacturer. Difficulty [change \| change source] In general, the traveling salesman problem is hard to solve. If there is a way to break this problem into smaller component problems, the components will be at least as complex as the original one. This is what computer scientists call NP-hard problems. Many people have studied this problem. The easiest (and most expensive solution) is to simply try all possibilities. The problem with this is that for N cities you have (N-1) factorial possibilities. This means that for only 10 cities there are over 180 thousand combinations to try (since the start city is defined, there can be permutations on the remaining nine). We only count half since each route has an equal route in reverse with the same length or cost. 9! / 2 = 181 440 Exact solutions to the problem can be found, using branch and bound algorithms. This is currently possible for up to 85,900 cities. Heuristics approaches use a set of guiding rules for selection of the next node. But since heuristics result in approximations, they will not always give the optimal solution, although high quality admissible heuristics can find a useful solution in a fraction of the time required for a full brute force of the problem. An example of a heuristic for a node would be a summation of how many unvisited nodes are "close by" a connected node. This could encourage the salesman to visit a group of close by nodes clustered together before moving onto another natural cluster in the graph. See Monte Carlo algorithms and Las Vegas algorithms Other websites [change \| change source] TSPLIB Archived 2017-03-25 at the Wayback Machine References [change \| change source] ↑ "Travelling Salesman Problem, Operations Research". www.universalteacherpublications.com. Archived from the original on 2017-11-21. Retrieved 2017-11-13. ↑ A discussion of the early work of Hamilton and Kirkman can be found in Graph Theory 1736–1936 ↑ A detailed treatment of the connection between Menger and Whitney as well as the growth in the study of TSP can be found in Alexander Schrijver's 2005 paper "On the history of combinatorial optimization (till 1960). Handbook of Discrete Optimization (K. Aardal, G.L. Nemhauser, R. Weismantel, eds.), Elsevier, Amsterdam, 2005, pp. 1–68.PS,PDF ↑ Traveling Salesman Problem - Branch and Bound on YouTube. How to cut unfruitful branches using reduced rows and columns as in Hungarian matrix algorithm ↑ "Optimal TSP tours". Retrieved from " Categories: Theoretical computer science Combinatorial optimization Hidden category: Webarchive template wayback links
1802	https://pre-med.jumedicine.com/wp-content/uploads/sites/6/2019/01/ganong-pdf.pdf	Ranges of Normal Values in Human Whole Blood (B), Plasma (P), or Serum (S)a Normal Value (Varies with Procedure Used) Determination Traditional Units SI Units Normal Value (Varies with Procedure Used) Determination Traditional Units SI Units Acetoacetate plus acetone (S) 0.3–2.0 mg/dL 3–20 mg/L Aldosterone (supine) (P) 3.0–10 ng/dL 83–227 pmol/L Alpha-amino nitrogen (P) 3.0–5.5 mg/dL 2.1–3.9 mmol/L Aminotransferases Alanine aminotransferase 3–48 units/L Aspartate aminotransferase 0–55 units/L Ammonia (B) 12–55 μmol/L 12–55 μmol/L Amylase (S) 53–123 units/L 884–2050 nmol s–1/L Ascorbic acid (B) 0.4–1.5 mg/dL (fasting) 23–85 μmol/L Bilirubin (S) Conjugated (direct): up to 0.4 mg/dL Up to 7 μmol/L Total (conjugated plus free): up to 1.0 mg/dL Up to 17 μmol/L Calcium (S) 8.5–10.5 mg/dL; 4.3–5.3 meq/L 2.1–2.6 mmol/L Carbon dioxide content (S) 24–30 meq/L 24–30 mmol/L Carotenoids (S) 0.8–4.0 μg/mL 1.5–7.4 μmol/L Ceruloplasmin (S) 23–43 mg/dL 240–430 mg/L Chloride (S) 100–108 meq/L 100–108 mmol/L Cholesterol (S) < 200 mg/dL < 5.17 mmol/L Cholesteryl esters (S) 60–70% of total cholesterol Copper (total) (S) 70–155 μg/dL 11.0–24.4 μmol/L Cortisol (P) (AM, fasting) 5–25 μg/dL 0.14–0.69 μmol/L Creatinine (P) 0.6–1.5 mg/dL 53–133 μmol/L Glucose, fasting (P) 70–110 mg/dL 3.9–6.1 mmol/L Iron (S) 50–150 μg/dL 9.0–26.9 μmol/L Lactic acid (B) 0.5–2.2 meq/L 0.5–2.2 mmol/L Lipase (S) 3–19 units/L Lipids, total (S) 450–1000 mg/dL 4.5–10 g/L Magnesium (S) 1.4–2.0 meq/L 0.7–1.0 mmol/L Osmolality (S) 280–296 mosm/kg H2O 280–296 mmol/kg H2O PCO2 (arterial) (B) 35–45 mm Hg 4.7–6.0 kPa Pepsinogen (P) 200–425 units/mL pH (B) 7.35–7.45 Phenylalanine (S) 0–2 mg/dL 0–120 μmol/L Phosphatase, acid (S) Males: 0–0.8 sigma unit/mL Females: 0.01–0.56 sigma unit/mL Phosphatase, alkaline (S) 13–39 units/L (adults) 0.22–0.65 μmol s–1/L Phospholipids (S) 9–16 mg/dL as lipid phosphorus 2.9–5.2 mmol/L Phosphorus, inorganic (S) 2.6–4.5 mg/dL (infants in first year: up to 6.0 mg/dL) 0.84–1.45 mmol/L PO2 (arterial) (B) 75–100 mm Hg 10.0–13.3 kPa Potassium (S) 3.5–5.0 meq/L 3.5–5.0 mmol/L Protein Total (S) 6.0–8.0 g/dL 60–80 g/L Albumin (S) 3.1–4.3 g/dL 31–43 g/L Globulin (S) 2.6–4.1 g/dL 26–41 g/L Pyruvic acid (P) 0–0.11 meq/L 0–110 μmol/L Sodium (S) 135–145 meq/L 135–145 mmol/L Urea nitrogen (S) 8–25 mg/dL 2.9–8.9 mmol/L Uric acid (S) Women 2.3–6.6 mg/dL 137–393 μmol/L Men 3.6–8.5 mg/dL 214–506 μmol/L aBased in part on Kratz A, et al. Laboratory reference values. N Engl J Med 2004;351:1548. Ranges vary somewhat from one laboratory to another depending on the details of the methods used, and specific values should be considered in the context of the range of values for the laboratory that made the determination. a LANGE medical book Ganong’s Review of Medical Physiology Twenty-Third Edition New York Chicago San Francisco Lisbon London Madrid Mexico City Milan New Delhi San Juan Seoul Singapore Sydney Toronto Kim E. Barrett, PhD Professor Department of Medicine Dean of Graduate Studies University of California, San Diego La Jolla, California Susan M. Barman, PhD Professor Department of Pharmacology/Toxicology Michigan State University East Lansing, Michigan Scott Boitano, PhD Associate Professor, Physiology Arizona Respiratory Center Bio5 Collaborative Research Institute University of Arizona Tucson, Arizona Heddwen L. Brooks, PhD Associate Professor Department of Physiology College of Medicine University of Arizona Tucson, Arizona Copyright © 2010 by The McGraw-Hill Companies, Inc. All rights reserved. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher. ISBN: 978-0-07-160568-7 MHID: 0-07-160568-1 The material in this eBook also appears in the print version of this title: ISBN: 978-0-07-160567-0, MHID: 0-07-160567-3. All trademarks are trademarks of their respective owners. Rather than put a trademark symbol after every occurrence of a trademarked name, we use names in an editorial fashion only, and to the benefit of the trademark owner, with no intention of infringement of the trademark. Where such designations appear in this book, they have been printed with initial caps. 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Under no circumstances shall McGraw-Hill and/or its licensors be liable for any indirect, incidental, special, punitive, consequential or similar damages that result from the use of or inability to use the work, even if any of them has been advised of the possibility of such damages. This limitation of liability shall apply to any claim or cause whatsoever whether such claim or cause arises in contract, tort or otherwise. iii WILLIAM FRANCIS GANONG William Francis (“Fran”) Ganong was an outstanding scien-tist, educator, and writer. He was completely dedicated to the field of physiology and medical education in general. Chair-man of the Department of Physiology at the University of Cal-ifornia, San Francisco, for many years, he received numerous teaching awards and loved working with medical students. Over the course of 40 years and some 22 editions, he was the sole author of the best selling Review of Medical Physiology, and a co-author of 5 editions of Pathophysiology of Disease: An Introduction to Clinical Medicine. He was one of the “deans” of the Lange group of authors who produced concise medical text and review books that to this day remain extraordinarily popu-lar in print and now in digital formats. Dr. Ganong made a gigantic impact on the education of countless medical students and clinicians. A general physiologist par excellence and a neuroendocrine physiologist by subspecialty, Fran developed and maintained a rare understanding of the entire field of physiology. This allowed him to write each new edition (every 2 years!) of the Review of Medical Physiology as a sole author, a feat remarked on and admired whenever the book came up for discussion among physiologists. He was an excellent writer and far ahead of his time with his objective of distilling a complex subject into a concise presentation. Like his good friend, Dr. Jack Lange, founder of the Lange series of books, Fran took great pride in the many different translations of the Review of Medical Physi-ology and was always delighted to receive a copy of the new edi-tion in any language. He was a model author, organized, dedicated, and enthusias-tic. His book was his pride and joy and like other best-selling authors, he would work on the next edition seemingly every day, updating references, rewriting as needed, and always ready and on time when the next edition was due to the publisher. He did the same with his other book, Pathophysiology of Disease: An Introduction to Clinical Medicine, a book that he worked on meticulously in the years following his formal retirement and appointment as an emeritus professor at UCSF. Fran Ganong will always have a seat at the head table of the greats of the art of medical science education and communi-cation. He died on December 23, 2007. All of us who knew him and worked with him miss him greatly. Dedication to iv • Thoroughly updated to reflect the latest research and developments in important areas such as the cellular basis of neurophysiology • Incorporates examples from clinical medicine throughout the chapters to illustrate important physiologic concepts • Delivers more detailed, clinically-relevant, high-yield information per page than any similar text or review • NEW full-color illustrations—the authors have worked with an outstanding team of medical illustrators, photographers, educators, and students to provide an unmatched collection of 600 illustrations and tables • NEW boxed clinical cases—featuring examples of diseases that illustrate important physiologic principles • NEW high-yield board review questions at the end of each chapter • NEW larger 8½ X 11” trim-size enhances the rich visual content • NEW companion online learning center (LangeTextbooks.com) offers a wealth of innovative learning tools and illustrations Key Features of the 23rd Edition of Ganong’s Review of Medical Physiology Full-color illustrations enrich the text NEW iPod-compatible review—Medical PodClass offers audio and text for study on the go KEY FEATURES v Clinical Cases illustrate essential physiologic principles Chapters conclude with Chapter Summaries and review questions Summary tables and charts encapsulate important information vi About the Authors KIM E. BARRETT Kim Barrett received her PhD in biological chemistry from University College London in 1982. Following postdoctoral training at the National Institutes of Health, she joined the faculty at the University of California, San Diego, School of Medicine in 1985, rising to her current rank of Professor of Medicine in 1996. Since 2006, she has also served the University as Dean of Graduate Studies. Her research interests focus on the physiology and pathophysiology of the intestinal epithelium, and how its function is altered by commensal, probiotics, and pathogenic bacteria as well as in specific disease states, such as inflammatory bowel diseases. She has published almost 200 articles, chapters, and reviews, and has received several honors for her research accomplishments including the Bowditch and Davenport Lectureships from the American Physiological Society and the degree of Doctor of Medical Sciences, honoris causa, from Queens University, Belfast. She is also a dedicated and award-winning instructor of medical, pharmacy, and graduate students, and has taught various topics in medical and systems physiology to these groups for more than 20 years. Her teaching experiences led her to author a prior volume (Gastrointestinal Physiology, McGraw-Hill, 2005) and she is honored to have been invited to take over the helm of Ganong. SUSAN M. BARMAN Susan Barman received her PhD in physiology from Loyola University School of Medicine in Maywood, Illinois. Afterward she went to Michigan State University (MSU) where she is currently a Professor in the Department of Pharmacology/ Toxicology and the Neuroscience Program. Dr Barman has had a career-long interest in neural control of cardiorespiratory function with an emphasis on the characterization and origin of the naturally occurring discharges of sympathetic and phrenic nerves. She was a recipient of a prestigious National Institutes of Health MERIT (Method to Extend Research in Time) Award. She is also a recipient of an Outstanding University Woman Faculty Award from the MSU Faculty Professional Women's Association and an MSU College of Human Medicine Distinguished Faculty Award. She has been very active in the American Physiological Society (APS) and recently served on its council. She has also served as Chair of the Central Nervous System Section of APS as well as Chair of both the Women in Physiology and Section Advisory Committees of APS. In her spare time, she enjoys daily walks, aerobic exercising, and mind-challenging activities like puzzles of various sorts. SCOTT BOITANO Scott Boitano received his PhD in genetics and cell biology from Washington State University in Pullman, Washington, where he acquired an interest in cellular signaling. He fostered this interest at University of California, Los Angeles, where he focused his research on second messengers and cellular physiology of the lung epithelium. He continued to foster these research interests at the University of Wyoming and at his current positions with the Department of Physiology and the Arizona Respiratory Center, both at the University of Arizona. HEDDWEN L. BROOKS Heddwen Brooks received her PhD from Imperial College, University of London and is an Associate Professor in the Department of Physiology at the University of Arizona (UA). Dr Brooks is a renal physiologist and is best known for her development of microarray technology to address in vivo signaling pathways involved in the hormonal regulation of renal function. Dr Brooks’ many awards include the American Physiological Society (APS) Lazaro J. Mandel Young Investigator Award, which is for an individual demonstrating outstanding promise in epithelial or renal physiology. She will receive the APS Renal Young Investigator Award at the 2009 annual meeting of the Federation of American Societies for Experimental Biology. Dr Brooks is a member of the APS Renal Steering Section and the APS Committee of Committees. She is on the Editorial Board of the American Journal of Physiology-Renal Physiology (since 2001), and she has also served on study sections of the National Institutes of Health and the American Heart Association. vii Contents Preface ix S E C T I O N I CELLULAR & MOLECULAR BASIS FOR MEDICAL PHYSIOLOGY 1 1. General Principles & Energy Production in Medical Physiology 1 2. Overview of Cellular Physiology in Medical Physiology 31 3. Immunity, Infection, & Inflammation 63 S E C T I O N II PHYSIOLOGY OF NERVE & MUSCLE CELLS 79 4. Excitable Tissue: Nerve 79 5. Excitable Tissue: Muscle 93 6. Synaptic & Junctional Transmission 115 7. Neurotransmitters & Neuromodulators 129 8. Properties of Sensory Receptors 149 9. Reflexes 157 S E C T I O N III CENTRAL & PERIPHERAL NEUROPHYSIOLOGY 167 10. Pain & Temperature 167 11. Somatosensory Pathways 173 12. Vision 181 13. Hearing & Equilibrium 203 14. Smell & Taste 219 15. Electrical Activity of the Brain, Sleep–Wake States, & Circadian Rhythms 229 16. Control of Posture & Movement 241 17. The Autonomic Nervous System 261 18. Hypothalamic Regulation of Hormonal Functions 273 19. Learning, Memory, Language, & Speech 289 S E C T I O N IV ENDOCRINE & REPRODUCTIVE PHYSIOLOGY 301 20. The Thyroid Gland 301 21. Endocrine Functions of the Pancreas & Regulation of Carbohydrate Metabolism 315 22. The Adrenal Medulla & Adrenal Cortex 337 23. Hormonal Control of Calcium and Phosphate Metabolism & the Physiology of Bone 363 24. The Pituitary Gland 377 25. The Gonads: Development & Function of the Reproductive System 391 S E C T I O N V GASTROINTESTINAL PHYSIOLOGY 429 26. Overview of Gastrointestinal Function & Regulation 429 viii CONTENTS 27. Digestion, Absorption, & Nutritional Principles 451 28. Gastrointestinal Motility 469 29. Transport & Metabolic Functions of the Liver 479 S E C T I O N VI CARDIOVASCULAR PHYSIOLOGY 489 30. Origin of the Heartbeat & the Electrical Activity of the Heart 489 31. The Heart as a Pump 507 32. Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 521 33. Cardiovascular Regulatory Mechanisms 555 34. Circulation Through Special Regions 569 S E C T I O N VII RESPIRATORY PHYSIOLOGY 587 35. Pulmonary Function 587 36. Gas Transport & pH in the Lung 609 37. Regulation of Respiration 625 S E C T I O N VIII RENAL PHYSIOLOGY 639 38. Renal Function & Micturition 639 39. Regulation of Extracellular Fluid Composition & Volume 665 40. Acidification of the Urine & Bicarbonate Excretion 679 Answers to Multiple Choice Questions 687 Index 689 ix Preface From the Authors We are very pleased to launch the 23rd edition of Ganong's Review of Medical Physiology. The current authors have at-tempted to maintain the highest standards of excellence, ac-curacy, and pedagogy developed by Fran Ganong over the 46 years in which he educated countless students worldwide with this textbook. At the same time, we have been attuned to the evolving needs of both students and professors in medical physiology. Thus, in addition to usual updates on the latest research and developments in areas such as the cellular basis of physiology and neurophysiology, this edition has added both outstanding pedagogy and learning aids for students. We are truly grateful for the many helpful insights, sugges-tions, and reviews from around the world that we received from colleagues and students. We hope you enjoy the new fea-tures and the 23rd edition! This edition is a revision of the original works of Dr. Francis Ganong. New 4 Color Illustrations • We have worked with a large team of medical illustrators, photographers, educators, and students to build an accurate, up-to-date, and visually appealing new illustration program. Full-color illustrations and tables are provided throughout, which also include detailed figure legends that tell a short sto-ry or describes the key point of the illustration. New 81 /2 x 11 Format • Based on student and instructor focus groups, we have in-creased the trim size, which will provide additional white space and allow our new art program to really show! New Boxed Clinical Cases • Highlighted in a shaded background, so students can recog-nize the boxed clinical cases, examples of diseases illustrat-ing important physiological principles are provided. New End of Chapter Board Review Questions • New to this edition, chapters now conclude with board re-view questions. New Media • This new edition has focused on creating new student con-tent that is built upon learning outcomes and assessing stu-dent performance. Free with every student copy is an iPod Review Tutorial Product. Questions and art based from each chapter tests students comprehension and is easy to navigate with a simple click of the scroll bar! • Online Learning Center will provide students and faculty with cases and art and board review questions on a dedicat-ed website. This page intentionally left blank 1 C H A P T E R SECTION I CELLULAR & MOLECULAR BASIS OF MEDICAL PHYSIOLOGY 1 General Principles & Energy Production in Medical Physiology O B J E C T I V E S After studying this chapter, you should be able to: ■Name the different fluid compartments in the human body. ■Define moles, equivalents, and osmoles. ■Define pH and buffering. ■Understand electrolytes and define diffusion, osmosis, and tonicity. ■Define and explain the resting membrane potential. ■Understand in general terms the basic building blocks of the cell: nucleotides, amino acids, carbohydrates, and fatty acids. ■Understand higher-order structures of the basic building blocks: DNA, RNA, proteins, and lipids. ■Understand the basic contributions of these building blocks to cell structure, function, and energy balance. INTRODUCTION In unicellular organisms, all vital processes occur in a single cell. As the evolution of multicellular organisms has progressed, various cell groups organized into tissues and organs have taken over particular functions. In humans and other verte-brate animals, the specialized cell groups include a gastrointes-tinal system to digest and absorb food; a respiratory system to take up O2 and eliminate CO2; a urinary system to remove wastes; a cardiovascular system to distribute nutrients, O2, and the products of metabolism; a reproductive system to perpetu-ate the species; and nervous and endocrine systems to coordi-nate and integrate the functions of the other systems. This book is concerned with the way these systems function and the way each contributes to the functions of the body as a whole. In this section, general concepts and biophysical and bio-chemical principles that are basic to the function of all the systems are presented. In the first chapter, the focus is on review of basic biophysical and biochemical principles and the introduction of the molecular building blocks that con-tribute to cellular physiology. In the second chapter, a review of basic cellular morphology and physiology is presented. In the third chapter, the process of immunity and inflammation, and their link to physiology, are considered. 2 SECTION I Cellular & Molecular Basis of Medical Physiology GENERAL PRINCIPLES THE BODY AS AN ORGANIZED “SOLUTION” The cells that make up the bodies of all but the simplest mul-ticellular animals, both aquatic and terrestrial, exist in an “in-ternal sea” of extracellular fluid (ECF) enclosed within the integument of the animal. From this fluid, the cells take up O2 and nutrients; into it, they discharge metabolic waste prod-ucts. The ECF is more dilute than present-day seawater, but its composition closely resembles that of the primordial oceans in which, presumably, all life originated. In animals with a closed vascular system, the ECF is divided into two components: the interstitial fluid and the circulating blood plasma. The plasma and the cellular elements of the blood, principally red blood cells, fill the vascular system, and together they constitute the total blood volume. The intersti-tial fluid is that part of the ECF that is outside the vascular system, bathing the cells. The special fluids considered together as transcellular fluids are discussed in the following text. About a third of the total body water is extracellular; the remaining two thirds is intracellular (intracellular fluid). In the average young adult male, 18% of the body weight is pro-tein and related substances, 7% is mineral, and 15% is fat. The remaining 60% is water. The distribution of this water is shown in Figure 1–1A. The intracellular component of the body water accounts for about 40% of body weight and the extracellular component for about 20%. Approximately 25% of the extracellular component is in the vascular system (plasma = 5% of body weight) and 75% outside the blood vessels (interstitial fluid = 15% of body weight). The total blood volume is about 8% of body weight. Flow between these compartments is tightly regulated. UNITS FOR MEASURING CONCENTRATION OF SOLUTES In considering the effects of various physiologically important substances and the interactions between them, the number of molecules, electric charges, or particles of a substance per unit volume of a particular body fluid are often more meaningful than simply the weight of the substance per unit volume. For this reason, physiological concentrations are frequently ex-pressed in moles, equivalents, or osmoles. Moles A mole is the gram-molecular weight of a substance, ie, the molecular weight of the substance in grams. Each mole (mol) consists of 6 × 1023 molecules. The millimole (mmol) is 1/1000 of a mole, and the micromole (μmol) is 1/1,000,000 of a mole. Thus, 1 mol of NaCl = 23 g + 35.5 g = 58.5 g, and 1 mmol = 58.5 mg. The mole is the standard unit for expressing the amount of substances in the SI unit system. The molecular weight of a substance is the ratio of the mass of one molecule of the substance to the mass of one twelfth the mass of an atom of carbon-12. Because molecular weight is a ratio, it is dimensionless. The dalton (Da) is a unit of mass equal to one twelfth the mass of an atom of carbon-12. The kilodalton (kDa = 1000 Da) is a useful unit for expressing the molecular mass of proteins. Thus, for example, one can speak of a 64-kDa protein or state that the molecular mass of the protein is 64,000 Da. However, because molecular weight is a dimensionless ratio, it is incorrect to say that the molecular weight of the protein is 64 kDa. Equivalents The concept of electrical equivalence is important in physiol-ogy because many of the solutes in the body are in the form of charged particles. One equivalent (eq) is 1 mol of an ionized substance divided by its valence. One mole of NaCl dissociates into 1 eq of Na+ and 1 eq of Cl–. One equivalent of Na+ = 23 g, but 1 eq of Ca2+ = 40 g/2 = 20 g. The milliequivalent (meq) is 1/1000 of 1 eq. Electrical equivalence is not necessarily the same as chemical equivalence. A gram equivalent is the weight of a substance that is chemically equivalent to 8.000 g of oxygen. The normality (N) of a solution is the number of gram equivalents in 1 liter. A 1 N solution of hydrochloric acid contains both H+ (1 g) and Cl– (35.5 g) equivalents, = (1 g + 35.5 g)/L = 36.5 g/L. WATER, ELECTROLYTES, & ACID/BASE The water molecule (H2O) is an ideal solvent for physiological reactions. H2O has a dipole moment where oxygen slightly pulls away electrons from the hydrogen atoms and creates a charge separation that makes the molecule polar. This allows water to dissolve a variety of charged atoms and molecules. It also allows the H2O molecule to interact with other H2O mol-ecules via hydrogen bonding. The resultant hydrogen bond network in water allows for several key properties in physiol-ogy: (1) water has a high surface tension, (2) water has a high heat of vaporization and heat capacity, and (3) water has a high dielectric constant. In layman’s terms, H2O is an excel-lent biological fluid that serves as a solute; it provides optimal heat transfer and conduction of current. Electrolytes (eg, NaCl) are molecules that dissociate in water to their cation (Na+) and anion (Cl–) equivalents. Because of the net charge on water molecules, these electro-lytes tend not to reassociate in water. There are many impor-tant electrolytes in physiology, notably Na+, K+, Ca2+, Mg2+, Cl–, and HCO3 –. It is important to note that electrolytes and other charged compounds (eg, proteins) are unevenly distrib-uted in the body fluids (Figure 1–1B). These separations play an important role in physiology. CHAPTER 1 General Principles & Energy Production in Medical Physiology 3 FIGURE 1–1 Organization of body fluids and electrolytes into compartments. A) Body fluids are divided into Intracellular and extracel-lular fluid compartments (ICF and ECF, respectively). Their contribution to percentage body weight (based on a healthy young adult male; slight variations exist with age and gender) emphasizes the dominance of fluid makeup of the body. Transcellular fluids, which constitute a very small percentage of total body fluids, are not shown. Arrows represent fluid movement between compartments. B) Electrolytes and proteins are un-equally distributed among the body fluids. This uneven distribution is crucial to physiology. Prot–, protein, which tends to have a negative charge at physiologic pH. Blood plasma: 5% body weight Interstitial fluid: 15% body weight Intracellular fluid: 40% body weight Skin Kidneys Intestines Stomach Lungs Extra-cellular fluid: 20% body weight A B 200 150 100 50 0 meq/L H2O K+ Na+ Cl− Prot− HCO3 − Plasma Extracellular fluid K+ Na+ Cl− HCO3 − Interstitial fluid K+ Na+ Cl− HCO3 − Intracellular fluid Capillaries Cell membrane Misc. phosphates Prot− 4 SECTION I Cellular & Molecular Basis of Medical Physiology pH AND BUFFERING The maintenance of a stable hydrogen ion concentration ([H+]) in body fluids is essential to life. The pH of a solution is defined as the logarithm to the base 10 of the reciprocal of the H+ concentration ([H+]), ie, the negative logarithm of the [H+]. The pH of water at 25 °C, in which H+ and OH– ions are present in equal numbers, is 7.0 (Figure 1–2). For each pH unit less than 7.0, the [H+] is increased tenfold; for each pH unit above 7.0, it is decreased tenfold. In the plasma of healthy in-dividuals, pH is slightly alkaline, maintained in the narrow range of 7.35 to 7.45. Conversely, gastric fluid pH can be quite acidic (on the order of 2.0) and pancreatic secretions can be quite alkaline (on the order of 8.0). Enzymatic activity and protein structure are frequently sensitive to pH; in any given body or cellular compartment, pH is maintained to allow for maximal enzyme/protein efficiency. Molecules that act as H+ donors in solution are considered acids, while those that tend to remove H+ from solutions are considered bases. Strong acids (eg, HCl) or bases (eg, NaOH) dissociate completely in water and thus can most change the [H+] in solution. In physiological compounds, most acids or bases are considered “weak,” that is, they contribute relatively few H+ or take away relatively few H+ from solution. Body pH is stabilized by the buffering capacity of the body fluids. A buffer is a substance that has the ability to bind or release H+ in solution, thus keeping the pH of the solution relatively con-stant despite the addition of considerable quantities of acid or base. Of course there are a number of buffers at work in bio-logical fluids at any given time. All buffer pairs in a homoge-nous solution are in equilibrium with the same [H+]; this is known as the isohydric principle. One outcome of this prin-ciple is that by assaying a single buffer system, we can under-stand a great deal about all of the biological buffers in that system. When acids are placed into solution, there is a dissociation of some of the component acid (HA) into its proton (H+) and free acid (A–). This is frequently written as an equation: HA → ← H+ + A–. According to the laws of mass action, a relationship for the dissociation can be defined mathematically as: Ka = [H+] [A–] / [HA] where Ka is a constant, and the brackets represent concentra-tions of the individual species. In layman’s terms, the product of the proton concentration ([H+]) times the free acid con-centration ([A–]) divided by the bound acid concentration ([HA]) is a defined constant (K). This can be rearranged to read: [H+] = Ka [HA]/[A–] If the logarithm of each side is taken: log [H+] = logKa + log[HA]/[A–] Both sides can be multiplied by –1 to yield: –log [H+] = –logKa + log[A–]/[HA] This can be written in a more conventional form known as the Henderson Hasselbach equation: pH = pKa + log [A–]/[HA] This relatively simple equation is quite powerful. One thing that we can discern right away is that the buffering capacity of a particular weak acid is best when the pKa of that acid is equal to the pH of the solution, or when: [A–] = [HA], pH = pKa Similar equations can be set up for weak bases. An impor-tant buffer in the body is carbonic acid. Carbonic acid is a weak acid, and thus is only partly dissociated into H+ and bicarbonate: H2CO3 → ← H+ + HCO3 – If H+ is added to a solution of carbonic acid, the equilib-rium shifts to the left and most of the added H+ is removed from solution. If OH– is added, H+ and OH– combine, taking H+ out of solution. However, the decrease is countered by more dissociation of H2CO3, and the decline in H+ concen-tration is minimized. A unique feature of bicarbonate is the linkage between its buffering ability and the ability for the lungs to remove carbon dioxide from the body. Other impor-tant biological buffers include phosphates and proteins. DIFFUSION Diffusion is the process by which a gas or a substance in a so-lution expands, because of the motion of its particles, to fill all the available volume. The particles (molecules or atoms) of a substance dissolved in a solvent are in continuous random movement. A given particle is equally likely to move into or FIGURE 1–2 Proton concentration and pH. Relative proton (H+) concentrations for solutions on a pH scale are shown. (Redrawn from Alberts B et al: Molecular Biology of the Cell, 4th ed. Garland Science, 2002.) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 10−1 10−2 10−3 10−4 10−5 10−6 10−7 10−8 10−9 10−10 10−11 10−12 10−13 10−14 pH H+ concentration (mol/L) ACIDIC ALKALINE For pure water, [H+] = 10−7 mol/L CHAPTER 1 General Principles & Energy Production in Medical Physiology 5 out of an area in which it is present in high concentration. However, because there are more particles in the area of high concentration, the total number of particles moving to areas of lower concentration is greater; that is, there is a net flux of sol-ute particles from areas of high to areas of low concentration. The time required for equilibrium by diffusion is proportion-ate to the square of the diffusion distance. The magnitude of the diffusing tendency from one region to another is directly proportionate to the cross-sectional area across which diffu-sion is taking place and the concentration gradient, or chem-ical gradient, which is the difference in concentration of the diffusing substance divided by the thickness of the boundary (Fick’s law of diffusion). Thus, J = –DA Δc Δx where J is the net rate of diffusion, D is the diffusion coeffi-cient, A is the area, and Δc/Δx is the concentration gradient. The minus sign indicates the direction of diffusion. When considering movement of molecules from a higher to a lower concentration, Δc/Δx is negative, so multiplying by –DA gives a positive value. The permeabilities of the boundaries across which diffusion occurs in the body vary, but diffusion is still a major force affecting the distribution of water and solutes. OSMOSIS When a substance is dissolved in water, the concentration of water molecules in the solution is less than that in pure water, because the addition of solute to water results in a solution that occupies a greater volume than does the water alone. If the so-lution is placed on one side of a membrane that is permeable to water but not to the solute, and an equal volume of water is placed on the other, water molecules diffuse down their con-centration (chemical) gradient into the solution (Figure 1–3). This process—the diffusion of solvent molecules into a region in which there is a higher concentration of a solute to which the membrane is impermeable—is called osmosis. It is an im-portant factor in physiologic processes. The tendency for movement of solvent molecules to a region of greater solute concentration can be prevented by applying pressure to the more concentrated solution. The pressure necessary to prevent solvent migration is the osmotic pressure of the solution. Osmotic pressure—like vapor pressure lowering, freezing-point depression, and boiling-point elevation—depends on the number rather than the type of particles in a solution; that is, it is a fundamental colligative property of solutions. In an ideal solution, osmotic pressure (P) is related to temperature and volume in the same way as the pressure of a gas: where n is the number of particles, R is the gas constant, T is the absolute temperature, and V is the volume. If T is held con-stant, it is clear that the osmotic pressure is proportional to the number of particles in solution per unit volume of solution. For this reason, the concentration of osmotically active parti-cles is usually expressed in osmoles. One osmole (Osm) equals the gram-molecular weight of a substance divided by the number of freely moving particles that each molecule lib-erates in solution. For biological solutions, the milliosmole (mOsm; 1/1000 of 1 Osm) is more commonly used. If a solute is a nonionizing compound such as glucose, the osmotic pressure is a function of the number of glucose mole-cules present. If the solute ionizes and forms an ideal solution, each ion is an osmotically active particle. For example, NaCl would dissociate into Na+ and Cl– ions, so that each mole in solution would supply 2 Osm. One mole of Na2SO4 would dissociate into Na+, Na+, and SO4 2– supplying 3 Osm. How-ever, the body fluids are not ideal solutions, and although the dissociation of strong electrolytes is complete, the number of particles free to exert an osmotic effect is reduced owing to interactions between the ions. Thus, it is actually the effective concentration (activity) in the body fluids rather than the number of equivalents of an electrolyte in solution that deter-mines its osmotic capacity. This is why, for example, 1 mmol of NaCl per liter in the body fluids contributes somewhat less than 2 mOsm of osmotically active particles per liter. The more concentrated the solution, the greater the deviation from an ideal solution. The osmolal concentration of a substance in a fluid is mea-sured by the degree to which it depresses the freezing point, with 1 mol of an ideal solution depressing the freezing point 1.86 °C. The number of milliosmoles per liter in a solution equals the freezing point depression divided by 0.00186. The osmolarity is the number of osmoles per liter of solution (eg, plasma), whereas the osmolality is the number of osmoles per kilogram of solvent. Therefore, osmolarity is affected by the volume of the various solutes in the solution and the tempera-ture, while the osmolality is not. Osmotically active substances in the body are dissolved in water, and the density of water is 1, so osmolal concentrations can be expressed as osmoles per P nRT V ----------= FIGURE 1–3 Diagrammatic representation of osmosis. Water molecules are represented by small open circles, solute molecules by large solid circles. In the diagram on the left, water is placed on one side of a membrane permeable to water but not to solute, and an equal volume of a solution of the solute is placed on the other. Water molecules move down their concentration (chemical) gradient into the solution, and, as shown in the diagram on the right, the volume of the solution increases. As indicated by the arrow on the right, the os-motic pressure is the pressure that would have to be applied to pre-vent the movement of the water molecules. Semipermeable membrane Pressure 6 SECTION I Cellular & Molecular Basis of Medical Physiology liter (Osm/L) of water. In this book, osmolal (rather than osmolar) concentrations are considered, and osmolality is expressed in milliosmoles per liter (of water). Note that although a homogeneous solution contains osmot-ically active particles and can be said to have an osmotic pres-sure, it can exert an osmotic pressure only when it is in contact with another solution across a membrane permeable to the sol-vent but not to the solute. OSMOLAL CONCENTRATION OF PLASMA: TONICITY The freezing point of normal human plasma averages –0.54 °C, which corresponds to an osmolal concentration in plasma of 290 mOsm/L. This is equivalent to an osmotic pressure against pure water of 7.3 atm. The osmolality might be expected to be higher than this, because the sum of all the cation and anion equivalents in plasma is over 300. It is not this high because plasma is not an ideal solution and ionic interactions reduce the number of particles free to exert an osmotic effect. Except when there has been insufficient time after a sudden change in composition for equilibrium to occur, all fluid compartments of the body are in (or nearly in) osmotic equilibrium. The term tonicity is used to describe the osmolality of a solution relative to plasma. Solutions that have the same osmolality as plasma are said to be isotonic; those with greater osmolality are hyper-tonic; and those with lesser osmolality are hypotonic. All solu-tions that are initially isosmotic with plasma (ie, that have the same actual osmotic pressure or freezing-point depression as plasma) would remain isotonic if it were not for the fact that some solutes diffuse into cells and others are metabolized. Thus, a 0.9% saline solution remains isotonic because there is no net movement of the osmotically active particles in the so-lution into cells and the particles are not metabolized. On the other hand, a 5% glucose solution is isotonic when initially in-fused intravenously, but glucose is metabolized, so the net ef-fect is that of infusing a hypotonic solution. It is important to note the relative contributions of the vari-ous plasma components to the total osmolal concentration of plasma. All but about 20 of the 290 mOsm in each liter of nor-mal plasma are contributed by Na+ and its accompanying anions, principally Cl– and HCO3 –. Other cations and anions make a relatively small contribution. Although the concentra-tion of the plasma proteins is large when expressed in grams per liter, they normally contribute less than 2 mOsm/L because of their very high molecular weights. The major nonelectro-lytes of plasma are glucose and urea, which in the steady state are in equilibrium with cells. Their contributions to osmolality are normally about 5 mOsm/L each but can become quite large in hyperglycemia or uremia. The total plasma osmolality is important in assessing dehydration, overhydration, and other fluid and electrolyte abnormalities (Clinical Box 1–1). NONIONIC DIFFUSION Some weak acids and bases are quite soluble in cell mem-branes in the undissociated form, whereas they cannot cross membranes in the charged (ie, dissociated) form. Conse-quently, if molecules of the undissociated substance diffuse from one side of the membrane to the other and then dissoci-ate, there is appreciable net movement of the undissociated substance from one side of the membrane to the other. This phenomenon is called nonionic diffusion. DONNAN EFFECT When an ion on one side of a membrane cannot diffuse through the membrane, the distribution of other ions to which the membrane is permeable is affected in a predictable way. For example, the negative charge of a nondiffusible anion hin-ders diffusion of the diffusible cations and favors diffusion of the diffusible anions. Consider the following situation, X Y m K+ K+ Cl– Cl– Prot– CLINICAL BOX 1–1 Plasma Osmolality & Disease Unlike plant cells, which have rigid walls, animal cell mem-branes are flexible. Therefore, animal cells swell when exposed to extracellular hypotonicity and shrink when exposed to ex-tracellular hypertonicity. Cells contain ion channels and pumps that can be activated to offset moderate changes in osmolality; however, these can be overwhelmed under certain pathologies. Hyperosmolality can cause coma (hyperosmolar coma). Because of the predominant role of the major solutes and the deviation of plasma from an ideal solution, one can or-dinarily approximate the plasma osmolality within a few mosm/liter by using the following formula, in which the con-stants convert the clinical units to millimoles of solute per liter: Osmolality (mOsm/L) = 2[Na+] (mEq/L) + 0.055[Glucose] (mg/dL) + 0.36[BUN] (mg/dL) BUN is the blood urea nitrogen. The formula is also useful in calling attention to abnormally high concentrations of other solutes. An observed plasma osmolality (measured by freez-ing-point depression) that greatly exceeds the value pre-dicted by this formula probably indicates the presence of a foreign substance such as ethanol, mannitol (sometimes in-jected to shrink swollen cells osmotically), or poisons such as ethylene glycol or methanol (components of antifreeze). CHAPTER 1 General Principles & Energy Production in Medical Physiology 7 in which the membrane (m) between compartments X and Y is impermeable to charged proteins (Prot–) but freely perme-able to K+ and Cl–. Assume that the concentrations of the an-ions and of the cations on the two sides are initially equal. Cl– diffuses down its concentration gradient from Y to X, and some K+ moves with the negatively charged Cl– because of its opposite charge. Therefore [K+ x] > [K+ y] Furthermore, [K+ x] + [Cl– x] + [Prot– x] > [K+ y] + [Cl– y] that is, more osmotically active particles are on side X than on side Y. Donnan and Gibbs showed that in the presence of a nondif-fusible ion, the diffusible ions distribute themselves so that at equilibrium their concentration ratios are equal: [K+ x] = [Cl– y] [K+ y] [Cl– x] Cross-multiplying, [K+ x] + [Cl– x] = [K+ y] + [Cl– y] This is the Gibbs–Donnan equation. It holds for any pair of cations and anions of the same valence. The Donnan effect on the distribution of ions has three effects in the body introduced here and discussed below. First, because of charged proteins (Prot–) in cells, there are more osmotically active particles in cells than in interstitial fluid, and because animal cells have flexible walls, osmosis would make them swell and eventually rupture if it were not for Na, K ATPase pumping ions back out of cells. Thus, normal cell volume and pressure depend on Na, K ATPase. Second, because at equilibrium the distribution of permeant ions across the membrane (m in the example used here) is asym-metric, an electrical difference exists across the membrane whose magnitude can be determined by the Nernst equation. In the example used here, side X will be negative relative to side Y. The charges line up along the membrane, with the con-centration gradient for Cl– exactly balanced by the oppositely directed electrical gradient, and the same holds true for K+. Third, because there are more proteins in plasma than in interstitial fluid, there is a Donnan effect on ion movement across the capillary wall. FORCES ACTING ON IONS The forces acting across the cell membrane on each ion can be analyzed mathematically. Chloride ions (Cl–) are present in higher concentration in the ECF than in the cell interior, and they tend to diffuse along this concentration gradient into the cell. The interior of the cell is negative relative to the exterior, and chloride ions are pushed out of the cell along this electrical gradient. An equilibrium is reached between Cl– influx and Cl– efflux. The membrane potential at which this equilibrium exists is the equilibrium potential. Its magnitude can be calculated from the Nernst equation, as follows: ECl = RT ln [Clo –] FZCl [Cli –] where ECl = equilibrium potential for Cl– R = gas constant T = absolute temperature F = the faraday (number of coulombs per mole of charge) ZCl = valence of Cl– (–1) [Clo –] = Cl– concentration outside the cell [Cli –] = Cl– concentration inside the cell Converting from the natural log to the base 10 log and replacing some of the constants with numerical values, the equation becomes: ECl = 61.5 log [Cli –] at 37 °C [Clo –] Note that in converting to the simplified expression the con-centration ratio is reversed because the –1 valence of Cl– has been removed from the expression. The equilibrium potential for Cl– (ECl), calculated from the standard values listed in Table 1–1, is –70 mV, a value identi-cal to the measured resting membrane potential of –70 mV. Therefore, no forces other than those represented by the chemical and electrical gradients need be invoked to explain the distribution of Cl– across the membrane. A similar equilibrium potential can be calculated for K+ (EK): EK = RT ln [Ko +] = 61.5log [Ko +] at 37 °C FZK [Ki +] [Ki +] where EK = equilibrium potential for K+ ZK = valence of K+ (+1) [Ko +] = K+ concentration outside the cell [Ki +] = K+ concentration inside the cell R, T, and F as above In this case, the concentration gradient is outward and the electrical gradient inward. In mammalian spinal motor neu-rons, EK is –90 mV (Table 1–1). Because the resting mem-brane potential is –70 mV, there is somewhat more K+ in the neurons than can be accounted for by the electrical and chem-ical gradients. The situation for Na+ is quite different from that for K+ and Cl–. The direction of the chemical gradient for Na+ is inward, to the area where it is in lesser concentration, and the electrical gradient is in the same direction. ENa is +60 mV (Table 1–1). Because neither EK nor ENa is equal to the membrane potential, 8 SECTION I Cellular & Molecular Basis of Medical Physiology one would expect the cell to gradually gain Na+ and lose K+ if only passive electrical and chemical forces were acting across the membrane. However, the intracellular concentration of Na+ and K+ remain constant because of the action of the Na, K ATPase that actively transports Na+ out of the cell and K+ into the cell (against their respective electrochemical gradients). GENESIS OF THE MEMBRANE POTENTIAL The distribution of ions across the cell membrane and the na-ture of this membrane provide the explanation for the mem-brane potential. The concentration gradient for K+ facilitates its movement out of the cell via K+ channels, but its electrical gradient is in the opposite (inward) direction. Consequently, an equilibrium is reached in which the tendency of K+ to move out of the cell is balanced by its tendency to move into the cell, and at that equilibrium there is a slight excess of cations on the outside and anions on the inside. This condition is maintained by Na, K ATPase, which uses the energy of ATP to pump K+ back into the cell and keeps the intracellular concentration of Na+ low. Because the Na, K ATPase moves three Na+ out of the cell for every two K+ moved in, it also contributes to the membrane potential, and thus is termed an electrogenic pump. It should be emphasized that the number of ions re-sponsible for the membrane potential is a minute fraction of the total number present and that the total concentrations of positive and negative ions are equal everywhere except along the membrane. ENERGY PRODUCTION ENERGY TRANSFER Energy is stored in bonds between phosphoric acid residues and certain organic compounds. Because the energy of bond formation in some of these phosphates is particularly high, relatively large amounts of energy (10–12 kcal/mol) are re-leased when the bond is hydrolyzed. Compounds containing such bonds are called high-energy phosphate compounds. Not all organic phosphates are of the high-energy type. Many, like glucose 6-phosphate, are low-energy phosphates that on hydrolysis liberate 2–3 kcal/mol. Some of the intermediates formed in carbohydrate metabolism are high-energy phos-phates, but the most important high-energy phosphate com-pound is adenosine triphosphate (ATP). This ubiquitous molecule (Figure 1–4) is the energy storehouse of the body. On hydrolysis to adenosine diphosphate (ADP), it liberates energy directly to such processes as muscle contraction, active transport, and the synthesis of many chemical compounds. Loss of another phosphate to form adenosine monophosphate (AMP) releases more energy. Another group of high-energy compounds are the thioesters, the acyl derivatives of mercaptans. Coenzyme A (CoA) is a widely distributed mercaptan-containing adenine, ribose, pan-tothenic acid, and thioethanolamine (Figure 1–5). Reduced CoA (usually abbreviated HS–CoA) reacts with acyl groups (R–CO–) to form R–CO–S–CoA derivatives. A prime example is the reaction of HS-CoA with acetic acid to form acetylcoen-zyme A (acetyl-CoA), a compound of pivotal importance in intermediary metabolism. Because acetyl-CoA has a much higher energy content than acetic acid, it combines readily with substances in reactions that would otherwise require out-side energy. Acetyl-CoA is therefore often called “active ace-tate.” From the point of view of energetics, formation of 1 mol of any acyl-CoA compound is equivalent to the formation of 1 mol of ATP. BIOLOGIC OXIDATIONS Oxidation is the combination of a substance with O2, or loss of hydrogen, or loss of electrons. The corresponding reverse pro-cesses are called reduction. Biologic oxidations are catalyzed by specific enzymes. Cofactors (simple ions) or coenzymes (or-ganic, nonprotein substances) are accessory substances that TABLE 1–1 Concentration of some ions inside and outside mammalian spinal motor neurons. Concentration (mmol/L of H2O) Ion Inside Cell Outside Cell Equilibrium Potential (mV) Na+ 15.0 150.0 +60 K+ 150.0 5.5 –90 Cl– 9.0 125.0 –70 Resting membrane potential = –70 mV FIGURE 1–4 Energy-rich adenosine derivatives. Adenosine triphosphate is broken down into its backbone purine base and sugar (at right) as well as its high energy phosphate derivatives (across bot-tom). (Reproduced, with permission, from Murray RK et al: Harper’s Biochemistry, 26th ed. McGraw-Hill, 2003.) NH2 N N C O N N HO OH CH2 C H H H H O Adenine Ribose — — P O O− O — — P O− O — — P O O− O −O Adenosine 5'-monophosphate (AMP) Adenosine 5'-diphosphate (ADP) Adenosine 5'-triphosphate (ATP) CHAPTER 1 General Principles & Energy Production in Medical Physiology 9 usually act as carriers for products of the reaction. Unlike the enzymes, the coenzymes may catalyze a variety of reactions. A number of coenzymes serve as hydrogen acceptors. One common form of biologic oxidation is removal of hydrogen from an R–OH group, forming R=O. In such dehydrogenation reactions, nicotinamide adenine dinucleotide (NAD+) and dihy-dronicotinamide adenine dinucleotide phosphate (NADP+) pick up hydrogen, forming dihydronicotinamide adenine dinu-cleotide (NADH) and dihydronicotinamide adenine dinucleo-tide phosphate (NADPH) (Figure 1–6). The hydrogen is then transferred to the flavoprotein–cytochrome system, reoxidizing the NAD+ and NADP+. Flavin adenine dinucleotide (FAD) is formed when riboflavin is phosphorylated, forming flavin mononucleotide (FMN). FMN then combines with AMP, forming the dinucleotide. FAD can accept hydrogens in a simi-lar fashion, forming its hydro (FADH) and dihydro (FADH2) derivatives. The flavoprotein–cytochrome system is a chain of enzymes that transfers hydrogen to oxygen, forming water. This process occurs in the mitochondria. Each enzyme in the chain is reduced FIGURE 1–5 Coenzyme A (CoA) and its derivatives. Left: Formula of reduced coenzyme A (HS-CoA) with its components highlighted. Right: Formula for reaction of CoA with biologically important compounds to form thioesters. R, remainder of molecule. NH2 N N O OH CH2 H H H H Adenine Ribose 3-phosphate P O O O O− P O O O− Pyrophosphate Coenzyme A P O O O− O− CH2 C H3C H3C CH OH H N CH2 CH2 H N CH2 CH2 SH Thioethanolamine β-Alanine Pantothenic acid OH + R CoA HS CoA C S R HOH O + C O C O C O N N FIGURE 1–6 Structures of molecules important in oxidation reduction reactions to produce energy. Top: Formula of the oxidized form of nicotinamide adenine dinucleotide (NAD+). Nicotinamide adenine dinucleotide phosphate (NADP+) has an additional phosphate group at the location marked by the asterisk. Bottom: Reaction by which NAD+ and NADP+ become reduced to form NADH and NADPH. R, remainder of molecule; R’, hydrogen donor. NH2 N N CONH2 CONH2 +N H R N+ N N CH2O OCH2 H H H H H H H OH H O OH OH OH — — P O OH O — — P O− O O + R'H2 CONH2 H H R N + H+ + R' Adenine Ribose Ribose Nicotinamide Diphosphate Oxidized coenzyme Reduced coenzyme 10 SECTION I Cellular & Molecular Basis of Medical Physiology and then reoxidized as the hydrogen is passed down the line. Each of the enzymes is a protein with an attached nonprotein prosthetic group. The final enzyme in the chain is cytochrome c oxidase, which transfers hydrogens to O2, forming H2O. It con-tains two atoms of Fe and three of Cu and has 13 subunits. The principal process by which ATP is formed in the body is oxidative phosphorylation. This process harnesses the energy from a proton gradient across the mitochondrial membrane to produce the high-energy bond of ATP and is briefly outlined in Figure 1–7. Ninety percent of the O2 consumption in the basal state is mitochondrial, and 80% of this is coupled to ATP syn-thesis. About 27% of the ATP is used for protein synthesis, and about 24% is used by Na, K ATPase, 9% by gluconeogenesis, 6% by Ca2+ ATPase, 5% by myosin ATPase, and 3% by ureagenesis. MOLECULAR BUILDING BLOCKS NUCLEOSIDES, NUCLEOTIDES, & NUCLEIC ACIDS Nucleosides contain a sugar linked to a nitrogen-containing base. The physiologically important bases, purines and pyrim-idines, have ring structures (Figure 1–8). These structures are bound to ribose or 2-deoxyribose to complete the nucleoside. When inorganic phosphate is added to the nucleoside, a nucleo-tide is formed. Nucleosides and nucleotides form the backbone for RNA and DNA, as well as a variety of coenzymes and regula-tory molecules (eg, NAD+, NADP+, and ATP) of physiological importance (Table 1–2). Nucleic acids in the diet are digested and their constituent purines and pyrimidines absorbed, but most of the purines and pyrimidines are synthesized from amino acids, principally in the liver. The nucleotides and RNA and DNA are then synthesized. RNA is in dynamic equilibrium with the amino acid pool, but DNA, once formed, is metabolically sta-ble throughout life. The purines and pyrimidines released by the breakdown of nucleotides may be reused or catabolized. Minor amounts are excreted unchanged in the urine. The pyrimidines are catabolized to the β-amino acids, β-alanine and β-aminoisobutyrate. These amino acids have their amino group on β-carbon, rather than the α-carbon typ-ical to physiologically active amino acids. Because β-ami-noisobutyrate is a product of thymine degradation, it can serve as a measure of DNA turnover. The β-amino acids are further degraded to CO2 and NH3. Uric acid is formed by the breakdown of purines and by direct synthesis from 5-phosphoribosyl pyrophosphate (5-PRPP) and glutamine (Figure 1–9). In humans, uric acid is excreted in the urine, but in other mammals, uric acid is fur-ther oxidized to allantoin before excretion. The normal blood uric acid level in humans is approximately 4 mg/dL (0.24 mmol/L). In the kidney, uric acid is filtered, reabsorbed, and secreted. Normally, 98% of the filtered uric acid is reabsorbed and the remaining 2% makes up approximately 20% of the amount excreted. The remaining 80% comes from the tubular secretion. The uric acid excretion on a purine-free diet is about 0.5 g/24 h and on a regular diet about 1 g/24 h. Excess uric acid in the blood or urine is a characteristic of gout (Clin-ical Box 1–2). FIGURE 1–7 Simplified diagram of transport of protons across the inner and outer lamellas of the inner mitochondrial membrane. The electron transport system (flavoprotein-cytochrome system) helps create H+ movement from the inner to the outer lamella. Return movement of protons down the proton gradient generates ATP. FIGURE 1–8 Principal physiologically important purines and pyrimidines. Purine and pyrimidine structures are shown next to repre-sentative molecules from each group. Oxypurines and oxypyrimidines may form enol derivatives (hydroxypurines and hydroxypyrimidines) by migration of hydrogen to the oxygen substituents. Ou te r l am ell a In ne r la me lla H+ ATP ADP N N N N C C C CH C H H H 1 2 3 4 2 1 6 6 5 N C C C C H H H 3 4 5 7 8 9 Purine nucleus Pyrimidine nucleus Adenine: Guanine: Hypoxanthine: Xanthine: 6-Aminopurine 1-Amino-6-oxypurine 6-Oxypurine 2,6-Dioxypurine Cytosine: Uracil: Thymine: 4-Amino-2-oxypyrimidine 2,4-Dioxypyrimidine 5-Methyl-2,4-dioxypyrimidine N TABLE 1–2 Purine- and pyrimidine-containing compounds. Type of Compound Components Nucleoside Purine or pyrimidine plus ribose or 2-deoxyribose Nucleotide (mononucleotide) Nucleoside plus phosphoric acid residue Nucleic acid Many nucleotides forming double-helical struc-tures of two polynucleotide chains Nucleoprotein Nucleic acid plus one or more simple basic proteins Contain ribose Ribonucleic acids (RNA) Contain 2-deoxyribose Deoxyribonucleic acids (DNA) CHAPTER 1 General Principles & Energy Production in Medical Physiology 11 DNA Deoxyribonucleic acid (DNA) is found in bacteria, in the nu-clei of eukaryotic cells, and in mitochondria. It is made up of two extremely long nucleotide chains containing the bases ad-enine (A), guanine (G), thymine (T), and cytosine (C) (Figure 1–10). The chains are bound together by hydrogen bonding between the bases, with adenine bonding to thymine and gua-nine to cytosine. This stable association forms a double-helical structure (Figure 1–11). The double helical structure of DNA is compacted in the cell by association with histones, and fur-ther compacted into chromosomes. A diploid human cell contains 46 chromosomes. A fundamental unit of DNA, or a gene, can be defined as the sequence of DNA nucleotides that contain the information for the production of an ordered amino acid sequence for a single polypeptide chain. Interestingly, the protein encoded by a sin-gle gene may be subsequently divided into several different physiologically active proteins. Information is accumulating at an accelerating rate about the structure of genes and their regu-lation. The basic structure of a typical eukaryotic gene is shown in diagrammatic form in Figure 1–12. It is made up of a strand of DNA that includes coding and noncoding regions. In eukaryotes, unlike prokaryotes, the portions of the genes that dictate the formation of proteins are usually broken into several segments (exons) separated by segments that are not translated (introns). Near the transcription start site of the gene is a pro-moter, which is the site at which RNA polymerase and its cofactors bind. It often includes a thymidine–adenine–thymi-dine–adenine (TATA) sequence (TATA box), which ensures that transcription starts at the proper point. Farther out in the 5' region are regulatory elements, which include enhancer and silencer sequences. It has been estimated that each gene has an average of five regulatory sites. Regulatory sequences are some-times found in the 3'-flanking region as well. Gene mutations occur when the base sequence in the DNA is altered from its original sequence. Such alterations can affect protein structure and be passed on to daughter cells after cell division. Point mutations are single base substitutions. A vari-ety of chemical modifications (eg, alkylating or intercalating agents, or ionizing radiation) can lead to changes in DNA sequences and mutations. The collection of genes within the full expression of DNA from an organism is termed its genome. An indication of the complexity of DNA in the human haploid genome (the total genetic message) is its size; it is made up of 3 × 109 base pairs that can code for approxi-mately 30,000 genes. This genetic message is the blueprint for FIGURE 1–9 Synthesis and breakdown of uric acid. Adeno-sine is converted to hypoxanthine, which is then converted to xanthine, and xanthine is converted to uric acid. The latter two reactions are both catalyzed by xanthine oxidase. Guanosine is converted directly to xan-thine, while 5-PRPP and glutamine can be converted to uric acid. An additional oxidation of uric acid to allantoin occurs in some mammals. C NH C C HN C O N H O O O C Uric acid (excreted in humans) NH NH C C H2N C O N H O C Allantoin (excreted in other mammals) NH H Guanosine 5-PRPP + Glutamine Hypoxanthine Adenosine Xanthine oxidase Xanthine oxidase Xanthine CLINICAL BOX 1–2 Gout Gout is a disease characterized by recurrent attacks of ar-thritis; urate deposits in the joints, kidneys, and other tis-sues; and elevated blood and urine uric acid levels. The joint most commonly affected initially is the metatarsopha-langeal joint of the great toe. There are two forms of “pri-mary” gout. In one, uric acid production is increased be-cause of various enzyme abnormalities. In the other, there is a selective deficit in renal tubular transport of uric acid. In “secondary” gout, the uric acid levels in the body fluids are elevated as a result of decreased excretion or increased production secondary to some other disease process. For example, excretion is decreased in patients treated with thiazide diuretics and those with renal disease. Production is increased in leukemia and pneumonia because of in-creased breakdown of uric acid-rich white blood cells. The treatment of gout is aimed at relieving the acute ar-thritis with drugs such as colchicine or nonsteroidal anti-in-flammatory agents and decreasing the uric acid level in the blood. Colchicine does not affect uric acid metabolism, and it apparently relieves gouty attacks by inhibiting the phagocytosis of uric acid crystals by leukocytes, a process that in some way produces the joint symptoms. Phenylb-utazone and probenecid inhibit uric acid reabsorption in the renal tubules. Allopurinol, which directly inhibits xan-thine oxidase in the purine degradation pathway, is one of the drugs used to decrease uric acid production. 12 SECTION I Cellular & Molecular Basis of Medical Physiology FIGURE 1–10 Basic structure of nucleotides and nucleic acids. A) At left, the nucleotide cytosine is shown with deoxyribose and at right with ribose as the principal sugar. B) Purine bases adenine and guanine are bound to each other or to pyrimidine bases, cytosine, thymine, or uracil via a phosphodiester backbone between 2'-deoxyribosyl moieties attached to the nucleobases by an N-glycosidic bond. Note that the backbone has a polarity (ie, a 5' and a 3' direction). Thymine is only found in DNA, while the uracil is only found in RNA. NH2 N N N N CH3 NH2 N N O O NH N O NH N NH2 O O Uracil (RNA only) Phosphate Sugar Nucleotide Adenine (DNA and RNA) Guanine (DNA and RNA) Cytosine (DNA and RNA) Thymine (DNA only) O N HN N N O O O– O P O CH2 O O– O P O CH2 O O O O– O P O CH2 O O O O– O P O CH2 O O O O– O P O CH2 O A B N N –O NH2 C O H C C OH H C H P H N O O N CH2 O Phosphate Base (cytosine) Sugar (ribose) Typical ribonucleotide NH2 C O H C C OH H C H P H H O O CH2 O Phosphate Base (cytosine) Sugar (deoxyribose) Typical deoxyribonucleotide OH O– –O O– CHAPTER 1 General Principles & Energy Production in Medical Physiology 13 the heritable characteristics of the cell and its descendants. The proteins formed from the DNA blueprint include all the enzymes, and these in turn control the metabolism of the cell. Each nucleated somatic cell in the body contains the full genetic message, yet there is great differentiation and special-ization in the functions of the various types of adult cells. Only small parts of the message are normally transcribed. Thus, the genetic message is normally maintained in a repressed state. However, genes are controlled both spatially and temporally. First, under physiological conditions, the double helix requires highly regulated interaction by proteins to unravel for replication, transcription, or both. REPLICATION: MITOSIS & MEIOSIS At the time of each somatic cell division (mitosis), the two DNA chains separate, each serving as a template for the syn-thesis of a new complementary chain. DNA polymerase cata-lyzes this reaction. One of the double helices thus formed goes to one daughter cell and one goes to the other, so the amount of DNA in each daughter cell is the same as that in the parent cell. The life cycle of the cell that begins after mitosis is highly regulated and is termed the cell cycle (Figure 1–13). The G1 (or Gap 1) phase represents a period of cell growth and divides the end of mitosis from the DNA synthesis (or S) phase. Fol-lowing DNA synthesis, the cell enters another period of cell growth, the G2 (Gap 2) phase. The ending of this stage is marked by chromosome condensation and the beginning of mitosis (M stage). In germ cells, reduction division (meiosis) takes place dur-ing maturation. The net result is that one of each pair of chro-mosomes ends up in each mature germ cell; consequently, each mature germ cell contains half the amount of chromoso-mal material found in somatic cells. Therefore, when a sperm unites with an ovum, the resulting zygote has the full comple-ment of DNA, half of which came from the father and half from the mother. The term “ploidy” is sometimes used to refer to the number of chromosomes in cells. Normal resting dip-loid cells are euploid and become tetraploid just before divi-sion. Aneuploidy is the condition in which a cell contains other than the haploid number of chromosomes or an exact multiple of it, and this condition is common in cancerous cells. RNA The strands of the DNA double helix not only replicate them-selves, but also serve as templates by lining up complementary bases for the formation in the nucleus of ribonucleic acids (RNA). RNA differs from DNA in that it is single-stranded, has uracil in place of thymine, and its sugar moiety is ribose rather than 2'-deoxyribose (Figure 1–13). The production of RNA from DNA is called transcription. Transcription can lead to several types of RNA including: messenger RNA (mRNA), transfer RNA (tRNA), ribosomal RNA (rRNA), and other RNAs. Transcription is catalyzed by various forms of RNA polymerase. FIGURE 1–11 Double-helical structure of DNA. The compact structure has an approximately 2.0 nm thickness and 3.4 nm between full turns of the helix that contains both major and minor grooves. The structure is maintained in the double helix by hydrogen bonding be-tween purines and pyrimidines across individual strands of DNA. Adenine (A) is bound to thymine (T) and cytosine (C) to guanine (G). (Reproduced with permission from Murray RK et al: Harper’s Biochemistry, 26th ed. McGraw-Hill, 2003.) 2.0 nm 3.4 nm Minor groove Major groove G C G G C C A T A A G C A A T T T T FIGURE 1–12 Diagram of the components of a typical eukaryotic gene. The region that produces introns and exons is flanked by non-coding regions. The 5'-flanking region contains stretches of DNA that interact with proteins to facilitate or inhibit transcription. The 3'-flanking re-gion contains the poly(A) addition site. (Modified from Murray RK et al: Harper’s Biochemistry, 26th ed. McGraw-Hill, 2003.) DNA 5' Regulatory region Basal promoter region Transcription start site 5' Noncoding region Intron Exon Exon Poly(A) addition site 3' Noncoding region 3' CAAT TATA AATAAA 14 SECTION I Cellular & Molecular Basis of Medical Physiology Typical transcription of an mRNA is shown in Figure 1–14. When suitably activated, transcription of the gene into a pre-mRNA starts at the cap site and ends about 20 bases beyond the AATAAA sequence. The RNA transcript is capped in the nucleus by addition of 7-methylguanosine triphosphate to the 5' end; this cap is necessary for proper binding to the ribosome. A poly(A) tail of about 100 bases is added to the untranslated segment at the 3' end to help maintain the stability of the mRNA. The pre-mRNA formed by capping and addition of the poly(A) tail is then processed by elimination of the introns, and once this posttranscriptional modification is complete, the mature mRNA moves to the cytoplasm. Posttranscriptional modification of the pre-mRNA is a regulated process where differential splicing can occur to form more than one mRNA from a single pre-mRNA. The introns of some genes are elimi-nated by spliceosomes, complex units that are made up of small RNAs and proteins. Other introns are eliminated by self-splicing by the RNA they contain. Because of introns and splic-ing, more than one mRNA can be formed from the same gene. Most forms of RNA in the cell are involved in translation, or protein synthesis. A brief outline of the transition from transcription to translation is shown in Figure 1–15. In the cytoplasm, ribosomes provide a template for tRNA to deliver specific amino acids to a growing polypeptide chain based on specific sequences in mRNA. The mRNA molecules are smaller than the DNA molecules, and each represents a FIGURE 1–13 Sequence of events during the cell cycle. Immediately following mitosis (M) the cell enters a gap phase (G1) before a DNA synthesis phase (S) a second gap phase (G2) and back to mitosis. Collectively G1, S, and G2 phases are referred to as interphase (I). Mitosis G2 Final growth and activity before mitosis S DNA replication Interphase Mitotic phase G1 Centrioles replicate Telophase Anaphase Metaphase Prophase Cytokinesis CHAPTER 1 General Principles & Energy Production in Medical Physiology 15 transcript of a small segment of the DNA chain. For comparison, the molecules of tRNA contain only 70–80 nitrogenous bases, compared with hundreds in mRNA and 3 billion in DNA. AMINO ACIDS & PROTEINS AMINO ACIDS Amino acids that form the basic building blocks for proteins are identified in Table 1–3. These amino acids are often re-ferred to by their corresponding three-letter, or single-letter abbreviations. Various other important amino acids such as ornithine, 5-hydroxytryptophan, L-dopa, taurine, and thy-roxine (T4) occur in the body but are not found in proteins. In higher animals, the L isomers of the amino acids are the only naturally occurring forms in proteins. The L isomers of hormones such as thyroxine are much more active than the D isomers. The amino acids are acidic, neutral, or basic in re-action, depending on the relative proportions of free acidic (–COOH) or basic (–NH2) groups in the molecule. Some of the amino acids are nutritionally essential amino acids, that is, they must be obtained in the diet, because they cannot be made in the body. Arginine and histidine must be provided through diet during times of rapid growth or recovery from illness and are termed conditionally essential. All others are nonessential amino acids in the sense that they can be syn-thesized in vivo in amounts sufficient to meet metabolic needs. FIGURE 1–14 Transcription of a typical mRNA. Steps in trans-cription from a typical gene to a processed mRNA are shown. Cap, cap site. (Modified from Baxter JD: Principles of endocrinology. In: Cecil Textbook of Medicine, 16th ed. Wyngaarden JB, Smith LH Jr (editors). Saunders, 1982.) Poly(A) Poly(A) Poly(A) Gene mRNA Pre-mRNA RNA processing Flanking DNA Introns Exons Cap Transcription Flanking DNA Translation FIGURE 1–15 Diagrammatic outline of transcription to translation. From the DNA molecule, a messenger RNA is produced and presented to the ribosome. It is at the ribosome where charged tRNA match up with their complementary codons of mRNA to position the amino acid for growth of the polypeptide chain. DNA and RNA are represented as lines with multiple short projections representing the individual bases. Small boxes labeled A represent individual amino acids. Posttranscriptional modification Posttranslational modification Translation DNA Chain separation Amino acid tRNA adenylate tRNA-amino acid-adenylate complex A3 A2 A1 Peptide chain Messenger RNA Coding triplets for A3 A4 A2 A4 A1 Ribosome Activating enzyme RNA strand formed on DNA strand (transcription) 16 SECTION I Cellular & Molecular Basis of Medical Physiology THE AMINO ACID POOL Although small amounts of proteins are absorbed from the gastrointestinal tract and some peptides are also absorbed, most ingested proteins are digested and their constituent ami-no acids absorbed. The body’s own proteins are being contin-uously hydrolyzed to amino acids and resynthesized. The turnover rate of endogenous proteins averages 80–100 g/d, be-ing highest in the intestinal mucosa and practically nil in the extracellular structural protein, collagen. The amino acids formed by endogenous protein breakdown are identical to those derived from ingested protein. Together they form a common amino acid pool that supplies the needs of the body (Figure 1–16). PROTEINS Proteins are made up of large numbers of amino acids linked into chains by peptide bonds joining the amino group of one amino acid to the carboxyl group of the next (Figure 1–17). In addition, some proteins contain carbohydrates (glycopro-teins) and lipids (lipoproteins). Smaller chains of amino acids are called peptides or polypeptides. The boundaries between peptides, polypeptides, and proteins are not well defined. For this text, amino acid chains containing 2–10 amino acid resi-dues are called peptides, chains containing more than 10 but fewer than 100 amino acid residues are called polypeptides, and chains containing 100 or more amino acid residues are called proteins. TABLE 1–3 Amino acids found in proteins. Amino acids with aliphatic side chains Amino acids with acidic side chains, or their amides Alanine (Ala, A) Aspartic acid (Asp, D) Valine (Val, V) Asparagine (Asn, N) Leucine (Leu, L) Glutamine (Gln, Q) Isoleucine (IIe, I) Glutamic acid (Glu, E) Hydroxyl-substituted amino acids γ-Carboxyglutamic acidb (Gla) Serine (Ser, S) Amino acids with side chains containing basic groups Threonine (Thr, T) Argininec (Arg, R) Sulfur-containing amino acids Lysine (Lys, K) Cysteine (Cys, C) Hydroxylysineb (Hyl) Methionine (Met, M) Histidinec (His, H) Selenocysteinea Imino acids (contain imino group but no amino group) Amino acids with aromatic ring side chains Proline (Pro, P) Phenylalanine (Phe, F) 4-Hydroxyprolineb (Hyp) Tyrosine (Tyr, Y) 3-Hydroxyprolineb Tryptophan (Trp, W) Those in bold type are the nutritionally essential amino acids. The generally accepted three-letter and one-letter abbreviations for the amino acids are shown in parentheses. aSelenocysteine is a rare amino acid in which the sulfur of cysteine is replaced by selenium. The codon UGA is usually a stop codon, but in certain situations it codes for selenocysteine. bThere are no tRNAs for these four amino acids; they are formed by post-translational modification of the corresponding unmodified amino acid in peptide linkage. There are tRNAs for selenocysteine and the remaining 20 amino acids, and they are incorporated into peptides and proteins under direct genetic control. cArginine and histidine are sometimes called “conditionally essential”—they are not necessary for maintenance of nitrogen balance, but are needed for normal growth. FIGURE 1–16 Amino acids in the body. There is an extensive network of amino acid turnover in the body. Boxes represent large pools of amino acids and some of the common interchanges are rep-resented by arrows. Note that most amino acids come from the diet and end up in protein, however, a large portion of amino acids are in-terconverted and can feed into and out of a common metabolic pool through amination reactions. Inert protein (hair, etc) Amino acid pool Body protein Diet Urea NH4 + Common metabolic pool Transamination Amination Deamination Purines, pyrimidines Hormones, neurotransmitters Creatine Urinary excretion CHAPTER 1 General Principles & Energy Production in Medical Physiology 17 The order of the amino acids in the peptide chains is called the primary structure of a protein. The chains are twisted and folded in complex ways, and the term secondary structure of a protein refers to the spatial arrangement produced by the twisting and folding. A common secondary structure is a regu-lar coil with 3.7 amino acid residues per turn (α-helix). Another common secondary structure is a β-sheet. An anti-parallel β-sheet is formed when extended polypeptide chains fold back and forth on one another and hydrogen bonding occurs between the peptide bonds on neighboring chains. Par-allel β-sheets between polypeptide chains also occur. The ter-tiary structure of a protein is the arrangement of the twisted chains into layers, crystals, or fibers. Many protein molecules are made of several proteins, or subunits (eg, hemoglobin), and the term quaternary structure is used to refer to the arrangement of the subunits into a functional structure. PROTEIN SYNTHESIS The process of protein synthesis, translation, is the conversion of information encoded in mRNA to a protein (Figure 1–15). As described previously, when a definitive mRNA reaches a ri-bosome in the cytoplasm, it dictates the formation of a polypep-tide chain. Amino acids in the cytoplasm are activated by combination with an enzyme and adenosine monophosphate (adenylate), and each activated amino acid then combines with a specific molecule of tRNA. There is at least one tRNA for each of the 20 unmodified amino acids found in large quantities in the body proteins of animals, but some amino acids have more than one tRNA. The tRNA–amino acid–adenylate complex is next attached to the mRNA template, a process that occurs in the ribosomes. The tRNA “recognizes” the proper spot to attach on the mRNA template because it has on its active end a set of three bases that are complementary to a set of three bases in a particular spot on the mRNA chain. The genetic code is made up of such triplets (codons), sequences of three purine, pyrimi-dine, or purine and pyrimidine bases; each codon stands for a particular amino acid. Translation typically starts in the ribosomes with an AUG (transcribed from ATG in the gene), which codes for methio-nine. The amino terminal amino acid is then added, and the chain is lengthened one amino acid at a time. The mRNA attaches to the 40S subunit of the ribosome during protein synthesis, the polypeptide chain being formed attaches to the 60S subunit, and the tRNA attaches to both. As the amino acids are added in the order dictated by the codon, the ribo-some moves along the mRNA molecule like a bead on a string. Translation stops at one of three stop, or nonsense, codons (UGA, UAA, or UAG), and the polypeptide chain is released. The tRNA molecules are used again. The mRNA molecules are typically reused approximately 10 times before being replaced. It is common to have more than one ribosome on a given mRNA chain at a time. The mRNA chain plus its collection of ribosomes is visible under the electron micro-scope as an aggregation of ribosomes called a polyribosome. POSTTRANSLATIONAL MODIFICATION After the polypeptide chain is formed, it “folds” into its biolog-ical form and can be further modified to the final protein by one or more of a combination of reactions that include hy-droxylation, carboxylation, glycosylation, or phosphorylation of amino acid residues; cleavage of peptide bonds that con-verts a larger polypeptide to a smaller form; and the further folding, packaging, or folding and packaging of the protein into its ultimate, often complex configuration. Protein folding is a complex process that is dictated primarily by the sequence of the amino acids in the polypeptide chain. In some instances, however, nascent proteins associate with other proteins called chaperones, which prevent inappropriate contacts with other proteins and ensure that the final “proper” conformation of the nascent protein is reached. Proteins also contain information that helps to direct them to individual cell compartments. Many proteins that are going to be secreted or stored in organelles and most transmembrane proteins have at their amino terminal a signal peptide (leader sequence) that guides them into the endoplasmic reticulum. The sequence is made up of 15 to 30 predominantly hydropho-bic amino acid residues. The signal peptide, once synthesized, binds to a signal recognition particle (SRP), a complex mole-cule made up of six polypeptides and 7S RNA, one of the small RNAs. The SRP stops translation until it binds to a translocon, a pore in the endoplasmic reticulum that is a heterotrimeric structure made up of Sec 61 proteins. The ribosome also binds, and the signal peptide leads the growing peptide chain into the cavity of the endoplasmic reticulum (Figure 1–18). The signal FIGURE 1–17 Amino acid structure and formation of peptide bonds. The dashed line shows where peptide bonds are formed be-tween two amino acids. The highlighted area is released as H2O. R, remainder of the amino acid. For example, in glycine, R = H; in glutamate, R = —(CH2)2—COO–. H H H C OH H–N R O R C H C H N O C H C R O N C H Amino acid Polypeptide chain 18 SECTION I Cellular & Molecular Basis of Medical Physiology peptide is next cleaved from the rest of the peptide by a signal peptidase while the rest of the peptide chain is still being syn-thesized. SRPs are not the only signals that help to direct pro-teins to their proper place in or out of the cell; other signal sequences, posttranslational modifications, or both (eg, glyco-sylation) can serve this function. PROTEIN DEGRADATION Like protein synthesis, protein degradation is a carefully regu-lated, complex process. It has been estimated that overall, up to 30% of newly produced proteins are abnormal, such as can oc-cur during improper folding. Aged normal proteins also need to be removed as they are replaced. Conjugation of proteins to the 74-amino-acid polypeptide ubiquitin marks them for degrada-tion. This polypeptide is highly conserved and is present in spe-cies ranging from bacteria to humans. The process of binding ubiquitin is called ubiquitination, and in some instances, mul-tiple ubiquitin molecules bind (polyubiquitination). Ubiquiti-nation of cytoplasmic proteins, including integral proteins of the endoplasmic reticulum, marks the proteins for degradation in multisubunit proteolytic particles, or proteasomes. Ubiquit-ination of membrane proteins, such as the growth hormone re-ceptors, also marks them for degradation, however these can be degraded in lysosomes as well as via the proteasomes. There is an obvious balance between the rate of production of a protein and its destruction, so ubiquitin conjugation is of major importance in cellular physiology. The rates at which individual proteins are metabolized vary, and the body has mechanisms by which abnormal proteins are recognized and degraded more rapidly than normal body constituents. For example, abnormal hemoglobins are metabolized rapidly in individuals with congenital hemoglobinopathies. CATABOLISM OF AMINO ACIDS The short-chain fragments produced by amino acid, carbohy-drate, and fat catabolism are very similar (see below). From this common metabolic pool of intermediates, carbohy-drates, proteins, and fats can be synthesized. These fragments can enter the citric acid cycle, a final common pathway of ca-tabolism, in which they are broken down to hydrogen atoms and CO2. Interconversion of amino acids involve transfer, re-moval, or formation of amino groups. Transamination reac-tions, conversion of one amino acid to the corresponding keto acid with simultaneous conversion of another keto acid to an amino acid, occur in many tissues: Alanine + α-Ketoglutarate → ← Pyruvate + Glutamate The transaminases involved are also present in the circula-tion. When damage to many active cells occurs as a result of a pathologic process, serum transaminase levels rise. An exam-ple is the rise in plasma aspartate aminotransferase (AST) following myocardial infarction. Oxidative deamination of amino acids occurs in the liver. An imino acid is formed by dehydrogenation, and this com-pound is hydrolyzed to the corresponding keto acid, with pro-duction of NH4 +: Amino acid + NAD+ → Imino acid + NADH + H+ Imino acid + H2O → Keto acid + NH4 + Interconversions between the amino acid pool and the common metabolic pool are summarized in Figure 1–19. Leucine, isoleucine, phenylalanine, and tyrosine are said to be ketogenic because they are converted to the ketone body ace-toacetate (see below). Alanine and many other amino acids are glucogenic or gluconeogenic; that is, they give rise to compounds that can readily be converted to glucose. UREA FORMATION Most of the NH4 + formed by deamination of amino acids in the liver is converted to urea, and the urea is excreted in the urine. The NH4 + forms carbamoyl phosphate, and in the mitochon-dria it is transferred to ornithine, forming citrulline. The en-zyme involved is ornithine carbamoyltransferase. Citrulline is converted to arginine, after which urea is split off and ornithine is regenerated (urea cycle; Figure 1–20). The overall reaction in the urea cycle consumes 3 ATP (not shown) and thus requires significant energy. Most of the urea is formed in the liver, and in severe liver disease the blood urea nitrogen (BUN) falls and blood NH3 rises (see Chapter 29). Congenital deficiency of or-nithine carbamoyltransferase can also lead to NH3 intoxication, even in individuals who are heterozygous for this deficiency. FIGURE 1–18 Translation of protein into endoplasmic reticulum according to the signal hypothesis. The ribosomes syn-thesizing a protein move along the mRNA from the 5' to the 3' end. When the signal peptide of a protein destined for secretion, the cell membrane, or lysosomes emerges from the large unit of the ribosome, it binds to a signal recognition particle (SRP), and this arrests further translation until it binds to the translocon on the endoplasmic reticu-lum. N, amino end of protein; C, carboxyl end of protein. (Reproduced, with permission, from Perara E, Lingappa VR: Transport of proteins into and across the endoplasmic reticulum membrane. In: Protein Transfer and Organelle Biogenesis. Das RC, Robbins PW (editors). Academic Press, 1988.) 5' 3' N N N N N N N N C C C C UAA SRP CHAPTER 1 General Principles & Energy Production in Medical Physiology 19 METABOLIC FUNCTIONS OF AMINO ACIDS In addition to providing the basic building blocks for proteins, amino acids also have metabolic functions. Thyroid hor-mones, catecholamines, histamine, serotonin, melatonin, and intermediates in the urea cycle are formed from specific ami-no acids. Methionine and cysteine provide the sulfur con-tained in proteins, CoA, taurine, and other biologically important compounds. Methionine is converted into S-ade-nosylmethionine, which is the active methylating agent in the synthesis of compounds such as epinephrine. CARBOHYDRATES Carbohydrates are organic molecules made of equal amounts of carbon and H2O. The simple sugars, or monosaccharides, including pentoses (5 carbons; eg, ribose) and hexoses (6 car-bons; eg, glucose) perform both structural (eg, as part of nu-cleotides discussed previously) and functional roles (eg, inositol 1,4,5 trisphosphate acts as a cellular signaling mole-cules) in the body. Monosaccharides can be linked together to form disaccharides (eg, sucrose), or polysaccharides (eg, gly-cogen). The placement of sugar moieties onto proteins (glyco-proteins) aids in cellular targeting, and in the case of some FIGURE 1–19 Involvement of the citric acid cycle in transamination and gluconeogenesis. The bold arrows indicate the main pathway of gluconeogenesis. Note the many entry positions for groups of amino acids into the citric acid cycle. (Reproduced with permission from Murray RK et al: Harper’s Biochemistry, 26th ed. McGraw-Hill, 2003.) Transaminase Transaminase Transaminase Phosphoenolpyruvate carboxykinase Oxaloacetate Aspartate Citrate α-Ketoglutarate Succinyl-CoA Fumarate Phosphoenolpyruvate CO2 CO2 Pyruvate Alanine Acetyl-CoA Glutamate Histidine Proline Glutamine Arginine Isoleucine Methionine Valine Hydroxyproline Serine Cysteine Threonine Glycine Tyrosine Phenylalanine Propionate Glucose Tryptophan Lactate FIGURE 1–20 Urea cycle. The processing of NH3 to urea for ex-cretion contains several coordinative steps in both the cytoplasm (Cy-to) and the mitochondria (Mito). The production of carbamoyl phosphate and its conversion to citrulline occurs in the mitochondria, whereas other processes are in the cytoplasm. NH2 + NH3 + NH3 + NH4 + NH3 H3N+ Argininosuccinate H2N C HN COO− COO− HC (CH2)3 (CH2)3 HC — — O NH3 + H2N C Pi HN COO− HC (CH2)3 — — Fumarate Aspartate Citrulline + NO Arginine Carbamoyl phosphate Urea Ornithine O NH2 C NH2 — — Cyto Mito 20 SECTION I Cellular & Molecular Basis of Medical Physiology receptors, recognition of signaling molecules. In this section we will discuss a major role for carbohydrates in physiology, the production and storage of energy. Dietary carbohydrates are for the most part polymers of hexoses, of which the most important are glucose, galactose, and fructose (Figure 1–21). Most of the monosaccharides occurring in the body are the D isomers. The principal prod-uct of carbohydrate digestion and the principal circulating sugar is glucose. The normal fasting level of plasma glucose in peripheral venous blood is 70 to 110 mg/dL (3.9–6.1 mmol/ L). In arterial blood, the plasma glucose level is 15 to 30 mg/ dL higher than in venous blood. Once it enters the cells, glucose is normally phosphorylated to form glucose 6-phosphate. The enzyme that catalyzes this reaction is hexokinase. In the liver, there is an additional enzyme called glucokinase, which has greater specificity for glucose and which, unlike hexokinase, is increased by insulin and decreased in starvation and diabetes. The glucose 6-phos-phate is either polymerized into glycogen or catabolized. The process of glycogen formation is called glycogenesis, and gly-cogen breakdown is called glycogenolysis. Glycogen, the stor-age form of glucose, is present in most body tissues, but the major supplies are in the liver and skeletal muscle. The break-down of glucose to pyruvate or lactate (or both) is called gly-colysis. Glucose catabolism proceeds via cleavage through fructose to trioses or via oxidation and decarboxylation to pentoses. The pathway to pyruvate through the trioses is the Embden–Meyerhof pathway, and that through 6-phospho-gluconate and the pentoses is the direct oxidative pathway (hexose monophosphate shunt). Pyruvate is converted to acetyl-CoA. Interconversions between carbohydrate, fat, and protein include conversion of the glycerol from fats to dihy-droxyacetone phosphate and conversion of a number of amino acids with carbon skeletons resembling intermediates in the Embden–Meyerhof pathway and citric acid cycle to these inter-mediates by deamination. In this way, and by conversion of lac-tate to glucose, nonglucose molecules can be converted to glucose (gluconeogenesis). Glucose can be converted to fats through acetyl-CoA, but because the conversion of pyruvate to acetyl-CoA, unlike most reactions in glycolysis, is irreversible, fats are not converted to glucose via this pathway. There is therefore very little net conversion of fats to carbohydrates in the body because, except for the quantitatively unimportant production from glycerol, there is no pathway for conversion. CITRIC ACID CYCLE The citric acid cycle (Krebs cycle, tricarboxylic acid cycle) is a sequence of reactions in which acetyl-CoA is metabolized to CO2 and H atoms. Acetyl-CoA is first condensed with the anion of a four-carbon acid, oxaloacetate, to form citrate and HS-CoA. In a series of seven subsequent reactions, 2CO2 mol-ecules are split off, regenerating oxaloacetate (Figure 1–22). Four pairs of H atoms are transferred to the flavoprotein– cytochrome chain, producing 12ATP and 4H2O, of which 2H2O is used in the cycle. The citric acid cycle is the common pathway for oxidation to CO2 and H2O of carbohydrate, fat, and some amino acids. The major entry into it is through acetyl-CoA, but a number of amino acids can be converted to citric acid cycle intermediates by deamination. The citric acid cycle requires O2 and does not function under anaerobic conditions. ENERGY PRODUCTION The net production of energy-rich phosphate compounds during the metabolism of glucose and glycogen to pyruvate depends on whether metabolism occurs via the Embden– Meyerhof pathway or the hexose monophosphate shunt. By oxidation at the substrate level, the conversion of 1 mol of phosphoglyceraldehyde to phosphoglycerate generates 1 mol of ATP, and the conversion of 1 mol of phosphoenolpyruvate to pyruvate generates another. Because 1 mol of glucose 6-phosphate produces, via the Embden–Meyerhof pathway, 2 mol of phosphoglyceraldehyde, 4 mol of ATP is generated per mole of glucose metabolized to pyruvate. All these reactions occur in the absence of O2 and consequently represent anaer-obic production of energy. However, 1 mol of ATP is used in forming fructose 1,6-diphosphate from fructose 6-phosphate and 1 mol in phosphorylating glucose when it enters the cell. Consequently, when pyruvate is formed anaerobically from glycogen, there is a net production of 3 mol of ATP per mole of glucose 6-phosphate; however, when pyruvate is formed from 1 mol of blood glucose, the net gain is only 2 mol of ATP. A supply of NAD+ is necessary for the conversion of phos-phoglyceraldehyde to phosphoglycerate. Under anaerobic conditions (anaerobic glycolysis), a block of glycolysis at the phosphoglyceraldehyde conversion step might be expected to develop as soon as the available NAD+ is converted to NADH. However, pyruvate can accept hydrogen from NADH, form-ing NAD+ and lactate: Pyruvate + NADH → ← Lactate + NAD+ In this way, glucose metabolism and energy production can continue for a while without O2. The lactate that accumulates is converted back to pyruvate when the O2 supply is restored, with NADH transferring its hydrogen to the flavoprotein– cytochrome chain. FIGURE 1–21 Structures of principal dietary hexoses. Glu-cose, galactose, and fructose are shown in their naturally occurring D isomers. — — — — C O H C H HO C OH H C OH H C OH H CH2OH C O — — C O H C H HO C OH H C H HO C OH H CH2OH C H HO C OH H C OH H CH2OH CH2OH D-Glucose D-Galactose D-Fructose CHAPTER 1 General Principles & Energy Production in Medical Physiology 21 During aerobic glycolysis, the net production of ATP is 19 times greater than the two ATPs formed under anaerobic con-ditions. Six ATPs are formed by oxidation via the flavopro-tein–cytochrome chain of the two NADHs produced when 2 mol of phosphoglyceraldehyde is converted to phosphoglyc-erate (Figure 1–22), six ATPs are formed from the two NADHs produced when 2 mol of pyruvate is converted to acetyl-CoA, and 24 ATPs are formed during the subsequent two turns of the citric acid cycle. Of these, 18 are formed by oxidation of six NADHs, 4 by oxidation of two FADH2s, and 2 by oxidation at the substrate level when succinyl-CoA is con-verted to succinate. This reaction actually produces GTP, but the GTP is converted to ATP. Thus, the net production of ATP per mol of blood glucose metabolized aerobically via the Embden–Meyerhof pathway and citric acid cycle is 2 + [2 × 3] + [2 × 3] + [2 × 12] = 38. Glucose oxidation via the hexose monophosphate shunt generates large amounts of NADPH. A supply of this reduced coenzyme is essential for many metabolic processes. The pentoses formed in the process are building blocks for nucleotides (see below). The amount of ATP generated depends on the amount of NADPH converted to NADH and then oxidized. “DIRECTIONAL-FLOW VALVES” Metabolism is regulated by a variety of hormones and other fac-tors. To bring about any net change in a particular metabolic process, regulatory factors obviously must drive a chemical re-action in one direction. Most of the reactions in intermediary metabolism are freely reversible, but there are a number of “di-rectional-flow valves,” ie, reactions that proceed in one direc-tion under the influence of one enzyme or transport mechanism and in the opposite direction under the influence of another. Five examples in the intermediary metabolism of carbohydrate are shown in Figure 1–23. The different pathways for fatty acid synthesis and catabolism (see below) are another example. Reg-ulatory factors exert their influence on metabolism by acting di-rectly or indirectly at these directional-flow valves. GLYCOGEN SYNTHESIS & BREAKDOWN Glycogen is a branched glucose polymer with two types of gly-coside linkages: 1:4α and 1:6α (Figure 1–24). It is synthesized on glycogenin, a protein primer, from glucose 1-phosphate via uridine diphosphoglucose (UDPG). The enzyme glycogen synthase catalyses the final synthetic step. The availability of FIGURE 1–22 Citric acid cycle. The numbers (6C, 5C, etc) indicate the number of carbon atoms in each of the intermediates. The conversion of pyruvate to acetyl-CoA and each turn of the cycle provide four NADH and one FADH2 for oxidation via the flavoprotein-cytochrome chain plus formation of one GTP that is readily converted to ATP. P Pyruvate 3C Acetyl-CoA 2C Oxaloacetate 4C Malate 4C Fumarate 4C Succinate 4C Succinyl-CoA 4C Citrate 6C Isocitrate 6C α-Ketoglutarate 5C NAD+ NAD+ NADH + H+ NADH + H+ FAD GTP GDP FADH2 CO2 CO2 CO2 NAD+ NADH + H+ NAD+ NADH + H+ 22 SECTION I Cellular & Molecular Basis of Medical Physiology glycogenin is one of the factors determining the amount of glycogen synthesized. The breakdown of glycogen in 1:4α linkage is catalyzed by phosphorylase, whereas another en-zyme catalyzes the breakdown of glycogen in 1:6α linkage. FACTORS DETERMINING THE PLASMA GLUCOSE LEVEL The plasma glucose level at any given time is determined by the balance between the amount of glucose entering the bloodstream and the amount leaving it. The principal deter-minants are therefore the dietary intake; the rate of entry into the cells of muscle, adipose tissue, and other organs; and the glucostatic activity of the liver (Figure 1–25). Five percent of ingested glucose is promptly converted into glycogen in the liver, and 30–40% is converted into fat. The remainder is me-tabolized in muscle and other tissues. During fasting, liver gly-cogen is broken down and the liver adds glucose to the bloodstream. With more prolonged fasting, glycogen is de-pleted and there is increased gluconeogenesis from amino ac-ids and glycerol in the liver. Plasma glucose declines modestly to about 60 mg/dL during prolonged starvation in normal in-dividuals, but symptoms of hypoglycemia do not occur be-cause gluconeogenesis prevents any further fall. FIGURE 1–23 Directional flow valves in energy production reactions. In carbohydrate metabolism there are several reactions that proceed in one direction by one mechanism and in the other direction by a different mechanism, termed “directional-flow valves.” Five examples of these reactions are illustrated (numbered at left). The double line in ex-ample 5 represents the mitochondrial membrane. Pyruvate is converted to malate in mitochondria, and the malate diffuses out of the mitochon-dria to the cytosol, where it is converted to phosphoenolpyruvate. Pyruvate Pyruvate Oxaloacetate Malate Malate Oxaloacetate 5. Phosphoenolpyruvate 4. Fructose 6-phosphate Phosphoenolpyruvate carboxykinase Fructose 1,6-biphosphate Fructose 1,6-biphosphatase Phospho-fructokinase 3. Glucose 1-phosphate Glycogen Phosphorylase Glycogen synthase 2. Glucose 1. Glucose entry into cells and glucose exit from cells Glucose 6-phosphate Glucose 6-phosphatase Hexokinase Pyruvate kinase ADP ATP FIGURE 1–24 Glycogen formation and breakdown. Glycogen is the main storage for glucose in the cell. It is cycled: built up from glucose 6-phosphate when energy is stored and broken down to glucose 6-phosphate when energy is required. Note the intermediate glucose 1-phosphate and enzymatic control by phosphorylase a and glycogen kinase. CH2OH O CH2OH O O CH2OH O O O CH2OH O CH2O O O CH2OH O CH2OH O O CH2 O CH2OH O CH2OH O O O 1:6α linkage 1:4α linkage P O O− O− P O O− O− Glucose 1-phosphate Glucose 6-phosphate Uridine diphospho-glucose Glycogen Phosphorylase a Glycogen synthase CHAPTER 1 General Principles & Energy Production in Medical Physiology 23 METABOLISM OF HEXOSES OTHER THAN GLUCOSE Other hexoses that are absorbed from the intestine include ga-lactose, which is liberated by the digestion of lactose and con-verted to glucose in the body; and fructose, part of which is ingested and part produced by hydrolysis of sucrose. After phosphorylation, galactose reacts with uridine diphosphoglu-cose (UDPG) to form uridine diphosphogalactose. The uri-dine diphosphogalactose is converted back to UDPG, and the UDPG functions in glycogen synthesis. This reaction is reversible, and conversion of UDPG to uridine diphospho-galactose provides the galactose necessary for formation of glycolipids and mucoproteins when dietary galactose intake is inadequate. The utilization of galactose, like that of glucose, depends on insulin. In the inborn error of metabolism known as galactosemia, there is a congenital deficiency of galactose 1-phosphate uridyl transferase, the enzyme responsible for the reaction between galactose 1-phosphate and UDPG, so that ingested galactose accumulates in the circulation. Serious dis-turbances of growth and development result. Treatment with galactose-free diets improves this condition without leading to galactose deficiency, because the enzyme necessary for the for-mation of uridine diphosphogalactose from UDPG is present. Fructose is converted in part to fructose 6-phosphate and then metabolized via fructose 1,6-diphosphate. The enzyme catalyzing the formation of fructose 6-phosphate is hexoki-nase, the same enzyme that catalyzes the conversion of glu-cose to glucose 6-phosphate. However, much more fructose is converted to fructose 1-phosphate in a reaction catalyzed by fructokinase. Most of the fructose 1-phosphate is then split into dihydroxyacetone phosphate and glyceraldehyde. The glyceraldehyde is phosphorylated, and it and the dihy-droxyacetone phosphate enter the pathways for glucose metabolism. Because the reactions proceeding through phos-phorylation of fructose in the 1 position can occur at a nor-mal rate in the absence of insulin, it has been recommended that fructose be given to diabetics to replenish their carbohy-drate stores. However, most of the fructose is metabolized in the intestines and liver, so its value in replenishing carbohy-drate elsewhere in the body is limited. Fructose 6-phosphate can also be phosphorylated in the 2 position, forming fructose 2,6-diphosphate. This compound is an important regulator of hepatic gluconeogenesis. When the fructose 2,6-diphosphate level is high, conversion of fruc-tose 6-phosphate to fructose 1,6-diphosphate is facilitated, and thus breakdown of glucose to pyruvate is increased. A decreased level of fructose 2,6-diphosphate facilitates the reverse reaction and consequently aids gluconeogenesis. FATTY ACIDS & LIPIDS The biologically important lipids are the fatty acids and their de-rivatives, the neutral fats (triglycerides), the phospholipids and related compounds, and the sterols. The triglycerides are made up of three fatty acids bound to glycerol (Table 1–4). Naturally occurring fatty acids contain an even number of carbon atoms. They may be saturated (no double bonds) or unsaturated (de-hydrogenated, with various numbers of double bonds). The phospholipids are constituents of cell membranes and provide structural components of the cell membrane, as well as an im-portant source of intra- and intercellular signaling molecules. Fatty acids also are an important source of energy in the body. FATTY ACID OXIDATION & SYNTHESIS In the body, fatty acids are broken down to acetyl-CoA, which enters the citric acid cycle. The main breakdown occurs in the mitochondria by β-oxidation. Fatty acid oxidation begins with activation (formation of the CoA derivative) of the fatty acid, a reaction that occurs both inside and outside the mitochon-dria. Medium- and short-chain fatty acids can enter the mito-chondria without difficulty, but long-chain fatty acids must be bound to carnitine in ester linkage before they can cross the inner mitochondrial membrane. Carnitine is β-hydroxy-γ-tri-methylammonium butyrate, and it is synthesized in the body from lysine and methionine. A translocase moves the fatty acid–carnitine ester into the matrix space. The ester is hydro-lyzed, and the carnitine recycles. β-oxidation proceeds by se-rial removal of two carbon fragments from the fatty acid (Figure 1–26). The energy yield of this process is large. For ex-ample, catabolism of 1 mol of a six-carbon fatty acid through the citric acid cycle to CO2 and H2O generates 44 mol of ATP, compared with the 38 mol generated by catabolism of 1 mol of the six-carbon carbohydrate glucose. KETONE BODIES In many tissues, acetyl-CoA units condense to form acetoacetyl-CoA (Figure 1–27). In the liver, which (unlike other tissues) contains a deacylase, free acetoacetate is formed. This β-keto acid is converted to β-hydroxybutyrate and acetone, and because these compounds are metabolized with difficulty in FIGURE 1–25 Plasma glucose homeostasis. Notice the gluco-static function of the liver, as well as the loss of glucose in the urine when the renal threshold is exceeded (dashed arrows). Kidney Brain Fat Muscle and other tissues Liver Amino acids Glycerol Diet Intestine Plasma glucose 70 mg/dL (3.9 mmol/L) Urine (when plasma glucose > 180 mg/dL) Lactate 24 SECTION I Cellular & Molecular Basis of Medical Physiology the liver, they diffuse into the circulation. Acetoacetate is also formed in the liver via the formation of 3-hydroxy-3-methyl-glutaryl-CoA, and this pathway is quantitatively more impor-tant than deacylation. Acetoacetate, β-hydroxybutyrate, and acetone are called ketone bodies. Tissues other than liver transfer CoA from succinyl-CoA to acetoacetate and metabo-lize the “active” acetoacetate to CO2 and H2O via the citric acid cycle. Ketone bodies are also metabolized via other path-ways. Acetone is discharged in the urine and expired air. An imbalance of ketone bodies can lead to serious health prob-lems (Clinical Box 1–3). CELLULAR LIPIDS The lipids in cells are of two main types: structural lipids, which are an inherent part of the membranes and other parts of cells; and neutral fat, stored in the adipose cells of the fat depots. Neutral fat is mobilized during starvation, but struc-tural lipid is preserved. The fat depots obviously vary in size, but in nonobese individuals they make up about 15% of body weight in men and 21% in women. They are not the inert structures they were once thought to be but, rather, active dy-namic tissues undergoing continuous breakdown and resyn-thesis. In the depots, glucose is metabolized to fatty acids, and neutral fats are synthesized. Neutral fat is also broken down, and free fatty acids are released into the circulation. A third, special type of lipid is brown fat, which makes up a small percentage of total body fat. Brown fat, which is some-what more abundant in infants but is present in adults as well, is located between the scapulas, at the nape of the neck, along the great vessels in the thorax and abdomen, and in other scattered locations in the body. In brown fat depots, the fat cells as well as the blood vessels have an extensive sympathetic innervation. This is in contrast to white fat depots, in which some fat cells may be innervated but the principal sympa-thetic innervation is solely on blood vessels. In addition, ordi-nary lipocytes have only a single large droplet of white fat, whereas brown fat cells contain several small droplets of fat. Brown fat cells also contain many mitochondria. In these mitochondria, an inward proton conductance that generates ATP takes places as usual, but in addition there is a second proton conductance that does not generate ATP. This “short-circuit” conductance depends on a 32-kDa uncoupling pro-tein (UCP1). It causes uncoupling of metabolism and genera-tion of ATP, so that more heat is produced. PLASMA LIPIDS & LIPID TRANSPORT The major lipids are relatively insoluble in aqueous solutions and do not circulate in the free form. Free fatty acids (FFAs) are bound to albumin, whereas cholesterol, triglycerides, and phospholipids are transported in the form of lipoprotein complexes. The complexes greatly increase the solubility of the lipids. The six families of lipoproteins (Table 1–5) are graded in size and lipid content. The density of these lipopro-teins is inversely proportionate to their lipid content. In general, the lipoproteins consist of a hydrophobic core of tri-glycerides and cholesteryl esters surrounded by phospholipids and protein. These lipoproteins can be transported from the intestine to the liver via an exogenous pathway, and between other tissues via an endogenous pathway. Dietary lipids are processed by several pancreatic lipases in the intestine to form mixed micelles of predominantly FFA, 2-monoglycerols, and cholesterol derivatives (see Chapter 27). These micelles additionally can contain important water-insoluble molecules such as vitamins A, D, E, and K. These mixed micelles are taken up into cells of the intestinal TABLE 1–4. Lipids. Typical fatty acids: Triglycerides (triacylglycerols): Esters of glycerol and three fatty acids. R = Aliphatic chain of various lengths and degrees of saturation. Phospholipids: A. Esters of glycerol, two fatty acids, and 1. Phosphate = phosphatidic acid 2. Phosphate plus inositol = phosphatidylinositol 3. Phosphate plus choline = phosphatidylcholine (lecithin) 4. Phosphate plus ethanolamine = phosphatidyl-ethanolamine (cephalin) 5. Phosphate plus serine = phosphatidylserine B. Other phosphate-containing derivatives of glycerol C. Sphingomyelins: Esters of fatty acid, phosphate, choline, and the amino alcohol sphingosine. Cerebrosides: Compounds containing galactose, fatty acid, and sphin-gosine. Sterols: Cholesterol and its derivatives, including steroid hormones, bile acids, and various vitamins. Palmitic acid: CH5(CH2)14—C—OH O Stearic acid: CH5(CH2)16—C—OH O Oleic acid: (Unsaturated) CH5(CH2)7CH=CH(CH2)7—C—OH O CH2—O—C—R CH2OH CHOH + 3HO—C—R CH2OH Glycerol O CH2—O—C—R + 3H2O CH2—O—C—R Triglyceride O O O CHAPTER 1 General Principles & Energy Production in Medical Physiology 25 FIGURE 1–26 Fatty acid oxidation. This process, splitting off two carbon fragments at a time, is repeated to the end of the chain. FIGURE 1–27 Formation and metabolism of ketone bodies. Note the two pathways for the formation of acetoacetate. OH + HS-CoA — — OH H α,β-Unsaturated fatty acid–CoA β-Keto fatty acid–CoA β-Hydroxy fatty acid–CoA "Active" fatty acid + Acetyl–CoA — — C O H2O + R — — O CH2CH2 CH2CH2 S R CoA + HS-CoA — — C S C O — — O R CH2 CoA — — C S C S CoA + CH3 O — — O R — — C O CH2 C R Mg2+ ATP ADP Fatty acid Oxidized flavoprotein Reduced flavoprotein "Active" fatty acid C S CoA H2O + R — — O CH CH C S CoA CoA NAD+ NADH + H+ R = Rest of fatty acid chain. S — — C O CH3 CoA + CH3 S — — C O CoA 2 Acetyl-CoA Acetoacetyl-CoA — — C O CH3 CH2 S — — C O CoA + HS-CoA β-Ketothiolase CH2 C CH3 S C CoA + H2O Acetoacetyl-CoA Acetyl-CoA + Acetoacetyl-CoA HMG-CoA Acetoacetate + H+ + Acetyl-CoA Acetoacetate Acetoacetate Acetone 3-Hydroxy-3-methylglutaryl-CoA (HMG-CoA) β-Hydroxybutyrate CH2 — C OH CH3 CH2 S C CoA + H+ CH2 –CO2 CO2 + ATP C CH3 O− C CH3 CH3 C + H+ CH2 CHOH CH3 O− — — C O + H+ C CH3 CH2 O− C COO− + H+ + HS-CoA Deacylase (liver only) +2H –2H Tissues except liver — — O — — O — — O — — O — — O — — O — — O — — O 26 SECTION I Cellular & Molecular Basis of Medical Physiology mucosa where large lipoprotein complexes, chylomicrons, are formed. The chylomicrons and their remnants constitute a transport system for ingested exogenous lipids (exogenous pathway). Chylomicrons can enter the circulation via the lymphatic ducts. The chylomicrons are cleared from the cir-culation by the action of lipoprotein lipase, which is located on the surface of the endothelium of the capillaries. The enzyme catalyzes the breakdown of the triglyceride in the chylomicrons to FFA and glycerol, which then enter adipose cells and are reesterified. Alternatively, the FFA can remain in the circulation bound to albumin. Lipoprotein lipase, which requires heparin as a cofactor, also removes triglycerides from circulating very low density lipoproteins (VLDL). Chylomicrons depleted of their triglyceride remain in the circulation as cholesterol-rich lipoproteins called chylomi-cron remnants, which are 30 to 80 nm in diameter. The rem-nants are carried to the liver, where they are internalized and degraded. CLINICAL BOX 1–3 Diseases Associated with Imbalance of β-oxidation of Fatty Acids Ketoacidosis and even fatal. Three conditions lead to deficient intracellular glucose supplies, and hence to ketoacidosis: starvation; diabetes mellitus; and a high-fat, low-carbohydrate diet. The acetone odor on the breath of children who have been vomiting is due to the ketosis of starvation. Parenteral administration of relatively small amounts of glucose abolishes the ketosis, and it is for this reason that carbohydrate is said to be antiketogenic. Carnitine Deficiency Deficient β-oxidation of fatty acids can be produced by carnitine deficiency or genetic defects in the translocase or other enzymes involved in the transfer of long-chain fatty acids into the mito-chondria. This causes cardiomyopathy. In addition, it causes hy-poketonemic hypoglycemia with coma, a serious and often fatal condition triggered by fasting, in which glucose stores are used up because of the lack of fatty acid oxidation to provide en-ergy. Ketone bodies are not formed in normal amounts because of the lack of adequate CoA in the liver. The normal blood ketone level in humans is low (about 1 mg/dL) and less than 1 mg is excreted per 24 h, because the ketones are normally metabolized as rapidly as they are formed. However, if the entry of acetyl-CoA into the citric acid cycle is depressed because of a decreased supply of the prod-ucts of glucose metabolism, or if the entry does not increase when the supply of acetyl-CoA increases, acetyl-CoA accumu-lates, the rate of condensation to acetoacetyl-CoA increases, and more acetoacetate is formed in the liver. The ability of the tissues to oxidize the ketones is soon exceeded, and they accu-mulate in the bloodstream (ketosis). Two of the three ketone bodies, acetoacetate and β-hydroxybutyrate, are anions of the moderately strong acids acetoacetic acid and β-hydroxybutyric acid. Many of their protons are buffered, reducing the decline in pH that would otherwise occur. However, the buffering capacity can be exceeded, and the metabolic acidosis that develops in conditions such as diabetic ketosis can be severe TABLE 1–5 The principal lipoproteins. Composition (%) Lipoprotein Size (nm) Protein Free Cholesteryl Cholesterol Esters Triglyceride Phospholipid Origin Chylomicrons 75–1000 2 2 3 90 3 Intestine Chylomicron remnants 30–80 … … … … … Capillaries Very low density lipoproteins (VLDL) 30–80 8 4 16 55 17 Liver and intestine Intermediate-density lipo-proteins (IDL) 25–40 10 5 25 40 20 VLDL Low-density lipoproteins (LDL) 20 20 7 46 6 21 IDL High-density lipoproteins (HDL) 7.5–10 50 4 16 5 25 Liver and intestine The plasma lipids include these components plus free fatty acids from adipose tissue, which circulate bound to albumin. CHAPTER 1 General Principles & Energy Production in Medical Physiology 27 The endogenous system, made up of VLDL, intermedi-ate-density lipoproteins (IDL), low-density lipoproteins (LDL), and high-density lipoproteins (HDL), also trans-ports triglycerides and cholesterol throughout the body. VLDL are formed in the liver and transport triglycerides formed from fatty acids and carbohydrates in the liver to extrahepatic tissues. After their triglyceride is largely removed by the action of lipoprotein lipase, they become IDL. The IDL give up phospholipids and, through the action of the plasma enzyme lecithin-cholesterol acyltransferase (LCAT), pick up cholesteryl esters formed from cholesterol in the HDL. Some IDL are taken up by the liver. The remain-ing IDL then lose more triglyceride and protein, probably in the sinusoids of the liver, and become LDL. LDL provide cholesterol to the tissues. The cholesterol is an essential con-stituent in cell membranes and is used by gland cells to make steroid hormones. FREE FATTY ACID METABOLISM In addition to the exogenous and endogenous pathways de-scribed above, FFA are also synthesized in the fat depots in which they are stored. They can circulate as lipoproteins bound to albumin and are a major source of energy for many organs. They are used extensively in the heart, but probably all tissues can oxidize FFA to CO2 and H2O. The supply of FFA to the tissues is regulated by two lipases. As noted above, lipoprotein lipase on the surface of the endothelium of the capillaries hydrolyzes the triglyc-erides in chylomicrons and VLDL, providing FFA and glyc-erol, which are reassembled into new triglycerides in the fat cells. The intracellular hormone-sensitive lipase of adipose tissue catalyzes the breakdown of stored triglycerides into glycerol and fatty acids, with the latter entering the circula-tion. Hormone-sensitive lipase is increased by fasting and stress and decreased by feeding and insulin. Conversely, feeding increases and fasting and stress decrease the activity of lipoprotein lipase. CHOLESTEROL METABOLISM Cholesterol is the precursor of the steroid hormones and bile ac-ids and is an essential constituent of cell membranes. It is found only in animals. Related sterols occur in plants, but plant sterols are not normally absorbed from the gastrointestinal tract. Most of the dietary cholesterol is contained in egg yolks and animal fat. Cholesterol is absorbed from the intestine and incorporated into the chylomicrons formed in the intestinal mucosa. After the chylomicrons discharge their triglyceride in adipose tissue, the chylomicron remnants bring cholesterol to the liver. The liver and other tissues also synthesize cholesterol. Some of the choles-terol in the liver is excreted in the bile, both in the free form and as bile acids. Some of the biliary cholesterol is reabsorbed from the intestine. Most of the cholesterol in the liver is incorporated into VLDL and circulates in lipoprotein complexes. The biosynthesis of cholesterol from acetate is summarized in Figure 1–28. Cholesterol feeds back to inhibit its own synthesis by inhibiting HMG-CoA reductase, the enzyme that con-verts 3-hydroxy-3-methylglutaryl-coenzyme A (HMG-CoA) to mevalonic acid. Thus, when dietary cholesterol intake is high, hepatic cholesterol synthesis is decreased, and vice versa. However, the feedback compensation is incomplete, because a diet that is low in cholesterol and saturated fat leads to only a modest decline in circulating plasma cholesterol. The most effective and most commonly used cholesterol-lowering drugs are lovastatin and other statins, which reduce cholesterol syn-thesis by inhibiting HMG-CoA. The relationship between cho-lesterol and vascular disease is discussed in Clinical Box 1–4. ESSENTIAL FATTY ACIDS Animals fed a fat-free diet fail to grow, develop skin and kidney lesions, and become infertile. Adding linolenic, linoleic, and arachidonic acids to the diet cures all the deficiency symptoms. These three acids are polyunsaturated fatty acids and because of their action are called essential fatty acids. Similar deficien-cy symptoms have not been unequivocally demonstrated in humans, but there is reason to believe that some unsaturated fats are essential dietary constituents, especially for children. FIGURE 1–28 Biosynthesis of cholesterol. Six mevalonic acid molecules condense to form squalene, which is then hydroxylated to cholesterol. The dashed arrow indicates feedback inhibition by cholesterol of HMG-CoA reductase, the enzyme that catalyzes meva-lonic acid formation. CH2 CH3 CH2 C OH HOOC CH2 OH HO Squalene (C30H50) Mevalonic acid Cholesterol (C27H46O) 3-Hydroxy-3-methylglutaryl-CoA Acetoacetyl-CoA Acetyl-CoA HMG-CoA reductase Acetoacetate Mevalonic acid Squalene Cholesterol Acetoacetate 28 SECTION I Cellular & Molecular Basis of Medical Physiology Dehydrogenation of fats is known to occur in the body, but there does not appear to be any synthesis of carbon chains with the ar-rangement of double bonds found in the essential fatty acids. EICOSANOIDS One of the reasons that essential fatty acids are necessary for health is that they are the precursors of prostaglandins, prosta-cyclin, thromboxanes, lipoxins, leukotrienes, and related com-pounds. These substances are called eicosanoids, reflecting their origin from the 20-carbon (eicosa-) polyunsaturated fat-ty acid arachidonic acid (arachidonate) and the 20-carbon derivatives of linoleic and linolenic acids. The prostaglandins are a series of 20-carbon unsaturated fatty acids containing a cyclopentane ring. They were first iso-lated from semen but are now known to be synthesized in most and possibly in all organs in the body. Prostaglandin H2 (PGH2) is the precursor for various other prostaglandins, thromboxanes, and prostacyclin. Arachidonic acid is formed from tissue phospholipids by phospholipase A2. It is converted to prostaglandin H2 (PGH2) by prostaglandin G/H synthases 1 and 2. These are bifunctional enzymes that have both cyclo-oxygenase and peroxidase activity, but they are more com-monly known by the names cyclooxygenase 1 (COX1) and cyclooxygenase 2 (COX2). Their structures are very similar, but COX1 is constitutive whereas COX2 is induced by growth factors, cytokines, and tumor promoters. PGH2 is converted to prostacyclin, thromboxanes, and prostaglandins by various tis-sue isomerases. The effects of prostaglandins are multitudinous and varied. They are particularly important in the female reproductive cycle, in parturition, in the cardiovascular system, in inflammatory responses, and in the causation of pain. Drugs that target production of prostaglandins are among the most common over the counter drugs available (Clinical Box 1–5). Arachidonic acid also serves as a substrate for the produc-tion of several physiologically important leukotrienes and lipoxins. The leukotrienes, thromboxanes, lipoxins, and CLINICAL BOX 1–4 Cholesterol & Atherosclerosis The interest in cholesterol-lowering drugs stems from the role of cholesterol in the etiology and course of athero-sclerosis. This extremely widespread disease predisposes to myocardial infarction, cerebral thrombosis, ischemic gangrene of the extremities, and other serious illnesses. It is characterized by infiltration of cholesterol and oxidized cholesterol into macrophages, converting them into foam cells in lesions of the arterial walls. This is followed by a complex sequence of changes involving platelets, macro-phages, smooth muscle cells, growth factors, and inflam-matory mediators that produces proliferative lesions which eventually ulcerate and may calcify. The lesions distort the vessels and make them rigid. In individuals with elevated plasma cholesterol levels, the incidence of atherosclerosis and its complications is increased. The normal range for plasma cholesterol is said to be 120 to 200 mg/dL, but in men, there is a clear, tight, positive correlation between the death rate from ischemic heart disease and plasma choles-terol levels above 180 mg/dL. Furthermore, it is now clear that lowering plasma cholesterol by diet and drugs slows and may even reverse the progression of atherosclerotic le-sions and the complications they cause. In evaluating plasma cholesterol levels in relation to athero-sclerosis, it is important to analyze the LDL and HDL levels as well. LDL delivers cholesterol to peripheral tissues, including atheromatous lesions, and the LDL plasma concentration cor-relates positively with myocardial infarctions and ischemic strokes. On the other hand, HDL picks up cholesterol from pe-ripheral tissues and transports it to the liver, thus lowering plasma cholesterol. It is interesting that women, who have a lower incidence of myocardial infarction than men, have higher HDL levels. In addition, HDL levels are increased in indi-viduals who exercise and those who drink one or two alco-holic drinks per day, whereas they are decreased in individuals who smoke, are obese, or live sedentary lives. Moderate drink-ing decreases the incidence of myocardial infarction, and obe-sity and smoking are risk factors that increase it. Plasma cho-lesterol and the incidence of cardiovascular diseases are increased in familial hypercholesterolemia, due to various loss-of-function mutations in the genes for LDL receptors. CLINICAL BOX 1–5 Pharmacology of Prostaglandins Because prostaglandins play a prominent role in the genesis of pain, inflammation, and fever, pharmacologists have long sought drugs to inhibit their synthesis. Glucocorticoids in-hibit phospholipase A2 and thus inhibit the formation of all eicosanoids. A variety of nonsteroidal anti-inflammatory drugs (NSAIDs) inhibit both cyclooxygenases, inhibiting the production of PGH2 and its derivatives. Aspirin is the best-known of these, but ibuprofen, indomethacin, and others are also used. However, there is evidence that prostaglandins synthesized by COX2 are more involved in the production of pain and inflammation, and prostaglandins synthesized by COX1 are more involved in protecting the gastrointestinal mucosa from ulceration. Drugs such as celecoxib and rofe-coxib that selectively inhibit COX2 have been developed, and in clinical use they relieve pain and inflammation, possi-bly with a significantly lower incidence of gastrointestinal ul-ceration and its complications than is seen with nonspecific NSAIDs. However, rofecoxib has been withdrawn from the market in the United States because of a reported increase of strokes and heart attacks in individuals using it. More re-search is underway to better understand all the effects of the COX enzymes, their products, and their inhibitors. CHAPTER 1 General Principles & Energy Production in Medical Physiology 29 prostaglandins have been called local hormones. They have short half-lives and are inactivated in many different tissues. They undoubtedly act mainly in the tissues at sites in which they are produced. The leukotrienes are mediators of allergic responses and inflammation. Their release is provoked when specific allergens combine with IgE antibodies on the surfaces of mast cells (see Chapter 3). They produce bronchoconstric-tion, constrict arterioles, increase vascular permeability, and attract neutrophils and eosinophils to inflammatory sites. Diseases in which they may be involved include asthma, pso-riasis, adult respiratory distress syndrome, allergic rhinitis, rheumatoid arthritis, Crohn’s disease, and ulcerative colitis. CHAPTER SUMMARY ■Cells contain approximately one third of the body fluids, while the remaining extracellular fluid is found between cells (intersti-tial fluid) or in the circulating blood plasma. ■The number of molecules, electrical charges, and particles of substances in solution are important in physiology. ■The high surface tension, high heat capacity, and high electrical ca-pacity allow H2O to function as an ideal solvent in physiology. ■Biological buffers including bicarbonate, proteins, and phos-phates can bind or release protons in solution to help maintain pH. Biological buffering capacity of a weak acid or base is great-est when pKa = pH. ■Fluid and electrolyte balance in the body is related to plasma os-molality. Isotonic solutions have the same osmolality as blood plasma, hypertonic have higher osmolality, while hypotonic have lower osmolality. ■Although the osmolality of solutions can be similar across a plasma membrane, the distribution of individual molecules and distribution of charge across the plasma membrane can be quite different. These are affected by the Gibbs-Donnan equilibrium and can be calculated using the Nernst potential equation. ■There is a distinct difference in concentration of ions in the extra-cellular and intracellular fluids (concentration gradient). The sep-aration of concentrations of charged species sets up an electrical gradient at the plasma membrane (inside negative). The electro-chemical gradient is in large part maintained by the Na, K ATPase. ■Cellular energy can be stored in high-energy phosphate com-pounds, including adenosine triphosphate (ATP). Coordinated oxidation-reduction reactions allow for production of a proton gradient at the inner mitochondrial membrane that ultimately yields to the production of ATP in the cell. ■Nucleotides made from purine or pyrimidine bases linked to ri-bose or 2-deoxyribose sugars with inorganic phosphates are the basic building blocks for nucleic acids, DNA, and RNA. ■DNA is a double-stranded structure that contains the funda-mental information for an organism. During cell division, DNA is faithfully replicated and a full copy of DNA is in every cell. The fundamental unit of DNA is the gene, which encodes infor-mation to make proteins in the cell. Genes are transcribed into messenger RNA, and with the help of ribosomal RNA and trans-fer RNAs, translated into proteins. ■Amino acids are the basic building blocks for proteins in the cell and can also serve as sources for several biologically active molecules. They exist in an “amino acid pool” that is derived from the diet, protein degradation, and de novo and resynthesis. ■Translation is the process of protein synthesis. After synthesis, proteins can undergo a variety of posttranslational modifica-tions prior to obtaining their fully functional cell state. ■Carbohydrates are organic molecules that contain equal amounts of C and H2O. Carbohydrates can be attached to pro-teins (glycoproteins) or fatty acids (glycolipids) and are critically important for the production and storage of cellular and body energy, with major supplies in the form of glycogen in the liver and skeletal muscle. The breakdown of glucose to generate en-ergy, or glycolysis, can occur in the presence or absence of O2 (aerobic or anaerobically). The net production of ATP during aerobic glycolysis is 19 times higher than anaerobic glycolysis. ■Fatty acids are carboxylic acids with extended hydrocarbon chains. They are an important energy source for cells and their derivatives, including triglycerides, phospholipids and sterols, and have additional important cellular applications. Free fatty acids can be bound to albumin and transported throughout the body. Triglycerides, phospholipids, and cholesterol are trans-ported as lipoprotein complexes. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. The membrane potential of a particular cell is at the K+ equilib-rium. The intracellular concentration for K+ is at 150 mmol/L and the extracellular concentration for K+ is at 5.5 mmol/L. What is the resting potential? A) –70 mv B) –90 mv C) +70 mv D) +90 mv 2. The difference in concentration of H+ in a solution of pH 2.0 compared with one of pH 7.0 is A) 5-fold. B) 1/5 as much. C) 105 fold. D) 10–5 as much. 3. Transcription refers to A) the process where an mRNA is used as a template for protein production. B) the process where a DNA sequence is copied into RNA for the purpose of gene expression. C) the process where DNA wraps around histones to form a nucleosome. D) the process of replication of DNA prior to cell division. 4. The primary structure of a protein refers to A) the twist, folds, or twist and folds of the amino acid sequence into stabilized structures within the protein (ie, α-helices and β-sheets). B) the arrangement of subunits to form a functional structure. C) the amino acid sequence of the protein. D) the arrangement of twisted chains and folds within a protein into a stable structure. 30 SECTION I Cellular & Molecular Basis of Medical Physiology 5. Fill in the blanks: Glycogen is a storage form of glucose. _ refers to the process of making glycogen and _ refers to the process of breakdown of glycogen. A) Glycogenolysis, glycogenesis B) Glycolysis, glycogenolysis C) Glycogenesis, glycogenolysis D) Glycogenolysis, glycolysis 6. The major lipoprotein source of the cholesterol used in cells is A) chylomicrons. B) intermediate-density lipoproteins (IDLs). C) albumin-bound free fatty acids. D) LDL. E) HDL. 7. Which of the following produces the most high-energy phos-phate compounds? A) aerobic metabolism of 1 mol of glucose B) anaerobic metabolism of 1 mol of glucose C) metabolism of 1 mol of galactose D) metabolism of 1 mol of amino acid E) metabolism of 1 mol of long-chain fatty acid 8. When LDL enters cells by receptor-mediated endocytosis, which of the following does not occur? A) Decrease in the formation of cholesterol from mevalonic acid. B) Increase in the intracellular concentration of cholesteryl esters. C) Increase in the transfer of cholesterol from the cell to HDL. D) Decrease in the rate of synthesis of LDL receptors. E) Decrease in the cholesterol in endosomes. CHAPTER RESOURCES Alberts B, et al: Molecular Biology of the Cell, 5th ed. Garland Science, 2007. Hille B: Ionic Channels of Excitable Membranes, 3rd ed. Sinauer Associates, 2001. Kandel ER, Schwartz JH, Jessell TM: Principles of Neural Science, 4th ed. McGraw-Hill, 2000. Macdonald RG, Chaney WG: USMLE Road Map, Biochemistry. McGraw- Hill, 2007. Murray RK, et al: Harper’s Biochemistry, 26th ed. McGraw-Hill, 2003. Pollard TD, Earnshaw WC: Cell Biology, 2nd ed. Saunders, Elsevier, 2008. Sack GH, Jr. USMLE Road Map, Genetics. McGraw Hill, 2008. Scriver CR, et al (editors): The Metabolic and Molecular Bases of Inherited Disease, 8th ed. McGraw-Hill, 2001. Sperelakis N (editor): Cell Physiology Sourcebook, 3rd ed. Academic Press, 2001. 31 C H A P T E R 2 Overview of Cellular Physiology in Medical Physiology O B J E C T I V E S After studying this chapter, you should be able to: ■Name the prominent cellular organelles and state their functions in cells. ■Name the building blocks of the cellular cytoskeleton and state their contributions to cell structure and function. ■Name the intercellular and cellular to extracellular connections. ■Define the processes of exocytosis and endocytosis, and describe the contribution of each to normal cell function. ■Define proteins that contribute to membrane permeability and transport. ■Describe specialized transport and filtration across the capillary wall. ■Recognize various forms of intercellular communication and describe ways in which chemical messengers (including second messengers) affect cellular physiology. ■Define cellular homeostasis. INTRODUCTION The cell is the fundamental working unit of all organisms. In humans, cells can be highly specialized in both structure and function; alternatively, cells from different organs can share features and function. In the previous chapter, we examined some basic principles of biophysics and the catabolism and metabolism of building blocks found in the cell. In some of those discussions, we examined how the building blocks could contribute to basic cellular physiology (eg, DNA repli-cation, transcription, and translation). In this chapter, we will briefly review more of the fundamental aspects of cellular and molecular physiology. Additional aspects that concern spe-cialization of cellular and molecular physiology are consid-ered in the next chapter concerning immune function and in the relevant chapters on the various organs. FUNCTIONAL MORPHOLOGY OF THE CELL A basic knowledge of cell biology is essential to an under-standing of the organ systems in the body and the way they function. A key tool for examining cellular constituents is the microscope. A light microscope can resolve structures as close as 0.2 μm, while an electron microscope can resolve structures as close as 0.002 μm. Although cell dimensions are quite vari-able, this resolution can give us a good look at the inner work-ings of the cell. The advent of common access to fluorescent, confocal, and other microscopy along with specialized probes for both static and dynamic cellular structures further expand-ed the examination of cell structure and function. Equally rev-olutionary advances in the modern biophysical, biochemical, and molecular biology techniques have also greatly contribut-ed to our knowledge of the cell. The specialization of the cells in the various organs is consid-erable, and no cell can be called “typical” of all cells in the body. However, a number of structures (organelles) are common to most cells. These structures are shown in Figure 2–1. Many of them can be isolated by ultracentrifugation combined with 32 SECTION I Cellular & Molecular Basis of Medical Physiology other techniques. When cells are homogenized and the result-ing suspension is centrifuged, the nuclei sediment first, fol-lowed by the mitochondria. High-speed centrifugation that generates forces of 100,000 times gravity or more causes a fraction made up of granules called the microsomes to sedi-ment. This fraction includes organelles such as the ribosomes and peroxisomes. CELL MEMBRANES The membrane that surrounds the cell is a remarkable struc-ture. It is made up of lipids and proteins and is semipermeable, allowing some substances to pass through it and excluding others. However, its permeability can also be varied because it contains numerous regulated ion channels and other trans-port proteins that can change the amounts of substances mov-ing across it. It is generally referred to as the plasma membrane. The nucleus and other organelles in the cell are bound by similar membranous structures. Although the chemical structures of membranes and their properties vary considerably from one location to another, they have certain common features. They are generally about 7.5 nm (75 Å) thick. The major lipids are phospholipids such as phosphatidylcholine and phosphatidylethanolamine. The shape of the phospholipid molecule reflects its solubility properties: the head end of the molecule contains the phos-phate portion and is relatively soluble in water (polar, hydro-philic) and the tails are relatively insoluble (nonpolar, hydrophobic). The possession of both hydrophilic and hydrophobic properties make the lipid an amphipathic mole-cule. In the membrane, the hydrophilic ends of the molecules are exposed to the aqueous environment that bathes the exte-rior of the cells and the aqueous cytoplasm; the hydrophobic ends meet in the water-poor interior of the membrane (Figure 2–2). In prokaryotes (ie, bacteria in which there is no nucleus), the membranes are relatively simple, but in eukary-otes (cells containing nuclei), cell membranes contain various glycosphingolipids, sphingomyelin, and cholesterol in addi-tion to phospholipids and phosphatidylcholine. Many different proteins are embedded in the membrane. They exist as separate globular units and many pass through the membrane (integral proteins), whereas others (peripheral proteins) stud the inside and outside of the membrane (Figure 2–2). The amount of protein varies significantly with the func-tion of the membrane but makes up on average 50% of the mass of the membrane; that is, there is about one protein molecule per 50 of the much smaller phospholipid molecules. The pro-teins in the membranes carry out many functions. Some are cell adhesion molecules that anchor cells to their neighbors or to basal laminas. Some proteins function as pumps, actively FIGURE 2–1 Diagram showing a hypothetical cell in the center as seen with the light microscope. Individual organelles are ex-panded for closer examination. (Adapted from Bloom and Fawcett. Reproduced with permission from Junqueira LC, Carneiro J, Kelley RO: Basic Histology, 9th ed. McGraw-Hill, 1998.) Secretory granules Centrioles Smooth endoplasmic reticulum Golgi apparatus Lipid droplets Rough endoplasmic reticulum Lysosomes Mitochondrion Globular heads Nuclear envelope Nucleolus CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 33 transporting ions across the membrane. Other proteins func-tion as carriers, transporting substances down electrochemical gradients by facilitated diffusion. Still others are ion channels, which, when activated, permit the passage of ions into or out of the cell. The role of the pumps, carriers, and ion channels in transport across the cell membrane is discussed below. Proteins in another group function as receptors that bind ligands or messenger molecules, initiating physiologic changes inside the cell. Proteins also function as enzymes, catalyzing reactions at the surfaces of the membrane. Examples from each of these groups are discussed later in this chapter. The uncharged, hydrophobic portions of the proteins are usually located in the interior of the membrane, whereas the charged, hydrophilic portions are located on the surfaces. Peripheral proteins are attached to the surfaces of the mem-brane in various ways. One common way is attachment to gly-cosylated forms of phosphatidylinositol. Proteins held by these glycosylphosphatidylinositol (GPI) anchors (Figure 2–3) include enzymes such as alkaline phosphatase, various antigens, a number of cell adhesion molecules, and three pro-teins that combat cell lysis by complement. Over 45 GPI-linked cell surface proteins have now been described in humans. Other proteins are lipidated, that is, they have spe-cific lipids attached to them (Figure 2–3). Proteins may be myristolated, palmitoylated, or prenylated (ie, attached to geranylgeranyl or farnesyl groups). The protein structure—and particularly the enzyme con-tent—of biologic membranes varies not only from cell to cell, but also within the same cell. For example, some of the enzymes embedded in cell membranes are different from those in mito-chondrial membranes. In epithelial cells, the enzymes in the cell membrane on the mucosal surface differ from those in the FIGURE 2–2 Organization of the phospholipid bilayer and associated proteins in a biological membrane. The phospholipid molecules each have two fatty acid chains (wavy lines) attached to a phos-phate head (open circle). Proteins are shown as irregular colored globules. Many are integral proteins, which extend into the membrane, but periph-eral proteins are attached to the inside or outside (not shown) of the membrane. Specific protein attachments and cholesterol commonly found in the bilayer are omitted for clarity. (Reproduced with permission from Widmaier EP, Raff H, Strang K: Vander’s Human Physiology: The Mechanisms of Body Function, 11th ed. McGraw-Hill, 2008.) Extracellular fluid Intracellular fluid Carbohydrate portion of glycoprotein Intregral proteins Polar regions Nonpolar regions Peripheral protein Phospholipids Transmembrane proteins Channel FIGURE 2–3 Protein linkages to membrane lipids. Some are linked by their amino terminals, others by their carboxyl terminals. Many are at-tached via glycosylated forms of phosphatidylinositol (GPI anchors). (Reproduced with permission from Fuller GM, Shields D: Molecular Basis of Medical Cell Biology. McGraw-Hill, 1998.) N H O Lipid membrane Cytoplasmic or external face of membrane Gly Protein Protein Protein Protein Protein COOH N-Myristoyl S-Cys O O O C CH2 CH C H2 O O O C Inositol P O O O C NH2 S-Palmitoyl S-Cys NH2 Geranylgeranyl S-Cys NH2 C C Farnesyl GPI anchor (Glycosylphosphatidylinositol) Hydrophobic domain Hydrophilic domain 34 SECTION I Cellular & Molecular Basis of Medical Physiology cell membrane on the basal and lateral margins of the cells; that is, the cells are polarized. Such polarization makes transport across epithelia possible. The membranes are dynamic struc-tures, and their constituents are being constantly renewed at different rates. Some proteins are anchored to the cytoskeleton, but others move laterally in the membrane. Underlying most cells is a thin, “fuzzy” layer plus some fibrils that collectively make up the basement membrane or, more properly, the basal lamina. The basal lamina and, more generally, the extracellular matrix are made up of many pro-teins that hold cells together, regulate their development, and determine their growth. These include collagens, laminins, fibronectin, tenascin, and various proteoglycans. MITOCHONDRIA Over a billion years ago, aerobic bacteria were engulfed by eu-karyotic cells and evolved into mitochondria, providing the eukaryotic cells with the ability to form the energy-rich com-pound ATP by oxidative phosphorylation. Mitochondria perform other functions, including a role in the regulation of apoptosis (programmed cell death), but oxidative phosphory-lation is the most crucial. Each eukaryotic cell can have hun-dreds to thousands of mitochondria. In mammals, they are generally depicted as sausage-shaped organelles (Figure 2–1), but their shape can be quite dynamic. Each has an outer mem-brane, an intermembrane space, an inner membrane, which is folded to form shelves (cristae), and a central matrix space. The enzyme complexes responsible for oxidative phosphory-lation are lined up on the cristae (Figure 2–4). Consistent with their origin from aerobic bacteria, the mitochondria have their own genome. There is much less DNA in the mitochondrial genome than in the nuclear genome, and 99% of the proteins in the mitochondria are the products of nuclear genes, but mitochondrial DNA is respon-sible for certain key components of the pathway for oxidative phosphorylation. Specifically, human mitochondrial DNA is a double-stranded circular molecule containing approximately 16,500 base pairs (compared with over a billion in nuclear DNA). It codes for 13 protein subunits that are associated with proteins encoded by nuclear genes to form four enzyme complexes plus two ribosomal and 22 transfer RNAs that are needed for protein production by the intramitochondrial ribosomes. The enzyme complexes responsible for oxidative phos-phorylation illustrate the interactions between the products of the mitochondrial genome and the nuclear genome. For example, complex I, reduced nicotinamide adenine dinucle-otide dehydrogenase (NADH), is made up of 7 protein sub-units coded by mitochondrial DNA and 39 subunits coded by nuclear DNA. The origin of the subunits in the other com-plexes is shown in Figure 2–4. Complex II, succinate dehydro-genase-ubiquinone oxidoreductase; complex III, ubiquinone-cytochrome c oxidoreductase; and complex IV, cytochrome c oxidase, act with complex I, coenzyme Q, and cytochrome c to convert metabolites to CO2 and water. Complexes I, III, and IV pump protons (H+) into the intermembrane space during this electron transfer. The protons then flow down their electrochemical gradient through complex V, ATP syn-thase, which harnesses this energy to generate ATP. As zygote mitochondria are derived from the ovum, their inheritance is maternal. This maternal inheritance has been used as a tool to track evolutionary descent. Mitochondria have an ineffective DNA repair system, and the mutation rate for mitochondrial DNA is over 10 times the rate for nuclear DNA. A large number of relatively rare diseases have now been traced to mutations in mitochondrial DNA. These include for the most part disorders of tissues with high meta-bolic rates in which energy production is defective as a result of abnormalities in the production of ATP. LYSOSOMES In the cytoplasm of the cell there are large, somewhat irregular structures surrounded by membrane. The interior of these structures, which are called lysosomes, is more acidic than the FIGURE 2–4 Components involved in oxidative phosphorylation in mitochondria and their origins. As enzyme complexes I through IV convert 2-carbon metabolic fragments to CO2 and H2O, protons (H+) are pumped into the intermembrane space. The proteins diffuse back to the matrix space via complex V, ATP synthase (AS), in which ADP is converted to ATP. The enzyme complexes are made up of subunits coded by mitochondrial DNA (mDNA) and nuclear DNA (nDNA), and the figures document the contribution of each DNA to the complexes. H+ CoQ H+ ADP ATP Cyt c Intramemb space Inner mito membrane Matrix space Complex Subunits from mDNA Subunits from nDNA I II III IV V 7 0 1 3 2 39 4 10 10 14 H+ H+ AS CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 35 rest of the cytoplasm, and external material such as endocy-tosed bacteria, as well as worn-out cell components, are di-gested in them. The interior is kept acidic by the action of a proton pump, or H+, ATPase. This integral membrane pro-tein uses the energy of ATP to move protons from the cytosol up their electrochemical gradient and keep the lysosome rela-tively acidic, near pH 5.0. Lysosomes can contain over 40 types of hydrolytic enzymes, some of which are listed in Table 2–1. Not surprisingly, these enzymes are all acid hydrolases, in that they function best at the acidic pH of the lysosomal compart-ment. This can be a safety feature for the cell; if the lysosome were to break open and release its contents, the enzymes would not be efficient at the near neutral cytosolic pH (7.2), and thus would be unable to digest cytosolic enzymes they may encounter. Diseases associated with lysosomal dysfunc-tion are discussed in Clinical Box 2–1. PEROXISOMES Peroxisomes are 0.5 μm in diameter, are surrounded by a membrane, and contain enzymes that can either produce H2O2 (oxidases) or break it down (catalases). Proteins are di-rected to the peroxisome by a unique signal sequence with the help of protein chaperones, peroxins. The peroxisome mem-brane contains a number of peroxisome-specific proteins that are concerned with transport of substances into and out of the matrix of the peroxisome. The matrix contains more than 40 enzymes, which operate in concert with enzymes outside the peroxisome to catalyze a variety of anabolic and catabolic re-actions (eg, breakdown of lipids). Peroxisomes can form by budding of endoplasmic reticulum, or by division. A number of synthetic compounds were found to cause proliferation of peroxisomes by acting on receptors in the nuclei of cells. These peroxisome proliferation activated receptors (PPARs) are members of the nuclear receptor superfamily. When activat-ed, they bind to DNA, producing changes in the production of mRNAs. The known effects for PPARs are extensive and can affect most tissues and organs. CYTOSKELETON All cells have a cytoskeleton, a system of fibers that not only maintains the structure of the cell but also permits it to change shape and move. The cytoskeleton is made up primarily of mi-crotubules, intermediate filaments, and microfilaments (Figure 2–5), along with proteins that anchor them and tie them together. In addition, proteins and organelles move along microtubules and microfilaments from one part of the cell to another, propelled by molecular motors. Microtubules (Figures 2–5 and 2–6) are long, hollow struc-tures with 5-nm walls surrounding a cavity 15 nm in diame-ter. They are made up of two globular protein subunits: α-and β-tubulin. A third subunit, γ-tubulin, is associated with the production of microtubules by the centrosomes. The α and β subunits form heterodimers, which aggregate to form long tubes made up of stacked rings, with each ring usually containing 13 subunits. The tubules interact with GTP to facilitate their formation. Although microtubule subunits can be added to either end, microtubules are polar with assembly predominating at the “+” end and disassembly predominating at the “–” end. Both processes occur simultaneously in vitro. The growth of microtubules is temperature sensitive (disas-sembly is favored under cold conditions) as well as under the control of a variety of cellular factors that can directly interact with microtubules in the cell. Because of their constant assembly and disassembly, micro-tubules are a dynamic portion of the cell skeleton. They provide the tracks along which several different molecular motors move transport vesicles, organelles such as secretory granules, and mitochondria, from one part of the cell to another. They also form the spindle, which moves the chromosomes in mitosis. Cargo can be transported in either direction on microtubules. There are several drugs available that disrupt cellular func-tion through interaction with microtubules. Microtubule assembly is prevented by colchicine and vinblastine. The anti-cancer drug paclitaxel (Taxol) binds to microtubules and TABLE 2–1 Some of the enzymes found in lysosomes and the cell components that are their substrates. Enzyme Substrate Ribonuclease RNA Deoxyribonuclease DNA Phosphatase Phosphate esters Glycosidases Complex carbohydrates; glycosides and polysaccharides Arylsulfatases Sulfate esters Collagenase Collagens Cathepsins Proteins CLINICAL BOX 2–1 Lysosomal Diseases When a lysosomal enzyme is congenitally absent, the lyso-somes become engorged with the material the enzyme normally degrades. This eventually leads to one of the lyso-somal storage diseases. For example, α-galactosidase A deficiency causes Fabry disease, and β-galactocerebrosi-dase deficiency causes Gaucher disease. These diseases are rare, but they are serious and can be fatal. Another example is the lysosomal storage disease called Tay–Sachs disease, which causes mental retardation and blindness. Tay–Sachs is caused by the loss of hexosaminidase A, a lysosomal en-zyme that catalyzes the biodegradation of gangliosides (fatty acid derivatives). 36 SECTION I Cellular & Molecular Basis of Medical Physiology makes them so stable that organelles cannot move. Mitotic spindles cannot form, and the cells die. Intermediate filaments (Figures 2–5 and 2–6) are 8 to 14 nm in diameter and are made up of various subunits. Some of these filaments connect the nuclear membrane to the cell membrane. They form a flexible scaffolding for the cell and help it resist external pressure. In their absence, cells rupture more easily, and when they are abnormal in humans, blistering of the skin is com-mon. The proteins that make up intermediate filaments are cell-type specific, and are thus frequently used as cellular markers. For example, vimentin is a major intermediate filament in fibro-blasts, whereas cytokeratin is expressed in epithelial cells. Microfilaments (Figures 2–5 and 2–6) are long solid fibers with a 4 to 6 nm diameter that are made up of actin. Although actin is most often associated with muscle contraction, it is present in all types of cells. It is the most abundant protein in mammalian cells, sometimes accounting for as much as 15% of the total protein in the cell. Its structure is highly conserved; for example, 88% of the amino acid sequences in yeast and rabbit actin are identical. Actin filaments polymerize and depolymerize in vivo, and it is not uncommon to find polymerization occurring at one end of the filament while depolymerization is occurring at the other end. Filamentous (F) actin refers to intact microfila-ments and globular (G) actin refers to the unpolymerized pro-tein actin subunits. F-actin fibers attach to various parts of the cytoskeleton and can interact directly or indirectly with mem-brane-bound proteins. They reach to the tips of the microvilli on the epithelial cells of the intestinal mucosa. They are also abundant in the lamellipodia that cells put out when they crawl along surfaces. The actin filaments interact with integrin FIGURE 2–5 Cytoskeletal elements of the cell. Artistic impressions that depict the major cytoskeletal elements are shown on the left, with basic properties of these elements on the right. (Reproduced with permission from Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology: The Mechanisms of Body Function, 11th ed. McGraw-Hill, 2008.) Cytoskeletal filaments Diameter (nm) Protein subunit Microfilament 7 Actin Intermediate filament 10 Several proteins Microtubule 25 Tubulin FIGURE 2–6 Microfilaments and microtubules. Electron micrograph (Left) of the cytoplasm of a fibroblast, displaying actin microfila-ments (MF) and microtubules (MT). (Reproduced, with permission, from Junqueira LC, Carneiro J: Basic Histology, 10th ed. McGraw-Hill, 2003.) Fluorescent micro-graphs of airway epithelial cells displaying actin microfilaments stained with phalloidin (Middle) and microtubules visualized with an antibody to β-tubulin (Right). Both fluorescent micrographs are counterstained with Hoechst dye (blue) to visualize nuclei. Note the distinct differences in cytoskeletal structure. CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 37 receptors and form focal adhesion complexes, which serve as points of traction with the surface over which the cell pulls itself. In addition, some molecular motors use microfilaments as tracks. MOLECULAR MOTORS The molecular motors that move proteins, organelles, and other cell parts (collectively referred to as “cargo”) to all parts of the cell are 100 to 500 kDa ATPases. They attach to their cargo at one end of the molecule and to microtubules or actin polymers with the other end, sometimes referred to as the “head.” They convert the energy of ATP into movement along the cytoskeleton, taking their cargo with them. There are three super families of molecular motors: kinesin, dynein, and my-osin. Examples of individual proteins from each superfamily are shown in Figure 2–7. It is important to note that there is extensive variation among superfamily members, allowing for specialization of function (eg, choice of cargo, cytoskeletal fil-ament type, and/or direction of movement). The conventional form of kinesin is a doubleheaded mole-cule that tends to move its cargo toward the “+” ends of microtubules. One head binds to the microtubule and then bends its neck while the other head swings forward and binds, producing almost continuous movement. Some kinesins are associated with mitosis and meiosis. Other kinesins perform different functions, including, in some instances, moving cargo to the “–” end of microtubules. Dyneins have two heads, with their neck pieces embedded in a complex of proteins. Cyto-plasmic dyneins have a function like that of conventional kinesin, except they tend to move particles and membranes to the “–” end of the microtubules. The multiple forms of myo-sin in the body are divided into 18 classes. The heads of myo-sin molecules bind to actin and produce motion by bending their neck regions (myosin II) or walking along microfila-ments, one head after the other (myosin V). In these ways, they perform functions as diverse as contraction of muscle and cell migration. CENTROSOMES Near the nucleus in the cytoplasm of eukaryotic animal cells is a centrosome. The centrosome is made up of two centrioles and surrounding amorphous pericentriolar material. The centri-oles are short cylinders arranged so that they are at right angles to each other. Microtubules in groups of three run longitudinally in the walls of each centriole (Figure 2–1). Nine of these triplets are spaced at regular intervals around the circumference. The centrosomes are microtubule-organizing centers (MTOCs) that contain γ-tubulin. The microtubules grow out of this γ-tubulin in the pericentriolar material. When a cell divides, the centrosomes duplicate themselves, and the pairs move apart to the poles of the mitotic spindle, where they monitor the steps in cell division. In multinucleate cells, a centrosome is near each nucleus. CILIA Cilia are specialized cellular projections that are used by unicellu-lar organisms to propel themselves through liquid and by multi-cellular organisms to propel mucus and other substances over the surface of various epithelia. Cilia are functionally indistinct from the eukaryotic flagella of sperm cells. Within the cilium there is an axoneme that comprises a unique arrangement of nine outer mi-crotubule doublets and two inner microtubules (“9+2” arrange-ment). Along this cytoskeleton is axonemal dynein. Coordinated dynein-microtubule interactions within the axoneme are the ba-sis of ciliary and sperm movement. At the base of the axoneme and just inside lies the basal body. It has nine circumferential triplet microtubules, like a centriole, and there is evidence that basal bodies and centrioles are interconvertible. FIGURE 2–7 Three examples of molecular motors. Conventional kinesin is shown attached to cargo, in this case a membrane-bound organelle. The way that myosin V “walks” along a microtubule is also shown. Note that the heads of the motors hydrolyze ATP and use the energy to produce motion. Cytoplasmic dynein 4 nm Cargo Light chains 80 nm Conventional kinesin Cargo-binding domain Head 2 Head 1 Head 1 Head 2 Actin ATP ADP ADP Myosin V 38 SECTION I Cellular & Molecular Basis of Medical Physiology CELL ADHESION MOLECULES Cells are attached to the basal lamina and to each other by cell adhesion molecules (CAMs) that are prominent parts of the intercellular connections described below. These adhesion proteins have attracted great attention in recent years because of their unique structural and signaling functions found to be important in embryonic development and formation of the nervous system and other tissues, in holding tissues together in adults, in inflammation and wound healing, and in the me-tastasis of tumors. Many CAMs pass through the cell mem-brane and are anchored to the cytoskeleton inside the cell. Some bind to like molecules on other cells (homophilic bind-ing), whereas others bind to nonself molecules (heterophilic binding). Many bind to laminins, a family of large cross-shaped molecules with multiple receptor domains in the ex-tracellular matrix. Nomenclature in the CAM field is somewhat chaotic, partly because the field is growing so rapidly and partly because of the extensive use of acronyms, as in other areas of modern biology. However, the CAMs can be divided into four broad families: (1) integrins, heterodimers that bind to various receptors; (2) adhesion molecules of the IgG superfamily of immunoglobulins; (3) cadherins, Ca2+-dependent molecules that mediate cell-to-cell adhesion by homophilic reactions; and (4) selectins, which have lectin-like domains that bind carbohydrates. Specific functions of some of these molecules are addressed in later chapters. The CAMs not only fasten cells to their neighbors, but they also transmit signals into and out of the cell. For example, cells that lose their contact with the extracellular matrix via integrins have a higher rate of apoptosis than anchored cells, and interactions between integrins and the cytoskeleton are involved in cell movement. INTERCELLULAR CONNECTIONS Intercellular junctions that form between the cells in tissues can be broadly split into two groups: junctions that fasten the cells to one another and to surrounding tissues, and junctions that permit transfer of ions and other molecules from one cell to another. The types of junctions that tie cells together and endow tissues with strength and stability include tight junc-tions, which are also known as the zonula occludens (Figure 2–8). The desmosome and zonula adherens also help to hold cells together, and the hemidesmosome and focal adhesions attach cells to their basal laminas. The gap junction forms a cytoplasmic “tunnel” for diffusion of small molecules (< 1000 Da) between two neighboring cells. Tight junctions characteristically surround the apical mar-gins of the cells in epithelia such as the intestinal mucosa, the walls of the renal tubules, and the choroid plexus. They are also important to endothelial barrier function. They are made up of ridges—half from one cell and half from the other—which adhere so strongly at cell junctions that they almost obliterate the space between the cells. There are three main families of transmembrane proteins that contribute to tight junctions: occludin, junctional adhesion molecules (JAMs), and claudins; and several more proteins that inter-act from the cytosolic side. Tight junctions permit the pas-sage of some ions and solute in between adjacent cells (paracellular pathway) and the degree of this “leakiness” varies, depending in part on the protein makeup of the tight junction. Extracellular fluxes of ions and solute across epithe-lia at these junctions are a significant part of overall ion and solute flux. In addition, tight junctions prevent the move-ment of proteins in the plane of the membrane, helping to maintain the different distribution of transporters and chan-nels in the apical and basolateral cell membranes that make transport across epithelia possible. In epithelial cells, each zonula adherens is usually a contin-uous structure on the basal side of the zonula occludens, and it is a major site of attachment for intracellular microfila-ments. It contains cadherins. Desmosomes are patches characterized by apposed thicken-ings of the membranes of two adjacent cells. Attached to the thickened area in each cell are intermediate filaments, some running parallel to the membrane and others radiating away from it. Between the two membrane thickenings the intercellu-lar space contains filamentous material that includes cadherins and the extracellular portions of several other transmembrane proteins. Hemidesmosomes look like half-desmosomes that attach cells to the underlying basal lamina and are connected FIGURE 2–8 Intercellular junctions in the mucosa of the small intestine. Tight junctions (zonula occludens), adherens junctions (zonula adherens), desmosomes, gap junctions, and hemidesmosomes are all shown in relative positions in a polarized epithelial cell. Tight junction (zonula occludens) Zonula adherens Desmosomes Gap junctions Hemidesmosome CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 39 intracellularly to intermediate filaments. However, they con-tain integrins rather than cadherins. Focal adhesions also attach cells to their basal laminas. As noted previously, they are labile structures associated with actin filaments inside the cell, and they play an important role in cell movement. GAP JUNCTIONS At gap junctions, the intercellular space narrows from 25 nm to 3 nm, and units called connexons in the membrane of each cell are lined up with one another (Figure 2–9). Each connexon is made up of six protein subunits called connexins. They sur-round a channel that, when lined up with the channel in the cor-responding connexon in the adjacent cell, permits substances to pass between the cells without entering the ECF. The diameter of the channel is normally about 2 nm, which permits the pas-sage of ions, sugars, amino acids, and other solutes with molec-ular weights up to about 1000. Gap junctions thus permit the rapid propagation of electrical activity from cell to cell, as well as the exchange of various chemical messengers. However, the FIGURE 2–9 Gap junction connecting the cytoplasm of two cells. A) A gap junction plaque, or collection of individual gap junctions, is shown to form multiple pores between cells that allow for the transfer of small molecules. Inset is electron micrograph from rat liver (N. Gilula). B) Topographical depiction of individual connexon and corresponding 6 connexin proteins that traverse the membrane. Note that each connexin traverses the membrane four times. (Reproduced with permission from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) 6 connexin subunits = 1 connexon (hemichannel) Presynaptic cytoplasm Extracellular loops for homophilic interactions Each of the 6 connexins has 4 membrane-spanning regions Cytoplasmic loops for regulation Extracellular space Channel formed by pores in each membrane B Postsynaptic cytoplasm Presynaptic cytoplasm A Normal extracellular space 3.5 nm 20 nm 40 SECTION I Cellular & Molecular Basis of Medical Physiology gap junction channels are not simply passive, nonspecific con-duits. At least 20 different genes code for connexins in humans, and mutations in these genes can lead to diseases that are highly selective in terms of the tissues involved and the type of com-munication between cells produced. For instance, X-linked Charcot–Marie–Tooth disease is a peripheral neuropathy as-sociated with mutation of one particular connexin gene. Exper-iments in mice in which particular connexins are deleted by gene manipulation or replaced with different connexins con-firm that the particular connexin subunits that make up con-nexons determine their permeability and selectivity. Recently it has been shown that connexons can be used as channels to re-lease small molecules from the cytosol into the ECF. NUCLEUS & RELATED STRUCTURES A nucleus is present in all eukaryotic cells that divide. If a cell is cut in half, the anucleate portion eventually dies without divid-ing. The nucleus is made up in large part of the chromosomes, the structures in the nucleus that carry a complete blueprint for all the heritable species and individual characteristics of the an-imal. Except in germ cells, the chromosomes occur in pairs, one originally from each parent. Each chromosome is made up of a giant molecule of DNA. The DNA strand is about 2 m long, but it can fit in the nucleus because at intervals it is wrapped around a core of histone proteins to form a nucleosome. There are about 25 million nucleosomes in each nucleus. Thus, the struc-ture of the chromosomes has been likened to a string of beads. The beads are the nucleosomes, and the linker DNA between them is the string. The whole complex of DNA and proteins is called chromatin. During cell division, the coiling around his-tones is loosened, probably by acetylation of the histones, and pairs of chromosomes become visible, but between cell divi-sions only clumps of chromatin can be discerned in the nucleus. The ultimate units of heredity are the genes on the chromo-somes). As discussed in Chapter 1, each gene is a portion of the DNA molecule. The nucleus of most cells contains a nucleolus (Figure 2–1), a patchwork of granules rich in RNA. In some cells, the nucleus contains several of these structures. Nucleoli are most promi-nent and numerous in growing cells. They are the site of syn-thesis of ribosomes, the structures in the cytoplasm in which proteins are synthesized. The interior of the nucleus has a skeleton of fine filaments that are attached to the nuclear membrane, or envelope (Fig-ure 2–1), which surrounds the nucleus. This membrane is a double membrane, and spaces between the two folds are called perinuclear cisterns. The membrane is permeable only to small molecules. However, it contains nuclear pore complexes. Each complex has eightfold symmetry and is made up of about 100 proteins organized to form a tunnel through which trans-port of proteins and mRNA occurs. There are many transport pathways, and proteins called importins and exportins have been isolated and characterized. Much current research is focused on transport into and out of the nucleus, and a more detailed understanding of these processes should emerge in the near future. ENDOPLASMIC RETICULUM The endoplasmic reticulum is a complex series of tubules in the cytoplasm of the cell (Figure 2–1). The inner limb of its membrane is continuous with a segment of the nuclear mem-brane, so in effect this part of the nuclear membrane is a cis-tern of the endoplasmic reticulum. The tubule walls are made up of membrane. In rough, or granular, endoplasmic reticulum, ribosomes are attached to the cytoplasmic side of the membrane, whereas in smooth, or agranular, endoplasmic reticulum, ri-bosomes are absent. Free ribosomes are also found in the cyto-plasm. The granular endoplasmic reticulum is concerned with protein synthesis and the initial folding of polypeptide chains with the formation of disulfide bonds. The agranular endoplas-mic reticulum is the site of steroid synthesis in steroid-secreting cells and the site of detoxification processes in other cells. A modified endoplasmic reticulum, the sarcoplasmic reticulum, plays an important role in skeletal and cardiac muscle. In partic-ular, the endoplasmic or sarcoplasmic reticulum can sequester Ca2+ ions and allow for their release as signaling molecules in the cytosol. RIBOSOMES The ribosomes in eukaryotes measure approximately 22 × 32 nm. Each is made up of a large and a small subunit called, on the basis of their rates of sedimentation in the ultracentrifuge, the 60S and 40S subunits. The ribosomes are complex structures, containing many different proteins and at least three ribosomal RNAs. They are the sites of protein synthesis. The ribosomes that become at-tached to the endoplasmic reticulum synthesize all transmem-brane proteins, most secreted proteins, and most proteins that are stored in the Golgi apparatus, lysosomes, and endosomes. These proteins typically have a hydrophobic signal peptide at one end (Figure 2–10). The polypeptide chains that form these proteins are extruded into the endoplasmic reticulum. The free ribosomes synthesize cytoplasmic proteins such as hemoglobin and the pro-teins found in peroxisomes and mitochondria. GOLGI APPARATUS & VESICULAR TRAFFIC The Golgi apparatus is a collection of membrane-enclosed sacs (cisterns) that are stacked like dinner plates (Figure 2–1). There are usually about six sacs in each apparatus, but there may be more. One or more Golgi apparati are present in all eu-karyotic cells, usually near the nucleus. Much of the organiza-tion of the Golgi is directed at proper glycosylation of proteins and lipids. There are more than 200 enzymes that function to add, remove, or modify sugars from proteins and lipids in the Golgi apparatus. CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 41 FIGURE 2–10 Rough endoplasmic reticulum and protein translation. Messenger RNA and ribosomes meet up in the cytosol for transla-tion. Proteins that have appropriate signal peptides begin translation, then associate with the endoplasmic reticulum (ER) to complete translation. The association of ribosomes is what gives the ER its “rough” appearance. (Reproduced with permission from Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology: The Mechanisms of Body Function, 11th ed. McGraw-Hill, 2008.) Cytoplasm Extracellular fluid mRNA from Gene A mRNA from Gene B Free ribosome Signal sequence Rough endoplasmic reticulum Growing polypeptide chain Carbohydrate group Cleaved signal sequences Vesicle Golgi apparatus Secretory vesicle Plasma membrane Exocytosis Digestive protein from Gene B Lysosome Secreted protein from Gene A 42 SECTION I Cellular & Molecular Basis of Medical Physiology The Golgi apparatus is a polarized structure, with cis and trans sides (Figure 2–11). Membranous vesicles containing newly synthesized proteins bud off from the granular endo-plasmic reticulum and fuse with the cistern on the cis side of the apparatus. The proteins are then passed via other vesicles to the middle cisterns and finally to the cistern on the trans side, from which vesicles branch off into the cytoplasm. From the trans Golgi, vesicles shuttle to the lysosomes and to the cell exterior via constitutive and nonconstitutive pathways, both involving exocytosis. Conversely, vesicles are pinched off from the cell membrane by endocytosis and pass to endo-somes. From there, they are recycled. Vesicular traffic in the Golgi, and between other membra-nous compartments in the cell, is regulated by a combination of common mechanisms along with special mechanisms that determine where inside the cell they will go. One prominent feature is the involvement of a series of regulatory proteins controlled by GTP or GDP binding (small G proteins) associ-ated with vesicle assembly and delivery. A second prominent feature is the presence of proteins called SNAREs (for solu-ble N-ethylmaleimide-sensitive factor attachment receptor). The v- (for vesicle) SNAREs on vesicle membranes interact in a lock-and-key fashion with t- (for target) SNAREs. Individual vesicles also contain structural protein or lipids in their mem-brane that help to target them for specific membrane compart-ments (eg, Golgi sacs, cell membranes). QUALITY CONTROL The processes involved in protein synthesis, folding, and migra-tion to the various parts of the cell are so complex that it is re-markable that more errors and abnormalities do not occur. The fact that these processes work as well as they do is because of mechanisms at each level that are responsible for “quality con-trol.” Damaged DNA is detected and repaired or bypassed. The various RNAs are also checked during the translation process. Finally, when the protein chains are in the endoplasmic reticu-lum and Golgi apparatus, defective structures are detected and the abnormal proteins are degraded in lysosomes and protea-somes. The net result is a remarkable accuracy in the produc-tion of the proteins needed for normal body function. APOPTOSIS In addition to dividing and growing under genetic control, cells can die and be absorbed under genetic control. This process is called programmed cell death, or apoptosis (Gr. apo “away” + ptosis “fall”). It can be called “cell suicide” in the sense that the cell’s own genes play an active role in its demise. It should be distinguished from necrosis (“cell murder”), in which healthy cells are destroyed by external processes such as inflammation. Apoptosis is a very common process during development and in adulthood. In the central nervous system, large numbers of neurons are produced and then die during the remodeling that occurs during development and synapse formation. In the immune system, apoptosis gets rid of inappropriate clones of immunocytes and is responsible for the lytic effects of gluco-corticoids on lymphocytes. Apoptosis is also an important fac-tor in processes such as removal of the webs between the fingers in fetal life and regression of duct systems in the course of sexual development in the fetus. In adults, it participates in the cyclic breakdown of the endometrium that leads to men-struation. In epithelia, cells that lose their connections to the basal lamina and neighboring cells undergo apoptosis. This is responsible for the death of the enterocytes sloughed off the tips of intestinal villi. Abnormal apoptosis probably occurs in autoimmune diseases, neurodegenerative diseases, and cancer. It is interesting that apoptosis occurs in invertebrates, including nematodes and insects. However, its molecular mechanism is much more complex than that in vertebrates. One final common pathway bringing about apoptosis is acti-vation of caspases, a group of cysteine proteases. Many of these have been characterized to date in mammals; 11 have been found in humans. They exist in cells as inactive proenzymes FIGURE 2–11 Cellular structures involved in protein processing. See text for details. Lysosome Nucleus Late endosome Early endosome Regulated secretion Constitutive secretion Endocytosis Recycling Secretory granules Golgi apparatus ER CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 43 until activated by the cellular machinery. The net result is DNA fragmentation, cytoplasmic and chromatin condensa-tion, and eventually membrane bleb formation, with cell breakup and removal of the debris by phagocytes (see Clinical Box 2–2). TRANSPORT ACROSS CELL MEMBRANES There are several mechanisms of transport across cellular membranes. Primary pathways include exocytosis, endocyto-sis, movement through ion channels, and primary and secon-dary active transport. Each of these are discussed below. EXOCYTOSIS Vesicles containing material for export are targeted to the cell membrane (Figure 2–11), where they bond in a similar man-ner to that discussed in vesicular traffic between Golgi stacks, via the v-SNARE/t-SNARE arrangement. The area of fusion then breaks down, leaving the contents of the vesicle outside and the cell membrane intact. This is the Ca2+-dependent pro-cess of exocytosis (Figure 2–12). Note that secretion from the cell occurs via two pathways (Figure 2–11). In the nonconstitutive pathway, proteins from the Golgi apparatus initially enter secretory granules, where processing of prohormones to the mature hormones occurs before exocytosis. The other pathway, the constitutive path-way, involves the prompt transport of proteins to the cell membrane in vesicles, with little or no processing or storage. The nonconstitutive pathway is sometimes called the regu-lated pathway, but this term is misleading because the output of proteins by the constitutive pathway is also regulated. ENDOCYTOSIS Endocytosis is the reverse of exocytosis. There are various types of endocytosis named for the size of particles being in-gested as well as the regulatory requirements for the particular process. These include phagocytosis, pinocytosis, clathrin-mediated endocytosis, caveolae-dependent uptake, and nonclathrin/noncaveolae endocytosis. Phagocytosis (“cell eating”) is the process by which bacteria, dead tissue, or other bits of microscopic material are engulfed by cells such as the polymorphonuclear leukocytes of the blood. The material makes contact with the cell membrane, which then invaginates. The invagination is pinched off, leaving the engulfed material in the membrane-enclosed vacuole and the cell membrane intact. Pinocytosis (“cell drinking”) is a similar process with the vesicles much smaller in size and the sub-stances ingested are in solution. The small size membrane that is ingested should not be misconstrued; cells undergoing active pinocytosis (eg, macrophages) can ingest the equivalent of their entire cell membrane in just 1 hour. Clathrin-mediated endocytosis occurs at membrane inden-tations where the protein clathrin accumulates. Clathrin mole-cules have the shape of triskelions, with three “legs” radiating from a central hub (Figure 2–13). As endocytosis progresses, CLINICAL BOX 2–2 Molecular Medicine Fundamental research on molecular aspects of genetics, regulation of gene expression, and protein synthesis has been paying off in clinical medicine at a rapidly accelerat-ing rate. One early dividend was an understanding of the mecha-nisms by which antibiotics exert their effects. Almost all act by inhibiting protein synthesis at one or another of the steps described previously. Antiviral drugs act in a similar way; for example, acyclovir and ganciclovir act by inhibiting DNA polymerase. Some of these drugs have this effect pri-marily in bacteria, but others inhibit protein synthesis in the cells of other animals, including mammals. This fact makes antibiotics of great value for research as well as for treat-ment of infections. Single genetic abnormalities that cause over 600 human diseases have now been identified. Many of the diseases are rare, but others are more common and some cause conditions that are severe and eventually fatal. Examples include the defectively regulated Cl– channel in cystic fibro-sis and the unstable trinucleotide repeats in various parts of the genome that cause Huntington’s disease, the fragile X syndrome, and several other neurologic diseases. Abnor-malities in mitochondrial DNA can also cause human dis-eases such as Leber’s hereditary optic neuropathy and some forms of cardiomyopathy. Not surprisingly, genetic aspects of cancer are probably receiving the greatest cur-rent attention. Some cancers are caused by oncogenes, genes that are carried in the genomes of cancer cells and are responsible for producing their malignant properties. These genes are derived by somatic mutation from closely related proto-oncogenes, which are normal genes that control growth. Over 100 oncogenes have been described. Another group of genes produce proteins that suppress tu-mors, and more than 10 of these tumor suppressor genes have been described. The most studied of these is the p53 gene on human chromosome 17. The p53 protein pro-duced by this gene triggers apoptosis. It is also a nuclear transcription factor that appears to increase production of a 21-kDa protein that blocks two cell cycle enzymes, slow-ing the cycle and permitting repair of mutations and other defects in DNA. The p53 gene is mutated in up to 50% of human cancers, with the production of p53 proteins that fail to slow the cell cycle and permit other mutations in DNA to persist. The accumulated mutations eventually cause cancer. 44 SECTION I Cellular & Molecular Basis of Medical Physiology the clathrin molecules form a geometric array that surrounds the endocytotic vesicle. At the neck of the vesicle, the GTP binding protein dynamin is involved, either directly or indi-rectly, in pinching off the vesicle. Once the complete vesicle is formed, the clathrin falls off and the three-legged proteins recy-cle to form another vesicle. The vesicle fuses with and dumps its contents into an early endosome (Figure 2–11). From the early endosome, a new vesicle can bud off and return to the cell membrane. Alternatively, the early endosome can become a late endosome and fuse with a lysosome (Figure 2–11) in which the contents are digested by the lysosomal proteases. Clathrin-mediated endocytosis is responsible for the internal-ization of many receptors and the ligands bound to them— including, for example, nerve growth factor and low-density lipoproteins. It also plays a major role in synaptic function. It is apparent that exocytosis adds to the total amount of membrane surrounding the cell, and if membrane were not removed elsewhere at an equivalent rate, the cell would enlarge. However, removal of cell membrane occurs by endocytosis, and such exocytosis–endocytosis coupling main-tains the surface area of the cell at its normal size. RAFTS & CAVEOLAE Some areas of the cell membrane are especially rich in choles-terol and sphingolipids and have been called rafts. These rafts are probably the precursors of flask-shaped membrane de-pressions called caveolae (little caves) when their walls be-come infiltrated with a protein called caveolin that resembles clathrin. There is considerable debate about the functions of rafts and caveolae, with evidence that they are involved in cho-lesterol regulation and transcytosis. It is clear, however, that cholesterol can interact directly with caveolin, effectively lim-iting the protein’s ability to move around in the membrane. Internalization via caveolae involves binding of cargo to cave-olin and regulation by dynamin. Caveolae are prominent in endothelial cells, where they help in the uptake of nutrients from the blood. FIGURE 2–12 Exocytosis and endocytosis. Note that in exocytosis the cytoplasmic sides of two membranes fuse, whereas in endocytosis two noncytoplasmic sides fuse. (Reproduced with permission from Alberts B et al: Molecular Biology of the Cell, 4th ed. Garland Science, 2002.) Exocytosis Endocytosis Cytoplasm FIGURE 2–13 Clathrin molecule on the surface of an endocytotic vesicle. Note the characteristic triskelion shape and the fact that with other clathrin molecules it forms a net supporting the vesicle. CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 45 COATS & VESICLE TRANSPORT It now appears that all vesicles involved in transport have pro-tein coats. In humans, 53 coat complex subunits have been identified. Vesicles that transport proteins from the trans Gol-gi to lysosomes have assembly protein 1 (AP-1) clathrin coats, and endocytotic vesicles that transport to endosomes have AP-2 clathrin coats. Vesicles that transport between the endoplasmic reticulum and the Golgi have coat proteins I and II (COPI and COPII). Certain amino acid sequences or at-tached groups on the transported proteins target the proteins for particular locations. For example, the amino acid sequence Asn–Pro–any amino acid–Tyr targets transport from the cell surface to the endosomes, and mannose-6-phosphate groups target transfer from the Golgi to mannose-6-phosphate recep-tors (MPR) on the lysosomes. Various small G proteins of the Rab family are especially important in vesicular traffic. They appear to guide and facili-tate orderly attachments of these vesicles. To illustrate the complexity of directing vesicular traffic, humans have 60 Rab proteins and 35 SNARE proteins. MEMBRANE PERMEABILITY & MEMBRANE TRANSPORT PROTEINS An important technique that has permitted major advances in our knowledge about transport proteins is patch clamping. A micropipette is placed on the membrane of a cell and forms a tight seal to the membrane. The patch of membrane under the pipette tip usually contains only a few transport proteins, al-lowing for their detailed biophysical study (Figure 2–14). The cell can be left intact (cell-attached patch clamp). Alterna-tively, the patch can be pulled loose from the cell, forming an inside-out patch. A third alternative is to suck out the patch with the micropipette still attached to the rest of the cell mem-brane, providing direct access to the interior of the cell (whole cell recording). Small, nonpolar molecules (including O2 and N2) and small uncharged polar molecules such as CO2 diffuse across the lipid membranes of cells. However, the membranes have very limited permeability to other substances. Instead, they cross the membranes by endocytosis and exocytosis and by passage through highly specific transport proteins, transmembrane proteins that form channels for ions or transport substances such as glucose, urea, and amino acids. The limited perme-ability applies even to water, with simple diffusion being sup-plemented throughout the body with various water channels (aquaporins). For reference, the sizes of ions and other bio-logically important substances are summarized in Table 2–2. Some transport proteins are simple aqueous ion channels, though many of these have special features that make them selective for a given substance such as Ca2+ or, in the case of aquaporins, for water. These membrane-spanning proteins (or collections of proteins) have tightly regulated pores that can be gated opened or closed in response to local changes (Figure 2–15). Some are gated by alterations in membrane potential (voltage-gated), whereas others are opened or closed in response to a ligand (ligand-gated). The ligand is FIGURE 2–14 Patch clamp to investigate transport. In a patch clamp experiment, a small pipette is carefully maneuvered to seal off a portion of a cell membrane. The pipette has an electrode bathed in an appropriate solution that allows for recording of electrical changes through any pore in the membrane (shown below). The illustrated setup is termed an “inside-out patch” because of the orientation of the mem-brane with reference to the electrode. Other configurations include cell attached, whole cell, and outside-out patches. (Modified from Ackerman MJ, Clapham DE: Ion channels: Basic science and clinical disease. N Engl J Med 1997;336:1575.) TABLE 2–2 Size of hydrated ions and other substances of biologic interest. Substance Atomic or Molecular Weight Radius (nm) Cl– 35 0.12 K+ 39 0.12 H2O 18 0.12 Ca2+ 40 0.15 Na+ 23 0.18 Urea 60 0.23 Li+ 7 0.24 Glucose 180 0.38 Sucrose 342 0.48 Inulin 5000 0.75 Albumin 69,000 7.50 Data from Moore EW: Physiology of Intestinal Water and Electrolyte Absorption. American Gastroenterological Association, 1976. Inside-out patch Cell membrane Pipette Electrode Closed Open ms pA 46 SECTION I Cellular & Molecular Basis of Medical Physiology often external (eg, a neurotransmitter or a hormone). How-ever, it can also be internal; intracellular Ca2+, cAMP, lipids, or one of the G proteins produced in cells can bind directly to channels and activate them. Some channels are also opened by mechanical stretch, and these mechanosensitive channels play an important role in cell movement. Other transport proteins are carriers that bind ions and other molecules and then change their configuration, moving the bound molecule from one side of the cell membrane to the other. Molecules move from areas of high concentration to areas of low concentration (down their chemical gradient), and cations move to negatively charged areas whereas anions move to positively charged areas (down their electrical gradi-ent). When carrier proteins move substances in the direction of their chemical or electrical gradients, no energy input is required and the process is called facilitated diffusion. A typi-cal example is glucose transport by the glucose transporter, which moves glucose down its concentration gradient from the ECF to the cytoplasm of the cell. Other carriers transport sub-stances against their electrical and chemical gradients. This form of transport requires energy and is called active trans-port. In animal cells, the energy is provided almost exclusively by hydrolysis of ATP. Not surprisingly, therefore, many carrier molecules are ATPases, enzymes that catalyze the hydrolysis of ATP. One of these ATPases is sodium–potassium adenosine triphosphatase (Na, K ATPase), which is also known as the Na, K pump. There are also H, K ATPases in the gastric mucosa and the renal tubules. Ca2+ATPase pumps Ca2+ out of cells. Proton ATPases acidify many intracellular organelles, including parts of the Golgi complex and lysosomes. Some of the transport proteins are called uniports because they transport only one substance. Others are called symports because transport requires the binding of more than one sub-stance to the transport protein and the substances are trans-ported across the membrane together. An example is the symport in the intestinal mucosa that is responsible for the cotransport by facilitated diffusion of Na+ and glucose from the intestinal lumen into mucosal cells. Other transporters are called antiports because they exchange one substance for another. ION CHANNELS There are ion channels specific for K+, Na+, Ca2+, and Cl–, as well as channels that are nonselective for cations or anions. Each type of channel exists in multiple forms with diverse properties. Most are made up of identical or very similar sub-units. Figure 2–16 shows the multiunit structure of various channels in diagrammatic cross-section. Most K+ channels are tetramers, with each of the four sub-units forming part of the pore through which K+ ions pass. Structural analysis of a bacterial voltage-gated K+ channel indicates that each of the four subunits have a paddle-like extension containing four charges. When the channel is closed, these extensions are near the negatively charged inte-rior of the cell. When the membrane potential is reduced, the paddles containing the charges bend through the membrane to its exterior surface, causing the channel to open. The bacte-rial K+ channel is very similar to the voltage-gated K+ chan-nels in a wide variety of species, including mammals. In the acetylcholine ion channel and other ligand-gated cation or anion channels, five subunits make up the pore. Members of the ClC family of Cl– channels are dimers, but they have two pores, one in each subunit. Finally, aquaporins are tetramers FIGURE 2–15 Regulation of gating in ion channels. Several types of gating are shown for ion channels. A) Ligand-gated channels open in response to ligand binding. B) Protein phosphorylation or de-phosphorylation regulate opening and closing of some ion channels. C) Changes in membrane potential alter channel openings. D) Me-chanical stretch of the membrane results in channel opening. (Repro-duced with permission from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Pi P Closed A Ligand-gated Open B Phosphorylation-gated C Voltage-gated D Stretch or pressure-gated Bind ligand Stretch Cytoskeleton Phosphorylate Change membrane potential Dephosphorylate + + + + + + + + – – – – – – – – CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 47 with a water pore in each of the subunits. Recently, a number of ion channels with intrinsic enzyme activity have been cloned. More than 30 different voltage-gated or cyclic nucleo-tide-gated Na+ and Ca2+ channels of this type have been described. Representative Na+, Ca2+, and K+ channels are shown in extended diagrammatic form in Figure 2–17. Another family of Na+channels with a different structure has been found in the apical membranes of epithelial cells in the kidneys, colon, lungs, and brain. The epithelial sodium channels (ENaCs) are made up of three subunits encoded by three different genes. Each of the subunits probably spans the membrane twice, and the amino terminal and carboxyl termi-nal are located inside the cell. The α subunit transports Na+, whereas the β and γ subunits do not. However, the addition of the β and γ subunits increases Na+ transport through the α subunit. ENaCs are inhibited by the diuretic amiloride, which binds to the α subunit, and they used to be called amiloride-inhibitable Na+channels. The ENaCs in the kidney play an important role in the regulation of ECF volume by aldoster-one. ENaC knockout mice are born alive but promptly die because they cannot move Na+, and hence water, out of their lungs. Humans have several types of Cl– channels. The ClC dimeric channels are found in plants, bacteria, and animals, and there are nine different ClC genes in humans. Other Cl– channels have the same pentameric form as the acetylcho-line receptor; examples include the γ-aminobutyric acid A (GABAA) and glycine receptors in the central nervous system (CNS). The cystic fibrosis transmembrane conductance regu-lator (CFTR) that is mutated in cystic fibrosis is also a Cl– channel. Ion channel mutations cause a variety of channelop-athies—diseases that mostly affect muscle and brain tissue and produce episodic paralyses or convulsions. Na, K ATPase As noted previously, Na, K ATPase catalyzes the hydrolysis of ATP to adenosine diphosphate (ADP) and uses the energy to extrude three Na+ from the cell and take two K+ into the cell for each molecule of ATP hydrolyzed. It is an electrogenic pump in that it moves three positive charges out of the cell for each two that it moves in, and it is therefore said to have a coupling ratio of 3:2. It is found in all parts of the body. Its activity is in-hibited by ouabain and related digitalis glycosides used in the treatment of heart failure. It is a heterodimer made up of an α subunit with a molecular weight of approximately 100,000 and a β subunit with a molecular weight of approximately 55,000. Both extend through the cell membrane (Figure 2–18). Separa-tion of the subunits eliminates activity. The β subunit is a gly-coprotein, whereas Na+ and K+ transport occur through the α subunit. The β subunit has a single membrane-spanning do-main and three extracellular glycosylation sites, all of which ap-pear to have attached carbohydrate residues. These residues account for one third of its molecular weight. The α subunit probably spans the cell membrane 10 times, with the amino and carboxyl terminals both located intracellularly. This sub-unit has intracellular Na+- and ATP-binding sites and a phos-phorylation site; it also has extracellular binding sites for K+ and ouabain. The endogenous ligand of the ouabain-binding site is unsettled. When Na+ binds to the α subunit, ATP also binds and is converted to ADP, with a phosphate being trans-ferred to Asp 376, the phosphorylation site. This causes a change in the configuration of the protein, extruding Na+ into the ECF. K+ then binds extracellularly, dephosphorylating the α subunit, which returns to its previous conformation, releas-ing K+ into the cytoplasm. The α and β subunits are heterogeneous, with α1, α2, and α3 subunits and β1, β2, and β3 subunits described so far. The α1 isoform is found in the membranes of most cells, whereas α2 is present in muscle, heart, adipose tissue, and brain, and α3 is present in heart and brain. The β1 subunit is widely distributed but is absent in certain astrocytes, vestibular cells of the inner ear, and glycolytic fast-twitch muscles. The fast-twitch muscles contain only β2 subunits. The different α and β subunit struc-tures of Na, K ATPase in various tissues probably represent spe-cialization for specific tissue functions. REGULATION OF Na, K ATPase ACTIVITY The amount of Na+ normally found in cells is not enough to saturate the pump, so if the Na+ increases, more is pumped out. Pump activity is affected by second messenger molecules (eg, cAMP and diacylglycerol [DAG]). The magnitude and di-rection of the altered pump effects vary with the experimental conditions. Thyroid hormones increase pump activity by a ge-nomic action to increase the formation of Na, K ATPase mol-ecules. Aldosterone also increases the number of pumps, although this effect is probably secondary. Dopamine in the kidney inhibits the pump by phosphorylating it, causing a natriuresis. Insulin increases pump activity, probably by a va-riety of different mechanisms. FIGURE 2–16 Different ways in which ion channels form pores. Many K+ channels are tetramers (A), with each protein subunit forming part of the channel. In ligand-gated cation and anion channels (B) such as the acetylcholine receptor, five identical or very similar sub-units form the channel. Cl– channels from the ClC family are dimers (C), with an intracellular pore in each subunit. Aquaporin water channels (D) are tetramers with an intracellular channel in each subunit. (Reproduced with permission from Jentsch TJ: Chloride channels are different. Nature 2002;415:276.) A B C D 48 SECTION I Cellular & Molecular Basis of Medical Physiology SECONDARY ACTIVE TRANSPORT In many situations, the active transport of Na+ is coupled to the transport of other substances (secondary active trans-port). For example, the luminal membranes of mucosal cells in the small intestine contain a symport that transports glu-cose into the cell only if Na+ binds to the protein and is trans-ported into the cell at the same time. From the cells, the glucose enters the blood. The electrochemical gradient for Na+ is maintained by the active transport of Na+ out of the mucosal cell into ECF. Other examples are shown in Figure 2– 19. In the heart, Na,K ATPase indirectly affects Ca2+ trans-port. An antiport in the membranes of cardiac muscle cells normally exchanges intracellular Ca2+ for extracellular Na+. Active transport of Na+ and K+ is one of the major energy-using processes in the body. On the average, it accounts for FIGURE 2–17 Diagrammatic representation of the pore-forming subunits of three ion channels. The α subunit of the Na+ and Ca2+ channels traverse the membrane 24 times in four repeats of six membrane-spanning units. Each repeat has a “P” loop between membrane spans 5 and 6 that does not traverse the membrane. These P loops are thought to form the pore. Note that span 4 of each repeat is colored in red, rep-resenting its net “+” charge. The K+ channel has only a single repeat of the six spanning regions and P loop. Four K+ subunits are assembled for a functional K+ channel. (Reproduced with permission from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Extracellular side Na+ channel I Cytoplasmic side P P P P 1 2 3 5 6 4 1 2 3 5 6 4 1 2 3 5 6 4 1 2 3 5 6 4 Ca2+ channel NH2 COOH NH2 NH2 COOH COOH 1 2 3 5 6 4 1 2 3 5 6 4 1 2 3 5 6 4 1 2 3 5 6 4 K+ channel 1 2 3 5 6 4 II III IV P P P P P CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 49 about 24% of the energy utilized by cells, and in neurons it accounts for 70%. Thus, it accounts for a large part of the basal metabolism. A major payoff for this energy use is the establishment of the electrochemical gradient in cells. TRANSPORT ACROSS EPITHELIA In the gastrointestinal tract, the pulmonary airways, the renal tubules, and other structures, substances enter one side of a cell and exit another, producing movement of the substance from one side of the epithelium to the other. For transepithelial trans-port to occur, the cells need to be bound by tight junctions and, obviously, have different ion channels and transport proteins in different parts of their membranes. Most of the instances of sec-ondary active transport cited in the preceding paragraph in-volve transepithelial movement of ions and other molecules. THE CAPILLARY WALL FILTRATION The capillary wall separating plasma from interstitial fluid is different from the cell membranes separating interstitial fluid from intracellular fluid because the pressure difference across it makes filtration a significant factor in producing movement of water and solute. By definition, filtration is the process by which fluid is forced through a membrane or other barrier be-cause of a difference in pressure on the two sides. ONCOTIC PRESSURE The structure of the capillary wall varies from one vascular bed to another. However, in skeletal muscle and many other organs, water and relatively small solutes are the only substances that cross the wall with ease. The apertures in the junctions between the endothelial cells are too small to permit plasma proteins and other colloids to pass through in significant quantities. The col-loids have a high molecular weight but are present in large amounts. Small amounts cross the capillary wall by vesicular transport, but their effect is slight. Therefore, the capillary wall behaves like a membrane impermeable to colloids, and these ex-ert an osmotic pressure of about 25 mm Hg. The colloid osmot-ic pressure due to the plasma colloids is called the oncotic pressure. Filtration across the capillary membrane as a result of the hydrostatic pressure head in the vascular system is opposed by the oncotic pressure. The way the balance between the hy-drostatic and oncotic pressures controls exchanges across the capillary wall is considered in detail in Chapter 32. TRANSCYTOSIS Vesicles are present in the cytoplasm of endothelial cells, and tagged protein molecules injected into the bloodstream have been found in the vesicles and in the interstitium. This indi-cates that small amounts of protein are transported out of cap-illaries across endothelial cells by endocytosis on the capillary side followed by exocytosis on the interstitial side of the cells. The transport mechanism makes use of coated vesicles that appear to be coated with caveolin and is called transcytosis, vesicular transport, or cytopempsis. FIGURE 2–18 Na+–K+ ATPase. The intracellular portion of the α subunit has a Na+-binding site (1), a phosphorylation site (4), and an ATP-binding site (5). The extracellular portion has a K+-binding site (2) and an ouabain-binding site (3). (From Horisberger J-D et al: Structure–function relationship of Na–K-ATPase. Annu Rev Physiol 1991;53:565. Reproduced with permission from the Annual Review of Physiology, vol. 53. Copyright © 1991 by Annual Reviews) 3 2 1 4 5 2K+ β α 3Na+ ECF Cytoplasm Ouabain FIGURE 2–19 Composite diagram of main secondary effects of active transport of Na+ and K+. Na,K ATPase converts the chemi-cal energy of ATP hydrolysis into maintenance of an inward gradient for Na+ and an outward gradient for K+. The energy of the gradients is used for countertransport, cotransport, and maintenance of the mem-brane potential. Some examples of cotransport and countertransport that use these gradients are shown. (Reproduced with permission from Skou JC: The Na–K pump. News Physiol Sci 1992;7:95.) Active transport 2K+ Ouabain 3Na+ Na+ Na+ H+ H+ K+ K+ Ca2+ ATP 3Na+ Na+ Na+ Na+ ADP + Pi Cl− K+, 2Cl− Cotransport Countertransport Sugars or amino acids Vm = −70 mV Na+ 140 meq/L K+ 4 − Cl− 105 − + + + + − − − − Na+ 15 meq/L K+ 150 − Cl− 7 − Cl− 50 SECTION I Cellular & Molecular Basis of Medical Physiology INTERCELLULAR COMMUNICATION Cells communicate with one another via chemical messen-gers. Within a given tissue, some messengers move from cell to cell via gap junctions without entering the ECF. In addition, cells are affected by chemical messengers secreted into the ECF, or by direct cell–cell contacts. Chemical messengers typ-ically bind to protein receptors on the surface of the cell or, in some instances, in the cytoplasm or the nucleus, triggering se-quences of intracellular changes that produce their physiolog-ic effects. Three general types of intercellular communication are mediated by messengers in the ECF: (1) neural communi-cation, in which neurotransmitters are released at synaptic junctions from nerve cells and act across a narrow synaptic cleft on a postsynaptic cell; (2) endocrine communication, in which hormones and growth factors reach cells via the circu-lating blood or the lymph; and (3) paracrine communication, in which the products of cells diffuse in the ECF to affect neighboring cells that may be some distance away (Figure 2–20). In addition, cells secrete chemical messengers that in some sit-uations bind to receptors on the same cell, that is, the cell that secreted the messenger (autocrine communication). The chemical messengers include amines, amino acids, steroids, polypeptides, and in some instances, lipids, purine nucleo-tides, and pyrimidine nucleotides. It is worth noting that in various parts of the body, the same chemical messenger can function as a neurotransmitter, a paracrine mediator, a hor-mone secreted by neurons into the blood (neural hormone), and a hormone secreted by gland cells into the blood. An additional form of intercellular communication is called juxtacrine communication. Some cells express multiple repeats of growth factors such as transforming growth factor alpha (TGFα) extracellularly on transmembrane proteins that provide an anchor to the cell. Other cells have TGFα receptors. Consequently, TGFα anchored to a cell can bind to a TGFα receptor on another cell, linking the two. This could be important in producing local foci of growth in tissues. RECEPTORS FOR CHEMICAL MESSENGERS The recognition of chemical messengers by cells typically be-gins by interaction with a receptor at that cell. There have been over 20 families of receptors for chemical messengers charac-terized. These proteins are not static components of the cell, but their numbers increase and decrease in response to vari-ous stimuli, and their properties change with changes in phys-iological conditions. When a hormone or neurotransmitter is present in excess, the number of active receptors generally de-creases (down-regulation), whereas in the presence of a defi-ciency of the chemical messenger, there is an increase in the number of active receptors (up-regulation). In its actions on the adrenal cortex, angiotensin II is an exception; it increases rather than decreases the number of its receptors in the adre-nal. In the case of receptors in the membrane, receptor-medi-ated endocytosis is responsible for down-regulation in some instances; ligands bind to their receptors, and the ligand– receptor complexes move laterally in the membrane to coated pits, where they are taken into the cell by endocytosis (inter-nalization). This decreases the number of receptors in the membrane. Some receptors are recycled after internalization, whereas others are replaced by de novo synthesis in the cell. Another type of down-regulation is desensitization, in which receptors are chemically modified in ways that make them less responsive. MECHANISMS BY WHICH CHEMICAL MESSENGERS ACT Receptor–ligand interaction is usually just the beginning of the cell response. This event is transduced into secondary responses within the cell that can be divided into four broad categories: (1) ion channel activation, (2) G-protein activation, (3) activa-tion of enzyme activity within the cell, or (4) direct activation of transcription. Within each of these groups, responses can be quite varied. Some of the common mechanisms by which FIGURE 2–20 Intercellular communication by chemical mediators. A, autocrine; P, paracrine. A P ENDOCRINE Message transmission Local or general Specificity depends on Anatomic location Local Local Anatomic location and receptors GAP JUNCTIONS SYNAPTIC PARACRINE AND AUTOCRINE By circulating body fluids General Receptors Receptors Locally diffuse By diffusion in interstitial fluid Across synaptic cleft Directly from cell to cell CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 51 chemical messengers exert their intracellular effects are sum-marized in Table 2–3. Ligands such as acetylcholine bind di-rectly to ion channels in the cell membrane, changing their conductance. Thyroid and steroid hormones, 1,25-dihydroxy-cholecalciferol, and retinoids enter cells and act on one or an-other member of a family of structurally related cytoplasmic or nuclear receptors. The activated receptor binds to DNA and in-creases transcription of selected mRNAs. Many other ligands in the ECF bind to receptors on the surface of cells and trigger the release of intracellular mediators such as cAMP, IP3, and DAG that initiate changes in cell function. Consequently, the extra-cellular ligands are called “first messengers” and the intra-cellular mediators are called “second messengers.” Second messengers bring about many short-term changes in cell func-tion by altering enzyme function, triggering exocytosis, and so on, but they also can lead to the alteration of transcription of various genes. A variety of enzymatic changes, protein–protein interactions or second messenger changes can be activated within a cell in an orderly fashion following receptor recogni-tion of the primary messenger. The resulting cell signaling pathway provides amplification of the primary signal and dis-tribution of the signal to appropriate targets within the cell. Ex-tensive cell signaling pathways also provide opportunities for feedback and regulation that can fine tune the signal for the cor-rect physiological response by the cell. The most predominant posttranslation modification of pro-teins, phosphorylation, is a common theme in cell signaling pathways. Cellular phosphorylation is under the control of two groups of proteins: kinases, enzymes that catalyze the phos-phorylation of tyrosine or serine and threonine residues in pro-teins (or in some cases, in lipids); and phosphatases, proteins that remove phosphates from proteins (or lipids). Some of the larger receptor families are themselves kinases. Tyrosine kinase receptors initiate phosphorylation on tyrosine residues on com-plementary receptors following ligand binding. Serine/threo-nine kinase receptors initiate phosphorylation on serines or threonines in complementary receptors following ligand bind-ing. Cytokine receptors are directly associated with a group of protein kinases that are activated following cytokine binding. Alternatively, second messengers changes can lead to phos-phorylation further downstream in the signaling pathway. More than 300 protein kinases have been described. Some of the principal ones that are important in mammalian cell signal-ing are summarized in Table 2–4. In general, addition of phos-phate groups changes the conformation of the proteins, altering their functions and consequently the functions of the cell. The close relationship between phosphorylation and dephosphory-lation of cellular proteins allows for a temporal control of acti-vation of cell signaling pathways. This is sometimes referred to as a “phosphate timer.” STIMULATION OF TRANSCRIPTION The activation of transcription, and subsequent translation, is a common outcome of cellular signaling. There are three TABLE 2–3 Common mechanisms by which chemical messengers in the ECF bring about changes in cell function. Mechanism Examples Open or close ion channels in cell membrane Acetylcholine on nicotinic cholin-ergic receptor; norepinephrine on K+ channel in the heart Act via cytoplasmic or nuclear re-ceptors to increase transcription of selected mRNAs Thyroid hormones, retinoic acid, steroid hormones Activate phospholipase C with in-tracellular production of DAG, IP3, and other inositol phosphates Angiotensin II, norepinephrine via α1-adrenergic receptor, vaso-pressin via V1 receptor Activate or inhibit adenylyl cyclase, causing increased or decreased intracellular production of cAMP Norepinephrine via β1-adrener-gic receptor (increased cAMP); norepinephrine via α2-adrener-gic receptor (decreased cAMP) Increase cGMP in cell Atrial natriuretic peptide; nitric oxide Increase tyrosine kinase activity of cytoplasmic portions of trans-membrane receptors Insulin, epidermal growth factor (EGF), platelet-derived growth factor (PDGF), monocyte colony-stimulating factor (M-CSF) Increase serine or threonine kinase activity TGFβ, activin, inhibin TABLE 2–4 Sample protein kinases. Phosphorylate serine or threonine residues, or both Calmodulin-dependent Myosin light-chain kinase Phosphorylase kinase Ca2+/calmodulin kinase I Ca2+/calmodulin kinase II Ca2+/calmodulin kinase III Calcium-phospholipid-dependent Protein kinase C (seven subspecies) Cyclic nucleotide-dependent cAMP-dependent kinase (protein kinase A; two subspecies) cGMP-dependent kinase Phosphorylate tyrosine residues Insulin receptor, EGF receptor, PDGF receptor, and M-CSF receptor 52 SECTION I Cellular & Molecular Basis of Medical Physiology distinct pathways for primary messengers to alter transcrip-tion of cells. First, as is the case with steroid or thyroid hor-mones, the primary messenger is able to cross the cell membrane and bind to a nuclear receptor, which then can di-rectly interact with DNA to alter gene expression. A second pathway to gene transcription is the activation of cytoplasmic protein kinases that can move to the nucleus to phosphorylate a latent transcription factor for activation. This pathway is a common endpoint of signals that go through the mitogen ac-tivated protein (MAP) kinase cascade. MAP kinases can be activated following a variety of receptor ligand interactions through second messenger signaling. They comprise a series of three kinases that coordinate a stepwise phosphorylation to activate each protein in series in the cytosol. Phosphorylation of the last MAP kinase in series allows it to migrate to the nu-cleus where it phosphorylates a latent transcription factor. A third common pathway is the activation of a latent transcrip-tion factor in the cytosol, which then migrates to the nucleus and alters transcription. This pathway is shared by a diverse set of transcription factors that include nuclear factor kappa B (NFκB; activated following tumor necrosis family receptor binding and others), and signal transducers of activated transcription (STATs; activated following cytokine receptor binding). In all cases the binding of the activated transcription factor to DNA increases (or in some cases, decreases) the transcription of mRNAs encoded by the gene to which it binds. The mRNAs are translated in the ribosomes, with the production of increased quantities of proteins that alter cell function. INTRACELLULAR Ca2+ AS A SECOND MESSENGER Ca2+ regulates a very large number of physiological processes that are as diverse as proliferation, neural signaling, learning, contraction, secretion, and fertilization, so regulation of intra-cellular Ca2+ is of great importance. The free Ca2+ concentra-tion in the cytoplasm at rest is maintained at about 100 nmol/ L. The Ca2+ concentration in the interstitial fluid is about 12,000 times the cytoplasmic concentration (ie, 1,200,000 nmol/L), so there is a marked inwardly directed concentration gradient as well as an inwardly directed electrical gradient. Much of the intracellular Ca2+ is stored at relatively high con-centrations in the endoplasmic reticulum and other organelles (Figure 2–21), and these organelles provide a store from which Ca2+ can be mobilized via ligand-gated channels to increase the concentration of free Ca2+ in the cytoplasm. Increased cy-toplasmic Ca2+ binds to and activates calcium-binding pro-teins. These proteins can have direct effects in cellular physiology, or can activate other proteins, commonly protein kinases, to further cell signaling pathways. Ca2+ can enter the cell from the extracellular fluid, down its electrochemical gradient, through many different Ca2+ chan-nels. Some of these are ligand-gated and others are voltage-gated. Stretch-activated channels exist in some cells as well. Many second messengers act by increasing the cytoplasmic Ca2+ concentration. The increase is produced by releasing Ca2+ from intracellular stores—primarily the endoplasmic reticu-lum—or by increasing the entry of Ca2+ into cells, or by both mechanisms. IP3 is the major second messenger that causes Ca2+ release from the endoplasmic reticulum through the direct activation of a ligand-gated channel, the IP3 receptor. In effect, the generation of one second messenger (IP3) can lead to the release of another second messenger (Ca2+). In many tissues, transient release of Ca2+ from internal stores into the cytoplasm triggers opening of a population of Ca2+ channels in the cell membrane (store-operated Ca2+ channels; SOCCs). The resulting Ca2+ influx replenishes the total intracellular Ca2+ supply and refills the endoplasmic reticulum. The exact identity of the SOCCs is still unknown, and there is debate about the signal from the endoplasmic reticulum that opens them. As with other second messenger molecules, the increase in Ca2+ within the cytosol is rapid, and is followed by a rapid decrease. Because the movement of Ca2+ outside of the cytosol (ie, across the plasma membrane or the membrane of the inter-nal store) requires that it move up its electrochemical gradient, it requires energy. Ca2+ movement out of the cell is facilitated by the plasma membrane Ca2+ ATPase. Alternatively, it can be transported by an antiport that exchanges three Na+ for each Ca2+ driven by the energy stored in the Na+ electrochemical gradient. Ca2+ movement into the internal stores is through the action of the sarcoplasmic or endoplasmic reticulum Ca2+ ATPase, also known as the SERCA pump. CALCIUM-BINDING PROTEINS Many different Ca2+-binding proteins have been described, in-cluding troponin, calmodulin, and calbindin. Troponin is the FIGURE 2–21 Ca2+ handling in mammalian cells. Ca2+ is stored in the endoplasmic reticulum and, to a lesser extent, mitochondria and can be released from them to replenish cytoplasmic Ca2+. Calcium-binding proteins (CaBP) bind cytoplasmic Ca2+ and, when activated in this fashion, bring about a variety of physiologic effects. Ca2+ enters the cells via voltage-gated (volt) and ligand-gated (lig) Ca2+ channels and store-operated calcium channels ( SOCCs). It is transported out of the cell by Ca, Mg ATPases (not shown), Ca, H ATPase and an Na, Ca antiport. It is also transported into the ER by Ca ATPases. CaBP Effects Ca2+ Ca2+ Ca2+ 2H+ 3Na+ ATP Mitochondrion Endoplasmic reticulum Ca2+ (volt) Ca2+ (lig) Ca2+ (SOCC) CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 53 Ca2+-binding protein involved in contraction of skeletal muscle (Chapter 5). Calmodulin contains 148 amino acid residues (Figure 2–22) and has four Ca2+-binding domains. It is unique in that amino acid residue 115 is trimethylated, and it is exten-sively conserved, being found in plants as well as animals. When calmodulin binds Ca2+, it is capable of activating five different calmodulin-dependent kinases (CaMKs; Table 2–4), among other proteins. One of the kinases is myosin light-chain kinase, which phosphorylates myosin. This brings about contraction in smooth muscle. CaMKI and CaMKII are concerned with syn-aptic function, and CaMKIII is concerned with protein synthe-sis. Another calmodulin-activated protein is calcineurin, a phosphatase that inactivates Ca2+ channels by dephosphorylat-ing them. It also plays a prominent role in activating T cells and is inhibited by some immunosuppressants. MECHANISMS OF DIVERSITY OF Ca2+ ACTIONS It may seem difficult to understand how intracellular Ca2+ can have so many varied effects as a second messenger. Part of the explanation is that Ca2+ may have different effects at low and at high concentrations. The ion may be at high concentration at the site of its release from an organelle or a channel (Ca2+ sparks) and at a subsequent lower concentration after it dif-fuses throughout the cell. Some of the changes it produces can outlast the rise in intracellular Ca2+ concentration because of the way it binds to some of the Ca2+-binding proteins. In ad-dition, once released, intracellular Ca2+ concentrations fre-quently oscillate at regular intervals, and there is evidence that the frequency and, to a lesser extent, the amplitude of those os-cillations codes information for effector mechanisms. Finally, increases in intracellular Ca2+ concentration can spread from cell to cell in waves, producing coordinated events such as the rhythmic beating of cilia in airway epithelial cells. G PROTEINS A common way to translate a signal to a biologic effect inside cells is by way of nucleotide regulatory proteins that are acti-vated after binding GTP (G proteins). When an activating sig-nal reaches a G protein, the protein exchanges GDP for GTP. The GTP–protein complex brings about the activating effect of the G protein. The inherent GTPase activity of the protein then converts GTP to GDP, restoring the G protein to an in-active resting state. G proteins can be divided into two princi-pal groups involved in cell signaling: small G proteins and heterotrimeric G proteins. Other groups that have similar regulation and are also important to cell physiology include elongation factors, dynamin, and translocation GTPases. There are six different families of small G proteins (or small GTPases) that are all highly regulated. GTPase activating proteins (GAPs) tend to inactivate small G proteins by encouraging hydrolysis of GTP to GDP in the central binding site. Guanine exchange factors (GEFs) tend to activate small G proteins by encouraging exchange of GDP for GTP in the active site. Some of the small G proteins contain lipid modifi-cations that help to anchor them to membranes, while others are free to diffuse throughout the cytosol. Small G proteins are involved in many cellular functions. Members of the Rab family regulate the rate of vesicle traffic between the endo-plasmic reticulum, the Golgi apparatus, lysosomes, endo-somes, and the cell membrane. Another family of small GTP-binding proteins, the Rho/Rac family, mediates interactions between the cytoskeleton and cell membrane; and a third family, the Ras family, regulates growth by transmitting sig-nals from the cell membrane to the nucleus. Another family of G proteins, the larger heterotrimeric G proteins, couple cell surface receptors to catalytic units that catalyze the intracellular formation of second messengers or couple the receptors directly to ion channels. Despite the knowledge of the small G proteins described above, the heter-omeric G proteins are frequently referred to in the shortened “G protein” form because they were the first to be identified. Heterotrimeric G proteins are made up of three subunits des-ignated α, β, and γ (Figure 2–23). Both the α and the γ sub-units have lipid modifications that anchor these proteins to plasma membrane. The α subunit is bound to GDP. When a ligand binds to a G protein-coupled receptor (GPCR), this GDP is exchanged for GTP and the α subunit separates from the combined β and γ subunits. The separated α subunit brings about many biologic effects. The β and γ subunits are tightly bound in the cell and together form a signaling mole-cule that can also activate a variety of effectors. The intrinsic FIGURE 2–22 Structure of calmodulin from bovine brain. Single-letter abbreviations are used for the amino acid residues. Note the four calcium domains (purple residues) flanked on either side by stretches of α helix. (Reproduced with permission from Cheung WY: Calmodulin: An overview. Fed Proc 1982;41:2253.) A D Q L T E E Q I A E F K E A F S L F E K D G N G T I T T K E G T V M S L G Q N P T E A E L Q D M I N E V D A D G N G T I D F P E F L T M M A R K M K D T D S E E E I R E A F R V F D K D G N G Y I S A A E L R H V M T N L G E K L T D E E V D E M I R E A N I D G D G E V N Y E E F V Q M M T A K R L COOH 10 Ca 20 30 NH Ac 140 Ca 130 120 100 110 (Me)3 N 90 80 60 40 Ca 70 50 Ca 54 SECTION I Cellular & Molecular Basis of Medical Physiology GTPase activity of the α subunit then converts GTP to GDP, and this leads to reassociation of the α with the βγ subunit and termination of effector activation. The GTPase activity of the α subunit can be accelerated by a family of regulators of G protein signaling (RGS). Heterotrimeric G proteins relay signals from over 1000 GPCRs, and their effectors in the cells include ion channels and enzymes (Table 2–5). There are 20 α, 6 β, and 12 γ genes, which allow for over 1400 α, β, and γ combinations. Not all combinations occur in the cell, but over 20 different heterotri-meric G proteins have been well documented in cell signaling. They can be divided into five families, each with a relatively characteristic set of effectors. G PROTEIN-COUPLED RECEPTORS All the heterotrimeric G protein-coupled receptors (GPCRs) that have been characterized to date are proteins that span the cell membrane seven times. Because of this structure they are alternatively referred to as seven-helix receptors or serpen-tine receptors. A very large number have been cloned, and their functions are multiple and diverse. The topological structures of two of them are shown in Figure 2–24. These re-ceptors further assemble into a barrel-like structure. Upon ligand binding, a conformational change activates a resting heterotrimeric G protein associated with the cytoplasmic leaf of the plasma membrane. Activation of a single receptor can result in 1, 10, or more active heterotrimeric G proteins, pro-viding amplification as well as transduction of the first mes-senger. Bound receptors can be inactivated to limit the amount of cellular signaling. This frequently occurs through phosphorylation of the cytoplasmic side of the receptor. INOSITOL TRISPHOSPHATE & DIACYLGLYCEROL AS SECOND MESSENGERS The link between membrane binding of a ligand that acts via Ca2+ and the prompt increase in the cytoplasmic Ca2+ con-centration is often inositol trisphosphate (inositol 1,4,5-tris-phosphate; IP3). When one of these ligands binds to its receptor, activation of the receptor produces activation of phospholipase C (PLC) on the inner surface of the membrane. Ligands bound to G protein-coupled receptor can do this through the Gq heterotrimeric G proteins, while ligands bound to tyrosine kinase receptors can do this through other cell signaling pathways. PLC has at least eight isoforms; PLCβ FIGURE 2–23 Heterotrimeric G proteins. Top Summary of overall reaction that occurs in the Gα subunit. Bottom: When the ligand (square) binds to the G protein-coupled receptor in the cell membrane, GTP replaces GDP on the α subunit. GTP-α separates from the βγ subunit and GTP-α and βγ both activate various effectors, pro-ducing physiologic effects. The intrinsic GTPase activity of GTP-α then converts GTP to GDP, and the α, β, and γ subunits reassociate. Nucleotide exchange Input GDP GTP GTPase activity Output Effectors β γ α β γ α TABLE 2–5 Some of the ligands for receptors coupled to heterotrimeric G proteins. Class Ligand Neurotransmitters Epinephrine Norepinephrine Dopamine 5-Hydroxytryptamine Histamine Acetylcholine Adenosine Opioids Tachykinins Substance P Neurokinin A Neuropeptide K Other peptides Angiotensin II Arginine vasopressin Oxytocin VIP, GRP, TRH, PTH Glycoprotein hormones TSH, FSH, LH, hCG Arachidonic acid derivatives Thromboxane A2 Other Odorants Tastants Endothelins Platelet-activating factor Cannabinoids Light CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 55 is activated by heterotrimeric G proteins, while PLCγ forms are activated through tyrosine kinase receptors. PLC isoforms can catalyze the hydrolysis of the membrane lipid phosphatidyl-inositol 4,5-diphosphate (PIP2) to form IP3 and diacylglycerol (DAG) (Figure 2–25). The IP3 diffuses to the endoplasmic reticulum, where it triggers the release of Ca2+ into the cyto-plasm by binding the IP3 receptor, a ligand-gated Ca2+ chan-nel (Figure 2–26). DAG is also a second messenger; it stays in the cell membrane, where it activates one of several isoforms of protein kinase C. FIGURE 2–24 Structures of two G protein-coupled receptors. The individual amino acid residues are identified by their single-letter codes, and the orange residues are sites of phosphorylation. The Y-shaped symbols identify glycosylation sites. Note the extracellular amino terminal, the intracellular carboxyl terminal, and the seven membrane-spanning portions of each protein. (Reproduced with permission from Benovic JL et al: Light-dependent phosphorylation of rhodopsin by β-adrenergic receptor kinase. Reprinted by permission from Nature 1986;321:869. Copyright © 1986 by Macmillan Magazines) M G P P G N D S D F L L T T N G S H V P D H D V T E E R D E A W V V G M A I L M S V I V F G N V L V I T A I L A I V L I N B A G F P V V V L D A C A L S T I A L G M NH2 A K F M W N F G N F W C K M E R L Q T V T N Y F E F W T S I D V L C I E T L C V I A V D V T A S R A T H Q K A I D C Y H K E T C C D F F T N Y W H M Q I P L F S V I W V M L I V M R T L Q S R Y I A I T S P F K Y Q S L L T K N K A Q A Y A I A S S I V P L V V M V F V Y S L S F Y V R V F Q V A K R Q G Q V E Q D G R S G L Q K I D K S E R R S S K F C L K E H K A L K T L G H G R F H S P N L Q D N L I P K I V H V I N V I F F L T F T G M I I G P L W C S P D F R I A F Q E L L C L R R E V Y I L L N W L G A F N P L I Y C R Y V N S G Y S S S K A Y G N N S S N G T K G Y D E G M E R L E S Q E K A S G C Q L G T E S F V N C Q G P G E D P S L S P V T D L Q S C S C N N T P S D L G R HOOC 2-Adrenergic receptor Extracellular surface Cytoplasmic surface G T E G P N F Y V P F S N K T C V V R S P F E A P Q Y Y L A E P W Q F S M L A A Y M F L G F P I N F L T L Y L I M L T Y L T T T F Q G F D A V A L N L L I Y V M F L M N NH2 V T V H G F Y F V G P L E C G T N G L S Q H K K L R T P L N F A T L G G E I A L V L A I E R Y V V V W S L A S R Y I M Q C P E G S C G I D Y Y T P H E E T N N W G V L P P A A C A W T F A V G M I A H L A M V C K P M S N F R F G E N E S F V I Y M F V V P L I V I F F C Y G M H F I I Q L V F T V A A K E Q E A Q Q T T A S Q K A E K E V T R T Q H G S F D G P F I Y F A V Q A Y P L F A I V M I I V L W C I M N K Q F R N C M V T T L C C G I F M T I P A F F A V Y N P V I Y I M K T S A D D N K P L G S A E V T T K S T E T Q S P A V A HOOC Rhodopsin Intradiskal surface Cytoplasmic surface F FIGURE 2–25 Metabolism of phosphatidylinositol in cell membranes. Phosphatidylinositol is successively phosphorylated to form phosphatidylinositol 4-phosphate (PIP), then phosphatidylinositol 4,5-bisphosphate (PIP2). Phospholipase Cβ and phospholipase Cγ catalyze the breakdown of PIP2 to inositol 1,4,5-trisphosphate (IP3) and diacylglycerol. Other inositol phosphates and phosphatidylinositol derivatives can also be formed. IP3 is dephosphorylated to inositol, and diacylglycerol is metabolized to cytosine diphosphate (CDP)-diacylglycerol. CDP-diacylglycerol and inositol then combine to form phosphatidylinositol, completing the cycle. (Modified from Berridge MJ: Inositol triphosphate and diacylglycerol as second messengers. Biochem J 1984;220:345.) P 1 4 P P 1 4 P P P 1 4 5 5 P P P 1 4 Phosphatidylinositol (PI) PIP Diacylglycerol PIP2 IP2 IP3 Phospholipase C Inositol IP Phosphatidic acid CDP-diacylglycerol + 56 SECTION I Cellular & Molecular Basis of Medical Physiology CYCLIC AMP Another important second messenger is cyclic adenosine 3',5'-monophosphate (cyclic AMP or cAMP; Figure 2–27). Cyclic AMP is formed from ATP by the action of the enzyme aden-ylyl cyclase and converted to physiologically inactive 5'AMP by the action of the enzyme phosphodiesterase. Some of the phosphodiesterase isoforms that break down cAMP are inhib-ited by methylxanthines such as caffeine and theophylline. Consequently, these compounds can augment hormonal and transmitter effects mediated via cAMP. Cyclic AMP activates one of the cyclic nucleotide-dependent protein kinases (pro-tein kinase A, PKA) that, like protein kinase C, catalyzes the phosphorylation of proteins, changing their conformation and altering their activity. In addition, the active catalytic sub-unit of PKA moves to the nucleus and phosphorylates the cAMP-responsive element-binding protein (CREB). This transcription factor then binds to DNA and alters transcrip-tion of a number of genes. PRODUCTION OF cAMP BY ADENLYL CYCLASE Adenylyl cyclase is a transmembrane protein, and it crosses the membrane 12 times. Ten isoforms of this enzyme have been described and each can have distinct regulatory proper-ties, permitting the cAMP pathway to be customized to specif-ic tissue needs. Notably, stimulatory heterotrimeric G proteins (Gs) activate, while inhibitory heterotrimeric G proteins (Gi) inactivate adenylyl cyclase (Figure 2–28). When the appropri-ate ligand binds to a stimulatory receptor, a Gs α subunit acti-vates one of the adenylyl cyclases. Conversely, when the appropriate ligand binds to an inhibitory receptor, a Gi α sub-unit inhibits adenylyl cyclase. The receptors are specific, re-sponding at low threshold to only one or a select group of related ligands. However, heterotrimeric G proteins mediate the stimulatory and inhibitory effects produced by many dif-ferent ligands. In addition, cross-talk occurs between the phospholipase C system and the adenylyl cyclase system, as several of the isoforms of adenylyl cyclase are stimulated by calmodulin. Finally, the effects of protein kinase A and protein kinase C are very widespread and can also affect directly, or in-directly, the activity at adenylyl cyclase. The close relationship between activation of G proteins and adenylyl cyclases also al-lows for spatial regulation of cAMP production. All of these events, and others, allow for fine-tuning the cAMP response for a particular physiological outcome in the cell. Two bacterial toxins have important effects on adenylyl cyclase that are mediated by G proteins. The A subunit of cholera toxin catalyzes the transfer of ADP ribose to an argi-nine residue in the middle of the α subunit of Gs. This inhibits FIGURE 2–26 Diagrammatic representation of release of inositol triphosphate (IP3) and diacylglycerol (DAG) as second messengers. Binding of ligand to G protein-coupled receptor acti-vates phospholipase C (PLC)β. Alternatively, activation of receptors with intracellular tyrosine kinase domains can activate PLCγ. The re-sulting hydrolysis of phosphatidylinositol 4,5-diphosphate (PIP2) pro-duces IP3, which releases Ca2+ from the endoplasmic reticulum (ER), and DAG, which activates protein kinase C (PKC). CaBP, Ca2+-binding proteins. ISF, interstitial fluid. Stimulatory receptor Tyrosine kinase Gq, etc ISF Phosphoproteins Physiologic effects ER Cytoplasm Ca2+ CaBP Physiologic effects IP3 PLC PIP2 DAG PKC α β γ FIGURE 2–27 Formation and metabolism of cAMP. The sec-ond messenger cAMP is a made from ATP by adenylyl cyclase and bro-ken down into cAMP by phosphodiesterase. OH HO H H OH O P O P O P O O O O H H OH OH OH CH2 Adenine OH H H OH O P O O HO H H OH CH2 Adenine OH OH H H O O O O H H CH2 P Adenine ATP AMP PP cAMP H2O Phosphodiesterase Adenylyl cyclase CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 57 its GTPase activity, producing prolonged stimulation of aden-ylyl cyclase. Pertussis toxin catalyzes ADP-ribosylation of a cysteine residue near the carboxyl terminal of the α subunit of Gi. This inhibits the function of Gi. In addition to the implica-tions of these alterations in disease, both toxins are used for fundamental research on G protein function. The drug for-skolin also stimulates adenylyl cyclase activity by a direct action on the enzyme. GUANYLYL CYCLASE Another cyclic nucleotide of physiologic importance is cyclic guanosine monophosphate (cyclic GMP or cGMP). Cyclic GMP is important in vision in both rod and cone cells. In ad-dition, there are cGMP-regulated ion channels, and cGMP ac-tivates cGMP-dependent kinase, producing a number of physiologic effects. Guanylyl cyclases are a family of enzymes that catalyze the formation of cGMP. They exist in two forms (Figure 2–29). One form has an extracellular amino terminal domain that is a receptor, a single transmembrane domain, and a cytoplas-mic portion with guanylyl cyclase catalytic activity. Three such guanylyl cyclases have been characterized. Two are receptors for atrial natriuretic peptide (ANP; also known as atrial natriuretic factor), and a third binds an Escherichia coli enterotoxin and the gastrointestinal polypeptide guanylin. The other form of guanylyl cyclase is soluble, contains heme, and is not bound to the membrane. There appear to be several isoforms of the intracellular enzyme. They are activated by nitric oxide (NO) and NO-containing compounds. GROWTH FACTORS Growth factors have become increasingly important in many different aspects of physiology. They are polypeptides and proteins that are conveniently divided into three groups. One group is made up of agents that foster the multiplication or de-velopment of various types of cells; nerve growth factor (NGF), insulin-like growth factor I (IGF-I), activins and in-hibins, and epidermal growth factor (EGF) are examples. More than 20 have been described. The cytokines are a second group. These factors are produced by macrophages and lym-phocytes, as well as other cells, and are important in regulation of the immune system (see Chapter 3). Again, more than 20 have been described. The third group is made up of the colo-ny-stimulating factors that regulate proliferation and matura-tion of red and white blood cells. Receptors for EGF, platelet-derived growth factor (PDGF), and many of the other factors that foster cell multiplication and growth have a single membrane-spanning domain with an intracellular tyrosine kinase domain (Figure 2–29). When ligand binds to a tyrosine kinase receptor, it first causes a dimerization of two similar receptors. The dimerization results in partial activation of the intracellular tyrosine kinase domains and a cross-phosphorylation to fully activate each other. One of the pathways activated by phosphorylation leads, through the small G protein Ras, to MAP kinases, and eventually to the production of transcription factors in the nucleus that alter gene expression (Figure 2–30). Receptors for cytokines and colony-stimulating factors differ from the other growth factors in that most of them do not have tyrosine kinase domains in their cytoplasmic portions and FIGURE 2–28 The cAMP system. Activation of adenylyl cyclase catalyzes the conversion of ATP to cAMP. Cyclic AMP activates protein kinase A, which phosphorylates proteins, producing physiologic ef-fects. Stimulatory ligands bind to stimulatory receptors and activate adenylyl cyclase via Gs. Inhibitory ligands inhibit adenylyl cyclase via inhibitory receptors and Gi. ISF, interstitial fluid. Stimulatory receptor Adenylyl cyclase Inhibitory receptor Protein kinase A ISF Phosphoproteins Cytoplasm Physiologic effects GS Gi ATP CAMP PDE 5' AMP α β γ α β γ FIGURE 2–29 Diagrammatic representation of guanylyl cyclases, tyrosine kinases, and tyrosine phosphatases. ANP, atrial natriuretic peptide; C, cytoplasm; cyc, guanylyl cyclase domain; EGF, epidermal growth factor; ISF, interstitial fluid; M, cell membrane; PDGF, platelet-derived growth factor; PTK, tyrosine kinase domain; PTP, ty-rosine phosphatase domain; ST, E. coli enterotoxin. (Modified from Koesling D, Böhme E, Schultz G: Guanylyl cyclases, a growing family of signal transducing enzymes. FASEB J 1991;5:2785.) ANP ISF M C PTK ST NH2 NH2 NH2 NH2 NH2 NH2 cyc cyc cyc COOH COOH COOHCOOH PTP PTP PTP PTK PTK Guanylyl cyclases Tyrosine kinases Tyrosine phosphatases COOH COOH COOH PDGF EGF 58 SECTION I Cellular & Molecular Basis of Medical Physiology some have little or no cytoplasmic tail. However, they initiate tyrosine kinase activity in the cytoplasm. In particular, they activate the so-called Janus tyrosine kinases (JAKs) in the cytoplasm (Figure 2–31). These in turn phosphorylate STAT proteins. The phosphorylated STATs form homo- and het-erodimers and move to the nucleus, where they act as trans-cription factors. There are four known mammalian JAKs and seven known STATs. Interestingly, the JAK–STAT pathway can also be activated by growth hormone and is another important direct path from the cell surface to the nucleus. However, it should be emphasized that both the Ras and the JAK–STAT pathways are complex and there is cross-talk between them and other signaling pathways discussed previously. Finally, note that the whole subject of second messengers and intracellular signaling has become immensely complex, with multiple pathways and interactions. It is only possible in a book such as this to list highlights and present general themes that will aid the reader in understanding the rest of physiology (see Clinical Box 2–3). HOMEOSTASIS The actual environment of the cells of the body is the intersti-tial component of the ECF. Because normal cell function FIGURE 2–30 One of the direct pathways by which growth factors alter gene activity. TK, tyrosine kinase domain; Grb2, Ras acti-vator controller; Sos, Ras activator; Ras, product of the ras gene; MAP K, mitogen-activated protein kinase; MAP KK, MAP kinase kinase; TF, transcription factors. There is cross-talk between this pathway and the cAMP pathway, as well as cross-talk with the IP3–DAG pathway. T K Grb2 SOS Growth factor Receptor Cell membrane Active Ras Inactive Ras GDP GTP Raf MAP KK MAP K TF Nucleus Altered gene activity Ras Ras FIGURE 2–31 Signal transduction via the JAK–STAT pathway. A) Ligand binding leads to dimerization of receptor. B) Acti-vation and tyrosine phosphorylation of JAKs. C) JAKs phosphorylate STATs. D) STATs dimerize and move to nucleus, where they bind to re-sponse elements on DNA. (Modified from Takeda K, Kishimoto T, Akira S: STAT6: Its role in interleukin 4-mediated biological functions. J Mol Med 1997;75:317.) Ligand C JAK STAT STAT P JAK P P P P P Ligand D JAK P JAK P P STAT P STAT P Ligand B JAK STAT P JAK P STAT P P Ligand A Cytoplasm ISF Receptor JAK JAK STAT P Nucleus DNA STAT CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 59 depends on the constancy of this fluid, it is not surprising that in multicellular animals, an immense number of regulatory mechanisms have evolved to maintain it. To describe “the var-ious physiologic arrangements which serve to restore the nor-mal state, once it has been disturbed,” W.B. Cannon coined the term homeostasis. The buffering properties of the body fluids and the renal and respiratory adjustments to the pres-ence of excess acid or alkali are examples of homeostatic mechanisms. There are countless other examples, and a large part of physiology is concerned with regulatory mechanisms that act to maintain the constancy of the internal environ-ment. Many of these regulatory mechanisms operate on the principle of negative feedback; deviations from a given normal set point are detected by a sensor, and signals from the sensor trigger compensatory changes that continue until the set point is again reached. CLINICAL BOX 2–3 Receptor & G Protein Diseases Many diseases are being traced to mutations in the genes for receptors. For example, loss-of-function receptor muta-tions that cause disease have been reported for the 1,25-dihydroxycholecalciferol receptor and the insulin receptor. Certain other diseases are caused by production of anti-bodies against receptors. Thus, antibodies against thyroid-stimulating hormone (TSH) receptors cause Graves’ dis-ease, and antibodies against nicotinic acetylcholine recep-tors cause myasthenia gravis. An example of loss of function of a receptor is the type of nephrogenic diabetes insipidus that is due to loss of the ability of mutated V2 vasopressin receptors to mediate con-centration of the urine. Mutant receptors can gain as well as lose function. A gain-of-function mutation of the Ca2+ receptor causes excess inhibition of parathyroid hormone secretion and familial hypercalciuric hypocalcemia. G proteins can also undergo loss-of-function or gain-of-func-tion mutations that cause disease (Table 2–6), In one form of pseudohypoparathyroidism, a mutated Gsα fails to re-spond to parathyroid hormone, producing the symptoms of hypoparathyroidism without any decline in circulating parathyroid hormone. Testotoxicosis is an interesting dis-ease that combines gain and loss of function. In this condi-tion, an activating mutation of Gsα causes excess testoster-one secretion and prepubertal sexual maturation. However, this mutation is temperature-sensitive and is active only at the relatively low temperature of the testes (33 °C). At 37 °C, the normal temperature of the rest of the body, it is replaced by loss of function, with the production of hypo-parathyroidism and decreased responsiveness to TSH. A different activating mutation in Gsα is associated with the rough-bordered areas of skin pigmentation and hypercorti-solism of the McCune–Albright syndrome. This mutation occurs during fetal development, creating a mosaic of nor-mal and abnormal cells. A third mutation in Gsα reduces its intrinsic GTPase activity. As a result, it is much more active than normal, and excess cAMP is produced. This causes hy-perplasia and eventually neoplasia in somatotrope cells of the anterior pituitary. Forty percent of somatotrope tumors causing acromegaly have cells containing a somatic muta-tion of this type. TABLE 2–6 Examples of abnormalities caused by loss- or gain-of-function mutations of heterotrimeric G protein-coupled receptors and G proteins. Site Type of Mutation Disease Receptor Cone opsins Loss Color blindness Rhodopsin Loss Congenital night blindness; two forms of retinitis pigmentosa V2 vasopressin Loss X-linked nephrogenic diabetes insipidus ACTH Loss Familial glucocorticoid deficiency LH Gain Familial male precocious puberty TSH Gain Familial nonautoimmune hyperthy-roidism TSH Loss Familial hypothyroidism Ca2+ Gain Familial hypercalciuric hypocalcemia Thromboxane A2 Loss Congenital bleeding Endothelin B Loss Hirschsprung disease G protein Gs α Loss Pseudohypothyroidism type 1a Gs α Gain/loss Testotoxicosis Gs α Gain (mosaic) McCune–Albright syndrome Gs α Gain Somatotroph adenomas with acro-megaly Gi α Gain Ovarian and adrenocortical tumors Modified from Lem J: Diseases of G-protein-coupled signal transduc-tion pathways: The mammalian visual system as a model. Semin Neuro-sci 1998;9:232. 60 SECTION I Cellular & Molecular Basis of Medical Physiology CHAPTER SUMMARY ■The cell and the intracellular organelles are surrounded by a semipermeable membrane. Biological membranes have a lipid bilayer with a hydrophobic core and hydrophilic outer regions that provide a barrier between inside and outside compartments as well as a template for biochemical reactions. The membrane is populated by structural and functional proteins that can be in-tegrated into the membrane or be associated with one side of the lipid bilayer. These proteins contribute greatly to the semiper-meable properties of biological membrane. ■Mitochondria are organelles that allow for oxidative phos-phorylation in eukaryotic cells. They contain their own DNA, however, proteins in the mitochondria are encoded by both mi-tochondrial and cellular DNA. Mitochondria also are important in specialized cellular signaling. ■Lysosomes and peroxisomes are membrane-bound organelles that contribute to protein and lipid processing. They do this in part by creating acidic (lysosomes) or oxidative (peroxisomes) contents relative to the cell cytosol. ■The cytoskeleton is a network of three types of filaments that provide structural integrity to the cell as well as a means for traf-ficking of organelles and other structures. Actin is the funda-mental building block for thin filaments and represents as much as 15% of cellular protein. Actin filaments are important in cel-lular contraction, migration, and signaling. Actin filaments also provide the backbone for muscle contraction. Intermediate fila-ments are primarily structural. Proteins that make up interme-diate filaments are cell-type specific. Microtubules are made up of tubulin subunits. Microtubules provide a dynamic structure in cells that allows for movement of cellular components around the cell. ■There are three superfamilies of molecular motor proteins in the cell that use the energy of ATP to generate force, movement, or both. Myosin is the force generator for muscle cell contraction. There are also cellular myosins that interact with the cytoskele-ton (primarily thin filaments) to participate in contraction as well as movement of cell contents. Kinesins and cellular dyneins are motor proteins that primarily interact with microtubules to move cargo around the cells. ■Cellular adhesion molecules aid in tethering cells to each other or to the extracellular matrix as well as providing for initiation of cellular signaling. There are four main families of these pro-teins: integrins, immunoglobulins, cadherins, and selectins. ■Cells contain distinct protein complexes that serve as cellular connections to other cells or the extracellular matrix. Tight junctions provide intercellular connections that link cells into a regulated tissue barrier. Tight junctions also provide a barrier to movement of proteins in the cell membrane and thus, are im-portant to cellular polarization. Gap junctions provide contacts between cells that allow for direct passage of small molecules be-tween two cells. Desmosomes and adherens junctions are spe-cialized structures that hold cells together. Hemidesmosomes and focal adhesions attach cells to their basal lamina. ■The nucleus is an organelle that contains the cellular DNA and is the site of transcription. There are several organelles that em-anate from the nucleus, including the endoplasmic reticulum and the Golgi apparatus. These two organelles are important in protein processing and the targeting of proteins to correct com-partments within the cell. ■Exocytosis and endocytosis are vesicular fusion events that al-low for movement of proteins and lipids between the cell interi-or, the plasma membrane, and the cell exterior. Exocytosis can be constitutive or nonconstitutive; both are regulated processes that require specialized proteins for vesicular fusion. Endocyto-sis is the formation of vesicles at the plasma membrane to take material from the extracellular space into the cell interior. Some endocytoses are defined in part by the size of the vesicles formed whereas others are defined by membrane structures that con-tribute to the endocytosis. All are tightly regulated processes. ■Membranes contain a variety of proteins and protein complexes that allow for transport of small molecules. Aqueous ion chan-nels are membrane-spanning proteins that can be gated open to allow for selective diffusion of ions across membranes and down their electrochemical gradient. Carrier proteins bind to small molecules and undergo conformational changes to deliver small molecules across the membrane. This facilitated transport can be passive or active. Active transport requires energy for trans-port and is typically provided by ATP hydrolysis. ■Cells can communicate with one another via chemical messen-gers. Individual messengers (or ligands) typically bind to a plas-ma membrane receptor to initiate intracellular changes that lead to physiologic changes. Plasma membrane receptor families in-clude ion channels, G protein-coupled receptors, or a variety of enzyme-linked receptors (eg, tyrosine kinase receptors). There are additional cytosolic receptors (eg, steroid receptors) that can bind membrane-permeant compounds. Activation of receptors lead to cellular changes that include changes in membrane po-tential, activation of heterotrimeric G proteins, increase in sec-ond messenger molecules, or initiation of transcription. ■Second messengers are molecules that undergo a rapid concen-tration changes in the cell following primary messenger recog-nition. Common second messenger molecules include Ca2+, cyclic adenosine monophosphate (cAMP), cyclic guanine monophosphate (cGMP), inositol trisphosphate (IP3) and nitric oxide (NO). CHAPTER 2 Overview of Cellular Physiology in Medical Physiology 61 MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. The electrogenic Na, K ATPase plays a critical role in cellular physiology by A) using the energy in ATP to extrude 3 Na+ out of the cell in exchange for taking two K+ into the cell. B) using the energy in ATP to extrude 3 K+ out of the cell in exchange for taking two Na+ into the cell. C) using the energy in moving Na+ into the cell or K+ outside the cell to make ATP. D) using the energy in moving Na+ outside of the cell or K+ inside the cell to make ATP. 2. Cell membranes A) contain relatively few protein molecules. B) contain many carbohydrate molecules. C) are freely permeable to electrolytes but not to proteins. D) have variable protein and lipid contents depending on their location in the cell. E) have a stable composition throughout the life of the cell. 3. Second messengers A) are substances that interact with first messengers outside cells. B) are substances that bind to first messengers in the cell membrane. C) are hormones secreted by cells in response to stimulation by another hormone. D) mediate the intracellular responses to many different hormones and neurotransmitters. E) are not formed in the brain. 4. The Golgi complex A) is an organelle that participates in the breakdown of proteins and lipids. B) is an organelle that participates in posttranslational processing of proteins. C) is an organelle that participates in energy production. D) is an organelle that participates in transcription and translation. E) is a subcellular compartment that stores proteins for trafficking to the nucleus. 5. Endocytosis A) includes phagocytosis and pinocytosis, but not clathrin-mediated or caveolae-dependent uptake of extracellular contents. B) refers to the merging of an intracellular vesicle with the plasma membrane to deliver intracellular contents to the extracellular milieu. C) refers to the invagination of the plasma membrane to uptake extracellular contents into the cell. D) refers to vesicular trafficking between Golgi stacks. 6. G protein-coupled receptors A) are intracellular membrane proteins that help to regulate movement within the cell. B) are plasma membrane proteins that couple the extracellular binding of primary signaling molecules to activation of small G proteins. C) are plasma membrane proteins that couple the extracellular binding of primary signaling molecules to the activation of heterotrimeric G proteins. D) are intracellular proteins that couple the binding of primary messenger molecules with transcription. 7. Gap junctions are intercellular connections that A) primarily serve to keep cells separated and allow for trans-port across a tissue barrier. B) serve as a regulated cytoplasmic bridge for sharing of small molecules between cells. C) serve as a barrier to prevent protein movement within the cellular membrane. D) are cellular components for constitutive exocytosis that occurs between adjacent cells. CHAPTER RESOURCES Alberts B et al: Molecular Biology of the Cell, 5th ed. Garland Science, 2007. Cannon WB: The Wisdom of the Body. Norton, 1932. Junqueira LC, Carneiro J, Kelley RO: Basic Histology, 9th ed. McGraw-Hill, 1998. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. Pollard TD, Earnshaw WC: Cell Biology, 2nd ed. Saunders, Elsevier, 2008. Sperelakis N (editor): Cell Physiology Sourcebook, 3rd ed. Academic Press, 2001. This page intentionally left blank 63 C H A P T E R 3 Immunity, Infection, & Inflammation O B J E C T I V E S After studying this chapter, you should be able to: ■Understand the significance of immunity, particularly with respect to defending the body against microbial invaders. ■Define the circulating and tissue cell types that contribute to immune and inflam-matory responses. ■Describe how phagocytes are able to kill internalized bacteria. ■Identify the functions of hematopoietic growth factors, cytokines, and chemokines. ■Delineate the roles and mechanisms of innate, acquired, humoral, and cellular im-munity. ■Understand the basis of inflammatory responses and wound healing. INTRODUCTION As an open system, the body is continuously called upon to defend itself from potentially harmful invaders such as bacte-ria, viruses, and other microbes. This is accomplished by the immune system, which is subdivided into innate and adaptive (or acquired) branches. The immune system is composed of specialized effector cells that sense and respond to foreign antigens and other molecular patterns not found in human tissues. Likewise, the immune system clears the body’s own cells that have become senescent or abnormal, such as cancer cells. Finally, occasionally, normal host tissues become the subject of inappropriate immune attack, such as in autoim-mune diseases or in settings where normal cells are harmed as innocent bystanders when the immune system mounts an inflammatory response to an invader. It is beyond the scope of this volume to provide a full treatment of all aspects of mod-ern immunology. Nevertheless, the student of physiology should have a working knowledge of immune functions and their regulation, due to a growing appreciation for the ways in which the immune system can contribute to normal physio-logical regulation in a variety of tissues, as well as contribu-tions of immune effectors to pathophysiology. IMMUNE EFFECTOR CELLS Many immune effector cells circulate in the blood as the white blood cells. In addition, the blood is the conduit for the pre-cursor cells that eventually develop into the immune cells of the tissues. The circulating immunologic cells include granu-locytes (polymorphonuclear leukocytes, PMNs), comprising neutrophils, eosinophils, and basophils; lymphocytes; and monocytes. Immune responses in the tissues are further am-plified by these cells following their extravascular migration, as well as tissue macrophages (derived from monocytes) and mast cells (related to basophils). Acting together, these cells provide the body with powerful defenses against tumors and viral, bacterial, and parasitic infections. 64 SECTION I Cellular & Molecular Basis for Medical Physiology GRANULOCYTES All granulocytes have cytoplasmic granules that contain bio-logically active substances involved in inflammatory and aller-gic reactions. The average half-life of a neutrophil in the circulation is 6 hours. To maintain the normal circulating blood level, it is therefore necessary to produce over 100 billion neutrophils per day. Many neutrophils enter the tissues, particularly if trig-gered to do so by an infection or by inflammatory cytokines. They are attracted to the endothelial surface by cell adhesion molecules known as selectins, and they roll along it. They then bind firmly to neutrophil adhesion molecules of the integrin family. They next insinuate themselves through the walls of the capillaries between endothelial cells by a process called dia-pedesis. Many of those that leave the circulation enter the gas-trointestinal tract and are eventually lost from the body. Invasion of the body by bacteria triggers the inflammatory response. The bone marrow is stimulated to produce and release large numbers of neutrophils. Bacterial products inter-act with plasma factors and cells to produce agents that attract neutrophils to the infected area (chemotaxis). The chemotac-tic agents, which are part of a large and expanding family of chemokines (see following text), include a component of the complement system (C5a); leukotrienes; and polypeptides from lymphocytes, mast cells, and basophils. Other plasma factors act on the bacteria to make them “tasty” to the phago-cytes (opsonization). The principal opsonins that coat the bacteria are immunoglobulins of a particular class (IgG) and complement proteins (see following text). The coated bacteria then bind to receptors on the neutrophil cell membrane. This triggers, via heterotrimeric G protein-mediated responses, increased motor activity of the cell, exocytosis, and the so-called respiratory burst. The increased motor activity leads to prompt ingestion of the bacteria by endocytosis (phagocyto-sis). By exocytosis, neutrophil granules discharge their con-tents into the phagocytic vacuoles containing the bacteria and also into the interstitial space (degranulation). The granules contain various proteases plus antimicrobial proteins called defensins. In addition, the cell membrane-bound enzyme NADPH oxidase is activated, with the production of toxic oxygen metabolites. The combination of the toxic oxygen metabolites and the proteolytic enzymes from the granules makes the neutrophil a very effective killing machine. Activation of NADPH oxidase is associated with a sharp increase in O2 uptake and metabolism in the neutrophil (the respiratory burst) and generation of O2 – by the following reaction: NADPH + H+ + 2O2 + → NADP+ + 2H+ + 2O2 – O2 – is a free radical formed by the addition of one electron to O2. Two O2 – react with two H+ to form H2O2 in a reaction catalyzed by the cytoplasmic form of superoxide dismutase (SOD-1): O2 –+ O2 – + H+ + H+ SOD-1 → H2O2 + O2 O2 – and H2O2 are both oxidants that are effective bacteri-cidal agents, but H2O2 is converted to H2O and O2 by the enzyme catalase. The cytoplasmic form of SOD contains both Zn and Cu. It is found in many parts of the body. It is defec-tive as a result of genetic mutation in a familial form of amyo-trophic lateral sclerosis (ALS; see Chapter 19). Therefore, it may be that O2 – accumulates in motor neurons and kills them in at least one form of this progressive, fatal disease. Two other forms of SOD encoded by at least one different gene are also found in humans. Neutrophils also discharge the enzyme myeloperoxidase, which catalyzes the conversion of Cl–, Br–, I–, and SCN– to the corresponding acids (HOCl, HOBr, etc). These acids are also potent oxidants. Because Cl– is present in greatest abun-dance in body fluids, the principal product is HOCl. In addition to myeloperoxidase and defensins, neutrophil granules contain an elastase, two metalloproteinases that attack collagen, and a variety of other proteases that help destroy invading organisms. These enzymes act in a cooperative fash-ion with the O2 –, H2O2, and HOCl formed by the action of the NADPH oxidase and myeloperoxidase to produce a killing zone around the activated neutrophil. This zone is effective in killing invading organisms, but in certain diseases (eg, rheuma-toid arthritis) the neutrophils may also cause local destruction of host tissue. The movements of the cell in phagocytosis, as well as migration to the site of infection, involve microtubules and microfilaments (see Chapter 1). Proper function of the microfilaments involves the interaction of the actin they con-tain with myosin-1 on the inside of the cell membrane (see Chapter 1). Like neutrophils, eosinophils have a short half-life in the circulation, are attracted to the surface of endothelial cells by selectins, bind to integrins that attach them to the vessel wall, and enter the tissues by diapedesis. Like neutrophils, they release proteins, cytokines, and chemokines that produce inflammation but are capable of killing invading organisms. However, eosinophils have some selectivity in the way in which they respond and in the killing molecules they secrete. Their maturation and activation in tissues is particularly stim-ulated by IL-3, IL-5, and GM-CSF (see below). They are espe-cially abundant in the mucosa of the gastrointestinal tract, where they defend against parasites, and in the mucosa of the respiratory and urinary tracts. Circulating eosinophils are increased in allergic diseases such as asthma and in various other respiratory and gastrointestinal diseases. Basophils also enter tissues and release proteins and cyto-kines. They resemble but are not identical to mast cells, and like mast cells they contain histamine (see below). They release histamine and other inflammatory mediators when activated by binding of specific antigens to cell-fixed IgE mol-ecules, and are essential for immediate-type hypersensitivity reactions. These range from mild urticaria and rhinitis to severe anaphylactic shock. The antigens that trigger IgE for-mation and basophil (and mast cell) activation are innocuous to most individuals, and are referred to as allergens. CHAPTER 3 Immunity, Infection, & Inflammation 65 MAST CELLS Mast cells are heavily granulated cells of the connective tissue that are abundant in tissues that come into contact with the ex-ternal environment, such as beneath epithelial surfaces. Their granules contain proteoglycans, histamine, and many proteas-es. Like basophils, they degranulate when allergens bind to IgE molecules directed against them that previously coat the mast cell surface. They are involved in inflammatory responses ini-tiated by immunoglobulins IgE and IgG (see below). The in-flammation combats invading parasites. In addition to this involvement in acquired immunity, they release TNF-α in re-sponse to bacterial products by an antibody-independent mechanism, thus participating in the nonspecific innate im-munity that combats infections prior to the development of an adaptive immune response (see following text). Marked mast cell degranulation produces clinical manifestations of allergy up to and including anaphylaxis. MONOCYTES Monocytes enter the blood from the bone marrow and circu-late for about 72 hours. They then enter the tissues and be-come tissue macrophages (Figure 3–1). Their life span in the tissues is unknown, but bone marrow transplantation data in humans suggest that they persist for about 3 months. It ap-pears that they do not reenter the circulation. Some of them end up as the multinucleated giant cells seen in chronic in-flammatory diseases such as tuberculosis. The tissue macro-phages include the Kupffer cells of the liver, pulmonary alveolar macrophages (see Chapter 35), and microglia in the brain, all of which come from the circulation. In the past, they have been called the reticuloendothelial system, but the gen-eral term tissue macrophage system seems more appropriate. Macrophages are activated by cytokines released from T lymphocytes, among others. Activated macrophages migrate in response to chemotactic stimuli and engulf and kill bacte-ria by processes generally similar to those occurring in neu-trophils. They play a key role in immunity (see below). They also secrete up to 100 different substances, including factors that affect lymphocytes and other cells, prostaglandins of the E series, and clot-promoting factors. GRANULOCYTE & MACROPHAGE COLONY-STIMULATING FACTORS The production of white blood cells is regulated with great precision in healthy individuals, and the production of granu-locytes is rapidly and dramatically increased in infections. The proliferation and self-renewal of hematopoietic stem cells (HSCs) depends on stem cell factor (SCF). Other factors specify particular lineages. The proliferation and maturation of the cells that enter the blood from the marrow are regulated by glycoprotein growth factors or hormones that cause cells in one or more of the committed cell lines to proliferate and ma-ture (Table 3–1). The regulation of erythrocyte production by erythropoietin is discussed in Chapter 39. Three additional factors are called colony-stimulating factors (CSFs), because they cause appropriate single stem cells to proliferate in soft agar, forming colonies in this culture medium. The factors stimulating the production of committed stem cells include granulocyte–macrophage CSF (GM-CSF), granulocyte CSF (G-CSF), and macrophage CSF (M-CSF). Interleukins IL-1 and IL-6 followed by IL-3 (Table 3–1) act in sequence to con-vert pluripotential uncommitted stem cells to committed pro-genitor cells. IL-3 is also known as multi-CSF. Each of the CSFs has a predominant action, but all the CSFs and interleu-kins also have other overlapping actions. In addition, they ac-tivate and sustain mature blood cells. It is interesting in this regard that the genes for many of these factors are located to-gether on the long arm of chromosome 5 and may have origi-nated by duplication of an ancestral gene. It is also interesting that basal hematopoiesis is normal in mice in which the GM-CSF gene is knocked out, indicating that loss of one factor can be compensated for by others. On the other hand, the absence of GM-CSF causes accumulation of surfactant in the lungs (see Chapter 35). As noted in Chapter 39, erythropoietin is produced in part by kidney cells and is a circulating hormone. The other factors are produced by macrophages, activated T cells, fibroblasts, and endothelial cells. For the most part, the factors act locally in the bone marrow (Clinical Box 3–1). LYMPHOCYTES Lymphocytes are key elements in the production of immunity (see below). After birth, some lymphocytes are formed in the bone marrow. However, most are formed in the lymph nodes (Figure 3–2), thymus, and spleen from precursor cells that originally came from the bone marrow and were processed in the thymus or bursal equivalent (see below). Lymphocytes en-ter the bloodstream for the most part via the lymphatics. At FIGURE 3–1 Macrophages contacting bacteria and preparing to engulf them. Figure is a colorized version of a scanning electron micrograph. Macrophages Pseudopods Bacteria 66 SECTION I Cellular & Molecular Basis for Medical Physiology any given time, only about 2% of the body lymphocytes are in the peripheral blood. Most of the rest are in the lymphoid or-gans. It has been calculated that in humans, 3.5 × 1010 lym-phocytes per day enter the circulation via the thoracic duct alone; however, this count includes cells that reenter the lym-phatics and thus traverse the thoracic duct more than once. The effects of adrenocortical hormones on the lymphoid or-gans, the circulating lymphocytes, and the granulocytes are discussed in Chapter 22. TABLE 3–1 Hematopoietic growth factors. Cytokine Cell Lines Stimulated Cytokine Source IL-1 Erythrocyte Multiple cell types Granulocyte Megakaryocyte Monocyte IL-3 Erythrocyte T lymphocytes Granulocyte Megakaryocyte Monocyte IL-4 Basophil T lymphocytes IL-5 Eosinophil T lymphocytes IL-6 Erythrocyte Endothelial cells Granulocyte Megakaryocyte Fibroblasts Monocyte Macrophages IL-11 Erythrocyte Fibroblasts Granulocyte Osteoblasts Megakaryocyte Erythropoietin Erythrocyte Kidney Kupffer cells of liver SCF Erythrocyte Multiple cell types Granulocyte Megakaryocyte Monocyte G-CSF Granulocyte Endothelial cells Fibroblasts Monocytes GM-CSF Erythrocyte Endothelial cells Fibroblasts Granulocyte Monocytes Megakaryocyte T lymphocytes M-CSF Monocyte Endothelial cells Fibroblasts Monocytes Thrombopoietin Megakaryocyte Liver, kidney Key: IL = interleukin; CSF = colony stimulating factor; G = granulocyte; M = macro-phage; SCF = stem cell factor. Reproduced with permission from McPhee SJ, Lingappa VR, Ganong WF (editors): Pathophysiology of Disease, 4th ed. McGraw-Hill, 2003. CLINICAL BOX 3–1 Disorders of Phagocytic Function More than 15 primary defects in neutrophil function have been described, along with at least 30 other conditions in which there is a secondary depression of the function of neutrophils. Patients with these diseases are prone to infec-tions that are relatively mild when only the neutrophil sys-tem is involved, but which can be severe when the mono-cyte-tissue macrophage system is also involved. In one syndrome (neutrophil hypomotility), actin in the neutro-phils does not polymerize normally, and the neutrophils move slowly. In another, there is a congenital deficiency of leukocyte integrins. In a more serious disease (chronic granulomatous disease), there is a failure to generate O2 – in both neutrophils and monocytes and consequent inability to kill many phagocytosed bacteria. In severe congenital glucose 6-phosphate dehydrogenase deficiency, there are multiple infections because of failure to generate the NADPH necessary for O2 – production. In congenital myelo-peroxidase deficiency, microbial killing power is reduced because hypochlorous acid is not formed. FIGURE 3–2 Anatomy of a normal lymph node. (After Chandrasoma. Reproduced with permission from McPhee SJ, Lingappa VR, Ganong WF [editors]: Pathophysiology of Disease, 4th ed. McGraw-Hill, 2003.) Cortical follicles, B cells Paracortex, T cells Medullary cords, plasma cells CHAPTER 3 Immunity, Infection, & Inflammation 67 IMMUNITY OVERVIEW Insects and other invertebrates have only innate immunity. This system is triggered by receptors that bind sequences of sugars, fats, or amino acids in common bacteria and activate various de-fense mechanisms. The receptors are coded in the germ line, and their fundamental structure is not modified by exposure to anti-gen. The activated defenses include, in various species, release of interferons, phagocytosis, production of antibacterial peptides, activation of the complement system, and several proteolytic cas-cades. Even plants release antibacterial peptides in response to in-fection. In vertebrates, innate immunity is also present, but is complemented by adaptive or acquired immunity, a system in which T and B lymphocytes are activated by very specific anti-gens. In both innate and acquired immunity, the receptors in-volved recognize the shape of antigens, not their specific chemical composition. In acquired immunity, activated B lymphocytes form clones that produce more antibodies which attack foreign proteins. After the invasion is repelled, small numbers persist as memory cells so that a second exposure to the same antigen pro-vokes a prompt and magnified immune attack. The genetic event that led to acquired immunity occurred 450 million years ago in the ancestors of jawed vertebrates and was probably insertion of a transposon into the genome in a way that made possible the generation of the immense repertoire of T cell receptors that are present in the body. In vertebrates, including humans, innate immunity provides the first line of defense against infections, but it also triggers the slower but more specific acquired immune response (Fig-ure 3–3). In vertebrates, natural and acquired immune mecha-nisms also attack tumors and tissue transplanted from other animals. Once activated, immune cells communicate by means of cytokines and chemokines. They kill viruses, bacteria, and other foreign cells by secreting other cytokines and activating the complement system. CYTOKINES Cytokines are hormonelike molecules that act—generally in a paracrine fashion—to regulate immune responses. They are secreted not only by lymphocytes and macrophages but by en-dothelial cells, neurons, glial cells, and other types of cells. Most of the cytokines were initially named for their actions, for example, B cell-differentiating factor, B cell-stimulating factor 2. However, the nomenclature has since been rational-ized by international agreement to that of the interleukins. For example, the name of B cell-differentiating factor was changed to interleukin-4. A number of cytokines selected for their biological and clinical relevance are listed in Table 3–2, but it would be beyond the scope of this text to list all cytokines, which now number more than 100. Many of the receptors for cytokines and hematopoietic growth factors (see above), as well as the receptors for prolactin FIGURE 3–3 How bacteria, viruses, and tumors trigger innate immunity and initiate the acquired immune response. Arrows indicate mediators/cytokines that act on the target cell shown and/or pathways of differentiation. APC, antigen-presenting cell; M, monocyte; N, neutrophil; TH1 and TH2, helper T cells type 1 and type 2, respectively. Naive T cell γδT cell TH 2 B M N Plasma cell Cytotoxic lymphocyte APC Bacteria Viruses Tumors IL-4 Chemokines TH 1 68 SECTION I Cellular & Molecular Basis for Medical Physiology (see Chapter 25), and growth hormone (see Chapter 24) are members of a cytokine-receptor superfamily that has three subfamilies (Figure 3–4). The members of subfamily 1, which includes the receptors for IL-4 and IL-7, are homodimers. The members of subfamily 2, which includes the receptors for IL-3, IL-5, and IL-6, are heterodimers. The receptor for IL-2 and several other cytokines is unique in that it consists of a het-erodimer plus an unrelated protein, the so-called Tac antigen. The other members of subfamily 3 have the same γ chain as IL-2R. The extracellular domain of the homodimer and het-erodimer subunits all contain four conserved cysteine residues plus a conserved Trp-Ser-X-Trp-Ser domain, and although the TABLE 3–2 Examples of cytokines and their clinical relevance. Cytokine Cellular Sources Major Activities Clinical Relevance Interleukin-1 Macrophages Activation of T cells and macrophages; promotion of inflammation Implicated in the pathogenesis of septic shock, rheumatoid arthritis, and atherosclerosis Interleukin-2 Type 1 (TH1) helper T cells Activation of lymphocytes, natural killer cells, and macrophages Used to induce lymphokine-activated killer cells; used in the treatment of metastatic renal-cell carci-noma, melanoma, and various other tumors Interleukin-4 Type 2 (TH2) helper T cells, mast cells, basophils, and eosinophils Activation of lymphocytes, monocytes, and IgE class switching As a result of its ability to stimulate IgE production, plays a part in mast-cell sensitization and thus in al-lergy and in defense against nematode infections Interleukin-5 Type 2 (TH2) helper T cells, mast cells, and eosinophils Differentiation of eosinophils Monoclonal antibody against interleukin-5 used to inhibit the antigen-induced late-phase eosinophil-ia in animal models of allergy Interleukin-6 Type 2 (TH2) helper T cells and macrophages Activation of lymphocytes; differentia-tion of B cells; stimulation of the produc-tion of acute-phase proteins Overproduced in Castleman’s disease; acts as an autocrine growth factor in myeloma and in mesan-gial proliferative glomerulonephritis Interleukin-8 T cells and macrophages Chemotaxis of neutrophils, basophils, and T cells Levels are increased in diseases accompanied by neutrophilia, making it a potentially useful marker of disease activity Interleukin-11 Bone marrow stromal cells Stimulation of the production of acute-phase proteins Used to reduce chemotherapy-induced thrombo-cytopenia in patients with cancer Interleukin-12 Macrophages and B cells Stimulation of the production of inter-feron γ by type 1 (TH1) helper T cells and by natural killer cells; induction of type 1 (TH1) helper T cells May be useful as an adjuvant for vaccines Tumor necrosis factor α Macrophages, natural killer cells, T cells, B cells, and mast cells Promotion of inflammation Treatment with antibodies against tumor necrosis factor α beneficial in rheumatoid arthritis Lymphotoxin (tumor necrosis factor β) Type 1 (TH1) helper T cells and B cells Promotion of inflammation Implicated in the pathogenesis of multiple sclero-sis and insulin-dependent diabetes mellitus Transforming growth factor β T cells, macrophages, B cells, and mast cells Immunosuppression May be useful therapeutic agent in multiple sclero-sis and myasthenia gravis Granulocyte-macrophage colony-stimulating factor T cells, macrophages, natu-ral killer cells, and B cells Promotion of the growth of granulo-cytes and monocytes Used to reduce neutropenia after chemotherapy for tumors and in ganciclovir-treated patients with AIDS; used to stimulate cell production after bone marrow transplantation Interferon-α Virally infected cells Induction of resistance of cells to viral infection Used to treat AIDS-related Kaposi sarcoma, mela-noma, chronic hepatitis B infection, and chronic hepatitis C infection Interferon-β Virally infected cells Induction of resistance of cells to viral infection Used to reduce the frequency and severity of relapses in multiple sclerosis Interferon-γ Type 1 (TH1) helper T cells and natural killer cells Activation of macrophages; inhibition of type 2 (TH2) helper T cells Used to enhance the killing of phagocytosed bacteria in chronic granulomatous disease Reproduced with permission from Delves PJ, Roitt IM: The immune system. First of two parts. N Engl J Med 2000;343:37. CHAPTER 3 Immunity, Infection, & Inflammation 69 intracellular portions do not contain tyrosine kinase catalytic domains, they activate cytoplasmic tyrosine kinases when ligand binds to the receptors. The effects of the principal cytokines are listed in Table 3–2. Some of them have systemic as well as local paracrine effects. For example, IL-1, IL-6, and tumor necrosis factor α cause fever, and IL-1 increases slow-wave sleep and reduces appetite. Another superfamily of cytokines is the chemokine family. Chemokines are substances that attract neutrophils (see pre-vious text) and other white blood cells to areas of inflamma-tion or immune response. Over 40 have now been identified, and it is clear that they also play a role in the regulation of cell growth and angiogenesis. The chemokine receptors are G protein-coupled receptors that cause, among other things, extension of pseudopodia with migration of the cell toward the source of the chemokine. THE COMPLEMENT SYSTEM The cell-killing effects of innate and acquired immunity are mediated in part by a system of more than 30 plasma proteins originally named the complement system because they “com-plemented” the effects of antibodies. Three different pathways or enzyme cascades activate the system: the classic pathway, triggered by immune complexes; the mannose-binding lectin pathway, triggered when this lectin binds mannose groups in bacteria; and the alternative or properdin pathway, triggered by contact with various viruses, bacteria, fungi, and tumor cells. The proteins that are produced have three functions: They help kill invading organisms by opsonization, chemo-taxis, and eventual lysis of the cells; they serve in part as a bridge from innate to acquired immunity by activating B cells and aiding immune memory; and they help dispose of waste products after apoptosis. Cell lysis, one of the principal ways the complement system kills cells, is brought about by insert-ing proteins called perforins into their cell membranes. These create holes, which permit free flow of ions and thus disrup-tion of membrane polarity. INNATE IMMUNITY The cells that mediate innate immunity include neutrophils, macrophages, and natural killer (NK) cells, large lympho-cytes that are not T cells but are cytotoxic. All these cells re-spond to lipid and carbohydrate sequences unique to bacterial cell walls and to other substances characteristic of tumor and transplant cells. Many cells that are not professional immuno-cytes may nevertheless also contribute to innate immune FIGURE 3–4 Members of one of the cytokine receptor superfamilies, showing shared structural elements. Note that all the subunits except the α subunit in subfamily 3 have four conserved cysteine residues (open boxes at top) and a Trp-Ser-X-Trp-Ser motif (pink). Many subunits also contain a critical regulatory domain in their cytoplasmic portions (green). CNTF, ciliary neurotrophic factor; LIF, leukemia inhibitory factor; OSM, oncostatin M; PRL, prolactin. (Modified from D’Andrea AD: Cytokine receptors in congenital hematopoietic disease. N Engl J Med 1994;330:839.) Subfamily 1 Subfamily 2 Subfamily 3 α α β β γ ECF Cytoplasm Erythropoietin G-CSF IL-4 IL-7 Growth hormone PRL IL-3 GM-CSF IL-5 IL-6 IL-11 LIF OSM CNTF Shared β subunit Shared gp130 subunit IL-2 IL-4 IL-7 IL-9 IL-15 70 SECTION I Cellular & Molecular Basis for Medical Physiology responses, such as endothelial and epithelial cells. The activat-ed cells produce their effects via the release of cytokines, as well as, in some cases, complement and other systems. An important link in innate immunity in Drosophila is a receptor protein named toll, which binds fungal antigens and triggers activation of genes coding for antifungal proteins. An expanding list of toll-like receptors (TLRs) have now been identified in humans. One of these, TLR4, binds bacterial lipopolysaccharide and a protein called CD14, and this initiates a cascade of intracellular events that activate transcription of genes for a variety of proteins involved in innate immune responses. This is important because bacterial lipopolysaccha-ride produced by gram-negative organisms is the cause of septic shock. TLR2 mediates the response to microbial lipoproteins, TLR6 cooperates with TLR2 in recognizing certain peptidogly-cans, and TLR9 recognizes the DNA of certain bacteria. ACQUIRED IMMUNITY As noted previously, the key to acquired immunity is the abil-ity of lymphocytes to produce antibodies (in the case of B cells) or cell-surface receptors (in the case of T cells) that are specific for one of the many millions of foreign agents that may invade the body. The antigens stimulating production of T cell receptors or antibodies are usually proteins and polypeptides, but antibodies can also be formed against nucle-ic acids and lipids if these are presented as nucleoproteins and lipoproteins, and antibodies to smaller molecules can be pro-duced experimentally if the molecules are bound to protein. Acquired immunity has two components: humoral immunity and cellular immunity. Humoral immunity is mediated by circulating immunoglobulin antibodies in the γ-globulin frac-tion of the plasma proteins. Immunoglobulins are produced by differentiated forms of B lymphocytes known as plasma cells, and they activate the complement system and attack and neutralize antigens. Humoral immunity is a major defense against bacterial infections. Cellular immunity is mediated by T lymphocytes. It is responsible for delayed allergic reactions and rejection of transplants of foreign tissue. Cytotoxic T cells attack and destroy cells that have the antigen which activated them. They kill by inserting perforins (see above) and by initi-ating apoptosis. Cellular immunity constitutes a major de-fense against infections due to viruses, fungi, and a few bacteria such as the tubercle bacillus. It also helps defend against tumors. DEVELOPMENT OF THE IMMUNE SYSTEM During fetal development, and to a much lesser extent during adult life, lymphocyte precursors come from the bone mar-row. Those that populate the thymus (Figure 3–5) become transformed by the environment in this organ into T lympho-cytes. In birds, the precursors that populate the bursa of Fabricius, a lymphoid structure near the cloaca, become trans-formed into B lymphocytes. There is no bursa in mammals, and the transformation to B lymphocytes occurs in bursal equivalents, that is, the fetal liver and, after birth, the bone marrow. After residence in the thymus or liver, many of the T and B lymphocytes migrate to the lymph nodes. T and B lymphocytes are morphologically indistinguishable but can be identified by markers on their cell membranes. B cells differentiate into plasma cells and memory B cells. There are three major types of T cells: cytotoxic T cells, helper T cells, and memory T cells. There are two subtypes of helper T cells: T helper 1 (TH1) cells secrete IL-2 and γ-interferon and are concerned primarily with cellular immu-nity; T helper 2 (TH2) cells secrete IL-4 and IL-5 and interact primarily with B cells in relation to humoral immunity. Cyto-toxic T cells destroy transplanted and other foreign cells, with their development aided and directed by helper T cells. Mark-ers on the surface of lymphocytes are assigned CD (clusters of differentiation) numbers on the basis of their reactions to a FIGURE 3–5 Development of the system mediating acquired immunity. Thymus T lymphocytes Memory T cells Bone marrow lymphocyte precursors Helper T cells (CD4 T cells) B lymphocytes Bursal equivalent (liver, bone marrow) Memory B cells Plasma cells IgG IgA IgM IgD IgE Humoral immunity Cellular immunity Cytotoxic T cells (mostly CD8 T cells) CHAPTER 3 Immunity, Infection, & Inflammation 71 panel of monoclonal antibodies. Most cytotoxic T cells display the glycoprotein CD8, and helper T cells display the glycopro-tein CD4. These proteins are closely associated with the T cell receptors and may function as coreceptors. On the basis of differences in their receptors and functions, cytotoxic T cells are divided into αβ and γδ types (see below). Natural killer cells (see above) are also cytotoxic lymphocytes, though they are not T cells. Thus, there are three main types of cytotoxic lymphocytes in the body: αβ T cells, γδ T cells, and NK cells. MEMORY B CELLS & T CELLS After exposure to a given antigen, a small number of activated B and T cells persist as memory B and T cells. These cells are readi-ly converted to effector cells by a later encounter with the same antigen. This ability to produce an accelerated response to a sec-ond exposure to an antigen is a key characteristic of acquired immunity. The ability persists for long periods of time, and in some instances (eg, immunity to measles) it can be lifelong. After activation in lymph nodes, lymphocytes disperse widely throughout the body and are especially plentiful in areas where invading organisms enter the body, for example, the mucosa of the respiratory and gastrointestinal tracts. This puts memory cells close to sites of reinfection and may account in part for the rapidity and strength of their response. Chemokines are involved in guiding activated lymphocytes to these locations. ANTIGEN RECOGNITION The number of different antigens recognized by lymphocytes in the body is extremely large. The repertoire develops initially without exposure to the antigen. Stem cells differentiate into many million different T and B lymphocytes, each with the ability to respond to a particular antigen. When the antigen first enters the body, it can bind directly to the appropriate re-ceptors on B cells. However, a full antibody response requires that the B cells contact helper T cells. In the case of T cells, the antigen is taken up by an antigen-presenting cell and partially digested. A peptide fragment of it is presented to the appropri-ate receptors on T cells. In either case, the cells are stimulated to divide, forming clones of cells that respond to this antigen (clonal selection). Effector cells are also subject to negative selection, during which lymphocyte precursors that are reac-tive with self antigens are normally deleted. This results in im-mune tolerance. It is this latter process that presumably goes awry in autoimmune diseases, where the body reacts to and destroys cells expressing normal proteins, with accompanying inflammation that may lead to tissue destruction. ANTIGEN PRESENTATION Antigen-presenting cells (APCs) include specialized cells called dendritic cells in the lymph nodes and spleen and the Langerhans dendritic cells in the skin. Macrophages and B cells themselves, and likely many other cell types, can also function as APCs. In APCs, polypeptide products of antigen digestion are coupled to protein products of the major histo-compatibility complex (MHC) genes and presented on the surface of the cell. The products of the MHC genes are called human leukocyte antigens (HLA). The genes of the MHC, which are located on the short arm of human chromosome 6, encode glycoproteins and are divided into two classes on the basis of structure and func-tion. Class I antigens are composed of a 45-kDa heavy chain associated noncovalently with β2-microglobulin encoded by a gene outside the MHC (Figure 3–6). They are found on all nucleated cells. Class II antigens are heterodimers made up of a 29- to 34-kDa α chain associated noncovalently with a 25-to 28-kDa β chain. They are present in antigen-presenting cells, including B cells, and in activated T cells. The class I MHC proteins (MHC-I proteins) are coupled primarily to peptide fragments generated from proteins syn-thesized within cells. The peptides to which the host is not tolerant (eg, those from mutant or viral proteins) are recog-nized by T cells. The digestion of these proteins occurs in FIGURE 3–6 Structure of human histocompatibility antigen HLA-A2. The antigen-binding pocket is at the top and is formed by the α1 and α2 parts of the molecule. The α3 portion and the associated β2-microglobulin (β2m) are close to the membrane. The extension of the C terminal from α3 that provides the transmembrane domain and the small cytoplasmic portion of the molecule have been omitted. (Reproduced with permission from Bjorkman PJ et al: Structure of the human histocompatibility antigen HLA-A2. Nature 1987;329:506.) N C C N α2 α1 2m β 3 α 72 SECTION I Cellular & Molecular Basis for Medical Physiology proteasomes, complexes of proteolytic enzymes that may be produced by genes in the MHC group, and the peptide frag-ments appear to bind to MHC proteins in the endoplasmic reticulum. The class II MHC proteins (MHC-II proteins) are concerned primarily with peptide products of extracellular antigens, such as bacteria, that enter the cell by endocytosis and are digested in the late endosomes. T CELL RECEPTORS The MHC protein–peptide complexes on the surface of the antigen-presenting cells bind to appropriate T cells. There-fore, receptors on the T cells must recognize a very wide vari-ety of complexes. Most of the receptors on circulating T cells are made up of two polypeptide units designated α and β. They form heterodimers that recognize the MHC proteins and the antigen fragments with which they are combined (Figure 3–7). These cells are called αβ T cells. About 10% of the circu-lating T cells have two different polypeptides designated γ and δ in their receptors, and they are called γδ T cells. These T cells are prominent in the mucosa of the gastrointestinal tract, and there is evidence that they form a link between the innate and acquired immune systems by way of the cytokines they secrete (Figure 3–3). CD8 occurs on the surface of cytotoxic T cells that bind MHC-I proteins, and CD4 occurs on the surface of helper T cells that bind MHC-II proteins (Figure 3–8). The CD8 and CD4 proteins facilitate the binding of the MHC proteins to the T cell receptors, and they also foster lymphocyte develop-ment, but how they produce these effects is unsettled. The activated CD8 cytotoxic T cells kill their targets directly, whereas the activated CD4 helper T cells secrete cytokines that activate other lymphocytes. The T cell receptors are surrounded by adhesion molecules and proteins that bind to complementary proteins in the anti-gen-presenting cell when the two cells transiently join to form the “immunologic synapse” that permits T cell activation to occur. It is now generally accepted that two signals are neces-sary to produce activation. One is produced by the binding of the digested antigen to the T cell receptor. The other is pro-duced by the joining of the surrounding proteins in the “syn-apse.” If the first signal occurs but the second does not, the T cell is inactivated and becomes unresponsive. B CELLS As noted above, B cells can bind antigens directly, but they must contact helper T cells to produce full activation and antibody formation. It is the TH2 subtype that is mainly involved. Helper T cells develop along the TH2 lineage in response to IL-4 (see below). On the other hand, IL-12 promotes the TH1 phenotype. IL-2 acts in an autocrine fashion to cause activated T cells to proliferate. The role of various cytokines in B cell and T cell ac-tivation is summarized in Figure 3–9. The activated B cells proliferate and transform into mem-ory B cells (see above) and plasma cells. The plasma cells secrete large quantities of antibodies into the general circu-lation. The antibodies circulate in the globulin fraction of the plasma and, like antibodies elsewhere, are called immunoglobulins. The immunoglobulins are actually the secreted form of antigen-binding receptors on the B cell membrane. FIGURE 3–7 Interaction between antigen-presenting cell (top) and αβ T lymphocyte (bottom). The MHC proteins (in this case, MHC-I) and their peptide antigen fragment bind to the α and β units that combine to form the T cell receptor. β α + + β2m α3 α1/α2 Antigen-presenting cell membrane Cytoplasm Cytoplasm ECF ECF Antigen fragment Variable regions Constant regions T cell membrane T cell receptor heterodimer (α:β) MHC molecular complex S–S FIGURE 3–8 Diagrammatic summary of the structure of CD4 and CD8, and their relation to MHC-I and MHC-II proteins. Note that CD4 is a single protein, whereas CD8 is a heterodimer. Class II MHC CD4 TCR CD8 TCR Class I MHC CHAPTER 3 Immunity, Infection, & Inflammation 73 IMMUNOGLOBULINS Circulating antibodies protect their host by binding to and neutralizing some protein toxins, by blocking the attachment of some viruses and bacteria to cells, by opsonizing bacteria (see above), and by activating complement. Five general types of immunoglobulin antibodies are produced by the lympho-cyte–plasma cell system. The basic component of each is a sym-metric unit containing four polypeptide chains (Figure 3–10). The two long chains are called heavy chains, whereas the two short chains are called light chains. There are two types of light chains, k and λ, and eight types of heavy chains. The chains are joined by disulfide bridges that permit mobility, and there are intrachain disulfide bridges as well. In addition, the heavy chains are flexible in a region called the hinge. Each heavy chain has a variable (V) segment in which the amino acid se-quence is highly variable, a diversity (D) segment in which the amino acid segment is also highly variable, a joining (J) seg-ment in which the sequence is moderately variable, and a con-stant (C) segment in which the sequence is constant. Each light chain has a V, a J, and a C segment. The V segments form part of the antigen-binding sites (Fab portion of the molecule [Fig-ure 3–10]). The Fc portion of the molecule is the effector por-tion, which mediates the reactions initiated by antibodies. Two of the classes of immunoglobulins contain additional polypeptide components (Table 3–3). In IgMs, five of the basic immunoglobulin units join around a polypeptide called the J chain to form a pentamer. In IgAs, the secretory immu-noglobulins, the immunoglobulin units form dimers and tri-mers around a J chain and a polypeptide that comes from epithelial cells, the secretory component (SC). In the intestine, bacterial and viral antigens are taken up by M cells (see Chapter 27) and passed on to underlying aggre-gates of lymphoid tissue (Peyer’s patches), where they acti-vate naive T cells. These lymphocytes then form B cells that infiltrate mucosa of the gastrointestinal, respiratory, geni-tourinary, and female reproductive tracts and the breast. There they secrete large amounts of IgAs when exposed again to the antigen that initially stimulated them. The epithelial cells produce the SC, which acts as a receptor for and binds the IgA. The resulting secretory immunoglobulin passes through the epithelial cell and is secreted by exocytosis. This system of secretory immunity is an important and effective defense mechanism. GENETIC BASIS OF DIVERSITY IN THE IMMUNE SYSTEM The genetic mechanism for the production of the immensely large number of different configurations of immunoglobulins produced by human B cells is a fascinating biologic problem. FIGURE 3–9 Summary of acquired immunity. (1) An antigen-presenting cell ingests and partially digests an antigen, then presents part of the antigen along with MHC peptides (in this case, MHC II pep-tides on the cell surface). (2) An “immune synapse” forms with a naive CD4 T cell, which is activated to produce IL-2. (3) IL-2 acts in an auto-crine fashion to cause the cell to multiply, forming a clone. (4) The ac-tivated CD4 cell may promote B cell activation and production of plasma cells or it may activate a cytotoxic CD8 cell. The CD8 cell can also be activated by forming a synapse with an MCH I antigen-present-ing cell. (Reproduced with permission from McPhee SJ, Lingappa VR, Ganong WF [editors]: Pathophysiology of Disease, 4th ed. McGraw-Hill, 2003.) MHC class II CD4 TCR IL-1 CD4 Activated T cell IL-2R IL-2 Inflammation and delayed hypersensitivity Cytokine-induced activation Activated B cell Antibody-producing cell 1 2 3 4 Macrophage (antigen-presenting cell) CD8 IL-2R MHC class I Cytotoxic T cell 4 4 Cell death FIGURE 3–10 Typical immunoglobulin G molecule. Fab, por-tion of the molecule that is concerned with antigen binding; Fc, effec-tor portion of the molecule. The constant regions are pink and purple, and the variable regions are orange. The constant segment of the heavy chain is subdivided into CH1, CH2, and CH3. SS lines indicate in-tersegmental disulfide bonds. On the right side, the C labels are omit-ted to show regions JH, D, and JL. SS SS SS SS Antigen-binding site Complement binding Macrophage binding Fab Fc Hinge VH VL VL VH JL JHD CL CH1 CH2 CH3 74 SECTION I Cellular & Molecular Basis for Medical Physiology Diversity is brought about in part by the fact that in immune globulin molecules there are two kinds of light chains and eight kinds of heavy chains. As noted previously, there are areas of great variability (hypervariable regions) in each chain. The vari-able portion of the heavy chains consists of the V, D, and J seg-ments. In the gene family responsible for this region, there are several hundred different coding regions for the V segment, about 20 for the D segment, and 4 for the J segment. During B cell development, one V coding region, one D coding region, and one J coding region are selected at random and recombined to form the gene that produces that particular variable portion. A similar variable recombination takes place in the coding regions respon-sible for the two variable segments (V and J) in the light chain. In addition, the J segments are variable because the gene segments join in an imprecise and variable fashion (junctional site diversi-ty) and nucleotides are sometimes added (junctional insertion di-versity). It has been calculated that these mechanisms permit the production of about 1015 different immunoglobulin molecules. Additional variability is added by somatic mutation. Similar gene rearrangement and joining mechanisms oper-ate to produce the diversity in T cell receptors. In humans, the α subunit has a V region encoded by 1 of about 50 different genes and a J region encoded by 1 of another 50 different genes. The β subunits have a V region encoded by 1 of about 50 genes, a D region encoded by 1 of 2 genes, and a J region encoded by 1 of 13 genes. These variable regions permit the generation of up to an estimated 1015 different T cell receptors (Clinical Box 3–2 and Clinical Box 3–3). A variety of immunodeficiency states can arise from defects in these various stages of B and T lymphocyte maturation. These are summarized in Figure 3–12. PLATELETS Platelets are circulating cells that are important mediators of hemostasis. While not immune cells, per se, they often partic-ipate in the response to tissue injury in cooperation with in-flammatory cell types (see below). They have a ring of microtubules around their periphery and an extensively in-vaginated membrane with an intricate canalicular system in contact with the ECF. Their membranes contain receptors for collagen, ADP, vessel wall von Willebrand factor (see below), and fibrinogen. Their cytoplasm contains actin, myosin, gly-cogen, lysosomes, and two types of granules: (1) dense gran-ules, which contain the nonprotein substances that are secreted in response to platelet activation, including seroto-nin, ADP, and other adenine nucleotides; and (2) α-granules, which contain secreted proteins other than the hydrolases in ly-sosomes. These proteins include clotting factors and platelet-derived growth factor (PDGF). PDGF is also produced by macrophages and endothelial cells. It is a dimer made up of A and B subunit polypeptides. Homodimers (AA and BB), as well as the heterodimer (AB), are produced. PDGF stimulates wound healing and is a potent mitogen for vascular smooth muscle. Blood vessel walls as well as platelets contain von Willebrand factor, which, in addition to its role in adhesion, regulates circulating levels of factor VIII (see below). When a blood vessel wall is injured, platelets adhere to the exposed collagen and von Willebrand factor in the wall via receptors on the platelet membrane. Von Willebrand factor is a very large circulating molecule that is produced by endothe-lial cells. Binding produces platelet activations which release the contents of their granules. The released ADP acts on the ADP receptors in the platelet membranes to produce further accumulation of more platelets (platelet aggregation). Humans have at least three different types of platelet ADP receptors: P2Y1, P2Y2, and P2X1. These are obviously attrac-tive targets for drug development, and several new inhibitors have shown promise in the prevention of heart attacks and strokes. Aggregation is also fostered by platelet-activating factor (PAF), a cytokine secreted by neutrophils and mono-cytes as well as platelets. This compound also has inflamma-tory activity. It is an ether phospholipid, 1-alkyl-2-acetylglyceryl-3-phosphorylcholine, which is produced from membrane lipids. It acts via a G protein-coupled receptor to TABLE 3–3 Human immunoglobulins.a Immunoglobulin Function Heavy Chain Additional Chain Structure Plasma Concentration (mg/dL) IgG Complement activation γ1, γ2, γ3, γ4 Monomer 1000 IgA Localized protection in external secretions (tears, intestinal secre-tions, etc) α1, α2 J, SC Monomer; dimer with J or SC chain; trimer with J chain 200 IgM Complement activation μ J Pentamer with J chain 120 IgD Antigen recognition by B cells δ Monomer 3 IgE Reagin activity; releases histamine from basophils and mast cells ε Monomer 0.05 aIn all instances, the light chains are k or γ. CHAPTER 3 Immunity, Infection, & Inflammation 75 increase the production of arachidonic acid derivatives, including thromboxane A2. The role of this compound in the balance between clotting and anticlotting activity at the site of vascular injury is discussed in Chapter 32. Platelet production is regulated by the colony-stimulating factors that control the production of megakaryocytes, plus thrombopoietin, a circulating protein factor. This factor, which facilitates megakaryocyte maturation, is produced con-stitutively by the liver and kidneys, and there are thrombopoi-etin receptors on platelets. Consequently, when the number of platelets is low, less is bound and more is available to stimulate production of platelets. Conversely, when the number of platelets is high, more is bound and less is available, produc-ing a form of feedback control of platelet production. The amino terminal portion of the thrombopoietin molecule has the platelet-stimulating activity, whereas the carboxyl termi-nal portion contains many carbohydrate residues and is con-cerned with the bioavailability of the molecule. When the platelet count is low, clot retraction is deficient and there is poor constriction of ruptured vessels. The resulting clinical syndrome (thrombocytopenic purpura) is character-ized by easy bruisability and multiple subcutaneous hemor-rhages. Purpura may also occur when the platelet count is normal, and in some of these cases, the circulating platelets are abnormal (thrombasthenic purpura). Individuals with throm-bocytosis are predisposed to thrombotic events. INFLAMMATION & WOUND HEALING LOCAL INJURY Inflammation is a complex localized response to foreign sub-stances such as bacteria or in some instances to internally pro-duced substances. It includes a sequence of reactions initially involving cytokines, neutrophils, adhesion molecules, com-plement, and IgG. PAF, an agent with potent inflammatory ef-fects, also plays a role. Later, monocytes and lymphocytes are involved. Arterioles in the inflamed area dilate, and capillary permeability is increased (see Chapters 33 and 34). When the inflammation occurs in or just under the skin (Figure 3–13), it CLINICAL BOX 3–2 Autoimmunity Sometimes the processes that eliminate antibodies against self antigens fail and a variety of different autoimmune dis-eases are produced. These can be B cell- or T cell-mediated and can be organ-specific or systemic. They include type 1 diabetes mellitus (antibodies against pancreatic islet B cells), myasthenia gravis (antibodies against nicotinic cholinergic receptors), and multiple sclerosis (antibodies against myelin basic protein and several other components of myelin). In some instances, the antibodies are against receptors and are capable of activating those receptors; for example, antibod-ies against TSH receptors increase thyroid activity and cause Graves’ disease (see Chapter 20). Other conditions are due to the production of antibodies against invading organisms that cross-react with normal body constituents (molecular mimicry). An example is rheumatic fever following a strep-tococcal infection; a portion of cardiac myosin resembles a portion of the streptococcal M protein, and antibodies in-duced by the latter attack the former and damage the heart. Some conditions may be due to bystander effects, in which inflammation sensitizes T cells in the neighborhood, causing them to become activated when otherwise they would not respond. However, much is still uncertain about the patho-genesis of autoimmune disease. CLINICAL BOX 3–3 Tissue Transplantation The T lymphocyte system is responsible for the rejection of transplanted tissue. When tissues such as skin and kidneys are transplanted from a donor to a recipient of the same spe-cies, the transplants “take” and function for a while but then become necrotic and are “rejected” because the recipient de-velops an immune response to the transplanted tissue. This is generally true even if the donor and recipient are close rela-tives, and the only transplants that are never rejected are those from an identical twin. A number of treatments have been developed to overcome the rejection of transplanted organs in humans. The goal of treatment is to stop rejection without leaving the patient vulnerable to massive infections. One approach is to kill T lymphocytes by killing all rapidly di-viding cells with drugs such as azathioprine, a purine antime-tabolite, but this makes patients susceptible to infections and cancer. Another is to administer glucocorticoids, which in-hibit cytotoxic T cell proliferation by inhibiting production of IL-2, but these cause osteoporosis, mental changes, and the other facets of Cushing syndrome (see Chapter 22). More re-cently, immunosuppressive drugs such as cyclosporine or tacrolimus (FK-506) have found favor. Activation of the T cell receptor normally increases intracellular Ca2+, which acts via calmodulin to activate calcineurin (Figure 3-11). Cal-cineurin dephosphorylates the transcription factor NF-AT, which moves to the nucleus and increases the activity of genes coding for IL-2 and related stimulatory cytokines. Cy-closporine and tacrolimus prevent the dephosphorylation of NF-AT. However, these drugs inhibit all T cell-mediated im-mune responses, and cyclosporine causes kidney damage and cancer. A new and promising approach to transplant re-jection is the production of T cell unresponsiveness by using drugs that block the costimulation that is required for normal activation (see text). Clinically effective drugs that act in this fashion could be of great value to transplant surgeons. 76 SECTION I Cellular & Molecular Basis for Medical Physiology is characterized by redness, swelling, tenderness, and pain. Elsewhere, it is a key component of asthma, ulcerative colitis, and many other diseases. Evidence is accumulating that a transcription factor, nuclear factor-κB, plays a key role in the inflammatory response. NF-κB is a heterodimer that normally exists in the cytoplasm of cells bound to IκBα, which renders it inactive. Stimuli such as cytokines, viruses, and oxidants separate NF-κB from IκBα, which is then degraded. NF-κB moves to the nucleus, where it binds to the DNA of the genes for numerous inflammatory mediators, resulting in their increased produc-tion and secretion. Glucocorticoids inhibit the activation of NF-κB by increasing the production of IκBα, and this is prob-ably the main basis of their anti-inflammatory action (see Chapter 22). SYSTEMIC RESPONSE TO INJURY Cytokines produced in response to inflammation and other injuries also produce systemic responses. These include alter-ations in plasma acute phase proteins, defined as proteins whose concentration is increased or decreased by at least 25% following injury. Many of the proteins are of hepatic origin. A number of them are shown in Figure 3–14. The causes of the changes in concentration are incompletely understood, but it can be said that many of the changes make homeostatic sense. Thus, for example, an increase in C-reactive protein activates monocytes and causes further production of cytokines. Other changes that occur in response to injury include somnolence, negative nitrogen balance, and fever. WOUND HEALING When tissue is damaged, platelets adhere to exposed matrix via integrins that bind to collagen and laminin (Figure 3–13). Blood coagulation produces thrombin, which promotes platelet aggregation and granule release. The platelet granules generate an inflammatory response. White blood cells are attracted by selectins and bind to integrins on endothelial cells, leading to their extravasation through the blood vessel walls. Cytokines re-leased by the white blood cells and platelets up-regulate inte-grins on macrophages, which migrate to the area of injury, and on fibroblasts and epithelial cells, which mediate wound healing FIGURE 3–11 Action of cyclosporine (CsA) and tacrolimus (TCL) in lymphocytes. BP, binding protein; CAM, calmodulin. T cell receptor Ca2+ CAM Calcineurin TCLBP CsABP P NF-AT IL-2 gene activation Nucleus FIGURE 3–12 Sites of congenital blockade of B and T lymphocyte maturation in various immunodeficiency states. SCID, severe combined immune deficiency. (Modified from Rosen FS, Cooper MD, Wedgwood RJP: The primary immunodeficiencies. N Engl J Med 1995;333:431.) Pluripotent stem cell Lymphoid progenitor Autosomal recessive SCID BONE MARROW pre-B cell THYMUS Immature T cell X-linked SCID X-linked agamma-globulinemia B cell Hyper-IgM syndrome IgM IgG IgA IgE MHC class I deficiency MHC class II deficiency CD8 cell CD4 cell CHAPTER 3 Immunity, Infection, & Inflammation 77 and scar formation. Plasmin aids healing by removing excess fi-brin. This aids the migration of keratinocytes into the wound to restore the epithelium under the scab. Collagen proliferates, producing the scar. Wounds gain 20% of their ultimate strength in 3 weeks and later gain more strength, but they never reach more than about 70% of the strength of normal skin. CHAPTER SUMMARY ■Immune and inflammatory responses are mediated by several different cell types—granulocytes, lymphocytes, monocytes, mast cells, tissue macrophages, and antigen presenting cells— that arise predominantly from the bone marrow and may circu-late or reside in connective tissues. ■Granulocytes mount phagocytic responses that engulf and de-stroy bacteria. These are accompanied by the release of reactive oxygen species and other mediators into adjacent tissues that may cause tissue injury. ■Mast cells and basophils underpin allergic reactions to substances that would be treated as innocuous by nonallergic individuals. ■A variety of soluble mediators orchestrate the development of immunologic effector cells and their subsequent immune and inflammatory reactions. ■Innate immunity represents an evolutionarily conserved, prim-itive response to stereotypical microbial components. ■Acquired immunity is slower to develop than innate immunity, but long-lasting and more effective. ■Genetic rearrangements endow B and T lymphocytes with a vast array of receptors capable of recognizing billions of foreign antigens. ■Self-reactive lymphocytes are normally deleted; a failure of this process leads to autoimmune disease. Disease can also result from abnormal function or development of granulocytes and lymphocytes. In these latter cases, deficient immune responses to microbial threats usually result. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. In normal human blood A) the eosinophil is the most common type of white blood cell. B) there are more lymphocytes than neutrophils. C) the iron is mostly in hemoglobin. D) there are more white cells than red cells. E) there are more platelets than red cells. 2. Lymphocytes A) all originate from the bone marrow after birth. B) are unaffected by hormones. C) convert to monocytes in response to antigens. D) interact with eosinophils to produce platelets. E) are part of the body’s defense against cancer. 3. The ability of the blood to phagocytose pathogens and mount a respiratory burst is increased by A) interleukin-2 (IL-2). B) granulocyte colony-stimulating factor (G-CSF). C) erythropoietin. D) interleukin-4 (IL-4). E) interleukin-5 (IL-5). 4. Cells responsible for innate immunity are activated most com-monly by A) glucocorticoids. B) pollen. C) carbohydrate sequences in bacterial cell walls. D) eosinophils. E) cytoplasmic proteins of bacteria. FIGURE 3–13 Cutaneous wound 3 days after injury, showing the multiple cytokines and growth factors affecting the repair process. VEGF, vascular endothelial growth factor. For other abbrevia-tions, see Appendix. Note the epidermis growing down under the fi-brin clot, restoring skin continuity. (Modified from Singer AJ, Clark RAF: Cutaneous wound healing. N Engl J Med 1999;341:738.) FIGURE 3–14 Time course of changes in some major acute phase proteins. C3, C3 component of complement. (Modified and reproduced with permission from Gitlin JD, Colten HR: Molecular biology of acute phase plasma proteins. In Pick F, et al [editors]: Lymphokines, vol 14, pages 123–153. Academic Press, 1987.) Platelet plug TGF-β1 TGF-β1 TGF-α FGF VEGF PDGF BB PDGF AB Macrophage Fibrin clot Neutrophil Blood vessel Neutrophil VEGF FGF-2 Fibroblast FGF-2 IGF 30,100 30,000 700 600 500 400 300 200 100 0 0 7 14 21 Time after inflammatory stimulus (d) Change in plasma concentration (%) C-reactive protein Serum amyloid A Haptoglobin Fibrinogen Transferrin Albumin C3 78 SECTION I Cellular & Molecular Basis for Medical Physiology CHAPTER RESOURCES Delibro G: The Robin Hood of antigen presentation. Science 2004;302:485. Delves PJ, Roitt IM: The immune system. (Two parts.) N Engl J Med 2000;343:37,108. Dhainaut J-K, Thijs LG, Park G (editors): Septic Shock. WB Saunders, 2000. Ganz T: Defensins and host defense. Science 1999;286:420. Samstein B, Emond JC: Liver transplant from living related donors. Annu Rev Med 2001;52:147. Singer AJ, Clark RAF: Cutaneous wound healing. N Engl J Med 1999;341:738 Tedder TF, et al: The selectins: Vascular adhesion molecules. FASEB J 1995;9:866. Tilney NL: Transplant: From Myth to Reality. Yale University Press, 2003. Walport MJ: Complement. (Two parts) N Engl J Med 2001;344:1058, 1140. 79 C H A P T E R SECTION II PHYSIOLOGY OF NERVE & MUSCLE CELLS 4 Excitable Tissue: Nerve O B J E C T I V E S After studying this chapter, you should be able to: ■Name the parts of a neuron and their functions. ■Name the various types of glia and their functions. ■Describe the chemical nature of myelin, and summarize the differences in the ways in which unmyelinated and myelinated neurons conduct impulses. ■Define orthograde and retrograde axonal transport and the molecular motors in-volved in each. ■Describe the changes in ionic channels that underlie electrotonic potentials, the action potential, and repolarization. ■List the various nerve fiber types found in the mammalian nervous system. ■Describe the function of neurotrophins. INTRODUCTION The human central nervous system (CNS) contains about 1011 (100 billion) neurons. It also contains 10–50 times this number of glial cells. The CNS is a complex organ; it has been calculated that 40% of the human genes participate, at least to a degree, in its formation. The neurons, the basic building blocks of the nervous system, have evolved from primitive neuroeffector cells that respond to various stimuli by con-tracting. In more complex animals, contraction has become the specialized function of muscle cells, whereas integration and transmission of nerve impulses have become the special-ized functions of neurons. This chapter describes the cellular components of the CNS and the excitability of neurons, which involves the genesis of electrical signals that enable neurons to integrate and transmit impulses (action potentials, receptor potentials, and synaptic potentials). 80 SECTION II Physiology of Nerve & Muscle Cells CELLULAR ELEMENTS IN THE CNS GLIAL CELLS For many years following their discovery, glial cells (or glia) were viewed as CNS connective tissue. In fact, the word glia is Greek for glue. However, today theses cells are recognized for their role in communication within the CNS in partnership with neurons. Unlike neurons, glial cells continue to undergo cell division in adulthood and their ability to proliferate is par-ticularly noticeable after brain injury (eg, stroke). There are two major types of glial cells in the vertebrate ner-vous system: microglia and macroglia. Microglia are scavenger cells that resemble tissue macrophages and remove debris resulting from injury, infection, and disease (eg, multiple scle-rosis, AIDS-related dementia, Parkinson disease, and Alzhei-mer disease). Microglia arise from macrophages outside of the nervous system and are physiologically and embryologically unrelated to other neural cell types. There are three types of macroglia: oligodendrocytes, Schwann cells, and astrocytes (Figure 4–1). Oligodendrocytes and Schwann cells are involved in myelin formation around axons in the CNS and peripheral nervous system, respectively. Astrocytes, which are found throughout the brain, are of two subtypes. Fibrous astrocytes, which contain many intermedi-ate filaments, are found primarily in white matter. Protoplas-mic astrocytes are found in gray matter and have a granular cytoplasm. Both types send processes to blood vessels, where they induce capillaries to form the tight junctions making up the blood–brain barrier. They also send processes that envelop synapses and the surface of nerve cells. Protoplasmic astrocytes have a membrane potential that varies with the external K+ concentration but do not generate propagated potentials. They produce substances that are tropic to neurons, and they help maintain the appropriate concentration of ions and neurotransmitters by taking up K+ and the neurotransmit-ters glutamate and γ-aminobutyrate (GABA). NEURONS Neurons in the mammalian central nervous system come in many different shapes and sizes. Most have the same parts as the typical spinal motor neuron illustrated in Figure 4–2. The cell body (soma) contains the nucleus and is the metabolic center of the neuron. Neurons have several processes called dendrites that extend outward from the cell body and arborize extensively. Particularly in the cerebral and cerebellar cortex, the dendrites have small knobby projections called dendritic spines. A typical neuron also has a long fibrous axon that orig-inates from a somewhat thickened area of the cell body, the axon hillock. The first portion of the axon is called the initial segment. The axon divides into presynaptic terminals, each ending in a number of synaptic knobs which are also called terminal buttons or boutons. They contain granules or vesi-cles in which the synaptic transmitters secreted by the nerves are stored. Based on the number of processes that emanate from their cell body, neurons can be classified as unipolar, bi-polar, and multipolar (Figure 4–3). FIGURE 4–1 The principal types of glial cells in the nervous system. A) Oligodendrocytes are small with relatively few processes. Those in the white matter provide myelin, and those in the gray matter support neurons. B) Schwann cells provide myelin to the peripheral nervous sys-tem. Each cell forms a segment of myelin sheath about 1 mm long; the sheath assumes its form as the inner tongue of the Schwann cell turns around the axon several times, wrapping in concentric layers. Intervals between segments of myelin are the nodes of Ranvier. C) Astrocytes are the most common glia in the CNS and are characterized by their starlike shape. They contact both capillaries and neurons and are thought to have a nutritive function. They are also involved in forming the blood–brain barrier. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Capillary End-foot Fibrous astrocyte Neuron End-foot Nodes of Ranvier Layers of myelin Inner tongue Nucleus Schwann cell Axon Neuron Axons A Oligodendrocyte B Schwann cell C Astrocyte Oligodendrocyte in white matter Perineural oligodendrocytes CHAPTER 4 Excitable Tissue: Nerve 81 FIGURE 4–2 Motor neuron with a myelinated axon. A motor neuron is comprised of a cell body (soma) with a nucleus, several processes called dendrites, and a long fibrous axon that originates from the axon hillock. The first portion of the axon is called the initial segment. A myelin sheath forms from Schwann cells and surrounds the axon except at its ending and at the nodes of Ranvier. Terminal buttons (boutons) are located at the terminal endings. FIGURE 4–3 Some of the types of neurons in the mammalian nervous system. A) Unipolar neurons have one process, with different seg-ments serving as receptive surfaces and releasing terminals. B) Bipolar neurons have two specialized processes: a dendrite that carries information to the cell and an axon that transmits information from the cell. C) Some sensory neurons are in a subclass of bipolar cells called pseudo-unipolar cells. As the cell develops, a single process splits into two, both of which function as axons—one going to skin or muscle and another to the spinal cord. D) Multipolar cells have one axon and many dendrites. Examples include motor neurons, hippocampal pyramidal cells with dendrites in the apex and base, and cerebellar Purkinje cells with an extensive dendritic tree in a single plane. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Cell body (soma) Initial segment of axon Node of Ranvier Schwann cell Terminal buttons Dendrites Axon hillock Nucleus A Unipolar cell C Pseudo-unipolar cell B Bipolar cell Dendrite Axon Cell body Invertebrate neuron Ganglion cell of dorsal root Bipolar cell of retina Central axon Single bifurcated process Cell body Peripheral axon to skin and muscle Dendrites Cell body Axon Dendrites Cell body Axon D Three types of multipolar cells Basal dendrite Cell body Axon Motor neuron of spinal cord Pyramidal cell of hippocampus Purkinje cell of cerebellum Dendrites Cell body Axon Apical dendrite Axon terminals 82 SECTION II Physiology of Nerve & Muscle Cells The conventional terminology used for the parts of a neuron works well enough for spinal motor neurons and interneurons, but there are problems in terms of “dendrites” and “axons” when it is applied to other types of neurons found in the ner-vous system. From a functional point of view, neurons generally have four important zones: (1) a receptor, or dendritic zone, where multiple local potential changes generated by synaptic connections are integrated; (2) a site where propagated action potentials are generated (the initial segment in spinal motor neurons, the initial node of Ranvier in cutaneous sensory neu-rons); (3) an axonal process that transmits propagated impulses to the nerve endings; and (4) the nerve endings, where action potentials cause the release of synaptic transmitters. The cell body is often located at the dendritic zone end of the axon, but it can be within the axon (eg, auditory neurons) or attached to the side of the axon (eg, cutaneous neurons). Its location makes no difference as far as the receptor function of the dendritic zone and the transmission function of the axon are concerned. The axons of many neurons are myelinated, that is, they acquire a sheath of myelin, a protein–lipid complex that is wrapped around the axon (Figure 4–2). In the peripheral ner-vous system, myelin forms when a Schwann cell wraps its membrane around an axon up to 100 times (Figure 4–1). The myelin is then compacted when the extracellular portions of a membrane protein called protein zero (P0) lock to the extracel-lular portions of P0 in the apposing membrane. Various muta-tions in the gene for P0 cause peripheral neuropathies; 29 different mutations have been described that cause symptoms ranging from mild to severe. The myelin sheath envelops the axon except at its ending and at the nodes of Ranvier, periodic 1-μm constrictions that are about 1 mm apart (Figure 4–2). The insulating function of myelin is discussed later in this chapter. Not all neurons are myelinated; some are unmyeli-nated, that is, simply surrounded by Schwann cells without the wrapping of the Schwann cell membrane that produces myelin around the axon. In the CNS of mammals, most neurons are myelinated, but the cells that form the myelin are oligodendrocytes rather than Schwann cells (Figure 4–1). Unlike the Schwann cell, which forms the myelin between two nodes of Ranvier on a single neuron, oligodendrocytes emit multiple processes that form myelin on many neighboring axons. In multiple sclero-sis, a crippling autoimmune disease, patchy destruction of myelin occurs in the CNS (see Clinical Box 4–1). The loss of myelin is associated with delayed or blocked conduction in the demyelinated axons. AXONAL TRANSPORT Neurons are secretory cells, but they differ from other secretory cells in that the secretory zone is generally at the end of the axon, far removed from the cell body. The apparatus for protein syn-thesis is located for the most part in the cell body, with transport of proteins and polypeptides to the axonal ending by axoplas-mic flow. Thus, the cell body maintains the functional and an-atomic integrity of the axon; if the axon is cut, the part distal to the cut degenerates (wallerian degeneration). Orthograde transport occurs along microtubules that run along the length of the axon and requires two molecular motors, dynein and ki-nesin (Figure 4–4). Orthograde transport moves from the cell body toward the axon terminals. It has both fast and slow com-ponents; fast axonal transport occurs at about 400 mm/day, and slow axonal transport occurs at 0.5 to 10 mm/day. Retro-grade transport, which is in the opposite direction (from the CLINICAL BOX 4–1 Demyelinating Diseases Normal conduction of action potentials relies on the insulat-ing properties of myelin. Thus, defects in myelin can have major adverse neurological consequences. One example is multiple sclerosis (MS), an autoimmune disease that af-fects over 3 million people worldwide, usually striking be-tween the ages of 20 and 50 and affecting women about twice as often as men. The cause of MS appears to include both genetic and environmental factors. It is most common among Caucasians living in countries with temperate cli-mates, including Europe, southern Canada, northern United States, and southeastern Australia. Environmental triggers include early exposure to viruses such as Epstein-Barr virus and those that cause measles, herpes, chicken pox, or influ-enza. In MS, antibodies and white blood cells in the immune system attack myelin, causing inflammation and injury to the sheath and eventually the nerves that it surrounds. Loss of myelin leads to leakage of K+ through voltage-gated channels, hyperpolarization, and failure to conduct action potentials. Typical physiological deficits range from muscle weakness, fatigue, diminished coordination, slurred speech, blurred or hazy vision, bladder dysfunction, and sensory dis-turbances. Symptoms are often exasperated by increased body temperature or ambient temperature. Progression of the disease is quite variable. In the most common form, transient episodes appear suddenly, last a few weeks or months, and then gradually disappear. Subsequent epi-sodes can appear years later, and eventually full recovery does not occur. Others have a progressive form of the dis-ease in which there are no periods of remission. Diagnosing MS is very difficult and generally is delayed until multiple episodes occur with deficits separated in time and space. Nerve conduction tests can detect slowed conduction in motor and sensory pathways. Cerebral spinal fluid analysis can detect the presence of oligoclonal bands indicative of an abnormal immune reaction against myelin. The most de-finitive assessment is magnetic resonance imaging (MRI) to visualize multiple scarred (sclerotic) areas in the brain. Although there is no cure for MS, some drugs (eg, β-inter-feron) that suppress the immune response reduce the se-verity and slow the progression of the disease. CHAPTER 4 Excitable Tissue: Nerve 83 nerve ending to the cell body), occurs along microtubules at about 200 mm/day. Synaptic vesicles recycle in the membrane, but some used vesicles are carried back to the cell body and de-posited in lysosomes. Some materials taken up at the ending by endocytosis, including nerve growth factor (NGF) and various viruses, are also transported back to the cell body. A potentially important exception to these principles seems to occur in some dendrites. In them, single strands of mRNA transported from the cell body make contact with appropriate ribosomes, and protein synthesis appears to create local protein domains. EXCITATION & CONDUCTION Nerve cells have a low threshold for excitation. The stimulus may be electrical, chemical, or mechanical. Two types of phys-icochemical disturbances are produced: local, nonpropagated potentials called, depending on their location, synaptic, gen-erator, or electrotonic potentials; and propagated potentials, the action potentials (or nerve impulses). These are the only electrical responses of neurons and other excitable tissues, and they are the main language of the nervous system. They are due to changes in the conduction of ions across the cell mem-brane that are produced by alterations in ion channels. The electrical events in neurons are rapid, being measured in mil-liseconds (ms); and the potential changes are small, being measured in millivolts (mV). The impulse is normally transmitted (conducted) along the axon to its termination. Nerves are not “telephone wires” that transmit impulses passively; conduction of nerve impulses, although rapid, is much slower than that of electricity. Nerve tissue is in fact a relatively poor passive conductor, and it would take a potential of many volts to produce a signal of a fraction of a volt at the other end of a meter-long axon in the absence of active processes in the nerve. Conduction is an active, self-propagating process, and the impulse moves along the nerve at a constant amplitude and velocity. The process is often compared to what happens when a match is applied to one end of a trail of gunpowder; by igniting the powder parti-cles immediately in front of it, the flame moves steadily down the trail to its end as it is extinguished in its progression. Mammalian neurons are relatively small, but giant unmyeli-nated nerve cells exist in a number of invertebrate species. Such cells are found, for example, in crabs (Carcinus), cuttle-fish (Sepia), and squid (Loligo). The fundamental properties of neurons were first determined in these species and then found to be similar in mammals. The neck region of the mus-cular mantle of the squid contains single axons up to 1 mm in diameter. The fundamental properties of these long axons are similar to those of mammalian axons. RESTING MEMBRANE POTENTIAL When two electrodes are connected through a suitable ampli-fier and placed on the surface of a single axon, no potential dif-ference is observed. However, if one electrode is inserted into the interior of the cell, a constant potential difference is FIGURE 4–4 Axonal transport along microtubules by dynein and kinesin. Fast and slow axonal orthograde transport occurs along mi-crotubules that run along the length of the axon from the cell body to the terminal. Retrograde transport occurs from the terminal to the cell body. (From Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology. McGraw-Hill, 2008.) 84 SECTION II Physiology of Nerve & Muscle Cells observed, with the inside negative relative to the outside of the cell at rest. A membrane potential results from separation of positive and negative charges across the cell membrane (Fig-ure 4–5). In neurons, the resting membrane potential is usu-ally about –70 mV, which is close to the equilibrium potential for K+ (Figure 4–6). In order for a potential difference to be present across a membrane lipid bilayer, two conditions must be met. First, there must be an unequal distribution of ions of one or more species across the membrane (ie, a concentration gradient). Two, the membrane must be permeable to one or more of these ion species. The permeability is provided by the exis-tence of channels or pores in the bilayer; these channels are usually permeable to a single species of ions. The resting membrane potential represents an equilibrium situation at which the driving force for the membrane-permeant ions down their concentration gradients across the membrane is equal and opposite to the driving force for these ions down their electrical gradients. FIGURE 4–5 This membrane potential results from separation of positive and negative charges across the cell membrane. The excess of positive charges (red circles) outside the cell and negative charges (blue circles) inside the cell at rest represents a small fraction of the total number of ions present. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.). – – – – – – – – – – – – – + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + – – – – – – – – – – – – – – – – – – – – – – Equal +,– Equal +,– Extracellular side Cytoplasmic side FIGURE 4–6 The changes in (a) membrane potential (mV) and (b) relative membrane permeability (P) to Na+ and K+ during an action po-tential. (From Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology. McGraw-Hill, 2008.) PK PNa Na+ K+ Membrane potential (mV) Time (ms) Relative membrane permeability (a) (b) +30 0 –70 600 300 50 1 0 1 2 3 4 3 2 4 5 6 7 1 K+ Gated Na+ channel Gated K+ channel CHAPTER 4 Excitable Tissue: Nerve 85 In neurons, the concentration of K+ is much higher inside than outside the cell, while the reverse is the case for Na+. This concentration difference is established by the Na+-K+ ATPase. The outward K+ concentration gradient results in passive movement of K+ out of the cell when K+-selective channels are open. Similarly, the inward Na+ concentration gradient results in passive movement of Na+ into the cell when Na+-selective channels are open. Because there are more open K+ channels than Na+ channels at rest, the membrane permeability to K+ is greater. Consequently, the intracellular and extracellular K+ concentrations are the prime determinants of the resting mem-brane potential, which is therefore close to the equilibrium potential for K+. Steady ion leaks cannot continue forever with-out eventually dissipating the ion gradients. This is prevented by the Na+-K+ ATPase, which actively moves Na+ and K+ against their electrochemical gradient. IONIC FLUXES DURING THE ACTION POTENTIAL The cell membranes of nerves, like those of other cells, contain many different types of ion channels. Some of these are volt-age-gated and others are ligand-gated. It is the behavior of these channels, and particularly Na+ and K+ channels, which explains the electrical events in nerves. The changes in membrane conductance of Na+ and K+ that occur during the action potentials are shown in Figure 4–6. The conductance of an ion is the reciprocal of its electrical resistance in the membrane and is a measure of the mem-brane permeability to that ion. In response to a depolarizing stimulus, some of the voltage-gated Na+ channels become active, and when the threshold potential is reached, the volt-age-gated Na+ channels overwhelm the K+ and other chan-nels and an action potential results (a positive feedback loop). The membrane potential moves toward the equilib-rium potential for Na+ (+60 mV) but does not reach it during the action potential, primarily because the increase in Na+ conductance is short-lived. The Na+ channels rapidly enter a closed state called the inactivated state and remain in this state for a few milliseconds before returning to the resting state, when they again can be activated. In addition, the direc-tion of the electrical gradient for Na+ is reversed during the overshoot because the membrane potential is reversed, and this limits Na+ influx. A third factor producing repolariza-tion is the opening of voltage-gated K+ channels. This open-ing is slower and more prolonged than the opening of the Na+ channels, and consequently, much of the increase in K+ con-ductance comes after the increase in Na+ conductance. The net movement of positive charge out of the cell due to K+ efflux at this time helps complete the process of repolariza-tion. The slow return of the K+ channels to the closed state also explains the after-hyperpolarization, followed by a return to the resting membrane potential. Thus, voltage-gated K+ channels bring the action potential to an end and cause closure of their gates through a negative feedback process. Figure 4–7 shows the sequential feedback control in voltage-gated K+ and Na+ channels during the action potential. Decreasing the external Na+ concentration reduces the size of the action potential but has little effect on the resting mem-brane potential. The lack of much effect on the resting mem-brane potential would be predicted, since the permeability of the membrane to Na+ at rest is relatively low. Conversely, increasing the external K+ concentration decreases the resting membrane potential. Although Na+ enters the nerve cell and K+ leaves it during the action potential, the number of ions involved is minute rela-tive to the total numbers present. The fact that the nerve gains Na+ and loses K+ during activity has been demonstrated exper-imentally, but significant differences in ion concentrations can be measured only after prolonged, repeated stimulation. Other ions, notably Ca2+, can affect the membrane potential through both channel movement and membrane interactions. A decrease in extracellular Ca2+ concentration increases the excitability of nerve and muscle cells by decreasing the amount of depolarization necessary to initiate the changes in the Na+ and K+ conductance that produce the action potential. Con-versely, an increase in extracellular Ca2+ concentration can sta-bilize the membrane by decreasing excitability. DISTRIBUTION OF ION CHANNELS IN MYELINATED NEURONS The spatial distribution of ion channels along the axon plays a key role in the initiation and regulation of the action potential. Voltage-gated Na+ channels are highly concentrated in the nodes of Ranvier and the initial segment in myelinated neu-rons. The initial segment and, in sensory neurons, the first node of Ranvier are the sites where impulses are normally gen-erated, and the other nodes of Ranvier are the sites to which the impulses jump during saltatory conduction. The number of Na+ channels per square micrometer of membrane in my-elinated mammalian neurons has been estimated to be 50–75 in the cell body, 350–500 in the initial segment, less than 25 on the surface of the myelin, 2000–12,000 at the nodes of Ranvier, and 20–75 at the axon terminals. Along the axons of unmyeli-nated neurons, the number is about 110. In many myelinated neurons, the Na+ channels are flanked by K+ channels that are involved in repolarization. “ALL-OR-NONE” LAW It is possible to determine the minimal intensity of stimulating current (threshold intensity) that, acting for a given duration, will just produce an action potential. The threshold intensity varies with the duration; with weak stimuli it is long, and with strong stimuli it is short. The relation between the strength and the duration of a threshold stimulus is called the strength–duration curve. Slowly rising currents fail to fire the nerve because the nerve adapts to the applied stimulus, a pro-cess called adaptation. 86 SECTION II Physiology of Nerve & Muscle Cells Once threshold intensity is reached, a full-fledged action potential is produced. Further increases in the intensity of a stimulus produce no increment or other change in the action potential as long as the other experimental conditions remain constant. The action potential fails to occur if the stimulus is subthreshold in magnitude, and it occurs with constant amplitude and form regardless of the strength of the stimulus if the stimulus is at or above threshold intensity. The action potential is therefore “all or none” in character and is said to obey the all-or-none law. ELECTROTONIC POTENTIALS, LOCAL RESPONSE, & FIRING LEVEL Although subthreshold stimuli do not produce an action po-tential, they do have an effect on the membrane potential. This can be demonstrated by placing recording electrodes within a few millimeters of a stimulating electrode and ap-plying subthreshold stimuli of fixed duration. Application of such currents leads to a localized depolarizing potential change that rises sharply and decays exponentially with time. The magnitude of this response drops off rapidly as the distance between the stimulating and recording elec-trodes is increased. Conversely, an anodal current produces a hyperpolarizing potential change of similar duration. These potential changes are called electrotonic potentials. As the strength of the current is increased, the response is greater due to the increasing addition of a local response of the membrane (Figure 4–8). Finally, at 7–15 mV of depolar-ization (potential of –55 mV), the firing level is reached and an action potential occurs. CHANGES IN EXCITABILITY DURING ELECTROTONIC POTENTIALS & THE ACTION POTENTIAL During the action potential, as well as during electrotonic po-tentials and the local response, the threshold of the neuron to stimulation changes. Hyperpolarizing responses elevate the threshold, and depolarizing potentials lower it as they move FIGURE 4–7 Feedback control in voltage-gated ion channels in the membrane. (a) Na+ channels exert positive feedback. (b) K+ chan-nels exert negative feedback. (From Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology. McGraw-Hill, 2008.) Depolarization of membrane potential Increased flow of Na+ into the cell Opening of voltage-gated Na+ channels Depolarizing stimulus Inactivation of Na+ channels Increased PNa Start Stop Positive feedback Repolarization of membrane potential Increased flow of K+ out of the cell Opening of voltage-gated K+ channels Depolarization of membrane by Na+ influx Increased PK Start Negative feedback (a) (b) + CHAPTER 4 Excitable Tissue: Nerve 87 the membrane potential closer to the firing level. During the local response, the threshold is lowered, but during the rising and much of the falling phases of the spike potential, the neu-ron is refractory to stimulation. This refractory period is di-vided into an absolute refractory period, corresponding to the period from the time the firing level is reached until repo-larization is about one-third complete, and a relative refrac-tory period, lasting from this point to the start of after-depolarization. During the absolute refractory period, no stimulus, no matter how strong, will excite the nerve, but dur-ing the relative refractory period, stronger than normal stim-uli can cause excitation. During after-depolarization, the threshold is again decreased, and during after-hyperpolariza-tion, it is increased. These changes in threshold are correlated with the phases of the action potential in Figure 4–9. ELECTROGENESIS OF THE ACTION POTENTIAL The nerve cell membrane is polarized at rest, with positive charges lined up along the outside of the membrane and neg-ative charges along the inside. During the action potential, this polarity is abolished and for a brief period is actually reversed (Figure 4–10). Positive charges from the membrane ahead of and behind the action potential flow into the area of negativity represented by the action potential (“current sink”). By draw-ing off positive charges, this flow decreases the polarity of the membrane ahead of the action potential. Such electrotonic de-polarization initiates a local response, and when the firing lev-el is reached, a propagated response occurs that in turn electrotonically depolarizes the membrane in front of it. SALTATORY CONDUCTION Conduction in myelinated axons depends on a similar pattern of circular current flow. However, myelin is an effective insu-lator, and current flow through it is negligible. Instead, depo-FIGURE 4–8 Electrotonic potentials and local response. The changes in the membrane potential of a neuron following application of stimuli of 0.2, 0.4, 0.6, 0.8, and 1.0 times threshold intensity are shown superimposed on the same time scale. The responses below the horizontal line are those recorded near the anode, and the re-sponses above the line are those recorded near the cathode. The stim-ulus of threshold intensity was repeated twice. Once it caused a propagated action potential (top line), and once it did not. −55 −70 −85 0.5 1.0 1.5 ms Resting membrane potential Propagated action potential Firing level Local response Membrane potential (mV) FIGURE 4–9 Relative changes in excitability of a nerve cell membrane during the passage of an impulse. Note that excitability is the reciprocal of threshold. (Modified from Morgan CT: Physiological Psychology. McGraw-Hill, 1943.) FIGURE 4–10 Local current flow (movement of positive charges) around an impulse in an axon. Top: Unmyelinated axon. Bottom: Myelinated axon. Positive charges from the membrane ahead of and behind the action potential flow into the area of negativ-ity represented by the action potential (“current sink”). In myelinated axons, depolarization jumps from one node of Ranvier to the next (sa-lutatory conduction). Spike potential After-depolarization After-hyperpolarization Local response Period of latent addition Supernormal period Refractory period Time Subnormal period Excitability Potential change Myelin Axon ECF _ + _ + Direction of propagation Active node Inactive node + + _ _ + + + + – – + + + – – – + + – – – – + + + + – – + + + – – – + + – – – – Axon ECF 88 SECTION II Physiology of Nerve & Muscle Cells larization in myelinated axons jumps from one node of Ranvier to the next, with the current sink at the active node serving to electrotonically depolarize the node ahead of the ac-tion potential to the firing level (Figure 4–10). This jumping of depolarization from node to node is called saltatory conduc-tion. It is a rapid process that allows myelinated axons to con-duct up to 50 times faster than the fastest unmyelinated fibers. ORTHODROMIC & ANTIDROMIC CONDUCTION An axon can conduct in either direction. When an action po-tential is initiated in the middle of it, two impulses traveling in opposite directions are set up by electrotonic depolarization on either side of the initial current sink. In the natural situa-tion, impulses pass in one direction only, ie, from synaptic junctions or receptors along axons to their termination. Such conduction is called orthodromic. Conduction in the oppo-site direction is called antidromic. Because synapses, unlike axons, permit conduction in one direction only, an antidrom-ic impulse will fail to pass the first synapse they encounter and die out at that point. BIPHASIC ACTION POTENTIALS The descriptions of the resting membrane potential and ac-tion potential outlined above are based on recording with two electrodes, one in the extracellular space and the other inside it. If both recording electrodes are placed on the sur-face of the axon, there is no potential difference between them at rest. When the nerve is stimulated and an impulse is conducted past the two electrodes, a characteristic sequence of potential changes results. As the wave of depolarization reaches the electrode nearest the stimulator, this electrode be-comes negative relative to the other electrode (Figure 4–11). When the impulse passes to the portion of the nerve between the two electrodes, the potential returns to zero, and then, as it passes the second electrode, the first electrode becomes positive relative to the second. It is conventional to connect the leads in such a way that when the first electrode becomes negative relative to the second, an upward deflection is re-corded. Therefore, the record shows an upward deflection followed by an isoelectric interval and then a downward de-flection. This sequence is called a biphasic action potential (Figure 4–11). PROPERTIES OF MIXED NERVES Peripheral nerves in mammals are made up of many axons bound together in a fibrous envelope called the epineurium. Potential changes recorded extracellularly from such nerves therefore represent an algebraic summation of the all-or-none action potentials of many axons. The thresholds of the indi-vidual axons in the nerve and their distance from the stimulat-ing electrodes vary. With subthreshold stimuli, none of the axons are stimulated and no response occurs. When the stim-uli are of threshold intensity, axons with low thresholds fire and a small potential change is observed. As the intensity of the stimulating current is increased, the axons with higher thresholds are also discharged. The electrical response in-creases proportionately until the stimulus is strong enough to excite all of the axons in the nerve. The stimulus that produces excitation of all the axons is the maximal stimulus, and appli-cation of greater, supramaximal stimuli produces no further increase in the size of the observed potential. NERVE FIBER TYPES & FUNCTION After a stimulus is applied to a nerve, there is a latent period before the start of the action potential. This interval corre-sponds to the time it takes the impulse to travel along the axon from the site of stimulation to the recording electrodes. Its du-ration is proportionate to the distance between the stimulating and recording electrodes and inversely proportionate to the speed of conduction. If the duration of the latent period and the distance between the stimulating and recording electrodes are known, axonal conduction velocity can be calculated. Erlanger and Gasser divided mammalian nerve fibers into A, B, and C groups, further subdividing the A group into α, β, γ, and δ fibers. In Table 4–1, the various fiber types are listed FIGURE 4–11 Biphasic action potential. Both recording elec-trodes are on the outside of the nerve membrane. It is conventional to connect the leads in such a way that when the first electrode becomes negative relative to the second, an upward deflection is recorded. Therefore, the record shows an upward deflection followed by an iso-electric interval and then a downward deflection. + _ + _ − + + _ + _ + _ + _ + _ + _ + _ + _ + _ + _ + _ + _ − + + _ + _ + _ + _ + _ + _ + _ + _ + _ + _ + _ + _ + _ + _ + _ + _ − + + _ + _ + _ + _ + − + _ + _ + _ − + + _ + _ Nerve Time mV CHAPTER 4 Excitable Tissue: Nerve 89 with their diameters, electrical characteristics, and functions. By comparing the neurologic deficits produced by careful dorsal root section and other nerve-cutting experiments with the histologic changes in the nerves, the functions and histo-logic characteristics of each of the families of axons responsi-ble for the various peaks of the compound action potential have been established. In general, the greater the diameter of a given nerve fiber, the greater its speed of conduction. The large axons are concerned primarily with proprioceptive sen-sation, somatic motor function, conscious touch, and pres-sure, while the smaller axons subserve pain and temperature sensations and autonomic function. The dorsal root C fibers conduct some impulses generated by touch and other cutane-ous receptors in addition to impulses generated by pain and temperature receptors. Further research has shown that not all the classically described lettered components are homogeneous, and a numer-ical system (Ia, Ib, II, III, IV) has been used by some physiolo-gists to classify sensory fibers. Unfortunately, this has led to confusion. A comparison of the number system and the letter system is shown in Table 4–2. In addition to variations in speed of conduction and fiber diameter, the various classes of fibers in peripheral nerves dif-fer in their sensitivity to hypoxia and anesthetics (Table 4–3). This fact has clinical as well as physiologic significance. Local anesthetics depress transmission in the group C fibers before they affect group A touch fibers. Conversely, pressure on a nerve can cause loss of conduction in large-diameter motor, touch, and pressure fibers while pain sensation remains rela-tively intact. Patterns of this type are sometimes seen in indi-viduals who sleep with their arms under their heads for long periods, causing compression of the nerves in the arms. Because of the association of deep sleep with alcoholic intoxi-cation, the syndrome is most common on weekends and has acquired the interesting name Saturday night or Sunday morning paralysis. NEUROTROPHINS TROPHIC SUPPORT OF NEURONS A number of proteins necessary for survival and growth of neu-rons have been isolated and studied. Some of these neurotro-phins are products of the muscles or other structures that the neurons innervate, but others are produced by astrocytes. These proteins bind to receptors at the endings of a neuron. They are internalized and then transported by retrograde transport to the neuronal cell body, where they foster the production of proteins associated with neuronal development, growth, and survival. Other neurotrophins are produced in neurons and transported in an anterograde fashion to the nerve ending, where they main-tain the integrity of the postsynaptic neuron. TABLE 4–1 Nerve fiber types in mammalian nerve.a Fiber Type Function Fiber Diameter (μm) Conduction Velocity (m/s) Spike Duration (ms) Absolute Refractory Period (ms) A α Proprioception; somatic motor 12–20 70–120 β Touch, pressure 5–12 30–70 0.4–0.5 0.4–1 γ Motor to muscle spindles 3–6 15–30 δ Pain, cold, touch 2–5 12–30 B Preganglionic autonomic <3 3–15 1.2 1.2 C Dorsal root Pain, temperature, some mechano-reception 0.4–1.2 0.5–2 2 2 Sympathetic Postganglionic sympathetic 0.3–1.3 0.7–2.3 2 2 aA and B fibers are myelinated; C fibers are unmyelinated. TABLE 4–2 Numerical classification sometimes used for sensory neurons. Number Origin Fiber Type Ia Muscle spindle, annulo-spiral ending A α Ib Golgi tendon organ A α II Muscle spindle, flower-spray ending; touch, pressure A β III Pain and cold receptors; some touch receptors A δ IV Pain, temperature, and other receptors Dorsal root C 90 SECTION II Physiology of Nerve & Muscle Cells RECEPTORS Four established neurotrophins and their three high-affinity receptors are listed in Table 4–4. Each of these trk receptors dimerizes, and this initiates autophosphorylation in the cyto-plasmic tyrosine kinase domains of the receptors. An addi-tional low-affinity NGF receptor that is a 75-kDa protein is called p75NTR. This receptor binds all four of the listed neu-rotrophins with equal affinity. There is some evidence that it can form a heterodimer with trk A monomer and that the dimer has increased affinity and specificity for NGF. However, it now appears that p75NTR receptors can form homodimers that in the absence of trk receptors cause apoptosis, an effect opposite to the usual growth-promoting and nurturing effects of neurotrophins. ACTIONS The first neurotrophin to be characterized was NGF, a protein growth factor that is necessary for the growth and maintenance of sympathetic neurons and some sensory neurons. It is present in a broad spectrum of animal species, including humans, and is found in many different tissues. In male mice, there is a par-ticularly high concentration in the submandibular salivary glands, and the level is reduced by castration to that seen in fe-males. The factor is made up of two α, two β, and two γ sub-units. The β subunits, each of which has a molecular mass of 13,200 Da, have all the nerve growth-promoting activity, the α subunits have trypsinlike activity, and the γ subunits are serine proteases. The function of the proteases is unknown. The struc-ture of the β unit of NGF resembles that of insulin. NGF is picked up by neurons and is transported in retro-grade fashion from the endings of the neurons to their cell bodies. It is also present in the brain and appears to be responsible for the growth and maintenance of cholinergic neurons in the basal forebrain and striatum. Injection of anti-serum against NGF in newborn animals leads to near total destruction of the sympathetic ganglia; it thus produces an immunosympathectomy. There is evidence that the mainte-nance of neurons by NGF is due to a reduction in apoptosis. Brain-derived neurotrophic factor (BDNF), neurotrophin 3 (NT-3), NT-4/5, and NGF each maintain a different pattern of neurons, although there is some overlap. Disruption of NT-3 by gene knockout causes a marked loss of cutaneous mechan-oreceptors, even in heterozygotes. BDNF acts rapidly and can TABLE 4–3 Relative susceptibility of mammalian A, B, and C nerve fibers to conduction block produced by various agents. Susceptibility to: Most Susceptible Intermediate Least Susceptible Hypoxia B A C Pressure A B C Local anesthetics C B A TABLE 4–4 Neurotrophins. Neurotrophin Receptor Nerve growth factor (NGF) trk A Brain-derived neurotrophic factor (BDNF) trk B Neurotrophin 3 (NT-3) trk C, less on trk A and trk B Neurotrophin 4/5 (NT-4/5) trk B CLINICAL BOX 4–2 Axonal Regeneration Peripheral nerve damage is often reversible. Although the axon will degenerate distal to the damage, connective ele-ments of the so-called distal stump often survive. Axonal sprouting occurs from the proximal stump, growing to-ward the nerve ending. This results from growth-promot-ing factors secreted by Schwann cells that attract axons toward the distal stump. Adhesion molecules of the immu-noglobulin superfamily (eg, NgCAM/L1) promote axon growth along cell membranes and extracellular matrices. Inhibitory molecules in the perineurium assure that the re-generating axons grow in a correct trajectory. Denervated distal stumps are able to upregulate production of neu-rotrophins that promote growth. Once the regenerated axon reaches its target, a new functional connection (eg, neuromuscular junction) is formed. Regeneration allows for considerable, although not full, recovery. For example, fine motor control may be permanently impaired because some motor neurons are guided to an inappropriate motor fiber. Nonetheless, recovery of peripheral nerves from dam-age far surpasses that of central nerve pathways. The proxi-mal stump of a damaged axon in the CNS will form short sprouts, but distant stump recovery is rare, and the dam-aged axons are unlikely to form new synapses. This is be-cause CNS neurons do not have the growth-promoting chemicals needed for regeneration. In fact, CNS myelin is a potent inhibitor of axonal growth. In addition, following CNS injury several events—astrocytic proliferation, acti-vation of microglia, scar formation, inflammation, and invasion of immune cells—provide an inappropriate envi-ronment for regeneration. Thus, treatment of brain and spi-nal cord injuries frequently focuses on rehabilitation rather than reversing the nerve damage. New research is aiming to identify ways to initiate and maintain axonal growth, to direct regenerating axons to reconnect with their target neurons, and to reconstitute original neuronal circuitry. CHAPTER 4 Excitable Tissue: Nerve 91 actually depolarize neurons. BDNF-deficient mice lose peripheral sensory neurons and have severe degenerative changes in their vestibular ganglia and blunted long-term potentiation. OTHER FACTORS AFFECTING NEURONAL GROWTH The regulation of neuronal growth is a complex process. Schwann cells and astrocytes produce ciliary neurotrophic factor (CNTF). This factor promotes the survival of damaged and embryonic spinal cord neurons and may prove to be of value in treating human diseases in which motor neurons de-generate. Glial cell line-derived neurotrophic factor (GD-NF) maintains midbrain dopaminergic neurons in vitro. However, GDNF knockouts have dopaminergic neurons that appear normal, but they have no kidneys and fail to develop an enteric nervous system. Another factor that enhances the growth of neurons is leukemia inhibitory factor (LIF). In ad-dition, neurons as well as other cells respond to insulinlike growth factor I (IGF-I) and the various forms of transform-ing growth factor (TGF), fibroblast growth factor (FGF), and platelet-derived growth factor (PDGF). Clinical Box 4–2 compares the ability to regenerate neurons after central and peripheral nerve injury. CHAPTER SUMMARY ■There are two main types of microglia and macroglia. Microglia are scavenger cells. Macroglia include oligodendrocytes, Schwann cells, and astrocytes. The first two are involved in my-elin formation; astrocytes produce substances that are tropic to neurons, and they help maintain the appropriate concentration of ions and neurotransmitters. ■Neurons are composed of a cell body (soma) which is the meta-bolic center of the neuron, dendrites that extend outward from the cell body and arborize extensively, and a long fibrous axon that originates from a somewhat thickened area of the cell body, the axon hillock. ■The axons of many neurons acquire a sheath of myelin, a pro-tein–lipid complex that is wrapped around the axon. Myelin is an effective insulator, and depolarization in myelinated axons jumps from one node of Ranvier to the next, with the current sink at the active node serving to electrotonically depolarize to the firing level the node ahead of the action potential. ■Orthograde transport occurs along microtubules that run the length of the axon and requires molecular motors, dynein, and kinesin. ■Two types of physicochemical disturbances occur in neurons: local, nonpropagated potentials (synaptic, generator, or electro-tonic potentials) and propagated potentials (action potentials). ■In response to a depolarizing stimulus, voltage-gated Na+ chan-nels become active, and when the threshold potential is reached, an action potential results. The membrane potential moves to-ward the equilibrium potential for Na+. The Na+ channels rap-idly enter a closed state (inactivated state) before returning to the resting state. The direction of the electrical gradient for Na+ is reversed during the overshoot because the membrane poten-tial is reversed, and this limits Na+ influx. Voltage-gated K+ channels open and the net movement of positive charge out of the cell helps complete the process of repolarization. The slow return of the K+ channels to the closed state explains after-hyperpolarization, followed by a return to the resting mem-brane potential. ■Nerve fibers are divided into different categories based on ax-onal diameter, conduction velocity, and function. ■Neurotrophins are produced by astrocytes and transported by retrograde transport to the neuronal cell body, where they foster the production of proteins associated with neuronal develop-ment, growth, and survival. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. The distance from between one stimulating electrode to record-ing electrode is 4.5 cm. When the axon is stimulated, the latent period is 1.5 ms. What is the conduction velocity of the axon? A) 15 m/s B) 30 m/s C) 40 m/s D) 67.5 m/s E) This cannot be determined from the information given. 2. Which of the following has the slowest conduction velocity? A) Aα fibers B) Aβ fibers C) Aγ fibers D) B fibers E) C fibers 3. A man falls into a deep sleep with one arm under his head. This arm is paralyzed when he awakens, but it tingles, and pain sensa-tion in it is still intact. The reason for the loss of motor function without loss of pain sensation is that in the nerves to his arm, A) A fibers are more susceptible to hypoxia than B fibers. B) A fibers are more sensitive to pressure than C fibers. C) C fibers are more sensitive to pressure than A fibers. D) motor nerves are more affected by sleep than sensory nerves. E) sensory nerves are nearer the bone than motor nerves and hence are less affected by pressure. 4. Which part of a neuron has the highest concentration of Na+ channels per square millimeter of cell membrane? A) dendrites B) cell body near dendrites C) initial segment D) axonal membrane under myelin E) none of the above 5. Which of the following statements about nerve growth factor is not true? A) It is made up of three polypeptide subunits. B) It facilitates the process of apoptosis. C) It is necessary for the growth and development of the sym-pathetic nervous system. D) It is picked up by nerves from the organs they innervate. E) It is present in the brain. 92 SECTION II Physiology of Nerve & Muscle Cells CHAPTER RESOURCES Aidley DJ: The Physiology of Excitable Cells, 4th ed. Cambridge University Press, 1998. Boron WF, Boulpaep EL: Medical Physiology, Elsevier, 2005. Bradbury EJ, McMahon SB: Spinal cord repair strategies: Why do they work? Nat Rev Neurosci 2006;7:644. Catterall WA: Structure and function of voltage-sensitive ion channels. Science 1988; 242:649. Hille B: Ionic Channels of Excitable Membranes, 3rd ed. Sinauer Associates, 2001. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science,4th ed. McGraw-Hill, 2000. Nicholls JG, Martin AR, Wallace BG: From Neuron to Brain: A Cellular and Molecular Approach to the Function of the Nervous System, 4th ed. Sinauer Associates, 2001. Thuret S, Moon LDF, Gage FH: Therapeutic interventions after spinal cord injury. Nat Rev Neurosci 2006;7:628. Volterra A, Meldolesi J: Astrocytes, from brain glue to communication elements: The revolution continues. Nat Rev Neurosci 2005;6:626. Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology. McGraw-Hill, 2008. 93 C H A P T E R 5 Excitable Tissue: Muscle O B J E C T I V E S After studying this chapter, you should be able to: ■Differentiate the major classes of muscle in the body. ■Describe the molecular and electrical makeup of muscle cell excitation– contraction coupling. ■Define thick and thick filaments and how they slide to create contraction. ■Differentiate the role(s) for Ca2+ in skeletal, cardiac, and smooth muscle contraction. ■Appreciate muscle cell diversity. INTRODUCTION Muscle cells, like neurons, can be excited chemically, electri-cally, and mechanically to produce an action potential that is transmitted along their cell membranes. Unlike neurons, they respond to stimuli by activating a contractile mechanism. The contractile protein myosin and the cytoskeletal protein actin are abundant in muscle, where they are the primary structural components that bring about contraction. Muscle is generally divided into three types: skeletal, cardiac, and smooth, although smooth muscle is not a homogeneous single category. Skeletal muscle makes up the great mass of the somatic musculature. It has well-developed cross-striations, does not normally contract in the absence of nervous stimulation, lacks anatomic and functional connections between individual muscle fibers, and is generally under voluntary control. Cardiac muscle also has cross-striations, but it is functionally syncytial and, although it can be modulated via the autonomic nervous system, it can contract rhythmically in the absence of external innervation owing to the presence in the myocardium of pace-maker cells that discharge spontaneously (see Chapter 30). Smooth muscle lacks cross-striations and can be further subdi-vided into two broad types: unitary (or visceral) smooth muscle and multiunit smooth muscle. The type found in most hollow viscera is functionally syncytial and contains pacemakers that discharge irregularly. The multiunit type found in the eye and in some other locations is not spontaneously active and resembles skeletal muscle in graded contractile ability. SKELETAL MUSCLE MORPHOLOGY ORGANIZATION Skeletal muscle is made up of individual muscle fibers that are the “building blocks” of the muscular system in the same sense that the neurons are the building blocks of the nervous system. Most skeletal muscles begin and end in tendons, and the mus-cle fibers are arranged in parallel between the tendinous ends, so that the force of contraction of the units is additive. Each muscle fiber is a single cell that is multinucleated, long, cylin-drical, and surrounded by a cell membrane, the sarcolemma (Figure 5–1). There are no syncytial bridges between cells. The muscle fibers are made up of myofibrils, which are divisible into individual filaments. These myofilaments contain several proteins that together make up the contractile machinery of the skeletal muscle. 94 SECTION II Physiology of Nerve & Muscle Cells The contractile mechanism in skeletal muscle largely depends on the proteins myosin-II, actin, tropomyosin, and troponin. Troponin is made up of three subunits: troponin I, troponin T, and troponin C. Other important proteins in muscle are involved in maintaining the proteins that partici-pate in contraction in appropriate structural relation to one another and to the extracellular matrix. FIGURE 5–1 Mammalian skeletal muscle. A single muscle fiber surrounded by its sarcolemma has been cut away to show individual myofibrils. The cut surface of the myofibrils shows the arrays of thick and thin filaments. The sarcoplasmic reticulum with its transverse (T) tubules and terminal cisterns surrounds each myofibril. The T tubules invaginate from the sarcolemma and contact the myofibrils twice in every sarcomere. Mitochondria are found between the myofibrils and a basal lamina surrounds the sarcolemma. (Reproduced with permission from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Myofibril Filaments Mitochondrion Terminal cistern Transverse tubules A Sarcoplasmic reticulum Sarcolemma (muscle fiber membrane) Sarcomere Z disk Thin filament (F-actin) Thick filament (myosin) Tropomyosin Troponin Actin Z disk B C CHAPTER 5 Excitable Tissue: Muscle 95 STRIATIONS Differences in the refractive indexes of the various parts of the muscle fiber are responsible for the characteristic cross-stria-tions seen in skeletal muscle when viewed under the micro-scope. The parts of the cross-striations are frequently identified by letters (Figure 5–2). The light I band is divided by the dark Z line, and the dark A band has the lighter H band in its center. A transverse M line is seen in the middle of the H band, and this line plus the narrow light areas on either side of it are sometimes called the pseudo-H zone. The area between two adjacent Z lines is called a sarcomere. The orderly arrange-ment of actin, myosin, and related proteins that produces this pattern is shown in Figure 5–3. The thick filaments, which are about twice the diameter of the thin filaments, are made up of myosin; the thin filaments are made up of actin, tropomyosin, and troponin. The thick filaments are lined up to form the A bands, whereas the array of thin filaments extends out of the A band and into the less dense staining I bands. The lighter H bands in the center of the A bands are the regions where, when the muscle is relaxed, the thin filaments do not overlap the thick filaments. The Z lines allow for anchoring of the thin fil-aments. If a transverse section through the A band is exam-ined under the electron microscope, each thick filament is seen to be surrounded by six thin filaments in a regular hexag-onal pattern. The form of myosin found in muscle is myosin-II, with two globular heads and a long tail. The heads of the myosin FIGURE 5–2 Electron micrograph of human gastrocnemius muscle. The various bands and lines are identified at the top (× 13,500). (Courtesy of Walker SM, Schrodt GR.) A band H band Z line M line I band FIGURE 5–3 A) Arrangement of thin (actin) and thick (myosin) filaments in skeletal muscle (compare to Figure 5–2). B) Sliding of actin on myosin during contraction so that Z lines move closer together. C) Detail of relation of myosin to actin in an individual sarcomere, the functional unit of the muscle. D) Diagrammatic representation of the arrangement of actin, tropomyosin, and troponin of the thin filaments in relation to a myosin thick filament. The globular heads of myosin interact with the thin filaments to create the contraction. Note that myosin thick filaments reverse polarity at the M line in the middle of the sarcomere, allowing for contraction. (C and D are modified with permision from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) A band Sarcomere Z line Thick filament Thin filament Myosin Relaxed Contracted Actin Z line Actin Z line Myosin M line Actin Tropomyosin Troponin I C T I C T Actin A B C D 96 SECTION II Physiology of Nerve & Muscle Cells molecules form cross-bridges with actin. Myosin contains heavy chains and light chains, and its heads are made up of the light chains and the amino terminal portions of the heavy chains. These heads contain an actin-binding site and a cata-lytic site that hydrolyzes ATP. The myosin molecules are arranged symmetrically on either side of the center of the sar-comere, and it is this arrangement that creates the light areas in the pseudo-H zone. The M line is the site of the reversal of polarity of the myosin molecules in each of the thick fila-ments. At these points, there are slender cross-connections that hold the thick filaments in proper array. Each thick fila-ment contains several hundred myosin molecules. The thin filaments are polymers made up of two chains of actin that form a long double helix. Tropomyosin molecules are long filaments located in the groove between the two chains in the actin (Figure 5–3). Each thin filament contains 300 to 400 actin molecules and 40 to 60 tropomyosin mole-cules. Troponin molecules are small globular units located at intervals along the tropomyosin molecules. Each of the three troponin subunits has a unique function: Troponin T binds the troponin components to tropomyosin; troponin I inhibits the interaction of myosin with actin; and troponin C contains the binding sites for the Ca2+ that helps to initiate contraction. Some additional structural proteins that are important in skeletal muscle function include actinin, titin, and desmin. Actinin binds actin to the Z lines. Titin, the largest known pro-tein (with a molecular mass near 3,000,000 Da), connects the Z lines to the M lines and provides scaffolding for the sarcomere. It contains two kinds of folded domains that provide muscle with its elasticity. At first when the muscle is stretched there is relatively little resistance as the domains unfold, but with fur-ther stretch there is a rapid increase in resistance that protects the structure of the sarcomere. Desmin adds structure to the Z lines in part by binding the Z lines to the plasma membrane. Although these proteins are important in muscle structure/ function, by no means do they represent an exhaustive list. SARCOTUBULAR SYSTEM The muscle fibrils are surrounded by structures made up of membranes that appear in electron photomicrographs as ves-icles and tubules. These structures form the sarcotubular sys-tem, which is made up of a T system and a sarcoplasmic reticulum. The T system of transverse tubules, which is con-tinuous with the sarcolemma of the muscle fiber, forms a grid perforated by the individual muscle fibrils (Figure 5–1). The space between the two layers of the T system is an extension of the extracellular space. The sarcoplasmic reticulum, which forms an irregular curtain around each of the fibrils, has en-larged terminal cisterns in close contact with the T system at the junctions between the A and I bands. At these points of contact, the arrangement of the central T system with a cistern of the sarcoplasmic reticulum on either side has led to the use of the term triads to describe the system. The T system, which is continuous with the sarcolemma, provides a path for the rapid transmission of the action potential from the cell mem-brane to all the fibrils in the muscle. The sarcoplasmic reticu-lum is an important store of Ca2+ and also participates in muscle metabolism. DYSTROPHIN–GLYCOPROTEIN COMPLEX The large dystrophin protein (molecular mass 427,000 Da) forms a rod that connects the thin actin filaments to the transmembrane protein β-dystroglycan in the sarcolemma by smaller proteins in the cytoplasm, syntrophins. β-dystro-glycan is connected to merosin (merosin refers to laminins that contain the α2 subunit in their trimeric makeup) in the extracellular matrix by α-dystroglycan (Figure 5–4). The dystroglycans are in turn associated with a complex of four transmembrane glycoproteins: α-, β-, γ-, and δ-sarcoglycan. This dystrophin–glycoprotein complex adds strength to the muscle by providing a scaffolding for the fibrils and con-necting them to the extracellular environment. Disruption of the tightly choreographed structure can lead to several different pathologies, or muscular dystrophies (see Clinical Box 5–1). ELECTRICAL PHENOMENA & IONIC FLUXES ELECTRICAL CHARACTERISTICS OF SKELETAL MUSCLE The electrical events in skeletal muscle and the ionic fluxes that underlie them share distinct similarities to those in nerve, with quantitative differences in timing and magnitude. The resting membrane potential of skeletal muscle is about –90 mV. The action potential lasts 2 to 4 ms and is conducted along the muscle fiber at about 5 m/s. The absolute refractory period is 1 to 3 ms long, and the after-polarizations, with their related changes in threshold to electrical stimulation, are rela-tively prolonged. The initiation of impulses at the myoneural junction is discussed in the next chapter. ION DISTRIBUTION & FLUXES The distribution of ions across the muscle fiber membrane is similar to that across the nerve cell membrane. Approximate values for the various ions and their equilibrium potentials are shown in Table 5–1. As in nerves, depolarization is largely a manifestation of Na+ influx, and repolarization is largely a manifestation of K+ efflux. CONTRACTILE RESPONSES It is important to distinguish between the electrical and me-chanical events in skeletal muscle. Although one response CHAPTER 5 Excitable Tissue: Muscle 97 does not normally occur without the other, their physiologic bases and characteristics are different. Muscle fiber membrane depolarization normally starts at the motor end plate, the spe-cialized structure under the motor nerve ending. The action potential is transmitted along the muscle fiber and initiates the contractile response. THE MUSCLE TWITCH A single action potential causes a brief contraction followed by relaxation. This response is called a muscle twitch. In Figure 5–5, the action potential and the twitch are plotted on the same time scale. The twitch starts about 2 ms after the start of depolarization of the membrane, before repolarization is com-plete. The duration of the twitch varies with the type of muscle being tested. “Fast” muscle fibers, primarily those concerned with fine, rapid, precise movement, have twitch durations as short as 7.5 ms. “Slow” muscle fibers, principally those in-volved in strong, gross, sustained movements, have twitch du-rations up to 100 ms. MOLECULAR BASIS OF CONTRACTION The process by which the contraction of muscle is brought about is a sliding of the thin filaments over the thick filaments. Note that this shortening is not due to changes in the actual lengths of the thick and thin filaments, rather, by their increased overlap within the muscle cell. The width of the A bands is con-stant, whereas the Z lines move closer together when the muscle contracts and farther apart when it relaxes (Figure 5–3). The sliding during muscle contraction occurs when the myo-sin heads bind firmly to actin, bend at the junction of the head with the neck, and then detach. This “power stroke” depends on the simultaneous hydrolysis of ATP. Myosin-II molecules are dimers that have two heads, but only one attaches to actin at any given time. The probable sequence of events of the power stroke is outlined in Figure 5–6. In resting muscle, troponin I is bound to actin and tropomyosin and covers the sites where myosin heads interact with actin. Also at rest, the myosin head contains tightly bound ADP. Following an action potential cytosolic Ca2+ is increased and free Ca2+ binds to troponin C. This binding results in a weakening of the troponin I interac-tion with actin and exposes the actin binding site for myosin to FIGURE 5–4 The dystrophin–glycoprotein complex. Dystrophin connects F-actin to the two members of the dystroglycan (DG) complex, α and β-dystroglycan, and these in turn connect to the merosin subunit of laminin 211 in the extracellular matrix. The sarcoglycan complex of four glycoproteins, α-, β-, γ-, and δ-sarcoglycan, sarcospan, and syntropins are all associated with the dystroglycan complex. There are muscle disorders associated with loss, abnormalities, or both of the sarcoglycans and merosin. (Reproduced with permission from Kandel ER, Scwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) α Dystroglycans Sarcoglycan complex Sarcospan Syntrophins β β γ δ Dystrophin F-Actin Functionally important carbohydrate side chains Laminin 2 α α2 β1 γ1 text continues on p. 100 98 SECTION II Physiology of Nerve & Muscle Cells CLINICAL BOX 5–1 Disease of Muscle have in common exercise intolerance and the possibility of muscle breakdown due to accumulation of toxic metabolites. Muscular Dystrophies The term muscular dystrophy is applied to diseases that cause progressive weakness of skeletal muscle. About 50 such diseas-es have been described, some of which include cardiac as well as skeletal muscle. They range from mild to severe and some are eventually fatal. They have multiple causes, but mutations in the genes for the various components of the dystrophin–glycopro-tein complex are a prominent cause. The dystrophin gene is one of the largest in the body, and mutations can occur at many dif-ferent sites in it. Duchenne muscular dystrophy is a serious form of dystrophy in which the dystrophin protein is absent from muscle. It is X-linked and usually fatal by the age of 30. In a milder form of the disease, Becker muscular dystrophy, dys-trophin is present but altered or reduced in amount. Limb-girdle muscular dystrophies of various types are associated with muta-tions of the genes coding for the sarcoglycans or other compo-nents of the dystrophin–glycoprotein complex. Ion Channel Myopathies In the various forms of clinical myotonia, muscle relaxation is prolonged after voluntary contraction. The molecular bases of myotonias are due to dysfunction of channels that shape the action potential. Myotonia dystrophy is caused by an autoso-mal dominant mutation that leads to overexpression of a K+ channel (although the mutation is not at the K+ channel). A variety of myotonias are associated with mutations in Na+ channels (eg, hyperkalemic periodic paralysis, paramyotonia congenita, or Na+ channel congenita) or Cl– channels (eg, dominant or recessive myotonia congenita). Malignant hyperthermia is another disease related to dys-functional muscle ion channels. Patients with malignant hy-perthermia can respond to general anesthetics such as hal-othane by eliciting rigidity in the muscles and a quick increase in body temperature. This disease has been traced to a mutation in RyR, the Ca2+ release channel in the sarco-plasmic reticulum. The mutation results in an inefficient feedback mechanism to shut down Ca2+ release after stimu-lation of the RyR, and thus, increased contractility and heat generation. Metabolic Myopathies Mutations in genes that code for enzymes involved in the me-tabolism of carbohydrates, fats, and proteins to CO2 and H2O in muscle and the production of ATP can cause metabolic my-opathies (eg, McArdle syndrome). Metabolic myopathies all TABLE 5–1 Steady-state distribution of ions in the intracellular and extracellular compartments of mammalian skeletal muscle, and the equilibrium potentials for these ions. Iona Concentration (mmol/L) Equilibrium Potential (mV) Intracellular Fluid Extracellular Fluid Na+ 12 145 +65 K+ 155 4 –95 H+ 13 × 10–5 3.8 × 10–5 –32 Cl– 3.8 120 –90 HCO3 – 8 27 –32 A– 155 0 … Membrane potential = –90 mV aA– represents organic anions. The value for intracellular Cl– is calculated from the membrane potential, using the Nernst equation. FIGURE 5–5 The electrical and mechanical responses of a mammalian skeletal muscle fiber to a single maximal stimulus. The electrical response (mV potential change) and the mechanical re-sponse (T, tension in arbitrary units) are plotted on the same abscissa (time). The mechanical response is relatively long-lived compared to the electrical response that initiates contraction. 0 5 10 15 20 25 ms 100 0 30 0 T mV CHAPTER 5 Excitable Tissue: Muscle 99 FIGURE 5–6 Power stroke of myosin in skeletal muscle. A) At rest, myosin heads are bound to adenosine diphosphate and are said to be in a “cocked” position in relation to the thin filament, which does not have Ca2+ bound to the troponin–tropomyosin complex. B) Ca2+ bound to the troponin–tropomyosin complex induced a conformational change in the thin filament that allows for myosin heads to cross-bridge with thin filament actin. C) Myosin heads rotate, move the attached actin and shorten the muscle fiber, forming the power stroke. D) At the end of the power stroke, ATP binds to a now exposed site, and causes a detachment from the actin filament. E) ATP is hydrolyzed into ADP and inorganic phosphate (Pi) and this chemical energy is used to “re-cock” the myosin head. (Modified with permission from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Ca2+ ADP Pi ADP ADP Actin Troponin Tropomyosin Exposed binding site Ca2+ Myosin Thin filament A Thick filament ADP B C D E Longitudinal force ATP 100 SECTION II Physiology of Nerve & Muscle Cells allow for formation of myosin/actin cross-bridges. Upon for-mation of the cross-bridge, ADP is released, causing a confor-mational change in the myosin head that moves the thin filament relative to the thick filament, comprising the cross-bridge “power stroke.” ATP quickly binds to the free site on the myosin, which leads to a detachment of the myosin head from the thin filament. ATP is hydrolyzed and inorganic phosphate (Pi) released, causing a “re-cocking” of the myosin head and completing the cycle. As long as Ca2+ remains elevated and suf-ficient ATP is available, this cycle repeats. Many myosin heads cycle at or near the same time, and they cycle repeatedly, pro-ducing gross muscle contraction. Each power stroke shortens the sarcomere about 10 nm. Each thick filament has about 500 myosin heads, and each head cycles about five times per second during a rapid contraction. The process by which depolarization of the muscle fiber initiates contraction is called excitation–contraction cou-pling. The action potential is transmitted to all the fibrils in the fiber via the T system (Figure 5–7). It triggers the release of Ca2+ from the terminal cisterns, the lateral sacs of the sar-coplasmic reticulum next to the T system. Depolarization of the T tubule membrane activates the sarcoplasmic reticulum via dihydropyridine receptors (DHPR), named for the drug dihydropyridine, which blocks them (Figure 5–8). DHPR are voltage-gated Ca2+ channels in the T tubule membrane. In cardiac muscle, influx of Ca2+ via these channels triggers the release of Ca2+ stored in the sarcoplasmic reticulum (calcium-induced calcium release) by activating the ryanodine recep-tor (RyR). The RyR is named after the plant alkaloid ryano-dine that was used in its discovery. It is a ligand-gated Ca2+ channel with Ca2+ as its natural ligand. In skeletal muscle, Ca2+ entry from the extracellular fluid (ECF) by this route is not required for Ca2+ release. Instead, the DHPR that serves as the voltage sensor unlocks release of Ca2+ from the nearby sarcoplasmic reticulum via physical interaction with the RyR. The released Ca2+ is quickly amplified through calcium-induced calcium release. Ca2+ is reduced in the muscle cell by the sarcoplasmic or endoplasmic reticulum Ca2+ ATPase (SERCA) pump. The SERCA pump uses energy from ATP hydrolysis to remove Ca2+ from the cytosol back into the ter-minal cisterns, where it is stored until released by the next action potential. Once the Ca2+ concentration outside the reticulum has been lowered sufficiently, chemical interaction between myosin and actin ceases and the muscle relaxes. Note that ATP provides the energy for both contraction (at the myosin head) and relaxation (via SERCA). If transport of Ca2+ into the reticulum is inhibited, relaxation does not occur even though there are no more action potentials; the resulting sustained contraction is called a contracture. TYPES OF CONTRACTION Muscular contraction involves shortening of the contractile elements, but because muscles have elastic and viscous ele-ments in series with the contractile mechanism, it is possible FIGURE 5–7 Flow of information that leads to muscle contraction. Release of transmitter (acetylcholine) at motor end-plate Discharge of motor neuron Generation of end-plate potential Steps in contractiona Binding of acetylcholine to nicotinic acetylcholine receptors Steps in relaxation Increased Na+ and K+ conductance in end-plate membrane Generation of action potential in muscle fibers Inward spread of depolarization along T tubules Binding of Ca2+ to troponin C, uncovering myosin-binding sites on actin Formation of cross-linkages between actin and myosin and sliding of thin on thick filaments, producing movement Release of Ca2+ from terminal cisterns of sarcoplasmic reticulum and diffusion to thick and thin filaments Ca2+ pumped back into sarcoplasmic reticulum Release of Ca2+ from troponin Cessation of interaction between actin and myosin aSteps 1–6 in contraction are discussed in Chapter 4. CHAPTER 5 Excitable Tissue: Muscle 101 for contraction to occur without an appreciable decrease in the length of the whole muscle (Figure 5–9). Such a contrac-tion is called isometric (“same measure” or length). Contrac-tion against a constant load with a decrease in muscle length is isotonic (“same tension”). Note that because work is the prod-uct of force times distance, isotonic contractions do work, whereas isometric contractions do not. In other situations, muscle can do negative work while lengthening against a con-stant weight. SUMMATION OF CONTRACTIONS The electrical response of a muscle fiber to repeated stimula-tion is like that of nerve. The fiber is electrically refractory only during the rising phase and part of the falling phase of the spike potential. At this time, the contraction initiated by the first stimulus is just beginning. However, because the contrac-tile mechanism does not have a refractory period, repeated stimulation before relaxation has occurred produces addition-al activation of the contractile elements and a response that is added to the contraction already present. This phenomenon is known as summation of contractions. The tension developed during summation is considerably greater than that during the single muscle twitch. With rapidly repeated stimulation, acti-vation of the contractile mechanism occurs repeatedly before any relaxation has occurred, and the individual responses fuse into one continuous contraction. Such a response is called a tetanus (tetanic contraction). It is a complete tetanus when no relaxation occurs between stimuli and an incomplete teta-nus when periods of incomplete relaxation take place between the summated stimuli. During a complete tetanus, the tension developed is about four times that developed by the individual twitch contractions. The development of an incomplete and a complete tetanus in response to stimuli of increasing frequen-cy is shown in Figure 5–10. FIGURE 5–8 Relation of the T tubule (TT) to the sarcoplasmic reticulum in Ca2+ transport. In skeletal muscle, the voltage-gated dihydropyridine receptor in the T tubule triggers Ca2+ release from the sarcoplasmic reticulum (SR) via the ryanodine receptor (RyR). Upon sensing a voltage change, there is a physical interaction between the sarcolemmal-bound DHPR and the SR-bound RyR. This interaction gates the RyR and allows for Ca2+ release from the SR. Dihydropyridine receptor Ryanodine receptor Extracellular space Cytoplasm COOH COOH Lumen of SR NH2 + + + + TT FIGURE 5–9 A) Muscle preparation arranged for recording iso-tonic contractions. B) Preparation arranged for recording isometric contractions. In A, the muscle is fastened to a writing lever that swings on a pivot. In B, it is attached to an electronic transducer that measures the force generated without permitting the muscle to shorten. A Pivot Recorder B Force transducer Recorder To stimulator FIGURE 5–10 Tetanus. Isometric tension of a single muscle fiber during continuously increasing and decreasing stimulation frequency. Dots at the top are at intervals of 0.2 s. Note the development of incomplete and then complete tetanus as stimulation is increased, and the return of incomplete tetanus, then full response, as stimulation frequency is decreased. 102 SECTION II Physiology of Nerve & Muscle Cells The stimulation frequency at which summation of contrac-tions occurs is determined by the twitch duration of the par-ticular muscle being studied. For example, if the twitch duration is 10 ms, frequencies less than 1/10 ms (100/s) cause discrete responses interrupted by complete relaxation, and frequencies greater than 100/s cause summation. RELATION BETWEEN MUSCLE LENGTH & TENSION & VELOCITY OF CONTRACTION Both the tension that a muscle develops when stimulated to contract isometrically (the total tension) and the passive ten-sion exerted by the unstimulated muscle vary with the length of the muscle fiber. This relationship can be studied in a whole skeletal muscle preparation such as that shown in Figure 5–9. The length of the muscle can be varied by changing the dis-tance between its two attachments. At each length, the passive tension is measured, the muscle is then stimulated electrically, and the total tension is measured. The difference between the two values at any length is the amount of tension actually gen-erated by the contractile process, the active tension. The records obtained by plotting passive tension and total tension against muscle length are shown in Figure 5–11. Similar curves are obtained when single muscle fibers are studied. The length of the muscle at which the active tension is maximal is usually called its resting length. The term comes originally from experiments demonstrating that the length of many of the muscles in the body at rest is the length at which they de-velop maximal tension. The observed length–tension relation in skeletal muscle is explained by the sliding filament mechanism of muscle con-traction. When the muscle fiber contracts isometrically, the tension developed is proportional to the number of cross-bridges between the actin and the myosin molecules. When muscle is stretched, the overlap between actin and myosin is reduced and the number of cross-linkages is therefore reduced. Conversely, when the muscle is appreciably shorter than resting length, the distance the thin filaments can move is reduced. The velocity of muscle contraction varies inversely with the load on the muscle. At a given load, the velocity is maximal at the resting length and declines if the muscle is shorter or longer than this length. FIBER TYPES Although skeletal muscle fibers resemble one another in a general way, skeletal muscle is a heterogeneous tissue made up of fibers that vary in myosin ATPase activity, contractile speed, and other properties. Muscles are frequently classified into two types, “slow” and “fast.” These muscles can contain a mixture of three fiber types: type I (or SO for slow-oxidative); type IIA (FOG for fast-oxidative-glycolytic); or type IIB (FG for fast glycolytic). Some of the properties associated with type I, type IIA, and type IIB fibers are summarized in Table 5–2. Although this classification scheme is valid for muscles across many mammalian species, there are significant variations of fibers within and between muscles. For example, type I fibers in a given muscle can be larger than type IIA fibers from a dif-ferent muscle in the same animal. Many of the differences in the fibers that make up muscles stem from differences in the proteins within them. Most of these are encoded by multigene families. Ten different isoforms of the myosin heavy chains (MHCs) have been characterized. Each of the two types of light chains also have isoforms. It appears that there is only one form of actin, but multiple isoforms of tropomyosin and all three components of troponin. ENERGY SOURCES & METABOLISM Muscle contraction requires energy, and muscle has been called “a machine for converting chemical energy into me-chanical work.” The immediate source of this energy is ATP, and this is formed by the metabolism of carbohydrates and lipids. PHOSPHORYLCREATINE ATP is resynthesized from ADP by the addition of a phos-phate group. Some of the energy for this endothermic reaction is supplied by the breakdown of glucose to CO2 and H2O, but there also exists in muscle another energy-rich phosphate compound that can supply this energy for short periods. This compound is phosphorylcreatine, which is hydrolyzed to cre-atine and phosphate groups with the release of considerable energy (Figure 5–12). At rest, some ATP in the mitochondria transfers its phosphate to creatine, so that a phosphorylcreatine FIGURE 5–11 Length–tension relationship for the human triceps muscle. The passive tension curve measures the tension exert-ed by this skeletal muscle at each length when it is not stimulated. The total tension curve represents the tension developed when the muscle contracts isometrically in response to a maximal stimulus. The active tension is the difference between the two. 30 20 10 0 0 1 2 3 4 5 Increase in muscle length (cm) Passive tension Active tension Total tension Resting length Tension (kg) CHAPTER 5 Excitable Tissue: Muscle 103 store is built up. During exercise, the phosphorylcreatine is hydrolyzed at the junction between the myosin heads and ac-tin, forming ATP from ADP and thus permitting contraction to continue. CARBOHYDRATE & LIPID BREAKDOWN At rest and during light exercise, muscles utilize lipids in the form of free fatty acids as their energy source. As the intensity of exercise increases, lipids alone cannot supply energy fast enough and so use of carbohydrate becomes the predominant component in the muscle fuel mixture. Thus, during exercise, much of the energy for phosphorylcreatine and ATP resynthe-sis comes from the breakdown of glucose to CO2 and H2O. Glucose in the bloodstream enters cells, where it is degraded through a series of chemical reactions to pyruvate. Another source of intracellular glucose, and consequently of pyruvate, is glycogen, the carbohydrate polymer that is especially abun-dant in liver and skeletal muscle. When adequate O2 is present, pyruvate enters the citric acid cycle and is metabo-lized—through this cycle and the so-called respiratory enzyme pathway—to CO2 and H2O. This process is called aerobic glycolysis. The metabolism of glucose or glycogen to CO2 and H2O forms large quantities of ATP from ADP. If O2 supplies are insufficient, the pyruvate formed from glucose does not enter the tricarboxylic acid cycle but is reduced to lactate. This process of anaerobic glycolysis is associated with the net pro-duction of much smaller quantities of energy-rich phosphate bonds, but it does not require the presence of O2. A brief over-view of the various reactions involved in supplying energy to skeletal muscle is shown in Figure 5–13. TABLE 5–2 Classification of fiber types in skeletal muscles. Type 1 Type IIA Type IIB Other names Slow, Oxidative (SO) Fast, Oxidative, Glycolytic (FOG) Fast, Glycolytic (FG) Color Red Red White Myosin ATPase Activity Slow Fast Fast Ca2+-pumping capacity of sarcoplasmic reticulum Moderate High High Diameter Small Large Large Glycolytic capacity Moderate High High Oxidative capacity High Moderate Low Associated Motor Unit Type Slow (S) Fast Resistant to Fatigue (FR) Fast Fatigable (FF) Membrane potential = –90 mV Oxidative capacity High Moderate Low FIGURE 5–12 Creatine, phosphorylcreatine, and creatinine cycling in muscle. During periods of high activity, cycling of phos-phorylcreatine allows for quick release of ATP to sustain muscle activity. Rest Exercise — — HN C O — — HN C CH2 PO3 CH3 N HN — — H2N+ C + ADP CH3 NCH2COO− Phosphorylcreatine Creatinine H2N — — H2N+ C + ATP CH3 NCH2COO− Creatine FIGURE 5–13 ATP turnover in muscle cells. Energy released by hydrolysis of 1 mol of ATP and reactions responsible for resynthesis of ATP. The amount of ATP formed per mole of free fatty acid (FFA) oxidized is large but varies with the size of the FFA. For example, com-plete oxidation of 1 mol of palmitic acid generates 140 mol of ATP. ATP + H2O ADP + H3PO4 + 7.3 kcal Phosphorylcreatine + ADP Creatine + ATP Glucose + 2 ATP (or glycogen + 1 ATP) Glucose + 2 ATP (or glycogen + 1 ATP) Anaerobic 2 Lactic acid + 4 ATP Oxygen 6 CO2 + 6 H2O + 40 ATP Oxygen CO2 + H2O + ATP FFA 104 SECTION II Physiology of Nerve & Muscle Cells THE OXYGEN DEBT MECHANISM During exercise, the muscle blood vessels dilate and blood flow is increased so that the available O2 supply is increased. Up to a point, the increase in O2 consumption is proportional to the energy expended, and all the energy needs are met by aerobic processes. However, when muscular exertion is very great, aerobic resynthesis of energy stores cannot keep pace with their utilization. Under these conditions, phosphoryl-creatine is still used to resynthesize ATP. In addition, some ATP synthesis is accomplished by using the energy released by the anaerobic breakdown of glucose to lactate. Use of the anaerobic pathway is self-limiting because in spite of rapid dif-fusion of lactate into the bloodstream, enough accumulates in the muscles to eventually exceed the capacity of the tissue buffers and produce an enzyme-inhibiting decline in pH. However, for short periods, the presence of an anaerobic path-way for glucose breakdown permits muscular exertion of a far greater magnitude than would be possible without it. For ex-ample, in a 100-m dash that takes 10 s, 85% of the energy con-sumed is derived anaerobically; in a 2-mi race that takes 10 min, 20% of the energy is derived anaerobically; and in a long-distance race that takes 60 min, only 5% of the energy comes from anaerobic metabolism. After a period of exertion is over, extra O2 is consumed to remove the excess lactate, replenish the ATP and phosphoryl-creatine stores, and replace the small amounts of O2 that were released by myoglobin. The amount of extra O2 consumed is proportional to the extent to which the energy demands during exertion exceeded the capacity for the aerobic synthesis of energy stores, ie, the extent to which an oxygen debt was incurred. The O2 debt is measured experimentally by determining O2 con-sumption after exercise until a constant, basal consumption is reached and subtracting the basal consumption from the total. The amount of this debt may be six times the basal O2 consump-tion, which indicates that the subject is capable of six times the exertion that would have been possible without it. RIGOR When muscle fibers are completely depleted of ATP and phos-phorylcreatine, they develop a state of rigidity called rigor. When this occurs after death, the condition is called rigor mortis. In rigor, almost all of the myosin heads attach to actin but in an abnormal, fixed, and resistant way. HEAT PRODUCTION IN MUSCLE Thermodynamically, the energy supplied to a muscle must equal its energy output. The energy output appears in work done by the muscle, in energy-rich phosphate bonds formed for later use, and in heat. The overall mechanical efficiency of skeletal muscle (work done/total energy expenditure) ranges up to 50% while lifting a weight during isotonic contraction and is essentially 0% during isometric contraction. Energy storage in phosphate bonds is a small factor. Consequently, heat production is considerable. The heat produced in muscle can be measured accurately with suitable thermocouples. Resting heat, the heat given off at rest, is the external man-ifestation of basal metabolic processes. The heat produced in excess of resting heat during contraction is called the initial heat. This is made up of activation heat, the heat that muscle produces whenever it is contracting, and shortening heat, which is proportionate in amount to the distance the muscle shortens. Shortening heat is apparently due to some change in the structure of the muscle during shortening. Following contraction, heat production in excess of resting heat continues for as long as 30 min. This recovery heat is the heat liberated by the metabolic processes that restore the mus-cle to its precontraction state. The recovery heat of muscle is approximately equal to the initial heat; that is, the heat pro-duced during recovery is equal to the heat produced during contraction. If a muscle that has contracted isotonically is restored to its previous length, extra heat in addition to recovery heat is pro-duced (relaxation heat). External work must be done on the muscle to return it to its previous length, and relaxation heat is mainly a manifestation of this work. PROPERTIES OF SKELETAL MUSCLES IN THE INTACT ORGANISM EFFECTS OF DENERVATION In the intact animal or human, healthy skeletal muscle does not contract except in response to stimulation of its motor nerve supply. Destruction of this nerve supply causes muscle atrophy. It also leads to abnormal excitability of the muscle and increases its sensitivity to circulating acetylcholine (de-nervation hypersensitivity; see Chapter 6). Fine, irregular con-tractions of individual fibers (fibrillations) appear. This is the classic picture of a lower motor neuron lesion. If the motor nerve regenerates, the fibrillations disappear. Usually, the contractions are not visible grossly, and they should not be confused with fasciculations, which are jerky, visible contrac-tions of groups of muscle fibers that occur as a result of patho-logic discharge of spinal motor neurons. THE MOTOR UNIT Because the axons of the spinal motor neurons supplying skel-etal muscle each branch to innervate several muscle fibers, the smallest possible amount of muscle that can contract in re-sponse to the excitation of a single motor neuron is not one muscle fiber but all the fibers supplied by the neuron. Each single motor neuron and the muscle fibers it innervates con-stitute a motor unit. The number of muscle fibers in a motor CHAPTER 5 Excitable Tissue: Muscle 105 unit varies. In muscles such as those of the hand and those concerned with motion of the eye (ie, muscles concerned with fine, graded, precise movement), each motor unit innervates very few (on the order of three to six) muscle fibers. On the other hand, values of 600 muscle fibers per motor unit can oc-cur in human leg muscles. The group of muscle fibers that contribute to a motor unit can be intermixed within a muscle. That is, although they contract as a unit, they are not necessar-ily “neighboring” fibers within the muscle. Each spinal motor neuron innervates only one kind of mus-cle fiber, so that all the muscle fibers in a motor unit are of the same type. On the basis of the type of muscle fiber they inner-vate, and thus on the basis of the duration of their twitch con-traction, motor units are divided into S (slow), FR (fast, resistant to fatigue), and FF (fast, fatigable) units. Interestingly, there is also a gradation of innervation of these fibers, with S fibers tending to have a low innervation ratio (ie, small units) and FF fibers tending to have a high innervation ratio (ie, large units). The recruitment of motor units during muscle contrac-tion is not random, rather it follows a general scheme, the size principle. In general, a specific muscle action is developed first by the recruitment of S muscle units that contract relatively slowly to produce controlled contraction. Next, FR muscle units are recruited, resulting in more powerful response over a shorter period of time. Lastly, FF muscle units are recruited for the most demanding tasks. For example, in muscles of the leg, the small, slow units are first recruited for standing. As walking motion is initiated, their recruitment of FR units increases. As this motion turns to running or jumping, the FF units are recruited. Of course, there is overlap in recruitment, but, in general, this principle holds true. The differences between types of muscle units are not inherent but are determined by, among other things, their activity. When the nerve to a slow muscle is cut and the nerve to a fast muscle is spliced to the cut end, the fast nerve grows and innervates the previously slow muscle. However, the mus-cle becomes fast and corresponding changes take place in its muscle protein isoforms and myosin ATPase activity. This change is due to changes in the pattern of activity of the mus-cle; in stimulation experiments, changes in the expression of MHC genes and consequently of MHC isoforms can be pro-duced by changes in the pattern of electrical activity used to stimulate the muscle. More commonly, muscle fibers can be altered by a change in activity initiated through exercise (or lack thereof). Increased activity can lead to muscle cell hyper-trophy, which allows for increase in contractile strength. Type IIA and IIB fibers are most susceptible to these changes. Alternatively, inactivity can lead to muscle cell atrophy and a loss of contractile strength. Type I fibers—that is, the ones used most often—are most susceptible to these changes. ELECTROMYOGRAPHY Activation of motor units can be studied by electromyography, the process of recording the electrical activity of muscle on an oscilloscope. This may be done in unanaesthetized humans by using small metal disks on the skin overlying the muscle as the pick-up electrodes or by using hypodermic needle electrodes. The record obtained with such electrodes is the electromyo-gram (EMG). With needle electrodes, it is usually possible to pick up the activity of single muscle fibers. The measured EMG depicts the potential difference between the two electrodes, which is altered by the activation of muscles in between the elec-trodes. A typical EMG is shown in Figure 5–14. It has been shown by electromyography that little if any spontaneous activity occurs in the skeletal muscles of normal individuals at rest. With minimal voluntary activity a few motor units discharge, and with increasing voluntary effort, more and more are brought into play to monitor the recruit-ment of motor units. Gradation of muscle response is there-fore in part a function of the number of motor units activated. In addition, the frequency of discharge in the individual nerve fibers plays a role, the tension developed during a tetanic con-traction being greater than that during individual twitches. The length of the muscle is also a factor. Finally, the motor units fire asynchronously, that is, out of phase with one another. This asynchronous firing causes the individual mus-cle fiber responses to merge into a smooth contraction of the whole muscle. In summary, EMGs can be used to quickly (and roughly) monitor abnormal electrical activity associated with muscle responses. THE STRENGTH OF SKELETAL MUSCLES Human skeletal muscle can exert 3 to 4 kg of tension per square centimeter of cross-sectional area. This figure is about the same as that obtained in a variety of experimental animals and seems to be constant for mammalian species. Because many of the muscles in humans have a relatively large cross-sectional area, the tension they can develop is quite large. The gastrocnemius, FIGURE 5–14 Electromyographic tracings from human biceps and triceps muscles during alternate flexion and extension of the elbow. Note the alternate activation and rest patterns as one muscle is used for flexion and the other for extension. Electrical activity of stimulated muscle can be recorded extracellularly, yielding typical excitation responses after stimulation. (Courtesy of Garoutte BC.) Biceps Triceps 0.5 s 500 μV 106 SECTION II Physiology of Nerve & Muscle Cells for example, not only supports the weight of the whole body during climbing but resists a force several times this great when the foot hits the ground during running or jumping. An even more striking example is the gluteus maximus, which can exert a tension of 1200 kg. The total tension that could be developed if all muscles in the body of an adult man pulled together is ap-proximately 22,000 kg (nearly 25 tons). BODY MECHANICS Body movements are generally organized in such a way that they take maximal advantage of the physiologic principles out-lined above. For example, the attachments of the muscles in the body are such that many of them are normally at or near their resting length when they start to contract. In muscles that ex-tend over more than one joint, movement at one joint may compensate for movement at another in such a way that rela-tively little shortening of the muscle occurs during contraction. Nearly isometric contractions of this type permit development of maximal tension per contraction. The hamstring muscles extend from the pelvis over the hip joint and the knee joint to the tibia and fibula. Hamstring contraction produces flexion of the leg on the thigh. If the thigh is flexed on the pelvis at the same time, the lengthening of the hamstrings across the hip joint tends to compensate for the shortening across the knee joint. In the course of various activities, the body moves in a way that takes advantage of this. Such factors as momentum and balance are integrated into body movement in ways that make possible maximal motion with minimal muscular exer-tion. One net effect is that the stress put on tendons and bones is rarely over 50% of their failure strength, protecting them from damage. In walking, each limb passes rhythmically through a sup-port or stance phase when the foot is on the ground and a swing phase when the foot is off the ground. The support phases of the two legs overlap, so that two periods of double support occur during each cycle. There is a brief burst of activity in the leg flexors at the start of each step, and then the leg is swung forward with little more active muscular contrac-tion. Therefore, the muscles are active for only a fraction of each step, and walking for long periods causes relatively little fatigue. A young adult walking at a comfortable pace moves at a velocity of about 80 m/min and generates a power output of 150 to 175 W per step. A group of young adults asked to walk at their most comfortable rate selected a velocity close to 80 m/min, and it was found that they had selected the velocity at which their energy output was minimal. Walking more rap-idly or more slowly took more energy. CARDIAC MUSCLE MORPHOLOGY The striations in cardiac muscle are similar to those in skeletal muscle, and Z lines are present. Large numbers of elongated mitochondria are in close contact with the muscle fibrils. The muscle fibers branch and interdigitate, but each is a complete unit surrounded by a cell membrane. Where the end of one muscle fiber abuts on another, the membranes of both fibers parallel each other through an extensive series of folds. These areas, which always occur at Z lines, are called intercalated disks (Figure 5–15). They provide a strong union between fi-bers, maintaining cell-to-cell cohesion, so that the pull of one contractile cell can be transmitted along its axis to the next. Along the sides of the muscle fibers next to the disks, the cell membranes of adjacent fibers fuse for considerable distances, forming gap junctions. These junctions provide low-resis-tance bridges for the spread of excitation from one fiber to an-other. They permit cardiac muscle to function as if it were a syncytium, even though no protoplasmic bridges are present between cells. The T system in cardiac muscle is located at the Z lines rather than at the A–I junction, where it is located in mammalian skeletal muscle. ELECTRICAL PROPERTIES RESTING MEMBRANE & ACTION POTENTIALS The resting membrane potential of individual mammalian cardiac muscle cells is about –80 mV. Stimulation produces a propagated action potential that is responsible for initiating contraction. Although action potentials vary among the car-diomyocytes in different regions of the heart (discussed in lat-er chapters), the action potential of a typical ventricular cardiomyocyte can be used as an example (Figure 5–16). De-polarization proceeds rapidly and an overshoot of the zero po-tential is present, as in skeletal muscle and nerve, but this is followed by a plateau before the membrane potential returns to the baseline. In mammalian hearts, depolarization lasts about 2 ms, but the plateau phase and repolarization last 200 ms or more. Repolarization is therefore not complete until the contraction is half over. As in other excitable tissues, changes in the external K+ concentration affect the resting membrane potential of car-diac muscle, whereas changes in the external Na+ concentra-tion affect the magnitude of the action potential. The initial rapid depolarization and the overshoot (phase 0) are due to opening of voltage-gated Na+ channels similar to that occur-ring in nerve and skeletal muscle (Figure 5–17). The initial rapid repolarization (phase 1) is due to closure of Na+ chan-nels and opening of one type of K+ channel. The subsequent prolonged plateau (phase 2) is due to a slower but prolonged opening of voltage-gated Ca2+ channels. Final repolarization (phase 3) to the resting membrane potential (phase 4) is due to closure of the Ca2+ channels and a slow, delayed increase of K+ efflux through various types of K+ channels. Cardiac myo-cytes contain at least two types of Ca2+ channels (T- and L-types), but the Ca2+ current is due mostly to opening of the slower L-type Ca2+ channels. CHAPTER 5 Excitable Tissue: Muscle 107 MECHANICAL PROPERTIES CONTRACTILE RESPONSE The contractile response of cardiac muscle begins just after the start of depolarization and lasts about 1.5 times as long as the action potential (Figure 5–16). The role of Ca2+ in excitation– contraction coupling is similar to its role in skeletal muscle (see above). However, it is the influx of extracellular Ca2+ through the voltage-sensitive DHPR in the T system that trig-gers calcium-induced calcium release through the RyR at the sarcoplasmic reticulum. Because there is a net influx of Ca2+ during activation, there is also a more prominent role for plas-ma membrane Ca2+ ATPases and the Na+/Ca2+ exchanger in recovery of intracellular Ca2+ concentrations. Specific effects FIGURE 5–15 Cardiac muscle. A) Electron photomicrograph of cardiac muscle. Note the similarity of the A-I regions seen in the skeletal muscle EM of Figure 3-2. The fuzzy thick lines are intercalated disks and function similarly to the Z-lines but occur at cell membranes (× 12,000). (Reproduced with permission from Bloom W, Fawcett DW: A Textbook of Histology, 10th ed. Saunders, 1975.) B) Artist interpretation of cardiac muscle as seen under the light microscope (top) and the electron microscope (bottom). Again, note the similarity to skeletal muscle structure. N, nucleus. (Reproduced with permission from Braunwald E, Ross J, Sonnenblick EH: Mechanisms of contraction of the normal and failing heart. N Engl J Med 1967;277:794. Courtesy of Little, Brown.) SARCOMERE FIBRIL Capillary FIBER Fibrils Sarcolemma 10 μm 2μm Sarcoplasmic reticulum T system Terminal cistern Intercalated disk Mitochondria Nucleus Intercalated disk N N N B A 108 SECTION II Physiology of Nerve & Muscle Cells of drugs that indirectly alter Ca2+ concentrations are dis-cussed in Clinical Box 5–2. During phases 0 to 2 and about half of phase 3 (until the membrane potential reaches approximately –50 mV during repolarization), cardiac muscle cannot be excited again; that is, it is in its absolute refractory period. It remains relatively refractory until phase 4. Therefore, tetanus of the type seen in skeletal muscle cannot occur. Of course, tetanization of car-diac muscle for any length of time would have lethal conse-quences, and in this sense, the fact that cardiac muscle cannot be tetanized is a safety feature. ISOFORMS Cardiac muscle is generally slow and has relatively low ATPase activity. Its fibers are dependent on oxidative metab-olism and hence on a continuous supply of O2. The human heart contains both the α and the β isoforms of the myosin heavy chain (α MHC and β MHC). β MHC has lower myosin ATPase activity than α MHC. Both are present in the atria, with the α isoform predominating, whereas the β isoform pre-dominates in the ventricle. The spatial differences in expres-sion contribute to the well-coordinated contraction of the heart. FIGURE 5–16 Comparison of action potentials and contractile response of a mammalian cardiac muscle fiber in a typical ventricular cell. In the top-most trace, the most commonly viewed surface action potential recording can be seen and it is broken down into four regions: Q, R, S, and T. In the middle trace, the intracellular re-cording of the action potential shows the quick depolarization and ex-tended recovery. In the bottom trace, the mechanical response is matched to the extracellular and intracellular electrical activities. Note that in the absolute refractory period (ARP), the cardiac myocyte can-not be excited, whereas in the relative refractory period (RRP) minimal excitation can occur. 0 100 200 300 ms ARP RRP Mechanical response Action potential recorded intra-cellularly Action potential recorded with surface electrode 0.5 g Q R S T 150 mV FIGURE 5–17 Dissection of the cardiac action potential. Top: The action potential of a cardiac muscle fiber can be broken down into several phases: 0, depolarization; 1, initial rapid repolariza-tion; 2, plateau phase; 3, late rapid repolarization; 4, baseline. Bottom: Diagrammatic summary of Na+, Ca2+, and cumulative K+ currents dur-ing the action potential. As is convention, inward currents are down-ward, and outward currents are upward. CLINICAL BOX 5–2 Glycolysidic Drugs & Cardiac Contractions Oubain and other digitalis glycosides are commonly used to treat failing hearts. These drugs have the effect of in-creasing the strength of cardiac contractions. Although there is discussion as to full mechanisms, a working hy-pothesis is based on the ability of these drugs to inhibit the Na, K ATPase in cell membranes of the cardiomyocytes. The block of the Na, K ATPase in cardiomyocytes would result in an increased intracellular Na+ concentration. Such an in-crease would result in a decreased Na+ influx and hence Ca2+ efflux via the Na+-Ca2+ exchange antiport during the Ca2+ recovery period. The resulting intracellular Ca2+ con-centration increase in turn increases the strength of con-traction of the cardiac muscle. With this mechanism in mind, these drugs can also be quite toxic. Overinhibition of the Na, K ATPase would result in a depolarized cell that could slow conduction, or even spontaneously activate. Al-ternatively, overly increased Ca2+ concentration could also have ill effects on cardiomyocyte physiology. INa IK ICa −90 +20 0 1 0 200 0 2 3 4 mV Time (ms) CHAPTER 5 Excitable Tissue: Muscle 109 CORRELATION BETWEEN MUSCLE FIBER LENGTH & TENSION The relation between initial fiber length and total tension in cardiac muscle is similar to that in skeletal muscle; there is a resting length at which the tension developed on stimulation is maximal. In the body, the initial length of the fibers is deter-mined by the degree of diastolic filling of the heart, and the pressure developed in the ventricle is proportionate to the vol-ume of the ventricle at the end of the filling phase (Starling’s law of the heart). The developed tension (Figure 5–18) in-creases as the diastolic volume increases until it reaches a max-imum, then tends to decrease. However, unlike skeletal muscle, the decrease in developed tension at high degrees of stretch is not due to a decrease in the number of cross-bridges between actin and myosin, because even severely dilated hearts are not stretched to this degree. The decrease is due in-stead to beginning disruption of the myocardial fibers. The force of contraction of cardiac muscle can be also increased by catecholamines, and this increase occurs without a change in muscle length. This positive ionotropic effect of catecholamines is mediated via innervated β1-adrenergic receptors, cyclic AMP, and their effects on Ca2+ homeostasis. The heart also contains noninnervated β2-adrenergic recep-tors, which also act via cyclic AMP, but their ionotropic effect is smaller and is maximal in the atria. Cyclic AMP activates protein kinase A, and this leads to phosphorylation of the voltage-dependent Ca2+ channels, causing them to spend more time in the open state. Cyclic AMP also increases the active transport of Ca2+ to the sarcoplasmic reticulum, thus accelerating relaxation and consequently shortening systole. This is important when the cardiac rate is increased because it permits adequate diastolic filling (see Chapter 31). METABOLISM Mammalian hearts have an abundant blood supply, numerous mitochondria, and a high content of myoglobin, a muscle pig-ment that can function as an O2 storage mechanism. Normally, less than 1% of the total energy liberated is provided by anaero-bic metabolism. During hypoxia, this figure may increase to nearly 10%; but under totally anaerobic conditions, the energy liberated is inadequate to sustain ventricular contractions. Un-der basal conditions, 35% of the caloric needs of the human heart are provided by carbohydrate, 5% by ketones and amino acids, and 60% by fat. However, the proportions of substrates utilized vary greatly with the nutritional state. After ingestion of large amounts of glucose, more lactate and pyruvate are used; during prolonged starvation, more fat is used. Circulating free fatty acids normally account for almost 50% of the lipid utilized. In untreated diabetics, the carbohydrate utilization of cardiac muscle is reduced and that of fat is increased. SMOOTH MUSCLE MORPHOLOGY Smooth muscle is distinguished anatomically from skeletal and cardiac muscle because it lacks visible cross-striations. Actin and myosin-II are present, and they slide on each other to produce contraction. However, they are not arranged in regular arrays, as in skeletal and cardiac muscle, and so the striations are absent. Instead of Z lines, there are dense bodies in the cytoplasm and attached to the cell membrane, and these are bound by α-actinin to actin filaments. Smooth muscle also contains tropomyosin, but troponin appears to be absent. The isoforms of actin and myosin differ from those in skeletal muscle. A sarcoplasmic reticulum is present, but it is less ex-tensive than those observed in skeletal or cardiac muscle. In general, smooth muscles contain few mitochondria and de-pend, to a large extent, on glycolysis for their metabolic needs. TYPES There is considerable variation in the structure and function of smooth muscle in different parts of the body. In general, smooth muscle can be divided into unitary (or visceral) smooth muscle and multiunit smooth muscle. Unitary smooth muscle occurs in large sheets, has many low-resis-tance gap junctional connections between individual muscle cells, and functions in a syncytial fashion. Unitary smooth muscle is found primarily in the walls of hollow viscera. The musculature of the intestine, the uterus, and the ureters are ex-amples. Multiunit smooth muscle is made up of individual units with few (or no) gap junctional bridges. It is found in structures such as the iris of the eye, in which fine, graded con-tractions occur. It is not under voluntary control, but it has many functional similarities to skeletal muscle. Each multi-unit smooth muscle cell has en passant endings of nerve fibers, but in unitary smooth muscle there are en passant junctions on fewer cells, with excitation spreading to other cells by gap FIGURE 5–18 Length–tension relationship for cardiac muscle. Comparison of the systolic intraventricular pressure (top trace) and diastolic intraventricular pressure (bottom trace) display the devel-oped tension in the cardiomyocyte. Values shown are for canine heart. Systolic intraventricular pressure Diastolic intraventricular pressure Diastolic volume (mL) Developed tension Pressure (mm Hg) 270 240 210 180 150 120 90 60 30 0 10 20 30 40 50 60 70 110 SECTION II Physiology of Nerve & Muscle Cells junctions. In addition, these cells respond to hormones and other circulating substances. Blood vessels have both unitary and multiunit smooth muscle in their walls. ELECTRICAL & MECHANICAL ACTIVITY Unitary smooth muscle is characterized by the instability of its membrane potential and by the fact that it shows continuous, irregular contractions that are independent of its nerve sup-ply. This maintained state of partial contraction is called to-nus, or tone. The membrane potential has no true “resting” value, being relatively low when the tissue is active and higher when it is inhibited, but in periods of relative quiescence val-ues for resting potential are on the order of –20 to –65 mV. Smooth muscle cells can display divergent electrical activity (eg, Figure 5–19). There are slow sine wave-like fluctuations a few millivolts in magnitude and spikes that sometimes over-shoot the zero potential line and sometimes do not. In many tissues, the spikes have a duration of about 50 ms, whereas in some tissues the action potentials have a prolonged plateau during repolarization, like the action potentials in cardiac muscle. As in the other muscle types, there are significant con-tributions of K+, Na+, and Ca2+ channels and Na, K ATPase to this electrical activity. However, discussion of contributions to individual smooth muscle types is beyond the scope of this text. Because of the continuous activity, it is difficult to study the relation between the electrical and mechanical events in uni-tary smooth muscle, but in some relatively inactive prepara-tions, a single spike can be generated. In such preparations the excitation–contraction coupling in unitary smooth muscle can occur with as much as a 500-ms delay. Thus, it is a very slow process compared with that in skeletal and cardiac mus-cle, in which the time from initial depolarization to initiation of contraction is less than 10 ms. Unlike unitary smooth mus-cle, multiunit smooth muscle is nonsyncytial and contrac-tions do not spread widely through it. Because of this, the contractions of multiunit smooth muscle are more discrete, fine, and localized than those of unitary smooth muscle. MOLECULAR BASIS OF CONTRACTION As in skeletal and cardiac muscle, Ca2+ plays a prominent role in the initiation of contraction of smooth muscle. How-ever, the source of Ca2+ increase can be much different in unitary smooth muscle. Depending on the activating stimu-lus, Ca2+ increase can be due to influx through voltage- or ligand-gated plasma membrane channels, efflux from intra-cellular stores through the RyR, efflux from intracellular stores through the inositol trisphosphate receptor (IP3R) Ca2+ channel, or via a combination of these channels. In ad-dition, the lack of troponin in smooth muscle prevents Ca2+ activation via troponin binding. Rather, myosin in smooth muscle must be phosphorylated for activation of the myosin ATPase. Phosphorylation and dephosphorylation of myosin also occur in skeletal muscle, but phosphorylation is not nec-essary for activation of the ATPase. In smooth muscle, Ca2+ binds to calmodulin, and the resulting complex activates cal-modulin-dependent myosin light chain kinase. This en-zyme catalyzes the phosphorylation of the myosin light chain on serine at position 19. The phosphorylation increases the ATPase activity. Myosin is dephosphorylated by myosin light chain phos-phatase in the cell. However, dephosphorylation of myosin light chain kinase does not necessarily lead to relaxation of the smooth muscle. Various mechanisms are involved. One appears to be a latch bridge mechanism by which myosin cross-bridges remain attached to actin for some time after the cytoplasmic Ca2+ concentration falls. This produces sustained contraction with little expenditure of energy, which is espe-cially important in vascular smooth muscle. Relaxation of the muscle presumably occurs when the Ca2+-calmodulin com-plex finally dissociates or when some other mechanism comes into play. The events leading to contraction and relaxation of unitary smooth muscle are summarized in Figure 5–20. The events in multiunit smooth muscle are generally similar. Unitary smooth muscle is unique in that, unlike other types of muscle, it contracts when stretched in the absence of any extrinsic innervation. Stretch is followed by a decline in mem-brane potential, an increase in the frequency of spikes, and a general increase in tone. If epinephrine or norepinephrine is added to a preparation of intestinal smooth muscle arranged for recording of intra-cellular potentials in vitro, the membrane potential usually becomes larger, the spikes decrease in frequency, and the muscle relaxes (Figure 5–21). Norepinephrine is the chemical mediator released at noradrenergic nerve endings, and stimu-lation of the noradrenergic nerves to the preparation pro-duces inhibitory potentials. Acetylcholine has an effect opposite to that of norepinephrine on the membrane poten-tial and contractile activity of intestinal smooth muscle. If acetylcholine is added to the fluid bathing a smooth muscle preparation in vitro, the membrane potential decreases and the spikes become more frequent. The muscle becomes more active, with an increase in tonic tension and the number of rhythmic contractions. The effect is mediated by phospholipase FIGURE 5–19 Electrical activity of individual smooth muscle cells in the guinea pig taenia coli. Left: Pacemaker-like activity with spikes firing at each peak. Right: Sinusoidal fluctuation of membrane potential with firing on the rising phase of each wave. In other fibers, spikes can occur on the falling phase of sinusoidal fluctuations and there can be mixtures of sinusoidal and pacemaker potentials in the same fiber. 4 s 50 mV CHAPTER 5 Excitable Tissue: Muscle 111 C, which produces IP3 and allows for Ca2+ release through IP3 receptors. In the intact animal, stimulation of cholinergic nerves causes release of acetylcholine, excitatory potentials, and increased intestinal contractions. Like unitary smooth muscle, multiunit smooth muscle is very sensitive to circulating chemical substances and is nor-mally activated by chemical mediators (acetylcholine and norepinephrine) released at the endings of its motor nerves. Norepinephrine in particular tends to persist in the muscle and to cause repeated firing of the muscle after a single stimu-lus rather than a single action potential. Therefore, the con-tractile response produced is usually an irregular tetanus rather than a single twitch. When a single twitch response is obtained, it resembles the twitch contraction of skeletal mus-cle except that its duration is 10 times as long. RELAXATION In addition to cellular mechanisms that increase contraction of smooth muscle, there are cellular mechanisms that lead to its relaxation (Clinical Box 5–3). This is especially important in smooth muscle that surrounds the blood vessels to increase blood flow. It was long known that endothelial cells that line the inside of blood cells could release a substance that relaxed smooth muscle (endothelial derived relaxation factor, FIGURE 5–20 Sequence of events in contraction and relaxation of smooth muscle. Flow chart illustrates many of the mo-lecular changes that occur from the initiation of contraction to its re-laxation. Note the distinct differences from skeletal and cardiac muscle excitation. FIGURE 5–21 Effects of various agents on the membrane potential of intestinal smooth muscle. Drugs and hormones can al-ter firing of smooth muscle action potentials by raising (top trace) or lowering (bottom trace) resting membrane potential. Binding of acetylcholine to muscarinic receptors Activation of calmodulin-dependent myosin light chain kinase Increased myosin ATPase activity and binding of myosin to actin Contraction Phosphorylation of myosin Increased influx of Ca2+ into the cell Dephosphorylation of myosin by myosin light chain phosphatase Relaxation, or sustained contraction due to the latch bridge and other mechanisms Acetylcholine, parasympathetic stimulation, cold, stretch Membrane potential Epinephrine, sympathetic stimulation mV 0 −50 CLINICAL BOX 5–3 Common Drugs That Act on Smooth Muscle Overexcitation of smooth muscle in the airways, such as that observed during an asthma attack, can lead to bron-choconstriction. Inhalers that deliver drugs to the conduct-ing airway are commonly used to offset this smooth muscle bronchoconstriction, as well as other symptoms in the asthmatic airways. The rapid effects of drugs in inhalers are related to smooth muscle relaxation. Rapid response in-haler drugs (eg, ventolin, albuterol, sambuterol) frequently target β-adrenergic receptors in the airway smooth muscle to elicit a relaxation. Although these β-adrenergic receptor agonists targeting the smooth muscle do not treat all symptoms associated with bronchial constriction (eg, in-flammation and increased mucus), they are quick and fre-quently allow for sufficient opening of the conducting air-way to restore airflow, and thus allow for other treatments to reduce airway obstruction. Smooth muscle is also a target for drugs developed to in-crease blood flow. As discussed in the text, NO is a natural signaling molecule that relaxes smooth muscle by raising cGMP. This signaling pathway is naturally down-regulated by the action of phosphodiesterase (PDE), which trans-forms cGMP into a nonsignaling form, GMP. The drugs sildenafil, tadalafil, and vardenafil are all specific inhibitors of PDE V, an isoform found mainly in the smooth muscle in the corpus cavernosum of the penis (see Chapter 25). Thus, oral administration of these drugs can block the action of PDE V, increasing blood flow in a very limited region in the body and offsetting erectile dysfunction. 112 SECTION II Physiology of Nerve & Muscle Cells EDRF). EDRF was later identified as the gaseous second mes-senger molecule, nitric oxide (NO). NO produced in endothe-lial cells is free to diffuse into the smooth muscle for its effects. Once in muscle, NO directly activates a soluble guanylate cycla-se to produce another second messenger molecule, cyclic guan-osine monophosphate (cGMP). This molecule can activate cGMP-specific protein kinases that can affect ion channels, Ca2+ homeostasis, or phosphatases, or all of those mentioned, that lead to smooth muscle relaxation (see Chapters 7 and 33). FUNCTION OF THE NERVE SUPPLY TO SMOOTH MUSCLE The effects of acetylcholine and norepinephrine on unitary smooth muscle serve to emphasize two of its important prop-erties: (1) its spontaneous activity in the absence of nervous stimulation, and (2) its sensitivity to chemical agents released from nerves locally or brought to it in the circulation. In mam-mals, unitary muscle usually has a dual nerve supply from the two divisions of the autonomic nervous system. The function of the nerve supply is not to initiate activity in the muscle but rather to modify it. Stimulation of one division of the auto-nomic nervous system usually increases smooth muscle activ-ity, whereas stimulation of the other decreases it. However, in some organs, noradrenergic stimulation increases and cholin-ergic stimulation decreases smooth muscle activity; in others, the reverse is true. FORCE GENERATION & PLASTICITY OF SMOOTH MUSCLE Smooth muscle displays a unique economy when compared to skeletal muscle. Despite approximately 20% of the myosin content and a 100-fold difference in ATP use when compared with skeletal muscle, they can generate similar force per cross-sectional area. One of the tradeoffs of obtaining force under these conditions is the noticeably slower contractions when compared to skeletal muscle. There are several known reasons for these noticeable changes, including unique isoforms of myosin and contractile-related proteins expressed in smooth muscle and their distinct regulation (discussed above). The unique architecture of the smooth cell and its coordinated units also likely contribute to these changes. Another special characteristic of smooth muscle is the vari-ability of the tension it exerts at any given length. If a unitary smooth muscle is stretched, it first exerts increased tension. However, if the muscle is held at the greater length after stretch-ing, the tension gradually decreases. Sometimes the tension falls to or below the level exerted before the muscle was stretched. It is consequently impossible to correlate length and developed tension accurately, and no resting length can be assigned. In some ways, therefore, smooth muscle behaves more like a vis-cous mass than a rigidly structured tissue, and it is this property that is referred to as the plasticity of smooth muscle. The consequences of plasticity can be demonstrated in humans. For example, the tension exerted by the smooth muscle walls of the bladder can be measured at different degrees of distention as fluid is infused into the bladder via a catheter. Initially, tension increases relatively little as volume is increased because of the plasticity of the bladder wall. How-ever, a point is eventually reached at which the bladder con-tracts forcefully (see Chapter 38). CHAPTER SUMMARY ■There are three main types of muscle cells: skeletal, cardiac, and smooth. ■Skeletal muscle is a true syncytium under voluntary control. Skeletal muscles receive electrical stimuli from neurons to elicit contraction: “excitation–contraction coupling.” Action poten-tials in muscle cells are developed largely through coordination of Na+, K+, and Ca2+ channels. Contraction in skeletal muscle cells is coordinated through Ca2+ regulation of the actomyosin system that gives the muscle its classic striated pattern under the microscope. ■There are several different types of skeletal muscle fibers (I, IIA, IIB) that have distinct properties in terms of protein makeup and force generation. Skeletal muscle fibers are arranged into motor units of like fibers within a muscle. Skeletal motor units are recruited in a specific pattern as the need for more force is increased. ■Cardiac muscle is a collection of individual cells (cardiomyo-cytes) that are linked as a syncytium by gap junctional commu-nication. Cardiac muscle cells also undergo excitation– contraction coupling. Pacemaker cells in the heart can initiate propagated action potentials. Cardiac muscle cells also have a striated, actomyosin system that underlies contraction. ■Smooth muscle cells are largely under control of the autonomic nervous system. ■There are two broad categories of smooth muscle cells: unitary and multiunit. Unitary smooth muscle contraction is synchro-nized by gap junctional communication to coordinate contrac-tion among many cells. Multiunit smooth muscle contraction is coordinated by motor units, functionally similar to skeletal muscle. ■Smooth muscle cells contract through an actomyosin system, but do not have well-organized striations. Unlike skeletal and cardiac muscle, Ca2+ regulation of contraction is primarily through phosphorylation–dephosphorylation reactions. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. The action potential of skeletal muscle A) has a prolonged plateau phase. B) spreads inward to all parts of the muscle via the T tubules. C) causes the immediate uptake of Ca2+ into the lateral sacs of the sarcoplasmic reticulum. D) is longer than the action potential of cardiac muscle. E) is not essential for contraction. CHAPTER 5 Excitable Tissue: Muscle 113 2. The functions of tropomyosin in skeletal muscle include A) sliding on actin to produce shortening. B) releasing Ca2+ after initiation of contraction. C) binding to myosin during contraction. D) acting as a “relaxing protein” at rest by covering up the sites where myosin binds to actin. E) generating ATP, which it passes to the contractile mechanism. 3. The cross-bridges of the sarcomere in skeletal muscle are made up of A) actin. B) myosin. C) troponin. D) tropomyosin. E) myelin. 4. The contractile response in skeletal muscle A) starts after the action potential is over. B) does not last as long as the action potential. C) produces more tension when the muscle contracts isometri-cally than when the muscle contracts isotonically. D) produces more work when the muscle contracts isometri-cally than when the muscle contracts isotonically. E) decreases in magnitude with repeated stimulation. 5. Gap junctions A) are absent in cardiac muscle. B) are present but of little functional importance in cardiac muscle. C) are present and provide the pathway for rapid spread of exci-tation from one cardiac muscle fiber to another. D) are absent in smooth muscle. E) connect the sarcotubular system to individual skeletal mus-cle cells. CHAPTER RESOURCES Alberts B, et al: Molecular Biology of the Cell, 5th ed. Garland Science, 2007. Fung YC: Biomechanics, 2nd ed. Springer, 1993. Hille B: Ionic Channels of Excitable Membranes, 3rd ed. Sinaver Associates, 2001. Horowitz A: Mechanisms of smooth muscle contraction. Physiol Rev 1996;76:967. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. Sperelakis N (editor): Cell Physiology Sourcebook, 3rd ed. Academic Press, 2001. This page intentionally left blank 115 C H A P T E R 6 Synaptic & Junctional Transmission O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the main morphologic features of synapses. ■Distinguish between chemical and electrical transmission at synapses. ■Define convergence and divergence in neural networks, and discuss their implications. ■Describe fast and slow excitatory and inhibitory postsynaptic potentials, outline the ionic fluxes that underlie them, and explain how the potentials interact to generate action potentials. ■Define and give examples of direct inhibition, indirect inhibition, presynaptic inhi-bition, and postsynaptic inhibition. ■Describe the neuromuscular junction, and explain how action potentials in the motor neuron at the junction lead to contraction of the skeletal muscle. ■Define and explain denervation hypersensitivity. INTRODUCTION The all-or-none type of conduction seen in axons and skeletal muscle has been discussed in Chapters 4 and 5. Impulses are transmitted from one nerve cell to another cell at synapses (Fig-ure 6–1). These are the junctions where the axon or some other portion of one cell (the presynaptic cell) terminates on the den-drites, soma, or axon of another neuron (Figure 6–2) or, in some cases, a muscle or gland cell (the postsynaptic cell). Cell-to-cell communication occurs across either a chemical or electrical syn-apse. At chemical synapses, a synaptic cleft separates the terminal of the presynaptic cell from the postsynaptic cell. An impulse in the presynaptic axon causes secretion of a chemical that diffuses across the synaptic cleft and binds to receptors on the surface of the postsynaptic cell. This triggers events that open or close chan-nels in the membrane of the postsynaptic cell. In electrical syn-apses, the membranes of the presynaptic and postsynaptic neurons come close together, and gap junctions form between the cells (see Chapter 2). Like the intercellular junctions in other tis-sues, these junctions form low-resistance bridges through which ions can pass with relative ease. There are also a few conjoint syn-apses in which transmission is both electrical and chemical. Regardless of the type of synapse, transmission is not a sim-ple jumping of an action potential from the presynaptic to the postsynaptic cell. The effects of discharge at individual synaptic endings can be excitatory or inhibitory, and when the postsyn-aptic cell is a neuron, the summation of all the excitatory and inhibitory effects determines whether an action potential is generated. Thus, synaptic transmission is a complex process that permits the grading and adjustment of neural activity nec-essary for normal function. Because most synaptic transmis-sion is chemical, consideration in this chapter is limited to chemical transmission unless otherwise specified. Transmission from nerve to muscle resembles chemical syn-aptic transmission from one neuron to another. The neuromus-cular junction, the specialized area where a motor nerve terminates on a skeletal muscle fiber, is the site of a stereotyped transmission process. The contacts between autonomic neurons and smooth and cardiac muscle are less specialized, and trans-mission in these locations is a more diffuse process. These forms of transmission are also considered in this chapter. 116 SECTION II Physiology of Nerve & Muscle Cells SYNAPTIC TRANSMISSION: FUNCTIONAL ANATOMY TYPES OF SYNAPSES The anatomic structure of synapses varies considerably in the different parts of the mammalian nervous system. The ends of the presynaptic fibers are generally enlarged to form terminal boutons (synaptic knobs) (Figure 6–2). In the cerebral and cerebellar cortex, endings are commonly located on dendrites and frequently on dendritic spines, which are small knobs projecting from dendrites (Figure 6–3). In some instances, the terminal branches of the axon of the presynaptic neuron form a basket or net around the soma of the postsynaptic cell (basket cells of the cerebellum and autonomic ganglia). In other loca-tions, they intertwine with the dendrites of the postsynaptic cell (climbing fibers of the cerebellum) or end on the dendrites di-rectly (apical dendrites of cortical pyramidal cells). Some end on axons of postsynaptic neurons (axoaxonal endings). On av-erage, each neuron divides to form over 2000 synaptic endings, and because the human central nervous system (CNS) has 1011 neurons, it follows that there are about 2 × 1014 synapses. Ob-viously, therefore, communication between neurons is ex-tremely complex. It should be noted as well that synapses are dynamic structures, increasing and decreasing in complexity and number with use and experience. It has been calculated that in the cerebral cortex, 98% of the synapses are on dendrites and only 2% are on cell bodies. In the spinal cord, the proportion of endings on dendrites is less; there are about 8000 endings on the dendrites of a typical spi-nal neuron and about 2000 on the cell body, making the soma appear encrusted with endings. PRESYNAPTIC & POSTSYNAPTIC STRUCTURE & FUNCTION Each presynaptic terminal of a chemical synapse is separated from the postsynaptic structure by a synaptic cleft that is 20 to 40 nm wide. Across the synaptic cleft are many neurotrans-mitter receptors in the postsynaptic membrane, and usually a postsynaptic thickening called the postsynaptic density (Fig-ures 6–2 and 6–3). The postsynaptic density is an ordered complex of specific receptors, binding proteins, and enzymes induced by postsynaptic effects. Inside the presynaptic terminal are many mitochondria, as well as many membrane-enclosed vesicles, which contain neu-rotransmitters. There are three kinds of synaptic vesicles: small, clear synaptic vesicles that contain acetylcholine, glycine, GABA, or glutamate; small vesicles with a dense core that contain cate-cholamines; and large vesicles with a dense core that contain FIGURE 6–1 Synapses on a typical motor neuron. The neuron has dendrites (1), an axon (2), and a prominent nucleus (3). Note that rough endoplasmic reticulum extends into the dendrites but not into the axon. Many different axons converge on the neuron, and their ter-minal boutons form axodendritic (4) and axosomatic (5) synapses. (6) Myelin sheath. (Reproduced with permission from Krstic RV: Ultrastructure of the Mammalian Cell. Springer, 1979.) 2 6 6 1 5 4 3 1 FIGURE 6–2 Electron photomicrograph of synaptic knob (S) ending on the shaft of a dendrite (D) in the central nervous system. P, postsynaptic density; M, mitochondrion. (×56,000). (Courtesy of DM McDonald.) M S S P P D CHAPTER 6 Synaptic & Junctional Transmission 117 neuropeptides. The vesicles and the proteins contained in their walls are synthesized in the neuronal cell body and transported along the axon to the endings by fast axoplasmic transport. The neuropeptides in the large dense-core vesicles must also be pro-duced by the protein-synthesizing machinery in the cell body. However, the small clear vesicles and the small dense-core vesi-cles recycle in the nerve ending. These vesicles fuse with the cell membrane and release transmitters through exocytosis and are then recovered by endocytosis to be refilled locally. In some instances, they enter endosomes and are budded off the endo-some and refilled, starting the cycle over again. The steps involved are shown in Figure 6–4. More commonly, however, the synaptic vesicle discharges its contents through a small hole in the cell membrane, then the opening reseals rapidly and the main vesicle stays inside the cell (kiss-and-run discharge). In this way, the full endocytotic process is short-circuited. The large dense-core vesicles are located throughout the pre-synaptic terminals that contain them and release their neu-ropeptide contents by exocytosis from all parts of the terminal. On the other hand, the small vesicles are located near the syn-aptic cleft and fuse to the membrane, discharging their contents very rapidly into the cleft at areas of membrane thickening called active zones (Figure 6–3). The active zones contain many proteins and rows of calcium channels. The Ca2+ that triggers exocytosis of transmitters enters the pre-synaptic neurons, and transmitter release starts within 200 μs. Therefore, it is not surprising that the voltage-gated Ca2+ channels are very close to the release sites at the active zones. In addition, for the transmitter to be effective on the postsynaptic neuron requires proximity of release to the postsynaptic receptors. This orderly organization of the synapse depends in part on neurexins, proteins bound to the membrane of the presynaptic neuron that bind neurexin receptors in the membrane of the postsynaptic neuron. In many vertebrates, neurexins are produced by a single gene that codes for the α isoform. However, in mice and humans they are encoded by three genes, and both α and β isoforms are produced. Each of the genes has two regulatory regions and extensive alternative splicing of their mRNAs. In this way, over 1000 different neurexins are produced. This raises the possibility that the neurexins not only hold synapses together, but also pro-vide a mechanism for the production of synaptic specificity. As noted in Chapter 2, vesicle budding, fusion, and dis-charge of contents with subsequent retrieval of vesicle mem-brane are fundamental processes occurring in most, if not all, cells. Thus, neurotransmitter secretion at synapses and the accompanying membrane retrieval are specialized forms of the general processes of exocytosis and endocytosis. The details of the processes by which synaptic vesicles fuse with the cell membrane are still being worked out. They involve the v-snare protein synaptobrevin in the vesicle membrane locking with the t-snare protein syntaxin in the cell membrane; a multipro-tein complex regulated by small GTPases such as rab3 is also involved in the process (Figure 6–5). The synapse begins in the presynaptic and not in the postsynaptic cell. The one-way gate at the synapses is necessary for orderly neural function. Clinical Box 6–1 describes the how neurotoxins can disrupt transmitter release in either the CNS or at the neuromuscular junction. ELECTRICAL EVENTS IN POSTSYNAPTIC NEURONS EXCITATORY & INHIBITORY POSTSYNAPTIC POTENTIALS Penetration of an α-motor neuron is a good example of the techniques used to study postsynaptic electrical activity. It is FIGURE 6–3 Axodendritic, axoaxonal, and axosomatic synapses. Many presynaptic neurons terminate on dendritic spines, as shown at the top, but some also end directly on the shafts of den-drites. Note the presence of clear and granulated synaptic vesicles in endings and clustering of clear vesicles at active zones. Presynaptic cell Postsynaptic cell Axo-axonal Microtubules Mitochondria Clear vesicles Active zone Axon Axodendritic Axodendritic Postsynaptic density Soma Axosomatic Dendrite Dendritic spine 118 SECTION II Physiology of Nerve & Muscle Cells achieved by advancing a microelectrode through the ventral portion of the spinal cord. Puncture of a cell membrane is sig-naled by the appearance of a steady 70-mV potential differ-ence between the microelectrode and an electrode outside the cell. The cell can be identified as a spinal motor neuron by stimulating the appropriate ventral root and observing the electrical activity of the cell. Such stimulation initiates an anti-dromic impulse (see Chapter 4) that is conducted to the soma and stops at this point. Therefore, the presence of an action potential in the cell after antidromic stimulation indicates that the cell that has been penetrated is an α-motor neuron. Stim-ulation of a dorsal root afferent (sensory neuron) can be used to study both excitatory and inhibitory events in α-motor neurons (Figure 6–6). When an impulse reaches the presynaptic terminals, an interval of at least 0.5 ms, the synaptic delay, occurs before a response is obtained in the postsynaptic neuron. It is due to the time it takes for the synaptic mediator to be released and to act on the membrane of the postsynaptic cell. Because of it, conduction along a chain of neurons is slower if many syn-apses are in the chain than if there are only a few. Because the minimum time for transmission across one synapse is 0.5 ms, it is also possible to determine whether a given reflex pathway is monosynaptic or polysynaptic (contains more than one synapse) by measuring the delay in transmission from the dorsal to the ventral root across the spinal cord. A single stimulus applied to the sensory nerves character-istically does not lead to the formation of a propagated action potential in the postsynaptic neuron. Instead, the stimulation produces either a transient partial depolarization or a tran-sient hyperpolarization. The initial depolarizing response produced by a single stimulus to the proper input begins about 0.5 ms after the afferent impulse enters the spinal cord. It reaches its peak 11.5 ms later and then declines exponentially. During this potential, the excitability of the neuron to other stimuli is increased, and consequently the potential is called an excitatory postsynaptic potential (EPSP) (Figure 6–6). FIGURE 6–4 Small synaptic vesicle cycle in presynaptic nerve terminals. Vesicles bud off the early endosome and then fill with neu-rotransmitter (NT; top left). They then move to the plasma membrane, dock, and become primed. Upon arrival of an action potential at the ending, Ca2+ influx triggers fusion and exocytosis of the granule contents to the synaptic cleft. The vesicle wall is then coated with clathrin and taken up by endocytosis. In the cytoplasm, it fuses with the early endosome, and the cycle is ready to repeat. (Reproduced with permission from Sdhof TC: The synaptic vesicle cycle: A cascade of proteinprotein interactions. Nature 1995;375:645. Copyright by Macmillan Magazines.) Early endosome ATP NT NT uptake Translocation Translocation Endosome fusion Budding Docking Priming Fusion/ exocytosis Endocytosis 4 Ca2+ H+ Ca2+ ? Plasma membrane Synaptic cleft Ca2+ FIGURE 6–5 Main proteins that interact to produce synaptic vesicle docking and fusion in nerve endings. (Reproduced with permission from Ferro-Novick S, John R: Vesicle fusion from yeast to man. Nature 1994;370:191. Copyright by Macmillan Magazines.) NSF rab3 GTP munc18/ rbSec1 α/γ SNAPs Syntaxin Synaptobrevin SNAP-25 Synaptic vesicle Plasma membrane Neuron: CHAPTER 6 Synaptic & Junctional Transmission 119 CLINICAL BOX 6–1 Botulinum and Tetanus Toxins Several deadly toxins which block neurotransmitter release are zinc endopeptidases that cleave and hence inactivate pro-teins in the fusion–exocytosis complex. Tetanus toxin and botulinum toxins B, D, F, and G act on synaptobrevin, and botulinum toxin C acts on syntaxin. Botulinum toxins A and B act on SNAP-25. Clinically, tetanus toxin causes spastic paraly-sis by blocking presynaptic transmitter release in the CNS, and botulism causes flaccid paralysis by blocking the release of acetylcholine at the neuromuscular junction. On the positive side, however, local injection of small doses of botulinum toxin (botox) has proved effective in the treatment of a wide variety of conditions characterized by muscle hyperactivity. Examples include injection into the lower esophageal sphinc-ter to relieve achalasia and injection into facial muscles to re-move wrinkles. FIGURE 6–6 Excitatory and inhibitory synaptic connections mediating the stretch reflex provide an example of typical circuits within the CNS. A) The stretch receptor sensory neuron of the quadriceps muscle makes an excitatory connection with the extensor motor neuron of the same muscle and an inhibitory interneuron projecting to flexor motor neurons supplying the antagonistic hamstring muscle. B) Experimental setup to study excitation and inhibition of the extensor motor neuron. Top panel shows two approaches to elicit an excitatory (depolarizing) postsynaptic potential or EPSP in the extensor motor neuron–electrical stimulation of the whole Ia afferent nerve using extracellular electrodes and intracellular current passing through an electrode in-serted into the cell body of a sensory neuron. Bottom panel shows that current passing through an inhibitory interneuron elicits an inhibitory (hyperpolariz-ing) postsynaptic potential or IPSP in the flexor motor neuron. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Quadriceps (extensor) Hamstring (flexor) Inhibitory interneuron Spinal cord Extensor motor neuron Current passing Recording Recording Extracellular stimulating electrodes Ia afferent fibers from muscle spindles of quadriceps Sensory neuron Extensor motor neuron Sensory neuron Muscle spindle A Stretch reflex circuit for knee jerk B Experimental setup for recording from cells in the circuit IPSP Motor neuron Interneuron Motor neuron Sensory neuron EPSP Current passing Recording EPSP EPSP IPSP Ia afferent fibers from muscle spindles of quadriceps Inhibitory interneurons Flexor motor neuron Recording Flexor motor neuron 120 SECTION II Physiology of Nerve & Muscle Cells The EPSP is produced by depolarization of the postsynaptic cell membrane immediately under the presynaptic ending. The excitatory transmitter opens Na+ or Ca2+ ion channels in the postsynaptic membrane, producing an inward current. The area of current flow thus created is so small that it does not drain off enough positive charge to depolarize the whole membrane. Instead, an EPSP is inscribed. The EPSP due to activity in one synaptic knob is small, but the depolarizations produced by each of the active knobs summate. EPSPs are produced by stimulation of some inputs, but stimulation of other inputs produces hyperpolarizing responses. Like the EPSPs, they peak 11.5 ms after the stim-ulus and decrease exponentially. During this potential, the excitability of the neuron to other stimuli is decreased; con-sequently, it is called an inhibitory postsynaptic potential (IPSP) (Figure 6–6). An IPSP can be produced by a localized increase in Cl– transport. When an inhibitory synaptic knob becomes active, the released transmitter triggers the opening of Cl– channels in the area of the postsynaptic cell membrane under the knob. Cl– moves down its concentration gradient. The net effect is the transfer of negative charge into the cell, so that the membrane potential increases. The decreased excitability of the nerve cell during the IPSP is due to movement of the membrane potential away from the firing level. Consequently, more excitatory (depo-larizing) activity is necessary to reach the firing level. The fact that an IPSP is mediated by Cl– can be demonstrated by repeating the stimulus while varying the resting membrane potential of the postsynaptic cell. When the membrane potential is at ECl, the potential disappears (Figure 6–7), and at more negative membrane potentials, it becomes positive (reversal potential). Because IPSPs are net hyperpolarizations, they can be pro-duced by alterations in other ion channels in the neuron. For example, they can be produced by opening of K+ channels, with movement of K+ out of the postsynaptic cell, or by clo-sure of Na+ or Ca2+ channels. TEMPORAL & SPATIAL SUMMATION Summation may be temporal or spatial (Figure 6–8). Tempo-ral summation occurs if repeated afferent stimuli cause new EPSPs before previous EPSPs have decayed. A longer time constant for the EPSP allows for a greater opportunity for summation. When activity is present in more than one synap-tic knob at the same time, spatial summation occurs and ac-tivity in one synaptic knob summates with activity in another to approach the firing level. The EPSP is therefore not an all-or-none response but is proportionate in size to the strength of the afferent stimulus. Spatial summation of IPSPs also occurs, as shown by the increasing size of the response, as the strength of an inhibi-tory afferent volley is increased. Temporal summation of IPSPs also occurs. FIGURE 6–7 IPSP is due to increased Cl influx during stimulation. This can be demonstrated by repeating the stimulus while varying the resting membrane potential (RMP) of the postsynap-tic cell. When the membrane potential is at ECl, the potential disap-pears, and at more negative membrane potentials, it becomes positive (reversal potential). FIGURE 6–8 Central neurons integrate a variety of synaptic inputs through temporal and spatial summation. A) The time con-stant of the postsynaptic neuron affects the amplitude of the depolar-ization caused by consecutive EPSPs produced by a single presynaptic neuron. B) The length constant of a postsynaptic cell affects the ampli-tude of two EPSPs produced by two presynaptic neurons, A and B. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) 5 mV 5 ms –100 mV –90 mV EK –70 mV ECl –60 mV RMP –40 mV Recording A Temporal summation Axon B Spatial summation Axon A Synaptic current A A B Synaptic potential Long time constant (100 ms) Short time constant (20 ms) Vm Vm Long length constant (1 mm) Short length constant (0.33 mm) Vm Vm 2 × 10–10 A 2 mV 2 mV 25 ms A Recording A B CHAPTER 6 Synaptic & Junctional Transmission 121 SLOW POSTSYNAPTIC POTENTIALS In addition to the EPSPs and IPSPs described previously, slow EPSPs and IPSPs have been described in autonomic ganglia, cardiac and smooth muscle, and cortical neurons. These postsynaptic potentials have a latency of 100 to 500 ms and last several seconds. The slow EPSPs are generally due to decreases in K+ conductance, and the slow IPSPs are due to increases in K+ conductance. GENERATION OF THE ACTION POTENTIAL IN THE POSTSYNAPTIC NEURON The constant interplay of excitatory and inhibitory activity on the postsynaptic neuron produces a fluctuating membrane potential that is the algebraic sum of the hyperpolarizing and depolarizing activity. The soma of the neuron thus acts as a sort of integrator. When the 10 to 15 mV of depolarization sufficient to reach the firing level is attained, a propagated spike results. However, the discharge of the neuron is slightly more complicated than this. In motor neurons, the portion of the cell with the lowest threshold for the production of a full-fledged action potential is the initial segment, the portion of the axon at and just beyond the axon hillock. This unmyelinat-ed segment is depolarized or hyperpolarized electrotonically by the current sinks and sources under the excitatory and in-hibitory synaptic knobs. It is the first part of the neuron to fire, and its discharge is propagated in two directions: down the axon and back into the soma. Retrograde firing of the soma in this fashion probably has value in wiping the slate clean for subsequent renewal of the interplay of excitatory and inhibito-ry activity on the cell. FUNCTION OF THE DENDRITES For many years, the standard view has been that dendrites are simply the sites of current sources or sinks that electrotonical-ly change the membrane potential at the initial segment; that is, they are merely extensions of the soma that expand the area available for integration. When the dendritic tree of a neuron is extensive and has multiple presynaptic knobs ending on it, there is room for a great interplay of inhibitory and excitatory activity. It is now well established that dendrites contribute to neural function in more complex ways. Action potentials can be recorded in dendrites. In many instances, these are initiated in the initial segment and conducted in a retrograde fashion, but propagated action potentials are initiated in some dendrites. Further research has demonstrated the malleability of den-dritic spines. Dendritic spines appear, change, and even disap-pear over a time scale of minutes and hours, not days and months. Also, although protein synthesis occurs mainly in the soma with its nucleus, strands of mRNA migrate into the den-drites. There, each can become associated with a single ribo-some in a dendritic spine and produce proteins, which alters the effects of input from individual synapses on the spine. Changes in dendritic spines have been implicated in motiva-tion, learning, and long-term memory. ELECTRICAL TRANSMISSION At synaptic junctions where transmission is electrical, the im-pulse reaching the presynaptic terminal generates an EPSP in the postsynaptic cell that, because of the low-resistance bridge between the two, has a much shorter latency than the EPSP at a synapse where transmission is chemical. In conjoint syn-apses, both a short-latency response and a longer-latency, chemically mediated postsynaptic response take place. INHIBITION & FACILITATION AT SYNAPSES DIRECT & INDIRECT INHIBITION Inhibition in the CNS can be postsynaptic or presynaptic. Postsynaptic inhibition during the course of an IPSP is called direct inhibition because it is not a consequence of previous discharges of the postsynaptic neuron. There are various forms of indirect inhibition, which is inhibition due to the ef-fects of previous postsynaptic neuron discharge. For example, the postsynaptic cell can be refractory to excitation because it has just fired and is in its refractory period. During after-hyperpolarization it is also less excitable. In spinal neurons, es-pecially after repeated firing, this after-hyperpolarization may be large and prolonged. POSTSYNAPTIC INHIBITION IN THE SPINAL CORD Various pathways in the nervous system are known to mediate postsynaptic inhibition, and one illustrative example is pre-sented here. Afferent fibers from the muscle spindles (stretch receptors) in skeletal muscle project directly to the spinal mo-tor neurons of the motor units supplying the same muscle (Figure 6–6). Impulses in this afferent fiber cause EPSPs and, with summation, propagated responses in the postsynaptic motor neurons. At the same time, IPSPs are produced in mo-tor neurons supplying the antagonistic muscles which have an inhibitory interneuron interposed between the afferent fiber and the motor neuron. Therefore, activity in the afferent fibers from the muscle spindles excites the motor neurons supplying the muscle from which the impulses come, and inhibits those supplying its antagonists (reciprocal innervation). 122 SECTION II Physiology of Nerve & Muscle Cells PRESYNAPTIC INHIBITION & FACILITATION Another type of inhibition occurring in the CNS is presynap-tic inhibition, a process mediated by neurons whose termi-nals are on excitatory endings, forming axoaxonal synapses (Figure 6–3). The neurons responsible for postsynaptic and presynaptic inhibition are compared in Figure 6–9. Three mechanisms of presynaptic inhibition have been described. First, activation of the presynaptic receptors increases Cl– con-ductance, and this has been shown to decrease the size of the action potentials reaching the excitatory ending (Figure 6–10). This in turn reduces Ca2+ entry and consequently the amount of excitatory transmitter released. Voltage-gated K+ channels are also opened, and the resulting K+ efflux also decreases the Ca2+ influx. Finally, there is evidence for direct inhibition of transmitter release independent of Ca2+ influx into the excita-tory ending. The first transmitter shown to produce presynaptic inhibi-tion was GABA. Acting via GABAA receptors, GABA increases Cl– conductance. GABAB receptors are also present in the spinal cord and appear to mediate presynaptic inhibition via a G protein that produces an increase in K+ conductance. Baclofen, a GABAB agonist, is effective in the treatment of the spasticity of spinal cord injury and multiple sclerosis, particu-larly when administered intrathecally via an implanted pump. Other transmitters also mediate presynaptic inhibition by G protein-mediated effects on Ca2+ channels and K+ channels. Conversely, presynaptic facilitation is produced when the action potential is prolonged (Figure 6–10) and the Ca2+ chan-nels are open for a longer period. The molecular events respon-sible for the production of presynaptic facilitation mediated by serotonin in the sea snail Aplysia have been worked out in detail. Serotonin released at an axoaxonal ending increases intraneuronal cAMP levels, and the resulting phosphorylation of one group of K+ channels closes the channels, slowing repo-larization and prolonging the action potential. ORGANIZATION OF INHIBITORY SYSTEMS Presynaptic and postsynaptic inhibition are usually produced by stimulation of certain systems converging on a given postsynaptic neuron (afferent inhibition). Neurons may also in-hibit themselves in a negative feedback fashion (negative feed-back inhibition). For instance, each spinal motor neuron regularly gives off a recurrent collateral that synapses with an inhibitory interneuron, which terminates on the cell body of the spinal neuron and other spinal motor neurons (Figure 6–11). This particular inhibitory neuron is sometimes called a Ren-shaw cell after its discoverer. Impulses generated in the motor neuron activate the inhibitory interneuron to secrete inhibitory mediators, and this slows or stops the discharge of the motor neuron. Similar inhibition via recurrent collaterals is seen in the cerebral cortex and limbic system. Presynaptic inhibition due to descending pathways that terminate on afferent pathways in the dorsal horn may be involved in the gating of pain transmission. Another type of inhibition is seen in the cerebellum. In this part of the brain, stimulation of basket cells produces IPSPs in the Purkinje cells. However, the basket cells and the Purkinje FIGURE 6–9 Arrangement of neurons producing presynaptic and postsynaptic inhibition. The neuron producing presynaptic inhi-bition is shown ending on an excitatory synaptic knob. Many of these neurons actually end higher up along the axon of the excitatory cell. Motor neuron Postsynaptic inhibition Presynaptic inhibition FIGURE 6–10 Effects of presynaptic inhibition and facilitation on the action potential and the Ca2+ current in the presynaptic neuron and the EPSP in the postsynaptic neuron. In each case, the solid lines are the controls and the dashed lines the records obtained during inhibition or facilitation. (Modified from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) EPSP in postsynaptic neuron Presynaptic action potential Ca2+ current in presynaptic neuron Presynaptic inhibition Presynaptic facilitation CHAPTER 6 Synaptic & Junctional Transmission 123 cells are excited by the same parallel-fiber excitatory input. This arrangement, which has been called feed-forward inhibi-tion, presumably limits the duration of the excitation pro-duced by any given afferent volley. SUMMATION & OCCLUSION As noted above, the axons of most neurons discharge onto many other neurons. Conversely, any given neuron receives input from many other neurons (convergence). In the hypothetical nerve net shown in Figure 6–12, neurons A and B converge on X, and neuron B diverges on X and Y. A stimulus applied to A or to B will set up an EPSP in X. If A and B are stimulated at the same time and action potentials are produced, two areas of depolarization will be produced in X and their actions will sum. The resultant EPSP in X will be twice as large as that produced by stimulation of A or B alone, and the membrane potential may well reach the firing level of X. The effect of the depolarization caused by the impulse in A adds to that due to activity in B, and vice versa; spatial summa-tion has taken place. In this case, Y has not fired, but its excit-ability has been increased, and it is easier for activity in neuron C to fire Y during the EPSP. Y is therefore said to be in the sub-liminal fringe of X. More generally stated, neurons are in the subliminal fringe if they are not discharged by an afferent vol-ley (not in the discharge zone) but do have their excitability increased. The neurons that have few active knobs ending on them are in the subliminal fringe, and those with many are in the discharge zone. Inhibitory impulses show similar temporal and spatial facilitation and subliminal fringe effects. If action potentials are produced repeatedly in neuron B, X and Y will discharge as a result of temporal summation of the EPSPs that are produced. If C is stimulated repeatedly, Y and Z will discharge. If B and C are fired repeatedly at the same time, X, Y, and Z will discharge. Thus, the response to stimu-lation of B and C together is not as great as the sum of responses to stimulation of B and C separately, because B and C both end on neuron Y. This decrease in expected response, due to presynaptic fibers sharing postsynaptic neurons, is called occlusion. NEUROMUSCULAR TRANSMISSION: NEUROMUSCULAR JUNCTION ANATOMY As the axon supplying a skeletal muscle fiber approaches its termination, it loses its myelin sheath and divides into a num-ber of terminal boutons, or endfeet (Figure 6–13). The endfeet contain many small, clear vesicles that contain acetylcholine, the transmitter at these junctions. The endings fit into junc-tional folds, which are depressions in the motor end plate, the thickened portion of the muscle membrane at the junc-tion. The space between the nerve and the thickened muscle membrane is comparable to the synaptic cleft at synapses. The whole structure is known as the neuromuscular, or myoneu-ral, junction. Only one nerve fiber ends on each end plate, with no convergence of multiple inputs. SEQUENCE OF EVENTS DURING TRANSMISSION The events occurring during transmission of impulses from the motor nerve to the muscle are somewhat similar to those occurring at neuron-to-neuron synapses (Figure 6–14). The impulse arriving in the end of the motor neuron increases the permeability of its endings to Ca2+. Ca2+ enters the endings and triggers a marked increase in exocytosis of the acetylcho-line-containing vesicles. The acetylcholine diffuses to the FIGURE 6–11 Negative feedback inhibition of a spinal motor neuron via an inhibitory interneuron (Renshaw cell). Motor neuron Motor neuron Inhibitory interneuron Axon FIGURE 6–12 Simple nerve net. Neurons A, B, and C have exci-tatory endings on neurons X, Y, and Z. A B C X Y Z 124 SECTION II Physiology of Nerve & Muscle Cells muscle-type nicotinic acetylcholine receptors, which are con-centrated at the tops of the junctional folds of the membrane of the motor end plate. Binding of acetylcholine to these re-ceptors increases the Na+ and K+ conductance of the mem-brane, and the resultant influx of Na+ produces a depolarizing potential, the end plate potential. The current sink created by this local potential depolarizes the adjacent muscle membrane to its firing level. Acetylcholine is then removed from the syn-aptic cleft by acetylcholinesterase, which is present in high concentration at the neuromuscular junction. Action poten-tials are generated on either side of the end plate and are con-ducted away from the end plate in both directions along the muscle fiber. The muscle action potential, in turn, initiates muscle contraction, as described in Chapter 5. END PLATE POTENTIAL An average human end plate contains about 15 to 40 million acetylcholine receptors. Each nerve impulse releases about 60 acetylcholine vesicles, and each vesicle contains about 10,000 molecules of the neurotransmitter. This amount is enough to activate about 10 times the number of acetylcholine receptors needed to produce a full end plate potential. Therefore, a propagated response in the muscle is regularly produced, and this large response obscures the end plate potential. However, the end plate potential can be seen if the tenfold safety factor is overcome and the potential is reduced to a size that is insuf-ficient to fire the adjacent muscle membrane. This can be ac-complished by administration of small doses of curare, a drug that competes with acetylcholine for binding to muscle-type nicotinic acetylcholine receptors. The response is then record-ed only at the end plate region and decreases exponentially away from it. Under these conditions, end plate potentials can be shown to undergo temporal summation. QUANTAL RELEASE OF TRANSMITTER Small quanta (packets) of acetylcholine are released randomly from the nerve cell membrane at rest. Each produces a minute depolarizing spike called a miniature end plate potential, which is about 0.5 mV in amplitude. The size of the quanta of acetylcholine released in this way varies directly with the Ca2+ concentration and inversely with the Mg2+ concentration at the end plate. When a nerve impulse reaches the ending, the number of quanta released increases by several orders of mag-nitude, and the result is the large end plate potential that ex-ceeds the firing level of the muscle fiber. Quantal release of acetylcholine similar to that seen at the myoneural junction has been observed at other cholinergic syn-apses, and quantal release of other transmitters probably occurs at noradrenergic, glutaminergic, and other synaptic junctions. Two diseases of the neuromuscular junction, myasthenia gravis and Lambert-Eaton syndrome, are described in Clini-cal Box 6–2 and Clinical Box 6–3, respectively. FIGURE 6–13 The neuromuscular junction. (a) Scanning electronmicrograph showing branching of motor axons with terminals embed-ded in grooves in the muscle fiber’s surface. (b) Structure of a neuromuscular junction. (From Widmaier EP, Raff H, Strang KT: Vanders Human Physiology. McGraw-Hill, 2008.) (a) (b) Myelin Motor nerve fiber Axon terminal Schwann cell Synaptic vesicles (containing ACh) Active zone Sarcolemma Region of sarcolemma with ACh receptors Junctional folds Nucleus of muscle fiber Synaptic cleft CHAPTER 6 Synaptic & Junctional Transmission 125 NERVE ENDINGS IN SMOOTH & CARDIAC MUSCLE ANATOMY The postganglionic neurons in the various smooth muscles that have been studied in detail branch extensively and come in close contact with the muscle cells (Figure 6–15). Some of these nerve fibers contain clear vesicles and are cholinergic, whereas others contain the characteristic dense-core vesicles that contain nor-epinephrine. There are no recognizable end plates or other postsynaptic specializations. The nerve fibers run along the membranes of the muscle cells and sometimes groove their sur-faces. The multiple branches of the noradrenergic and, presum-ably, the cholinergic neurons are beaded with enlargements (varicosities) and contain synaptic vesicles (Figure 6–15). In noradrenergic neurons, the varicosities are about 5 μm apart, with up to 20,000 varicosities per neuron. Transmitter is appar-ently liberated at each varicosity, that is, at many locations along each axon. This arrangement permits one neuron to innervate many effector cells. The type of contact in which a neuron forms a synapse on the surface of another neuron or a smooth muscle cell and then passes on to make similar contacts with other cells is called a synapse en passant. In the heart, cholinergic and noradrenergic nerve fibers end on the sinoatrial node, the atrioventricular node, and the bun-dle of His. Noradrenergic fibers also innervate the ventricular muscle. The exact nature of the endings on nodal tissue is not known. In the ventricle, the contacts between the noradrener-gic fibers and the cardiac muscle fibers resemble those found in smooth muscle. JUNCTIONAL POTENTIALS In smooth muscles in which noradrenergic discharge is exci-tatory, stimulation of the noradrenergic nerves produces dis-crete partial depolarizations that look like small end plate potentials and are called excitatory junction potentials (EJPs). These potentials summate with repeated stimuli. Sim-ilar EJPs are seen in tissues excited by cholinergic discharges. In tissues inhibited by noradrenergic stimuli, hyperpolarizing inhibitory junction potentials (IJPs) are produced by stimu-lation of the noradrenergic nerves. Junctional potentials spread electrotonically. FIGURE 6–14 Events at the neuromuscular junction that lead to an action potential in the muscle fiber plasma membrane. Al-though potassium exits the muscle cell when Ach receptors are open, sodium entry and depolarization dominate. (From Widmaier EP, Raff H, Strang KT: Vanders Human Physiology. McGraw-Hill, 2008.) 3 8 2 4 5 6 7 1 + + Acetylcholine release Motor neuron action potential Muscle fiber action potential initiation Local current between depolarized end plate and adjacent muscle plasma membrane Acetylcholine receptor Acetylcholine degradation Acetylcholinesterase Motor end plate Acetylcholine vesicle Voltage-gated Na+ channels + + + – – – + + + + – + – – – + + – – + – – + – + – + – + – + + – – + – + – Na+ entry Ca2+ enters voltage-gated channels Propagated action potential in muscle plasma membrane 126 SECTION II Physiology of Nerve & Muscle Cells DENERVATION HYPERSENSITIVITY When the motor nerve to skeletal muscle is cut and allowed to degenerate, the muscle gradually becomes extremely sensitive to acetylcholine. This denervation hypersensitivity or super-sensitivity is also seen in smooth muscle. Smooth muscle, un-like skeletal muscle, does not atrophy when denervated, but it becomes hyperresponsive to the chemical mediator that nor-mally activates it. A good example of denervation hypersensitiv-ity is the response of the denervated iris. If the postganglionic sympathetic nerves to one iris are cut in an experimental animal and, after several weeks, norepinephrine (the transmitter re-leased by sympathetic postganglionic neurons) is injected intra-venously, the denervated pupil dilates widely. A much smaller, less prolonged response is observed on the intact side. The reactions triggered by section of an axon are summa-rized in Figure 6–16. Hypersensitivity of the postsynaptic structure to the transmitter previously secreted by the axon endings is a general phenomenon, largely due to the synthesis or activation of more receptors. There is in addition ortho-grade degeneration (wallerian degeneration) and retrograde degeneration of the axon stump to the nearest collateral (sus-taining collateral). A series of changes occur in the cell body that include a decrease in Nissl substance (chromatolysis). The nerve then starts to regrow, with multiple small branches projecting along the path the axon previously fol-lowed (regenerative sprouting). Axons sometimes grow back to their original targets, especially in locations like the neuro-muscular junction. However, nerve regeneration is generally limited because axons often become entangled in the area of tissue damage at the site where they were disrupted. This CLINICAL BOX 6–2 Myasthenia Gravis Myasthenia gravis is a serious and sometimes fatal disease in which skeletal muscles are weak and tire easily. It occurs in 25 to 125 of every 1 million people worldwide and can occur at any age but seems to have a bimodal distribution, with peak occurrences in individuals in their 20s (mainly women) and 60s (mainly men). It is caused by the formation of circu-lating antibodies to the muscle type of nicotinic acetylcho-line receptors. These antibodies destroy some of the recep-tors and bind others to neighboring receptors, triggering their removal by endocytosis. Normally, the number of quanta released from the motor nerve terminal declines with successive repetitive stimuli. In myasthenia gravis, neuro-muscular transmission fails at these low levels of quantal re-lease. This leads to the major clinical feature of the disease– muscle fatigue with sustained or repeated activity. There are two major forms of the disease. In one form, the extraocular muscles are primarily affected. In the second form, there is a generalized weakness of skeletal muscles. Weakness im-proves after a period of rest or after administration of acetyl-cholinesterase inhibitors. Cholinesterase inhibitors pre-vent metabolism of acetylcholine and can thus compensate for the normal decline in released neurotransmitters during repeated stimulation. In severe cases, all muscles, including the diaphragm, can become weak and respiratory failure and death can ensue. The major structural abnormality in myas-thenia gravis is the appearance of sparse, shallow, and ab-normally wide or absent synaptic clefts in the motor end plate. Studies show that the postsynaptic membrane has a reduced response to acetylcholine and a 70–90% decrease in the number of receptors per end plate in affected muscles. Patients with mysathenia gravis have a greater than normal tendency to also have rheumatoid arthritis, systemic lupus erythematosus, and polymyositis. About 30% of mysathenia gravis patients have a maternal relative with an autoimmune disorder. These associations suggest that individuals with myasthenia gravis share a genetic predisposition to autoim-mune disease. The thymus may play a role in the pathogene-sis of the disease by supplying helper T cells sensitized against thymic proteins that cross-react with acetylcholine receptors. In most patients, the thymus is hyperplastic, and 10–15% have thymomas. Thymectomy is indicated if a thy-moma is suspected. Even in those without thymoma, thy-mectomy induces remission in 35% and improves symptoms in another 45% of patients. CLINICAL BOX 6–3 Lambert–Eaton Syndrome Another condition that resembles myasthenia gravis is the relatively rare condition called Lambert–Eaton Syndrome (LEMS). In this condition, muscle weakness is caused by an autoimmune attack against one of the Ca2+ channels in the nerve endings at the neuromuscular junction. This de-creases the normal Ca2+ influx that causes acetylcholine re-lease. Proximal muscles of the lower extremities are primar-ily affected, producing a waddling gait and difficulty raising the arms. Repetitive stimulation of the motor nerve facili-tates accumulation of Ca2+ in the nerve terminal and in-creases acetylcholine release, leading to an increase in muscle strength. This is in contrast to myasthenia gravis in which symptoms are exasperated by repetitive stimulation. About 40% of patients with LEMS also have cancer, espe-cially small cell cancer of the lung. One theory is that anti-bodies that have been produced to attack the cancer cells may also attack Ca2+ channels, leading to LEMS. LEMS has also been associated with lymphosarcoma, malignant thy-moma, and cancer of the breast, stomach, colon, prostate, bladder, kidney, or gall bladder. Clinical signs usually pre-cede the diagnosis of cancer. A syndrome similar to LEMS can occur after the use of aminoglycoside antibiotics, which also impair Ca2+channel function. CHAPTER 6 Synaptic & Junctional Transmission 127 difficulty has been reduced by administration of neurotro-phins. For example, sensory neurons torn when dorsal nerve roots are avulsed from the spinal cord regrow and form func-tional connections in the spinal cord if experimental animals are treated with NGF, neurotrophin 3, or GDNF. Hypersensitivity is limited to the structures immediately innervated by the destroyed neurons and fails to develop in neu-rons and muscle farther downstream. Suprasegmental spinal cord lesions do not lead to hypersensitivity of the paralyzed skeletal muscles to acetylcholine, and destruction of the preganglionic autonomic nerves to visceral structures does not cause hypersensitivity of the denervated viscera. This fact has practical implications in the treatment of diseases due to spasm of the blood vessels in the extremities. For example, if the upper extremity is sympathectomized by removing the upper part of the ganglionic chain and the stellate ganglion, the hypersensitive smooth muscle in the vessel walls is stimu-lated by circulating norepinephrine, and episodic vasospasm continues to occur. However, if preganglionic sympathectomy of the arm is performed by cutting the ganglion chain below the third ganglion (to interrupt ascending preganglionic fibers) and the white rami of the first three thoracic nerves, no hypersensitivity results. Denervation hypersensitivity has multiple causes. As noted in Chapter 2, a deficiency of a given chemical messenger gen-erally produces an upregulation of its receptors. Another fac-tor is lack of reuptake of secreted neurotransmitters. CHAPTER SUMMARY ■Presynaptic terminals are separated from the postsynaptic structure by a synaptic cleft. The postsynaptic membrane con-tains many neurotransmitter receptors and usually a postsynap-tic thickening called the postsynaptic density. ■At chemical synapses, an impulse in the presynaptic axon causes secretion of a chemical that diffuses across the synaptic cleft and binds to postsynaptic receptors, triggering events that open or close channels in the membrane of the postsynaptic cell. At elec-trical synapses, the membranes of the presynaptic and postsyn-aptic neurons come close together, and gap junctions form FIGURE 6–15 Endings of postganglionic autonomic neurons on smooth muscle. Neurotransmitter, released from varicosities along the branched axon, diffuses to receptors on smooth muscle cell plasma membranes. (From Widmaier EP , Raff H, Strang KT: Vanders Human Physiology. McGraw-Hill, 2008.) Varicosity Synaptic vesicles Mitochondrion Autonomic nerve fiber Varicosities Sheet of cells FIGURE 6–16 Summary of changes occurring in a neuron and the structure it innervates when its axon is crushed or cut at the point marked X. Hypersensitivity of the postsynaptic structure to the transmitter previously secreted by the axon occurs largely due to the synthesis or activation of more receptors. There is both orthograde (wallerian) degeneration from the point of damage to the terminal and retrograde degeneration of the axon stump to the nearest collateral (sustaining collateral). Changes also occur in the cell body, including chromatolysis. The nerve starts to regrow, with multiple small branch-es projecting along the path the axon previously followed (regenera-tive sprouting). Axon branch (sustaining collateral) Receptor Receptor hypersensitive Retrograde degeneration Site of injury Retrograde reaction: chromatolysis Regenerative sprouting Orthograde (wallerian) degeneration X 128 SECTION II Physiology of Nerve & Muscle Cells low-resistance bridges through which ions pass with relative ease from one neuron to the next. ■A neuron receives input from many other neurons (conver-gence), and a neuron branches to innervate many other neurons (divergence). ■An EPSP is produced by depolarization of the postsynaptic cell af-ter a latency of 0.5 ms; the excitatory transmitter opens Na+ or Ca2+ ion channels in the postsynaptic membrane, producing an inward current. An IPSP is produced by a hyperpolarization of the postsynaptic cell; it can be produced by a localized increase in Cl– transport. Slow EPSPs and IPSPs occur after a latency of 100 to 500 ms in autonomic ganglia, cardiac and smooth muscle, and cortical neurons. The slow EPSPs are due to decreases in K+ conductance, and the slow IPSPs are due to increases in K+ conductance. ■Postsynaptic inhibition during the course of an IPSP is called di-rect inhibition. Indirect inhibition is due to the effects of previous postsynaptic neuron discharge; for example, the postsynaptic cell cannot be activated during its refractory period. Presynaptic inhi-bition is a process mediated by neurons whose terminals are on excitatory endings, forming axoaxonal synapses; in response to activation of the presynaptic terminal. Activation of the presyn-aptic receptors can increase Cl– conductance, decreasing the size of the action potentials reaching the excitatory ending, and reduc-ing Ca2+ entry and the amount of excitatory transmitter released. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Fast inhibitory postsynaptic potentials (IPSPs) A) are a consequence of decreased Cl– conductance. B) occur in skeletal muscle. C) can be produced by an increase in Na+ conductance. D) can be produced by an increase in Ca2+ conductance. E) interact with other fast and slow potentials to move the membrane potential of the postsynaptic neuron toward or away from the firing level. 2. Fast excitatory postsynaptic potentials (EPSPs) A) are a consequence of decreased Cl– conductance. B) occur in skeletal muscle. C) can be produced by an increase in Na+ conductance. D) can be produced by a decrease in Ca2+ conductance. E) all of the above 3. Initiation of an action potential in skeletal muscle by stimulating its motor nerve A) requires spatial facilitation. B) requires temporal facilitation. C) is inhibited by a high concentration of Ca2+ at the neuro-muscular junction. D) requires the release of norepinephrine. E) requires the release of acetylcholine. 4. A 35-year-old woman sees her physician to report muscle weak-ness in the extraocular eye muscles and muscles of the extremi-ties. She states that she feels fine when she gets up in the morning, but the weakness begins soon after she becomes active. The weakness is improved by rest. Sensation appears normal. The physician treats her with an anticholinesterase inhibitor, and she notes immediate return of muscle strength. Her physician diag-noses her with A) Lambert–Eaton syndrome. B) myasthenia gravis. C) multiple sclerosis. D) Parkinson disease. E) muscular dystrophy. CHAPTER RESOURCES Boron WF, Boulpaep EL: Medical Physiology, Elsevier, 2005. Hille B: Ionic Channels of Excitable Membranes, 3rd ed. Sinauer Associates, 2001. Jessell TM, Kandel ER: Synaptic transmission: A bidirectional and a self-modifiable form of cellcell communication. Cell 1993;72(Suppl):1. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. McPhee SJ, Ganong WF: Pathophysiology of Disease. An Introduction to Clinical Medicine, 5th ed. McGraw-Hill, 2006. Squire LR, et al (editors): Fundamental Neuroscience, 3rd ed., Academic Press, 2008. Unwin N: Neurotransmitter action: Opening of ligand-gated ion channels. Cell 1993; 72(Suppl):31. Van der Kloot W, Molg J: Quantal acetylcholine release at the vertebrate neuromuscular junction. Physiol Rev 1994;74:899. 129 C H A P T E R 7 Neurotransmitters & Neuromodulators O B J E C T I V E S After studying this chapter, you should be able to: ■List neurotransmitters and the principal sites in the nervous system at which they are released. ■Describe the receptors for catecholamines, acetylcholine, 5-HT, amino acids, and opioids. ■Summarize the steps involved in the biosynthesis, release, action, and removal from the synaptic cleft of the various synaptic transmitters. ■Define opioid peptide, list the principal opioid peptides in the body, and name the precursor molecules from which they originate. INTRODUCTION The fact that transmission at most synapses is chemical is of great physiologic and pharmacologic importance. Nerve endings have been called biological transducers that convert electrical energy into chemical energy. In broad terms, this conversion process involves the synthesis of the neurotransmitters, their storage in synaptic vesicles, and their release by the nerve impulses into the synaptic cleft. The secreted transmitters then act on appropriate receptors on the membrane of the postsynaptic cell and are rap-idly removed from the synaptic cleft by diffusion, metabolism, and, in many instances, reuptake into the presynaptic neuron. Some chemicals released by neurons have little or no direct effects on their own but can modify the effects of neurotransmit-ters. These chemicals are called neuromodulators. All these pro-cesses, plus the postreceptor events in the postsynaptic neuron, are regulated by many physiologic factors and at least in theory can be altered by drugs. Therefore, pharmacologists (in theory) should be able to develop drugs that regulate not only somatic and visceral motor activity but also emotions, behavior, and all the other complex functions of the brain. CHEMICAL TRANSMISSION OF SYNAPTIC ACTIVITY CHEMISTRY OF TRANSMITTERS One suspects that a substance is a neurotransmitter if it is un-evenly distributed in the nervous system and its distribution parallels that of its receptors and synthesizing and catabolizing enzymes. Additional evidence includes demonstration that it is released from appropriate brain regions in vitro and that it pro-duces effects on single target neurons when applied to their membranes by means of a micropipette (microiontophoresis). Many transmitters and enzymes involved in their synthesis and catabolism have been localized in nerve endings by immuno-histochemistry, a technique in which antibodies to a given sub-stance are labeled and applied to brain and other tissues. The antibodies bind to the substance, and the location of the sub-stance is then determined by locating the label with the light mi-croscope or electron microscope. In situ hybridization histochemistry, which permits localization of the mRNAs for particular synthesizing enzymes or receptors, has also been a valuable tool. Identified neurotransmitters and neuromodulators can be divided into two major categories: small-molecule transmitters and large-molecule transmitters. Small-molecule transmitters 130 SECTION II Physiology of Nerve & Muscle Cells include monoamines (eg, acetylcholine, serotonin, hista-mine), catecholamines (dopamine, norepinephrine, and epi-nephrine), and amino acids (eg, glutamate, GABA, glycine). Large-molecule transmitters include a large number of pep-tides called neuropeptides including substance P, enkephalin, vasopressin, and a host of others. In general, neuropeptides are colocalized with one of the small-molecule neurotransmitters (Table 7–1). There are also other substances thought to be released into the synaptic cleft to act as either a transmitter or modulator of synaptic transmission. These include purine derivatives like adenosine and adenosine triphosphate (ATP) and a gaseous molecule, nitric oxide (NO). Figure 7–1 shows the biosynthesis of some common small-molecule transmitters released by neurons in the central or peripheral nervous system. Figure 7–2 shows the location of major groups of neurons that contain norepinephrine, epi-nephrine, dopamine, and acetylcholine. These are some of the major neuromodulatory systems. RECEPTORS Cloning and other molecular biology techniques have permit-ted spectacular advances in knowledge about the structure and function of receptors for neurotransmitters and other chemi-cal messengers. The individual receptors, along with their ligands, are discussed in the following parts of this chapter. However, five themes have emerged that should be mentioned in this introductory discussion. First, in every instance studied in detail to date, it has become clear that each ligand has many subtypes of receptors. Thus, for example, norepinephrine acts on α1 and α2 recep-tors, and three of each subtype have been cloned. In addition, there are β1, β2, and β3 receptors. Obviously, this multiplies the possible effects of a given ligand and makes its effects in a given cell more selective. Second, there are receptors on the presynaptic as well as the postsynaptic elements for many secreted transmitters. These presynaptic receptors, or autoreceptors, often inhibit fur-ther secretion of the ligand, providing feedback control. For example, norepinephrine acts on α2 presynaptic receptors to inhibit norepinephrine secretion. However, autoreceptors can also facilitate the release of neurotransmitters. Third, although there are many ligands and many subtypes of receptors for each ligand, the receptors tend to group in large families as far as structure and function are concerned. Many receptors act via trimeric G proteins and protein kinases to produce their effects. Others are ion channels. The receptors for a group of selected, established neurotransmitters and neu-romodulators are listed in Table 7–2, along with their princi-pal second messengers and, where established, their net effect on ion channels. It should be noted that this table is an over-simplification. For example, activation of α2-adrenergic recep-tors decreases intracellular cAMP concentrations, but there is evidence that the G protein activated by α2-adrenergic presyn-aptic receptors also acts directly on Ca2+ channels to inhibit norepinephrine release by decreasing Ca2+ increases. Fourth, receptors are concentrated in clusters in postsynaptic structures close to the endings of neurons that secrete the neu-rotransmitters specific for them. This is generally due to the presence of specific binding proteins for them. In the case of nic-otinic acetylcholine receptors at the neuromuscular junction, the protein is rapsyn, and in the case of excitatory glutamatergic receptors, a family of PB2-binding proteins is involved. GABAA receptors are associated with the protein gephyrin, which also binds glycine receptors, and GABAC receptors are bound to the cytoskeleton in the retina by the protein MAP-1B. At least in the case of GABAA receptors, the binding protein gephyrin is located in clumps in the postsynaptic membrane. With activity, the free receptors move rapidly to the gephyrin and bind to it, creating membrane clusters. Gephyrin binding slows and restricts their further movement. Presumably, during neural inactivity, the receptors are unbound and move again. Fifth, prolonged exposure to their ligands causes most receptors to become unresponsive, that is, to undergo desensi-tization. This can be of two types: homologous desensitiza-tion, with loss of responsiveness only to the particular ligand and maintained responsiveness of the cell to other ligands; and heterologous desensitization, in which the cell becomes unresponsive to other ligands as well. Desensitization in β-adrenergic receptors has been studied in considerable detail. One form involves phosphorylation of the carboxyl terminal TABLE 7–1 Examples of colocalization of small-molecule transmitters with neuropeptides. Small-Molecule Transmitter Neuropeptide Monoamines Acetylcholine Enkephalin, calcitonin-gene-related peptide, galanin, gonadotropin-releasing hormone, neurotensin, somatostatin, substance P, vaso-active intestinal polypeptide Serotonin Cholecystokinin, enkephalin, neuropeptide Y, substance P, vasoactive intestinal polypeptide Catecholamines Dopamine Cholecystokinin, enkephalin, neurotensin Norepinephrine Enkephalin, neuropeptide Y, neurotensin, so-matostatin, vasopressin Epinephrine Enkephalin, neuropeptide Y, neurotensin, substance P Amino Acids Glutamate Substance P Glycine Neurotensin GABA Cholecystokinin, enkephalin, somatostatin, substance P, thyrotropin-releasing hormone CHAPTER 7 Neurotransmitters & Neuromodulators 131 FIGURE 7–1 Biosynthesis of some common small molecule transmitters. (Reproduced with permission from Boron WF, Boulpaep EL: Medical Physiology. Elsevier, 2005.) 132 SECTION II Physiology of Nerve & Muscle Cells region of the receptor by a specific β-adrenergic receptor kinase (β-ARK) or binding β-arrestins. Four β-arrestins have been described in mammals. Two are expressed in rods and cones of the retina and inhibit visual responses. The other two, β-arrestin 1 and β-arrestin 2, are more ubiquitous. They desensitize β-adrenegic receptors, but they also inhibit other heterotrimeric G protein-coupled receptors. In addition, they foster endocytosis of ligands, adding to desensitization. REUPTAKE Neurotransmitters are transported from the synaptic cleft back into the cytoplasm of the neurons that secreted them, a process referred to as reuptake (Figure 7–3). The high-affinity reuptake systems employ two families of transporter proteins. One family has 12 transmembrane domains and cotransports the transmitter with Na+ and Cl–. Members of this family in-clude transporters for norepinephrine, dopamine, serotonin, GABA, and glycine, as well as transporters for proline, taurine, and the acetylcholine precursor choline. In addition, there FIGURE 7–2 Four diffusely connected systems of central neurons using modulatory transmitters. A) Norepinephrine-containing neu-rons. B) Serotonin-containing neurons. C) Dopamine-containing neurons. D) Acetylcholine-containing neurons. (Reproduced with permission from Boron WF, Boulpaep EL: Medical Physiology. Elsevier, 2005.) CHAPTER 7 Neurotransmitters & Neuromodulators 133 may be an epinephrine transporter. The other family is made up of at least three transporters that mediate glutamate uptake by neurons and two that transport glutamate into astrocytes. These glutamate transporters are coupled to the cotransport of Na+ and the countertransport of K+, and they are not depen-dent on Cl– transport. There is a debate about their structure, and they may have 6, 8, or 10 transmembrane domains. One of them transports glutamate into glia rather than neurons (see Chapter 4). There are in addition two vesicular monoamine trans-porters, VMAT1 and VMAT2, that transport neurotrans-mitters from the cytoplasm to synaptic vesicles. They are coded by different genes but have extensive homology. Both have a broad specificity, moving dopamine, norepinephrine, epinephrine, serotonin, and histamine from the cytoplasm into secretory granules. Both are inhibited by reserpine, which accounts for the marked monoamine depletion pro-duced by this drug. Like the neurotransmitter membrane transporter family, they have 12 transmembrane domains, but they have little homology to the other transporters. There is also a vesicular GABA transporter (VGAT) that moves GABA and glycine into vesicles and a vesicular ace-tylcholine transporter. Reuptake is a major factor in terminating the action of transmitters, and when it is inhibited, the effects of transmit-ter release are increased and prolonged. This has clinical con-sequences. For example, several effective antidepressant drugs are inhibitors of the reuptake of amine transmitters, and cocaine is believed to inhibit dopamine reuptake. Glutamate uptake into neurons and glia is important because glutamate is an excitotoxin that can kill cells by overstimulating them (see Clinical Box 7–1). There is evidence that during ischemia TABLE 7–2 Mechanism of action of selected small-molecule transmitters. Transmitter Receptor Second Messenger Net Channel Effects Monoamines Acetylcholine Nicotinic ↑Na+, K+ M1, M3, M5 ↑IP3, DAG ↑Ca2+ M2, M4 ↓Cyclic AMP ↑K+ Serotonin 5HT1A ↓Cyclic AMP ↑K+ 5HT1B ↓Cyclic AMP 5HT1D ↓Cyclic AMP ↓K+ 5HT2A ↑IP3, DAG ↓K+ 5HT2C ↑IP3, DAG 5HT3 ↑Na+ 5HT4 ↑Cyclic AMP Catecholamines Dopamine D1, D5 ↑Cyclic AMP D2 ↓Cyclic AMP ↑K+, ↓Ca2+ D3, D4 ↓Cyclic AMP Norepinephrine α1 ↑IP3, DAG ↓K+ α2 ↓Cyclic AMP ↑K+, ↓Ca2+ β1 ↑Cyclic AMP β2 ↑Cyclic AMP β3 ↑Cyclic AMP Amino Acids Glutamate Metabotropica Ionotropic AMPA, Kainate ↑Na+, K+ NMDA ↑Na+, K+,Ca2+ GABA GABAA ↑Cl– GABAB ↑IP3, DAG ↑K+,↓Ca2+ Glycine Glycine ↑Cl– aEleven subtypes identified; all decrease cAMP or increase IP3 and DAG, except one, which increases cAMP. 134 SECTION II Physiology of Nerve & Muscle Cells and anoxia, loss of neurons is increased because glutamate reuptake is inhibited. SMALL-MOLECULE TRANSMITTERS Synaptic physiology is a rapidly expanding, complex field that cannot be covered in detail in this book. However, it is appro-priate to summarize information about the principal neu-rotransmitters and their receptors. MONOAMINES Acetylcholine Acetylcholine, which is the acetyl ester of choline, is largely enclosed in small, clear synaptic vesicles in high concentra-tion in the terminal boutons of neurons that release acetyl-choline (cholinergic neurons). Synthesis of acetylcholine involves the reaction of choline with acetate (Figure 7–1). Acetylcholine is the transmitter at the neuromuscular junc-tion, in autonomic ganglia, and in postganglionic parasympa-thetic nerve-target organ junctions and some postganglionic sympathetic nerve-target junctions. It is also found within the brain, including the basal forebrain complex and pon-tomesencephalic cholinergic complex (Figure 7–2). These systems may be involved in regulation of sleep-wake states, learning, and memory. Cholinergic neurons actively take up choline via a trans-porter (Figure 7–4). Choline is also synthesized in neurons. The acetate is activated by the combination of acetate groups with reduced coenzyme A. The reaction between active ace-tate (acetyl-coenzyme A, acetyl-CoA) and choline is catalyzed by the enzyme choline acetyltransferase. This enzyme is found in high concentration in the cytoplasm of cholinergic FIGURE 7–3 Fate of monoamines secreted at synaptic junctions. In each monoamine-secreting neuron, the monoamine is synthesized in the cytoplasm and the secretory granules (1) and its concentration in secretory granules is maintained (2) by the two vesicular monoamine transporters (VMAT). The monoamine is secreted by exocytosis of the granules (3), and it acts (4) on receptors (Y-shaped structures labeled R). Many of these receptors are postsynaptic, but some are presynaptic and some are located on glia. In addition, there is extensive reuptake into the cytoplasm of the presynaptic terminal (5) via the monoamine neurotransmitter transporter (NTT) for the monoamine that is synthesized in the neuron. (Reproduced with permission from Hoffman BJ, et al: Distribution of monoamine neurotransmitter transporters in the rat brain. Front Neuroendocrinol 1998;19:187.) 2 1 3 5 4 4 Transmitter synthesis NTT Amino acid precursor Presynaptic terminal K+ Na+ R R Second messengers + Other receptors Postsynaptic terminal Neuron Neuron Glial cell R VMAT H+ Cl− CHAPTER 7 Neurotransmitters & Neuromodulators 135 nerve endings. Acetylcholine is then taken up into synaptic vesicles by a vesicular transporter, VAChT. Cholinesterases Acetylcholine must be rapidly removed from the synapse if re-polarization is to occur. The removal occurs by way of hydrol-ysis of acetylcholine to choline and acetate, a reaction catalyzed by the enzyme acetylcholinesterase. This enzyme is also called true or specific cholinesterase. Its greatest affinity is for acetyl-choline, but it also hydrolyzes other choline esters. There are a variety of esterases in the body. One found in plasma is capable of hydrolyzing acetylcholine but has different properties from acetylcholinesterase. It is therefore called pseudocholinest-erase or nonspecific cholinesterase. The plasma moiety is partly under endocrine control and is affected by variations in liver function. On the other hand, the specific cholinesterase molecules are clustered in the postsynaptic membrane of cho-linergic synapses. Hydrolysis of acetylcholine by this enzyme is rapid enough to explain the observed changes in Na+ conduc-tance and electrical activity during synaptic transmission. Acetylcholine Receptors Historically, acetylcholine receptors have been divided into two main types on the basis of their pharmacologic proper-ties. Muscarine, the alkaloid responsible for the toxicity of toadstools, has little effect on the receptors in autonomic gan-glia but mimics the stimulatory action of acetylcholine on smooth muscle and glands. These actions of acetylcholine are therefore called muscarinic actions, and the receptors in-volved are muscarinic cholinergic receptors. They are blocked by the drug atropine. In sympathetic ganglia, small amounts of acetylcholine stimulate postganglionic neurons and large amounts block transmission of impulses from preganglionic to postganglionic neurons. These actions are unaffected by atropine but mimicked by nicotine. Conse-quently, these actions of acetylcholine are nicotinic actions and the receptors are nicotinic cholinergic receptors. Nico-tinic receptors are subdivided into those found in muscle at neuromuscular junctions and those found in autonomic gan-glia and the central nervous system. Both muscarinic and nic-otinic acetylcholine receptors are found in large numbers in the brain. The nicotinic acetylcholine receptors are members of a superfamily of ligand-gated ion channels that also includes the GABAA and glycine receptors and some of the glutamate receptors. They are made up of multiple subunits coded by dif-ferent genes. Each nicotinic cholinergic receptor is made up of five subunits that form a central channel which, when the receptor is activated, permits the passage of Na+ and other cat-ions. The 5 subunits come from a menu of 16 known subunits, α1–α9, β2–β5, γ, δ, and ε, coded by 16 different genes. Some of the receptors are homomeric—for example, those that contain five α7 subunits—but most are heteromeric. The muscle type nicotinic receptor found in the fetus is made up of two α1 sub-units, a β1 subunit, a γ subunit, and a δ subunit (Figure 7–5). In adult mammals, the γ subunit is replaced by a δ subunit, which decreases the channel open time but increases its CLINICAL BOX 7–1 Excitotoxins Glutamate is usually cleared from the brain’s extracellular fluid by Na+-dependent uptake systems in neurons and glia, keeping only micromolar levels of the chemical in the extracellular fluid despite millimolar levels inside neurons. However, excessive levels of glutamate occur in response to ischemia, anoxia, hypoglycemia, or trauma. Glutamate and some of its synthetic congeners are unique in that when they act on neuronal cell bodies, they can produce so much Ca2+ influx that neurons die. This is the reason why microinjection of these excitotoxins is used in research to produce discrete lesions that destroy neuronal cell bodies without affecting neighboring axons. Evidence is accumu-lating that excitotoxins play a significant role in the dam-age done to the brain by a stroke. When a cerebral artery is occluded, the cells in the severely ischemic area die. Sur-rounding partially ischemic cells may survive but lose their ability to maintain the transmembrane Na+ gradient. The elevated levels of intracellular Na+ prevent the ability of as-trocytes to remove glutamate from the brain’s extracellu-lar fluid. Therefore, glutamate accumulates to the point that excitotoxic damage and cell death occurs in the pen-umbra, the region around the completely infarcted area. FIGURE 7–4 Biochemical events at cholinergic endings. ACh, acetylcholine; ASE, acetylcholinesterase; X, receptor. Cholinergic neuron Postsynaptic tissue Acetyl-CoA + Choline ACh ACh ASE Choline 136 SECTION II Physiology of Nerve & Muscle Cells conductance. The nicotinic cholinergic receptors in autonomic ganglia are heteromers that usually contain α3 subunits in combination with others, and the nicotinic receptors in the brain are made up of many other subunits. Many of the nico-tinic cholinergic receptors in the brain are located presynapti-cally on glutamate-secreting axon terminals, and they facilitate the release of this transmitter. However, others are postsynap-tic. Some are located on structures other than neurons, and some seem to be free in the interstitial fluid, that is, they are perisynaptic in location. Each α subunit has a binding site for acetylcholine, and when an acetylcholine molecule binds to each of them, they induce a confirmational change in the protein so that the channel opens. This increases the conductance of Na+ and other cations, and the resulting influx of Na+ produces a depolarizing potential. A prominent feature of neuronal nico-tinic cholinergic receptors is their high permeability to Ca2+. Muscarinic cholinergic receptors are very different from nicotinic cholinergic receptors. Five types, encoded by five separate genes, have been cloned. The exact status of M5 is uncertain, but the remaining four receptors are coupled via G proteins to adenylyl cyclase, K+ channels, and/or phospholi-pase C (Table 7–2). The nomenclature of these receptors has not been standardized, but the receptor designated M1 in Table 7–2 is abundant in the brain. The M2 receptor is found in the heart. The M4 receptor is found in pancreatic acinar and islet tissue, where it mediates increased secretion of pan-creatic enzymes and insulin. The M3 and M4 receptors are associated with smooth muscle. Serotonin Serotonin (5-hydroxytryptamine; 5-HT) is present in highest concentration in blood platelets and in the gastrointestinal tract, where it is found in the enterochromaffin cells and the myenteric plexus. It is also found within the brain stem in the midline raphé nuclei which project to portions of the hypo-thalamus, the limbic system, the neocortex, the cerebellum, and the spinal cord (Figure 7–2). Serotonin is formed in the body by hydroxylation and decarboxylation of the essential amino acid tryptophan (Fig-ures 7–1 and 7–6). After release from serotonergic neurons, much of the released serotonin is recaptured by an active reuptake mechanism and inactivated by monoamine oxidase (MAO) to form 5-hydroxyindoleacetic acid (5-HIAA). This substance is the principal urinary metabolite of serotonin, and urinary output of 5-HIAA is used as an index of the rate of serotonin metabolism in the body. Tryptophan hydroxylase in the human CNS is slightly differ-ent from the tryptophan hydroxylase in peripheral tissues, and is coded by a different gene. This is presumably why knockout of the TPH1 gene, which codes for tryptophan hydroxylase in peripheral tissues, has much less effect on brain serotonin pro-duction than on peripheral serotonin production. As described in Clinical Box 7–2, there is evidence for a relationship between behavior and brain serotonin content. Serotonergic Receptors The number of cloned and characterized serotonin receptors has increased rapidly. There are at least seven types of 5-HT FIGURE 7–5 Three-dimensional model of the nicotinic acetylcholine-gated ion channel. The receptor–channel complex consists of five subunits, all of which contribute to forming the pore. When two molecules of acetylcholine bind to portions of the α-subunits exposed to the membrane surface, the receptor–channel changes conformation. This opens the pore in the portion of the channel emnbedded in the lipid bilayer, and both K+ and Na+ flow through the open channel down their electrochemical gradient. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Na+ K+ ACh No ACh bound: Channel closed Two ACh molecules bound: Channel open CHAPTER 7 Neurotransmitters & Neuromodulators 137 receptors (from 5-HT1 through 5-HT7 receptors). Within the 5-HT1 group are the 5-HT1A, 5-HT1B, 5-HT1D, 5-HT1E, and 5-HT1F subtypes. Within the 5-HT2 group there are 5-HT2A, 5-HT2B, and 5-HT2C subtypes. There are two 5-HT5 sub-types: 5-HT5A and 5-HT5B. Most of these are G protein-cou-pled receptors and affect adenylyl cyclase or phospholipase C (Table 7–2). However, the 5-HT3 receptors, like nicotinic cholinergic receptors, are ligand-gated ion channels. Some of the serotonin receptors are presynaptic, and others are postsynaptic. 5-HT2A receptors mediate platelet aggregation and smooth muscle contraction. Mice in which the gene for 5-HT2C receptors has been knocked out are obese as a result of increased food intake despite normal responses to leptin, and they are prone to fatal seizures. 5-HT3 receptors are present in the gastrointestinal tract and the area postrema and are related to vomiting. 5-HT4 receptors are also present in the gastrointestinal tract, where they facilitate secretion and peristalsis, and in the brain. 5-HT6 and 5-HT7 receptors in the brain are distributed throughout the limbic system, and the 5-HT6 receptors have a high affinity for antidepressant drugs. Histamine Histaminergic neurons have their cell bodies in the tubero-mammillary nucleus of the posterior hypothalamus, and their axons project to all parts of the brain, including the ce-rebral cortex and the spinal cord. Histamine is also found in cells in the gastric mucosa and in heparin-containing cells called mast cells that are plentiful in the anterior and poster-ior lobes of the pituitary gland as well as at body surfaces. Histamine is formed by decarboxylation of the amino acid histidine (Figure 7–1). Histamine is converted to methylhista-mine or, alternatively, to imidazoleacetic acid. The latter reac-tion is quantitatively less important in humans. It requires the enzyme diamine oxidase (histaminase) rather than MAO, even though MAO catalyzes the oxidation of methylhistamine to methylimidazoleacetic acid. The three known types of histamine receptors—H1, H2, and H3—are all found in both peripheral tissues and the brain. Most, if not all, of the H3 receptors are presynaptic, and they mediate inhibition of the release of histamine and other transmitters via a G protein. H1 receptors activate phospholi-pase C, and H2 receptors increase the intracellular cAMP con-centration. The function of this diffuse histaminergic system is unknown, but evidence links brain histamine to arousal, FIGURE 7–6 Biochemical events at serotonergic synapses. 5-HTP, 5-hydroxytryptophan; 5-HT, 5-hydroxytryptamine (serotonin); 5-HIAA, 5-hydroxyindoleacetic acid; X, serotonin receptor. For clarity, the presynaptic receptors have been omitted. Serotonergic neuron Postsynaptic tissue MAO 5-HT 5-HT 5-HTP 5-HIAA L-Tryptophan Reuptake CLINICAL BOX 7–2 Role of Serotonin in Mood & Behavior The hallucinogenic agent lysergic acid diethylamide (LSD) is a serotonin agonist that produces its effects by activating 5-HT2 receptors in the brain. The transient hallucinations and other mental aberrations produced by this drug were discov-ered when the chemist who synthesized it inhaled some by accident. Its discovery called attention to the correlation between behavior and variations in brain serotonin con-tent. Psilocin (and its phosphorylated form, psilocybin), a substance found in certain mushrooms, and N,N-dimethyl-tryptamine (DMT) are also hallucinogenic and, like seroto-nin, are derivatives of tryptamine. 2,5-Dimethoxy-4-methyl-amphetamine (DOM) and mescaline and its congeners, the other true hallucinogens, are phenylethylamines rather than indolamines. However, all these hallucinogens appear to exert their effects by binding to 5-HT2 receptors. 3,4-Meth-ylenedioxymethamphetamine, a drug known as MDMA or ecstasy, is a popular drug of abuse. It produces euphoria, but this is followed by difficulty in concentrating, depression, and, in monkeys, insomnia. The drug causes release of sero-tonin followed by serotonin depletion; the euphoria may be due to the release and the later symptoms to the depletion. Drugs that increase extracellular norepinephrine levels in the brain elevate mood, and drugs that decrease extracellu-lar norepinephrine levels cause depression. However, indi-viduals with congenital dopamine β-hydroxylase (DBH) defi-ciency are normal as far as mood is concerned. Drugs that inhibit norepinephrine reuptake were of considerable value in the treatment of depression, but these drugs also inhibit serotonin reuptake. It is also known that the primary seroto-nin metabolite 5-HIAA is low in CSF of depressed individu-als. Drugs such as fluoxetine (Prozac), which inhibit seroto-nin reuptake without affecting norepinephrine reuptake, are effective as antidepressants. Thus, the focus in treating clini-cal depression has shifted from norepinephrine to serotonin. 138 SECTION II Physiology of Nerve & Muscle Cells sexual behavior, blood pressure, drinking, pain thresholds, and regulation of the secretion of several anterior pituitary hormones. CATECHOLAMINES Norepinephrine & Epinephrine The chemical transmitter present at most sympathetic post-ganglionic endings is norepinephrine. It is stored in the syn-aptic knobs of the neurons that secrete it in characteristic small vesicles that have a dense core (granulated vesicles; see above). Norepinephrine and its methyl derivative, epineph-rine, are secreted by the adrenal medulla, but epinephrine is not a mediator at postganglionic sympathetic endings. As discussed in Chapter 6, each sympathetic postganglionic neuron has multiple varicosities along its course, and each of these varicosities appears to be a site at which norepineph-rine is secreted. There are also norepinephrine-secreting and epinephrine-secreting neurons in the brain. Norepinephrine-secreting neu-rons are properly called noradrenergic neurons, although the term adrenergic neurons is also applied. However, it seems appropriate to reserve the latter term for epinephrine-secreting neurons. The cell bodies of the norepinephrine-containing neurons are located in the locus ceruleus and other medullary and pontine nuclei (Figure 7–2). From the locus ceruleus, the axons of the noradrenergic neurons form the locus ceruleus system. They descend into the spinal cord, enter the cerebel-lum, and ascend to innervate the paraventricular, supraoptic, and periventricular nuclei of the hypothalamus, the thalamus, the basal telencephalon, and the entire neocortex. Biosynthesis & Release of Catecholamines The principal catecholamines found in the body—norepi-nephrine, epinephrine, and dopamine—are formed by hy-droxylation and decarboxylation of the amino acid tyrosine (Figure 7–1). Some of the tyrosine is formed from phenylala-nine, but most is of dietary origin. Phenylalanine hydroxylase is found primarily in the liver (see Clinical Box 7–3). Tyrosine is transported into catecholamine-secreting neurons and ad-renal medullary cells by a concentrating mechanism. It is con-verted to dopa and then to dopamine in the cytoplasm of the cells by tyrosine hydroxylase and dopa decarboxylase. The decarboxylase, which is also called aromatic L-amino acid de-carboxylase, is very similar but probably not identical to 5-hy-droxytryptophan decarboxylase. The dopamine then enters the granulated vesicles, within which it is converted to norepi-nephrine by dopamine β-hydroxylase (DBH). L-Dopa is the isomer involved, but the norepinephrine that is formed is in the D configuration. The rate-limiting step in synthesis is the conversion of tyrosine to dopa. Tyrosine hydroxylase, which catalyzes this step, is subject to feedback inhibition by dopa-mine and norepinephrine, thus providing internal control of the synthetic process. The cofactor for tyrosine hydroxylase is tetrahydrobiopterin, which is converted to dihydrobiopterin when tyrosine is converted to dopa. Some neurons and adrenal medullary cells also contain the cytoplasmic enzyme phenylethanolamine-N-methyltrans-ferase (PNMT), which catalyzes the conversion of norepineph-rine to epinephrine. In these cells, norepinephrine apparently leaves the vesicles, is converted to epinephrine, and then enters other storage vesicles. In granulated vesicles, norepinephrine and epinephrine are bound to ATP and associated with a protein called chromo-granin A. In some but not all noradrenergic neurons, the large granulated vesicles also contain neuropeptide Y. Chro-mogranin A is a 49-kDa acid protein that is also found in many other neuroendocrine cells and neurons. Six related chromogranins have been identified. They have been claimed to have multiple intracellular and extracellular func-tions. Their level in the plasma is elevated in patients with a variety of tumors and in essential hypertension, in which they probably reflect increased sympathetic activity. However, their specific functions remain unsettled. The catecholamines are transported into the granulated vesicles by two vesicular transporters, and these transporters are inhibited by the drug reserpine. Catecholamines are released from autonomic neurons and adrenal medullary cells by exocytosis. Because they are present in the granulated vesicles, ATP, chromogranin A, and the CLINICAL BOX 7–3 Phenylketonuria Phenylketonuria is a disorder characterized by severe men-tal deficiency and the accumulation in the blood, tissues, and urine of large amounts of phenylalanine and its keto acid derivatives. It is usually due to decreased function resulting from mutation of the gene for phenylalanine hydroxylase. This gene is located on the long arm of chromosome 12. Cat-echolamines are still formed from tyrosine, and the cognitive impairment is largely due to accumulation of phenylalanine and its derivatives in the blood. Therefore, it can be treated with considerable success by markedly reducing the amount of phenylalanine in the diet. The condition can also be caused by tetrahydrobiopterin (BH4) deficiency. Because BH4 is a cofactor for tyrosine hydroxylase and tryptophan hydroxylase, as well as phenylalanine hydroxylase, cases due to tetrahydrobiopterin deficiency have catecholamine and serotonin deficiencies in addition to hyperphenylalaninemia. These cause hypotonia, inactivity, and developmental prob-lems. They are treated with tetrahydrobiopterin, levodopa, and 5-hydroxytryptophan in addition to a low-phenylalanine diet. BH4 is also essential for the synthesis of nitric oxide (NO) by nitric oxide synthase. Severe BH4 deficiency can lead to impairment of NO formation, and the CNS may be subjected to increased oxidative stress. CHAPTER 7 Neurotransmitters & Neuromodulators 139 dopamine β hydroxylase that is not membrane-bound are released with norepinephrine and epinephrine. The half-life of circulating dopamine β-hydroxylase is much longer than that of the catecholamines, and circulating levels of this substance are affected by genetic and other factors in addition to the rate of sympathetic activity. Catabolism of Catecholamines Norepinephrine, like other amine and amino acid transmit-ters, is removed from the synaptic cleft by binding to postsyn-aptic receptors, binding to presynaptic receptors (Figure 7–3), reuptake into the presynaptic neurons, or catabolism. Reup-take is a major mechanism in the case of norepinephrine, and the hypersensitivity of sympathetically denervated structures is explained in part on this basis. After the noradrenergic neurons are cut, their endings degenerate with loss of reup-take in them. Consequently, more norepinephrine from oth-er sources is available to stimulate the receptors on the autonomic effectors. Epinephrine and norepinephrine are metabolized to biologi-cally inactive products by oxidation and methylation. The former reaction is catalyzed by MAO and the latter by cate-chol-O-methyltransferase (COMT). MAO is located on the outer surface of the mitochondria. It has two isoforms, MAO-A and MAO-B, which differ in substrate specificity and sensitivity to drugs. Both are found in neurons. MAO is widely distrib-uted, being particularly plentiful in the nerve endings at which catecholamines are secreted. COMT is also widely distributed, particularly in the liver, kidneys, and smooth muscles. In the brain, it is present in glial cells, and small amounts are found in postsynaptic neurons, but none is found in presynaptic norad-renergic neurons. Consequently, catecholamine metabolism has two different patterns. Extracellular epinephrine and norepinephrine are for the most part O-methylated, and measurement of the concentra-tions of the O-methylated derivatives normetanephrine and metanephrine in the urine is a good index of the rate of secretion of norepinephrine and epinephrine. The O-methyl-ated derivatives that are not excreted are largely oxidized, and 3-methoxy-4-hydroxymandelic acid (vanillylmandelic acid, VMA) is the most plentiful catecholamine metabolite in the urine. Small amounts of the O-methylated derivatives are also conjugated to sulfate and glucuronide. In the noradrenergic nerve terminals, on the other hand, some of the norepinephrine is constantly being converted by intracellular MAO (Figure 7–7) to the physiologically inactive deaminated derivatives, 3,4-dihydroxymandelic acid (DOMA) and its corresponding glycol (DHPG). These are subsequently converted to their corresponding O-methyl derivatives, VMA and 3-methoxy-4-hydroxyphenylglycol (MHPG). α & β Receptors Epinephrine and norepinephrine both act on α and β receptors, with norepinephrine having a greater affinity for α-adrenergic receptors and epinephrine for β-adrenergic receptors. As noted previously, the α and β receptors are typical G protein-coupled receptors, and each has multiple forms. They are closely related to the cloned receptors for dopamine and serotonin and to mus-carinic acetylcholine receptors. Clonidine lowers blood pressure when administered centrally. It is an α2 agonist and was initially thought to act on presynaptic α2 receptors, reducing central norepinephrine discharge. How-ever, its structure resembles that of imidazoline, and it binds to imidazoline receptors with higher affinity than to α2 adrenergic receptors. A subsequent search led to the discovery that imidazo-line receptors occur in the nucleus tractus solitarius and the vent-rolateral medulla. Administration of imidazolines lowers blood pressure and has a depressive effect. However, the full signifi-cance of these observations remains to be explored. Dopamine In certain parts of the brain, catecholamine synthesis stops at dopamine (Figure 7–1) which can then be secreted into the synaptic cleft. Active reuptake of dopamine occurs via a Na+-and Cl–-dependent dopamine transporter. Dopamine is me-tabolized to inactive compounds by MAO and COMT in a manner analogous to the inactivation of norepinephrine. 3,4-Dihydroxyphenylacetic acid (DOPAC) and homovanillic acid (HVA) are conjugated, primarily to sulfate. Dopaminergic neurons are located in several brain regions including the nigrostriatal system, which projects from the substantia nigra to the striatum and is involved in FIGURE 7–7 Biochemical events at noradrenergic endings. NE, norepinephrine; COMT, catechol-O-methyltransferase; MAO, monoamine oxidase; X, receptor. For clarity, the presynaptic receptors have been omitted. Note that MAO is intracellular, so that norepineph-rine is being constantly deaminated in noradrenergic endings. COMT acts primarily on secreted norepinephrine. Noradrenergic neuron Postsynaptic tissue Dopamine Dopa NE NE COMT MAO Reuptake Tyrosine Deaminated derivatives Normetanephrine 140 SECTION II Physiology of Nerve & Muscle Cells motor control, and the mesocortical system, which arises primarily in the ventral tegmental area (Figure 7–2). The mesocortical system projects to the nucleus accumbens and limbic subcortical areas, and it is involved in reward behav-ior and addiction. Studies by PET scanning in normal humans show that a steady loss of dopamine receptors occurs in the basal ganglia with age. The loss is greater in men than in women. Dopamine Receptors Five different dopamine receptors have been cloned, and several of these exist in multiple forms. This provides for va-riety in the type of responses produced by dopamine. Most, but perhaps not all, of the responses to these receptors are mediated by heterotrimeric G proteins. One of the two forms of D2 receptors can form a heterodimer with the somatosta-tin SST5 receptor, further increasing the dopamine response menu. Overstimulation of D2 receptors is thought to be re-lated to schizophrenia (see Clinical Box 7–4). D3 receptors are highly localized, especially to the nucleus accumbens (Figure 7–2). D4 receptors have a greater affinity than the other dopamine receptors for the “atypical” antipsychotic drug clozapine, which is effective in schizophrenia but pro-duces fewer extrapyramidal side effects than the other major tranquilizers do. EXCITATORY & INHIBITORY AMINO ACIDS Glutamate The amino acid glutamate is the main excitatory transmitter in the brain and spinal cord, and it has been calculated that it is the transmitter responsible for 75% of the excitatory trans-mission in the brain. Glutamate is formed by reductive amina-tion of the Krebs cycle intermediate α-ketoglutarate in the cytoplasm. The reaction is reversible, but in glutaminergic neurons, glutamate is concentrated in synaptic vesicles by the vesicle-bound transporter BPN1. The cytoplasmic store of glutamine is enriched by three transporters that import glutamate from the interstitial fluid, and two additional trans-porters carry glutamate into astrocytes, where it is converted to glutamine and passed on to glutaminergic neurons. The in-teraction of astrocytes and glutaminergic neurons is shown in Figure 7–8. Released glutamate is taken up by astrocytes and converted to glutamine, which passes back to the neurons and is converted back to glutamate, which is released as the synap-tic transmitter. Uptake into neurons and astrocytes is the main mechanism for removal of glutamate from synapses. Glutamate Receptors Glutamate receptors are of two types: metabotropic receptors and ionotropic receptors. The metabotropic receptors are G protein-coupled receptors that increase intracellular IP3 and CLINICAL BOX 7-4 Schizophrenia Schizophrenia is an illness that involves deficits of multiple brain systems that alter an individual’s inner thoughts as well as their interactions with others. Individuals with schizophre-nia suffer from hallucinations, delusions, and racing thoughts (positive symptoms); and they experience apathy, difficulty dealing with novel situations, and little spontaneity or motivation (negative symptoms). Worldwide, about 1–2% of the population lives with schizophrenia. A combination of genetic, biological, cultural, and psychological factors con-tributes to the illness. A large amount of evidence indicates that a defect in the mesocortical system is responsible for the development of at least some of the symptoms of schizophrenia. Attention was initially focused on overstimu-lation of limbic D2 dopamine receptors. Amphetamine, which causes release of dopamine as well as norepinephrine in the brain, causes a schizophrenialike psychosis; brain le-vels of D2 receptors are said to be elevated in schizophrenics; and there is a clear positive correlation between the antis-chizophrenic activity of many drugs and their ability to block D2 receptors. However, several recently developed drugs are effective antipsychotic agents but bind D2 receptors to a lim-ited degree. Instead, they bind to D4 receptors, and there is active ongoing research into the possibility that these recep-tors are abnormal in individuals with schizophrenia. FIGURE 7–8 The glutamate–glutamine cycle through glutaminergic neurons and astrocytes. Glutamate released into the synaptic cleft is taken up by a Na+-dependent glutamate transporter, and in the astrocyte it is converted to glutamine. The glutamine enters the neuron and is converted to glutamate. Glucose is transported out of capillaries and enters astrocytes and neurons. In astrocytes, it is me-tabolized to lactate, producing two ATPs. One of these powers the conversion of glutamate to glutamine, and the other is used by Na+– K+ ATPase to transport three Na+ out of the cell in exchange for two K+. In neurons, the glucose is metabolized further through the citric acid cycle, producing 34 ATPs. Glucose Capillary Lactate Lactate ATP ATP 34 ATP Gln Glu 3Na+ 3Na+ 2K+ Glutaminergic synapse Astrocyte Gln Glu Glu CHAPTER 7 Neurotransmitters & Neuromodulators 141 DAG levels or decrease intracellular cAMP levels. Eleven sub-types have been identified (Table 7–2). They are both presyn-aptic and postsynaptic, and they are widely distributed in the brain. They appear to be involved in the production of synap-tic plasticity, particularly in the hippocampus and the cerebel-lum. Knockout of the gene for one of these receptors, one of the forms of mGluR1, causes severe motor incoordination and deficits in spatial learning. The ionotropic receptors are ligand-gated ion channels that resemble nicotinic cholinergic receptors and GABA and glycine receptors. There are three general types, each named for the congeners of glutamate to which they respond in maximum fashion. These are the kainate receptors (kainate is an acid iso-lated from seaweed), AMPA receptors (for α-amino-3-hydroxy-5-methylisoxazole-4-propionate), and NMDA recep-tors (for N-methyl-D-aspartate). Four AMPA, five kainate, and six NMDA subunits have been identified, and each is coded by a different gene. The receptors were initially thought to be pen-tamers, but some may be tetramers, and their exact stoichiome-try is unsettled. The kainate receptors are simple ion channels that, when open, permit Na+ influx and K+ efflux. There are two popula-tions of AMPA receptors: one is a simple Na+ channel and one also passes Ca2+. The balance between the two in a given syn-apse can be shifted by activity. The NMDA receptor is also a cation channel, but it per-mits passage of relatively large amounts of Ca2+, and it is unique in several ways (Figure 7–9). First, glycine facilitates its function by binding to it, and glycine appears to be essen-tial for its normal response to glutamate. Second, when glutamate binds to it, it opens, but at normal membrane potentials, its channel is blocked by a Mg2+ ion. This block is removed only when the neuron containing the receptor is partially depolarized by activation of AMPA or other chan-nels that produce rapid depolarization via other synaptic cir-cuits. Third, phencyclidine and ketamine, which produce amnesia and a feeling of dissociation from the environment, bind to another site inside the channel. Most target neurons for glutamate have both AMPA and NMDA receptors. Kain-ate receptors are located presynaptically on GABA-secreting nerve endings and postsynaptically at various localized sites in the brain. Kainate and AMPA receptors are found in glia as well as neurons, but it appears that NMDA receptors occur only in neurons. The concentration of NMDA receptors in the hippocampus is high, and blockade of these receptors prevents long-term potentiation, a long-lasting facilitation of transmission in neural pathways following a brief period of high-frequency stimulation. Thus, these receptors may well be involved in memory and learning. GABA GABA is the major inhibitory mediator in the brain, includ-ing being responsible for presynaptic inhibition. GABA, which exists as β-aminobutyrate in the body fluids, is formed by decarboxylation of glutamate (Figure 7–1). The enzyme that catalyzes this reaction is glutamate decarboxylase (GAD), which is present in nerve endings in many parts of the brain. GABA is metabolized primarily by transamination to succinic semialdehyde and thence to succinate in the citric acid cycle. GABA transaminase (GABA-T) is the enzyme that catalyzes the transamination. Pyridoxal phosphate, a de-rivative of the B complex vitamin pyridoxine, is a cofactor for GAD and GABA-T. In addition, there is an active reuptake of GABA via the GABA transporter. A vesicular GABA trans-porter (VGAT) transports GABA and glycine into secretory vesicles. GABA Receptors Three subtypes of GABA receptors have been identified: GABAA, GABAB, and GABAC. The GABAA and GABAB re-ceptors are widely distributed in the CNS, whereas in adult ver-tebrates the GABAC receptors are found almost exclusively in the retina. The GABAA and GABAC receptors are ion channels made up of five subunits surrounding a pore, like the nicotinic acetylcholine receptors and many of the glutamate receptors. In this case, the ion is Cl– (Figure 7–10). The GABAB receptors are metabotropic and are coupled to heterotrimeric G proteins that increase conductance in K+ channels, inhibit adenylyl cyclase, and inhibit Ca2+ influx. Increases in Cl– influx and K+ efflux and decreases in Ca2+ influx all hyperpolarize neurons, produc-ing an IPSP. The G protein mediation of GABAB receptor ef-fects is unique in that a G protein heterodimer, rather than a single protein, is involved. The GABAC receptors are relatively simple in that they are pentamers of three ρ subunits in various combinations. On FIGURE 7–9 Diagrammatic representation of the NMDA receptor. When glycine and glutamate bind to the receptor, the closed ion channel (left) opens, but at the resting membrane poten-tial, the channel is blocked by Mg2+ (right). This block is removed if partial depolarization is produced by other inputs to the neuron con-taining the receptor, and Ca2+ and Na+ enter the neuron. Blockade can also be produced by the drug dizocilpine maleate (MK-801). L-Glutamate Glycine Extracellular Intracellular Ca2+ Na+ K+ Open ion channel Closed ion channel Channel blocker Mg2+ MK-801 142 SECTION II Physiology of Nerve & Muscle Cells the other hand, the GABAA receptors are pentamers made up of various combinations of six α subunits, four β, four γ, one δ, and one ε. This endows them with considerably different properties from one location to another. An observation of considerable interest is that there is a chronic low-level stimulation of GABAA receptors in the CNS that is aided by GABA in the interstitial fluid. This back-ground stimulation cuts down on the “noise” caused by inci-dental discharge of the billions of neural units and greatly improves the signal-to-noise ratio in the brain. It may be that this GABA discharge declines with advancing age, resulting in a loss of specificity of responses of visual neurons. Support for this hypothesis comes from studies in which microinjection of GABA in older monkeys resulted in restoration of the spec-ificity of visual neurons. The increase in Cl– conductance produced by GABAA recep-tors is potentiated by benzodiazepines, drugs that have marked anti-anxiety activity and are also effective muscle relaxants, anticonvulsants, and sedatives. Benzodiazepines bind to the α subunits. Diazepam and other benzodiazepines are used throughout the world. At least in part, barbiturates and alcohol also act by facilitating Cl– conductance through the Cl– chan-nel. Metabolites of the steroid hormones progesterone and deoxycorticosterone bind to GABAA receptors and increase Cl– conductance. It has been known for many years that progester-one and deoxycorticosterone are sleep-inducing and anesthetic in large doses, and these effects are due to their action on GABAA receptors. A second class of benzodiazepine receptors is found in ster-oid-secreting endocrine glands and other peripheral tissues, and hence these receptors are called peripheral benzodiaze-pine receptors. They may be involved in steroid biosynthesis, possibly performing a function like that of the StAR protein in moving steroids into the mitochondria. Another possibility is a role in the regulation of cell proliferation. Peripheral-type benzodiazepine receptors are also present in astrocytes in the brain, and they are found in brain tumors. Glycine Glycine has both excitatory and inhibitory effects in the CNS. When it binds to NMDA receptors, it makes them more sen-sitive. It appears to spill over from synaptic junctions into the interstitial fluid, and in the spinal cord, for example, this gly-cine may facilitate pain transmission by NMDA receptors in the dorsal horn. However, glycine is also responsible in part for direct inhibition, primarily in the brain stem and spinal cord. Like GABA, it acts by increasing Cl– conductance. Its ac-tion is antagonized by strychnine. The clinical picture of con-vulsions and muscular hyperactivity produced by strychnine emphasizes the importance of postsynaptic inhibition in nor-mal neural function. The glycine receptor responsible for in-hibition is a Cl– channel. It is a pentamer made up of two subunits: the ligand-binding α subunit and the structural β subunit. Recently, solid evidence has been presented that three kinds of neurons are responsible for direct inhibition in the spinal cord: neurons that secrete glycine, neurons that secrete GABA, and neurons that secrete both. Presumably, neurons that secrete only glycine have the glycine transporter GLYT2, those that secrete only GABA have GAD, and those that se-crete glycine and GABA have both. This third type of neuron is of special interest because the neurons seem to have glycine and GABA in the same vesicles. FIGURE 7–10 Diagram of GABAA and GABAB receptors, showing their principal actions. The G protein that mediates the effects of GABAB receptors is a heterodimer. (Reproduced with permission from Bowery NG, Brown DA: The cloning of GABAB receptors. Nature 1997;386:223. Copyright © 1997 by Macmillan Magazines.) AC α β α γ β γ − GABAA Extracellular Intracellular Cl− GABAB NH2 Ca2+ K+ COOH CHAPTER 7 Neurotransmitters & Neuromodulators 143 Anesthesia Although general anesthetics have been used for millennia, lit-tle has been understood about their mechanisms of action. However, it now appears that alcohols, barbiturates, and many volatile inhaled anesthetics as well act on ion channel recep-tors and specifically on GABAA and glycine receptors to in-crease Cl– conductance. Regional variation in anesthetic actions in the CNS seems to parallel the variation in subtypes of GABAA receptors. Other inhaled anesthetics do not act by increasing GABA receptor activity, but appear to act by inhib-iting NMDA and AMPA receptors instead. In contrast to general anesthetics, local anesthetics produce anesthesia by blocking conduction in peripheral nerves via reversibly binding to and inactivating Na+ channels. Na+ influx through these channels normally causes depolarization of nerve cell membranes and propagation of impulses toward the nerve terminal. When depolarization and propagation are interrupted, the individual loses sensation in the area supplied by the nerve. LARGE-MOLECULE TRANSMITTERS: NEUROPEPTIDES Substance P & Other Tachykinins Substance P is a polypeptide containing 11 amino acid residues that is found in the intestine, various peripheral nerves, and many parts of the CNS. It is one of a family of six mammalian polypeptides called tachykinins that differ at the amino terminal end but have in common the carboxyl terminal sequence of Phe-X-Gly-LeuMet-NH2, where X is Val, His, Lys, or Phe. The members of the family are listed in Table 7–3. There are many related tachykinins in other vertebrates and in invertebrates. The mammalian tachykinins are encoded by two genes. The neurokinin B gene encodes only one known polypep-tide, neurokinin B. The substance P/neurokinin A gene encodes the remaining five polypeptides. Three are formed by alternative processing of the primary RNA and two by post-translational processing. There are three neurokinin receptors. Two of these, the sub-stance P and the neuropeptide K receptors, are G protein-cou-pled receptors. Activation of the substance P receptor causes activation of phospholipase C and increased formation of IP3 and DAG. Substance P is found in high concentration in the endings of primary afferent neurons in the spinal cord, and it is proba-bly the mediator at the first synapse in the pathways for pain transmission in the dorsal horn. It is also found in high con-centrations in the nigrostriatal system, where its concentra-tion is proportional to that of dopamine, and in the hypothalamus, where it may play a role in neuroendocrine regulation. Upon injection into the skin, it causes redness and swelling, and it is probably the mediator released by nerve fibers that is responsible for the axon reflex. In the intestine, it is involved in peristalsis. It has recently been reported that a centrally active NK-1 receptor antagonist has antidepressant activity in humans. This antidepressant effect takes time to develop, like the effect of the antidepressants that affect brain monoamine metabolism, but the NK-1 inhibitor does not alter brain monoamines in experimental animals. The func-tions of the other tachykinins are unsettled. Opioid Peptides The brain and the gastrointestinal tract contain receptors that bind morphine. The search for endogenous ligands for these receptors led to the discovery of two closely related pentapeptides (enkephalins; Table 7–4) that bind to these opioid receptors. TABLE 7–3 Mammalian tachykinins. Gene Polypeptide Products Receptors SP/NKA Substance P Substance P (NK-1) Neurokinin A Neuropeptide K Neuropeptide K (NK-2) Neuropeptide α Neurokinin A (3–10) NKB Neurokinin B Neurokinin B (NK-3) TABLE 7–4 Opioid peptides and their precursors. Precursor Opioid Peptides Structures Proenkephalin Met-enkephalin Tyr-Gly-Gly-Phe-Met Leu-enkephalin Tyr-Gly-Gly-Phe-Leu Octapeptide Tyr-Gly-Gly-Phe-Met-Arg-Gly-Leu Heptapeptide Tyr-Gly-Gly-Phe-Met-Arg-Phe Proopiomel-anocortin β-Endorphin Tyr-Gly-Glu-Phe-Met-Thr-Ser-Lys-Ser-Gln-Thr-Pro-Leu-Val-Thr-Leu-Phe-Lys-Asn-Ala-Ile-Val-Lys-Asn-Ala-His-Lys-Lys-Gly-Gln Prodynorphin Dynorphin 1–8 Tyr-Gly-Gly-Phe-Leu-Arg-Arg-lle Dynorphin 1–17 Tyr-Gly-Gly-Phe-Leu-Arg-Arg-lle-Arg-Pro-Lys-Leu-Lys-Trp-Asp-Asn-Gln α-Neoendorphin Tyr-Gly-Gly-Phe-Leu-Arg-Lys-Tyr-Pro-Lys β-Neoendorphin Tyr-Gly-Gly-Phe-Leu-Arg-Lys-Tyr-Pro 144 SECTION II Physiology of Nerve & Muscle Cells One contains methionine (met-enkephalin), and one con-tains leucine (leu-enkephalin). These and other peptides that bind to opioid receptors are called opioid peptides. The en-kephalins are found in nerve endings in the gastrointestinal tract and many different parts of the brain, and they appear to function as synaptic transmitters. They are found in the sub-stantia gelatinosa and have analgesic activity when injected into the brain stem. They also decrease intestinal motility. Like other small peptides, the opioid peptides are synthe-sized as part of larger precursor molecules. More than 20 active opioid peptides have been identified. Unlike other pep-tides, however, the opioid peptides have a number of different precursors. Each has a prepro form and a pro form from which the signal peptide has been cleaved. The three precur-sors that have been characterized, and the opioid peptides they produce, are shown in Table 7–4. Proenkephalin was first identified in the adrenal medulla, but it is also the precur-sor for met-enkephalin and leu-enkephalin in the brain. Each proenkephalin molecule contains four met-enkephalins, one leuenkephalin, one octapeptide, and one heptapeptide. Proo-piomelanocortin, a large precursor molecule found in the anterior and intermediate lobes of the pituitary gland and the brain, contains β-endorphin, a polypeptide of 31 amino acid residues that has metenkephalin at its amino terminal. There are separate enkephalin-secreting and β endorphin-secreting systems of neurons in the brain. β-Endorphin is also secreted into the bloodstream by the pituitary gland. A third precursor molecule is prodynorphin, a protein that contains three leu-enkephalin residues associated with dynorphin and neoen-dorphin. Dynorphin 1-17 is found in the duodenum and dynorphin 1-8 in the posterior pituitary and hypothalamus. Alpha- and β-neoendorphins are also found in the hypothala-mus. The reasons for the existence of multiple opioid peptide precursors and for the presence of the peptides in the circula-tion as well as in the brain and the gastrointestinal tract are presently unknown. Enkephalins are metabolized primarily by two peptidases: enkephalinase A, which splits the Gly-Phe bond, and enkephali-nase B, which splits the Gly-Gly bond. Aminopeptidase, which splits the Tyr-Gly bond, also contributes to their metabolism. Opioid receptors have been studied in detail, and three are now established: μ, κ, and δ. They differ in physiologic effects (Table 7–5), distribution in the brain and elsewhere, and affinity for various opioid peptides. All three are G protein-coupled receptors, and all inhibit adenylyl cyclase. In mice in which the μ receptors have been knocked out, morphine fails to produce analgesia, withdrawal symptoms, and self-admin-istration of nicotine. Selective knockout of the other system fails to produce this blockade. Activation of μ receptors increases K+ conductance, hyperpolarizing central neurons and primary afferents. Activation of κ receptors and δ recep-tors closes Ca2+ channels. The affinities of individual ligands for the three types of receptors are summarized in Figure 7–11. Endorphins bind only to μ receptors, the main receptors that mediate analgesia. Other opioid peptides bind to multiple opioid receptors. Other Polypeptides Numerous other polypeptides are found in the brain. For ex-ample, somatostatin is found in various parts of the brain, where it apparently functions as a neurotransmitter with ef-fects on sensory input, locomotor activity, and cognitive func-tion. In the hypothalamus, this growth hormone-inhibiting hormone is secreted into the portal hypophysial vessels; in the endocrine pancreas, it inhibits insulin secretion and the secre-tion of other pancreatic hormones; and in the gastrointestinal tract, it is an important inhibitory gastrointestinal regulator. A family of five different somatostatin receptors have been TABLE 7–5 Physiologic effects produced by stimulation of opiate receptors. Receptor Effect μ Analgesia Site of action of morphine Respiratory depression Constipation Euphoria Sedation Increased secretion of growth hormone and prolactin Meiosis κ Analgesia Diuresis Sedation Meiosis Dysphoria δ Analgesia FIGURE 7–11 Opioid receptors. The ligands for the κ, μ, and δ receptors are shown with the width of the arrows proportionate to the affinity of the receptor for each ligand. (Reproduced with permission from Julius DJ: Another spark for the masses? Nature 1997;386:442. Copyright © 1997 by Macmillan Magazines.) Dynorphins β-Endorphin Enkephalins Endomorphins κ δ CHAPTER 7 Neurotransmitters & Neuromodulators 145 identified (SSTR1 through SSTR5). All are G protein-coupled receptors. They inhibit adenylyl cyclase and exert various oth-er effects on intracellular messenger systems. It appears that SSTR2 mediates cognitive effects and inhibition of growth hormone secretion, whereas SSTR5 mediates the inhibition of insulin secretion. Vasopressin and oxytocin are not only secreted as hormones but also are present in neurons that project to the brain stem and spinal cord. The brain contains bradykinin, angiotensin II, and endothelin. The gastrointestinal hormones VIP, CCK-4, and CCK-8 are also found in the brain. There are two kinds of CCK receptors in the brain, CCK-A and CCK-B. CCK-8 acts at both binding sites, whereas CCK-4 acts at the CCK-B sites. Gastrin, neurotensin, galanin, and gastrin-releasing peptide are also found in the gastrointestinal tract and brain. Neurotensin and VIP receptors have been cloned and shown to be G pro-tein-coupled receptors. The hypothalamus contains both gas-trin 17 and gastrin 34. VIP produces vasodilation and is found in vasomotor nerve fibers. The functions of these peptides in the nervous system are unknown. Calcitonin gene-related peptide (CGRP) is a polypeptide that exists in two forms in rats and humans: CGRPα and CGRPβ. In humans, these two forms differ by only three amino acid residues, yet they are encoded by different genes. In rats, and presumably in humans, CGRPβ is present in the gastrointestinal tract, whereas CGRPβ is found in primary afferent neurons, neurons that project which taste impulses to the thalamus, and neurons in the medial forebrain bundle. It is also present along with substance P in the branches of primary afferent neurons that end near blood vessels. CGRP-like immunoreactivity is present in the circulation, and injection of CGRP causes vasodilation. CGRPα and the calcium-lowering hormone calcitonin are both products of the calcitonin gene. In the thyroid gland, splicing produces the mRNA that codes for calcitonin, whereas in the brain, alternative splicing pro-duces the mRNA that codes for CGRPα. CGRP has little effect on Ca2+ metabolism, and calcitonin is only a weak vasodilator. Neuropeptide Y is a polypeptide containing 36 amino acid residues that acts on at least two of the four known G protein-coupled receptors: Y1, Y2, Y4, and Y5. Neuropeptide Y is found throughout the brain and the autonomic nervous system. When injected into the hypothalamus, this polypeptide increases food intake, and inhibitors of neuropeptide Y syn-thesis decrease food intake. Neuropeptide Y-containing neu-rons have their cell bodies in the arcuate nuclei and project to the paraventricular nuclei. OTHER CHEMICAL TRANSMITTERS Purine & Pyrimidine Transmitters After extended debate, it now seems clear that ATP, uridine, adenosine, and adenosine metabolites are neurotransmitters or neuromodulators. Adenosine is a neuromodulator that acts as a general CNS depressant and has additional widespread ef-fects throughout the body. It acts on four receptors: A1, A2A, A2B, and A3. All are G protein-coupled receptors and increase (A2A and A2B) or decrease (A1 and A3) cAMP concentrations. The stimulatory effects of coffee and tea are due to blockade of adenosine receptors by caffeine and theophylline. Currently, there is considerable interest in the potential use of A1 antag-onists to decrease excessive glutamate release and thus to min-imize the effects of strokes. ATP is also becoming established as a transmitter, and it has widespread receptor-mediated effects in the body. It appears that soluble nucleotidases are released with ATP, and these accelerate its removal after it has produced its effects. ATP has now been shown to mediate rapid synaptic responses in the autonomic nervous system and a fast response in the habenula. ATP binds to P2X receptors which are ligand-gated ion chan-nel receptors; seven subtypes (P2X1–P2X7) have been identi-fied. P2X receptors have widespread distributions throughout the body; for example, P2X1 and P2X2 receptors are present in the dorsal horn, indicating a role for ATP in sensory transmis-sion. ATP also binds to P2Y receptors which are G protein-coupled receptors. There are eight subtypes of P2Y receptors: P2Y1, P2Y2, P2Y4, P2Y6, P2Y11, P2Y12, P2Y13, and P2Y14. Cannabinoids Two receptors with a high affinity for Δ9-tetrahydrocannab-inol (THC), the psychoactive ingredient in marijuana, have been cloned. The CB1 receptor triggers a G protein-mediated decrease in intracellular cAMP levels and is common in cen-tral pain pathways as well as in parts of the cerebellum, hippo-campus, and cerebral cortex. The endogenous ligand for the receptor is anandamide, a derivative of arachidonic acid. This compound mimics the euphoria, calmness, dream states, drowsiness, and analgesia produced by marijuana. There are also CB1 receptors in peripheral tissues, and blockade of these receptors reduces the vasodilator effect of anandamide. How-ever, it appears that the vasodilator effect is indirect. A CB2 re-ceptor has also been cloned, and its endogenous ligand may be palmitoylethanolamide (PEA). However, the physiologic role of this compound is unsettled. Gases Nitric oxide (NO), a compound released by the endothelium of blood vessels as endothelium-derived relaxing factor (EDRF), is also produced in the brain. It is synthesized from arginine, a reaction catalyzed in the brain by one of the three forms of NO synthase. It activates guanylyl cyclase and, unlike other transmitters, it is a gas, which crosses cell membranes with ease and binds directly to guanylyl cyclase. It may be the signal by which postsynaptic neurons communicate with pre-synaptic endings in long-term potentiation and long-term de-pression. NO synthase requires NADPH, and it is now known that NADPH-diaphorase (NDP), for which a histochemical stain has been available for many years, is NO synthase. 146 SECTION II Physiology of Nerve & Muscle Cells Other Substances Prostaglandins are derivatives of arachidonic acid found in the nervous system. They are present in nerve-ending frac-tions of brain homogenates and are released from neural tis-sue in vitro. A putative prostaglandin transporter with 12 membrane-spanning domains has been described. However, prostaglandins appear to exert their effects by modulating re-actions mediated by cAMP rather than by functioning as syn-aptic transmitters. Many steroids are neuroactive steroids; that is, they affect brain function, although they are not neurotransmitters in the usual sense. Circulating steroids enter the brain with ease, and neurons have numerous sex steroid and glucocorticoid recep-tors. In addition to acting in the established fashion by bind-ing to DNA (genomic effects), some steroids seem to act rapidly by a direct effect on cell membranes (nongenomic effects). The effects of steroids on GABA receptors have been discussed previously. Evidence has now accumulated that the brain can produce some hormonally active steroids from sim-pler steroid precursors, and the term neurosteroids has been coined to refer to these products. Progesterone facilitates the formation of myelin, but the exact role of most steroids in the regulation of brain function remains to be determined. CHAPTER SUMMARY ■Neurotransmitters and neuromodulators are divided into two major categories: small-molecule transmitters (monoamines, catecholamines, and amino acids) and large-molecule transmit-ters (neuropeptides). Usually neuropeptides are colocalized with one of the small-molecule neurotransmitters. ■Monoamines include acetylcholine, serotonin, and histamine. Catecholamines include norepinephrine, epinephrine, and do-pamine. Amino acids include glutamate, GABA, and glycine. ■Acetylcholine is found at the neuromuscular junction, in auto-nomic ganglia, and in postganglionic parasympathetic nerve-target organ junctions and some postganglionic sympathetic nerve-target junctions. It is also found in the basal forebrain complex and pontomesencephalic cholinergic complex. There are two major types of cholinergic receptors: muscarinic (G pro-tein-coupled receptors) and nicotinic (ligand-gated ion channel receptors). ■Serotonin (5-HT) is found within the brain stem in the midline raphé nuclei which project to portions of the hypothalamus, the limbic system, the neocortex, the cerebellum, and the spinal cord. There are at least seven types of 5-HT receptors, and many of these contain subtypes. Most are G protein-coupled receptors. ■Norepinephrine-containing neurons are in the locus ceruleus and other medullary and pontine nuclei. Some neurons also contain PNMT, which catalyzes the conversion of norepineph-rine to epinephrine. Epinephrine and norepinephrine act on α and β receptors, with norepinephrine having a greater affinity for α-adrenergic receptors and epinephrine for β-adrenergic re-ceptors. They are G protein-coupled receptors, and each has multiple forms. ■The amino acid glutamate is the main excitatory transmitter in the CNS. There are two major types of gluatamate receptors: metabotropic (G protein-coupled receptors) and ionotropic (ligand-gated ion channels receptors, including kainite, AMPA, and NMDA). ■GABA is the major inhibitory mediator in the brain. Three sub-types of GABA receptors have been identified: GABAA and GABAC (ligand-gated ion channel) and GABAB (G protein-coupled). The GABAA and GABAB receptors are widely distrib-uted in the CNS. ■There are three types of G protein-coupled opioid receptors (μ, κ, and δ) that differ in physiological effects, distribution in the brain and elsewhere, and affinity for various opioid peptides. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Which of the following is a ligand-gated ion channel? A) VIP receptor B) norepinephrine receptor C) GABAA receptor D) GABAB receptor E) metabotropic glutamate receptor 2. Which of the following synaptic transmitters is not a peptide, polypeptide, or protein? A) substance P B) met-enkephalin C) β-endorphin D) serotonin E) dynorphin 3. Activation of which of the following receptors would be expected to decrease anxiety? A) nicotinic cholinergic receptors B) glutamate receptors C) GABAA receptors D) glucocorticoid receptors E) α1-adrenergic receptors 4. Which of the following receptors is coupled to a heterotrimeric G protein? A) glycine receptor B) GABAB receptor C) nicotinic acetylcholine receptor at myoneural junction D) 5-HT3 receptor E) ANP receptor 5. Which of the following would not be expected to enhance nor-adrenergic transmission? A) A drug that increases the entry of arginine into neurons. B) A drug that enhances tyrosine hydroxylase activity. C) A drug that enhances dopamine β-hydroxylase activity. D) A drug that inhibits monoamine oxidase. E) A drug that inhibits norepinephrine reuptake. CHAPTER RESOURCES Boron WF, Boulpaep EL: Medical Physiology. Elsevier, 2005. Cooper JR, Bloom FE, Roth RH: The Biochemical Basis of Neuropharmacology, 8th ed. Oxford University Press, 2002. Fink KB, Göthert M: 5-HT receptor regulation of neurotransmitter release. Pharmacol Rev 2007;59:360. CHAPTER 7 Neurotransmitters & Neuromodulators 147 Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. Monaghan DT, Bridges RJ, Cotman CW: The excitatory amino acid receptors: Their classes, pharmacology, and distinct properties in the function of the central nervous system. Ann Rev Pharmacol Toxicol 1989;29:365. Nadeau SE, et al: Medical Neuroscience, Sauders, 2004. Olsen RW: The molecular mechanism of action of general anesthetics: Structural aspects of interactions with GABAA receptors. Toxicol Lett 1998;100:193. Owens DF, Kriegstein AR: Is there more to GABA than synaptic inhibition? Nat Rev Neurosci 2002;3:715. Squire LR, et al (editors): Fundamental Neuroscience, 3rd ed. Academic Press, 2008. This page intentionally left blank 149 C H A P T E R 8 Properties of Sensory Receptors O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the classification of sensory receptors. ■Name the types of sensory receptors found in the skin, and discuss their relation to touch, cold, warmth, and pain. ■Define generator potential. ■Explain the essential elements of sensory coding. INTRODUCTION Information about the internal and external environment activates the CNS via a variety of sensory receptors. These receptors are transducers that convert various forms of energy in the environment into action potentials in neurons. The characteristics of these receptors, the way they generate impulses in afferent neurons, and the general principles or “laws” that apply to sensation are considered in this chapter. Emphasis is placed on receptors mediating the sensation of touch, and later chapters focus on other sensory processes. We learn in elementary school that there are “five senses,” but this dictum takes into account only some of the senses that reach our consciousness. In addition, some sensory receptors relay information that does not reach consciousness. For exam-ple, the muscle spindles provide information about muscle length, and other receptors provide information about arterial blood pressure, the temperature of the blood in the head, and the pH of the cerebrospinal fluid. The list of senses in Table 8–1 is somewhat simplified. The rods and cones, for example, respond maximally to light of different wavelengths, and three different types of cones are present, one for each of the three primary colors. There are five different modalities of taste: sweet, salt, sour, bitter, and umami. Sounds of different pitches are heard primarily because different groups of hair cells in the cochlea are activated maximally by sound waves of different frequencies. Whether these various responses to light, taste, and sound should be considered separate senses is a semantic question that in the present context is largely academic. SENSE RECEPTORS & SENSE ORGANS It is worth noting that the term receptor is used in physiology to refer not only to sensory receptors but also, in a very differ-ent sense, to proteins that bind neurotransmitters, hormones, and other substances with great affinity and specificity as a first step in initiating specific physiologic responses. CLASSIFICATION OF SENSORY RECEPTORS Numerous attempts have been made to classify sensory recep-tors, but none has been entirely successful. One classification divides them into (1) teleceptors (“distance receivers”), which are concerned with events at a distance; (2) exteroceptors, which are concerned with the external environment near at hand; (3) interoceptors, which are concerned with the internal 150 SECTION II Physiology of Nerve & Muscle Cells environment; and (4) proprioceptors, which provide informa-tion about the position of the body in space at any given in-stant. However, the conscious component of proprioception (“body image”) is actually synthesized from information com-ing not only from receptors in and around joints but also from cutaneous touch and pressure receptors. Other special terms are frequently used to identify sensory receptors. The cutaneous receptors for touch and pressure are mechanoreceptors. Potentially harmful stimuli such as pain, extreme heat, and extreme cold are said to be mediated by nociceptors. The term chemoreceptor is used to refer to receptors stimulated by a change in the chemical composition of the environment in which they are located. These include receptors for taste and smell as well as visceral receptors such as those sensitive to changes in the plasma level of O2, pH, and osmolality. Photoreceptors are those in the rods and cones in the retina that respond to light. SENSE ORGANS Sensory receptors can be specialized dendritic endings of af-ferent nerve fibers, and they are often associated with nonneu-ral cells that surround it, forming a sense organ. Touch and pressure are sensed by four types of mechanoreceptors (Figure 8–1). Meissner corpuscles are dendrites encapsulated in con-nective tissue and respond to changes in texture and slow vi-brations. Merkel cells are expanded dendritic endings, and they respond to sustained pressure and touch. Ruffini corpus-cles are enlarged dendritic endings with elongated capsules, and they respond to sustained pressure. Pacinian corpuscles consist of unmyelinated dendritic endings of a sensory nerve fiber, 2 μm in diameter, encapsulated by concentric lamellae of connective tissue that give the organ the appearance of a cocktail onion. Theses receptors respond to deep pressure and fast vibration. The Na+ channel BNC1 is closely associated with touch receptors. This channel is one of the degenerins, so called because when they are hyperexpressed, they cause the neu-rons they are in to degenerate. However, it is not known if BNC1 is part of the receptor complex or the neural fiber at the point of initiation of the spike potential. The receptor may be opened mechanically by pressure on the skin. Some sensory receptors are not specialized organs but rather are free nerve endings. Pain and temperature sensa-tions arise from unmyelinated dendrites of sensory neurons located around hair follicles throughout the glaborous and hairy skin as well as deep tissue. TABLE 8–1 Principle sensory modalities. Sensory System Modality Stimulus Energy Receptor Class Receptor Cell Types Somatosensory Touch Tap, flutter 5–40 Hz Cutaneous mechanoreceptor Meissner corpuscles Somatosensory Touch Motion Cutaneous mechanoreceptor Hair follicle receptors Somatosensory Touch Deep pressure, vibration 60–300 Hz Cutaneous mechanoreceptor Pacinian corpuscles Somatosensory Touch Touch, pressure Cutaneous mechanoreceptor Merkel cells Somatosensory Touch Sustained pressure Cutaneous mechanoreceptor Ruffini corpuscles Somatosensory Proprioception Stretch Mechanoreceptor Muscle spindles Somatosensory Proprioception Tension Mechanoreceptor Golgi tendon organ Somatosensory Temperature Thermal Thermoreceptor Cold and warm receptors Somatosensory Pain Chemical, thermal, and mechanical Chemoreceptor, thermorecep-tor, and mechanoreceptor Polymodal receptors or chemical, thermal, and mechanical nociceptors Somatosensory Itch Chemical Chemoreceptor Chemical nociceptor Visual Vision Light Photoreceptor Rods, cones Auditory Hearing Sound Mechanoreceptor Hair cells (cochlea) Vestibular Balance Angular acceleration Mechanoreceptor Hair cells (semicircular canals) Vestibular Balance Linear acceleration, gravity Mechanoreceptor Hair cells (otolith organs) Olfactory Smell Chemical Chemoreceptor Olfactory sensory neuron Gustatory Taste Chemical Chemoreceptor Taste buds CHAPTER 8 Properties of Sensory Receptors 151 GENERATION OF IMPULSES IN CUTANEOUS RECEPTORS PACINIAN CORPUSCLES The way receptors generate action potentials in the sensory nerves that innervate them varies with the complexity of the sense organ. In the skin, the Pacinian corpuscle has been studied in some detail. As noted above, the Pacinian corpus-cles are touch receptors. Because of their relatively large size and accessibility, they can be isolated, studied with microelec-trodes, and subjected to microdissection. The myelin sheath of the sensory nerve begins inside the corpuscle (Figure 8–2). The first node of Ranvier is also located inside, whereas the second is usually near the point at which the nerve fiber leaves the corpuscle. GENERATOR POTENTIALS Recording electrodes can be placed on the sensory nerve as it leaves a Pacinian corpuscle and graded pressure applied to the corpuscle. When a small amount of pressure is applied, a non-propagated depolarizing potential resembling an EPSP is re-corded. This is called the generator potential or receptor potential (Figure 8–2). As the pressure is increased, the mag-nitude of the receptor potential increases. When the magni-tude of the generator potential is about 10 mV, an action potential is generated in the sensory nerve. As the pressure is further increased, the generator potential becomes even larger and the sensory nerve fires repetitively. SOURCE OF THE GENERATOR POTENTIAL By microdissection techniques, it has been shown that removal of the connective tissue lamellas from the unmyelinated nerve FIGURE 8–1 Sensory systems encode four elementary attributes of stimuli: modality, location (receptive field), intensity, and duration (timing). A) The human hand has four types of mechanoreceptors; their combined activation produces the sensation of contact with an object. Selective activation of Merkel cells and Ruffini endings causes sensation of steady pressure; selective activation of Meissner’s and Pacinian corpuscles causes tingling and vibratory sensation. B) Location of a stimulus is encoded by spatial distribution of the population of receptors ac-tivated. A receptor fires only when the skin close to its sensory terminals is touched. These receptive fields of mechanoreceptors (shown as red areas on fingertips) differ in size and response to touch. Merkel cells and Meissner’s corpuscles provide the most precise localization as they have the smallest receptive fields and are most sensitive to pressure applied by a small probe. C) Stimulus intensity is signaled by firing rates of individ-ual receptors; duration of stimulus is signaled by time course of firing. The spike trains indicate action potentials elicited by pressure from a small probe at the center of each receptive field. Meissner’s and Pacinian corpuscles adapt rapidly, the others adapt slowly. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Touch Meissner’s corpuscle Stimulus Neural spike train Receptors A Modality B Location C Intensity and time course Receptive field Merkel cells Pacinian corpuscle Ruffini endings 152 SECTION II Physiology of Nerve & Muscle Cells ending in a Pacinian corpuscle does not abolish the generator po-tential. When the first node of Ranvier is blocked by pressure or narcotics, the generator potential is unaffected but conducted im-pulses are abolished (Figure 8–2). When the sensory nerve is sec-tioned and the nonmyelinated terminal is allowed to degenerate, no generator potential is formed. These and other experiments have established that the generator potential is produced in the unmyelinated nerve terminal. The receptor therefore converts mechanical energy into an electrical response, the magnitude of which is proportionate to the intensity of the stimulus. The gen-erator potential in turn depolarizes the sensory nerve at the first node of Ranvier. Once the firing level is reached, an action poten-tial is produced and the membrane repolarizes. If the generator potential is great enough, the neuron fires again as soon as it re-polarizes, and it continues to fire as long as the generator poten-tial is large enough to bring the membrane potential of the node to the firing level. Thus, the node converts the graded response of the receptor into action potentials, the frequency of which is pro-portionate to the magnitude of the applied stimuli. SENSORY CODING Converting a receptor stimulus to a recognizable sensation is termed sensory coding. All sensory systems code for four ele-mentary attributes of a stimulus: modality, location, intensity, and duration. Modality is the type of energy transmitted by the stimulus. Location is the site on the body or space where the stimulus originated. Intensity is signaled by the response am-plitude or frequency of action potential generation. Duration refers to the time from start to end of a response in the receptor. These attributes of sensory coding are shown for the modality of touch in Figure 8–1. MODALITY Humans have four basic classes of receptors based on their sen-sitivity to one predominant form of energy: mechanical, ther-mal, electromagnetic, or chemical. The particular form of energy to which a receptor is most sensitive is called its adequate stim-ulus. The adequate stimulus for the rods and cones in the eye, for example, is light (an example of electromagnetic energy). Recep-tors do respond to forms of energy other than their adequate stimuli, but the threshold for these nonspecific responses is much higher. Pressure on the eyeball will stimulate the rods and cones, for example, but the threshold of these receptors to pres-sure is much higher than the threshold of the pressure receptors in the skin. LOCATION The term sensory unit is applied to a single sensory axon and all its peripheral branches. These branches vary in number but may be numerous, especially in the cutaneous senses. The receptive field of a sensory unit is the spatial distribution from which a stimulus produces a response in that unit (Figure 8–1). Repre-sentation of the senses in the skin is punctate. If the skin is care-fully mapped, millimeter by millimeter, with a fine hair, a sensation of touch is evoked from spots overlying these touch re-ceptors. None is evoked from the intervening areas. Similarly, temperature sensations and pain are produced by stimulation of FIGURE 8–2 Demonstration that the generator potential in a Pacinian corpuscle originates in the unmyelinated nerve terminal. (1) The electrical responses to a pressure of 1× (record a), 2× (b), 3× (c), and 4× (d) were recorded. The strongest stimulus produced an action po-tential in the sensory nerve (e). (2) Similar responses persisted after removal of the connective tissue capsule, except that the responses were more prolonged because of partial loss of adaptation. (3) The generator responses persisted but the action potential was absent when the first node of Ranvier was blocked by pressure or with narcotics (arrow). (4) All responses disappeared when the sensory nerve was cut and allowed to degen-erate before the experiment. 1 2 4 3 d e e cb a dcb a dcb a abcd CHAPTER 8 Properties of Sensory Receptors 153 the skin only over the spots where the receptors for these modal-ities are located. In the cornea and adjacent sclera of the eye, the surface area supplied by a single sensory unit is 50–200 mm2. Generally, the areas supplied by one unit overlap and interdigi-tate with the areas supplied by others. One of the most important mechanisms that enable local-ization of a stimulus site is lateral inhibition. Information from sensory neurons whose receptors are at the peripheral edge of the stimulus is inhibited compared to information from the sensory neurons at the center of the stimulus. Thus, lateral inhibition enhances the contrast between the center and periphery of a stimulated area and increases the ability of the brain to localize a sensory input. Lateral inhibition under-lies two-point discrimination (see Clinical Box 8–1). INTENSITY The intensity of sensation is determined by the amplitude of the stimulus applied to the receptor. This is illustrated in Figure 8–3. As a greater pressure is applied to the skin, the receptor poten-tial in the mechanoreceptor increases (not shown), and the fre-quency of the action potentials in a single axon transmitting information to the CNS is also increased. In addition to increas-ing the firing rate in a single axon, the greater intensity of stim-ulation also will recruit more receptors into the receptive field. It has long been taught that the magnitude of the sensation felt is proportional to the log of the intensity of the stimulus (Weber–Fechner law). It now appears, however, that a power function more accurately describes this relation. In other words, R = KSA where R is the sensation felt, S is the intensity of the stimulus, and, for any specific sensory modality, K and A are constants. The frequency of the action potentials generated in a sensory nerve fiber is also related to the intensity of the initiating stimulus by a power function. An example of this relation is shown is shown in Figure 8–4, in which the calculated expo-nent is 0.52. However, the relation between direct stimulation of a sensory nerve and the sensation felt is linear. Conse-quently, it appears that for any given sensory modality, the relation between sensation and stimulus intensity is deter-mined primarily by the properties of the peripheral receptors. DURATION When a maintained stimulus of constant strength is applied to a receptor, the frequency of the action potentials in its sensory nerve declines over time. This phenomenon is known as adap-tation or desensitization. The degree to which adaptation oc-curs varies from one sense to another. Receptors can be classified into rapidly adapting (phasic) receptors and slowly adapting (tonic) receptors. This is illustrated for different types of touch receptors in Figure 8–1. Meissner and Pacinian corpuscles are examples of rapidly adapting receptors, and Merkel cells and Ruffini endings are examples of slowly adapt-ing receptors. Other examples of slowly adapting receptors are muscle spindles and nociceptors. Different types of sensory adaptation appear to have some value to the individual. Light touch would be distracting if it were persistent; and, converse-ly, slow adaptation of spindle input is needed to maintain pos-ture. Similarly, input from nociceptors provides a warning that would lose its value if it adapted and disappeared. SENSORY INFORMATION The speed of conduction and other characteristics of sensory nerve fibers vary, but action potentials are similar in all nerves. The action potentials in the nerve from a touch receptor, for example, are essentially identical to those in the nerve from a warmth receptor. This raises the question of why stimulation of a touch receptor causes a sensation of touch and not of warmth. It also raises the question of how it is possible to tell whether the touch is light or heavy. LAW OF SPECIFIC NERVE ENERGIES The sensation evoked by impulses generated in a receptor de-pends in part on the specific part of the brain they ultimately activate. The specific sensory pathways are discrete from sense CLINICAL BOX 8–1 Two-Point Discrimination The size of the receptive fields for light touch can be mea-sured by the two-point threshold test. In this procedure, the two points on a pair of calipers are simultaneously posi-tioned on the skin and one determines the minimum dis-tance between the two caliper points that can be perceived as separate points of stimulation. This is called the two-point discrimination threshold. If the distance is very small, each caliper point is touching the receptive field of only one sen-sory neuron. If the distance between stimulation points is less than this threshold, only one point of stimulation can be felt. Thus, the two-point discrimination threshold is a meas-ure of tactile acuity. The magnitude of two-point discrimi-nation thresholds varies from place to place on the body and is smallest where touch receptors are most abundant. Stimu-lus points on the back, for instance, must be separated by at least 65 mm before they can be distinguished as separate, whereas on the fingertips two stimuli are recognized if they are separated by as little as 2 mm. Blind individuals benefit from the tactile acuity of fingertips to facilitate the ability to read Braille; the dots forming Braille symbols are separated by 2.5 mm. Two-point discrimination is used to test the in-tegrity of the dorsal column (medial lemniscus) system, the central pathway for touch and proprioception. 154 SECTION II Physiology of Nerve & Muscle Cells FIGURE 8–3 Relationship between stimulus and impulse frequency in an afferent fiber. Action potentials in an afferent fiber from a mechanoreceptor of a single sensory unit increase in frequency as branches of the afferent neuron are stimulated by pressure of increasing mag-nitude. (From Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology. McGraw-Hill, 2008.) FIGURE 8–4 Relation between magnitude of touch stimulus (S) and frequency of action potentials in sensory nerve fibers (R). Dots are individual values from cats and are plotted on linear coordinates (left) and log–log coordinates (right). The equation shows the calculated power function relationship between R and S. (Reproduced, with permission, from Werner G, Mountcastle VB: Neural activity in mechanoreceptive cutaneous afferents. Stimulus–response relations, Weber functions, and information transmission. J Neurophysiol 1965;28:359.) Action potentials Skin Time Pressure (mmHg) 180 120 60 Glass probe Afferent neuron 100 90 80 70 60 50 40 30 20 10 S (% maximum stimulus) R = 9.4 S0.52 R (% response to maximum stimulus) 10 20 30 40 50 60 70 80 90 100 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9 2.1 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 .9 .8 .7 .6 .5 .4 .3 .2 Log S Log R CHAPTER 8 Properties of Sensory Receptors 155 organ to cortex. Therefore, when the nerve pathways from a particular sense organ are stimulated, the sensation evoked is that for which the receptor is specialized no matter how or where along the pathway the activity is initiated. This princi-ple, first enunciated by Müller in 1835, has been given the rather cumbersome name of the law of specific nerve ener-gies. For example, if the sensory nerve from a Pacinian cor-puscle in the hand is stimulated by pressure at the elbow or by irritation from a tumor in the brachial plexus, the sensation evoked is touch. Similarly, if a fine enough electrode could be inserted into the appropriate fibers of the dorsal columns of the spinal cord, the thalamus, or the postcentral gyrus of the cerebral cortex, the sensation produced by stimulation would be touch. The general principle of specific nerve energies re-mains one of the cornerstones of sensory physiology. LAW OF PROJECTION No matter where a particular sensory pathway is stimulated along its course to the cortex, the conscious sensation pro-duced is referred to the location of the receptor. This principle is called the law of projection. Cortical stimulation experi-ments during neurosurgical procedures on conscious patients illustrate this phenomenon. For example, when the cortical re-ceiving area for impulses from the left hand is stimulated, the patient reports sensation in the left hand, not in the head. RECRUITMENT OF SENSORY UNITS As the strength of a stimulus is increased, it tends to spread over a large area and generally not only activates the sense organs im-mediately in contact with it but also “recruits” those in the sur-rounding area. Furthermore, weak stimuli activate the receptors with the lowest thresholds, and stronger stimuli also activate those with higher thresholds. Some of the receptors activated are part of the same sensory unit, and impulse frequency in the unit therefore increases. Because of overlap and interdigitation of one unit with another, however, receptors of other units are also stimulated, and consequently more units fire. In this way, more afferent pathways are activated, which is interpreted in the brain as an increase in intensity of the sensation. NEUROLOGICAL EXAM The sensory component of a neurological exam includes an assessment of various sensory modalities including touch, proprioception, vibratory sense, and pain. Clinical Box 8–2 describes the test for vibratory sensibility. Cortical sensory function can be tested by placing familiar objects in a patient’s hands and asking him or her to identify it with the eyes closed (see Clinical Box 8–3). CHAPTER SUMMARY ■Sensory receptors are commonly classified as mechanorecep-tors, nociceptors, chemoreceptors, or photoreceptors. ■Touch and pressure are sensed by four types of mechanoreceptors: Meissner’s corpuscles (respond to changes in texture and slow vi-brations), Merkel’s cells (respond to sustained pressure and touch), Ruffini corpuscles (respond to sustained pressure), and Pa-cinian corpuscles (respond to deep pressure and fast vibrations). ■Nociceptors and thermoreceptors are free nerve endings on un-myelinated or lightly myelinated fibers in hairy and glaborous skin and deep tissues. ■The generator or receptor potential is the nonpropagated depo-larizing potential recorded in a sensory organ after an adequate CLINICAL BOX 8–2 Vibratory Sensibility Vibratory sensibility is tested by applying a vibrating (128-Hz) tuning fork to the skin on the fingertip, tip of the toe, or bony prominences of the toes. The normal response is a “buzzing” sensation. The sensation is most marked over bones. The term pallesthesia is also used to describe this ability to feel mechanical vibrations. The receptors involved are the receptors for touch, especially Pacinian corpuscles, but a time factor is also necessary. A pattern of rhythmic pressure stimuli is interpreted as vibration. The impulses re-sponsible for the vibrating sensation are carried in the dor-sal columns. Degeneration of this part of the spinal cord oc-curs in poorly controlled diabetes, pernicious anemia, vitamin B12 deficiencies, or early tabes dorsalis. Elevation of the threshold for vibratory stimuli is an early symptom of this degeneration. Vibratory sensation and proprioception are closely related; when one is diminished, so is the other. CLINICAL BOX 8–3 Stereognosis Stereognosis is the perception of the form and nature of an object without looking at it. Normal persons can readily identify objects such as keys and coins of various denomina-tions. This ability depends on relatively intact touch and pressure sensation and is compromised when the dorsal col-umns are damaged. The inability to identify an object by touch is called tactile agnosia. It also has a large cortical component; impaired stereognosis is an early sign of dam-age to the cerebral cortex and sometimes occurs in the ab-sence of any detectable defect in touch and pressure sensa-tion when there is a lesion in the parietal lobe posterior to the postcentral gyrus. Stereoagnosia can also be expressed by the failure to identify an object by sight (visual agnosia), the inability to identify sounds or words (auditory agnosia) or color (color agnosia), or the inability to identify the loca-tion or position of an extremity (position agnosia). 156 SECTION II Physiology of Nerve & Muscle Cells stimulus is applied. As the stimulus is increased, the magnitude of the receptor potential increases. When it reaches a critical threshold, an action potential is generated in the sensory nerve. ■Converting a receptor stimulus to a recognizable sensation is termed sensory coding. All sensory systems code for four ele-mentary attributes of a stimulus: modality, location, intensity, and duration. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Pacinian corpuscles are A) a type of thermoreceptor. B) usually innervated by Aδ nerve fibers. C) rapidly adapting touch receptors. D) slowly adapting touch receptors. E) nociceptors. 2. Adaptation to a sensory stimulus produces A) a diminished sensation when other types of sensory stimuli are withdrawn. B) a more intense sensation when a given stimulus is applied repeatedly. C) a sensation localized to the hand when the nerves of the bra-chial plexus are stimulated. D) a diminished sensation when a given stimulus is applied repeatedly over time. E) a decreased firing rate in the sensory nerve from the recep-tor when one’s attention is directed to another matter. 3. Sensory systems code for the following attributes of a stimulus: A) modality, location, intensity, and duration B) threshold, receptive field, adaptation, and discrimination C) touch, taste, hearing, and smell D) threshold, laterality, sensation, and duration E) sensitization, discrimination, energy, and projection 4. In which of the following is the frequency of stimulation not lin-early related to the strength of the sensation felt? A) sensory area of the cerebral cortex B) specific projection nuclei of the thalamus C) lateral spinothalamic tract D) dorsal horn E) cutaneous receptors 5. Which of the following receptors and sense organs are incorrectly paired? A) rods and cones : eye B) receptors sensitive to sodium : taste buds C) hair cells : olfactory epithelium D) receptors sensitive to stretch : carotid sinus E) glomus cells : carotid body 6. Which best describes the law of specific nerve energies? A) No matter where a particular sensory pathway is stimulated along its course to the cortex, the conscious sensation pro-duced is referred to the location of the receptor. B) A nerve can only be stimulated by electrical energy. C) Receptors can respond to forms of energy other than their adequate stimuli, but the threshold for these nonspecific responses is much higher. D) For any given sensory modality, the specific relationship between sensation and stimulus intensity is determined by the properties of the peripheral receptors. E) The sensation evoked by impulses generated in a receptor depends in part on the specific part of the brain they ulti-mately activate. 7. Which of the following does not contain cation channels that are activated by mechanical distortion, producing depolarization? A) olfactory receptors B) Pacinian corpuscles C) hair cells in cochlea D) hair cells in semicircular canals E) hair cells in utricle CHAPTER RESOURCES Barlow HB, Mollon JD (editors): The Senses. Cambridge University Press, 1982. Bell J, Bolanowski S, Holmes MH: The structure and function of Pacinian corpuscles: A review. Prog Neurobiol 1994;42:79. Haines DE (editor): Fundamental Neuroscience for Basic and Clinical Applications, 3rd ed. Elsevier, 2006. Iggo A (editor): Handbook of Sensory Physiology. Vol 2, Somatosensory System. Springer-Verlag, 1973. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. Mountcastle VB: Perceptual Neuroscience. Harvard University Press, 1999. Squire LR, et al (editors): Fundamental Neuroscience, 3rd ed. Academic Press, 2008. 157 C H A P T E R 9 Reflexes O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the components of a reflex arc. ■Describe the muscle spindles and their role in the stretch reflex. ■Describe the Golgi tendon organs and analyze their function as part of a feedback system that maintains muscle force. ■Define reciprocal innervation, inverse stretch reflex, clonus, and lengthening reaction. INTRODUCTION The basic unit of integrated reflex activity is the reflex arc. This arc consists of a sense organ, an afferent neuron, one or more synapses within a central integrating station, an efferent neuron, and an effector. In mammals, the connection between afferent and efferent somatic neurons is generally in the brain or spinal cord. The afferent neurons enter via the dorsal roots or cranial nerves and have their cell bodies in the dorsal root ganglia or in the homologous ganglia on the cranial nerves. The efferent fibers leave via the ventral roots or correspond-ing motor cranial nerves. The principle that in the spinal cord the dorsal roots are sensory and the ventral roots are motor is known as the Bell–Magendie law. Activity in the reflex arc starts in a sensory receptor with a receptor potential whose magnitude is proportional to the strength of the stimulus (Figure 9–1). This generates all-or-none action potentials in the afferent nerve, the number of action potentials being proportional to the size of the genera-tor potential. In the central nervous system (CNS), the responses are again graded in terms of excitatory postsynaptic potentials (EPSPs) and inhibitory postsynaptic potentials (IPSPs) at the synaptic junctions. All-or-none responses are generated in the efferent nerve. When these reach the effector, they again set up a graded response. When the effector is smooth muscle, responses summate to produce action poten-tials in the smooth muscle, but when the effector is skeletal muscle, the graded response is always adequate to produce action potentials that bring about muscle contraction. The connection between the afferent and efferent neurons is usu-ally in the CNS, and activity in the reflex arc is modified by the multiple inputs converging on the efferent neurons or at any synaptic station within the reflex loop. The simplest reflex arc is one with a single synapse between the afferent and efferent neurons. Such arcs are monosynaptic, and reflexes occurring in them are called monosynaptic reflexes. Reflex arcs in which one or more interneuron is inter-posed between the afferent and efferent neurons are called polysynaptic reflexes. There can be anywhere from two to hundreds of synapses in a polysynaptic reflex arc. 158 SECTION II Physiology of Nerve & Muscle Cells MONOSYNAPTIC REFLEXES: THE STRETCH REFLEX When a skeletal muscle with an intact nerve supply is stretched, it contracts. This response is called the stretch reflex. The stimu-lus that initiates the reflex is stretch of the muscle, and the re-sponse is contraction of the muscle being stretched. The sense organ is a small encapsulated spindlelike or fusiform shaped structure called the muscle spindle, located within the fleshy part of the muscle. The impulses originating from the spindle are transmitted to the CNS by fast sensory fibers that pass directly to the motor neurons which supply the same muscle. The neu-rotransmitter at the central synapse is glutamate. The stretch re-flex is the best known and studied monosynaptic reflex and is typified by the knee jerk reflex (see Clinical Box 9–1). STRUCTURE OF MUSCLE SPINDLES Each muscle spindle has three essential elements: (1) a group of specialized intrafusal muscle fibers with contractile polar ends and a noncontractile center, (2) large diameter myelinat-ed afferent nerves (types Ia and II) originating in the central portion of the intrafusal fibers, and (3) small diameter myeli-nated efferent nerves supplying the polar contractile regions of the intrafusal fibers (Figure 9–2A). It is important to under-stand the relationship of these elements to each other and to the muscle itself to appreciate the role of this sense organ in signaling changes in the length of the muscle in which it is lo-cated. Changes in muscle length are associated with changes in joint angle; thus muscle spindles provide information on po-sition (ie, proprioception). The intrafusal fibers are positioned in parallel to the extrafusal fibers (the regular contractile units of the muscle) with the ends of the spindle capsule attached to the tendons at either end of the muscle. Intrafusal fibers do not contribute to the overall contractile force of the muscle, but rather serve a pure sensory function. There are two types of intrafusal fibers in mammalian muscle spindles. The first type contains many nuclei in a dilated central area and is called a nuclear bag fiber (Figure 9–2B). There are two subtypes of nuclear bag fibers, dynamic and static. Typically, there are two or three nuclear bag fibers per spindle. The second intrafusal fiber type, the nuclear chain fiber, is thinner and shorter and lacks a definite bag. Each spindle has about five of these fibers. There are two kinds of sensory endings in each spindle, a single primary (group Ia) ending and up to eight secondary (group II) endings. The Ia afferent fiber wraps around the cen-ter of the dynamic and static nuclear bag fibers and nuclear chain fibers. Group II sensory fibers are located adjacent to the centers of the static nuclear bag and nuclear chain fibers; these fibers do not innervate the dynamic nuclear bag fibers. Ia affer-ents are very sensitive to the velocity of the change in muscle length during a stretch (dynamic response); thus they provide information about the speed of movements and allow for quick corrective movements. The steady-state (tonic) activity of group Ia and II afferents provide information on steady-state length of the muscle (static response). The top trace in Figure 9–2C shows the dynamic and static components of activity in a Ia afferent during muscle stretch. Note that they discharge most rapidly while the muscle is being stretched (shaded area of graphs) and less rapidly during sustained stretch. The spindles have a motor nerve supply of their own. These nerves are 3–6 μm in diameter, constitute about 30% of the fibers in the ventral roots, and are called γ-motor neurons. There are two types of γ-motor neurons: dynamic, which sup-ply the dynamic nuclear bag fibers and static, which supply the static nuclear bag fibers and the nuclear chain fibers. Acti-vation of dynamic γ-motor neurons increases the dynamic sensitivity of the group Ia endings. Activation of the static γ-motor neurons increases the tonic level of activity in both group Ia and II endings, decreases the dynamic sensitivity of group Ia afferents, and can prevent silencing of Ia afferents during muscle stretch (Figure 9–2C). FIGURE 9–1 The reflex arc. Note that at the receptor and in the CNS a nonpropagated graded response occurs that is proportionate to the magnitude of the stimulus. The response at the neuromuscular junction is also graded, though under normal conditions it is always large enough to produce a response in skeletal muscle. On the other hand, in the portions of the arc specialized for transmission (afferent and efferent axons, muscle membrane), the responses are all-or-none action potentials. Sense organ Afferent neuron Efferent neuron Neuromuscular junction Synapse Muscle Generator potential Action potentials EPSPs (and IPSPs) Action potentials Endplate potentials Action potentials CHAPTER 9 Reflexes 159 CLINICAL BOX 9–1 Knee Jerk Reflex Tapping the patellar tendon elicits the knee jerk, a stretch reflex of the quadriceps femoris muscle, because the tap on the tendon stretches the muscle. A similar contraction is ob-served if the quadriceps is stretched manually. Stretch re-flexes can also be elicited from most of the large muscles of the body. Tapping on the tendon of the triceps brachii, for example, causes an extensor response at the elbow as a re-sult of reflex contraction of the triceps; tapping on the Achil-les tendon causes an ankle jerk due to reflex contraction of the gastrocnemius; and tapping on the side of the face causes a stretch reflex in the masseter. The knee jerk reflex is an example of a deep tendon reflex (DTR) in a neurological exam and is graded on the following scale: 0 (absent), 1+ (hy-poactive), 2+ (brisk, normal), 3+ (hyperactive without clonus), 4+ (hyperactive with mild clonus), and 5+ (hyperactive with sustained clonus). Absence of the knee jerk can signify an ab-normality anywhere within the reflex arc, including the mus-cle spindle, the Ia afferent nerve fibers, or the motor neurons to the quadriceps muscle. The most common cause is a pe-ripheral neuropathy from such things as diabetes, alcoholism, and toxins. A hyperactive reflex can signify an interruption of corticospinal and other descending pathways that influence the reflex arc. FIGURE 9–2 Mammalian muscle spindle. A) Diagrammatic representation of the main components of mammalian muscle spindle includ-ing intrafusal muscle fibers, afferent sensory fiber endings, and efferent motor fibers (γ-motor neurons). B) Three types of intrafusal muscle fibers: dynamic nuclear bag, static nuclear bag, and nuclear chain fibers. A single Ia afferent fiber innervates all three types of fibers to form a primary sensory ending. A group II sensory fiber innervates nuclear chain and static bag fibers to form a secondary sensory ending. Dynamic γ-motor neu-rons innervate dynamic bag fibers; static γ-motor neurons innervate combinations of chain and static bag fibers. C) Comparison of discharge pat-tern of Ia afferent activity during stretch alone and during stimulation of static or dynamic γ-motor neurons. Without γ-stimulation, Ia fibers show a small dynamic response to muscle stretch and a modest increase in steady-state firing. When static γ-motor neurons are activated, the steady-state response increases and the dynamic response decreases. When dynamic γ-motor neurons are activated, the dynamic response is markedly increased but the steady-state response gradually returns to its original level. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Afferent axons Efferent axons Gamma motor endings Sensory endings Capsule Intrafusal muscle fibers Ia II Static Dynamic A Muscle spindle Dynamic nuclear bag fiber Static nuclear bag fiber Nuclear chain fiber B Intrafusal fibers of the muscle spindle C Response of Ia sensory fiber to selective activation of motor neurons Stretch alone Dynamic response Steady-state response 0 Stimulate static gamma fiber 200 0 Stimulate dynamic gamma fiber Stretch Imp/s 200 0 6 0.2 s Imp/s 0 200 Imp/s 160 SECTION II Physiology of Nerve & Muscle Cells CENTRAL CONNECTIONS OF AFFERENT FIBERS Ia fibers end directly on motor neurons supplying the extrafus-al fibers of the same muscle (Figure 9–3). The time between the application of the stimulus and the response is called the reac-tion time. In humans, the reaction time for a stretch reflex such as the knee jerk is 19–24 ms. Weak stimulation of the sen-sory nerve from the muscle, known to stimulate only Ia fibers, causes a contractile response with a similar latency. Because the conduction velocities of the afferent and efferent fiber types are known and the distance from the muscle to the spinal cord can be measured, it is possible to calculate how much of the reac-tion time was taken up by conduction to and from the spinal cord. When this value is subtracted from the reaction time, the remainder, called the central delay, is the time taken for the re-flex activity to traverse the spinal cord. In humans, the central delay for the knee jerk is 0.6–0.9 ms, and figures of similar mag-nitude have been found in experimental animals. Because the minimal synaptic delay is 0.5 ms, only one synapse could have been traversed. Muscle spindles also make connections that cause muscle contraction via polysynaptic pathways, and the afferents involved are probably those from the secondary endings. However, group II fibers also make monosynaptic connec-tions to the motor neurons and make a small contribution to the stretch reflex. FUNCTION OF MUSCLE SPINDLES When the muscle spindle is stretched, its sensory endings are distorted and receptor potentials are generated. These in turn set up action potentials in the sensory fibers at a frequency pro-portional to the degree of stretching. Because the spindle is in parallel with the extrafusal fibers, when the muscle is passively stretched, the spindles are also stretched, referred to as “loading the spindle.” This initiates reflex contraction of the extrafusal fi-bers in the muscle. On the other hand, the spindle afferents characteristically stop firing when the muscle is made to con-tract by electrical stimulation of the α-motor neurons to the ex-trafusal fibers because the muscle shortens while the spindle is unloaded (Figure 9–4). Thus, the spindle and its reflex connections constitute a feedback device that operates to maintain muscle length; if the muscle is stretched, spindle discharge increases and reflex shortening is produced, whereas if the muscle is shortened without a change in γ-motor neuron discharge, spindle affer-ent activity decreases and the muscle relaxes. Dynamic and static responses of muscle spindle afferents influence physio-logical tremor (see Clinical Box 9–2). EFFECTS OF γ-MOTOR NEURON DISCHARGE Stimulation of γ-motor neurons produces a very different pic-ture from that produced by stimulation of the extrafusal fibers. Such stimulation does not lead directly to detectable contrac-tion of the muscles because the intrafusal fibers are not strong enough or plentiful enough to cause shortening. However, stimulation does cause the contractile ends of the intrafusal fi-bers to shorten and therefore stretches the nuclear bag portion of the spindles, deforming the endings and initiating impulses in the Ia fibers (Figure 9–4). This in turn can lead to reflex con-traction of the muscle. Thus, muscle can be made to contract via stimulation of the α-motor neurons that innervate the ex-trafusal fibers or the γ-motor neurons that initiate contraction indirectly via the stretch reflex. If the whole muscle is stretched during stimulation of the γ-motor neurons, the rate of discharge in the Ia fibers is further increased (Figure 9–4). Increased γ-motor neuron activity thus increases spindle sensitivity during stretch. In response to descending excitatory input to spinal motor circuits, both α- and γ-motor neurons are activated. Because of this “α–γ coactivation,” intrafusal and extrafusal fibers shorten together, and spindle afferent activity can occur throughout the period of muscle contraction. In this way, the spindle remains capable of responding to stretch and reflexly adjusting α-motor neuron discharge. FIGURE 9–3 Diagram illustrating the pathways responsible for the stretch reflex and the inverse stretch reflex. Stretch stimu-lates the muscle spindle, which activates Ia fibers that excite the motor neuron. Stretch also stimulates the Golgi tendon organ, which acti-vates Ib fibers that excite an interneuron that releases the inhibitory mediator glycine. With strong stretch, the resulting hyperpolarization of the motor neuron is so great that it stops discharging. Motor endplate on extrafusal fiber Ventral root Motor neuron Dorsal root Interneuron releasing inhibitory mediator Ib fiber from Golgi tendon organ from muscle spindle a fiber I CHAPTER 9 Reflexes 161 CONTROL OF γ-MOTOR NEURON DISCHARGE The γ-motor neurons are regulated to a large degree by de-scending tracts from a number of areas in the brain. Via these pathways, the sensitivity of the muscle spindles and hence the threshold of the stretch reflexes in various parts of the body can be adjusted and shifted to meet the needs of postural control. Other factors also influence γ-motor neuron discharge. Anxi-ety causes an increased discharge, a fact that probably explains the hyperactive tendon reflexes sometimes seen in anxious patients. In addition, unexpected movement is associated with a greater efferent discharge. Stimulation of the skin, especially by noxious agents, increases γ-motor neuron discharge to ipsilateral flexor muscle spindles while decreasing that to extensors and produces the opposite pattern in the opposite limb. It is well known that trying to pull the hands apart when the flexed fingers are hooked together facilitates the knee jerk reflex (Jendrassik’s maneuver), and this may also be due to increased γ-motor neu-ron discharge initiated by afferent impulses from the hands. RECIPROCAL INNERVATION When a stretch reflex occurs, the muscles that antagonize the ac-tion of the muscle involved (antagonists) relax. This phenome-non is said to be due to reciprocal innervation. Impulses in the Ia fibers from the muscle spindles of the protagonist muscle cause postsynaptic inhibition of the motor neurons to the antag-onists. The pathway mediating this effect is bisynaptic. A collat-eral from each Ia fiber passes in the spinal cord to an inhibitory interneuron that synapses on a motor neuron supplying the FIGURE 9–4 Effect of various conditions on muscle spindle discharge. When the whole muscle is stretched, the muscle spindle is also stretched and its sensory endings are activated at a frequency pro-portional to the degree of stretching (“loading the spindle”). Spindle af-ferents stop firing when the muscle contracts (“unloading the spindle”). Stimulation of γ-motor neurons cause the contractile ends of the in-trafusal fibers to shorten. This stretches the nuclear bag region, initiating impulses in sensory fibers. If the whole muscle is stretched during stim-ulation of the γ-motor neurons, the rate of discharge in sensory fibers is further increased. Increased γ efferent discharge—muscle stretched Increased γ efferent discharge Muscle contracted Muscle stretched Muscle at rest Tendon Spindle Extrafusal fiber Sensory nerve Impulses in sensory nerve CLINICAL BOX 9–2 Physiological Tremor The response of the Ia sensory fiber endings to the dynamic (phasic) as well as the static events in the muscle is impor-tant because the prompt, marked phasic response helps to dampen oscillations caused by conduction delays in the feed-back loop regulating muscle length. Normally a small oscilla-tion occurs in this feedback loop. This physiologic tremor has a low amplitude (barely visible to the naked eye) and a fre-quency of approximately 10 Hz. Physiological tremor is a nor-mal phenomenon which affects everyone while maintaining posture or during movements. However, the tremor would be worse if it were not for the sensitivity of the spindle to velocity of stretch. It can become exaggerated in some situations such as when we are anxious or tired or because of drug toxicity. Numerous factors contribute to the genesis of physiological tremor. It is likely dependent on not only central (inferior ol-ive) sources but also from peripheral factors including motor unit firing rates, reflexes, and mechanical resonance. 162 SECTION II Physiology of Nerve & Muscle Cells antagonist muscles. This example of postsynaptic inhibition is dis-cussed in Chapter 6, and the pathway is illustrated in Figure 6–6. INVERSE STRETCH REFLEX Up to a point, the harder a muscle is stretched, the stronger is the reflex contraction. However, when the tension becomes great enough, contraction suddenly ceases and the muscle re-laxes. This relaxation in response to strong stretch is called the inverse stretch reflex or autogenic inhibition. The receptor for the inverse stretch reflex is in the Golgi ten-don organ (Figure 9–5). This organ consists of a netlike collec-tion of knobby nerve endings among the fascicles of a tendon. There are 3–25 muscle fibers per tendon organ. The fibers from the Golgi tendon organs make up the Ib group of myelinated, rapidly conducting sensory nerve fibers. Stimulation of these Ib fibers leads to the production of IPSPs on the motor neurons that supply the muscle from which the fibers arise. The Ib fibers end in the spinal cord on inhibitory interneurons that in turn terminate directly on the motor neurons (Figure 9–3). They also make excitatory connections with motor neurons supply-ing antagonists to the muscle. Because the Golgi tendon organs, unlike the spindles, are in series with the muscle fibers, they are stimulated by both pas-sive stretch and active contraction of the muscle. The threshold of the Golgi tendon organs is low. The degree of stimulation by passive stretch is not great because the more elastic muscle fibers take up much of the stretch, and this is why it takes a strong stretch to produce relaxation. However, discharge is reg-ularly produced by contraction of the muscle, and the Golgi tendon organ thus functions as a transducer in a feedback cir-cuit that regulates muscle force in a fashion analogous to the spindle feedback circuit that regulates muscle length. The importance of the primary endings in the spindles and the Golgi tendon organs in regulating the velocity of the muscle contraction, muscle length, and muscle force is illustrated by the fact that that section of the afferent nerves to an arm causes the limb to hang loosely in a semiparalyzed state. The organiza-tion of the system is shown in Figure 9–6. The interaction of FIGURE 9–5 Golgi tendon organ. (Reproduced, with permission, from Goss CM [editor]: Gray’s Anatomy of the Human Body, 29th ed. Lea & Febiger, 1973.) Organ of Golgi, showing ramification of nerve fibrils Muscular fibers Tendon bundles Nerve fiber FIGURE 9–6 Block diagram of peripheral motor control system. The dashed line indicates the nonneural feedback from muscle that lim-its length and velocity via the inherent mechanical properties of muscle. γd, dynamic γ-motor neurons; γs, static γ-motor neurons. (Reproduced, with permission, from Houk J in: Medical Physiology, 13th ed. Mount-Castle VB [editor]. Mosby, 1974.) Inter-neurons Force feedback Efferent signal Length and velocity feedback Tendon organs Muscle Muscular force Muscle length Load Spindles α − − + + + γd γs Interneuronal control signal α Control signal γ-Dynamic control signal γ-Static control signal Internal disturbances External forces Length and velocity CHAPTER 9 Reflexes 163 spindle discharge, tendon organ discharge, and reciprocal innervation determines the rate of discharge of α-motor neu-rons (see Clinical Box 9–3). MUSCLE TONE The resistance of a muscle to stretch is often referred to as its tone or tonus. If the motor nerve to a muscle is cut, the muscle offers very little resistance and is said to be flaccid. A hyper-tonic (spastic) muscle is one in which the resistance to stretch is high because of hyperactive stretch reflexes. Somewhere be-tween the states of flaccidity and spasticity is the ill-defined area of normal tone. The muscles are generally hypotonic when the rate of γ-motor neuron discharge is low and hyper-tonic when it is high. When the muscles are hypertonic, the sequence of moderate stretch → muscle contraction, strong stretch → muscle relax-ation is clearly seen. Passive flexion of the elbow, for example, meets immediate resistance as a result of the stretch reflex in the triceps muscle. Further stretch activates the inverse stretch reflex. The resistance to flexion suddenly collapses, and the arm flexes. Continued passive flexion stretches the muscle again, and the sequence may be repeated. This sequence of resistance followed by give when a limb is moved passively is known as the clasp-knife effect because of its resemblance to the closing of a pocket knife. It is also known as the lengthen-ing reaction because it is the response of a spastic muscle to lengthening. POLYSYNAPTIC REFLEXES: THE WITHDRAWAL REFLEX Polysynaptic reflex paths branch in a complex fashion (Figure 9–7). The number of synapses in each of their branches varies. Because of the synaptic delay at each synapse, activity in the branches with fewer synapses reaches the motor neurons first, followed by activity in the longer pathways. This causes pro-longed bombardment of the motor neurons from a single stim-ulus and consequently prolonged responses. Furthermore, some of the branch pathways turn back on themselves, permit-ting activity to reverberate until it becomes unable to cause a propagated transsynaptic response and dies out. Such reverber-ating circuits are common in the brain and spinal cord. WITHDRAWAL REFLEX The withdrawal reflex is a typical polysynaptic reflex that occurs in response to a usually painful stimulation of the skin or sub-cutaneous tissues and muscle. The response is flexor muscle contraction and inhibition of extensor muscles, so that the body part stimulated is flexed and withdrawn from the stimulus. When a strong stimulus is applied to a limb, the response in-cludes not only flexion and withdrawal of that limb but also CLINICAL BOX 9–3 Clonus A characteristic of states in which increased γ-motor neu-ron discharge is present is clonus. This neurologic sign is the occurrence of regular, repetitive, rhythmic contractions of a muscle subjected to sudden, maintained stretch. Only sustained clonus with five or more beats is considered ab-normal. Ankle clonus is a typical example. This is initiated by brisk, maintained dorsiflexion of the foot, and the re-sponse is rhythmic plantar flexion at the ankle. The stretch reflex–inverse stretch reflex sequence may contribute to this response. However, it can occur on the basis of syn-chronized motor neuron discharge without Golgi tendon organ discharge. The spindles of the tested muscle are hy-peractive, and the burst of impulses from them discharges all the motor neurons supplying the muscle at once. The consequent muscle contraction stops spindle discharge. However, the stretch has been maintained, and as soon as the muscle relaxes it is again stretched and the spindles stimulated. Clonus may also occur after disruption of de-scending cortical input to a spinal glycinergic inhibitory in-terneuron called the Renshaw cell. This cell receives exci-tatory input from α-motor neurons via an axon collateral (and in turn it inhibits the same). In addition, cortical fibers activating ankle flexors contact Renshaw cells (as well as type Ia inhibitory interneurons) that inhibit the antagonis-tic ankle extensors. This circuitry prevents reflex stimulation of the extensors when flexors are active. Therefore, when the descending cortical fibers are damaged (upper motor neuron lesion), the inhibition of antagonists is absent. The result is repetitive, sequential contraction of ankle flexors and extensors (clonus). Clonus may be seen in patients with amyotrophic lateral sclerosis, stroke, multiple sclerosis, spi-nal cord damage, and hepatic encephalopathy. FIGURE 9–7 Diagram of polysynaptic connections between afferent and efferent neurons in the spinal cord. The dorsal root fi-ber activates pathway A with three interneurons, pathway B with four interneurons, and pathway C with four interneurons. Note that one of the interneurons in pathway C connects to a neuron that doubles back to other interneurons, forming reverberating circuits. Sensory neuron A C Motor neuron B 164 SECTION II Physiology of Nerve & Muscle Cells extension of the opposite limb. This crossed extensor response is properly part of the withdrawal reflex. Strong stimuli in ex-perimental animals generate activity in the interneuron pool that spreads to all four extremities. This is difficult to demon-strate in normal animals but is easily demonstrated in an animal in which the modulating effects of impulses from the brain have been abolished by prior section of the spinal cord (spinal ani-mal). For example, when the hind limb of a spinal cat is pinched, the stimulated limb is withdrawn, the opposite hind limb extended, the ipsilateral forelimb extended, and the con-tralateral forelimb flexed. This spread of excitatory impulses up and down the spinal cord to more and more motor neurons is called irradiation of the stimulus, and the increase in the num-ber of active motor units is called recruitment of motor units. IMPORTANCE OF THE WITHDRAWAL REFLEX Flexor responses can be produced by innocuous stimulation of the skin or by stretch of the muscle, but strong flexor responses with withdrawal are initiated only by stimuli that are noxious or at least potentially harmful to the animal. These stimuli are therefore called nociceptive stimuli. Sherrington pointed out the survival value of the withdrawal response. Flexion of the stimulated limb gets it away from the source of irritation, and extension of the other limb supports the body. The pattern as-sumed by all four extremities puts the animal in position to run away from the offending stimulus. Withdrawal reflexes are prepotent; that is, they preempt the spinal pathways from any other reflex activity taking place at the moment. Many of the characteristics of polysynaptic reflexes can be demonstrated by studying the withdrawal reflex. A weak nox-ious stimulus to one foot evokes a minimal flexion response; stronger stimuli produce greater and greater flexion as the stimulus irradiates to more and more of the motor neuron pool supplying the muscles of the limb. Stronger stimuli also cause a more prolonged response. A weak stimulus causes one quick flexion movement; a strong stimulus causes prolonged flexion and sometimes a series of flexion movements. This prolonged response is due to prolonged, repeated firing of the motor neu-rons. The repeated firing is called after-discharge and is due to continued bombardment of motor neurons by impulses arriv-ing by complicated and circuitous polysynaptic paths. As the strength of a noxious stimulus is increased, the reac-tion time is shortened. Spatial and temporal facilitation occurs at synapses in the polysynaptic pathway. Stronger stimuli pro-duce more action potentials per second in the active branches and cause more branches to become active; summation of the EPSPs to the firing level therefore occurs more rapidly. FRACTIONATION & OCCLUSION Another characteristic of the withdrawal response is the fact that supramaximal stimulation of any of the sensory nerves from a limb never produces as strong a contraction of the flex-or muscles as that elicited by direct electrical stimulation of the muscles themselves. This indicates that the afferent inputs fractionate the motor neuron pool; that is, each input goes to only part of the motor neuron pool for the flexors of that par-ticular extremity. On the other hand, if all the sensory inputs are dissected out and stimulated one after the other, the sum of the tension developed by stimulation of each is greater than that produced by direct electrical stimulation of the muscle or stimulation of all inputs at once. This indicates that the vari-ous afferent inputs share some of the motor neurons and that occlusion occurs when all inputs are stimulated at once. GENERAL PROPERTIES OF REFLEXES It is apparent from the preceding description of the properties of monosynaptic and polysynaptic reflexes that reflex activity is stereotyped and specific in terms of both the stimulus and the response; a particular stimulus elicits a particular re-sponse. The fact that reflex responses are stereotyped does not exclude the possibility of their being modified by experience. Reflexes are adaptable and can be modified to perform motor tasks and maintain balance. Descending inputs from higher brain regions play an important role in modulating and adapt-ing spinal reflexes. ADEQUATE STIMULUS The stimulus that triggers a reflex is generally very precise. This stimulus is called the adequate stimulus for the particular re-flex. A dramatic example is the scratch reflex in the dog. This spinal reflex is adequately stimulated by multiple linear touch stimuli such as those produced by an insect crawling across the skin. The response is vigorous scratching of the area stimulated. If the multiple touch stimuli are widely separated or not in a line, the adequate stimulus is not produced and no scratching occurs. Fleas crawl, but they also jump from place to place. This jumping separates the touch stimuli so that an adequate stimu-lus for the scratch reflex is not produced. It is doubtful if the flea population would survive long without the ability to jump. FINAL COMMON PATH The motor neurons that supply the extrafusal fibers in skeletal muscles are the efferent side of many reflex arcs. All neural influ-ences affecting muscular contraction ultimately funnel through them to the muscles, and they are therefore called the final common paths. Numerous inputs converge on them. Indeed, the surface of the average motor neuron and its dendrites ac-commodates about 10,000 synaptic knobs. At least five inputs go from the same spinal segment to a typical spinal motor neuron. In addition to these, there are excitatory and inhibito-ry inputs, generally relayed via interneurons, from other levels of the spinal cord and multiple long-descending tracts from CHAPTER 9 Reflexes 165 the brain. All of these pathways converge on and determine the activity in the final common paths. CENTRAL EXCITATORY & INHIBITORY STATES The spread up and down the spinal cord of subliminal fringe effects from excitatory stimulation has already been men-tioned. Direct and presynaptic inhibitory effects can also be widespread. These effects are generally transient. However, the spinal cord also shows prolonged changes in excitability, pos-sibly because of activity in reverberating circuits or prolonged effects of synaptic mediators. The terms central excitatory state and central inhibitory state have been used to describe prolonged states in which excitatory influences overbalance in-hibitory influences and vice versa. When the central excitatory state is marked, excitatory impulses irradiate not only to many somatic areas of the spinal cord but also to autonomic areas. In chronically paraplegic humans, for example, a mild noxious stimulus may cause, in addition to prolonged withdrawal-ex-tension patterns in all four limbs, urination, defecation, sweat-ing, and blood pressure fluctuations (mass reflex). CHAPTER SUMMARY ■A reflex arc consists of a sense organ, an afferent neuron, one or more synapses within a central integrating station, an efferent neuron, and an effector response. ■A muscle spindle is a group of specialized intrafusal muscle fi-bers with contractile polar ends and a noncontractile center that is located in parallel to the extrafusal muscle fibers and is inner-vated by types Ia and II afferent fibers and γ-motor neurons. Muscle stretch activates the muscle spindle to initiate reflex con-traction of the extrafusal muscle fibers in the same muscle (stretch reflex). ■A Golgi tendon organ is a netlike collection of knobby nerve endings among the fascicles of a tendon that is located in series with extrafusal muscle fibers and innervated by type Ib afferents. They are stimulated by both passive stretch and active contrac-tion of the muscle to relax the muscle (inverse stretch reflex) and function as a transducer to regulate muscle force. ■A collateral from an Ia afferent branches to terminate on an inhib-itory interneuron that synapses on an antagonistic muscle (recip-rocal innervation) to relax that muscle when the agonist contracts. Clonus is the occurrence of regular, rhythmic contractions of a muscle subjected to sudden, maintained stretch. A sequence of in-creased resistance followed by reduced resistance when a limb is moved passively is known as the lengthening reaction. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. The inverse stretch reflex A) has a lower threshold than the stretch reflex. B) is a monosynaptic reflex. C) is a disynaptic reflex with a single interneuron inserted between the afferent and efferent limbs. D) is a polysynaptic reflex with many interneurons inserted between the afferent and efferent limbs. E) requires the discharge of central neurons that release acetyl-choline. 2. When γ-motor neuron discharge increases at the same time as α-motor neuron discharge to muscle, A) prompt inhibition of discharge in spindle Ia afferents takes place. B) the contraction of the muscle is prolonged. C) the muscle will not contract. D) the number of impulses in spindle Ia afferents is smaller than when α discharge alone is increased. E) the number of impulses in spindle Ia afferents is greater than when α discharge alone is increased. 3. Which of the following is not characteristic of a reflex? A) Modification by impulses from various parts of the CNS B) May involve simultaneous contraction of some muscles and relaxation of others C) Chronically suppressed after spinal cord transection D) Always involves transmission across at least one synapse E) Frequently occurs without conscious perception 4. Withdrawal reflexes are not A) initiated by nociceptive stimuli. B) prepotent. C) prolonged if the stimulus is strong. D) an example of a flexor reflex. E) accompanied by the same response on both sides of the body. CHAPTER RESOURCES Haines DE (editor): Fundamental Neuroscience for Basic and Clinical Applications, 3rd ed. Elsevier, 2006. Hulliger M: The mammalian muscle spindle and its central control. Rev Physiol Biochem Pharmacol 1984;101:1. Hunt CC: Mammalian muscle spindle: Peripheral mechanisms. Physiol Rev 1990;70: 643. Jankowska E: Interneuronal relay in spinal pathways from proprioceptors. Prog Neurobiol 1992;38:335. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. Lundberg A: Multisensory control of spinal reflex pathways. Prog Brain Res 1979;50:11. Matthews PBC: Mammalian Muscle Receptors and Their Central Actions, Williams & Wilkins, 1972. This page intentionally left blank 167 C H A P T E R SECTION III CENTRAL & PERIPHERAL NEUROPHYSIOLOGY 10 Pain & Temperature O B J E C T I V E S After studying this chapter, you should be able to: ■Name the types of peripheral nerve fibers and receptor types that mediate warmth, cold, and nociception. ■Explain the difference between pain and nociception. ■Explain the differences between fast and slow pain and acute and chronic pain. ■Explain hyperalgesia and allodynia. ■Describe and explain referred pain. INTRODUCTION One of the most common reasons an individual seeks the advice of a physician is because he or she is in pain. Pain was called by Sherrington, “the physical adjunct of an imperative protective reflex.” Painful stimuli generally initiate potent withdrawal and avoidance responses. Pain differs from other sensations in that it sounds a warning that something is wrong, preempts other signals, and is associated with an unpleasant affect. It turns out to be immensely complex because when pain is prolonged and tissue is damaged, cen-tral nociceptor pathways are sensitized and reorganized. NOCICEPTORS & THERMORECEPTORS Pain and temperature sensations arise from unmyelinated dendrites of sensory neurons located around hair follicles throughout the glabrous and hairy skin as well as deep tissue. Impulses from nociceptors (pain) are transmitted via two fi-ber types. One system comprises thinly myelinated Aδ fibers (2–5 μm in diameter) which conduct at rates of 12–30 m/s. The other is unmyelinated C fibers (0.4–1.2 μm in diameter) which conduct at low rates of 0.5–2 m/s. Thermoreceptors also span these two fiber types. Cold receptors are on dendritic endings of Aδ fibers and C fibers, whereas warmth (heat) re-ceptors are on C fibers. Mechanical nociceptors respond to strong pressure (eg, from a sharp object). Thermal nociceptors are activated by skin temperatures above 45 °C or by severe cold. Chemically sensitive nociceptors respond to various agents like bradyki-nin, histamine, high acidity, and environmental irritants. Poly-modal nociceptors respond to combinations of these stimuli. 168 SECTION III Central & Peripheral Neurophysiology Mapping experiments show that the skin has discrete cold-sensitive and heat-sensitive spots. There are 4 to 10 times as many cold-sensitive as heat-sensitive spots. The threshold for activation of warmth receptors is 30 °C, and they increase their firing rate up to 46 °C. Cold receptors are inactive at tempera-tures of 40 °C, but then steadily increase their firing rate as skin temperature falls to about 24 °C. As skin temperature further decreases, the firing rate of cold receptors decreases until the temperature reaches 10 °C. Below that temperature, they are inactive and the cold becomes an effective local anesthetic. Because the sense organs are located subepithelially, it is the temperature of the subcutaneous tissues that determines the responses. Cool metal objects feel colder than wooden objects of the same temperature because the metal conducts heat away from the skin more rapidly, cooling the subcutaneous tissues to a greater degree. A major advance in this field has been the cloning of three thermoreceptors and nociceptors. The receptor for moderate cold is the cold- and menthol-sensitive receptor 1 (CMR 1). Two types of vanilloid receptors respond to noxious heat (VR1 and VRL-1). Vanillins are a group of compounds, including capsaicin, that cause pain. The VR1 receptors respond not only to capsaicin but also to protons and to potentially harmful tem-peratures above 43 °C. VRL-1, which responds to temperatures above 50 °C but not to capsaicin, has been isolated from C fibers. There may be many types of receptors on single periph-eral C fiber endings, so single fibers can respond to many dif-ferent noxious stimuli. However, the different properties of the VR1 and the VRL-1 receptors make it likely that there are many different nociceptor C fibers systems as well. CMR1, VR1, and VRL1 are members of the transient receptor potential (TRP) family of excitatory ion channels. VR1 has a PIP2 binding site, and when the amount of PIP2 bound is decreased, the sensitivity of the receptors is increased. Aside from the fact that activation of the cool receptor causes an influx of Ca2+, little is known about the ionic basis of the initial depolarization they produce. In the cutaneous receptors in general, depolarization could be due to inhibition of K+ channels, activation of Na+ channels, or inhi-bition of the Na+–K+ pump, but the distinction between these possibilities has not been made. CLASSIFICATION OF PAIN For scientific and clinical purposes, pain is defined by the In-ternational Association for the Study of Pain (IASP) as, “an unpleasant sensory and emotional experience associated with actual or potential tissue damage, or described in terms of such damage.” This is to be distinguished from the term noci-ception which the IASP defines as the unconscious activity in-duced by a harmful stimulus applied to sense receptors. Pain is sometimes classified as fast and slow pain. A painful stimulus causes a “bright,” sharp, localized sensation (fast pain) followed by a dull, intense, diffuse, and unpleasant feeling (slow pain). Evidence suggests that fast pain is due to activity in the Aδ pain fibers, whereas slow pain is due to activity in the C pain fibers. Itch and tickle are related to pain sensation (see Clinical Box 10–1). Pain is frequently classified as physiologic or acute pain and pathologic or chronic pain, which includes inflamma-tory pain and neuropathic pain. Acute pain typically has a sudden onset and recedes during the healing process. Acute pain can be considered as “good pain” as it serves an impor-tant protective mechanism. The withdrawal reflex is an exam-ple of this protective role of pain. Chronic pain can be considered “bad pain” because it persists long after recovery from an injury and is often refractory to common analgesic agents, including nonsteroidal anti-inflam-matory drugs (NSAIDs) and opiates. Chronic pain can result from nerve injury (neuropathic pain) including diabetic neu-ropathy, toxin-induced nerve damage, and ischemia. Causalgia is a type of neuropathic pain (see Clinical Box 10–2). Pain is often accompanied by hyperalgesia and allodynia. Hyperalgesia is an exaggerated response to a noxious stimulus, whereas allodynia is a sensation of pain in response to an innoc-uous stimulus. An example of the latter is the painful sensation from a warm shower when the skin is damaged by sunburn. CLINICAL BOX 10–1 Itch & Tickle Itching (pruritus) is not much of a problem for normal indi-viduals, but severe itching that is difficult to treat occurs in diseases such as chronic renal failure, some forms of liver dis-ease, atopic dermatitis, and HIV infection. Especially in areas where many naked endings of unmyelinated nerve fibers oc-cur, itch spots can be identified on the skin by careful map-ping. In addition, itch-specific fibers have been demon-strated in the ventrolateral spinothalamic tract. This and other evidence implicate the existence of an itch-specific path. Relatively mild stimulation, especially if produced by something that moves across the skin, produces itch and tickle. Scratching relieves itching because it activates large, fast-conducting afferents that gate transmission in the dorsal horn in a manner analogous to the inhibition of pain by stim-ulation of similar afferents. It is interesting that a tickling sen-sation is usually regarded as pleasurable, whereas itching is annoying and pain is unpleasant. Itching can be produced not only by repeated local mechanical stimulation of the skin but also by a variety of chemical agents. Histamine pro-duces intense itching, and injuries cause its liberation in the skin. However, in most instances of itching, endogenous his-tamine does not appear to be the responsible agent; doses of histamine that are too small to produce itching still pro-duce redness and swelling on injection into the skin, and se-vere itching frequently occurs without any visible change in the skin. The kinins cause severe itching. CHAPTER 10 Pain & Temperature 169 Hyperalgesia and allodynia signify increased sensitivity of nociceptive afferent fibers. Figure 10–1 shows how chemicals released at the site of injury can further activate nociceptors lead-ing to inflammatory pain. Injured cells release chemicals such as K+ that depolarize nerve terminals, making nociceptors more responsive. Injured cells also release bradykinin and Substance P , which can further sensitize nociceptive terminals. Histamine is released from mast cells, serotonin (5-HT) from platelets, and prostaglandins from cell membranes, all contributing to the inflammatory process and they activate or sensitize the nocicep-tors. Some released substances act by releasing another one (eg, bradykinin activates both Aδ and C fibers and increases synthesis and release of prostaglandins). Prostaglandin E2 (a cyclooxygen-ase metabolite of arachidonic acid) is released from damaged cells and produces hyperalgesia. This is why aspirin and other NSAIDs (inhibitors of cyclooxygenase) alleviate pain. DEEP PAIN The main difference between superficial and deep sensibility is the different nature of the pain evoked by noxious stimuli. This is probably due to a relative deficiency of Aδ nerve fibers in deep structures, so there is little rapid, bright pain. In addi-tion, deep pain and visceral pain are poorly localized, nauseat-ing, and frequently associated with sweating and changes in blood pressure. Pain can be elicited experimentally from the periosteum and ligaments by injecting hypertonic saline into them. The pain produced in this fashion initiates reflex con-traction of nearby skeletal muscles. This reflex contraction is similar to the muscle spasm associated with injuries to bones, tendons, and joints. The steadily contracting muscles become ischemic, and ischemia stimulates the pain receptors in the muscles (see Clinical Box 10–3). The pain in turn initiates more spasm, setting up a vicious cycle. VISCERAL PAIN In addition to being poorly localized, unpleasant, and associ-ated with nausea and autonomic symptoms, visceral pain of-ten radiates or is referred to other areas. CLINICAL BOX 10–2 Neuropathic Pain Neuropathic pain may occur when nerve fibers are injured. Commonly, it is excruciating and a difficult condition to treat. It occurs in various forms in humans. For example, in causal-gia, spontaneous burning pain occurs long after seemingly trivial injuries. The pain is often accompanied by hyperalge-sia and allodynia. Reflex sympathetic dystrophy is often present as well. In this condition, the skin in the affected area is thin and shiny, and there is increased hair growth. Research in animals indicates that nerve injury leads to sprouting and eventual overgrowth of noradrenergic sympathetic nerve fi-bers into the dorsal root ganglia of the sensory nerves from the injured area. Sympathetic discharge then brings on pain. Thus, it appears that the periphery has been short-circuited and that the relevant altered fibers are being stimulated by norepinephrine at the dorsal root ganglion level. Alpha-adre-nergic blockade produces relief of causalgia-type pain in hu-mans, though for unknown reasons α1-adrenergic blockers are more effective than α2-adrenergic blocking agents. Treat-ment of painful sensory neuropathy is a major challenge and current therapies are often inadequate. FIGURE 10–1 In response to tissue injury, chemical mediators can sensitize and activate nociceptors. These factors contribute to hy-peralgesia and allodynia. Tissue injury releases bradykinin and prostaglandins that sensitize or activate nociceptors, which in turn releases sub-stance P and calcitonin gene-related peptide (CGRP). Substance P acts on mast cells to cause degranulation and release histamine, which activates nociceptors. Substance P causes plasma extravasation and CGRP dilates blood vessels; the resulting edema causes additional release of bradykinin. Serotonin (5-HT) is released from platelets and activates nociceptors. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science. McGraw-Hill, 2000.) Lesion 5-HT Bradykinin Prostaglandin K+ Mast cell Spinal cord Histamine CGRP CGRP Blood vessel Substance P Dorsal root ganglion neuron Substance P 170 SECTION III Central & Peripheral Neurophysiology The autonomic nervous system, like the somatic, has afferent components, central integrating stations, and effector path-ways. The receptors for pain and the other sensory modalities present in the viscera are similar to those in skin, but there are marked differences in their distribution. There are no proprio-ceptors in the viscera, and few temperature and touch recep-tors. Nociceptors are present, although they are more sparsely distributed than in somatic structures. Afferent fibers from visceral structures reach the CNS via sympathetic and parasympathetic nerves. Their cell bodies are located in the dorsal roots and the homologous cranial nerve ganglia. Specifically, there are visceral afferents in the facial, glossopharyngeal, and vagus nerves; in the thoracic and upper lumbar dorsal roots; and in the sacral roots (Figure 10–2). There may also be visceral afferent fibers from the eye in the trigeminal nerve. As almost everyone knows from personal experience, visceral pain can be very severe. The receptors in the walls of the hollow viscera are especially sensitive to distention of these organs. Such distention can be produced experimentally in the gastrointesti-nal tract by inflation of a swallowed balloon attached to a tube. This produces pain that waxes and wanes (intestinal colic) as the intestine contracts and relaxes on the balloon. Similar colic is produced in intestinal obstruction by the contractions of the dilated intestine above the obstruction. When a visceral organ is inflamed or hyperemic, relatively minor stimuli cause severe pain. This is probably a form of hyperalgesia. REFERRED PAIN Irritation of a visceral organ frequently produces pain that is felt not at that site but in some somatic structure that may be a considerable distance away. Such pain is said to be referred to the somatic structure. Obviously, knowledge of referred pain and the common sites of pain referral from each of the viscera is of great importance to the physician. Perhaps the best-known example is referral of cardiac pain to the inner aspect of the left arm. Other examples include pain in the tip of the shoulder caused by irritation of the central portion of the dia-phragm and pain in the testicle due to distention of the ureter. Additional instances abound in the practices of medicine, sur-gery, and dentistry. However, sites of reference are not ster-eotyped, and unusual reference sites occur with considerable frequency. Cardiac pain, for instance, may be referred to the right arm, the abdominal region, or even the back and neck. When pain is referred, it is usually to a structure that devel-oped from the same embryonic segment or dermatome as the structure in which the pain originates. This principle is called the dermatomal rule. For example, the heart and the arm have the same segmental origin, and the testicle has migrated with its nerve supply from the primitive urogenital ridge from which the kidney and ureter have developed. The basis for referred pain may be convergence of somatic and visceral pain fibers on the same second-order neurons in the dorsal horn that project to the thalamus and then to the somatosensory cortex (Figure 10–3). This is called the conver-gence–projection theory. Somatic and visceral neurons con-verge in lamina I–VI of the ipsilateral dorsal horn, but neurons in lamina VII receive afferents from both sides of the body—a requirement if convergence is to explain referral to the side opposite that of the source of pain. The somatic nociceptive fibers normally do not activate the second-order neurons, but when the visceral stimulus is prolonged, facilitation of the somatic fiber endings occurs. They now stimulate the second-order neurons, and of course the brain cannot determine whether the stimulus came from the viscera or from the area of referral. CLINICAL BOX 10–3 Muscle Pain If a muscle contracts rhythmically in the presence of an ade-quate blood supply, pain does not usually result. However, if the blood supply to a muscle is occluded, contraction soon causes pain. The pain persists after the contraction until blood flow is reestablished. These observations are difficult to interpret except in terms of the release during contraction of a chemical agent (Lewis’s “P factor”) that causes pain when its local concentration is high enough. When the blood supply is restored, the material is washed out or metabo-lized. The identity of the P factor is not settled, but it could be K+. Clinically, the substernal pain that develops when the myocardium becomes ischemic during exertion (angina pectoris) is a classic example of the accumulation of P factor in a muscle. Angina is relieved by rest because this decreases the myocardial O2 requirement and permits the blood sup-ply to remove the factor. Intermittent claudication, the pain produced in the leg muscles of persons with occlusive vascular disease, is another example. It characteristically comes on while the patient is walking and disappears on stopping. Visceral pain, like deep somatic pain, initiates re-flex contraction of nearby skeletal muscle. This reflex spasm is usually in the abdominal wall and makes the abdominal wall rigid. It is most marked when visceral inflammatory pro-cesses involve the peritoneum. However, it can occur with-out such involvement. The spasm protects the underlying in-flamed structures from inadvertent trauma. Indeed, this reflex spasm is sometimes called “guarding.” CHAPTER 10 Pain & Temperature 171 CHAPTER SUMMARY ■Pain impulses are transmitted via lightly myelinated Aδ and un-myelinated C fibers. Cold receptors are on dendritic endings of Aδ fibers and C fibers, whereas heat receptors are on C fibers. ■Pain is an unpleasant sensory and emotional experience associat-ed with actual or potential tissue damage, or described in terms of such damage, whereas nociception is the unconscious activity in-duced by a harmful stimulus applied to sense receptors. ■Fast pain is mediated by Aδ fibers and causes sharp, localized sensation. Slow pain is mediated by C fibers and causes a dull, intense, diffuse, and unpleasant feeling. ■Acute pain has a sudden onset, recedes during the healing pro-cess, and serves as an important protective mechanism. Chronic pain is persistent and caused by nerve damage; it is often refrac-tory to NSAIDs and opiates. ■Hyperalgesia is an exaggerated response to a noxious stimulus; al-lodynia is a sensation of pain in response to an innocuous stimulus. ■Referred pain is pain that originates in a visceral organ but is sensed at a somatic site. It may be due to convergence of somatic and visceral nociceptive afferent fibers on the same second-order neurons in the spinal dorsal horn that project to the thal-amus and then to the somatosensory cortex. FIGURE 10–2 Pain innervation of the viscera. Pain afferents from structures between the pain lines reach the CNS via sympathetic path-ways, whereas, they traverse parasympathetic pathways from structures above the thoracic pain line and below the pelvic pain line. (After White JC. Reproduced with permission from Ruch TC: In Physiology and Biophysics, 19th ed. Ruch TC, Patton HD [editors]. Saunders, 1965.) PARASYMPATHETIC SYMPATHETIC PARASYMPATHETIC Glossopharyngeal nerve Superior laryngeal nerve Upper thoracic vagal rami Brachial plexus THORACIC PAIN LINE PELVIC PAIN LINE Vagus A p i c a l p l e u r a Pa rie ta l pl e u r a (Intercostal nerves) V i s c e ra l pl e ur a (i n s e n si ti ve ) Central diaphragm (phrenic nerve) Peripheral diaphragm (intercostal nerves) Lower splanch-nic nerves (T10–L1) Ureter (T11–L1) Somatic nerves (T11–L1) P a ri e t a l p e r it o n e u m Fundus (T11–L1) Cervix and upper vagina (S2–4) Trigone Prostate Urethra (pelvic nerves, S2–4) Testicle (sacral nerves, S2–4) (genitofemoral nerves, L1–2) (spermatic plexus, T10) Parasympathetic rami (S2–4) Colon (T11–L1) Splanchnic nerves (T7–9) lleum Small intestine splanchnic nerves (T9–11) Duodenum and jejunum (splanchnic nerves) R splanchnic nerves (T7–9) FIGURE 10–3 Diagram of the way in which convergence of somatic and visceral nociceptive fibers in lamina VII of the dorsal horn may cause referred pain. When a visceral stimulus is pro-longed, somatic fiber facilitation occurs. This leads to activation of spi-nothalamic tract neurons, and of course the brain cannot determine whether the stimulus came from the viscera or from the somatic area. To brain Somatic structure Spinothalamic tract Viscus 172 SECTION III Central & Peripheral Neurophysiology MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. The distance by which two touch stimuli must be separated to be perceived as two separate stimuli is greatest on the A) lips. B) palm of the hand. C) back of the scapula. D) dorsum of the hand. E) tips of the fingers. 2. Visceral pain A) shows relatively rapid adaptation. B) is mediated by B fibers in the dorsal roots of the spinal nerves. C) is poorly localized. D) resembles "fast pain" produced by noxious stimulation of the skin. E) causes relaxation of nearby skeletal muscles. 3. Nociceptors A) are activated by strong pressure, severe cold, severe heat, and chemicals. B) are absent in visceral organs. C) are specialized structures located in the skin and joints. D) are innervated by group II afferents. E) all of the above 4. Thermoreceptors A) are activated only by severe cold or severe heat. B) are located on superficial layers of the skin. C) are a subtype of nociceptors. D) are on dendritic endings of Aδ fibers and C fibers. E) all of the above CHAPTER RESOURCES Boron WF, Boulpaep EL: Medical Physiology, Elsevier, 2005. Craig AD: How do you feel? Interoception: The sense of the physiological condition of the body. Nat Rev Neurosci 2002;3:655. Garry EM, Jones E, Fleetwood-Walker SM: Nociception in vertebrates: Key receptors participating in spinal mechanisms of chronic pain in animals. Brain Res Rev 2004;46: 216. Haines DE (editor): Fundamental Neuroscience for Basic and Clinical Applications, 3rd ed. Elsevier, 2006. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. Marchand F, Perretti M, McMahon SB: Role of the immune system in chronic pain. Nat Rev Neurosci 2005;6:521. Mendell JR, Sahenk Z: Painful sensory neuropathy. N Engl J Med 2003;348:1243. 173 C H A P T E R 11 Somatosensory Pathways O B J E C T I V E S After studying this chapter, you should be able to: ■Compare the pathway that mediates sensory input from touch, proprioceptive, and vibratory senses to that mediating information from pain and thermoreceptors. ■Describe the somatotopic organization of ascending sensory pathways. ■Describe descending pathways that modulate transmission in pain pathways. ■List some drugs that have been used for relief of pain, and give the rationale for their use and their clinical effectiveness. INTRODUCTION Earlier chapters have described properties of receptors medi-ating the sensations of touch, vibration, proprioception, tem-perature, and pain. This chapter will review central ascending pathways that transmit and process the information from peripheral receptors to the cerebral cortex as well as describe some deficits in sensation resulting from lesions at various steps within the ascending systems. Also, various ways to modulate pain transmission will be described. DORSAL HORN The dorsal horns are divided on the basis of histologic charac-teristics into laminae I–VII, with I being the most superficial and VII the deepest. Lamina VII receives afferents from both sides of the body, whereas the other laminae receive only uni-lateral input. Lamina II and part of lamina III make up the substantia gelatinosa, a lightly stained area near the top of each dorsal horn. Three types of primary afferent fibers (with cell bodies in the dorsal root ganglia) mediate cutaneous sen-sation: (1) large myelinated Aα and Aβ fibers that transmit impulses generated by mechanical stimuli; (2) small myelinat-ed Aδ fibers, some of which transmit impulses from cold re-ceptors and nociceptors that mediate pain and some of which transmit impulses from mechanoreceptors; and (3) small un-myelinated C fibers that are concerned primarily with pain and temperature. However, a few C fibers also transmit im-pulses from mechanoreceptors. The orderly distribution of these fibers in various layers of the dorsal horn is shown in Figure 11–1. DORSAL COLUMN PATHWAY The principal direct pathways to the cerebral cortex for touch, vi-bratory sense, and proprioception (position sense) are shown in Figure 11–2. Fibers mediating these sensations ascend ipsilater-ally in the dorsal columns to the medulla, where they synapse in the gracilus and cuneate nuclei. The second-order neurons from these nuclei cross the midline and ascend in the medial lemnis-cus to end in the contralateral ventral posterior lateral (VPL) nucleus and related specific sensory relay nuclei of the thalamus. This ascending system is called the dorsal column or medial lemniscal system. The fibers within the dorsal column pathway are joined in the brain stem by fibers mediating sensation from the head. Touch and proprioception are relayed mostly via the main sensory and mesencephalic nuclei of the trigeminal nerve. SOMATOTOPIC ORGANIZATION Within the dorsal columns, fibers arising from different levels of the cord are somatotopically organized. Specifically, fibers 174 SECTION III Central & Peripheral Neurophysiology from the sacral cord are positioned most medially and those from the cervical cord are most lateral. This arrangement con-tinues in the medulla with lower body (eg, foot) representation in the gracilus nucleus and upper body (eg, finger) representa-tion in cuneate nucleus. The medial lemniscus is organized dorsal to ventral representing from neck to foot. Somatotopic organization continues through the thalamus and cortex. VPL thalamic neurons carrying sensory informa-tion project in a highly specific way to the two somatic sen-sory areas of the cortex: somatic sensory area I (SI) in the postcentral gyrus and somatic sensory area II (SII) in the wall of the sylvian fissure. In addition, SI projects to SII. SI corresponds to Brodmann’s areas 3, 2, and 1. Brodmann was a histologist who painstakingly divided the cerebral cortex into numbered areas based on their histologic characteristics. The arrangement of projections to SI is such that the parts of the body are represented in order along the postcentral gyrus, with the legs on top and the head at the foot of the gyrus FIGURE 11–1 Schematic representation of the terminations of the three types of primary afferent neurons in the various layers of the dorsal horn of the spinal cord. I II III IV V VI VII Mechanoreceptors Mechanoreceptors Nociceptors Cold receptors Nociceptors Thermoreceptors Mechanoreceptors Aδ C Aβ To dorsal columns FIGURE 11–2 Ascending tracts carrying sensory information from peripheral receptors to the cerebral cortex. (a) Dorsal-column pathway mediating touch, vibratory sense, and proprioception. (b) Ventrolateral spinothalamic tract mediating pain and temperature. (From Fox SI, Human Physiology. McGraw-Hill, 2008.) Axons of third-order neurons Cerebral cortex Medial lemniscal tract (axons of second-order neurons) Fasciculus cuneatus (axons of first-order sensory neurons) Joint stretch receptor (proprioceptor) Spinal cord Fasciculus gracilis (axons of first-order sensory neurons) Medulla oblongata Touch receptor Temperature receptor Pain receptor Axons of first-order neurons (not part of spinothalamic tract) Lateral spinothalamic tract (axons of second-order neurons) Thalamus Postcentral gyrus (a) (b) CHAPTER 11 Somatosensory Pathways 175 (Figure 11–3). Not only is there detailed localization of the fibers from the various parts of the body in the postcentral gyrus, but also the size of the cortical receiving area for impulses from a particular part of the body is proportionate to the use of the part. The relative sizes of the cortical receiving areas are shown dramatically in Figure 11–4, in which the pro-portions of the homunculus have been distorted to correspond to the size of the cortical receiving areas for each. Note that the cortical areas for sensation from the trunk and back are small, whereas very large areas are concerned with impulses from the hand and the parts of the mouth concerned with speech. Studies of the sensory receiving area emphasize the very dis-crete nature of the point-for-point localization of peripheral areas in the cortex and provide further evidence for the general validity of the law of specific nerve energies (see Chapter 8). Stimulation of the various parts of the postcentral gyrus gives rise to sensations projected to appropriate parts of the body. The sensations produced are usually numbness, tingling, or a sense of movement, but with fine enough electrodes it has been possible to produce relatively pure sensations of touch, warmth, and cold. The cells in the postcentral gyrus are orga-nized in vertical columns, like cells in the visual cortex. The cells in a given column are all activated by afferents from a given part of the body, and all respond to the same sensory modality. SII is located in the superior wall of the sylvian fissure, the fissure that separates the temporal from the frontal and pari-etal lobes. The head is represented at the inferior end of the postcentral gyrus, and the feet at the bottom of the sylvian fis-sure. The representation of the body parts is not as complete or detailed as it is in the postcentral gyrus. Conscious awareness of the positions of the various parts of the body in space depends in part on impulses from sense organs in and around the joints. The organs involved are slowly adapting spray endings, structures that resemble Golgi tendon organs, and probably Pacinian corpuscles in the syn-ovia and ligaments. Impulses from these organs, touch recep-tors in the skin and other tissues, and muscle spindles are synthesized in the cortex into a conscious picture of the posi-tion of the body in space. Microelectrode studies indicate that many of the neurons in the sensory cortex respond to particu-lar movements, not just to touch or static position. VENTROLATERAL SPINOTHALAMIC TRACT Fibers from nociceptors and thermoreceptors synapse on neu-rons in the dorsal horn (Figure 11–1). Aδ fibers terminate pri-marily on neurons in laminae I and V, whereas the dorsal root C fibers terminate on neurons in laminae I and II. The synap-tic transmitter secreted by afferent fibers subserving fast mild pain is glutamate, and the transmitter subserving slow severe pain is substance P. The axons from these neurons cross the midline and ascend in the ventrolateral quadrant of the spinal cord, where they form the ventrolateral spinothalamic tract (Figure 11–2). Fibers within this tract synapse in the VPL. Other dorsal horn neurons that receive nociceptive input synapse in the reticular formation of the brain stem (spinoreticular pathway) and then project to the centrolateral nucleus of the thalamus. Positron emission tomographic (PET) and functional mag-netic resonance imaging (fMRI) studies in normal humans indicate that pain activates cortical areas SI, SII, and the cingu-late gyrus on the side opposite the stimulus. In addition, the mediofrontal cortex, the insular cortex, and the cerebellum are activated. These technologies were important in distinguishing FIGURE 11–3 Brain areas concerned with somatic sensation, and some of the cortical receiving areas for other sensory modalities in the human brain. The numbers are those of Brod-mann’s cortical areas. The primary auditory area is actually located in the sylvian fissure on the top of the superior temporal gyrus and is not normally visible in a lateral view of the cortex. FIGURE 11–4 Sensory homunculus, drawn overlying a coronal section through the postcentral gyrus. Gen., genitalia. (Reproduced, with permission, from Penfield W, Rasmussen G: The Cerebral Cortex of Man. Macmillan, 1950.) Hand Face Tongue Sll Trunk Auditory Sl Posterior parietal cortex Visual 9 8 6 4 3 1 2 Intra-abdominal Pharynx Tongue Gen. Toes Foot Trunk Neck Head Shoulder Leg Hip Teeth, gums, and jaw Lower lip Lips Upper lip Face Nose Eye Thumb Index Ring Little Hand Wrist Forearm Elbow Arm Middle 176 SECTION III Central & Peripheral Neurophysiology two components of pain pathways. From VPL nuclei in the thalamus, fibers project to SI and SII. This is the pathway responsible for the discriminative aspect of pain, and is also called the neospinothalamic tract. In contrast, the pathway that includes synapses in the brain stem reticular formation and centrolateral thalamic nucleus projects to the frontal lobe, lim-bic system, and insula. This pathway mediates the motiva-tional-affect component of pain and is called the paleospinothalamic tract. In the central nervous system (CNS), visceral sensation travels along the same pathways as somatic sensation in the spinothalamic tracts and thalamic radiations, and the cortical receiving areas for visceral sensation are intermixed with the somatic receiving areas. CORTICAL PLASTICITY It is now clear that the extensive neuronal connections de-scribed above are not innate and immutable but can be changed relatively rapidly by experience to reflect the use of the represented area. Clinical Box 11–1 describes remarkable changes in cortical and thalamic organization that occur in re-sponse to limb amputation to lead to the phenomenon of phantom limb pain. Numerous animal studies point to dramatic reorganization of cortical structures. If a digit is amputated in a monkey, the cortical representation of the neighboring digits spreads into the cortical area that was formerly occupied by the represen-tation of the amputated digit. Conversely, if the cortical area representing a digit is removed, the somatosensory map of the digit moves to the surrounding cortex. Extensive, long-term deafferentation of limbs leads to even more dramatic shifts in somatosensory representation in the cortex, with, for exam-ple, the limb cortical area responding to touching the face. The explanation of these shifts appears to be that cortical con-nections of sensory units to the cortex have extensive conver-gence and divergence, with connections that can become weak with disuse and strong with use. Plasticity of this type occurs not only with input from cuta-neous receptors but also with input in other sensory systems. For example, in cats with small lesions of the retina, the corti-cal area for the blinded spot begins to respond to light striking other areas of the retina. Development of the adult pattern of retinal projections to the visual cortex is another example of this plasticity. At a more extreme level, experimentally routing visual input to the auditory cortex during development cre-ates visual receptive fields in the auditory system. PET scanning in humans also documents plastic changes, sometimes from one sensory modality to another. Thus, for example, tactile and auditory stimuli increase metabolic activ-ity in the visual cortex in blind individuals. Conversely, deaf individuals respond faster and more accurately than normal individuals to moving stimuli in the visual periphery. Plastic-ity also occurs in the motor cortex. These findings illustrate the malleability of the brain and its ability to adapt. EFFECTS OF CNS LESIONS Ablation of SI in animals causes deficits in position sense and in the ability to discriminate size and shape. Ablation of SII causes deficits in learning based on tactile discrimination. Ablation of SI causes deficits in sensory processing in SII, whereas ablation of SII has no gross effect on processing in SI. Thus, it seems clear that SI and SII process sensory information in series rather than in parallel and that SII is concerned with further elaboration of sensory data. SI also projects to the posterior parietal cortex (Figure 11–3), and lesions of this association area produce com-plex abnormalities of spatial orientation on the contralateral side of the body. In experimental animals and humans, cortical lesions do not abolish somatic sensation. Proprioception and fine touch are most affected by cortical lesions. Temperature sensibility is less affected, and pain sensibility is only slightly altered. Only very extensive lesions completely interrupt touch sensa-tion. When the dorsal columns are destroyed, vibratory sensation CLINICAL BOX 11–1 Phantom Limb Pain In 1551, a military surgeon, Ambroise Pare, wrote, ”. . . the patients, long after the amputation is made, say they still feel pain in the amputated part. Of this they complain strongly, a thing worthy of wonder and almost incredible to people who have not experienced this.” This is perhaps the earliest description of phantom limb pain. Between 50% and 80% of amputees experience phantom sensations, usu-ally pain, in the region of their amputated limb. Phantom sensations may also occur after the removal of body parts other than the limbs, for example, after amputation of the breast, extraction of a tooth (phantom tooth pain), or re-moval of an eye (phantom eye syndrome). Numerous the-ories have been evoked to explain this phenomenon. The current theory is based on evidence that the brain can reor-ganize if sensory input is cut off. The ventral posterior tha-lamic nucleus is one example where this change can occur. In patients who have had their leg amputated, single neu-ron recordings show that the thalamic region that once re-ceived input from the leg and foot now respond to stimula-tion of the stump (thigh). Others have demonstrated remapping of the somatosensory cortex. For example, in some individuals who have had an arm amputated, stroking different parts of the face can lead to the feeling of being touched in the area of the missing limb. Spinal cord stimula-tion has been shown to be an effective therapy for phantom pain. Electric current is passed through an electrode that is placed next to the spinal cord to stimulate spinal pathways. This interferes with the impulses ascending to the brain and lessens the pain felt in the phantom limb. Instead, amputees feel a tingling sensation in the phantom limb. CHAPTER 11 Somatosensory Pathways 177 and proprioception are reduced, the touch threshold is ele-vated, and the number of touch-sensitive areas in the skin is decreased. In addition, localization of touch sensation is impaired. An increase in touch threshold and a decrease in the number of touch spots in the skin are also observed after interrupting the spinothalamic tract, but the touch deficit is slight and touch localization remains normal. The informa-tion carried in the lemniscal system is concerned with the detailed localization, spatial form, and temporal pattern of tactile stimuli. The information carried in the spinothalamic tracts, on the other hand, is concerned with poorly localized, gross tactile sensations. Clinical Box 11–2 describes the char-acteristic changes in sensory (and motor) functions that occur in response to spinal hemisection. Proprioceptive information is transmitted up the spinal cord in the dorsal columns. A good deal of the proprioceptive input goes to the cerebellum, but some passes via the medial lemniscus and thalamic radiations to the cortex. Diseases of the dorsal columns produce ataxia because of the interruption of proprioceptive input to the cerebellum. MODULATION OF PAIN TRANSMISSION STRESS-INDUCED ANALGESIA It is well known that soldiers wounded in the heat of battle of-ten feel no pain until the battle is over (stress-induced analge-sia). Many people have learned from practical experience that touching or shaking an injured area decreases the pain due to the injury. Stimulation with an electric vibrator at the site of pain also gives some relief. The relief may result from inhibi-tion of pain pathways in the dorsal horn gate by stimulation of large-diameter touch-pressure afferents. Figure 11–1 shows that collaterals from these myelinated afferent fibers synapse in the dorsal horn. These collaterals may modify the input from nociceptive afferent terminals that also synapse in the dorsal horn. This is called the gate-control hypothesis. The same mechanism is probably responsible for the effi-cacy of counterirritants. Stimulation of the skin over an area of visceral inflammation produces some relief of the pain due to the visceral disease. The old-fashioned mustard plaster works on this principle. Surgical procedures undertaken to relieve severe pain include cutting the nerve from the site of injury or ventrolat-eral cordotomy, in which the spinothalamic tracts are care-fully cut. However, the effects of these procedures are transient at best if the periphery has been short-circuited by sympathetic or other reorganization of the central pathways. MORPHINE & ENKEPHALINS Pain can often be handled by administration of analgesic drugs in adequate doses, though this is not always the case. The most effective of these agents is morphine. Morphine is particularly effective when given intrathecally. The receptors that bind mor-phine and the body’s own morphines, the opioid peptides, are found in the midbrain, brain stem, and spinal cord. There are at least three nonmutually exclusive sites at which opioids can act to produce analgesia: peripherally, at the site of an injury; in the dorsal horn, where nociceptive fibers synapse on dorsal root ganglion cells; and at more rostral sites in the brain stem. Figure 11–5 shows various modes of action of opi-ates to decrease transmission in pain pathways. Opioid recep-tors are produced in dorsal root ganglion cells and migrate both peripherally and centrally along their nerve fibers. In the periphery, inflammation causes the production of opioid pep-tides by immune cells, and these presumably act on the recep-tors in the afferent nerve fibers to reduce the pain that would otherwise be felt. The opioid receptors in the dorsal horn region could act presynaptically to decrease release of substance P, although presynaptic nerve endings have not been identified. Finally, injections of morphine into the periaqueductal gray matter of the midbrain relieve pain by activating descending pathways that produce inhibition of primary afferent transmis-sion in the dorsal horn. There is evidence that this activation occurs via projections from the periaqueductal gray matter to the nearby raphé magnus nucleus and that descending seroto-nergic fibers from this nucleus mediate the inhibition. Chronic use of morphine to relieve pain can cause patients to develop resistance to the drug, requiring progressively higher doses for pain relief. This acquired tolerance is differ-ent from addiction, which refers to a psychological craving. Psychological addiction rarely occurs when morphine is used to treat chronic pain, provided the patient does not have a his-tory of drug abuse. Clinical Box 11–3 describes mechanisms involved in motivation and addiction. CLINICAL BOX 11–2 Brown–Séquard Syndrome A functional hemisection of the spinal cord causes a char-acteristic and easily recognized clinical picture that reflects damage to ascending sensory (dorsal-column pathway, ventrolateral spinothalamic tract) and descending motor (corticospinal tract) pathways, which is called the Brown– Séquard syndrome. The lesion to fasciculus gracilus or fas-ciculus cuneatus leads to ipsilateral loss of discriminative touch, vibration, and proprioception below the level of le-sion. The loss of the spinothalamic tract leads to contralat-eral loss of pain and temperature sensation beginning one or two segments below the lesion. Damage to the cortico-spinal tract produces weakness and spasticity in certain muscle groups on the same side of the body. Although a precise spinal hemisection is rare, the syndrome is fairly common because it can be caused by spinal cord tumor, trauma, degenerative disc disease, and ischemia. 178 SECTION III Central & Peripheral Neurophysiology Despite intensive study, relatively little is known about the brain mechanisms that cause tolerance and dependence. However, the two can be separated. Absence of β-arrestin-2 blocks tolerance but has no effect on dependence. β-Arrestin-2 is a member of a family of proteins that inhibit heterotrim-eric G proteins by phosphorylating them. Acupuncture at a location distant from the site of a pain may act by releasing endorphins. Acupuncture at the site of the pain appears to act primarily in the same way as touching or shaking (gate-control mechanism). A component of stress-induced analgesia appears to be mediated by endogenous opi-oids, because in experimental animals some forms of stress-induced analgesia are prevented by naloxone, a morphine antagonist. However, other forms are unaffected, and so other components are also involved. ACETYLCHOLINE Epibatidine, a cholinergic agonist first isolated from the skin of a frog, is a potent nonopioid analgesic agent, and even more FIGURE 11–5 Local-circuit interneurons in the superficial dorsal horn of the spinal cord integrate descending and afferent pathways. A) Possible interactions of nociceptive afferent fibers, interneurons, and descending fibers in the dorsal horn. Nociceptive fibers termi-nate on second-order spinothalamic projection neurons. Enkephalin (ENK)-containing interneurons exert both presynaptic and postsynaptic in-hibitory actions. Serotonergic and noradrenergic neurons in the brain stem activate opioid interneurons and suppress the activity spinothalamic projection neurons. B1) Activation of nociceptors releases glutamate and neuropeptides from sensory terminals, depolarizing and activating pro-jection neurons. B2) Opiates decrease Ca2+ influx leading to a decrease in the duration of nociceptor action potentials and a decreased release of transmitter. Also, opiates hyperpolarize the membrane of dorsal horn neurons by activating K+ conductance and decrease the amplitude of the EPSP produced by stimulation of nociceptors. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Nociceptor Nociceptor Projection neuron Morphine Morphine Glutamate Neuropeptides Control No input No input + opiates Sensory input Sensory input + opiates Opiate Neuropeptides Ca2+ Glutamate ENK Projection neuron Norepinephrine Serotonin B1 Sensory input A B2 Sensory input + opiates/opioids Control Control Enkephalin Enkephalin Control Ca2+ Enkephalin Enkephalin CHAPTER 11 Somatosensory Pathways 179 potent synthetic congeners of this compound have been devel-oped. Their effects are blocked by cholinergic blocking drugs, and as yet there is no evidence that they are addictive. Con-versely, the analgesic effect of nicotine is reduced in mice lack-ing the α4 and β2 nicotine cholinergic receptor subunits. These observations make it clear that a nicotinic cholinergic mechanism is involved in the regulation of pain, although its exact role remains to be determined. CANNABINOIDS The cannabinoids anandamide and palmitoylethanolamide (PEA) are produced endogenously and bind to CB1 and CB2 receptors, respectively. Anandamide has been shown to have an analgesic effect, and there are anandamide-containing neu-rons in the periaqueductal gray and other areas concerned with pain. When PEA is administered, it acts peripherally to augment the analgesic effects of anandamide. CHAPTER SUMMARY ■Discriminitive touch, proprioception, and vibratory sensations are relayed via the dorsal column (medial lemniscus) pathway to SI. Pain and temperature sensations are mediated via the ven-trolateralspinothalmic tract to SI. ■The ascending pathways mediating sensation are organized so-matotopically all the way from the spinal cord to SI. ■Descending pathways from the mesencephalic periaqueductal gray inhibit transmission in nociceptive pathways. This de-scending pathway includes a synapse in the ventromedial me-dulla (raphé nucleus) and the release of endogenous opiates. ■Morphine is an effective antinociceptive agent that binds to en-dogenous opiate receptors in the midbrain, brain stem, and spi-nal cord. ■Anandamide is an endogenous cannabinoid that binds to CB1 receptors and acts centrally as an analgesic agent. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. A ventrolateral cordotomy is performed that produces relief of pain in the right leg. It is effective because it interrupts the A) left dorsal column. B) left ventral spinothalamic tract. C) right lateral spinothalamic tract. D) left lateral spinothalamic tract. E) right corticospinal tract. 2. Which of the following does not exert an analgesic effect? A) morphine B) cholinergic antagonists C) adrenergic antagonists D) substance P antagonists E) anandamide 3. A 40-year-old man loses his right hand in a farm accident. Four years later, he has episodes of severe pain in the missing hand (phantom limb pain). A detailed PET scan study of his cerebral cortex might be expected to show A) expansion of the right hand area in his right somatic sensory area I (SI). B) expansion of the right-hand area in his left SI. C) a metabolically inactive spot where his hand area in his left SI would normally be. D) projection of fibers from neighboring sensory areas into the right-hand area of his right SI. E) projection of fibers from neighboring sensory areas into the right-hand area of his left SI. CLINICAL BOX 11–3 Motivation & Addiction Forebrain neurons in the ventral tegmental area and nu-cleus acumbens are thought to be involved in motivated behaviors such as reward, laughter, pleasure, addiction, and fear. These areas have been referred to as the brain’s reward center or pleasure center. Addiction, defined as the re-peated compulsive use of a substance despite negative health consequences, can be produced by a variety of differ-ent drugs. According to the World Health Organization, over 76 million people worldwide suffer from alcohol abuse, and over 15 million suffer from drug abuse. Not surprisingly, alco-hol and drug addiction are associated with the reward sys-tem. The mesocortical dopaminergic neurons that project from the midbrain to the nucleus accumbens and the fron-tal cortex are also involved. The best studied addictive drugs are opiates such as morphine and heroin, cocaine, ampheta-mine, ethyl alcohol, cannabinoids from marijuana, and nico-tine. These drugs affect the brain in different ways, but all have in common the fact that they increase the amount of dopamine available to act on D3 receptors in the nucleus ac-cumbens. Thus, acutely they stimulate the reward system of the brain. On the other hand, long-term addiction involves the development of tolerance, that is, the need for increas-ing amounts of a drug to produce a high. In addition, with-drawal produces psychologic and physical symptoms. Injec-tions of β-noradrenergic antagonists or α2-noradrenergic agonists in the bed nucleus of the stria terminalis reduce the symptoms of opioid withdrawal, and so do bilateral lesions of the lateral tegmental noradrenergic fibers. One of the characteristics of addiction is the tendency of addicts to re-lapse after treatment. For opiate addicts, for example, the re-lapse rate in the first year is about 80%. Relapse often occurs on exposure to sights, sounds, and situations that were previ-ously associated with drug use. An interesting observation that may be relevant in this regard is that as little as a single dose of an addictive drug facilitates release of excitatory neu-rotransmitters in brain areas concerned with memory. The medial frontal cortex, the hippocampus, and the amygdala are concerned with memory, and they all project via excita-tory glutamatergic pathways to the nucleus accumbens. 180 SECTION III Central & Peripheral Neurophysiology 4. A 50-year-old woman undergoes a neurological exam that indi-cates loss of pain and temperature sensitivity, vibratory sense, and proprioception in both legs. These symptoms could be explained by A) a tumor on the medial lemniscal pathway in the sacral spinal cord. B) a peripheral neuropathy. C) a large tumor in the sacral dorsal horn. D) a large tumor affecting the posterior paracentral gyri. E) a large tumor in the ventral posterolateral and posterome-dial thalamic nuclei. CHAPTER RESOURCES Baron R, Maier C: Phantom limb pain: Are cutaneous nociceptors and spinothalamic neurons involved in the signaling and maintenance of spontaneous and touch-evoked pain? A case report. Pain 1995;60:223. Blumenfeld H: Neuroanatomy Through Clinical Cases. Sinauer Associates, 2002. Haines DE (editor): Fundamental Neuroscience for Basic and Clinical Applications, 3rd ed. Elsevier, 2006. Herman J: Phantom limb: From medical knowledge to folk wisdom and back. Ann Int Med 1998;128:76. Hopkins K: Show me where it hurts: Tracing the pathways of pain. J NIH Res 1997;9:37. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. Melzack R: The tragedy of needless pain. Sci Am 1991;262:27. Penfield W, Rasmussen T: The Cerebral Cortex of Man: A Clinical Study of Localization of Function. Macmillan, 1950. Willis WD: The somatosensory system, with emphasis on structures important for pain. Brain Res Rev 2007;55:297. 181 C H A P T E R 12 Vision O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the various parts of the eye and list the functions of each. ■Trace the neural pathways that transmit visual information from the rods and cones to the visual cortex. ■Explain how light rays in the environment are brought to a focus on the retina and the role of accommodation in this process. ■Define hyperopia, myopia, astigmatism, presbyopia, and strabismus. ■Describe the electrical responses produced by rods and cones, and explain how these responses are produced. ■Describe the electrical responses and function of bipolar, horizontal, amacrine, and ganglion cells. ■Describe the responses of cells in the visual cortex and the functional organization of the dorsal and ventral pathways to the parietal cortex. ■Define and explain dark adaptation and visual acuity. ■Describe the neural pathways involved in color vision. ■Name the four types of eye movements and the function of each. INTRODUCTION The eyes are complex sense organs that have evolved from primitive light-sensitive spots on the surface of invertebrates. Within its protective casing, each eye has a layer of receptors, a lens system that focuses light on these receptors, and a sys-tem of nerves that conducts impulses from the receptors to the brain. The way these components operate to set up con-scious visual images is the subject of this chapter. ANATOMIC CONSIDERATIONS The principal structures of the eye are shown in Figure 12–1. The outer protective layer of the eyeball, the sclera, is modified anteriorly to form the transparent cornea, through which light rays enter the eye. Inside the sclera is the choroid, a layer that contains many of the blood vessels that nourish the structures in the eyeball. Lining the posterior two thirds of the choroid is the retina, the neural tissue containing the receptor cells. The crystalline lens is a transparent structure held in place by a circular lens suspensary ligament (zonule). The zonule is attached to the thickened anterior part of the choroid, the ciliary body. The ciliary body contains circular muscle fibers and longitudinal muscle fibers that attach near the corneo-scleral junction. In front of the lens is the pigmented and opaque iris, the colored portion of the eye. The iris contains circular muscle fibers that constrict and radial fibers that dilate the pupil. Variations in the diameter of the pupil can 182 SECTION III Central & Peripheral Neurophysiology produce up to fivefold changes in the amount of light reach-ing the retina. The space between the lens and the retina is filled primarily with a clear gelatinous material called the vitreous (vitreous humor). Aqueous humor, a clear liquid that nourishes the cor-nea and lens, is produced in the ciliary body by diffusion and active transport from plasma. It flows through the pupil and fills the anterior chamber of the eye. It is normally reabsorbed through a network of trabeculae into the canal of Schlemm, a venous channel at the junction between the iris and the cornea (anterior chamber angle). Obstruction of this outlet leads to increased intraocular pressure (see Clinical Box 12–1). RETINA The retina extends anteriorly almost to the ciliary body. It is orga-nized in 10 layers and contains the rods and cones, which are the visual receptors, plus four types of neurons: bipolar cells, gangli-on cells, horizontal cells, and amacrine cells (Figure 12–2). There are many different synaptic transmitters. The rods and cones, which are next to the choroid, synapse with bipolar cells, and the bipolar cells synapse with ganglion cells. About 12 differ-ent types of bipolar cells occur, based on morphology and func-tion. The axons of the ganglion cells converge and leave the eye as the optic nerve. Horizontal cells connect receptor cells to the other receptor cells in the outer plexiform layer. Amacrine cells connect ganglion cells to one another in the inner plexiform lay-er via processes of varying length and patterns. At least 29 types of amacrine cells have been described on the basis of their con-nections. Gap junctions also connect retinal neurons to one an-other, and the permeability of these gap junctions is regulated. Because the receptor layer of the retina rests on the pig-ment epithelium next to the choroid, light rays must pass through the ganglion cell and bipolar cell layers to reach the rods and cones. The pigment epithelium absorbs light rays, preventing the reflection of rays back through the retina. Such reflection would produce blurring of the visual images. The neural elements of the retina are bound together by glial cells called Müller cells. The processes of these cells form an internal limiting membrane on the inner surface of the ret-ina and an external limiting membrane in the receptor layer. The optic nerve leaves the eye and the retinal blood vessels enter it at a point 3 mm medial to and slightly above the pos-terior pole of the globe. This region is visible through the oph-thalmoscope as the optic disk (Figure 12–3). There are no visual receptors over the disk, and consequently this spot is blind (the blind spot). Near the posterior pole of the eye is a yellowish pigmented spot, the macula lutea. This marks the location of the fovea FIGURE 12–1 The internal anatomy of the eye. (From Fox SI, Human Physiology. McGraw-Hill, 2008.) Superior rectus muscle Conjunctiva Cornea Pupil Lens Iris Posterior chamber Vitreous chamber (posterior cavity) Zonular fibers of suspensory ligament Ciliary body Anterior cavity Posterior chamber Anterior chamber Inferior rectus muscle Sclera Choroid Retina Fovea centralis Optic nerve Central artery Central vein CLINICAL BOX 12–1 Glaucoma Increased intraocular pressure does not cause glaucoma, a degenerative disease in which there is loss of retinal ganglia cells. In fact, a substantial minority of the patients with this disease have normal intraocular pressure (10–20 mm Hg). However, increased pressure makes glaucoma worse, and treatment is aimed at lowering the pressure. One cause of increased pressure is decreased permeability through the trabeculae (open-angle glaucoma), and another is forward movement of the iris, obliterating the angle (angle-closure glaucoma). Glaucoma can be treated with β-adrenergic blocking drugs or carbonic anhydrase inhibitors, both of which decrease the production of aqueous humor, or with cholinergic agonists, which increase aqueous outflow. CHAPTER 12 Vision 183 FIGURE 12–2 Neural components of the extrafoveal portion of the retina. C, cone; R, rod; MB, RB, and FB, midget, rod, and flat bipolar cells; DG and MG, diffuse and midget ganglion cells; H, horizontal cells; A, amacrine cells. (Modified from Dowling JE, Boycott BB: Organization of the primate retina: Electron microscopy. Proc R Soc Lond Ser B [Biol] 1966;166:80.) FIGURE 12–3 Retina seen through the ophthalmoscope in a normal human. (a) A photograph and (b) an illustration of the optic fundus (back of the eye). Optic nerve fibers leave the eyeball at the optic disc to form the optic nerve. The arteries, arterioles, and veins in the superficial layers of the retina near its vitreous surface can be seen through the ophthalmoscope. (From Fox SI, Human Physiology. McGraw-Hill, 2008.) C C C C R R R R R H MB MG MG DG DG MB RB RB FB FB A A Pigment epithelium Rod and cone Outer nuclear layer Inner nuclear layer Outer plexiform layer Inner plexiform layer Ganglion cell layer Optic nerve fibers Outer segments Inner segments Arteriole Venule Optic disc Macula lutea Fovea centralis (a) (b) 184 SECTION III Central & Peripheral Neurophysiology centralis, a thinned-out, rod-free portion of the retina that is present in humans and other primates. In it, the cones are densely packed, and each synapses to a single bipolar cell, which, in turn, synapses on a single ganglion cell, providing a direct pathway to the brain. There are very few overlying cells and no blood vessels. Consequently, the fovea is the point where visual acuity is greatest (see Clinical Box 12–2). When attention is attracted to or fixed on an object, the eyes are normally moved so that light rays coming from the object fall on the fovea. The arteries, arterioles, and veins in the superficial layers of the retina near its vitreous surface can be seen through the ophthalmoscope. Because this is the one place in the body where arterioles are readily visible, ophthalmoscopic exami-nation is of great value in the diagnosis and evaluation of dia-betes mellitus, hypertension, and other diseases that affect blood vessels. The retinal vessels supply the bipolar and gan-glion cells, but the receptors are nourished, for the most part, by the capillary plexus in the choroid. This is why retinal detachment is so damaging to the receptor cells. NEURAL PATHWAYS The axons of the ganglion cells pass caudally in the optic nerve and optic tract to end in the lateral geniculate body in the thalamus (Figure 12–4). The fibers from each nasal hemiretina decussate in the optic chiasm. In the geniculate body, the fi-bers from the nasal half of one retina and the temporal half of the other synapse on the cells whose axons form the geniculo-calcarine tract. This tract passes to the occipital lobe of the ce-rebral cortex. The effects of lesions in these pathways on visual function are discussed below. The primary visual receiving area (primary visual cortex, Brodmann’s area 17; also known as V1), is located principally on the sides of the calcarine fissure (Figure 12–5). The organi-zation of the primary visual cortex is discussed below. Some ganglion cell axons pass from the lateral geniculate nucleus to the pretectal region of the midbrain and the supe-rior colliculus, where they form connections that mediate pupillary reflexes and eye movements. The frontal cortex is also concerned with eye movement, and especially its refine-ment. The bilateral frontal eye fields in this part of the cortex are concerned with control of saccades, and an area just ante-rior to these fields is concerned with vergence and the near response. The frontal areas concerned with vision probably project to the nucleus reticularis tegmentalis pontinus, and from there to the other brain stem nuclei mentioned above. Other axons pass directly from the optic chiasm to the suprachiasmatic nuclei in the hypothalamus, where they form connections that synchronize a variety of endocrine and other circadian rhythms with the light–dark cycle. The brain areas activated by visual stimuli have been investi-gated in monkeys and humans by positron emission tomogra-phy (PET) and other imaging techniques. Activation occurs not only in the occipital lobe but also in parts of the inferior tempo-ral cortex, the posteroinferior parietal cortex, portions of the frontal lobe, and the amygdala. The subcortical structures acti-vated in addition to the lateral geniculate body include the superior colliculus, pulvinar, caudate nucleus, putamen, and claustrum. RECEPTORS Each rod and cone is divided into an outer segment, an inner segment that includes a nuclear region, and a synaptic zone (Figure 12–6). The outer segments are modified cilia and are made up of regular stacks of flattened saccules or disks com-posed of membrane. These saccules and disks contain the photosensitive compounds that react to light, initiating action potentials in the visual pathways. The inner segments are rich in mitochondria. The rods are named for the thin, rodlike ap-pearance of their outer segments. Cones generally have thick inner segments and conical outer segments, although their morphology varies from place to place in the retina. In cones, the saccules are formed in the outer segments by infoldings of the cell membrane, but in rods the disks are separated from the cell membrane. Rod outer segments are being constantly renewed by for-mation of new disks at the inner edge of the segment and phagocytosis of old disks from the outer tip by cells of the CLINICAL BOX 12–2 Visual Acuity Visual acuity is the degree to which the details and contours of objects are perceived, and it is usually defined in terms of the shortest distance by which two lines can be separated and still be perceived as two lines. Clinically, visual acuity is often determined by the use of the familiar Snellen letter charts viewed at a distance of 20 ft (6 m). The individual being tested reads aloud the smallest line distinguishable. The results are expressed as a fraction. The numerator of the fraction is 20, the distance at which the subject reads the chart. The denomina-tor is the greatest distance from the chart at which a normal individual can read the smallest line. Normal visual acuity is 20/20; a subject with 20/15 visual acuity has better than nor-mal vision (not farsightedness); and one with 20/100 visual acuity has subnormal vision. The Snellen charts are designed so that the height of the letters in the smallest line a normal in-dividual can read at 20 ft subtends a visual angle of 5 minutes. Each of the lines is separated by 1 minute of arc. Thus, the min-imum separable in a normal individual corresponds to a visual angle of about 1 minute. Visual acuity is a complex phenome-non and is influenced by a large variety of factors, including optical factors (eg, the state of the image-forming mecha-nisms of the eye), retinal factors (eg, the state of the cones), and stimulus factors (eg, illumination, brightness of the stimu-lus, contrast between the stimulus and the background, length of time the subject is exposed to the stimulus). CHAPTER 12 Vision 185 pigment epithelium. Cone renewal is a more diffuse process and appears to occur at multiple sites in the outer segments. In the extrafoveal portions of the retina, rods predominate (Figure 12–7), and there is a good deal of convergence. Flat bipolar cells (Figure 12–2) make synaptic contact with several cones, and rod bipolar cells make synaptic contact with several rods. Because there are approximately 6 million cones and 120 million rods in each human eye but only 1.2 million nerve fibers in each optic nerve, the overall convergence of receptors through bipolar cells on ganglion cells is about 105:1. However, there is divergence from this point on. There are twice as many fibers in the geniculocalcarine tracts as in the optic nerves, and in the visual cortex the number of neurons concerned with vision is 1000 times the number of fibers in the optic nerves. PROTECTION The eye is well protected from injury by the bony walls of the orbit. The cornea is moistened and kept clear by tears that FIGURE 12–4 Visual pathways. Transection of the pathways at the locations indicated by the letters causes the visual field defects shown in the diagrams on the right. The fibers from the nasal half of each retina decussate in the optic chiasm, so that the fibers in the optic tracts are those from the temporal half of one retina and the nasal half of the other. A lesion that interrupts one optic nerve causes blindness in that eye (A). A lesion in one optic tract causes blindness in half of the visual field (C) and is called homonymous (same side of both visual fields) hemianopia (half-blindness). Lesions affecting the optic chiasm destroy fibers from both nasal hemiretinas and produce a heteronymous (opposite sides of the visual fields) hemianopia (B). Occipital lesions may spare the fibers from the macula (as in D) because of the separation in the brain of these fibers from the others subserving vision (see Figure 12–5). Temporal field Ganglion cell Geniculocalcarine tract LEFT EYE LEFT RIGHT A B C D Optic nerve Optic chiasm Pretectal region Nasal field Lateral geniculate body A B C D Optic tract Occipital cortex RIGHT EYE FIGURE 12–5 Medial view of the human right cerebral hemisphere showing projection of the retina on the primary visual cortex (Brodmann’s area 17; also known as V1) in the occipital cortex around the calcarine fissure. The geniculocalcarine fibers from the medial half of the lateral geniculate terminate on the superior lip of the calcarine fissure, and those from the lateral half ter-minate on the inferior lip. Also, the fibers from the lateral geniculate body that relay macular vision separate from those that relay periph-eral vision and end more posteriorly on the lips of the calcarine fissure. Upper peripheral quadrant of retina Upper quadrant of macula Lower quadrant of macula Lower peripheral quadrant of retina 186 SECTION III Central & Peripheral Neurophysiology course from the lacrimal gland in the upper portion of each orbit across the surface of the eye to empty via the lacrimal duct into the nose. Blinking helps keep the cornea moist. One of the most important characteristics of the visual sys-tem is its ability to function over a wide range of light inten-sity. When one goes from near darkness to bright sunlight, light intensity increases by 10 log units, that is, by a factor of 10 billion. One factor reducing the fluctuation in intensity is the diameter of the pupil; when this is reduced from 8 mm to 2 mm, its area decreases by a factor of 16 and light intensity at the retina is reduced by more than 1 log unit. Another factor in reacting to fluctuations in intensity is the presence of two types of receptors. The rods are extremely sensitive to light and are the receptors for night vision (scotopic vision). The scotopic visual apparatus is incapable of resolving the details and boundaries of objects or deter-mining their color. The cones have a much higher threshold, but the cone system has a much greater acuity and is the sys-tem responsible for vision in bright light (photopic vision) and for color vision. There are thus two kinds of inputs to the central nervous system (CNS) from the eye: input from the rods and input from the cones. The existence of these two kinds of input, each working maximally under different con-ditions of illumination, is called the duplicity theory. THE IMAGE-FORMING MECHANISM The eyes convert energy in the visible spectrum into action po-tentials in the optic nerve. The wavelengths of visible light range from approximately 397–723 nm. The images of objects in the environment are focused on the retina. The light rays striking the retina generate potentials in the rods and cones. Impulses initiated in the retina are conducted to the cerebral cortex, where they produce the sensation of vision. PRINCIPLES OF OPTICS Light rays are bent when they pass from a medium of one den-sity into a medium of a different density, except when they strike perpendicular to the interface (Figure 12–8). The bend-ing of light rays is called refraction and is the mechanism that FIGURE 12–6 Schematic diagram of a rod and a cone. Each rod and cone is divided into an outer segment, an inner segment with a nuclear region, and a synaptic zone. The saccules and disks in the outer segment contain photosensitive compounds that react to light to initiate action potentials in the visual pathways. (Reproduced with permission from Lamb TD: Electrical responses of photoreceptors. In: Recent Advances in Physiology. No.10. Baker PF [editor]. Churchill Livingstone, 1984.) Ciliary neck Mitochondria Nucleus Plasma membrane 30 nm Disks Sacs Rod Cone Outer segment Inner segment Synaptic terminal FIGURE 12–7 Rod and cone density along the horizontal meridian through the human retina. A plot of the relative acuity of vision in the various parts of the light-adapted eye would parallel the cone density curve; a similar plot of relative acuity of the dark-adapted eye would parallel the rod density curve. Cones Rods Blind spot 2000 1600 1200 800 400 0 100° 80° 60° 40° Nasal retina Temporal retina 20° 20° 40° 60° 80° 0° Fovea Distance from the fovea Number of rods or cones in an area of 0.0069 mm2 CHAPTER 12 Vision 187 allows one to focus an accurate image onto the retina. Parallel light rays striking a biconvex lens are refracted to a point (principal focus) behind the lens. The principal focus is on a line passing through the centers of curvature of the lens, the principal axis. The distance between the lens and the princi-pal focus is the principal focal distance. For practical purpos-es, light rays from an object that strike a lens more than 6 m (20 ft) away are considered to be parallel. The rays from an ob-ject closer than 6 m are diverging and are therefore brought to a focus farther back on the principal axis than the principal fo-cus. Biconcave lenses cause light rays to diverge. Refractive power is greatest when the curvature of a lens is greatest. The refractive power of a lens is conveniently mea-sured in diopters, the number of diopters being the reciprocal of the principal focal distance in meters. For example, a lens with a principal focal distance of 0.25 m has a refractive power of 1/0.25, or 4 diopters. The human eye has a refractive power of approximately 60 diopters at rest. In the eye, light is actually refracted at the anterior surface of the cornea and at the anterior and posterior surfaces of the lens. The process of refraction can be represented diagram-matically, however, without introducing any appreciable error, by drawing the rays of light as if all refraction occurs at the anterior surface of the cornea (Figure 12–8). It should be noted that the retinal image is inverted. The connections of the retinal receptors are such that from birth any inverted image on the retina is viewed right side up and projected to the visual field on the side opposite to the retinal area stimu-lated. This perception is present in infants and is innate. If ret-inal images are turned right side up by means of special lenses, the objects viewed look as if they are upside down. COMMON DEFECTS OF THE IMAGE-FORMING MECHANISM In some individuals, the eyeball is shorter than normal and the parallel rays of light are brought to a focus behind the retina. This abnormality is called hyperopia or farsightedness (Figure 12–9). Sustained accommodation, even when viewing distant objects, can partially compensate for the defect, but the pro-longed muscular effort is tiring and may cause headaches and blurring of vision. The prolonged convergence of the visual axes associated with the accommodation may lead eventually to FIGURE 12–8 Focusing point sources of light. (a) When diverging light rays enter a dense medium at an angle to its convex surface, re-fraction bends them inward. (b) Refraction of light by the lens system. For simplicity, refraction is shown only at the corneal surface (site of greatest refraction) although it also occurs in the lens and elsewhere. Incoming light from a (above) and b (below) is bent in opposite directions, resulting in b' being above a' on the retina. (From Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology, 11th ed. McGraw-Hill, 2008.) b' a' a b Glass Air Refraction Refraction No refraction Point source of light (a) (b) 188 SECTION III Central & Peripheral Neurophysiology squint (strabismus; see Clinical Box 12–3). The defect can be corrected by using glasses with convex lenses, which aid the refractive power of the eye in shortening the focal distance. In myopia (nearsightedness), the anteroposterior diameter of the eyeball is too long (Figure 12–9). Myopia is said to be genetic in origin. However, there is a positive correlation between sleep-ing in a lighted room before the age of 2 and the subsequent development of myopia. Thus, the shape of the eye appears to be determined in part by the refraction presented to it. In young adult humans the extensive close work involved in activities such as studying accelerates the development of myopia. This defect can be corrected by glasses with biconcave lenses, which make parallel light rays diverge slightly before they strike the eye. Astigmatism is a common condition in which the curvature of the cornea is not uniform (Figure 12–9). When the curva-ture in one meridian is different from that in others, light rays in that meridian are refracted to a different focus, so that part of the retinal image is blurred. A similar defect may be produced if the lens is pushed out of alignment or the curvature of the lens is not uniform, but these conditions are rare. Astigmatism can usually be corrected with cylindric lenses placed in such a way that they equalize the refraction in all meridians. FIGURE 12–9 Common defects of the optical system of the eye. In hyperopia (farsightedness), the eyeball is too short and light rays come to a focus behind the retina. A biconvex lens corrects this by adding to the refractive power of the lens of the eye. In myopia (near-sightedness), the eyeball is too long and light rays focus in front of the retina. Placing a biconcave lens in front of the eye causes the light rays to diverge slightly before striking the eye, so that they are brought to a focus on the retina. (From Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology, 11th ed. McGraw-Hill, 2008.) Normal sight (faraway object is clear) Normal sight (near object is clear) Nearsighted (eyeball too long) Nearsightedness corrected Farsighted (eyeball too short) Farsightedness corrected (b) (a) CLINICAL BOX 12–3 Strabismus & Amblyopia Strabismus is a misalignment of the eyes and one of the most common eye problems in children, affecting about 4% of children under 6 years of age. It is characterized by one or both eyes turning inward (crossed-eyes), outward (wall eyes), upward, or downward. In some cases, more than one of these conditions is present. Strabismus is also commonly called “wandering eye” or “crossed-eyes.” It occurs when visual im-ages do not fall on corresponding retinal points. When visual images chronically fall on noncorresponding points in the two retinas in young children, one is eventually suppressed (sup-pression scotoma). This suppression is a cortical phenome-non, and it usually does not develop in adults. It is important to institute treatment before age 6 in affected children, be-cause if the suppression persists, the loss of visual acuity in the eye generating the suppressed image is permanent. A similar suppression with subsequent permanent loss of visual acuity can occur in children in whom vision in one eye is blurred or distorted owing to a refractive error. The loss of vision in these cases is called amblyopia ex anopsia, a term that refers to uncorrectable loss of visual acuity that is not directly due to organic disease of the eye. Typically, an af-fected child has one weak eye with poor vision and one strong eye with normal vision. It affects about 3% of the general population. Amblyopia is also referred to as “lazy eye,” and it often co-exists with strabismus. Some types of strabismus can be corrected by surgical shortening of some of the eye muscles, by eye muscle training exercises, and by the use of glasses with prisms that bend the light rays suffi-ciently to compensate for the abnormal position of the eye-ball. However, subtle defects in depth perception persist. It has been suggested that congenital abnormalities of the vi-sual tracking mechanisms may cause both strabismus and the defective depth perception. In infant monkeys, covering one eye with a patch for 3 months causes a loss of ocular dominance columns; input from the remaining eye spreads to take over all the cortical cells, and the patched eye be-comes functionally blind. Comparable changes may occur in children with strabismus. CHAPTER 12 Vision 189 ACCOMMODATION When the ciliary muscle is relaxed, parallel light rays striking the optically normal (emmetropic) eye are brought to a focus on the retina. As long as this relaxation is maintained, rays from objects closer than 6 m from the observer are brought to a focus behind the retina, and consequently the objects appear blurred. The problem of bringing diverging rays from close objects to a focus on the retina can be solved by increasing the distance between the lens and the retina or by increasing the curvature or refractive power of the lens. In bony fish, the problem is solved by increasing the length of the eyeball, a so-lution analogous to the manner in which the images of objects closer than 6 m are focused on the film of a camera by moving the lens away from the film. In mammals, the problem is solved by increasing the curvature of the lens. The process by which the curvature of the lens is increased is called accommodation. At rest, the lens is held under ten-sion by the lens ligaments. Because the lens substance is mal-leable and the lens capsule has considerable elasticity, the lens is pulled into a flattened shape. When the gaze is directed at a near object, the ciliary muscle contracts. This decreases the distance between the edges of the ciliary body and relaxes the lens ligaments, so that the lens springs into a more convex shape (Figure 12–10). The change is greatest in the anterior surface of the lens. In young individuals, the change in shape may add as many as 12 diopters to the refractive power of the eye. The relaxation of the lens ligaments produced by contrac-tion of the ciliary muscle is due partly to the sphincterlike action of the circular muscle fibers in the ciliary body and partly to the contraction of longitudinal muscle fibers that attach anteriorly, near the corneoscleral junction. When these fibers contract, they pull the whole ciliary body forward and inward. This motion brings the edges of the ciliary body closer together. Changes in accommodation with age are described in Clinical Box 12–4. In addition to accommodation, the visual axes converge and the pupil constricts when an individual looks at a near object. This three-part response—accommodation, conver-gence of the visual axes, and pupillary constriction—is called the near response. OTHER PUPILLARY REFLEXES When light is directed into one eye, the pupil constricts (pu-pillary light reflex). The pupil of the other eye also con-stricts (consensual light reflex). The optic nerve fibers that carry the impulses initiating these pupillary responses leave the optic nerves near the lateral geniculate bodies. On each side, they enter the midbrain via the brachium of the superi-or colliculus and terminate in the pretectal nucleus. From this nucleus, the second-order neurons project to the ipsilat-eral and contralateral Edinger–Westphal nucleus. The third-order neurons pass from this nucleus to the ciliary ganglion in the oculomotor nerve, and the fourth-order neurons pass from this ganglion to the ciliary body. This pathway is dorsal to the pathway for the near response. Con-sequently, the light response is sometimes lost while the re-sponse to accommodation remains intact (Argyll Robertson pupil). One cause of this abnormality is CNS syphilis, but the Argyll Robertson pupil is also seen in other diseases pro-ducing selective lesions in the midbrain. FIGURE 12–10 Accommodation. The solid lines represent the shape of the lens, iris, and ciliary body at rest, and the dashed lines rep-resent the shape during accommodation. When gaze is directed at a near object, ciliary muscles contract. This decreases the distance be-tween the edges of the ciliary body and relaxes the lens ligaments, and the lens becomes more convex. CLINICAL BOX 12–4 Accommodation & Aging Accommodation is an active process, requiring muscular effort, and can therefore be tiring. Indeed, the ciliary mus-cle is one of the most used muscles in the body. The de-gree to which the lens curvature can be increased is lim-ited, and light rays from an object very near the individual cannot be brought to a focus on the retina, even with the greatest of effort. The nearest point to the eye at which an object can be brought into clear focus by accommodation is called the near point of vision. The near point recedes throughout life, slowly at first and then rapidly with ad-vancing age, from approximately 9 cm at age 10 to ap-proximately 83 cm at age 60. This recession is due princi-pally to increasing hardness of the lens, with a resulting loss of accommodation due to the steady decrease in the degree to which the curvature of the lens can be in-creased. By the time a normal individual reaches age 40– 45, the loss of accommodation is usually sufficient to make reading and close work difficult. This condition, which is known as presbyopia, can be corrected by wear-ing glasses with convex lenses. 190 SECTION III Central & Peripheral Neurophysiology THE PHOTORECEPTOR MECHANISM ELECTRICAL RESPONSES The potential changes that initiate action potentials in the ret-ina are generated by the action of light on photosensitive com-pounds in the rods and cones. When light is absorbed by these substances, their structure changes, and this triggers a se-quence of events that initiates neural activity. The eye is unique in that the receptor potentials of the photo-receptors and the electrical responses of most of the other neural elements in the retina are local, graded potentials, and it is only in the ganglion cells that all-or-none action potentials transmitted over appreciable distances are generated. The responses of the rods, cones, and horizontal cells are hyperpolarizing (Figure 12–11), and the responses of the bipolar cells are either hyperpo-larizing or depolarizing, whereas amacrine cells produce depo-larizing potentials and spikes that may act as generator potentials for the propagated spikes produced in the ganglion cells. The cone receptor potential has a sharp onset and offset, whereas the rod receptor potential has a sharp onset and slow offset. The curves relating the amplitude of receptor potentials to stimulus intensity have similar shapes in rods and cones, but the rods are much more sensitive. Therefore, rod responses are pro-portionate to stimulus intensity at levels of illumination that are below the threshold for cones. On the other hand, cone responses are proportionate to stimulus intensity at high levels of illumina-tion when the rod responses are maximal and cannot change. This is why cones generate good responses to changes in light intensity above background but do not represent absolute illumi-nation well, whereas rods detect absolute illumination. IONIC BASIS OF PHOTORECEPTOR POTENTIALS Na+ channels in the outer segments of the rods and cones are open in the dark, so current flows from the inner to the outer segment (Figure 12–12). Current also flows to the synaptic ending of the photoreceptor. The Na+–K+ pump in the inner segment maintains ionic equilibrium. Release of synaptic transmitter is steady in the dark. When light strikes the outer segment, the reactions that are initiated close some of the Na+ channels, and the result is a hyperpolarizing receptor poten-tial. The hyperpolarization reduces the release of synaptic transmitter, and this generates a signal in the bipolar cells that ultimately leads to action potentials in ganglion cells. The ac-tion potentials are transmitted to the brain. PHOTOSENSITIVE COMPOUNDS The photosensitive compounds in the rods and cones of the eyes of humans and most other mammals are made up of a protein called an opsin, and retinene1, the aldehyde of vitamin A1. The term retinene1 is used to distinguish this compound from retinene2, which is found in the eyes of some animal spe-cies. Because the retinenes are aldehydes, they are also called retinals. The A vitamins themselves are alcohols and are therefore called retinols (see Clinical Box 12–5). RHODOPSIN The photosensitive pigment in the rods is called rhodopsin (visual purple). Its opsin is called scotopsin. Rhodopsin has a peak sensitivity to light at a wavelength of 505 nm. Human rhodopsin has a molecular weight of 41,000. It is found in the membranes of the rod disks and makes up 90% of the total protein in these membranes. It is one of the many recep-tors coupled to G proteins. Retinene1 is parallel to the surface of FIGURE 12–11 Intracellularly recorded responses of cells in the retina to light. The synaptic connections of the cells are also indi-cated. The eye is unique in that the receptor potentials of the photore-ceptors and the electrical responses of most of the other neural elements in the retina are local, graded potentials. The rod (R) on the left is receiving a light flash, whereas the rod on the right is receiving steady, low-intensity illumination. The responses of rods and horizon-tal cells (H) are hyperpolarizing, responses of bipolar cells (B) are either hyperpolarizing or depolarizing, and amacrine (A) cells produce depo-larizing potentials and spikes that may act as generator potentials for propagated spikes of ganglion cells (G). (Reproduced with permission from Dowling JE: Organization of vertebrate retinas. Invest Ophthalmol 1970;9:655.) R R G B A B G G H CHAPTER 12 Vision 191 the membrane (Figure 12–13) and is attached to a lysine resi-due at position 296 in the seventh transmembrane domain. In the dark, the retinene1 in rhodopsin is in the 11-cis con-figuration. The only action of light is to change the shape of the retinene, converting it to the all-trans isomer. This, in turn, alters the configuration of the opsin, and the opsin change activates the associated heterotrimeric G protein, which in this case is called transducin or Gt1. The G protein exchanges GDP for GTP, and the α subunit separates. This subunit remains active until its intrinsic GTPase activity hydrolyzes the GTP. Termination of the activity of transducin is also accelerated by its binding of β-arrestin. The α subunit activates cGMP phosphodiesterase, which converts cGMP to 5'-GMP (Figure 12–14). cGMP normally acts directly on Na+ channels to maintain them in the open position, so the decline in the cytoplasmic cGMP concentra-tion causes some Na+ channels to close. This produces the hyperpolarizing potential. This cascade of reactions occurs very rapidly and amplifies the light signal. The amplification helps explain the remarkable sensitivity of rod photorecep-tors; these receptors are capable of producing a detectable response to as little as one photon of light. After retinene1 is converted to the all-trans configuration, it separates from the opsin (bleaching). Some of the all-trans retinene is converted back to the 11-cis retinene by retinal isomerase, and then again associates with scotopsin, replen-ishing the rhodopsin supply. Some 11-cis retinene is also syn-thesized from vitamin A. All of these reactions, except the formation of the all-trans isomer of retinene1, are indepen-dent of the light intensity, proceeding equally well in light or FIGURE 12–12 Effect of light on current flow in visual receptors. In the dark, Na+ channels in the outer segment are held open by cGMP. Light leads to increased conversion of cGMP to 5'-GMP, and some of the channels close. This produces hyperpolarization of the synaptic terminal of the photoreceptor. K+ Na+ Na+ Dark K+ Na+ Light CLINICAL BOX 12–5 Vitamin Deficiencies In view of the importance of vitamin A in the synthesis of retinene1, it is not surprising that a deficiency in this vita-min produces visual abnormalities. Among these, one of the earliest to appear is night blindness (nyctalopia). Vita-min A deficiency also contributes to blindness by causing the eye to become very dry, which damages the cornea (xerophthalmia) and retina. Vitamin A first alters rod func-tion, but concomitant cone degeneration occurs as vitamin A deficiency develops. Vitamin A deficiency is due to inade-quate intake of foods high in vitamin A (liver, kidney, whole eggs, milk, cream, and cheese) or beta-carotene, a precur-sor of vitamin A, found in dark green leafy vegetables and yellow or orange fruits and vegetables. Vitamin A defi-ciency is rare in the United States, but it is still a major pub-lic health problem in the developing world. Annually, about 80,000 individuals worldwide (mostly children in un-derdeveloped countries) lose their sight from severe vita-min A deficiency. Prolonged deficiency is associated with anatomic changes in the rods and cones followed by de-generation of the neural layers of the retina. Treatment with vitamin A can restore retinal function if given before the receptors are destroyed. Other vitamins, especially those of the B complex, are also necessary for the normal functioning of the retina and other neural tissues. FIGURE 12–13 Diagrammatic representation of the structure of rhodopsin, showing the position of retinene1 (R) in the rod disk membrane. Retinene1 is parallel to the surface of the membrane and is attached to a lysine residue at position 296 in the seventh transmembrane domain. Rod disk membrane Cytoplasmic surface Intradiskal surface R N C OH OH OH OH OH OH 192 SECTION III Central & Peripheral Neurophysiology darkness. The amount of rhodopsin in the receptors therefore varies inversely with the incident light level. CONE PIGMENTS Primates have three different kinds of cones. These receptors subserve color vision and respond maximally to light at wave-lengths of 440, 535, and 565 nm. Each contains retinene1 and an opsin. The opsin resembles rhodopsin and spans the cone membrane seven times but has a characteristic structure in each type of cone. The cell membrane of cones is invaginated to form the saccules, but the cones have no separate intracel-lular disks like those in rods. The details of the responses of cones to light are probably similar to those in rods. Light ac-tivates retinene1, and this activates Gt2, a G protein that dif-fers somewhat from rod transducin. Gt2 in turn activates phosphodiesterase, catalyzing the conversion of cGMP to 5'-GMP. This results in closure of Na+ channels between the ex-tracellular fluid and the cone cytoplasm, a decrease in intra-cellular Na+ concentration, and hyperpolarization of the cone synaptic terminals. The sequence of events in photoreceptors by which incident light leads to production of a signal in the next succeeding neural unit in the retina is summarized in Figure 12–15. RESYNTHESIS OF CYCLIC GMP Light reduces the concentration of Ca2+ as well as that of Na+ in photoreceptors. The resulting decrease in Ca2+ concentra-tion activates guanylyl cyclase, which generates more cGMP. It also inhibits the light-activated phosphodiesterase. Both ac-tions speed recovery, restoring the Na+ channels to their open position. MELANOPSIN A small number of photoreceptors contain melanopsin rather than rhodopsin or cone pigments. The axons of these neurons project to the suprachiasmatic nuclei and the part of the lateral geniculate nuclei that controls the pupillary responses to light. When the gene for melanopsin is knocked out, circadian photo-entrainment is abolished. The papillary light responses are re-duced, and they are abolished when the rods and cones are also inactivated. Thus, a part of the pupillary responses and all the cir-cadian entrainment responses to light–dark changes are con-trolled by a system separate from the rod and cone systems. PROCESSING OF VISUAL INFORMATION IN THE RETINA In a sense, the processing of visual information in the retina in-volves the formation of three images. The first image, formed by the action of light on the photoreceptors, is changed to a second image in the bipolar cells, and this in turn is converted to a third image in the ganglion cells. In the formation of the second im-age, the signal is altered by the horizontal cells, and in the for-mation of the third, it is altered by the amacrine cells. There is little change in the impulse pattern in the lateral geniculate bod-ies, so the third image reaches the occipital cortex. FIGURE 12–14 Initial steps in phototransduction in rods. Light activates rhodopsin, which activates transducin to bind GTP. This activates phosphodiesterase, which catalyzes the conversion of cGMP to 5'-GMP. The resulting decrease in the cytoplasmic cGMP concentra-tion causes cGMP-gated ion channels to close. Outer segment membrane Rhodopsin Transducin cGMP phospho-diesterase Disk Light GTP 5'-GMP cGMP-gated channel Na+ ECF Rod outer segment cGMP FIGURE 12–15 Sequence of events involved in phototransduction in rods and cones. Incident light Structural change in the retinene1 of photopigment Conformational change of photopigment Activation of transducin Activation of phosphodiesterase Decreased intracellular cGMP Closure of Na+ channels Hyperpolarization Decreased release of synaptic transmitter Response in bipolar cells and other neural elements CHAPTER 12 Vision 193 A characteristic of the bipolar and ganglion cells (as well as the lateral geniculate cells and the cells in layer 4 of the visual cortex) is that they respond best to a small, circular stimulus and that, within their receptive field, an annulus of light around the center (surround illumination) inhibits the response to the central spot (Figure 12–16). The center can be excitatory with an inhibitory surround (an “on-center” cell) or inhibitory with an excitatory surround (an “off-center” cell). The inhibition of the center response by the surround is probably due to inhibitory feedback from one photoreceptor to another mediated via horizontal cells. Thus, activation of nearby photoreceptors by addition of the annulus triggers horizontal cell hyperpolarization, which in turn inhibits the response of the centrally activated photoreceptors. The inhibi-tion of the response to central illumination by an increase in surrounding illumination is an example of lateral inhibi-tion—that form of inhibition in which activation of a particu-lar neural unit is associated with inhibition of the activity of nearby units. It is a general phenomenon in mammalian sen-sory systems and helps to sharpen the edges of a stimulus and improve discrimination. A remarkable degree of processing of visual input occurs in the retina, largely via amacrine cells. For example, movement of an object within the visual field is separated from movement of the background caused by changes in posture and movement of the eyes. This was demonstrated by recording from optic neu-rons. When an object moved at a different speed or in a differ-ent direction than the background, an impulse was generated. However, when the object moved like the background, inhibi-tion occurred and no optic nerve signal was generated. At least in some vertebrates, dopamine secreted between the inner nuclear and the inner plexiform layers of the retina (Fig-ure 12–2) diffuses throughout the retina and affects the struc-ture of gap junctions. These junctions allow current to pass freely through horizontal cells in the dark, enlarging the recep-tive fields of the photoreceptors. Light reduces the current flow, decoupling the horizontal cells, and this decoupling appears to be due to increased release of dopamine in daylight. RESPONSES IN THE VISUAL PATHWAYS & CORTEX PATHWAYS TO THE CORTEX The axons of retinal ganglion cells project a detailed spatial rep-resentation of the retina on the lateral geniculate body. Each ge-niculate body contains six well-defined layers (Figure 12–17). Layers 3–6 have small cells and are called parvocellular, whereas layers 1 and 2 have large cells and are called magnocellular. On each side, layers 1, 4, and 6 receive input from the contralateral eye, whereas layers 2, 3, and 5 receive input from the ipsilateral eye. In each layer, there is a precise point-for-point representa-tion of the retina, and all six layers are in register so that along a line perpendicular to the layers, the receptive fields of the cells in each layer are almost identical. It is worth noting that only 10–20% of the input to the lateral geniculate nucleus comes from the retina. Major inputs also occur from the visual cortex and other brain regions. The feedback pathway from the visual cortex has been shown to be involved in visual processing relat-ed to the perception of orientation and motion. Two kinds of ganglion cells can be distinguished in the ret-ina: large ganglion cells (magno, or M cells), which add responses from different kinds of cones and are concerned with movement and stereopsis; and small ganglion cells (parvo, or P cells), which subtract input from one type of cone from input from another and are concerned with color, texture, and shape. The M ganglion cells project to the magnocellular portion of the lateral geniculate, whereas the P ganglion cells project to the parvocellular portion. From the lateral geniculate nucleus, a magnocellular pathway and a parvocellular pathway project to the visual cortex. The magnocellular pathway, from layers 1 and 2, carries signals for detection of movement, depth, and flicker. The parvocellular pathway, from layers 3–6, carries signals for color vision, texture, shape, and fine detail. Cells in the interlaminar region of the lateral geniculate nucleus also receive input from P ganglion cells, probably via FIGURE 12–16 Responses of retinal ganglion cells to light on the portions of their receptive fields indicated in white. Beside each recep-tive field diagram is a diagram of the ganglion cell response, indicated by extracellularly recorded action potentials. Note that in three of the four situations, there is increased discharge when the light is turned off. (Modified from Kandel E, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) On-center field Off-center field Light Light Central illumination Surround illumination 194 SECTION III Central & Peripheral Neurophysiology dendrites of interlaminar cells that penetrate the parvocellular layers. They project via a separate component of the P path-way to the blobs in the visual cortex. PRIMARY VISUAL CORTEX Just as the ganglion cell axons project a detailed spatial repre-sentation of the retina on the lateral geniculate body, the lateral geniculate body projects a similar point-for-point representa-tion on the primary visual cortex (Figure 12–5). In the visual cortex, many nerve cells are associated with each incoming fi-ber. Like the rest of the neocortex, the visual cortex has six lay-ers. The axons from the lateral geniculate nucleus that form the magnocellular pathway end in layer 4, specifically in its deepest part, layer 4C. Many of the axons that form the parvocellular pathway also end in layer 4C. However, the axons from the in-terlaminar region end in layers 2 and 3. Layers 2 and 3 of the cortex contain clusters of cells about 0.2 mm in diameter that, unlike the neighboring cells, contain a high concentration of the mitochondrial enzyme cyto-chrome oxidase. The clusters have been named blobs. They are arranged in a mosaic in the visual cortex and are con-cerned with color vision. However, the parvocellular pathway also carries color opponent data to the deep part of layer 4. Like the ganglion cells, the lateral geniculate neurons and the neurons in layer 4 of the visual cortex respond to stimuli in their receptive fields with on centers and inhibitory sur-rounds or off centers and excitatory surrounds. A bar of light covering the center is an effective stimulus for them because it stimulates the entire center and relatively little of the sur-round. However, the bar has no preferred orientation and, as a stimulus, is equally effective at any angle. The responses of the neurons in other layers of the visual cortex are strikingly different. So-called simple cells respond to bars of light, lines, or edges, but only when they have a par-ticular orientation. When, for example, a bar of light is rotated as little as 10 degrees from the preferred orientation, the firing rate of the simple cell is usually decreased, and if the stimulus is rotated much more, the response disappears. There are also complex cells, which resemble simple cells in requiring a pre-ferred orientation of a linear stimulus but are less dependent upon the location of a stimulus in the visual field than the simple cells and the cells in layer 4. They often respond maxi-mally when a linear stimulus is moved laterally without a change in its orientation. They probably receive input from the simple cells. The visual cortex, like the somatosensory cortex, is arranged in vertical columns that are concerned with orientation (orien-tation columns). Each is about 1 mm in diameter. However, the orientation preferences of neighboring columns differ in a systematic way; as one moves from column to column across the cortex, sequential changes occur in orientation preference of 5–10 degrees. Thus, it seems likely that for each ganglion cell receptive field in the visual field, there is a collection of columns in a small area of visual cortex representing the possible pre-ferred orientations at small intervals throughout the full 360 degrees. The simple and complex cells have been called feature detectors because they respond to and analyze certain features of the stimulus. Feature detectors are also found in the cortical areas for other sensory modalities. The orientation columns can be mapped with the aid of radioactive 2-deoxyglucose. The uptake of this glucose deriv-ative is proportionate to the activity of neurons. When this technique is employed in animals exposed to uniformly ori-ented visual stimuli such as vertical lines, the brain shows a remarkable array of intricately curved but evenly spaced ori-entation columns over a large area of the visual cortex. Another feature of the visual cortex is the presence of ocular dominance columns. The geniculate cells and the cells in layer 4 receive input from only one eye, and the layer 4 cells alternate with cells receiving input from the other eye. If a large amount of a radioactive amino acid is injected into one eye, the amino acid is incorporated into protein and transported by axoplas-mic flow to the ganglion cell terminals, across the geniculate synapses, and along the geniculocalcarine fibers to the visual cortex. In layer 4, labeled endings from the injected eye alter-nate with unlabeled endings from the uninjected eye. The result, when viewed from above, is a vivid pattern of stripes that FIGURE 12–17 Ganglion cell projections from the right hemiretina of each eye to the right lateral geniculate body and from this nucleus to the right primary visual cortex. Note the six layers of the geniculate. P ganglion cells project to layers 3–6, and M ganglion cells project to layers 1 and 2. The ipsilateral (I) and contralat-eral (C) eyes project to alternate layers. Not shown are the interlaminar area cells, which project via a separate component of the P pathway to blobs in the visual cortex. (Modified from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Primary visual cortex (area 17) Parvocellular pathway 1 2 3 4 5 6 Ventral Magnocellular pathway Optic tracts Optic nerves Optic chiasm Lateral geniculate nucleus Dorsal C C C I I I CHAPTER 12 Vision 195 covers much of the visual cortex (Figure 12–18) and is separate from and independent of the grid of orientation columns. About half the simple and complex cells receive an input from both eyes. The inputs are identical or nearly so in terms of the portion of the visual field involved and the preferred orientation. However, they differ in strength, so that between the cells to which the input comes totally from the ipsilateral or the contralateral eye, there is a spectrum of cells influenced to different degrees by both eyes. Thus, the primary visual cortex segregates information about color from that concerned with form and movement, combines the input from the two eyes, and converts the visual world into short line segments of various orientations. OTHER CORTICAL AREAS CONCERNED WITH VISION As mentioned above, the primary visual cortex (V1) projects to many other parts of the occipital lobes and other parts of the brain. These are often identified by number (V2, V3, etc) or by letters (LO, MT, etc). The distribution of some of these in the human brain is shown in Figure 12–19, and their putative func-tions are listed in Table 12–1. Studies of these areas have been carried out in monkeys trained to do various tasks and then fit-ted with implanted microelectrodes. In addition, the availability of PET and functional magnetic resonance imaging (fMRI) scanning has made it possible to conduct sophisticated experi-ments on visual cognition and other cortical visual functions in normal, conscious humans. The visual projections from V1 can be divided roughly into a dorsal or parietal pathway, con-cerned primarily with motion, and a ventral or temporal path-way, concerned with shape and recognition of forms and faces. In addition, connections to the sensory areas are important. For example, in the occipital cortex, visual responses to an object are better if the object is felt at the same time. There are many other relevant connections to other systems. It is apparent from the preceding paragraphs that parallel pro-cessing of visual information occurs along multiple paths. In some as yet unknown way, all the information is eventually pulled together into what we experience as a conscious visual image. COLOR VISION CHARACTERISTICS OF COLOR Colors have three attributes: hue, intensity, and saturation (de-gree of freedom from dilution with white). For any color there is FIGURE 12–18 Reconstruction of ocular dominance columns in a subdivision of layer 4 of a portion of the right visual cortex of a rhesus monkey. Dark stripes represent one eye, light stripes the other. (Reproduced with permission from LeVay S, Hubel DH, Wiesel TN: The pattern of ocular dominance columns in macaque visual cortex revealed by a reduced silver stain. J Comp Neurol 1975;159:559.) 5 mm FIGURE 12–19 Some of the main areas to which the primary visual cortex (V1) projects in the human brain. Lateral and medial views. See also Table 8–1. (Modified from Logothetis N: Vision: A window on consciousness. Sci Am [Nov] 1999;281:99.) LO V7 MT/V5 Parietal lobe Occipital lobe V3A V3 V2 V1 VP V8 V4v Cerebellum V3 V1 V2 VP V4v V8 V3A V7 196 SECTION III Central & Peripheral Neurophysiology a complementary color that, when properly mixed with it, pro-duces a sensation of white. Black is the sensation produced by the absence of light, but it is probably a positive sensation because the blind eye does not “see black;” rather, it “sees nothing.” Another observation of basic importance is the demonstra-tion that the sensation of white, any spectral color, and even the extraspectral color, purple, can be produced by mixing various proportions of red light (wavelength 723–647 nm), green light (575–492 nm), and blue light (492–450 nm). Red, green, and blue are therefore called the primary colors. A third important point is that the color perceived depends in part on the color of other objects in the visual field. Thus, for example, a red object is seen as red if the field is illuminated with green or blue light, but as pale pink or white if the field is illuminated with red light. Clinical Box 12–6 describes color blindness. RETINAL MECHANISMS The Young–Helmholtz theory of color vision in humans postu-lates the existence of three kinds of cones, each containing a differ-ent photopigment and that are maximally sensitive to one of the three primary colors, with the sensation of any given color being determined by the relative frequency of the impulses from each of these cone systems. The correctness of this theory has been dem-onstrated by the identification and chemical characterization of each of the three pigments (Figure 12–20). One pigment (the blue-sensitive, or short-wave, pigment) absorbs light maximally in the blue-violet portion of the spectrum. Another (the green-sensi-tive, or middle-wave, pigment) absorbs maximally in the green portion. The third (the red-sensitive, or long-wave, pigment) ab-sorbs maximally in the yellow portion. Blue, green, and red are the primary colors, but the cones with their maximal sensitivity in the yellow portion of the spectrum are sensitive enough in the red por-tion to respond to red light at a lower threshold than green. This is all the Young–Helmholtz theory requires. The gene for human rhodopsin is on chromosome 3, and the gene for the blue-sensitive S cone pigment is on chromo-some 7. The other two cone pigments are encoded by genes arranged in tandem on the q arm of the X chromosome. The green-sensitive M and red-sensitive L pigments are very simi-lar in structure; their opsins show 96% homology of amino acid sequences, whereas each of these pigments has only about 43% homology with the opsin of blue-sensitive pig-ment, and all three have about 41% homology with rhodop-sin. Many mammals are dichromats; that is, they have only two cone pigments, a short-wave and a long-wave pigment. TABLE 12–1 Functions of visual projection areas in the human brain. V1 Primary visual cortex; receives input from lateral genicu-late nucleus, begins processing in terms of orientation, edges, etc V2, V3, VP Continued processing, larger visual fields V3A Motion V4v Unknown MT/V5 Motion; control of movement LO Recognition of large objects V7 Unknown V8 Color vision Modified from Logothetis N: Vision: a window on consciousness. Sci Am (Nov) 1999;281:99. CLINICAL BOX 12–6 Color Blindness The most common test for color blindness uses the Ishi-hara charts, which are plates containing figures made up of colored spots on a background of similarly shaped colored spots. The figures are intentionally made up of colors that are liable to look the same as the background to an individ-ual who is color blind. Some color-blind individuals are un-able to distinguish certain colors, whereas others have only a color weakness. The prefixes “prot-,” “deuter-,” and “trit-” refer to defects of the red, green, and blue cone systems, re-spectively. Individuals with normal color vision are called trichromats. Dichromats are individuals with only two cone systems; they may have protanopia, deuteranopia, or tritan-opia. Monochromats have only one cone system. Dichro-mats can match their color spectrum by mixing only two pri-mary colors, and monochromats match theirs by varying the intensity of only one. Abnormal color vision is present as an inherited abnormality in Caucasian populations in about 8% of the males and 0.4% of the females. Tritanopia is rare and shows no sexual selectivity. However, about 2% of the color-blind males are dichromats who have protanopia or deutera-nopia, and about 6% are anomalous trichromats in whom the red-sensitive or the green-sensitive pigment is shifted in its spectral sensitivity. These abnormalities are inherited as recessive and X-linked characteristics. Color blindness is present in males if the X chromosome has the abnormal gene. Females show a defect only when both X chromo-somes contain the abnormal gene. However, female children of a man with X-linked color blindness are carriers of the color blindness and pass the defect on to half of their sons. Therefore, X-linked color blindness skips generations and ap-pears in males of every second generation. Color blindness can also occur in individuals with lesions of area V8 of the vi-sual cortex since this region appears to be uniquely con-cerned with color vision in humans. This deficit is called ach-romatopsia. Transient blue-green color weakness occurs as a side effect in individuals taking sildenafil (Viagra) for the treatment of erectile dysfunction because the drug inhibits the retinal as well as the penile form of phosphodiesterase. CHAPTER 12 Vision 197 Old World monkeys, apes, and humans are trichromats, with separate middle- and long-wave pigments—in all probability because there was duplication of the ancestral long-wave gene followed by divergence. There are variations in the red, long-wave pigment in humans. It has been known for some time that responses to the Rayleigh match, the amounts of red and green light that a subject mixes to match a monochromatic orange, are bimodal. This correlates with new evidence that 62% of otherwise color-normal individu-als have serine at site 180 of their long-wave cone opsin, whereas 38% have alanine. The absorption curve of the subjects with serine at position 180 peaks at 556.7 nm, and they are more sen-sitive to red light, whereas the absorption curve of the subjects with alanine at position 180 peaks at 552.4 nm. NEURAL MECHANISMS Color is mediated by ganglion cells that subtract or add input from one type of cone to input from another type. Processing in the ganglion cells and the lateral geniculate nucleus produces im-pulses that pass along three types of neural pathways that project to V1: a red–green pathway that signals differences between L-and M-cone responses, a blue–yellow pathway that signals dif-ferences between S-cone and the sum of L- and M-cone respons-es, and a luminance pathway that signals the sum of L- and M-cone responses. These pathways project to the blobs and the deep portion of layer 4C of V1. From the blobs and layer 4, color information is projected to V8. However, it is not known how V8 converts color input into the sensation of color. OTHER ASPECTS OF VISUAL FUNCTION DARK ADAPTATION If a person spends a considerable length of time in brightly lighted surroundings and then moves to a dimly lighted envi-ronment, the retinas slowly become more sensitive to light as the individual becomes “accustomed to the dark.” This decline in visual threshold is known as dark adaptation. It is nearly maximal in about 20 minutes, although some further decline occurs over longer periods. On the other hand, when one pass-es suddenly from a dim to a brightly lighted environment, the light seems intensely and even uncomfortably bright until the eyes adapt to the increased illumination and the visual thresh-old rises. This adaptation occurs over a period of about 5 min-utes and is called light adaptation, although, strictly speaking, it is merely the disappearance of dark adaptation. The dark adaptation response actually has two components. The first drop in visual threshold, rapid but small in magni-tude, is known to be due to dark adaptation of the cones because when only the foveal, rod-free portion of the retina is tested, the decline proceeds no further. In the peripheral por-tions of the retina, a further drop occurs as a result of adapta-tion of the rods. The total change in threshold between the light-adapted and the fully dark-adapted eye is very great. Radiologists, aircraft pilots, and others who need maximal visual sensitivity in dim light can avoid having to wait 20 min-utes in the dark to become dark-adapted if they wear red gog-gles when in bright light. Light wavelengths in the red end of the spectrum stimulate the rods to only a slight degree while permitting the cones to function reasonably well. Therefore, a person wearing red glasses can see in bright light during the time it takes for the rods to become dark-adapted. The time required for dark adaptation is determined in part by the time required to build up the rhodopsin stores. In bright light, much of the pigment is continuously being bro-ken down, and some time is required in dim light for accumu-lation of the amounts necessary for optimal rod function. However, dark adaptation also occurs in the cones, and addi-tional factors are undoubtedly involved. CRITICAL FUSION FREQUENCY The time-resolving ability of the eye is determined by measur-ing the critical fusion frequency (CFF), the rate at which stimuli can be presented and still be perceived as separate stimuli. Stimuli presented at a higher rate than the CFF are perceived as continuous stimuli. Motion pictures move be-cause the frames are presented at a rate above the CFF, and movies begin to flicker when the projector slows down. VISUAL FIELDS & BINOCULAR VISION The visual field of each eye is the portion of the external world visible out of that eye. Theoretically, it should be circular, but actually it is cut off medially by the nose and superiorly by the roof of the orbit (Figure 12–21). Mapping the visual fields is important in neurologic diagnosis. The peripheral portions of the visual fields are mapped with an instrument called a pe-rimeter, and the process is referred to as perimetry. One eye is covered while the other is fixed on a central point. A small FIGURE 12–20 Absorption spectra of the three cone pigments in the human retina. The S pigment that peaks at 440 nm senses blue, and the M pigment that peaks at 535 nm senses green. The remaining L pigment peaks in the yellow portion of the spectrum, at 565 nm, but its spectrum extends far enough into the long wave-lengths to sense red. (Reproduced with permission from Michael CR: Color vision. N Engl J Med 1973;288:724.) 700 nm 600 500 Wavelength 400 Absorption 198 SECTION III Central & Peripheral Neurophysiology target is moved toward this central point along selected merid-ians, and, along each, the location where the target first be-comes visible is plotted in degrees of arc away from the central point (Figure 12–21). The central visual fields are mapped with a tangent screen, a black felt screen across which a white target is moved. By noting the locations where the target dis-appears and reappears, the blind spot and any objective sco-tomas (blind spots due to disease) can be outlined. The central parts of the visual fields of the two eyes coin-cide; therefore, anything in this portion of the field is viewed with binocular vision. The impulses set up in the two retinas by light rays from an object are fused at the cortical level into a single image (fusion). The points on the retina on which the image of an object must fall if it is to be seen binocularly as a single object are called corresponding points. If one eye is gently pushed out of the line while staring fixedly at an object in the center of the visual field, double vision (diplopia) results; the image on the retina of the eye that is displaced no longer falls on the corresponding point. When visual images no longer fall on corresponding retinal points, strabismus occurs (see Clinical Box 12–3). Binocular vision has an important role in the perception of depth. However, depth perception also has numerous monoc-ular components, such as the relative sizes of objects, the degree one looks down at them, their shadows, and, for mov-ing objects, their movement relative to one another (move-ment parallax). EFFECT OF LESIONS IN THE OPTIC PATHWAYS The anatomy of the pathways from the eyes to the brain is shown in Figure 12–4. Lesions along these pathways can be lo-calized with a high degree of accuracy by the effects they pro-duce in the visual fields. The fibers from the nasal half of each retina decussate in the optic chiasm, so that the fibers in the optic tracts are those from the temporal half of one retina and the nasal half of the other. In other words, each optic tract subserves half of the field of vision. Therefore, a lesion that interrupts one optic nerve causes blindness in that eye, but a lesion in one optic tract causes blindness in half of the visual field (Figure 12–4). This defect is classified as a homonymous (same side of both visual fields) hemianopia (half-blindness). Lesions affecting the optic chiasm, such as pituitary tumors expand-ing out of the sella turcica, cause destruction of the fibers from both nasal hemiretinas and produce a heteronymous (opposite sides of the visual fields) hemianopia. Because the fibers from the maculas are located posteriorly in the optic chiasm, hemianopic scotomas develop before vision in the two hemiretinas is completely lost. Selective visual field defects are further classified as bitemporal, binasal, and right or left. The optic nerve fibers from the upper retinal quadrants subserving vision in the lower half of the visual field termi-nate in the medial half of the lateral geniculate body, whereas the fibers from the lower retinal quadrants terminate in the lateral half. The geniculocalcarine fibers from the medial half of the lateral geniculate terminate on the superior lip of the calcarine fissure, and those from the lateral half terminate on the inferior lip. Furthermore, the fibers from the lateral genic-ulate body that subserve macular vision separate from those that subserve peripheral vision and end more posteriorly on the lips of the calcarine fissure (Figure 12–5). Because of this anatomic arrangement, occipital lobe lesions may produce discrete quadrantic visual field defects (upper and lower quadrants of each half visual field). Macular sparing, that is, loss of peripheral vision with intact macular vision, is also common with occipital lesions (Figure 12–4), because the macular representation is separate from that of the peripheral fields and very large relative to that of the peripheral fields. Therefore, occipital lesions must extend considerable dis-tances to destroy macular as well as peripheral vision. Bilat-eral destruction of the occipital cortex in humans causes subjective blindness. However, there is appreciable blind-sight, that is, residual responses to visual stimuli even though they do not reach consciousness. For example, when these individuals are asked to guess where a stimulus is located dur-ing perimetry, they respond with much more accuracy than can be explained by chance. They are also capable of consider-able discrimination of movement, flicker, orientation, and even color. Similar biasing of responses can be produced by stimuli in the blind areas in patients with hemianopia due to lesions in the visual cortex. The fibers to the pretectal region that subserve the reflex pupillary constriction produced by shining a light into the eye leave the optic tracts near the geniculate bodies. Therefore, blindness with preservation of the pupillary light reflex is usu-ally due to bilateral lesions behind the optic tract. FIGURE 12–21 Monocular and binocular visual fields. The dashed line encloses the visual field of the left eye; the solid line, that of the right eye. The common area (heart-shaped clear zone in the cen-ter) is viewed with binocular vision. The colored areas are viewed with monocular vision. 90° 90° 0° 180° CHAPTER 12 Vision 199 EYE MOVEMENTS The eye is moved within the orbit by six ocular muscles (Fig-ure 12–22). These are innervated by the oculomotor, trochle-ar, and abducens nerves. Because the oblique muscles pull medially, their actions vary with the position of the eye. When the eye is turned nasally, the inferior oblique elevates it and the superior oblique depresses it. When it is turned laterally, the superior rectus elevates it and the inferior rectus depresses it. Because much of the visual field is binocular, it is clear that a very high order of coordination of the movements of the two eyes is necessary if visual images are to fall at all times on corresponding points in the two retinas and diplopia is to be avoided. There are four types of eye movements, each controlled by a different neural system but sharing the same final common path, the motor neurons that supply the external ocular mus-cles. Saccades, sudden jerky movements, occur as the gaze shifts from one object to another. They bring new objects of interest onto the fovea and reduce adaptation in the visual pathway that would occur if gaze were fixed on a single object for long periods. Smooth pursuit movements are tracking movements of the eyes as they follow moving objects. Vestib-ular movements, adjustments that occur in response to stim-uli initiated in the semicircular canals, maintain visual fixation as the head moves. Convergence movements bring the visual axes toward each other as attention is focused on objects near the observer. The similarity to a human-made tracking system on an unstable platform such as a ship is apparent: saccadic movements seek out visual targets, pursuit movements follow them as they move about, and vestibular movements stabilize the tracking device as the platform on which the device is mounted (ie, the head) moves about. In primates, these eye movements depend on an intact visual cortex. Saccades are programmed in the frontal cortex and the superior colliculi and pursuit movements in the cerebellum. SUPERIOR COLLICULI The superior colliculi, which regulate saccades, are innervated by M fibers from the retina. They also receive extensive inner-vation from the cerebral cortex. Each superior colliculus has a map of visual space plus a map of the body surface and a map for sound in space. A motor map projects to the regions of the brain stem that control eye movements. There are also projec-tions via the tectopontine tract to the cerebellum and via the tectospinal tract to areas concerned with reflex movements of the head and neck. The superior colliculi are constantly active positioning the eyes, and they have one of the highest rates of blood flow and metabolism of any region in the brain. CHAPTER SUMMARY ■The major parts of the eye are the sclera (protective covering), cornea (transfer light rays), choroids (nourishment), retina (re-ceptor cells), lens, and iris. ■The visual pathway is from the rods and cones to bipolar cells to ganglion cells then via the optic tract to the thalamic lateral ge-niculate body to the occipital lobe of the cerebral cortex. The fi-bers from each nasal hemiretina decussate in the optic chiasm; the fibers from the nasal half of one retina and the temporal half of the other synapse on the cells whose axons form the genicu-localcarine tract. ■The bending of light rays (refraction) allows one to focus an ac-curate image onto the retina. Light is refracted at the anterior surface of the cornea and at the anterior and posterior surfaces of the lens. To bring diverging rays from close objects to a focus on the retina, the curvature of the lens is increased, a process called accommodation. ■In hyperopia (farsightedness), the eyeball is too short and light rays come to a focus behind the retina. In myopia (nearsighted-ness), the anteroposterior diameter of the eyeball is too long. Astigmatism is a common condition in which the curvature of the cornea is not uniform. Presbyopia is the loss of accommoda-tion for near vision. Strabismus is squinting in an attempt to correct visual acuity. ■Na+ channels in the outer segments of the rods and cones are open in the dark, so current flows from the inner to the outer segment. When light strikes the outer segment, some of the Na+ channels are closed and the cells are hyperpolarized. ■In response to light, horizontal cells are hyperpolarized, bipolar cells are either hyperpolarized or depolarized, and amacrine cells are depolarized and develop spikes that may act as generator FIGURE 12–22 Extraocular muscles subserving the six cardinal positions of gaze. The eye is adducted by the medial rectus and abducted by the lateral rectus. The adducted eye is elevated by the in-ferior oblique and depressed by the superior oblique; the abducted eye is elevated by the superior rectus and depressed by the inferior rectus. (From Squire LR, et al [editors]: Fundamental Neuroscience, 3rd ed. Academic Press, 2008.) 200 SECTION III Central & Peripheral Neurophysiology potentials for the propagated spikes produced in the ganglion cells. ■Neurons in layer 4 of the visual cortex respond to stimuli in their receptive fields with on centers and inhibitory surrounds or off centers and excitatory surrounds. Neurons in other layers are called simple cells if they respond to bars of light, lines, or edges, but only when they have a particular orientation. Complex cells also require a preferred orientation of a linear stimulus but are less dependent on the location of a stimulus in the visual field. ■Projections from V1 can be divided into a dorsal or parietal pathway (concerned primarily with motion) and a ventral or temporal pathway (concerned with shape and recognition of forms and faces). ■The decline in visual threshold after spending long periods of time in a dimly lit room is called dark adaptation. ■The Young–Helmholtz theory of color vision postulates the ex-istence of three kinds of cones, each containing a different pho-topigment and that are maximally sensitive to one of the three primary colors, with the sensation of any given color being de-termined by the relative frequency of the impulses from each of these cone systems. ■Saccades (sudden jerky movements) occur as the gaze shifts from one object to another, and they reduce adaptation in the visual pathway that would occur if gaze were fixed on a single object for long periods. Smooth pursuit movements are tracking movements of the eyes as they follow moving objects. Vestibular movements occur in response to stimuli in the semicircular ca-nals to maintain visual fixation as the head moves. Convergence movements bring the visual axes toward each other as attention is focused on objects near the observer. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. A visual exam in an 80-year-old man shows he has a reduced ability to see objects in the upper and lower quadrants of the left visual fields of both eyes but some vision remains in the central regions of the visual field. The diagnosis is A) central scotoma. B) heteronymous hemianopia with macular sparing. C) lesion of the optic chiasm. D) homonymous hemianopia with macular sparing. E) retinopathy. 2. Visual accommodation involves A) increased tension on the lens ligaments. B) a decrease in the curvature of the lens. C) relaxation of the sphincter muscle of the iris. D) contraction of the ciliary muscle. E) increased intraocular pressure. 3. The fovea of the eye A) has the lowest light threshold. B) is the region of highest visual acuity. C) contains only red and green cones. D) contains only rods. E) is situated over the head of the optic nerve. 4. Which of the following parts of the eye has the greatest concen-tration of rods? A) ciliary body B) iris C) optic disk D) fovea E) parafoveal region 5. The correct sequence of events involved in phototransduction in rods and cones in response to light is: A) activation of transducin, decreased release of glutamate, structural changes in rhodopsin, closure of Na+ channels, and decrease in intracellular cGMP. B) decreased release of glutamate, activation of transducin, clo-sure of Na+ channels, decrease in intracellular cGMP, and structural changes in rhodopsin. C) structural changes in rhodopsin, decrease in intracellular cGMP, decreased release of glutamate, closure of Na+ chan-nels, and activation of transducin. D) structural changes in rhodopsin, activation of transducin, decrease in intracellular cGMP, closure of Na+ channels, and decreased release of glutamate. E) activation of transducin, structural changes in rhodopsin, closure of Na+ channels, decrease in intracellular cGMP, and decreased release of glutamate. 6. Vitamin A is a precursor for the synthesis of A) somatostatin. B) retinene1. C) the pigment of the iris. D) scotopsin. E) aqueous humor. 7. Abnormal color vision is 20 times more common in men than women because most cases are caused by an abnormal A) dominant gene on the Y chromosome. B) recessive gene on the Y chromosome. C) dominant gene on the X chromosome. D) recessive gene on the X chromosome. E) recessive gene on chromosome 22. 8. Which of the following is not involved in color vision? A) activation of a pathway that signals differences between S cone responses and the sum of L and M cone responses B) geniculate layers 3–6 C) P pathway D) area V3A of visual cortex E) area V8 of visual cortex CHAPTER RESOURCES Chiu C, Weliky M: Synaptic modification by vision. Science 2003;300:1890. Dowling JE: Organization of vertebrate retinas. Invest Ophthalmol 1970;9:655 Dowling JE: The Retina: An Approachable Part of the Brain. Belknap, 1987. Gegenfurtner KR, Kiper DC: Color vision. Annu Rev Neurosci 2003;26:181. Hubel DH: Eye, Brain, and Vision. Scientific American Library, 1988. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. CHAPTER 12 Vision 201 Lamb TD: Electrical responses of photoreceptors. In: Recent Advances in Physiology. No.10. Baker PF (editor). Churchill Livingstone, 1984. LeVay S, Hubel DH, Wiesel TN: The pattern of ocular dominance columns in macaque visual cortex revealed by a reduced silver stain. J Comp Neurol 1975;159:559. Logothetis N: Vision: A window on consciousness. Sci Am 1999;281:99. Oyster CW: The Human Eye: Structure and Function. Sinauer, 1999. Squire LR, et al (editors): Fundamental Neuroscience, 3rd ed. Academic Press, 2008. Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology, 11th ed. McGraw-Hill, 2008. This page intentionally left blank 203 C H A P T E R 13 Hearing & Equilibrium O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the components and functions of the external, middle, and inner ear. ■Describe the way that movements of molecules in the air are converted into im-pulses generated in hair cells in the cochlea. ■Trace the path of auditory impulses in the neural pathways from the cochlear hair cells to the auditory cortex, and discuss the function of the auditory cortex. ■Explain how pitch, loudness, and timbre are coded in the auditory pathways. ■Describe the various forms of deafness. ■Explain how the receptors in the semicircular canals detect rotational acceleration and how the receptors in the saccule and utricle detect linear acceleration. ■List the major sensory inputs that provide the information which is synthesized in the brain into the sense of position in space. INTRODUCTION Receptors for two sensory modalities, hearing and equilib-rium, are housed in the ear. The external ear, the middle ear, and the cochlea of the inner ear are concerned with hearing. The semicircular canals, the utricle, and the saccule of the inner ear are concerned with equilibrium. Receptors in the semicircular canals detect rotational acceleration, receptors in the utricle detect linear acceleration in the horizontal direc-tion, and receptors in the saccule detect linear acceleration in the vertical direction. The receptors for hearing and equilib-rium are hair cells, six groups of which are present in each inner ear: one in each of the three semicircular canals, one in the utricle, one in the saccule, and one in the cochlea. ANATOMIC CONSIDERATIONS EXTERNAL & MIDDLE EAR The external ear funnels sound waves to the external auditory meatus (Figure 13–1). In some animals, the ears can be moved like radar antennas to seek out sound. From the external audi-tory meatus, sound waves pass inward to the tympanic mem-brane (eardrum). The middle ear is an air-filled cavity in the temporal bone that opens via the auditory (eustachian) tube into the nasopharynx and through the nasopharynx to the exterior. The tube is usually closed, but during swallowing, chewing, and yawning it opens, keeping the air pressure on the two sides of the eardrum equalized. The three auditory ossicles, the malleus, incus, and stapes, are located in the middle ear (Fig-ure 13–2). The manubrium (handle of the malleus) is attached to the back of the tympanic membrane. Its head is attached to the wall of the middle ear, and its short process is attached to the incus, which in turn articulates with the head of the stapes. The stapes is named for its resemblance to a stirrup. Its foot plate is attached by an annular ligament to the walls of the oval window. Two small skeletal muscles, the tensor tympani and the stapedius, are also located in the middle ear. Contraction of 204 SECTION III Central & Peripheral Neurophysiology FIGURE 13–1 The structures of the outer, middle, and inner portions of the human ear. To make the relationships clear, the cochlea has been turned slightly and the middle ear muscles have been omitted. (From Fox SI, Human Physiology. McGraw-Hill, 2008.) FIGURE 13–2 The medial view of the middle ear. The locations of auditory muscles attached to the middle-ear ossicles are indicated. (From Fox SI, Human Physiology. McGraw-Hill, 2008.) CHAPTER 13 Hearing & Equilibrium 205 the former pulls the manubrium of the malleus medially and decreases the vibrations of the tympanic membrane; contrac-tion of the latter pulls the foot plate of the stapes out of the oval window. The functions of the ossicles and the muscles are con-sidered in more detail below. INNER EAR The inner ear (labyrinth) is made up of two parts, one within the other. The bony labyrinth is a series of channels in the petrous portion of the temporal bone. Inside these channels, surrounded by a fluid called perilymph, is the membranous labyrinth (Fig-ure 13–3). This membranous structure more or less duplicates the shape of the bony channels. It is filled with a K+-rich fluid called endolymph, and there is no communication between the spaces filled with endolymph and those filled with perilymph. COCHLEA The cochlear portion of the labyrinth is a coiled tube which in humans is 35 mm long and makes a two and three quarter turns. Throughout its length, the basilar membrane and Reissner’s membrane divide it into three chambers or scalae (Figure 13–4). The upper scala vestibuli and the lower scala tympani contain perilymph and communicate with each other at the apex of the cochlea through a small opening called the helicotrema. At the base of the cochlea, the scala vestibuli ends at the oval window, which is closed by the footplate of the stapes. The scala tympani ends at the round window, a foramen on the medial wall of the middle ear that is closed by the flexible secondary tympanic membrane. The scala media, the middle cochlear chamber, is continuous with the membranous labyrinth and does not com-municate with the other two scalae. ORGAN OF CORTI Located on the basilar membrane is the organ of Corti, the structure that contains the hair cells, which are the auditory receptors. This organ extends from the apex to the base of the cochlea and consequently has a spiral shape. The processes of the hair cells pierce the tough, membrane-like reticular lami-na that is supported by the pillar cells or rods of Corti (Figure FIGURE 13–3 Schematic of the human inner ear showing the membranous labyrinth with enlargements of the structures in which hair cells are embedded. The membranous labyrinth is suspended in perilymph and filled with K+-rich endolymph which bathes the receptors. Hair cells (darkened for emphasis) occur in different arrays characteristic of the receptor organs. The three semicircular canals are sensitive to an-gular accelerations which deflect the gelatinous cupula and associated hair cells. In the cochlea, hair cells spiral along the basilar membrane in the organ of Corti. Airborne sounds set the eardrum in motion, which is conveyed to the cochlea by bones of the middle ear. This flexes the membrane up and down. Hair cells in the organ of Corti are stimulated by shearing motion. The otolithic organs (saccule and utricle) are sensitive to linear acceleration in vertical and horizontal planes. Hair cells are attached to the otolithic membrane. VIII, eighth cranial nerve, with auditory and ves-tibular divisions. (Reproduced with permission from Hudspeth AJ: How the ear’s works work. Nature 1989;341:397. Copyright © 1989 by Macmillan Magazines.) Cupula VIII Semicircular canal Sacculus Otolithic membrane Tectorial membrane Basilar membrane Cochlea 206 SECTION III Central & Peripheral Neurophysiology 13–4). The hair cells are arranged in four rows: three rows of outer hair cells lateral to the tunnel formed by the rods of Corti, and one row of inner hair cells medial to the tunnel. There are 20,000 outer hair cells and 3500 inner hair cells in each human cochlea. Covering the rows of hair cells is a thin, viscous, but elastic tectorial membrane in which the tips of the hairs of the outer but not the inner hair cells are embedded. The cell bodies of the sensory neurons that arborize around the bases of the hair cells are located in the spiral ganglion within the modiolus, the bony core around which the cochlea is wound. Ninety to 95% of these sensory neurons innervate the inner hair cells; only 5–10% innervate the more numerous outer hair cells, and each sensory neuron innervates several outer hair cells. By contrast, most of the efferent fibers in the auditory nerve terminate on the outer rather than inner hair cells. The axons of the afferent neurons that innervate the hair cells form the auditory (cochlear) division of the eighth cra-nial nerve. In the cochlea, tight junctions between the hair cells and the adjacent phalangeal cells prevent endolymph from reaching the bases of the cells. However, the basilar membrane is relatively permeable to perilymph in the scala tympani, and consequently, the tunnel of the organ of Corti and the bases of the hair cells are bathed in perilymph. Because of similar tight junctions, the arrangement is similar for the hair cells in other parts of the inner ear; that is, the processes of the hair cells are bathed in endolymph, whereas their bases are bathed in perilymph. SEMICIRCULAR CANALS On each side of the head, the semicircular canals are perpendic-ular to each other, so that they are oriented in the three planes of space. Inside the bony canals, the membranous canals are suspended in perilymph. A receptor structure, the crista amp-ullaris, is located in the expanded end (ampulla) of each of the membranous canals. Each crista consists of hair cells and sup-porting (sustentacular) cells surmounted by a gelatinous parti-tion (cupula) that closes off the ampulla (Figure 13–3). The processes of the hair cells are embedded in the cupula, and the bases of the hair cells are in close contact with the afferent fibers of the vestibular division of the eighth cranial nerve. UTRICLE & SACCULE Within each membranous labyrinth, on the floor of the utri-cle, is an otolithic organ (macula). Another macula is located on the wall of the saccule in a semivertical position. The mac-ulae contain supporting cells and hair cells, surmounted by an otolithic membrane in which are embedded crystals of calci-um carbonate, the otoliths (Figure 13–3). The otoliths, which are also called otoconia or ear dust, range from 3 to 19 μm in length in humans and are more dense than the endolymph. The processes of the hair cells are embedded in the membrane. The nerve fibers from the hair cells join those from the cristae in the vestibular division of the eighth cranial nerve. HAIR CELLS STRUCTURE As noted above, the sensory receptors in the ear consist of six patches of hair cells in the membranous labyrinth. The hair cells in the organ of Corti signal hearing; the hair cells in the utricle signal horizontal acceleration; the hair cells in the sac-cule signal vertical acceleration; and a patch in each of the three semicircular canals signal rotational acceleration. These hair cells have a common structure (Figure 13–5). Each is em-bedded in an epithelium made up of supporting cells, with the basal end in close contact with afferent neurons. Projecting from the apical end are 30 to 150 rod-shaped processes, or hairs. FIGURE 13–4 Top: Cross-section of the cochlea, showing the organ of Corti and the three scalae of the cochlea. Bottom: Structure of the organ of Corti, as it appears in the basal turn of the cochlea. DC, outer phalangeal cells (Deiters’ cells) supporting out-er hair cells; IPC, inner phalangeal cell supporting inner hair cell. (Reproduced with permission from Pickels JO: An Introduction to the Physiology of Hearing, 2nd ed. Academic Press, 1988.) Arch DCs Inner hair cell IPC Tectorial membrane Reticular lamina Outer hair cells Basilar membrane Pillar cell (rod of Corti) Habenula perforata Spiral lamina Nerve fibers Spiral ganglion Modiolus Tectorial membrane Reissner’s membrane Limbus Scala vestibuli Scala media Spiral prominence Stria vascularis Spiral ligament Spiral lamina Organ of Corti Basilar membrane Spiral ligament Scala tympani Tunnel CHAPTER 13 Hearing & Equilibrium 207 Except in the cochlea, one of these, the kinocilium, is a true but nonmotile cilium with nine pairs of microtubules around its circumference and a central pair of microtubules. It is one of the largest processes and has a clubbed end. The kinocilium is lost from the hair cells of the cochlea in adult mammals. However, the other processes, which are called stereocilia, are present in all hair cells. They have cores composed of parallel filaments of actin. The actin is coated with various isoforms of myosin. Within the clump of processes on each cell there is an orderly structure. Along an axis toward the kinocilium, the stereocilia increase progressively in height; along the perpen-dicular axis, all the stereocilia are the same height. ELECTRICAL RESPONSES The resting membrane potential of the hair cells is about –60 mV. When the stereocilia are pushed toward the kinocilium, the membrane potential is decreased to about –50 mV. When the bundle of processes is pushed in the opposite direction, the cell is hyperpolarized. Displacing the processes in a direction perpendicular to this axis provides no change in membrane potential, and displacing the processes in directions that are intermediate between these two directions produces depolar-ization or hyperpolarization that is proportionate to the de-gree to which the direction is toward or away from the kinocilium. Thus, the hair processes provide a mechanism for generating changes in membrane potential proportional to the direction and distance the hair moves. GENESIS OF ACTION POTENTIALS IN AFFERENT NERVE FIBERS Very fine processes called tip links (Figure 13–6) tie the tip of each stereocilium to the side of its higher neighbor, and at the junction are cation channels in the higher process that appear to be mechanically sensitive. When the shorter stereocilia are FIGURE 13–5 Left: Structure of a hair cell in the saccule. Hair cells in the membranous labyrinth of the ear have a common structure, and each is within an epithelium of supporting cells (SC) surmounted by an otolithic membrane (OM) embedded with crystals of calcium carbonate, the otoliths (OT). Projecting from the apical end are rod-shaped processes, or hair cells (RC), in contact with afferent (A) and efferent (E) nerve fibers. Except in the cochlea, one of these, kinocilium (K), is a true but nonmotile cilium with nine pairs of microtubules around its circumference and a central pair of microtubules. The other processes, stereocilia (S), are found in all hair cells; they have cores of actin filaments coated with isoforms of myosin. Within the clump of processes on each cell there is an orderly structure. Along an axis toward the kinocilium, the stereocilia increase progressively in height; along the perpendicular axis, all the stereocilia are the same height. (Reproduced with permission from Hillman DE: Morphology of peripheral and central vestibular systems. In: Llinas R, Precht W [editors]: Frog Neurobiology. Springer, 1976.) Right: Scanning electron photomicrograph of process-es on a hair cell in the saccule. The otolithic membrane has been removed. The small projections around the hair cell are microvilli on supporting cells. (Courtesy of AJ Hudspeth.) OM K S OL OM RC SC A E 208 SECTION III Central & Peripheral Neurophysiology pushed toward the higher, the open time of these channels in-creases. K+—the most abundant cation in endolymph—and Ca2+ enter via the channel and produce depolarization. There is still considerable uncertainty about subsequent events. Howev-er, one hypothesis is that a molecular motor in the higher neigh-bor next moves the channel toward the base, releasing tension in the tip link (Figure 13–6). This causes the channel to close and permits restoration of the resting state. The motor appar-ently is myosin-based. Depolarization of hair cells causes them to release a neurotransmitter, probably glutamate, which ini-tiates depolarization of neighboring afferent neurons. The K+ that enters hair cells via the mechanically sensitive cation channels is recycled (Figure 13–7). It enters supporting cells and then passes on to other supporting cells by way of tight junctions. In the cochlea, it eventually reaches the stria vascularis and is secreted back into the endolymph, complet-ing the cycle. The processes of the hair cells project into the endolymph whereas the bases are bathed in perilymph. This arrangement is necessary for the normal production of generator poten-tials. The perilymph is formed mainly from plasma. On the other hand, endolymph is formed in the scala media by the stria vascularis and has a high concentration of K+ and a low concentration of Na+ (Figure 13–7). Cells in the stria vascu-laris have a high concentration of Na+–K+ pump. In addition, it appears that a unique electrogenic K+ pump in the stria vascularis accounts for the fact that the scala media is electri-cally positive by 85 mV relative to the scala vestibuli and scala tympani. HEARING SOUND WAVES Sound is the sensation produced when longitudinal vibrations of the molecules in the external environment—that is, alter-nate phases of condensation and rarefaction of the mole-cules—strike the tympanic membrane. A plot of these movements as changes in pressure on the tympanic mem-brane per unit of time is a series of waves (Figure 13–8); such movements in the environment are generally called sound waves. The waves travel through air at a speed of approximate-ly 344 m/s (770 mph) at 20 °C at sea level. The speed of sound increases with temperature and with altitude. Other media in which humans occasionally find themselves also conduct sound waves but at different speeds. For example, the speed of sound is 1450 m/s at 20 °C in fresh water and is even greater in salt water. It is said that the whistle of the blue whale is as loud as 188 decibels and is audible for 500 miles. FIGURE 13–6 Schematic representation of the role of tip links in the responses of hair cells. When a stereocilium is pushed toward a taller stereocilium, the tip line is stretched and opens an ion channel in its taller neighbor. The channel next is presumably moved down the taller stereocilium by a molecular motor, so the tension on the tip link is released. When the hairs return to the resting position, the motor moves back up the stereocilium. (Modified from Kandel ER, Schwartz JH, Jessel TM [editors]: Principles of Neuroscience, 4th ed. McGraw-Hill, 2000.) Myosin Ca2+ K+ Tip link FIGURE 13–7 Ionic composition of perilymph in the scala vestibuli, endolymph in the scala media, and perilymph in the scala tympani. SL, spiral ligament. SV, stria vascularis. The dashed ar-row indicates the path by which K+ recycles from the hair cells to the supporting cells to the spiral ligament and is then secreted back into the endolymph by cells in the stria vascularis. Organ of Corti Scala vestibuli Cl− 125 Na+ 150 K+ 5 Scala tympani Cl− 130 Na+ 1 K+ 150 SL and SV Cl− 125 Na+ 150 K+ 3 CHAPTER 13 Hearing & Equilibrium 209 Generally speaking, the loudness of a sound is correlated with the amplitude of a sound wave and its pitch with the fre-quency (number of waves per unit of time). The greater the amplitude, the louder the sound; and the greater the fre-quency, the higher the pitch. Sound waves that have repeating patterns, even though the individual waves are complex, are perceived as musical sounds; aperiodic nonrepeating vibra-tions cause a sensation of noise. Most musical sounds are made up of a wave with a primary frequency that determines the pitch of the sound plus a number of harmonic vibrations (overtones) that give the sound its characteristic timbre (quality). Variations in timbre permit us to identify the sounds of the various musical instruments even though they are playing notes of the same pitch. The amplitude of a sound wave can be expressed in terms of the maximum pressure change at the eardrum, but a relative scale is more convenient. The decibel scale is such a scale. The intensity of a sound in bels is the logarithm of the ratio of the intensity of that sound and a standard sound. A decibel (dB) is 0.1 bel. The standard sound reference level adopted by the Acoustical Society of America corresponds to 0 dB at a pressure level of 0.000204 × dyne/cm2, a value that is just at the auditory threshold for the average human. A value of 0 dB does not mean the absence of sound but a sound level of an intensity equal to that of the standard. The 0- to 140-dB range from threshold pressure to a pressure that is potentially dam-aging to the organ of Corti actually represents a 107 (10 mil-lion)-fold variation in sound pressure. Put another way, atmospheric pressure at sea level is 15 lb/in2 or 1 bar, and the range from the threshold of hearing to potential damage to the cochlea is 0.0002 to 2000 μbar. A range of 120 to 160 dB (eg, firearms, jackhammer, jet plane on take off) is classified as painful; 90 to 110 dB (eg, subway, bass drum, chain saw, lawn mower) is classified as extremely high; 60 to 80 dB (eg, alarm clock, busy traffic, dishwasher, conversation) is classified as very loud; 40 to 50 dB (eg, moderate rainfall, normal room noise) is moderate; and 30 dB (eg, whisper, library) is faint. The sound frequencies audible to humans range from about 20 to a maximum of 20,000 cycles per second (cps, Hz). In bats and dogs, much higher frequencies are audible. The threshold of the human ear varies with the pitch of the sound (Figure 13–9), the greatest sensitivity being in the 1000- to 4000-Hz range. The pitch of the average male voice in conver-sation is about 120 Hz and that of the average female voice about 250 Hz. The number of pitches that can be distin-guished by an average individual is about 2000, but trained musicians can improve on this figure considerably. Pitch dis-crimination is best in the 1000- to 3000-Hz range and is poor at high and low pitches. The presence of one sound decreases an individual’s ability to hear other sounds, a phenomenon known as masking. It is believed to be due to the relative or absolute refractoriness of previously stimulated auditory receptors and nerve fibers to other stimuli. The degree to which a given tone masks others is related to its pitch. The masking effect of the background noise in all but the most carefully soundproofed environ-ments raises the auditory threshold by a definite and measur-able amount. SOUND TRANSMISSION The ear converts sound waves in the external environment into action potentials in the auditory nerves. The waves are transformed by the eardrum and auditory ossicles into move-ments of the foot plate of the stapes. These movements set up FIGURE 13–8 Characteristics of sound waves. A is the record of a pure tone. B has a greater amplitude and is louder than A. C has the same amplitude as A but a greater frequency, and its pitch is high-er. D is a complex wave form that is regularly repeated. Such patterns are perceived as musical sounds, whereas waves like that shown in E, which have no regular pattern, are perceived as noise. Time Frequency (cycles per unit time) Pressure change Amplitude 1 cycle A B C D E FIGURE 13–9 Human audibility curve. The middle curve is that obtained by audiometry under the usual conditions. The lower curve is that obtained under ideal conditions. At about 140 decibels (top curve), sounds are felt as well as heard. 140 100 0 10 (Tickle in ear) Threshold of feeling Frequency (Hz) 102 103 104 2 × 104 Intensity level (decibels) Threshold of hearing–audiometer Threshold of hearing–ideal 210 SECTION III Central & Peripheral Neurophysiology waves in the fluid of the inner ear (Figure 13–10). The action of the waves on the organ of Corti generates action potentials in the nerve fibers. In response to the pressure changes produced by sound waves on its external surface, the tympanic membrane moves in and out. The membrane therefore functions as a resonator that reproduces the vibrations of the sound source. It stops vibrating almost immediately when the sound wave stops. The motions of the tympanic membrane are imparted to the manubrium of the malleus. The malleus rocks on an axis through the junction of its long and short processes, so that the short process transmits the vibrations of the manubrium to the incus. The incus moves in such a way that the vibra-tions are transmitted to the head of the stapes. Movements of the head of the stapes swing its foot plate to and fro like a door hinged at the posterior edge of the oval window. The auditory ossicles thus function as a lever system that converts the reso-nant vibrations of the tympanic membrane into movements of the stapes against the perilymph-filled scala vestibuli of the cochlea (Figure 13–10). This system increases the sound pres-sure that arrives at the oval window, because the lever action of the malleus and incus multiplies the force 1.3 times and the area of the tympanic membrane is much greater than the area of the foot plate of the stapes. Some sound energy is lost as a result of resistance, but it has been calculated that at frequen-cies below 3000 Hz, 60% of the sound energy incident on the tympanic membrane is transmitted to the fluid in the cochlea. TYMPANIC REFLEX When the middle ear muscles (tensor tympani and stapedius) contract, they pull the manubrium of the malleus inward and the footplate of the stapes outward (Figure 13–2). This de-creases sound transmission. Loud sounds initiate a reflex con-traction of these muscles called the tympanic reflex. Its function is protective, preventing strong sound waves from causing excessive stimulation of the auditory receptors. How-ever, the reaction time for the reflex is 40 to 160 ms, so it does not protect against brief intense stimulation such as that pro-duced by gunshots. BONE & AIR CONDUCTION Conduction of sound waves to the fluid of the inner ear via the tympanic membrane and the auditory ossicles, the main path-way for normal hearing, is called ossicular conduction. Sound waves also initiate vibrations of the secondary tympanic mem-brane that closes the round window. This process, unimpor-tant in normal hearing, is air conduction. A third type of conduction, bone conduction, is the transmission of vibra-tions of the bones of the skull to the fluid of the inner ear. Con-siderable bone conduction occurs when tuning forks or other vibrating bodies are applied directly to the skull. This route also plays a role in transmission of extremely loud sounds. TRAVELING WAVES The movements of the foot plate of the stapes set up a series of traveling waves in the perilymph of the scala vestibuli. A dia-gram of such a wave is shown in Figure 13–11. As the wave moves up the cochlea, its height increases to a maximum and then drops off rapidly. The distance from the stapes to this point of maximum height varies with the frequency of the vibrations initiating the wave. High-pitched sounds generate waves that reach maximum height near the base of the cochlea; low-pitched sounds generate waves that peak near the apex. The FIGURE 13–10 Schematic representation of the auditory ossicles and the way their movement translates movements of the tympanic membrane into a wave in the fluid of the inner ear. The wave is dissipated at the round window. The movements of the ossicles, the membranous labyrinth, and the round window are indi-cated by dashed lines. Stapes Oval window Reissner's membrane Basilar membrane Round window Auditory tube Organ of Corti Pivot Malleus Incus FIGURE 13–11 Traveling waves. Top: The solid and the short-dashed lines represent the wave at two instants of time. The long-dashed line shows the “envelope” of the wave formed by connecting the wave peaks at successive instants. Bottom: Displacement of the basilar membrane by the waves generated by stapes vibration of the frequencies shown at the top of each curve. 22 0 10 20 30 24 26 Distance from stapes (mm) 800 Hz 400 Hz 50 Hz 1600 Hz Distance from stapes (mm) 28 30 32 Relative amplitude Displace-ment of basilar membrane CHAPTER 13 Hearing & Equilibrium 211 bony walls of the scala vestibuli are rigid, but Reissner’s mem-brane is flexible. The basilar membrane is not under tension, and it also is readily depressed into the scala tympani by the peaks of waves in the scala vestibuli. Displacements of the fluid in the scala tympani are dissipated into air at the round window. Therefore, sound produces distortion of the basilar membrane, and the site at which this distortion is maximal is determined by the frequency of the sound wave. The tops of the hair cells in the organ of Corti are held rigid by the reticular lamina, and the hairs of the outer hair cells are embedded in the tectorial mem-brane (Figure 13–4). When the stapes moves, both membranes move in the same direction, but they are hinged on different ax-es, so a shearing motion bends the hairs. The hairs of the inner hair cells are not attached to the tectorial membrane, but they are apparently bent by fluid moving between the tectorial mem-brane and the underlying hair cells. FUNCTIONS OF THE INNER & OUTER HAIR CELLS The inner hair cells are the primary sensory cells that generate action potentials in the auditory nerves, and presumably they are stimulated by the fluid movements noted above. The outer hair cells, on the other hand, have a different func-tion. These respond to sound, like the inner hair cells, but depo-larization makes them shorten and hyperpolarization makes them lengthen. They do this over a very flexible part of the basal membrane, and this action somehow increases the amplitude and clarity of sounds. These changes in outer hair cells occur in parallel with changes in prestin, a membrane protein, and this protein may well be the motor protein of outer hair cells. The outer hair cells receive cholinergic innervation via an efferent component of the auditory nerve, and acetylcholine hyperpolarizes the cells. However, the physiologic function of this innervation is unknown. ACTION POTENTIALS IN AUDITORY NERVE FIBERS The frequency of the action potentials in single auditory nerve fi-bers is proportional to the loudness of the sound stimuli. At low sound intensities, each axon discharges to sounds of only one frequency, and this frequency varies from axon to axon depend-ing on the part of the cochlea from which the fiber originates. At higher sound intensities, the individual axons discharge to a wid-er spectrum of sound frequencies, particularly to frequencies lower than that at which threshold simulation occurs. The major determinant of the pitch perceived when a sound wave strikes the ear is the place in the organ of Corti that is maximally stimulated. The traveling wave set up by a tone pro-duces peak depression of the basilar membrane, and conse-quently maximal receptor stimulation, at one point. As noted above, the distance between this point and the stapes is inversely related to the pitch of the sound, with low tones pro-ducing maximal stimulation at the apex of the cochlea and high tones producing maximal stimulation at the base. The path-ways from the various parts of the cochlea to the brain are dis-tinct. An additional factor involved in pitch perception at sound frequencies of less than 2000 Hz may be the pattern of the action potentials in the auditory nerve. When the frequency is low enough, the nerve fibers begin to respond with an impulse to each cycle of a sound wave. The importance of this volley effect, however, is limited; the frequency of the action potentials in a given auditory nerve fiber determines princi-pally the loudness, rather than the pitch, of a sound. Although the pitch of a sound depends primarily on the fre-quency of the sound wave, loudness also plays a part; low tones (below 500 Hz) seem lower and high tones (above 4000 Hz) seem higher as their loudness increases. Duration also affects pitch to a minor degree. The pitch of a tone cannot be perceived unless it lasts for more than 0.01 s, and with durations between 0.01 and 0.1 s, pitch rises as duration increases. Finally, the pitch of complex sounds that include harmonics of a given fre-quency is still perceived even when the primary frequency (missing fundamental) is absent. CENTRAL PATHWAY The afferent fibers in the auditory division of the eighth cranial nerve end in dorsal and ventral cochlear nuclei (Figure 13–12). From there, auditory impulses pass by various routes to the infe-rior colliculi, the centers for auditory reflexes, and via the medi-al geniculate body in the thalamus to the auditory cortex. Other impulses enter the reticular formation. Information from both ears converges on each superior olive, and beyond this, most of the neurons respond to inputs from both sides. The primary au-ditory cortex is Brodmann’s area 41 (see Figure 13–13). In hu-mans, low tones are represented anterolaterally and high tones posteromedially in the auditory cortex. In the primary auditory cortex, most neurons respond to inputs from both ears, but strips of cells are stimulated by input from the contralateral ear and inhibited by input from the ipsi-lateral ear. There are several additional auditory receiving areas, just as there are several receiving areas for cutaneous sensation. The auditory association areas adjacent to the primary auditory receiving areas are widespread. The olivocochlear bundle is a prominent bundle of efferent fibers in each auditory nerve that arises from both ipsilateral and contralateral superior olivary complexes and ends pri-marily around the bases of the outer hair cells of the organ of Corti. AUDITORY RESPONSES OF NEURONS IN THE MEDULLA OBLONGATA The responses of individual second-order neurons in the cochlear nuclei to sound stimuli are like those of the individu-al auditory nerve fibers. The frequency at which sounds of the 212 SECTION III Central & Peripheral Neurophysiology lowest intensity evoke a response varies from unit to unit; with in-creased sound intensities, the band of frequencies to which a re-sponse occurs becomes wider. The major difference between the responses of the first- and second-order neurons is the presence of a sharper “cutoff” on the low-frequency side in the medullary neurons. This greater specificity of the second-order neurons is probably due to an inhibitory process in the brain stem. OTHER CORTICAL AREAS CONCERNED WITH AUDITION The increasing availability of positron emission tomography (PET) scanning and functional magnetic resonance imaging (fMRI) has led to rapid increases in knowledge about auditory association areas in humans. The auditory pathways in the cor-tex resemble the visual pathways in that increasingly complex processing of auditory information takes place along them. An interesting observation is that although the auditory areas look very much the same on the two sides of the brain, there is marked hemispheric specialization. For example, Brodmann’s area 22 is concerned with the processing of auditory signals re-lated to speech. During language processing, it is much more active on the left side than on the right side. Area 22 on the right side is more concerned with melody, pitch, and sound intensity. The auditory pathways are also very plastic, and, like the visual and somasthetic pathways, they are modified by experience. Ex-amples of auditory plasticity in humans include the observation that in individuals who become deaf before language skills are fully developed, viewing sign language activates auditory asso-ciation areas. Conversely, individuals who become blind early in life are demonstrably better at localizing sound than individ-uals with normal eyesight. Musicians provide additional examples of cortical plasticity. In these individuals, the size of the auditory areas activated by musical tones is increased. In addition, violinists have altered somatosensory representation of the areas to which the fin-gers they use in playing their instruments project. Musicians also have larger cerebellums than nonmusicians, presumably because of learned precise finger movements. A portion of the posterior superior temporal gyrus known as the planum temporale (Figure 13–13) is regularly larger in the left than in the right cerebral hemisphere, particularly in right-handed individuals. This area appears to be involved in language-related auditory processing. A curious observation, which is presently unexplained, is that the planum temporale is even larger than normal on the left side in musicians and others who have perfect pitch. FIGURE 13–12 Simplified diagram of main auditory (left) and vestibular (right) pathways superimposed on a dorsal view of the brain stem. Cerebellum and cerebral cortex have been removed. Medulla Thalamus To cortex (superior temporal gyrus) From cochlea Spiral ganglion Inferior colliculus Pineal Reticular formation Medial geniculate body Superior olives IV ventricle Dorsal and ventral cochlear nuclei From utricle, semicircular canals Anterior vestibulo-spinal tracts Lateral vestibulo-spinal tract Vestibular ganglion Vestibular nuclei: superior, lateral (Deiters’), medial, spinal To cerebellum To somatosensory cortex Medial longitudinal fasciculus Thalamus III IV VI AUDITORY VESTIBULAR CHAPTER 13 Hearing & Equilibrium 213 SOUND LOCALIZATION Determination of the direction from which a sound emanates in the horizontal plane depends on detecting the difference in time between the arrival of the stimulus in the two ears and the consequent difference in phase of the sound waves on the two sides; it also depends on the fact that the sound is louder on the side closest to the source. The detectable time difference, which can be as little as 20 μs, is said to be the most important factor at frequencies below 3000 Hz and the loudness differ-ence the most important at frequencies above 3000 Hz. Neu-rons in the auditory cortex that receive input from both ears respond maximally or minimally when the time of arrival of a stimulus at one ear is delayed by a fixed period relative to the time of arrival at the other ear. This fixed period varies from neuron to neuron. Sounds coming from directly in front of the individual dif-fer in quality from those coming from behind because each pinna (the visible portion of the exterior ear) is turned slightly forward. In addition, reflections of the sound waves from the pinnal surface change as sounds move up or down, and the change in the sound waves is the primary factor in locating sounds in the vertical plane. Sound localization is markedly disrupted by lesions of the auditory cortex. AUDIOMETRY Auditory acuity is commonly measured with an audiometer. This device presents the subject with pure tones of various fre-quencies through earphones. At each frequency, the threshold intensity is determined and plotted on a graph as a percentage of normal hearing. This provides an objective measurement of the degree of deafness and a picture of the tonal range most affected. DEAFNESS Hearing loss is the most common sensory defect in humans. According to the World Health Organization, over 270 mil-lion people worldwide have moderate to profound hearing loss, with one fourth of these cases beginning in childhood. Presbycusis, the gradual hearing loss associated with aging, affects more than one-third of those over 75 and is probably due to gradual cumulative loss of hair cells and neurons. In most cases, hearing loss is a multifactorial disorder caused by both genetic and environmental factors. Genetic factors con-tributing to deafness are described in Clinical Box 13–1. Deafness can be divided into two major categories: conduc-tive (or conduction) and sensorineural hearing loss. Conduc-tive deafness refers to impaired sound transmission in the external or middle ear and impacts all sound frequencies. Among the causes of conduction deafness are plugging of the external auditory canals with wax (cerumen) or foreign bod-ies, otitis externa (inflammation of the outer ear, “swimmer’s ear”) and otitis media (inflammation of the middle ear) caus-ing fluid accumulation, perforation of the eardrum, and osteosclerosis in which bone is resorbed and replaced with sclerotic bone that grows over the oval window. Sensorineural deafness is most commonly the result of loss of cochlear hair cells but can also be due to problems with the eighth cranial nerve or within central auditory pathways. It often impairs the ability to hear certain pitches while others are unaffected. Aminoglycoside antibiotics such as streptomy-cin and gentamicin obstruct the mechanosensitive channels in the stereocilia of hair cells and can cause the cells to degen-erate, producing sensorineural hearing loss and abnormal vestibular function. Damage to the outer hair cells by pro-longed exposure to noise is associated with hearing loss. Other causes include tumors of the eighth cranial nerve and cerebellopontine angle and vascular damage in the medulla. Conduction and sensorineural deafness can be differenti-ated by simple tests with a tuning fork. Three of these tests, named for the individuals who developed them, are outlined in Table 13–1. The Weber and Schwabach tests demonstrate the important masking effect of environmental noise on the auditory threshold. VESTIBULAR SYSTEM The vestibular system can be divided into the vestibular appa-ratus and central vestibular nuclei. The vestibular apparatus FIGURE 13–13 Left and right planum temporale in a brain sectioned horizontally along the plane of the sylvian fissure. Plane of section shown in the insert at the bottom. (Reproduced with permission from Kandel ER, Schwartz JH, Jessel TM [editors]: Principles of Neural Science, 3rd ed. McGraw-Hill, 1991.) Left planum temporale Right planum temporale Occipital pole Frontal pole Heschl’s sulcus 214 SECTION III Central & Peripheral Neurophysiology within the inner ear detects head motion and position and transduces this information to a neural signal (Figure 13–3). The vestibular nuclei are primarily concerned with maintain-ing the position of the head in space. The tracts that descend from these nuclei mediate head-on-neck and head-on-body adjustments. CENTRAL PATHWAY The cell bodies of the 19,000 neurons supplying the cristae and maculae on each side are located in the vestibular ganglion. Each vestibular nerve terminates in the ipsilateral four-part ves-tibular nucleus and in the flocculonodular lobe of the cerebel-lum (Figure 13–12). Fibers from the semicircular canals end primarily in the superior and medial divisions of the vestibular nucleus and project mainly to nuclei controlling eye movement. Fibers from the utricle and saccule end predominantly in the lateral division (Deiters nucleus), which projects to the spinal cord. They also end on neurons that project to the cerebellum and the reticular formation. The vestibular nuclei also project to the thalamus and from there to two parts of the primary soma-tosensory cortex. The ascending connections to cranial nerve nuclei are largely concerned with eye movements. RESPONSES TO ROTATIONAL ACCELERATION Rotational acceleration in the plane of a given semicircular ca-nal stimulates its crista. The endolymph, because of its inertia, is displaced in a direction opposite to the direction of rotation. The fluid pushes on the cupula, deforming it. This bends the processes of the hair cells (Figure 13–3). When a constant speed of rotation is reached, the fluid spins at the same rate as the body and the cupula swings back into the upright position. When rotation is stopped, deceleration produces displace-ment of the endolymph in the direction of the rotation, and the cupula is deformed in a direction opposite to that during acceleration. It returns to mid position in 25 to 30 s. Move-ment of the cupula in one direction commonly causes an CLINICAL BOX 13–1 Genetic Mutations Contributing to Deafness Single-gene mutations have been shown to cause hearing loss. This type of hearing loss is a monogenic disorder with an autosomal dominant, autosomal recessive, X-linked, or mito-chondrial mode of inheritance. Monogenic forms of deafness can be defined as syndromic (hearing loss associated with other abnormalities) or nonsyndromic (only hearing loss). About 0.1% of newborns have genetic mutations leading to deafness. Nonsyndromic deafness due to genetic mutations can first appear in adults rather than in children and may ac-count for many of the 16% of all adults who have significant hearing impairment. It is now estimated that the products of 100 or more genes are essential for normal hearing, and deaf-ness loci have been described in all but 5 of the 24 human chromosomes. The most common mutation leading to con-genital hearing loss is that of the protein connexin 26. This de-fect prevents the normal recycling of K+ through the sustenac-ular cells. Mutations in three nonmuscle myosins also cause deafness. These are myosin-VIIa, associated with the actin in the hair cell processes; myosin-Ib, which is probably part of the “adaptation motor” that adjusts tension on the tip links; and myosin-VI, which is essential in some way for the formation of normal cilia. Deafness is also associated with mutant forms of α-tectin, one of the major proteins in the tectorial membrane. An example of syndromic deafness is Pendred syndrome, in which a mutant sulfate transport protein causes deafness and goiter. Another example is one form of the long QT syn-drome in which one of the K+ channel proteins, KVLQT1, is mutated. In the stria vascularis, the normal form of this protein is essential for maintaining the high K+ concentration in en-dolymph, and in the heart it helps maintain a normal QT inter-val. Individuals who are homozygous for mutant KVLQT1 are deaf and predisposed to the ventricular arrhythmias and sud-den death that characterize the long QT syndrome. Mutations of the membrane protein barttin can cause deafness as well as the renal manifestations of Bartter syndrome. TABLE 13–1 Common tests with a tuning fork to distinguish between sensorineural and conduction deafness. Weber Rinne Schwabach Method Base of vibrating tuning fork placed on vertex of skull. Base of vibrating tuning fork placed on mastoid process until subject no longer hears it, then held in air next to ear. Bone conduction of patient com-pared with that of normal subject. Normal Hears equally on both sides. Hears vibration in air after bone conduction is over. Conduction deaf-ness (one ear) Sound louder in diseased ear because masking effect of environmental noise is absent on diseased side. Vibrations in air not heard after bone con-duction is over. Bone conduction better than nor-mal (conduction defect excludes masking noise). Sensorineural deaf-ness (one ear) Sound louder in normal ear. Vibration heard in air after bone conduction is over, as long as nerve deafness is partial. Bone conduction worse than normal. CHAPTER 13 Hearing & Equilibrium 215 increase in the firing rate of single nerve fibers from the crista, whereas movement in the opposite direction commonly in-hibits neural activity (Figure 13–14). Rotation causes maximal stimulation of the semicircular canals most nearly in the plane of rotation. Because the canals on one side of the head are a mirror image of those on the other side, the endolymph is displaced toward the ampulla on one side and away from it on the other. The pattern of stimu-lation reaching the brain therefore varies with the direction as well as the plane of rotation. Linear acceleration probably fails to displace the cupula and therefore does not stimulate the cristae. However, there is considerable evidence that when one part of the labyrinth is destroyed, other parts take over its functions. Clinical Box 13–2 describes the characteristic eye movements that occur during a period of rotation. RESPONSES TO LINEAR ACCELERATION In mammals, the utricular and saccular maculae respond to linear acceleration. In general, the utricle responds to horizon-tal acceleration and the saccule to vertical acceleration. The otoliths are more dense than the endolymph, and acceleration in any direction causes them to be displaced in the opposite di-rection, distorting the hair cell processes and generating activ-ity in the nerve fibers. The maculae also discharge tonically in the absence of head movement, because of the pull of gravity on the otoliths. The impulses generated from these receptors are partly responsible for labyrinth righting reflexes. These reflexes are a series of responses integrated for the most part in the nuclei of the midbrain. The stimulus for the reflex is tilting of the head, which stimulates the otolithic organs; the response is compensatory contraction of the neck muscles to keep the head level. In cats, dogs, and primates, visual cues can initiate optical righting reflexes that right the animal in the absence of labyrinthine or body stimulation. In humans, the operation of these reflexes maintains the head in a stable position and the eyes fixed on visual targets despite movements of the body and the jerks and jolts of everyday life. The responses are initi-ated by vestibular stimulation, stretching of neck muscles, and movement of visual images on the retina, and the responses are the vestibulo-ocular reflex and other remarkably precise reflex contractions of the neck and extraocular muscles. FIGURE 13–14 Ampullary responses to rotation. Average time course of impulse discharge from the ampulla of two semicircular canals during rotational acceleration, steady rotation, and decelera-tion. Movement of the cupula in one direction increases the firing rate of single nerve fibers from the crista, and movement in the opposite direction inhibits neural activity. (Reproduced with permission from Adrian ED: Discharge from vestibular receptors in the cat. J Physiol [Lond] 1943;101:389.) 0 10 20 30 40 50 60 Time (s) Rotation Angular velocity Frequency of impulses/s 60 40 20 0 60 40 20 0 CLINICAL BOX 13–2 Nystagmus The characteristic jerky movement of the eye observed at the start and end of a period of rotation is called nystagmus. It is actually a reflex that maintains visual fixation on stationary points while the body rotates, although it is not initiated by visual impulses and is present in blind individuals. When rota-tion starts, the eyes move slowly in a direction opposite to the direction of rotation, maintaining visual fixation (vestibulo-ocular reflex, VOR). When the limit of this movement is reached, the eyes quickly snap back to a new fixation point and then again move slowly in the other direction. The slow component is initiated by impulses from the vestibular laby-rinths; the quick component is triggered by a center in the brain stem. Nystagmus is frequently horizontal (ie, the eyes move in the horizontal plane), but it can also be vertical (when the head is tipped sidewise during rotation) or rotatory (when the head is tipped forward). By convention, the direc-tion of eye movement in nystagmus is identified by the direc-tion of the quick component. The direction of the quick com-ponent during rotation is the same as that of the rotation, but the postrotatory nystagmus that occurs owing to displace-ment of the cupula when rotation is stopped is in the oppo-site direction. Clinically, nystagmus is seen at rest in patients with lesions of the brain stem. Nystagmus can persist for hours at rest in patients with acute temporal bone fracture af-fecting semicircular canals or after damage to the flocculon-odular lobe or midline structures such as the fastigial nucleus. Nystagmus can be used as a diagnostic indicator of the integ-rity of the vestibular system. Caloric stimulation can be used to test the function of the vestibular labyrinth. The semicircu-lar canals are stimulated by instilling warm (40 °C) or cold (30 °C) water into the external auditory meatus. The tempera-ture difference sets up convection currents in the endolymph, with consequent motion of the cupula. In normal subjects, warm water causes nystagmus that bears toward the stimulus, whereas cold water induces nystagmus that bears toward the opposite ear. This test is given the mnemonic COWS (Cold water nystagmus is Opposite sides, Warm water nystagmus is Same side). In the case of a unilateral lesion in the vestibular pathway, nystagmus is reduced or absent on the side of the le-sion. To avoid nystagmus, vertigo, and nausea when irrigating the ear canals in the treatment of ear infections, it is important to be sure that the fluid used is at body temperature. 216 SECTION III Central & Peripheral Neurophysiology Although most of the responses to stimulation of the macu-lae are reflex in nature, vestibular impulses also reach the cerebral cortex. These impulses are presumably responsible for conscious perception of motion and supply part of the information necessary for orientation in space. Vertigo is the sensation of rotation in the absence of actual rotation and is a prominent symptom when one labyrinth is inflamed. SPATIAL ORIENTATION Orientation in space depends in part on input from the vestib-ular receptors, but visual cues are also important. Pertinent in-formation is also supplied by impulses from proprioceptors in joint capsules, which supply data about the relative position of the various parts of the body, and impulses from cutaneous exteroceptors, especially touch and pressure receptors. These four inputs are synthesized at a cortical level into a continuous picture of the individual’s orientation in space. Clinical Box 13–3 describes some common vestibular disorders. CHAPTER SUMMARY ■The external ear funnels sound waves to the external auditory meatus and tympanic membrane. From there, sound waves pass through three auditory ossicles (malleus, incus, and stapes) in the middle ear. The inner ear, or labyrinth, contains the cochlea and organ of Corti. ■The hair cells in the organ of Corti signal hearing. The stereocil-ia provide a mechanism for generating changes in membrane potential proportional to the direction and distance the hair moves. Sound is the sensation produced when longitudinal vi-brations of air molecules strike the tympanic membrane. ■The activity within the auditory pathway passes from the eighth cranial nerve afferent fibers to the dorsal and ventral cochlear nuclei to the inferior colliculi to the thalamic medial geniculate body and then to the auditory cortex. ■Loudness is correlated with the amplitude of a sound wave, pitch with the frequency, and timbre with harmonic vibrations. ■Conductive deafness is due to impaired sound transmission in the external or middle ear and impacts all sound frequencies. Sensorineural deafness is usually due to loss of cochlear hair cells but can also occur after damage to the eighth cranial nerve or central auditory pathways. ■Rotational acceleration stimulates the crista in the semicircular, displacing the endolymph in a direction opposite to the direc-tion of rotation, deforming the cupula and bending the hair cell. The utricle responds to horizontal acceleration and the saccule to vertical acceleration. Acceleration in any direction displaces the otoliths, distorting the hair cell processes and generating neural. ■Spatial orientation is dependent on input from vestibular recep-tors, visual cues, proprioceptors in joint capsules, and cutaneous touch and pressure receptors. CLINICAL BOX 13–3 Vestibular Disorders Vestibular balance disorders are the ninth most common reason for visits to a primary care physician. It is one of the most common reasons elderly people seek medical advice. Patients often describe balance problems in terms of ver-tigo, dizziness, lightheadedness, and motion sickness. Nei-ther lightheadedness nor dizziness is necessarily a symp-tom of vestibular problems, but vertigo is a prominent symptom of a disorder of the inner ear or vestibular system, especially when one labyrinth is inflamed. Benign parox-ysmal positional vertigo is the most common vestibular disorder characterized by episodes of vertigo that occur with particular changes in body position (eg, turning over in bed, bending over). One possible cause is that otoconia from the utricle separate from the otolith membrane and become lodged in the cupula of the posterior semicircular canal. This causes abnormal deflections when the head changes position relative to gravity. Ménière disease is an abnormality of the inner ear caus-ing vertigo or severe dizziness, tinnitus, fluctuating hearing loss, and the sensation of pressure or pain in the affected ear lasting several hours. Symptoms can occur suddenly and recur daily or very rarely. The hearing loss is initially transient but can become permanent. The pathophysiology likely in-volves an immune reaction. An inflammatory response can increase fluid volume within the membranous labyrinth, causing it to rupture and allowing the endolymph and peri-lymph to mix together. There is no cure for Ménière disease but the symptoms can be controlled by reducing the fluid re-tention through dietary changes (low-salt or salt-free diet, no caffeine, no alcohol) or medication. The nausea, blood pressure changes, sweating, pallor, and vomiting that are the well-known symptoms of motion sick-ness are produced by excessive vestibular stimulation and occurs when conflicting information is fed into the vestibular and other sensory systems. The symptoms are probably due to reflexes mediated via vestibular connections in the brain stem and the flocculonodular lobe of the cerebellum. Space motion sickness— the nausea, vomiting, and vertigo expe-rienced by astronauts—develops when they are first ex-posed to microgravity and often wears off after a few days of space flight. It can then recur with reentry, as the force of gravity increases again. It is believed to be due to mis-matches in neural input created by changes in the input from some parts of the vestibular apparatus and other grav-ity sensors without corresponding changes in the other spa-tial orientation inputs. CHAPTER 13 Hearing & Equilibrium 217 MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. A 40-year-old male, employed as a road construction worker for nearly 20 years, went to his physician to report that he recently began to notice difficulty hearing during normal conversations. A Weber test showed that sound from a vibrating tuning fork was localized to the right ear. A Schwabach test showed that bone conduction was below normal. A Rinne test showed that both air and bone conduction were abnormal, but air conduc-tion lasted longer than bone conduction. The diagnosis was: A) sensorial deafness in both ears B) conduction deafness in the right ear C) sensorial deafness in the right ear D) conduction deafness in the left ear E) sensorineural deafness in the left ear 2. What would the diagnosis be if a patient had the following test results? Weber test showed that sound from a vibrating tuning fork was louder than normal; Schwabach test showed that bone conduction was better than normal; and Rinne test showed that air conduction did not outlast bone conduction. A sensorial deafness in both ears B) conduction deafness in both ears C) normal hearing D) both sensorial and conduction deafness E) a possible tumor on the eighth cranial nerve 3. Postrotatory nystagmus is caused by continued movement of A) aqueous humor over the ciliary body in the eye. B) cerebrospinal fluid over the parts of the brain stem that con-tain the vestibular nuclei. C) endolymph in the semicircular canals, with consequent bending of the cupula and stimulation of hair cells. D) endolymph toward the helicotrema. E) perilymph over hair cells that have their processes embed-ded in the tectorial membrane. 4. Some diseases damage the hair cells in the ear. When the damage to the outer hair cells is greater than the damage to the inner hair cells, A) perception of vertical acceleration is disrupted. B) K+ concentration in endolymph is decreased. C) K+ concentration in perilymph is decreased. D) there is severe hearing loss. E) affected hair cells fail to shorten when exposed to sound. 5. Which of the following are incorrectly paired? A) tympanic membrane : manubrium of malleus B) helicotrema : apex of cochlea C) footplate of stapes : oval window D) otoliths : semicircular canals E) basilar membrane : organ of Corti 6. The direction of nystagmus is vertical when a subject is rotated A) after warm water is put in one ear. B) with the head tipped backward. C) after cold water is put in both ears. D) with the head tipped sideways. E) after section of one vestibular nerve. 7. In the utricle, tip links in hair cells are involved in A) formation of perilymph. B) depolarization of the stria vascularis. C) movements of the basement membrane. D) perception of sound. E) regulation of distortion-activated ion channels. 8. A patient enters the hospital for evaluation of deafness. He is found to also have an elevated plasma renin, although his blood pressure is 118/75 mm Hg. Mutation of what single gene may explain these findings? A) the gene for barttin B) the gene for Na+ channel C) the gene for renin D) the gene for cystic fibrosis transmembrane conductance reg-ulator E) the gene for tyrosine hydroxylase CHAPTER RESOURCES Baloh RW, Halmagyi M: Disorders of the Vestibular System. Oxford University Press, 1996. Fox SI: Human Physiology. McGraw-Hill, 2008. Haines DE (editor): Fundamental Neuroscience for Basic and Clinical Applications, 3rd ed. Elsevier, 2006. Highstein SM, Fay RR, Popper AN (editors): The Vestibular System. Springer, 2004. Hudspeth AJ: The cellular basis of hearing: The biophysics of hair cells. Science 1985;230:745. Hudspeth AJ: How the ear’s works work. Nature 1989;341:397. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. Oertel D, Fay RR, Popper AN (editors): Integrative Functions in the Mammalian Auditory Pathway. Springer, 2002. Pickles JO: An Introduction to the Physiology of Hearing, 2nd ed. Academic Press, 1988. Squire LR, et al (editors): Fundamental Neuroscience, 3rd ed. Academic Press, 2008. Willems PJ: Genetic causes of hearing loss. NE J Med 2000;342:1101. This page intentionally left blank 219 C H A P T E R 14 Smell & Taste O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the basic features of the neural elements in the olfactory epithelium and olfactory bulb. ■Describe signal transduction in odorant receptors. ■Outline the pathway by which impulses generated in the olfactory epithelium reach the olfactory cortex. ■Describe the location and cellular composition of taste buds. ■Name the five major taste receptors and signal transduction mechanisms in these receptors. ■Outline the pathways by which impulses generated in taste receptors reach the in-sular cortex. INTRODUCTION Smell and taste are generally classified as visceral senses because of their close association with gastrointestinal func-tion. Physiologically, they are related to each other. The fla-vors of various foods are in large part a combination of their taste and smell. Consequently, food may taste “different” if one has a cold that depresses the sense of smell. Both smell and taste receptors are chemoreceptors that are stimulated by molecules in solution in mucus in the nose and saliva in the mouth. Because stimuli arrive from an external source, they are also classified as exteroceptors. SMELL OLFACTORY EPITHELIUM The olfactory sensory neurons are located in a specialized por-tion of the nasal mucosa, the yellowish pigmented olfactory epi-thelium. In dogs and other animals in which the sense of smell is highly developed (macrosmatic animals), the area covered by this membrane is large; in microsmatic animals, such as humans, it is small. In humans, it covers an area of 5 cm2 in the roof of the nasal cavity near the septum (Figure 14–1). The human olfactory epi-thelium contains 10 to 20 million bipolar olfactory sensory neu-rons interspersed with glia-like supporting (sustentacular) cells and basal stem cells. The olfactory epithelium is said to be the place in the body where the nervous system is closest to the exter-nal world. Each neuron has a short, thick dendrite that projects into the nasal cavity where it terminates in a knob containing 10 to 20 cilia (Figure 14–2). The cilia are unmyelinated processes about 2 μm long and 0.1 μm in diameter and contain specific re-ceptors for odorants (odorant receptors). The axons of the olfac-tory sensory neurons pass through the cribriform plate of the ethmoid bone and enter the olfactory bulbs (Figure 14–1). New olfactory sensory neurons are generated by basal stem cells as needed to replace those damaged by exposure to the environment. The olfactory renewal process is carefully regu-lated, and there is evidence that in this situation a bone mor-phogenic protein (BMP) exerts an inhibitory effect. BMPs are a large family of growth factors originally described as promoters of bone growth but now known to act on most tissues in the body during development, including many types of nerve cells. 220 SECTION III Central & Peripheral Neurophysiology OLFACTORY BULBS In the olfactory bulbs, the axons of the olfactory sensory neu-rons (first cranial nerve) contact the primary dendrites of the mitral cells and tufted cells (Figure 14–3) to form anatomi-cally discrete synaptic units called olfactory glomeruli. The tufted cells are smaller than the mitral cells and have thinner axons, but both types send axons into the olfactory cortex, and they appear to be similar from a functional point of view. In addition to mitral and tufted cells, the olfactory bulbs contain FIGURE 14–1 Olfactory sensory neurons embedded within the olfactory epithelium in the dorsal posterior recess of the nasal cavity. These neurons project axons to the olfactory bulb of the brain, a small ovoid structure that rests on the cribriform plate of the ethmoid bone. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Olfactory bulb Olfactory bulb Olfactory sensory neurons Cribriform plate Olfactory epithelium FIGURE 14–2 Structure of the olfactory epithelium. There are three cell types: olfactory sensory neurons, supporting cells, and bas-al stem cells at the base of the epithelium. Each sensory neuron has a dendrite that projects to the epithelial surface. Numerous cilia protrude into the mucosal layer lining the nasal lumen. A single axon projects from each neuron to the olfactory bulb. Odorants bind to specific odor-ant receptors on the cilia and initiate a cascade of events leading to gen-eration of action potentials in the sensory axon. (Modified from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Axon Basal cells Mucus Supporting cell Olfactory sensory neuron To olfactory bulb Dendrite Cilia FIGURE 14–3 Basic neural circuits in the olfactory bulb. Note that olfactory receptor cells with one type of odorant receptor project to one olfactory glomerulus (OG) and olfactory receptor cells with another type of receptor project to a different olfactory glomeru-lus. CP, cribriform plate; PG, periglomerular cell; M, mitral cell; T, tufted cell; Gr, granule cell. (Modified from Mori K, Nagao H, Yoshihara Y: The olfactory bulb: Coding and processing of odor molecular information. Science 1999;286:711.) Gr M T PG OG To olfactory cortex CP CHAPTER 14 Smell & Taste 221 periglomerular cells, which are inhibitory neurons connect-ing one glomerulus to another, and granule cells, which have no axons and make reciprocal synapses with the lateral den-drites of the mitral and tufted cells (Figure 14–3). At these syn-apses, the mitral or tufted cell excites the granule cell by releasing glutamate, and the granule cell in turn inhibits the mitral or tufted cell by releasing GABA. OLFACTORY CORTEX The axons of the mitral and tufted cells pass posteriorly through the lateral olfactory stria to terminate on apical den-drites of pyramidal cells in five regions of the olfactory cortex: anterior olfactory nucleus, olfactory tubercle, piriform cor-tex, amygdala, and entorhinal cortex (Figure 14–4). From these regions, information travels directly to the frontal cortex or via the thalamus to the orbitofrontal cortex. Conscious dis-crimination of odors is dependent on the pathway to the orb-itofrontal cortex. The orbitofrontal activation is generally greater on the right side than the left; thus, cortical represen-tation of olfaction is asymmetric. The pathway to the amygda-la is probably involved with the emotional responses to olfactory stimuli, and the pathway to the entorhinal cortex is concerned with olfactory memories. OLFACTORY THRESHOLDS & DISCRIMINATION The olfactory epithelium is covered by a thin layer of mucus secreted by the supporting cells and Bowman glands, which lie beneath the epithelium. The mucus bathes the odorant recep-tors on the cilia and provides the appropriate molecular and ionic environment for odor detection. The olfactory thresholds for substances shown in Table 14–1 illustrate the remarkable sensitivity of the odorant receptors. For FIGURE 14–4 Diagram of the olfactory pathway. Information is transmitted from the olfactory bulb by axons of mitral and tufted relay neurons in the lateral olfactory tract. Mitral cells project to five regions of the olfactory cortex: anterior olfactory nucleus, olfactory tubercle, piri-form cortex, and parts of the amygdala and entorhinal cortex. Tufted cells project to anterior olfactory nucleus and olfactory tubercle; mitral cells in the accessory olfactory bulb project only to the amygdala. Conscious discrimination of odor depends on the neocortex (orbitofrontal and frontal cortices). Emotive aspects of olfaction derive from limbic projections (amygdala and hypothalamus). (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Olfactory epithelium Vomeronasal organ Lateral olfactory tract Anterior olfactory nucleus Olfactory tubercle Piriform cortex Amygdala Entorhinal cortex Contralateral olfactory bulb Hypothalamus Hippocampus Thalamus Orbitofrontal cortex Frontal cortex Mitral cell Tufted cell Olfactory bulb Accessory olfactory bulb Mitral cell TABLE 14–1 Some olfactory thresholds. Substance mg/L of Air Ethyl ether 5.83 Chloroform 3.30 Pyridine 0.03 Oil of peppermint 0.02 Lodoform 0.02 Butyric acid 0.009 Propyl mercaptan 0.006 Artificial musk 0.00004 Methyl mercaptan 0.0000004 222 SECTION III Central & Peripheral Neurophysiology example, methyl mercaptan, one of the substances in garlic, can be smelled at a concentration of less than 500 pg/L of air. In addition, olfactory discrimination is remarkable; for example, humans can recognize more than 10,000 different odors. On the other hand, determination of differences in the intensity of any given odor is poor. The concentration of an odor-producing substance must be changed by about 30% before a difference can be detected. The comparable visual discrimination threshold is a 1% change in light intensity. The direction from which a smell comes may be indicated by the slight difference in the time of arrival of odoriferous molecules in the two nostrils. Odor-producing molecules are generally small, containing from 3 to 20 carbon atoms, and molecules with the same num-ber of carbon atoms but different structural configurations have different odors. Relatively high water and lipid solubility are characteristic of substances with strong odors. Some common abnormalities in odor detection are described in Clinical Box 14–1. SIGNAL TRANSDUCTION The olfactory system has received considerable attention in re-cent years because of the intriguing biologic question of how a simple sense organ such as the olfactory epithelium and its brain representation, which apparently lacks a high degree of complexity, can mediate discrimination of more than 10,000 different odors. One part of the answer to this question is that there are many different odorant receptors. The genes that code for about 1000 different types of odor-ant receptors make up the largest gene family so far described in mammals—larger than the immunoglobulin and T-cell receptor gene families combined. The amino acid sequences of odorant receptors are very diverse, but all the odorant receptors are coupled to heterotrimeric G proteins. When an odorant molecule binds to its receptor, the G protein subunits (α, β, γ) dissociate (Figure 14–5). The α-subunit activates adenylate cyclase to catalyze the production of cAMP, which acts as a second messenger to open cation channels, causing an inward-directed Ca2+ current. This produces the graded receptor potential, which then leads to an action potential in the olfactory nerve. A second part of the answer to the question of how 10,000 different odors can be detected lies in the neural organization of the olfactory pathway. Although there are millions of olfactory sensory neurons, each expresses only one of the 1000 different odorant receptors. Each neuron projects to one or two glomer-uli (Figure 14–3). This provides a distinct two-dimensional map in the olfactory bulb that is unique to the odorant. The mitral cells with their glomeruli project to different parts of the olfactory cortex. The olfactory glomeruli demonstrate lateral inhibition medi-ated by periglomerular cells and granule cells. This sharpens and focuses olfactory signals. In addition, the extracellular field potential in each glomerulus oscillates, and the granule cells appear to regulate the frequency of the oscillation. The exact function of the oscillation is unknown, but it probably also helps to focus the olfactory signals reaching the cortex. CLINICAL BOX 14–1 Abnormalities in Odor Detection Anosmia (inability to smell) and hyposmia or hypesthesia (diminished olfactory sensitivity) can result from simple nasal congestion or be a sign of a more serious problem in-cluding damage to the olfactory nerves due to fractures of the cribriform plate, tumors such as neuroblastomas or meningiomas, or infections (such as abscesses). Alzheimer disease can also damage the olfactory nerves. Aging is also associated with abnormalities in smell sensation; more than 75% of humans over the age of 80 have an impaired ability to identify smells. Hyperosmia (enhanced olfactory sensitivity) is less common than loss of smell, but pregnant women commonly become oversensitive to smell. Dysos-mia (distorted sense of smell) can be caused by several dis-orders including sinus infections, partial damage to the ol-factory nerves, and poor dental hygiene. FIGURE 14–5 Signal transduction in an odorant receptor. Olfactory receptors are G protein-coupled receptors that dissociate upon binding to the odorant. The α-subunit of G proteins activates adenylate cyclase to catalyze production of cAMP. cAMP acts as a sec-ond messenger to open cation channels. Inward diffusion of Na+ and Ca2+ produces depolarization. (From Fox SI: Human Physiology. McGraw-Hill, 2008.) cAMP cAMP (a) (b) Odorant Odorant Odorant receptor Odorant receptor G-proteins Adenylate cyclase Adenylate cyclase Na+/Ca2+ channel Na+/Ca2+ channel Ca2+ Na+ ATP CHAPTER 14 Smell & Taste 223 ODORANT-BINDING PROTEINS In contrast to the low threshold for olfactory stimulation when the olfactory epithelium is intact, single olfactory receptors that have been patch-clamped have a relatively high threshold and a long latency. In addition, lipophilic odor-producing molecules must traverse the hydrophilic mucus in the nose to reach the re-ceptors. These facts led to the suggestion that the olfactory mu-cus might contain one or more odorant-binding proteins (OBP) that concentrate the odorants and transfer them to the re-ceptors. An 18-kDa OBP that is unique to the nasal cavity has been isolated, and other related proteins probably exist. The pro-tein has considerable homology to other proteins in the body that are known to be carriers for small lipophilic molecules. A similar binding protein appears to be associated with taste. VOMERONASAL ORGAN In rodents and various other mammals, the nasal cavity con-tains another patch of olfactory epithelium located along the nasal septum in a well-developed vomeronasal organ. This structure is concerned with the perception of odors that act as pheromones. Vomeronasal sensory neurons project to the ac-cessory olfactory bulb and from there primarily to areas in the amygdala and hypothalamus that are concerned with re-production and ingestive behavior. Vomeronasal input has major effects on these functions. An example is pregnancy block in mice; the pheromones of a male from a different strain prevent pregnancy as a result of mating with that male, but mating with a mouse of the same strain does not produce blockade. The vomeronasal organ has about 100 G protein-coupled odorant receptors that differ in structure from those in the rest of the olfactory epithelium. The organ is not well developed in humans, but an anatomi-cally separate and biochemically unique area of olfactory epithe-lium occurs in a pit in the anterior third of the nasal septum, which appears to be a homologous structure. There is evidence for the existence of pheromones in humans, and there is a close relationship between smell and sexual function. Perfume adver-tisements bear witness to this. The sense of smell is said to be more acute in women than in men, and in women it is most acute at the time of ovulation. Smell, and to a lesser extent, taste, have a unique ability to trigger long-term memories, a fact noted by novelists and documented by experimental psychologists. SNIFFING The portion of the nasal cavity containing the olfactory recep-tors is poorly ventilated in humans. Most of the air normally moves smoothly over the turbinates with each respiratory cy-cle, although eddy currents pass some air over the olfactory epithelium. These eddy currents are probably set up by con-vection as cool air strikes the warm mucosal surfaces. The amount of air reaching this region is greatly increased by sniff-ing, an action that includes contraction of the lower part of the nares on the septum, deflecting the airstream upward. Sniffing is a semireflex response that usually occurs when a new odor attracts attention. ROLE OF PAIN FIBERS IN THE NOSE Naked endings of many trigeminal pain fibers are found in the olfactory epithelium. They are stimulated by irritating sub-stances and leads to the characteristic “odor” of such substanc-es as peppermint, menthol, and chlorine. Activation of these endings by nasal irritants also initiates sneezing, lacrimation, respiratory inhibition, and other reflexes. ADAPTATION It is common knowledge that when one is continuously ex-posed to even the most disagreeable odor, perception of the odor decreases and eventually ceases. This sometimes benefi-cent phenomenon is due to the fairly rapid adaptation, or de-sensitization, that occurs in the olfactory system. It is mediated by Ca2+ acting via calmodulin on cyclic nucleotide-gated (CNG) ion channels. When the CNG A4 subunit is knocked out, adaptation is slowed. TASTE TASTE BUDS The specialized sense organ for taste (gustation) consists of ap-proximately 10,000 taste buds, which are ovoid bodies measur-ing 50–70 μm. There are four morphologically distinct types of cells within each taste bud: basal cells, dark cells, light cells, and intermediate cells (Figure 14–6). The latter three cell types are all referred to as Type I, II, and III taste cells. They are the sen-sory neurons that respond to taste stimuli or tastants. The three cell types may represent various stages of differentiation of de-veloping taste cells, with the light cells being the most mature. Alternatively, each cell type might represent different cell lin-eages. The apical ends of taste cells have microvilli that project into the taste pore, a small opening on the dorsal surface of the tongue where tastes cells are exposed to the oral contents. Each taste bud is innervated by about 50 nerve fibers, and conversely, each nerve fiber receives input from an average of five taste buds. The basal cells arise from the epithelial cells surrounding the taste bud. They differentiate into new taste cells, and the old cells are continuously replaced with a half-time of about 10 days. If the sensory nerve is cut, the taste buds it innervates de-generate and eventually disappear. In humans, the taste buds are located in the mucosa of the epiglottis, palate, and pharynx and in the walls of papillae of the tongue (Figure 14–6). The fungiform papillae are rounded structures most numerous near the tip of the tongue; the circumvallate papillae are prominent structures arranged in a V on the back of the tongue; the foliate papillae are on 224 SECTION III Central & Peripheral Neurophysiology the posterior edge of the tongue. Each fungiform papilla has up to five taste buds, mostly located at the top of the papilla, while each vallate and foliate papilla contain up to 100 taste buds, mostly located along the sides of the papillae. TASTE PATHWAYS The sensory nerve fibers from the taste buds on the anterior two-thirds of the tongue travel in the chorda tympani branch of the facial nerve, and those from the posterior third of the tongue reach the brain stem via the glossopharyngeal nerve (Figure 14–7). The fibers from areas other than the tongue (eg, pharynx) reach the brain stem via the vagus nerve. On each side, the myelinated but relatively slowly conducting taste fibers in these three nerves unite in the gustatory portion of the nucleus of the solitary tract (NTS) in the medulla oblongata (Figure 14–7). From there, axons of second-order neurons ascend in the ipsilateral medial lemniscus and, in primates, pass directly to the ventral posteromedial nucleus of the thalamus. From the thalamus, the axons of the third-order neurons pass to neurons in the anterior insula and the frontal operculum in the ipsilater-al cerebral cortex. This region is rostral to the face area of the postcentral gyrus, which is probably the area that mediates con-scious perception of taste and taste discrimination. BASIC TASTE MODALITIES Humans have five established basic tastes: sweet, sour, bitter, salt, and umami. It used to be thought that the surface of the tongue had special areas for each of the first four of these sen-sations, but it is now clear that all tastants are sensed from all parts of the tongue and adjacent structures. Afferent nerves to the NTS contain fibers from all types of taste receptors, with-out any clear localization of types. The fifth taste sense, umami, was recently added to the four classic tastes. This taste has actually been known for almost 100 years, and it became established once its receptor was identified. It is triggered by glutamate and particularly by the monosodium glutamate (MSG) used so extensively in Asian cooking. The taste is pleasant and sweet but differs from the standard sweet taste. TASTE RECEPTORS & TRANSDUCTION The putative receptors for taste are shown diagrammatically in Figure 14–8. The salty taste is triggered by NaCl. Salt-sensi-tive taste is mediated by a Na+-selective channel known as ENaC, the amiloride-sensitive epithelial sodium channel. The entry of Na+ into the salt receptors depolarizes the membrane, generating the receptor potential. In humans, the amiloride sensitivity of salt taste is less pronounced than in some species, FIGURE 14–6 Taste buds located in papillae of the human tongue. A) Taste buds on the anterior two-thirds of the tongue are innervated by the chorda tympani branch of the facial nerve; those on the posterior one-third of the tongue are innervated by the lingual branch of the glos-sopharyngeal nerve. B) The three major types of papillae (circumvallate, foliate, and fungiform) are located on specific parts of the tongue. C) Taste buds are composed of basal stem cells and three types of taste cells (dark, light, and intermediate). Taste cells extend from the base of the taste bud to the taste pore, where microvilli contact tastants dissolved in saliva and mucus. (Modified from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Chorda tympani nerve (VII) Circumvallate Foliate Taste bud Fungiform Serous gland Glossopharyngeal nerve (IX) A B Taste pore Saliva Epithelial cell Taste cell Basal cell To sensory ganglion Gustatory afferent nerve C CHAPTER 14 Smell & Taste 225 FIGURE 14–7 Diagram of taste pathways. Signals from the taste buds travel via different nerves to gustatory areas of the nucleus of the solitary tract which relays information to the thalamus; the thalamus projects to the gustatory cortex. (Modified from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) FIGURE 14–8 Signal transduction in taste receptors. Salt-sensitive taste is mediated by a Na+-selective channel (ENaC); sour taste is me-diated by H+ ions permeable to ENaCs; umami taste is mediated by glutamate acting on a metabotropic receptor, mGluR4; bitter taste is mediated by the T2R family of G protein-coupled receptors; sweet taste may be dependent on the T1R3 family of G protein-coupled receptors which couple to the G protein gustducin. (Modified from Lindemann B: Receptors and transduction in taste. Nature 2001;413:219.) Tongue Pharynx Nucleus of solitary tract Gustatory area Nodose ganglion Petrosal ganglion N. X N. IX N. VII Geniculate ganglion Ventral posterior medial nucleus of thalamus Gustatory cortex (anterior insula-frontal operculum) Chorda tympani Glossopharyngeal N C C C N N C C N N C C X N I O-γ α Salty ENaC, others ENaC, HCN, others Taste mGluR4 T2R family, others T1R3 (sac locus) Sour Umami (L-glutamate) Bitter Predicted sweet receptor 226 SECTION III Central & Peripheral Neurophysiology suggesting that there are additional mechanisms to activate salt-sensitive receptors. The sour taste is triggered by protons (H+ ions). ENaCs permit the entry of protons and may contribute to the sensa-tion of sour taste. The H+ ions can also bind to and block a K+-sensitive channel. The fall in K+ permeability can depolar-ize the membrane. Also, HCN, a hyperpolarization-activated cyclic nucleotide-gated cation channel, and other mecha-nisms may contribute to sour transduction. Umami taste is due to activation of a truncated metabotro-pic glutamate receptor, mGluR4, in the taste buds. The way activation of the receptor produces depolarization is unset-tled. Glutamate in food may also activate ionotropic glutamate receptors to depolarize umami receptors. Bitter taste is produced by a variety of unrelated compounds. Many of these are poisons, and bitter taste serves as a warning to avoid them. Some bitter compounds bind to and block K+-selective channels. Many G protein-linked receptors in the human genome are taste receptors (T2R family) and are stimu-lated by bitter substances such as strychnine. In some cases, these receptors couple to the heterotrimeric G protein, gustdu-cin. Gustducin lowers cAMP and increases the formation of inositol phosphates which could lead to depolarization. Some bitter compounds are membrane permeable and may not involve G proteins; quinine is an example. Substances that taste sweet also act via the G protein gustdu-cin. The T1R3 family of G protein-coupled receptors is expressed by about 20% of taste cells, some of which also express gustducin. Sugars taste sweet, but so do compounds such as saccharin that have an entirely different structure. It appears at present that natural sugars such as sucrose and syn-thetic sweeteners act via different receptors on gustducin. Like the bitter-responsive receptors, sweet-responsive receptors act via cyclic nucleotides and inositol phosphate metabolism. TASTE THRESHOLDS & INTENSITY DISCRIMINATIONS The ability of humans to discriminate differences in the intensity of tastes, like intensity discrimination in olfaction, is relatively crude. A 30% change in the concentration of the substance being tasted is necessary before an intensity difference can be detected. The threshold concentrations of substances to which the taste buds respond vary with the particular substance (Table 14–2). A protein that binds taste-producing molecules has been cloned. It is produced by Ebner gland that secretes mucus into the cleft around vallate papillae (Figure 14–6) and proba-bly has a concentrating and transport function similar to that of the OBP described for olfaction. Some common abnormal-ities in taste detection are described in Clinical Box 14–2. VARIATION & AFTER EFFECTS Taste exhibits after reactions and contrast phenomena that are similar in some ways to visual after images and contrasts. Some of these are chemical “tricks,” but others may be true central phenomena. A taste modifier protein, miraculin, has been discovered in a plant. When applied to the tongue, this protein makes acids taste sweet. Animals, including humans, form particularly strong aver-sions to novel foods if eating the food is followed by illness. The survival value of such aversions is apparent in terms of avoiding poisons. CHAPTER SUMMARY ■Olfactory sensory neurons, supporting (sustentacular) cells, and basal stem cells are located in the olfactory epithelium within the upper portion of the nasal cavity. ■The cilia located on the dendritic knob of the olfactory sensory neuron contain odorant receptors which are coupled to heterot-rimeric G proteins. ■Axons of olfactory sensory neurons contact the dendrites of mitral and tufted cells in the olfactory bulbs to form olfactory glomeruli. TABLE 14–2 Some taste thresholds. Substance Taste Threshold Concentration (μmol/L) Hydrochloric acid Sour 100 Sodium chloride Salt 2000 Strychnine hydrochloride Bitter 1.6 Glucose Sweet 80,000 Sucrose Sweet 10,000 Saccharin Sweet 23 CLINICAL BOX 14–2 Abnormalities in Taste Detection Ageusia (absence of the sense of taste) and hypogeusia (di-minished taste sensitivity) can be caused by damage to the lingual or glossopharyngeal nerve. Neurological disorders such as vestibular schwannoma, Bell palsy, familial dysauto-nomia, multiple sclerosis, and certain infections (eg, primary amoeboid meningoencephalopathy) can also cause prob-lems with taste sensitivity. Ageusia can also be an adverse side effect of various drugs including cisplatin and captopril or vitamin B3 or zinc deficiencies. Aging and tobacco abuse also contribute to diminished taste. Dysgeusia or parageu-sia (unpleasant perception of taste) causes a metallic, salty, foul, or rancid taste. In many cases, dysgeusia is a temporary problem. Factors contributing to ageusia or hypogeusia can also lead to abnormal taste sensitivity. CHAPTER 14 Smell & Taste 227 ■Information from the olfactory bulb travels via the lateral olfac-tory stria directly to the olfactory cortex, including the anterior olfactory nucleus, olfactory tubercle, piriform cortex, amygdala, and entorhinal cortex. ■Taste buds are the specialized sense organs for taste and are comprised of basal stem cells and three types of taste cells (dark cells, light cells, and intermediate cells). The three types of taste cells may represent various stages of differentiation of develop-ing taste cells, with the light cells being the most mature. Taste buds are located in the mucosa of the epiglottis, palate, and pharynx and in the walls of papillae of the tongue. ■There are taste receptors for sweet, sour, bitter, salt, and umami. Signal transduction mechanisms include passage through ion channels, binding to and blocking ion channels, and second messenger systems. ■The afferents from taste buds in the tongue travel via the sev-enth, ninth, and tenth cranial nerves to synapse in the nucleus of the tractus solitarius. From there, axons ascend via the ipsilater-al medial lemniscus to the ventral posteromedial nucleus of the thalamus, and on to the anterior insula and frontal operculum in the ipsilateral cerebral cortex. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Odorant receptors are A) located in the olfactory bulb. B) located on dendrites of mitral and tufted cells. C) located on neurons that project directly to the olfactory cortex. D) located on neurons in the olfactory epithelium that project to mitral cells and from there directly to the olfactory cortex. E) located on sustentacular cells that project to the olfactory bulb. 2. Taste receptors A) for sweet, sour, bitter, salt, and umami are spatially separated on the surface of the tongue. B) are synonymous with taste buds. C) are a type of chemoreceptor. D) are innervated by afferents in the facial, trigeminal, and glossopharyngeal nerves. E) all of the above 3. Which of the following does not increase the ability to discrimi-nate many different odors? A) many different receptors B) pattern of olfactory receptors activated by a given odorant C) projection of different mitral cell axons to different parts of the brain D) high β-arrestin content in olfactory neurons E) sniffing 4. As a result of an automobile accident, a 10-year-old boy suffered damage to the brain including the periamygdaloid, piriform, and entorhinal cortices. Which of the following sensory deficits is he most likely to experience? A) visual disturbance B) hyperosmia C) auditory problems D) taste and odor abnormalities E) no major sensory deficits 5. Which of the following are incorrectly paired? A) ENaC : sour B) α-gustducin : bitter taste C) nucleus tractus solitarius : blood pressure D) Heschel sulcus : smell E) Ebner glands : taste acuity 6. Which of the following is true about olfactory transmission? A) An olfactory sensory neuron expresses a wide range of odor-ant receptors. B) Lateral inhibition within the olfactory glomeruli reduces the ability to distinguish between different types of odorant receptors. C) Conscious discrimination of odors is dependent on the pathway to the orbitofrontal cortex. D) Olfaction is closely related to gustation because odorant and gustatory receptors use the same central pathways. E) all of the above 7. Which of the following is not true about gustatory sensation? A) The sensory nerve fibers from the taste buds on the anterior two-thirds of the tongue travel in the chorda tympani branch of the facial nerve. B) The sensory nerve fibers from the taste buds on the poster-ior third of the tongue travel in the petrosal branch of the glossopharyngeal nerve. C) The pathway from taste buds on the left side of the tongue is transmitted ipsilaterally to the cerebral cortex. D) Sustentacular cells in the taste buds serve as stem cells to permit growth of new taste buds. E) The pathway from taste receptors includes synapses in the nucleus of the tractus solitarius in the brain stem and ventral posterior medial nucleus in the thalamus. 8. A 20-year-old woman was diagnosed with Bell palsy (damage to facial nerve). Which of the following symptoms is she likely to exhibit? A) loss of sense of taste B) facial twitching C) droopy eyelid D) ipsilateral facial paralysis E) all of the above CHAPTER RESOURCES Adler E, et al: A novel family of mammalian taste receptors. Cell 2000;100:693. Anholt RRH: Odor recognition and olfactory transduction: The new frontier. Chem Senses 1991;16:421. Boron WF, Boulpaep EL: Medical Physiology. Elsevier, 2005. Gilbertson TA, Damak S, Margolskee RF: The molecular physiology of taste transduction. Curr Opin Neurobiol 2000;10:519. Gold GH: Controversial issues in vertebrate olfactory transduction. Annu Rev Physiol 1999;61:857. Herness HM, Gilbertson TA: Cellular mechanisms of taste transduction. Annu Rev Physiol 1999;61:873. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. Lindemann B: Receptors and transduction in taste. Nature 2001;413:219. Mombaerts P: Genes and ligands for odorant, vomeronasal and taste receptors. Nature Rev Neurosci 2004;5:263. 228 SECTION III Central & Peripheral Neurophysiology Ronnett GV, Moon C: G proteins and olfactory signal transduction. Annu Rev Physiol 2002;64:189. Shepherd GM, Singer MS, Greer CA: Olfactory receptors: A large gene family with broad affinities and multiple functions (Review). Neuroscientist 1996;2:262. Stern P, Marks J (editors): Making sense of scents. Science 1999;286:703. 229 C H A P T E R 15 Electrical Activity of the Brain, Sleep–Wake States, & Circadian Rhythms O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the primary types of rhythms that make up the electroencephalogram (EEG). ■List the main clinical uses of the EEG. ■Summarize the behavioral and EEG characteristics of each of the stages of nonrap-id eye movement (NREM) and rapid eye movement (REM) sleep and the mecha-nisms responsible for their production. ■Describe the pattern of normal nighttime sleep in adults and the variations in this pattern from birth to old age. ■Discuss the circadian rhythm and the role of the suprachiasmatic nuclei (SCN) in its regulation. ■Describe the diurnal regulation of synthesis of melatonin from serotonin in the pi-neal gland and its secretion into the bloodstream. INTRODUCTION Most of the various sensory pathways described in Chapters 11–14 relay impulses from sense organs via three- and four-neuron chains to particular loci in the cerebral cortex. The impulses are responsible for perception and localization of individual sensations. However, they must be processed in the awake brain to be perceived. At least in mammals, there is a spectrum of behavioral states ranging from deep sleep through light sleep, REM sleep, and the two awake states: relaxed awareness and awareness with concentrated attention. Discrete patterns of brain electrical activity correlate with each of these states. Feedback oscillations within the cerebral cortex and between the thalamus and the cortex serve as producers of this activity and possible determinants of the behavioral state. Arousal can be produced by sensory stimulation and by impulses ascending in the reticular core of the midbrain. Many of these activities have rhythmic fluctuations that are approxi-mately 24 h in length; that is, they are circadian. THALAMUS, CEREBRAL CORTEX, & RETICULAR FORMATION THALAMIC NUCLEI The thalamus is a large collection of neuronal groups within the diencephalons; it participates in sensory, motor, and lim-bic functions. Virtually all information that reaches the cortex is processed by the thalamus, leading to its being called the “gateway” to the cerebral cortex. The thalamus can be divided into nuclei that project dif-fusely to wide regions of the neocortex and nuclei that project to specific discrete portions of the neocortex and limbic sys-tem. The nuclei that project to wide regions of the neocortex are the midline and intralaminar nuclei. The nuclei that 230 SECTION III Central & Peripheral Neurophysiology project to specific areas include the specific sensory relay nuclei and the nuclei concerned with efferent control mecha-nisms. The specific sensory relay nuclei include the medial and lateral geniculate bodies, which relay auditory and visual impulses to the auditory and visual cortices; and the ventral posterior lateral (VPL) and ventral posteromedial, which relay somatosensory information to the postcentral gyrus. The ven-tral anterior and ventral lateral nuclei are concerned with motor function. They receive input from the basal ganglia and the cerebellum and project to the motor cortex. The anterior nuclei receive afferents from the mamillary bodies and project to the limbic cortex, which may be involved in memory and emotion. Most of the thalamic nuclei described are excitatory neurons that release glutamate. The thalamus also contains inhibitory neurons in the thalamic reticular nucleus. These neurons release GABA, and unlike the other thalamic neurons just described, their axons do not project to the cortex. Rather, they are thalamic interneurons and modulate the responses of other thalamic neurons to input coming from the cortex. CORTICAL ORGANIZATION The neocortex is generally arranged in six layers (Figure 15–1). The most common neuronal type is the pyramidal cell with an extensive vertical dendritic tree (Figures 15–1 and 15–2) that may reach to the cortical surface. Their cell bodies can be found in all cortical layers except layer I. The axons of these cells usu-ally give off recurrent collaterals that turn back and synapse on the superficial portions of the dendritic trees. Afferents from the specific nuclei of the thalamus terminate primarily in cor-tical layer IV, whereas the nonspecific afferents are distributed to layers I–IV. Pyramidal neurons are the only projection neu-rons of the cortex, and they are excitatory neurons that release glutamate at their terminals. The other cortical cell types are lo-cal circuit neurons (interneurons) which have been classified based on their shape, pattern of projection, and neurotransmit-ter. Inhibitory interneurons (basket cells and chandelier cells) release GABA as their neurotransmitter. Basket cells have long axonal endings that surround the soma of pyramidal neurons; FIGURE 15–1 Structure of the cerebral cortex. The cortical layers are indicated by the numbers. Golgi stain shows neuronal cell bodies and dendrites, Nissl stain shows cell bodies, and Weigert myelin sheath stain shows myelinated nerve fibers. (Modified from Ranson SW, Clark SL: The Anatomy of the Nervous System, 10th ed. Saunders, 1959.) Multiform layer White matter Molecular layer Internal pyramidal cell layer Internal granule cell layer External granule cell layer External pyramidal cell layer VI I V IV II III Golgi stain Pial surface Nissl stain Weigert stain CHAPTER 15 Electrical Activity of the Brain, Sleep–Wake States, & Circadian Rhythms 231 they account for most inhibitory synapses on the pyramidal soma and dendrites. Chandelier cells are a powerful source of inhibition of pyramidal neurons because they have axonal end-ings that terminate exclusively on the initial segment of the py-ramidal cell axon. Their terminal boutons form short vertical rows that resemble candlesticks, thus accounting for their name. Spiny stellate cells are excitatory interneurons that re-lease glutamate as a neurotransmitter. These cells are located primarily in layer IV and are a major recipient of sensory infor-mation arising from the thalamus; they are an example of a multipolar neuron (Chapter 4) with local dendritic and axonal arborizations. In addition to being organized into layers, the cortex is also organized into columns. Neurons within a column have simi-lar response properties, suggesting they comprise a local pro-cessing network (eg, orientation and ocular dominance columns in the visual cortex). RETICULAR FORMATION & RETICULAR ACTIVATING SYSTEM The reticular formation, the phylogenetically old reticular core of the brain, occupies the midventral portion of the me-dulla and midbrain. It is primarily an anatomic area made up of various neural clusters and fibers with discrete functions. For example, it contains the cell bodies and fibers of many of the serotonergic, noradrenergic, adrenergic, and cholinergic systems. It also contains many of the areas concerned with regulation of heart rate, blood pressure, and respiration. Some of the descending fibers in it inhibit transmission in sensory and motor pathways in the spinal cord; various reticular areas and the pathways from them are concerned with spasticity and adjustment of stretch reflexes. The reticular activating sys-tem (RAS) and related components of the brain concerned with consciousness and sleep are considered in this chapter. The RAS is a complex polysynaptic pathway arising from the brain stem reticular formation with projections to the intralaminar and reticular nuclei of the thalamus which, in turn, project diffusely and nonspecifically to wide regions of the cortex (Figure 15–3). Collaterals funnel into it not only from the long ascending sensory tracts but also from the tri-geminal, auditory, visual, and olfactory systems. The com-plexity of the neuron net and the degree of convergence in it abolish modality specificity, and most reticular neurons are activated with equal facility by different sensory stimuli. The system is therefore nonspecific, whereas the classic sensory pathways are specific in that the fibers in them are activated by only one type of sensory stimulation. EVOKED CORTICAL POTENTIALS The electrical events that occur in the cortex after stimulation of a sense organ can be monitored with an exploring electrode connected to another electrode at an indifferent point some distance away. A characteristic response is seen in animals un-der barbiturate anesthesia, which eliminates much of the background electrical activity. If the exploring electrode is over the primary receiving area for a particular sense, a sur-face-positive wave appears with a latency of 5 to 12 ms. This is followed by a small negative wave, and then a larger, more prolonged positive deflection frequently occurs with a latency of 20 to 80 ms. The first positive–negative wave sequence is the FIGURE 15–2 Neocortical pyramidal cell, showing the distribution of neurons that terminate on it. A denotes nonspecific afferents from the reticular formation and the thalamus; B denotes re-current collaterals of pyramidal cell axons; C denotes commissural fibers from mirror image sites in the contralateral hemisphere; D denotes spe-cific afferents from thalamic sensory relay nuclei. (Modified from Chow KL, Leiman AL: The structural and functional organization of the neocortex. Neurosci Res Program Bull 1970;8:157.) A B B C D Axon FIGURE 15–3 Diagram showing the ascending reticular system in the human midbrain, its projections to the intralaminar nuclei of the thalamus, and the output from the intralaminar nuclei to many parts of the cerebral cortex. Activation of these ar-eas is shown by PET scans when subjects shift from a relaxed awake state to an attention-demanding task. Intralaminar nuclei of thalamus Midbrain reticular formation Cortex 232 SECTION III Central & Peripheral Neurophysiology primary evoked potential; the second is the diffuse secon-dary response. The primary evoked potential is highly specific in its loca-tion and can be observed only where the pathways from a par-ticular sense organ end. An electrode on the pial surface of the cortex samples activity to a depth of only 0.3–0.6 mm. The primary response is negative rather than positive when it is recorded with a microelectrode inserted in layers II–VI of the underlying cortex, and the negative wave within the cortex is followed by a positive wave. The negative–positive sequence indicates depolarization on the dendrites and somas of the cells in the cortex, followed by hyperpolarization. The posi-tive–negative wave sequence recorded from the surface of the cortex occurs because the superficial cortical layers are posi-tive relative to the initial negativity, then negative relative to the deep hyperpolarization. In unanesthetized animals or humans, the primary evoked potential is largely obscured by the spontaneous activity of the brain, but it can be demon-strated by superimposing multiple traces so that the back-ground activity is averaged out. It is somewhat more diffuse in unanesthetized animals but still well localized compared with the diffuse secondary response. The surface-positive diffuse secondary response, unlike the primary, is not highly localized. It appears at the same time over most of the cortex and is due to activity in projections from the midline and related thalamic nuclei. PHYSIOLOGIC BASIS OF THE ELECTROENCEPHALOGRAM The background electrical activity of the brain in unanesthetized animals was first described in the 19th century. Subsequently, it was analyzed in systematic fashion by the German psychiatrist Hans Berger, who introduced the term electroencephalogram (EEG) to denote the record of the variations in brain potential. The EEG can be recorded with scalp electrodes through the un-opened skull or with electrodes on or in the brain. The term electrocorticogram (ECoG) is used for the record obtained with electrodes on the pial surface of the cortex. EEG records may be bipolar or unipolar. Bipolar records show fluctuations in the potential difference between two corti-cal electrodes; unipolar records show the potential difference between a cortical electrode and a theoretically indifferent elec-trode on some part of the body distant from the cortex. CORTICAL DIPOLES The EEG recorded from the scalp is a measure of the summa-tion of dendritic postsynaptic potentials rather than action po-tentials (Figure 15–4). The dendrites of the cortical cells are a forest of similarly oriented, densely packed units in the superfi-cial layers of the cerebral cortex (Figure 15–1). Propagated po-tentials can be generated in dendrites. In addition, recurrent axon collaterals end on dendrites in the superficial layers. As ex-citatory and inhibitory endings on the dendrites of each cell be-come active, current flows into and out of these current sinks and sources from the rest of the dendritic processes and the cell body. The cell–dendrite relationship is therefore that of a con-stantly shifting dipole. Current flow in this dipole produces wave-like potential fluctuations in a volume conductor (Figure 15–4). When the sum of the dendritic activity is negative rela-tive to the cell, the cell is depolarized and hyperexcitable; when it is positive, the cell is hyperpolarized and less excitable. The cerebellar cortex and the hippocampus are two other parts of the central nervous system (CNS) where many complex, paral-lel dendritic processes are located subpially over a layer of cells. In both areas, characteristic rhythmic fluctuations occur in sur-face potential similar to that observed in the cortical EEG. CLINICAL USES OF THE EEG The EEG is sometimes of value in localizing pathologic pro-cesses. When a collection of fluid overlies a portion of the cor-tex, activity over this area may be damped. This fact may aid in diagnosing and localizing conditions such as subdural he-matomas. Lesions in the cerebral cortex cause local formation of irregular or slow waves that can be picked up in the EEG leads. Epileptogenic foci sometimes generate high-voltage waves that can be localized. Epilepsy is a syndrome with multiple causes. In some forms, characteristic EEG patterns occur during seizures; between attacks, however, abnormalities are often difficult to demon-strate. Seizures are divided into partial (focal) seizures and generalized seizures. FIGURE 15–4 Diagrammatic comparison of the electrical responses of the axon and the dendrites of a large cortical neuron. Current flow to and from active synaptic knobs on the den-drites produces wave activity, while all-or-none action potentials are transmitted along the axon. 100 μV 200 mV Wave activity Dendritic tree Axon Action potentials CHAPTER 15 Electrical Activity of the Brain, Sleep–Wake States, & Circadian Rhythms 233 Partial seizures originate in a small group of neurons and can result from head injury, brain infection, stroke, or tumor, but often the cause is unknown. Symptoms depend on the sei-zure focus. They are further subdivided into simple partial seizures (without loss of consciousness) and complex partial seizures (with altered consciousness). An example of a partial seizure is localized jerking movements in one hand progress-ing to clonic movements of the entire arm. Auras typically precede the onset of a partial seizure and include abnormal sensations. The time after the seizure until normal neurologi-cal function returns is called the postictal period. Generalized seizures are associated with widespread elec-trical activity and involve both hemispheres simultaneously. They are further subdivided into convulsive and nonconvul-sive categories depending on whether tonic or clonic move-ments occur. Absence seizures (formerly called petit mal seizures) are one of the forms of nonconvulsive generalized seizures characterized by a momentary loss of consciousness. They are associated with 3/s doublets, each consisting of a typical spike and rounded wave, and lasting about 10 s (Fig-ure 15–5). They are not accompanied by auras or postictal periods. The most common convulsive generalized seizure is tonic– clonic seizure (formerly called grand mal seizure). This is associated with sudden onset of contraction of limb muscles (tonic phase) lasting about 30 s, followed by a clonic phase with symmetric jerking of the limbs as a result of alternating contraction and relaxation (clonic phase) lasting 1–2 min. There is fast EEG activity during the tonic phase. Slow waves, each preceded by a spike, occur at the time of each clonic jerk. For a while after the attack, slow waves are present. Recent research provides insight into a possible role of release of glutamate from astrocytes in the pathophysiology of epilepsy. Also, there is evidence to support the view that reor-ganization of astrocytes along with dendritic sprouting and new synapse formation form the structural basis for recurrent excitation in the epileptic brain. Clinical Box 15–1 describes information regarding the role of genetic mutations in some forms of epilepsy. SLEEP–WAKE CYCLE ALPHA, BETA, & GAMMA RHYTHMS In adult humans who are awake but at rest with the mind wan-dering and the eyes closed, the most prominent component of the EEG is a fairly regular pattern of waves at a frequency of 8– 13 Hz and amplitude of 50–100 μV when recorded from the scalp. This pattern is the alpha rhythm (Figure 15–6). It is most marked in the parietal and occipital lobes and is associated with decreased levels of attention. A similar rhythm has been ob-served in a wide variety of mammalian species. There are some minor variations from species to species, but in all mammals the pattern is remarkably similar (see Clinical Box 15–2). When attention is focused on something, the alpha rhythm is replaced by an irregular 13–30 Hz low-voltage activity, the beta rhythm (Figure 15–6). This phenomenon is called alpha block and can be produced by any form of sensory stimula-tion or mental concentration, such as solving arithmetic prob-lems. Another term for this phenomenon is the arousal or alerting response, because it is correlated with the aroused, alert state. It has also been called desynchronization, because it represents breaking up of the obviously synchronized neural FIGURE 15–5 Absence seizures. Record of four cortical EEG leads from a 6-year-old boy who, during the recording, had one of his “blank spells” in which he was transiently unaware of his surroundings and blinked his eyelids. Time is indicated by the horizontal calibration line. (Reproduced with permission from Waxman SG: Neuroanatomy with Clinical Correlations, 25th ed. McGraw-Hill, 2003.) 1 s CLINICAL BOX 15–1 Genetic Mutations & Epilepsy Epilepsy has no geographical, racial, gender, or social bias. It can occur at any age, but is most often diagnosed in in-fancy, childhood, adolescence, and old age. According to the World Health Organization, it is estimated that 50 mil-lion people worldwide (8.2 per 1000 individuals) experi-ence epileptic seizures. The prevalence in developing countries (such as Colombia, Ecuador, India, Liberia, Nige-ria, Panama, United Republic of Tanzania, and Venezuela) is more than 10 per 1000. Many affected individuals experi-ence unprovoked seizures, for no apparent reason, and without any other neurological abnormalities. These are called idiopathic epilepsies and are assumed to be ge-netic in origin. Mutations in voltage-gated potassium, so-dium, and chloride channels have been linked to some forms of idiopathic epilepsy. Mutated ion channels can lead to neuronal hyperexcitability via various pathogenic mech-anisms. Scientists have recently identified the mutated gene responsible for development of childhood absence epilepsy (CAE). Several patients with CAE were found to have mutations in a subunit gene of the GABA receptor called GABRB3. Also, SCN1A and SCN1B mutations have been identified in an inherited form of epilepsy called gen-eralized epilepsy with febrile seizures. SCN1A and SCN1B are sodium channel subunit genes that are widely expressed within the nervous system. SCN1A mutations are suspected in several forms of epilepsy. 234 SECTION III Central & Peripheral Neurophysiology activity necessary to produce regular waves. However, the rapid EEG activity seen in the alert state is also synchronized, but at a higher rate. Therefore, the term desynchronization is misleading. Gamma oscillations at 30–80 Hz are often seen when an individual is aroused and focuses attention on some-thing. This is often replaced by irregular fast activity as the individual initiates motor activity in response to the stimulus. SLEEP STAGES There are two kinds of sleep: rapid eye movement (REM) sleep and non-REM (NREM), or slow-wave sleep. REM sleep is so named because of the characteristic eye movements that occur during this stage of sleep. NREM sleep is divided into four stag-es (Figure 15–7). A person falling asleep first enters stage 1, the EEG begins to show a low-voltage, mixed frequency pattern. A theta rhythm (4–7 Hz) can be seen at this early stage of slow-wave sleep. Throughout NREM sleep, there is some activity of skeletal muscle but no eye movements occur. Stage 2 is marked by the appearance of sinusoidal waves called sleep spindles (12–14 Hz) and occasional high voltage biphasic waves called K complexes. In stage 3, a high-amplitude delta rhythm (0.5–4 Hz) dominates the EEG waves. Maximum slowing with large waves is seen in stage 4. Thus, the characteristic of deep sleep is a pattern of rhythmic slow waves, indicating marked synchro-nization; it is sometimes referred to as slow-wave sleep. Whereas theta and delta rhythms are normal during sleep, their appearance during wakefulness is a sign of brain dysfunction. REM SLEEP The high-amplitude slow waves seen in the EEG during sleep are periodically replaced by rapid, low-voltage EEG activity, which resembles that seen in the awake, aroused state and in stage 1 sleep (Figure 15–7). For this reason, REM sleep is also called paradoxical sleep. However, sleep is not interrupted; indeed, the threshold for arousal by sensory stimuli and by FIGURE 15–6 EEG records showing the alpha and beta rhythms. When attention is focused on something, the 8–13 Hz alpha rhythm is replaced by an irregular 13–30 Hz low-voltage activity, the beta rhythm. (From Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology, 11th ed. McGraw-Hill, 2008.) (a) (b) Alpha rhythm (relaxed with eyes closed) Beta rhythm (alert) Time CLINICAL BOX 15–2 Variations in the Alpha Rhythm In humans, the frequency of the dominant EEG rhythm at rest varies with age. In infants, there is fast, beta-like activ-ity, but the occipital rhythm is a slow 0.5- to 2-Hz pattern. During childhood this latter rhythm speeds up, and the adult alpha pattern gradually appears during adolescence. The frequency of the alpha rhythm is decreased by low blood glucose levels, low body temperature, low levels of adrenal glucocorticoid hormones, and high arterial partial pressure of CO2 (PaCO2). It is increased by the reverse con-ditions. Forced over-breathing to lower the PaCO2 is some-times used clinically to bring out latent EEG abnormalities. FIGURE 15–7 EEG and muscle activity during various stages of the sleep–wake cycle. NREM sleep has four stages. Stage 1 is character-ized by a slight slowing of the EEG. Stage 2 has high-amplitude K complexes and spindles. Stages 3 and 4 have slow, high-amplitude delta waves. REM sleep is characterized by eye movements, loss of muscle tone, and a low-amplitude, high-frequency activity pattern. The higher voltage ac-tivity in the EOG tracings during stages 2 and 3 reflect high amplitude EEG activity in the prefrontal areas rather than eye movements. EOG, electro-oculogram registering eye movements; EMG, electromyogram registering skeletal muscle activity. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) EOG Awake Sleep stage 1 2 3 4 REM EMG EEG 1 s 50 μV CHAPTER 15 Electrical Activity of the Brain, Sleep–Wake States, & Circadian Rhythms 235 stimulation of the reticular formation is elevated. Rapid, roving movements of the eyes occur during paradoxical sleep, and it is for this reason that it is also called REM sleep. Another charac-teristic of REM sleep is the occurrence of large phasic potentials that originate in the cholinergic neurons in the pons and pass rapidly to the lateral geniculate body and from there to the oc-cipital cortex. They are called pontogeniculo-occipital (PGO) spikes. The tone of the skeletal muscles in the neck is markedly reduced during REM sleep. Humans aroused at a time when they show the EEG charac-teristics of REM sleep generally report that they were dream-ing, whereas individuals awakened from slow-wave sleep do not. This observation and other evidence indicate that REM sleep and dreaming are closely associated. Positron emission tomography (PET) scans of humans in REM sleep show increased activity in the pontine area, amygdala, and anterior cingulate gyrus, but decreased activity in the prefrontal and parietal cortex. Activity in visual associ-ation areas is increased, but there is a decrease in the primary visual cortex. This is consistent with increased emotion and operation of a closed neural system cut off from the areas that relate brain activity to the external world. DISTRIBUTION OF SLEEP STAGES In a typical night of sleep, a young adult first enters NREM sleep, passes through stages 1 and 2, and spends 70–100 min-utes in stages 3 and 4. Sleep then lightens, and a REM period follows. This cycle is repeated at intervals of about 90 minutes throughout the night (Figure 15–8). The cycles are similar, though there is less stage 3 and 4 sleep and more REM sleep to-ward morning. Thus, four to six REM periods occur per night. REM sleep occupies 80% of total sleep time in premature in-fants (Figure 15–9) and 50% in full-term neonates. Thereafter, the proportion of REM sleep falls rapidly and plateaus at about 25% until it falls further in old age. Children have more total sleep time and stage 4 sleep than adults. THALAMOCORTICAL LOOP A circuit linking the cortex and thalamus is thought to be im-portant in generating patterns of brain activity in sleep–wake states. Figure 15–10 shows properties of activity in such a FIGURE 15–8 Normal sleep cycles at various ages. REM sleep is indicated by the darker colored areas. (Reproduced with permission from Kales AM, Kales JD: Sleep disorders. N Engl J Med 1974;290:487.) Awake REM 1 2 3 4 1 2 3 4 5 6 7 Young Adults Sleep stages Awake REM 1 2 3 4 1 2 3 4 5 6 7 Children Sleep stages 1 2 3 4 5 6 7 Awake REM 1 2 3 4 Elderly Sleep stages Hours of sleep FIGURE 15–9 Changes in human sleep pattern with age. Each plot shows data points for the ages of 6, 10, 21, 30, 69, and 84 years. (Data from Kandel ER, Schwartz JH, Jessel TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) 10 9 8 7 6 Total sleep time Hr/day 25 20 15 10 Percent of total sleep time in stage 4 sleep Percent 30 25 20 Percent of total sleep time in REM sleep Percent 0 20 40 60 80 100 Age (years) 236 SECTION III Central & Peripheral Neurophysiology thalamocortical circuit hypothesized to be involved in gener-ating rhythmic activity. Although not shown, inhibitory tha-lamic reticular neurons are elements of this network. The EEG shows the characteristic awake, light sleep, and deep sleep pat-terns of activity described above. Likewise, recordings from individual thalamic and cortical neurons show different pat-terns of rhythmic activity. In the waking state, corticocortical and thalamocortical networks generate higher-frequency rhythmic activity (30–80 Hz; gamma rhythm). This rhythm may be generated within the cells and networks of the cerebral cortex or within thalamocortical loops. The gamma rhythm has been suggested as a mechanism to “bind” together diverse sensory information into a single percept and action, but this theory is still controversial. In fact, disturbances in the integ-rity of this thalamocortical loop and its interaction with other brain structures may underlie some neurological disorders, including seizure activity. IMPORTANCE OF SLEEP Sleep has persisted throughout evolution of mammals and birds, so it is likely that it is functionally important. Indeed, if humans are awakened every time they show REM sleep, then permitted to sleep without interruption, they show a great deal more than the normal amount of REM sleep for a few nights. Relatively prolonged REM deprivation does not seem to have adverse psychological effects. Rats deprived of sleep for long periods lose weight in spite of increased caloric intake and eventually die. Various studies imply that sleep is needed to maintain metabolic-caloric balance, thermal equilibrium, and immune competence. In experimental animals, sleep is necessary for learning and memory consolidation. Learning sessions do not improve per-formance until a period of slow-wave or slow-wave plus REM sleep has occurred. Clinical Box 15–3 describes several com-mon sleep disorders. CIRCADIAN RHYTHMS & THE SLEEP–WAKE CYCLE CIRCADIAN RHYTHMS Most, if not all, living cells in plants and animals have rhyth-mic fluctuations in their function on a circadian cycle. Nor-mally they become entrained, that is, synchronized to the day– night light cycle in the environment. If they are not entrained, they become progressively more out of phase with the light– dark cycle because they are longer or shorter than 24 hours. The entrainment process in most cases is dependent on the su-prachiasmatic nuclei (SCN) located bilaterally above the op-tic chiasm (Figure 15–11). These nuclei receive information about the light–dark cycle via a special neural pathway, the retinohypothalamic fibers. Efferents from the SCN initiate FIGURE 15–10 Correlation between behavioral states, EEG, and single-cell responses in the cerebral cortex and thalamus. The EEG is characterized by high-frequency oscillations in the awake state and low-frequency rhythms during sleep. Thalamic and cortical neurons can also show different patterns of rhythmic activity. Thalamocortical neurons show slow rhythmic oscillations during deep sleep, and fire tonic trains of action potentials in the awake state. Most pyramidal neurons in the cortex generate only tonic trains of action potentials, although others may participate in the generation of high frequency rhythms through activation of rhythmic bursts of spikes. The thalamus and cerebral cortex are con-nected together in a loop. (Modified from McCormick DA: Are thalamocortical rhythms the Rosetta stone of a subset of neurological disorders? Nat Med 1999;5:1349.) Thalamocortical loop Cerebral cortex Thalamus EEG Single cell properties Awake Pyramidal cells Tonic firing 30–50 Hz gamma oscillations Light sleep Deep sleep 20–80 Hz rhythms 7–15 Hz rhythms 0.5–4 Hz rhythms Thalamocortical cell 0.5–4 Hz burst firing Tonic firing Transition from sleep to waking CHAPTER 15 Electrical Activity of the Brain, Sleep–Wake States, & Circadian Rhythms 237 neural and humoral signals that entrain a wide variety of well-known circadian rhythms including the sleep–wake cycle and the secretion of the pineal hormone melatonin. Evidence suggests that the SCN have two peaks of circadian activity. This may correlate with the observation that expo-sure to bright light can either advance, delay, or have no effect on the sleep–wake cycle in humans depending on the time of day when it is experienced. During the usual daytime it has no effect, but just after dark it delays the onset of the sleep period, and just before dawn it accelerates the onset of the next sleep period. Injections of melatonin have similar effects. In experimental animals, exposure to light turns on immedi-ate-early genes in the SCN, but only at times during the circa-dian cycle when light is capable of influencing entrainment. Stimulation during the day is ineffective. NEUROCHEMICAL MECHANISMS PROMOTING SLEEP & AROUSAL Transitions between sleep and wakefulness manifest a circadi-an rhythm consisting of an average of 8 h of sleep and 16 h of wakefulness. Nuclei in both the brain stem and hypothalamus are critical for the transitions between these states of con-sciousness. A classic study by Moruzzi and Magoun in 1949 showed that high-frequency stimulation of the midbrain retic-ular formation (the RAS) produces the EEG alerting response and arouses a sleeping animal. Damage to the area produces a comatose state. Electrical stimulation of the posterior hypo-thalamus also produces arousal similar to that elicited by stim-ulation of the midbrain, while electrical stimulation of the anterior hypothalamus and adjacent basal forebrain region in-duces sleep. As described above, the brainstem RAS is composed of several groups of neurons which release norepinephrine, serotonin, or acetylcholine. The locations of these neuronal populations are shown in Figure 7–2. In the case of the forebrain neurons CLINICAL BOX 15–3 Sleep Disorders Narcolepsy is a chronic neurological disorder caused by the brain’s inability to regulate sleep–wake cycles normally in which there is a sudden loss of voluntary muscle tone (cataplexy), an eventual irresistible urge to sleep during daytime, and possibly also brief episodes of total paralysis at the beginning or end of sleep. Narcolepsy is character-ized by a sudden onset of REM sleep, unlike normal sleep which begins with NREM, slow-wave sleep. The prevalence of narcolepsy ranges from 1 in 600 in Japan to 1 in 500,000 in Israel, with 1 in 1000 Americans being affected. Narco-lepsy has a familial incidence strongly associated with a class II antigen of the major histocompatibility complex on chromosome 6 at the HLA-DR2 or HLA-DQW1 locus, imply-ing a genetic susceptibility to narcolepsy. The HLA com-plexes are interrelated genes that regulate the immune sys-tem. Brains from humans with narcolepsy often contain fewer hypocretin (orexin)-producing neurons in the hypo-thalamus. It is thought that the HLA complex may increase susceptibility to an immune attack on these neurons, lead-ing to their degeneration. Obstructive sleep apnea (OSA) is the most common cause of daytime sleepiness due to fragmented sleep at night and affects about 24% of middle-aged men and 9% of women in the United States. Breathing ceases for more than 10 s during frequent episodes of obstruction of the upper air-way (especially the pharynx) due to reduction in muscle tone. The apnea causes brief arousals from sleep in order to rees-tablish upper airway tone. Snoring is a common patient com-plaint. There is actually not a reduction in total sleep time, but individuals with OSA experience a much greater time in stage 1 NREM sleep (from an average of 10% of total sleep to 30–50%) and a marked reduction in slow-wave sleep (stages 3 and 4 NREM sleep). The pathophysiology of OSA includes both a re-duction in neuromuscular tone at the onset of sleep and a change in the central respiratory drive. Periodic limb movement disorder (PLMD) is a stereotypical rhythmic extension of the big toe and dorsiflexion of the ankle and knee during sleep lasting for about 0.5 to 10 s and recurring at intervals of 20 to 90 s. Movements can actually range from shallow, continual movement of the ankle or toes, to wild and strenuous kicking and flailing of the legs and arms. Electromyo-graph (EMG) recordings show bursts of activity during the first hours of NREM sleep associated with brief EEG signs of arousal. The duration of stage 1 NREM sleep may be increased and that of stages 3 and 4 may be decreased compared to age-matched controls. PLMD is reported to occur in 5% of individuals between the ages of 30 and 50 years and increases to 44% of those over the age of 65. PLMD is similar to restless leg syndrome in which individuals have an irresistible urge to move their legs while at rest all day long. Sleepwalking (somnambulism), bed-wetting (nocturnal enuresis), and night terrors are referred to as parasomnias, which are sleep disorders associated with arousal from NREM and REM sleep. Episodes of sleepwalking are more common in children than in adults and occur predominantly in males. They may last several minutes. Somnambulists walk with their eyes open and avoid obstacles, but when awakened they can-not recall the episodes. 238 SECTION III Central & Peripheral Neurophysiology involved in control of the sleep–wake cycles, preoptic neu-rons in the hypothalamus release GABA and posterior hypo-thalamic neurons release histamine. One theory regarding the basis for transitions from sleep to wakefulness involves alternating reciprocal activity of different groups of RAS neurons. In this model (Figure 15–12), wake-fulness and REM sleep are at opposite extremes. When the activity of norepinephrine- and serotonin-containing neurons (locus coeruleus and raphé nuclei) is dominant, there is a reduced level of activity in acetylcholine-containing neurons in the pontine reticular formation. This pattern of activity con-tributes to the appearance of the awake state. The reverse of this pattern leads to REM sleep. When there is a more even balance in the activity of the aminergic and cholinergic neu-rons, NREM sleep occurs. In addition, an increased release of GABA and reduced release of histamine increase the likelihood of NREM sleep via deactivation of the thalamus and cortex. Wakefulness occurs when GABA release is reduced and histamine release is increased. MELATONIN AND THE SLEEP–WAKE STATE In addition to the previously described neurochemical mech-anisms promoting changes in the sleep–wake state, melatonin release from the richly vascularized pineal gland plays a role in sleep mechanisms (Figure 15–11). The pineal arises from the roof of the third ventricle in the diencephalon and is en-capsulated by the meninges. The pineal stroma contains glial cells and pinealocytes with features suggesting that they have a secretory function. Like other endocrine glands, it has highly permeable fenestrated capillaries. In infants, the pineal is large and the cells tend to be arranged in alveoli. It begins to invo-lute before puberty and small concretions of calcium phos-phate and carbonate (pineal sand) appear in the tissue. Because the concretions are radiopaque, the pineal is often vis-ible on x-ray films of the skull in adults. Displacement of a cal-cified pineal from its normal position indicates the presence of a space-occupying lesion such as a tumor in the brain. Melatonin and the enzymes responsible for its synthesis from serotonin by N-acetylation and O-methylation are present in pineal pinealocytes, and the hormone is secreted by them into the blood and the cerebrospinal fluid (Figure 15–13). Two FIGURE 15–11 Secretion of melatonin. Retinohypothalamic fibers synapse in the suprachiasmatic nuclei (SCN), and there are connections from the SCN to sympathetic preganglionic neurons in the spinal cord that project to the superior cervical ganglion. Postganglionic neurons project from this ganglion to the pineal gland that secretes melatonin. The cyclic activity of SCN sets up a circadian rhythm for melatonin release. This rhythm is entrained to light/dark cycles by neurons in the retina. (From Fox SI: Human Physiology. McGraw-Hill, 2008.) Inhibition Retinohypothalamic tract Suprachiasmatic nucleus (the "biological clock") Superior cervical ganglion Sympathetic neurons Pineal gland Stimulation CH3O H C H H C H H N O C N H CH3 Melatonin Day Night CHAPTER 15 Electrical Activity of the Brain, Sleep–Wake States, & Circadian Rhythms 239 melatonin-binding sites have been characterized: a high-affin-ity ML1 site and a low affinity ML2 site. Two subtypes of the ML1 receptor have been cloned: Mel 1a and Mel 1b. All the receptors are coupled to G proteins, with ML1 receptors inhib-iting adenylyl cyclase and ML2 receptors stimulating phospho-inositide hydrolysis. The diurnal change in melatonin secretion may function as a timing signal to coordinate events with the light–dark cycle in the environment. Melatonin synthesis and secretion are increased during the dark period of the day and maintained at a low level during daylight hours (Figure 15–13). This diurnal variation in secretion is brought about by norepinephrine secreted by the postganglionic sympathetic nerves that inner-vate the pineal gland (Figure 15–11). Norepinephrine acts via β-adrenergic receptors to increase intracellular cAMP, and the cAMP in turn produces a marked increase in N-acetyltrans-ferase activity. This results in increased melatonin synthesis and secretion. Circulating melatonin is rapidly metabolized in the liver by 6-hydroxylation followed by conjugation, and over 90% of the melatonin that appears in the urine is in the form of 6-hydroxy conjugates and 6-sulfatoxymelatonin. The pathway by which the brain metabolizes melatonin is unset-tled but may involve cleavage of the indole nucleus. The discharge of the sympathetic nerves to the pineal is entrained to the light–dark cycle in the environment via the retinohypothalamic nerve fibers to the SCN. From the hypo-thalamus, descending pathways converge onto preganglionic sympathetic neurons that in turn innervate the superior cer-vical ganglion, the site of origin of the postganglionic neurons to the pineal gland. CHAPTER SUMMARY ■The major rhythms in the EEG are alpha (8–13 Hz), beta (13–30 Hz), theta (4–7 Hz), delta (0.5–4 Hz), and gamma (30–80 Hz) oscillations. ■The EEG is of some value in localizing pathologic processes, and it is useful in characterizing different types of epilepsy. ■Throughout NREM sleep, there is some activity of skeletal mus-cle. A theta rhythm can be seen during stage 1 of sleep. Stage 2 is marked by the appearance of sleep spindles and occasional K complexes. In stage 3, a delta rhythm is dominant. Maximum slowing with slow waves is seen in stage 4. ■REM sleep is characterized by low-voltage, high-frequency EEG activity and rapid, roving movements of the eyes. ■A young adult typically passes through stages 1 and 2, and spends 70–100 min in stages 3 and 4. Sleep then lightens, and a REM period follows. This cycle repeats at 90-min intervals throughout the night. REM sleep occupies 50% of total sleep time in full-term neonates; this proportion declines rapidly and plateaus at about 25% until it falls further in old age. ■Transitions from sleep to wakefulness may involve alternating reciprocal activity of different groups of RAS neurons. When the activity of norepinephrine- and serotonin-containing FIGURE 15–12 A model of how alternating activity of brain stem and hypothalamic neurons may influence the different states of consciousness. (From Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology, 11th ed. McGraw-Hill, 2008.) Hypothalamus with circadian and homeostatic centers NREM sleep REM sleep Waking Activation of the thalamus and cortex Histamine GABA Histamine GABA Activation of the thalamus and cortex Acetylcholine Brainstem nuclei that are part of the reticular activating system Norepinephrine and serotonin Norepinephrine and serotonin Acetylcholine FIGURE 15–13 Diurnal rhythms of compounds involved in melatonin synthesis in the pineal. Melatonin and the enzymes re-sponsible for its synthesis from serotonin are found in pineal pinealo-cytes; melatonin is secreted into the bloodstream. Melatonin synthesis and secretion are increased during the dark period (shaded area) and maintained at a low level during the light period. Pineal paren-chymal cells Blood Melatonin N-Acetylserotonin Serotonin N-Acetyl-transferase Hydroxyindole-O-methyltransferase Melatonin 0 12 24 Time (hrs) 240 SECTION III Central & Peripheral Neurophysiology neurons is dominant, the activity in acetylcholine-containing neurons is reduced, leading to the appearance of wakefulness. The reverse of this pattern leads to REM sleep. Also, wakefulness occurs when GABA release is reduced and histamine release is increased. ■The entrainment of biological processes to the light–dark cycle is regulated by the SCN. ■The diurnal change in melatonin secretion from serotonin in the pineal gland may function as a timing signal to coordinate events with the light–dark cycle, including the sleep–wake cycle. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. In a healthy, alert adult sitting with the eyes closed, the dominant EEG rhythm observed with electrodes over the occipital lobes is A) delta (0.5–4 Hz). B) theta (4–7 Hz). C) alpha (8–13 Hz). D) beta (18–30 Hz). E) fast, irregular low-voltage activity. 2. Which of the following pattern of changes in central neurotrans-mitters/neuromodulators are associated with the transition from NREM to wakefulness? A) decrease in norepinephrine, increase in epinephrine, increase in acetylcholine, decrease in histamine, and decrease in GABA B) decrease in norepinephrine, increase in epinephrine, increase in acetylcholine, decrease in histamine, and increase in GABA C) decrease in norepinephrine, decrease in epinephrine, increase in acetylcholine, increase in histamine, and increase in GABA D) increase in norepinephrine, increase in epinephrine, decrease in acetylcholine, increase in histamine, and decrease in GABA E) increase in norepinephrine, decrease in epinephrine, decrease in acetylcholine, increase in histamine, and decrease in GABA 3. A gamma rhythm (30–80 Hz) A) is characteristic of seizure activity. B) is seen in an individual who is awake but not focused. C) may be a mechanism to bind together sensory information into a single percept and action. D) is independent of thalamocortical loops. E) is generated in the hippocampus. 4. Melatonin secretion would probably not be increased by A) stimulation of the superior cervical ganglia. B) intravenous infusion of tryptophan. C) intravenous infusion of epinephrine. D) stimulation of the optic nerve. E) induction of pineal hydroxyindole-O-methyltransferase. 5. Absence seizures A) are a form of nonconvulsive generalized seizures accompa-nied by momentary loss of consciousness. B) are a form of complex partial seizures accompanied by momentary loss of consciousness. C) are a form of nonconvulsive generalized seizures without a loss of consciousness. D) are a form of simple partial seizures without a loss of con-sciousness. E) are a form of convulsive generalized seizures accompanied by momentary loss of consciousness. 6. Narcolepsy is triggered by abnormalities in the A) skeletal muscles. B) medulla oblongata. C) hypothalamus. D) olfactory bulb. E) neocortex. CHAPTER RESOURCES Blackman S: Consciousness: An Introduction. Oxford University Press, 2004. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. McCormick DA, Contreras D: Of the cellular and network bases of epileptic seizures. Annu Rev Physiol 2001;63:815. Moruzzi G, Magoun HW: Brain stem reticular formation and activation of the EEG. Electroencephalogr Clin Neurophysiol 1949;1:455. Oberheim NA, et al: Loss of astrocytic domain organization in the epileptic brain. J Neurosci 2008;28:3264. Siegel JM: Narcolepsy. Sci Am 2000;282:76. Squire LR, et al (editors): Fundamental Neuroscience, 3rd ed. Academic Press, 2008. Steinlein O: Genetic mechanisms that underlie epilepsy. Nat Rev Neurosci 2004;5:400. Steriade M, McCarley RW: Brain Stem Control of Wakefulness and Sleep. Plenum, 1990. Steriade M, Paré D: Gating in Cerebral Networks. Cambridge University Press, 2007. Thorpy M (editor): Handbook of Sleep Disorders. Marcel Dekker, 1990. Waxman SG: Neuroanatomy with Clinical Correlations, 25th ed. McGraw-Hill, 2003. 241 C H A P T E R 16 Control of Posture & Movement O B J E C T I V E S After studying this chapter, you should be able to: ■Describe how skilled movements are planned and carried out. ■Name the posture-regulating parts of the central nervous system and discuss the role of each. ■Define spinal shock and describe the initial and long-term changes in spinal reflex-es that follow transection of the spinal cord. ■Define decerebrate and decorticate rigidity, and comment on the cause and phys-iologic significance of each. ■Describe the basal ganglia and list the pathways that interconnect them, along with the neurotransmitters in each pathway. ■Describe and explain the symptoms of Parkinson disease and Huntington disease. ■List the pathways to and from the cerebellum and the connections of each within the cerebellum. ■Discuss the functions of the cerebellum and the neurologic abnormalities pro-duced by diseases of this part of the brain. INTRODUCTION Somatic motor activity depends ultimately on the pattern and rate of discharge of the spinal motor neurons and homologous neurons in the motor nuclei of the cranial nerves. These neu-rons, the final common paths to skeletal muscle, are bom-barded by impulses from an immense array of descending pathways, other spinal neurons, and peripheral afferents. Some of these inputs end directly on α-motor neurons, but many exert their effects via interneurons or via γ-motor neu-rons to the muscle spindles and back through the Ia afferent fibers to the spinal cord. It is the integrated activity of these multiple inputs from spinal, medullary, midbrain, and cortical levels that regulates the posture of the body and makes coor-dinated movement possible. The inputs converging on motor neurons subserve three functions: they bring about voluntary activity, they adjust body posture to provide a stable background for movement, and they coordinate the action of the various muscles to make movements smooth and precise. The patterns of voluntary activity are planned within the brain, and the commands are sent to the muscles primarily via the corticospinal and corti-cobulbar systems. Posture is continually adjusted not only before but also during movement by descending brain stem pathways and peripheral afferents. Movement is smoothed and coordinated by the medial and intermediate portions of the cerebellum (spinocerebellum) and its connections. The basal ganglia and the lateral portions of the cerebellum (cere-brocerebellum) are part of a feedback circuit to the premotor and motor cortex that is concerned with planning and orga-nizing voluntary movement. 242 SECTION III Central & Peripheral Neurophysiology GENERAL PRINCIPLES ORGANIZATION There are two types of motor output: reflexive (involuntary) and voluntary. A subdivision of reflex responses includes some rhythmic movements such as swallowing, chewing, scratching, and walking, which are largely involuntary but subject to voluntary adjustment and control. To move a limb, the brain must plan a movement, arrange appropriate motion at many different joints at the same time, and adjust the motion by comparing plan with performance. The motor system “learns by doing” and performance improves with repetition. This involves synaptic plasticity. There is considerable evidence for the general motor control scheme shown in Figure 16–1. Commands for voluntary movement originate in cortical association areas. The move-ments are planned in the cortex as well as in the basal ganglia and the lateral portions of the cerebellar hemispheres, as indi-cated by increased electrical activity before the movement. The basal ganglia and cerebellum funnel information to the pre-motor and motor cortex by way of the thalamus. Motor com-mands from the motor cortex are relayed in large part via the corticospinal tracts to the spinal cord and the corresponding corticobulbar tracts to motor neurons in the brain stem. How-ever, collaterals from these pathways and a few direct connec-tions from the motor cortex end on brain stem nuclei, which also project to motor neurons in the brain stem and spinal cord. These pathways can also mediate voluntary movement. Movement sets up alterations in sensory input from the special senses and from muscles, tendons, joints, and the skin. This feedback information, which adjusts and smoothes movement, is relayed directly to the motor cortex and to the spinocerebel-lum. The spinocerebellum projects in turn to the brain stem. The main brain stem pathways that are concerned with pos-ture and coordination are the rubrospinal, reticulospinal, tectospinal, and vestibulospinal tracts. CONTROL OF AXIAL & DISTAL MUSCLES Within the brain stem and spinal cord, pathways and neurons that are concerned with the control of skeletal muscles of the trunk and proximal portions of the limbs are located medially or ventrally, whereas pathways and neurons that are con-cerned with the control of skeletal muscles in the distal por-tions of the limbs are located laterally. The axial muscles are concerned with postural adjustments and gross movements, whereas the distal limb muscles mediate fine, skilled move-ments. Thus, for example, neurons in the medial portion of the ventral horn innervate proximal limb muscles, particularly the flexors, whereas lateral ventral horn neurons innervate distal limb muscles. Similarly, the ventral corticospinal tract and medial descending brain stem pathways (tectospinal, reticulospinal, and vestibulospinal tracts) are concerned with adjustments of proximal muscles and posture, whereas the lat-eral corticospinal and rubrospinal tracts are concerned with distal limb muscles and, particularly in the case of the lateral corticospinal tract, with skilled voluntary movements. Phylo-genetically, the lateral pathways are newer. More details about these motor pathways are provided below. CORTICOSPINAL & CORTICOBULBAR TRACTS DESCENDING PROJECTIONS The axons of neurons from the motor cortex that project to spinal motor neurons form the corticospinal tracts, a large bundle of about 1 million fibers. About 80% of these fibers cross the midline in the medullary pyramids to form the later-al corticospinal tract (Figure 16–2). The remaining 20% make up the ventral corticospinal tract, which does not cross the midline until it reaches the level of the spinal cord at which it terminates. Lateral corticospinal tract neurons make mono-synaptic connections to motor neurons, especially those con-cerned with skilled movements. Corticospinal tract neurons also synapse on spinal interneurons antecedent to motor FIGURE 16–1 Control of voluntary movement. Commands for voluntary movement originate in cortical association areas. The cortex, basal ganglia, and cerebellum work cooperatively to plan movements. Movement executed by the cortex is relayed via the corticospinal tracts and corticobulbar tracts to motor neurons. The cerebellum provides feedback to adjust and smooth movement. Idea Cortical association areas Basal ganglia Lateral cerebellum Premotor and motor cortex Intermediate cerebellum Movement Plan Execute CHAPTER 16 Control of Posture & Movement 243 neurons; this indirect pathway is important in coordinating groups of muscles. The trajectory from the cortex to the spinal cord passes through the corona radiata to the posterior limb of the inter-nal capsule. Within the midbrain they traverse the cerebral peduncle and the basilar pons until they reach the medullary pyramids on their way to the spinal cord. The corticobulbar tract is composed of the fibers that pass from the motor cortex to motor neurons in the trigeminal, facial, and hypoglossal nuclei. Corticobulbar neurons end either directly on the cranial nerve nuclei or on their anteced-ent interneurons within the brain stem. Their axons traverse through the genu of the internal capsule, the cerebral pedun-cle (medial to corticospinal tract neurons), to descend with corticospinal tract fibers in the pons and medulla. The motor system can be divided into lower and upper motor neurons. Lower motor neurons refer to the spinal and cranial motor neurons that directly innervate skeletal mus-cles. Upper motor neurons are those in the cortex and brain stem that activate the lower motor neurons. The pathophysio-logical responses to damage to lower and upper motor neu-rons are very distinctive (see Clinical Box 16–1). ORIGINS OF CORTICOSPINAL & CORTICOBULBAR TRACTS Corticospinal and corticobulbar tract neurons are pyramidal shaped and located in layer V of the cerebral cortex (see Chap-ter 11). The cortical areas from which these tracts originate were identified on the basis of electrical stimulation that pro-duced prompt discrete movement. Figure 16–3 shows the ma-jor cortical regions involved in motor control. About 31% of the corticospinal tract neurons are from the primary motor cortex (M1; Brodmann’s area 4). This region is in the precen-tral gyrus of the frontal lobe, extending into the central sulcus. The premotor cortex and supplementary motor cortex (Brodmann’s area 6) account for 29% of the corticospinal tract neurons. The premotor area is anterior to the precentral gyrus, on the lateral and medial cortical surface; and the sup-plementary motor area is on and above the superior bank of the cingulate sulcus on the medial side of the hemisphere. The other 40% of corticospinal tract neurons originate in the pari-etal lobe (Brodmann’s area 5, 7) and primary somatosenso-ry area (Brodmann’s area 3, 1, 2) in the postcentral gyrus. MOTOR CORTEX & VOLUNTARY MOVEMENT PRIMARY MOTOR CORTEX By means of stimulation experiments in patients undergoing craniotomy under local anesthesia, it has been possible to out-line most of the motor projections from M1. These have been confirmed in unanesthetized unoperated humans by PET scan and fMRI (Figure 16–4). The various parts of the body are rep-resented in the precentral gyrus, with the feet at the top of the gyrus and the face at the bottom (Figure 16–5). The facial area is represented bilaterally, but the rest of the representation is generally unilateral, with the cortical motor area controlling the musculature on the opposite side of the body. The cortical representation of each body part is proportionate in size to the skill with which the part is used in fine, voluntary movement. The areas involved in speech and hand movements are espe-cially large in the cortex; use of the pharynx, lips, and tongue to form words and of the fingers and apposable thumbs to ma-nipulate the environment are activities in which humans are especially skilled. A somatotopic organization continues throughout the corticospinal and corticobulbar pathways. FIGURE 16–2 The corticospinal tracts. This tract originates in the precentral gyrus and passes through the internal capsule. Most fi-bers decussate in the pyramids and descend in the lateral white matter of the spinal cord to form the lateral division of the tract which can make monosynaptic connections with spinal motor neurons. The ven-tral division of the tract remains uncrossed until reaching the spinal cord where axons terminate on spinal interneurons antecedent to mo-tor neurons. Corticospinal tract Decussation of the pyramids Anterior horn cell Internal capsule Lateral cortico-spinal tract (80% of fibers) Spinal nerve Pyramids Ventral cortico-spinal tract (20% of fibers) Interneuron Precentral gyrus (area 4, etc) Distal muscle Proximal muscle 244 SECTION III Central & Peripheral Neurophysiology The conditions under which the human stimulation studies were performed precluded stimulation of the banks of the sulci and other inaccessible areas. Meticulous study has shown that in monkeys there is a regular representation of the body, with the axial musculature and the proximal portions of the limbs represented along the anterior edge of the precentral gyrus and the distal part of the limbs along the posterior edge. The cells in the cortical motor areas are arranged in col-umns. The ability to elicit discrete movements of a single muscle by electrical stimulation of a column within M1 led to the view that this area was responsible for control of individ-ual muscles. More recent work has shown that neurons in sev-eral cortical columns project to the same muscle; in fact, most stimuli activate more than one muscle. Moreover, the cells in each column receive fairly extensive sensory input from the CLINICAL BOX 16–1 Lower versus Upper Motor Neuron Damage Lower motor neurons are those whose axons terminate on skeletal muscles. Damage to these neurons is associated with flaccid paralysis, muscular atrophy, fasciculations (visible muscle twitches that appear as flickers under the skin), hypotonia (decreased muscle tone), and hyporeflexia or areflexia. An example of a disease that leads to lower motor neuron damage is amyotrophic lateral sclerosis (ALS). “Amyotrophic” means “no muscle nourishment” and describes the atrophy that muscles undergo because of dis-use. “Sclerosis” refers to the hardness felt when a pathologist examines the spinal cord on autopsy; the hardness is due to proliferation of astrocytes and scarring of the lateral columns of the spinal cord. ALS is a selective, progressive degenera-tion of α-motor neurons. This fatal disease is also known as Lou Gehrig disease because Gehrig, a famous American baseball player, died of it. The worldwide annual incidence of ALS has been estimated to be 0.5–3 cases per 100,000 people. Most cases are sporadic, but 5–10% of the cases are familial. Forty percent of the familial cases have a mutation in the gene for Cu/Zn superoxide dismutase (SOD-1) on chro-mosome 21. SOD is a free radical scavenger that reduces oxi-dative stress. A defective SOD-1 gene permits free radicals to accumulate and kill neurons. The disease has no racial, socio-economic, or ethnic boundaries. The life expectancy of ALS patients is usually 3–5 years after diagnosis. ALS is most com-monly diagnosed in middle age and affects men more often than women. The worldwide incidence of ALS is 2 per 100,000 of total population. The causes of ALS are unclear, but possibilities include viruses, neurotoxins, heavy metals, DNA defects (especially in familial ALS), immune system ab-normalities, and enzyme abnormalities. Upper motor neurons typically refer to corticospinal tract neurons that innervate spinal motor neurons, but they can also include brain stem neurons that control spinal motor neurons. Damage to these neurons initially causes muscles to become weak and flaccid but eventually leads to spasticity, hypertonia (increased resistance to passive movement), hyperactive stretch reflexes, and abnormal plantar extensor reflex (Babinski sign). The Babinski sign is dorsiflexion of the great toe and fanning of the other toes when the lateral aspect of the sole of the foot is scratched. In adults, the normal response to this stimulation is plantar flexion in all the toes. The Babinski sign is believed to be a flexor withdrawal reflex that is normally held in check by the lateral corticospinal system. It is of value in the localiza-tion of disease processes, but its physiologic significance is unknown. However, in infants whose corticospinal tracts are not well developed, dorsiflexion of the great toe and fanning of the other toes is the natural response to stimuli applied to the sole of the foot. FIGURE 16–3 A view of the human cerebral cortex, showing the motor cortex (Brodmann’s area 4) and other areas concerned with control of voluntary movement, along with the numbers assigned to the regions by Brodmann. (Reproduced with permission from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) FIGURE 16–4 Hand area of motor cortex demonstrated by functional magnetic resonance imaging (fMRI) in a 7-year-old boy. Changes in activity associated with squeezing a rubber ball with the right hand are shown in white and with the left hand in black. (Reproduced with permission from Waxman SG: Neuroanatomy with Clinical Correlations, 25th ed. McGraw-Hill, 2003.) 7 3,1,2 Posterior parietal cortex Primary somatic sensory cortex Motor cortex Supplementary motor area Premotor cortex Prefrontal cortex 7 5 4 6 CHAPTER 16 Control of Posture & Movement 245 peripheral area in which they produce movement, providing the basis for feedback control of movement. Some of this input may be direct and some is relayed from other parts of the cortex. The current view is that M1 neurons represent movements of groups of muscles for different tasks. SUPPLEMENTARY MOTOR AREA For the most part, the supplementary motor area projects to the motor cortex. This region also contains a map of the body, but it is less precise than in M1. It appears to be involved pri-marily in organizing or planning motor sequences, while M1 executes the movements. Lesions of this area in monkeys pro-duce awkwardness in performing complex activities and diffi-culty with bimanual coordination. When human subjects count to themselves without speak-ing, the motor cortex is quiescent, but when they speak the numbers aloud as they count, blood flow increases in M1 and the supplementary motor area. Thus, the supplementary motor area as well as M1 is involved in voluntary movement when the movements being performed are complex and involve planning. Blood flow increases whether or not a planned movement is carried out. The increase occurs whether the movement is performed by the contralateral or the ipsilateral hand. PREMOTOR CORTEX The premotor cortex, which also contains a somatotopic map, receives input from sensory regions of the parietal cortex and projects to M1, the spinal cord, and the brain stem reticular for-mation. Its function is still incompletely understood, but it may be concerned with setting posture at the start of a planned move-ment and with getting the individual prepared to move. It is most involved in control of proximal limb muscles needed to orient the body for movement. POSTERIOR PARIETAL CORTEX In addition to providing fibers that run in the corticospinal and corticobulbar tracts, the somatic sensory area and related portions of the posterior parietal lobe project to the premotor area. Lesions of the somatic sensory area cause defects in mo-tor performance that are characterized by inability to execute learned sequences of movements such as eating with a knife and fork. Some of the neurons in area 5 (Figure 16–3) are con-cerned with aiming the hands toward an object and manipu-lating it, whereas some of the neurons in area 7 are concerned with hand–eye coordination. ROLE IN MOVEMENT The corticospinal and corticobulbar system is the primary pathway for the initiation of skilled voluntary movement. This does not mean that movement—even skilled movement—is impossible without it. Nonmammalian vertebrates have es-sentially no corticospinal and corticobulbar system, but they move with great agility. Cats and dogs stand, walk, and run af-ter complete destruction of this system. Only in primates are relatively marked deficits produced. Careful section of the pyramids producing highly selective destruction of the lateral corticospinal tract in laboratory pri-mates produces prompt and sustained loss of the ability to grasp small objects between two fingers and to make isolated movements of the wrists. However, the animal can still use the hand in a gross fashion and can stand and walk. These deficits are consistent with loss of control of the distal musculature of the limbs, which is concerned with fine-skilled movements. On the other hand, lesions of the ventral corticospinal tract produce axial muscle deficits that cause difficulty with bal-ance, walking, and climbing. PLASTICITY A striking discovery made possible by PET and fMRI is that the motor cortex shows the same kind of plasticity as the sensory cortex (Chapter 11). For example, the finger areas of the contra-lateral motor cortex enlarge as a pattern of rapid finger move-ment is learned with the fingers of one hand; this change is detectable at 1 week and maximal at 4 weeks. Cortical areas of output to other muscles also increase in size when motor learning involves these muscles. When a small focal ischemic lesion is pro-duced in the hand area of the motor cortex of monkeys, the hand area may reappear, with return of motor function, in an adjacent FIGURE 16–5 Motor homunculus. The figure represents, on a coronal section of the precentral gyrus, the location of the cortical rep-resentation of the various parts. The size of the various parts is propor-tionate to the cortical area devoted to them. Compare with Figure 11–4. (Reproduced with permission from Penfield W, Rasmussen G: The Cere-bral Cortex of Man. Macmillan, 1950.) Toes Ankle Knee Hip Trunk Shoulder Elbow Wrist Hand Little Ring Middle Index Thumb Neck Brow Eyelid and eyeball Face Lips VO CAL IZA TIO N S A LI V A TI O N M A S TI C A T I O N Jaw Tongue Swallowing 246 SECTION III Central & Peripheral Neurophysiology undamaged part of the cortex. Thus, the maps of the motor cor-tex are not immutable, and they change with experience. BRAIN STEM PATHWAYS INVOLVED IN POSTURE AND VOLUNTARY MOVEMENT As mentioned above, spinal motor neurons are organized such that those innervating the most proximal muscles are lo-cated most medially and those innervating the more distal muscles are located more laterally. This organization is also reflected in descending brain stem pathways (Figure 16–6). MEDIAL BRAIN STEM PATHWAYS The medial brain stem pathways, which work in concert with the ventral corticospinal tract, are the pontine and medullary reticulospinal, vestibulospinal, and tectospinal tracts. These pathways descend in the ipsilateral ventral columns of the spi-nal cord and terminate predominantly on interneurons and long propriospinal neurons in the ventromedial part of the ventral horn to control axial and proximal muscles. A few me-dial pathway neurons synapse directly on motor neurons con-trolling axial muscles. The medial and lateral vestibulospinal tracts were briefly described in Chapter 13. The medial tract originates in the medial and inferior vestibular nuclei and projects bilaterally to cervical spinal motor neurons that control neck muscula-ture. The lateral tract originates in the lateral vestibular nuclei and projects ipsilaterally to neurons at all spinal levels. It acti-vates motor neurons to antigravity muscles (eg, proximal limb extensors) to control posture and balance. The pontine and medullary reticulospinal tracts project to all spinal levels. They are involved in the maintenance of pos-ture and in modulating muscle tone, especially via an input to γ-motor neurons. Pontine reticulospinal neurons are primar-ily excitatory and medullary reticulospinal neurons are pri-marily inhibitory. FIGURE 16–6 Medial and lateral descending brain stem pathways involved in motor control. A) Medial pathways (reticulospinal, ves-tibulospinal, and tectospinal) terminate in ventromedial area of spinal gray matter and control axial and proximal muscles. B) Lateral pathway (ru-brospinal) terminates in dorsolateral area of spinal gray matter and controls distal muscles. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Rubrospinal tract B Lateral brain stem pathways Red nucleus (magnocellular part) A Medial brain stem pathways Tectum Medial reticular formation Reticulospinal tract Tectospinal tract Vestibulospinal tracts Lateral and medial vestibular nuclei CHAPTER 16 Control of Posture & Movement 247 The tectospinal tract originates in the superior colliculus of the midbrain. It projects to the contralateral cervical spinal cord to control head and eye movements. LATERAL BRAIN STEM PATHWAY The main control of distal muscles arise from the lateral corti-cospinal tract, but neurons within the red nucleus of the mid-brain cross the midline and project to interneurons in the dorsolateral part of the spinal ventral horn to also influence motor neurons that control distal limb muscles. This ru-brospinal tract excites flexor motor neurons and inhibits ex-tensor motor neurons. This pathway is not very prominent in humans, but it may play a role in the posture typical of decor-ticate rigidity (see below). POSTURE-REGULATING SYSTEMS INTEGRATION In the intact animal, individual motor responses are submerged in the total pattern of motor activity. When the neural axis is transected, the activities integrated below the section are cut off, or released, from the control of higher brain centers and often appear to be accentuated. Release of this type, long a cardinal principle in neurology, may be due in some situations to remov-al of an inhibitory control by higher neural centers. A more im-portant cause of the apparent hyperactivity is loss of differentiation of the reaction so that it no longer fits into the broader pattern of motor activity. An additional factor may be denervation hypersensitivity of the centers below the transec-tion, but the role of this component remains to be determined. Animal experimentation has led to information on the role of cortical and brain stem mechanisms involved in control of voluntary movement and posture. The deficits in motor con-trol seen after various lesions mimic those seen in humans with damage in the same structures. DECEREBRATION A complete transection of the brain stem between the superior and inferior colliculi permits the brain stem pathways to func-tion independent of their input from higher brain structures. This is called a midcollicular decerebration and is diagramed in Figure 16–7 by the dashed line labeled A. This lesion inter-rupts all input from the cortex (corticospinal and corticobul-bar tracts) and red nucleus (rubrospinal tract), primarily to distal muscles of the extremities. The excitatory and inhibitory reticulospinal pathways (primarily to postural extensor mus-cles) remain intact. The dominance of drive from ascending sensory pathways to the excitatory reticulospinal pathway leads to hyperactivity in extensor muscles in all four extremi-ties which is called decerebrate rigidity. This resembles what ensues after supratentorial lesions in humans cause uncal herniation. Uncal herniation can occur in patients with large tumors or a hemorrhage in the cerebral hemisphere. Figure 16–8A shows the posture typical of such a patient. Clinical Box 16–2 describes complications related to uncal herniation. In midcollicular decerebrate cats, section of dorsal roots to a limb (dashed line labeled B in Figure 16–7) immediately eliminates the hyperactivity of extensor muscles. This sug-gests that decerebrate rigidity is spasticity due to facilitation of the myotatic stretch reflex. That is, the excitatory input from the reticulospinal pathway activates γ-motor neurons which indirectly activate α-motor neurons (via Ia spindle afferent activity). This is called the gamma loop. The exact site of origin within the cerebral cortex of the fibers that inhibit stretch reflexes is unknown. Under certain condi-tions, stimulation of the anterior edge of the precentral gyrus can cause inhibition of stretch reflexes and cortically evoked movements. This region, which also projects to the basal gan-glia, has been named area 4s, or the suppressor strip. There is also evidence that decerebrate rigidity leads to direct activation of α-motor neurons. If the anterior lobe of the cere-bellum is removed in a decerebrate animal (dashed line labeled C in Figure 16–7), extensor muscle hyperactivity is exaggerated (decerebellate rigidity). This cut eliminates cortical inhibition of the cerebellar fastigial nucleus and secondarily increases excitation to vestibular nuclei. Subsequent dorsal root section does not reverse the rigidity, thus it was due to activation of α-motor neurons independent of the gamma loop. DECORTICATION Removal of the cerebral cortex (decortication; dashed line la-beled D in Figure 16–7) produces decorticate rigidity which is characterized by flexion of the upper extremities at the el-bow and extensor hyperactivity in the lower extremities (Fig-ure 16–8B). The flexion can be explained by rubrospinal excitation of flexor muscles in the upper extremities; the hy-perextension of lower extremities is due to the same changes that occur after midcollicular decerebration. Decorticate rigidity is seen on the hemiplegic side in humans after hemorrhages or thromboses in the internal capsule. Prob-ably because of their anatomy, the small arteries in the internal capsule are especially prone to rupture or thrombotic obstruc-tion, so this type of decorticate rigidity is fairly common. Sixty percent of intracerebral hemorrhages occur in the internal cap-sule, as opposed to 10% in the cerebral cortex, 10% in the pons, 10% in the thalamus, and 10% in the cerebellum. SPINAL INTEGRATION The responses of animals and humans to spinal cord injury (SCI) illustrate the integration of reflexes at the spinal level. The deficits seen after SCI vary, of course, depending on the level of the injury. Clinical Box 16–3 provides information on long-term problems related to SCI and recent advancements in treatment options. 248 SECTION III Central & Peripheral Neurophysiology FIGURE 16–7 A circuit drawing representing lesions produced in experimental animals to replicate decerebrate and decorticate deficits seen in humans. Bilateral transections are indicated by dashed lines A, B, C, and D. Decerebration is at a midcollicular level (A), decor-tication is rostral to the superior colliculus, dorsal roots sectioned for one extremity (B), and removal of anterior lobe of cerebellum (C). The objective was to identify anatomic substrates responsible for decerebrate or decorticate rigidity/posturing seen in humans with lesions that either isolate the forebrain from the brain stem or separate rostral from caudal brain stem and spinal cord. (Reproduced with permission from Haines DE [editor]: Fundamental Neuroscience for Basic and Clinical Applications, 3rd ed. Elsevier, 2006.) CHAPTER 16 Control of Posture & Movement 249 SPINAL SHOCK In all vertebrates, transection of the spinal cord is followed by a period of spinal shock during which all spinal reflex responses are profoundly depressed. Subsequently, reflex responses re-turn and become hyperactive. The duration of spinal shock is proportionate to the degree of encephalization of motor func-tion in the various species. In frogs and rats it lasts for minutes; in dogs and cats it lasts for 1 to 2 h; in monkeys it lasts for days; and in humans it usually lasts for a minimum of 2 wk. The cause of spinal shock is uncertain. Cessation of tonic bom-bardment of spinal neurons by excitatory impulses in descending pathways undoubtedly plays a role, but the subsequent return of reflexes and their eventual hyperactivity also have to be explained. The recovery of reflex excitability may be due to the development of denervation hypersensitivity to the mediators released by the remaining spinal excitatory endings. Another possibility for which there is some evidence is the sprouting of collaterals from existing neurons, with the formation of addi-tional excitatory endings on interneurons and motor neurons. The first reflex response to appear as spinal shock wears off in humans is often a slight contraction of the leg flexors and adductors in response to a noxious stimulus. In some patients, the knee jerk reflex recovers first. The interval between cord transection and the return of reflex activity is about 2 weeks in the absence of any complications, but if complications are present it is much longer. It is not known why infection, mal-nutrition, and other complications of SCI inhibit spinal reflex activity. Once the spinal reflexes begin to reappear after spinal shock, their threshold steadily drops. LOCOMOTION GENERATOR Circuits intrinsic to the spinal cord can produce walking movements when stimulated in a suitable fashion even after spinal cord transection in cats and dogs. There are two loco-motor pattern generators in the spinal cord: one in the cervi-cal region and one in the lumbar region. However, this does not mean that spinal animals or humans can walk without stimulation; the pattern generator has to be turned on by tonic discharge of a discrete area in the midbrain, the mesencephalic CLINICAL BOX 16–2 Uncal Herniation Space-occupying lesions from large tumors, hemorrhages, strokes, or abscesses in the cerebral hemisphere can drive the uncus of the temporal lobe over the edge of the cere-bellar tentorium, compressing the ipsilateral cranial nerve III (uncal herniation). Before the herniation these patients experience a decreased level of consciousness, lethargy, poorly reactive pupils, deviation of the eye to a “down and out” position, hyperactive reflexes, and a bilateral Babinski sign (due to compression of the ipsilateral corticospinal tract). After the brain herniates, the patients are decere-brate and comatose, have fixed and dilated pupils, and eye movements are absent. Once damage extends to the mid-brain, a Cheyne–Stokes respiratory pattern develops, characterized by a pattern of waxing-and-waning depth of respiration with interposed periods of apnea. Eventually, medullary function is lost, breathing ceases, and recovery is unlikely. Hemispheric masses closer to the midline com-press the thalamic reticular formation and can cause coma before eye findings develop (central herniation). As the mass enlarges, midbrain function is affected, the pupils di-late, and a decerebrate posture ensues. With progressive herniation, pontine vestibular and then medullary respira-tory functions are lost. FIGURE 16–8 Decerebrate and decorticate postures. A) Damage to lower midbrain and upper pons causes decerebrate posturing in which lower extremities are extended with toes pointed inward and upper extremities extended with fingers flexed and forearms pronate. Neck and head are extended. B) Damage to upper midbrain may cause decorticate posturing in which upper limbs are flexed, lower limbs are extended with toes pointed slightly inward, and head is extended. (Modified from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) B Upper midbrain damage A Upper pontine damage 250 SECTION III Central & Peripheral Neurophysiology locomotor region, and, of course, this is only possible in pa-tients with incomplete spinal cord transection. Interestingly, the generators can also be turned on in experimental animals by administration of the norepinephrine precursor L-dopa (levodopa) after complete section of the spinal cord. Progress is being made in teaching humans with SCI to take a few steps by placing them, with support, on a treadmill. BASAL GANGLIA ANATOMIC CONSIDERATIONS The term basal ganglia (or basal nuclei) is generally applied to five interactive structures on each side of the brain (Figure 16–9). These are the caudate nucleus, putamen, and globus pallidus (three large nuclear masses underlying the cortical mantle), the subthalamic nucleus, and substantia nigra. The globus pallidus is divided into external and internal segments (GPe and GPi). The substantia nigra is divided into a pars compacta and a pars reticulata. The caudate nucleus and putamen are commonly called the striatum; the putamen and globus pallidus are sometimes called the lenticular nucleus. The main inputs to the basal ganglia terminate in the stria-tum (Figure 16–10). They include the excitatory corticostri-ate pathway from M1 and premotor cortex. There is also a projection from intralaminar nuclei of the thalamus to the striatum (thalamostriatal pathway). The connections between the parts of the basal ganglia include a dopaminergic nigrostriatal projection from the sub-stantia nigra pars compacta to the striatum and a correspond-ing GABAergic projection from the striatum to substantia nigra pars reticulata. The striatum projects to both GPe and GPi. GPe projects to the subthalamic nucleus, which in turn projects to both GPe and GPi. The principal output from the basal ganglia is from GPi via the thalamic fasciculus to the ventral lateral, ventral anterior, and centromedian nuclei of the thalamus. From the thalamic nuclei, fibers project to the prefrontal and premotor cortex. The substantia nigra also projects to the thalamus. These con-nections, along with the probable synaptic transmitters involved, are summarized in Figure 16–10. CLINICAL BOX 16–3 Spinal Cord Injury It has been estimated that the worldwide annual incidence of sustaining spinal cord injury (SCI) is between 10 and 83 per million of the population. Leading causes are vehicle ac-cidents, violence, and sports injuries. The mean age of pa-tients who sustain an SCI is 33 years old, and men outnum-ber women with a nearly 4 to 1 ratio. Approximately 52% of SCI cases result in quadriplegia and about 42% lead to para-plegia. In quadriplegic humans, the threshold of the with-drawal reflex is very low; even minor noxious stimuli may cause not only prolonged withdrawal of one extremity but marked flexion–extension patterns in the other three limbs. Stretch reflexes are also hyperactive. Afferent stimuli irradi-ate from one reflex center to another after SCI. When even a relatively minor noxious stimulus is applied to the skin, it may activate autonomic neurons and produce evacuation of the bladder and rectum, sweating, pallor, and blood pres-sure swings in addition to the withdrawal response. This dis-tressing mass reflex can sometimes be used to give para-plegic patients a degree of bladder and bowel control. They can be trained to initiate urination and defecation by strok-ing or pinching their thighs, thus producing an intentional mass reflex. If the cord section is incomplete, the flexor spasms initiated by noxious stimuli can be associated with bursts of pain that are particularly bothersome. They can be treated with considerable success with baclofen, a GABAB receptor agonist that crosses the blood–brain barrier and fa-cilitates inhibition. Treatment of SCI patients presents complex problems. Ad-ministration of large doses of glucocorticoids has been shown to foster recovery and minimize loss of function after SCI. They need to be given soon after the injury and then discontinued because of the well-established deleterious effects of long-term steroid treatment. Their immediate value is likely due to reduction of the inflammatory response in the damaged tissue. Due to immobilization, SCI patients develop a negative nitro-gen balance and catabolize large amounts of body protein. Their body weight compresses the circulation to the skin over bony prominences, causing decubitus ulcers to form. The ul-cers heal poorly and are prone to infection because of body protein depletion. The tissues that are broken down include the protein matrix of bone and this, plus the immobilization, cause Ca2+ to be released in large amounts, leading to hyper-calcemia, hypercalciuria, and formation of calcium stones in the urinary tract. The combination of stones and bladder paral-ysis cause urinary stasis, which predisposes to urinary tract in-fection, the most common complication of SCI. The search continues for ways to get axons of neurons in the spinal cord to regenerate. Administration of neurotrophins shows some promise in experimental animals, and so does implantation of embryonic stem cells at the site of injury. Another possibility being explored is bypassing the site of SCI with brain–com-puter interface devices. However, these novel approaches are a long way from routine clinical use. CHAPTER 16 Control of Posture & Movement 251 The main feature of the connections of the basal ganglia is that the cerebral cortex projects to the striatum, the striatum to GPi, GPi to the thalamus, and the thalamus back to the cor-tex, completing a loop. The output from GPi to the thalamus is inhibitory, whereas the output from the thalamus to the cerebral cortex is excitatory. The striatum is made up of two parts that differ histologi-cally: a unique mosaic of patches or striosomes (an area with little acetylcholinesterase) and a matrix (an area high in ace-tylcholinesterase). The neurons of the corticostriate projec-tion that originate in the deep portion of layer V of the cortex terminate in the patches, whereas the neurons that originate in layers II and III and the superficial part of layer V end pri-marily in the matrix. Neurons with their cell bodies in patches project in large part to dopaminergic neurons in the substan-tia nigra pars compacta, whereas many of the neurons with their cell bodies in the matrix project to GABAergic neurons in the substantia nigra pars reticulata. FUNCTION Neurons in the basal ganglia, like those in the lateral portions of the cerebellar hemispheres, discharge before movements begin. This observation, plus careful analysis of the effects of diseases of the basal ganglion in humans and the effects of drugs that destroy dopaminergic neurons in animals, have led to the idea that the basal ganglia are involved in the planning and programming of movement or, more broadly, in the pro-cesses by which an abstract thought is converted into volun-tary action (Figure 16–1). They influence the motor cortex via the thalamus, and the corticospinal pathways provide the final common pathway to motor neurons. In addition, GPi projects to nuclei in the brain stem, and from there to motor neurons in the brain stem and spinal cord. The basal ganglia, particularly FIGURE 16–9 The basal ganglia. The basal ganglia are composed of the caudate nucleus, putamen, and globus pallidus and the function-ally related subthalamic nucleus and substantia nigra. The frontal (coronal) section shows the location of the basal ganglia in relation to surround-ing structures. Caudate nucleus Thalamus Putamen and globus pallidus Amygdaloid nucleus Lateral view Horizontal section Caudate nucleus Globus pallidus Thalamus Tail of caudate nucleus Putamen Amygdala Substantia nigra Subthalamic nucleus Internal segment External segment Putamen Globus pallidus: Caudate nucleus Lateral ventricle Internal capsule Thalamus Frontal section Internal capsule FIGURE 16–10 Diagrammatic representation of the principal connections of the basal ganglia. Solid lines indicate excitatory path-ways, dashed lines inhibitory pathways. The transmitters are indicated in the pathways, where they are known. Glu, glutamate; DA, dopamine. Acetylcholine is the transmitter produced by interneurons in the stria-tum. SNPR, substantia nigra, pars reticulata; SNPC, substantia nigra, pars compacta; ES, external segment; IS, internal segment; PPN, peduncu-lopontine nuclei. The subthalamic nucleus also projects to the pars com-pacta of the substantia nigra; this pathway has been omitted for clarity. Cerebral cortex Striatum (acetylcholine) Globus pallidus, ES Subthalamic nucleus Brain stem and spinal cord GABA GABA GABA GABA GABA GABA GABA Glu Glu Glu DA Globus pallidus, IS Thalamus SNPR SNPC PPN 252 SECTION III Central & Peripheral Neurophysiology the caudate nuclei, also play a role in some cognitive processes. Possibly because of the interconnections of this nucleus with the frontal portions of the neocortex, lesions of the caudate nuclei disrupt performance on tests involving object reversal and delayed alternation. In addition, lesions of the head of the left but not the right caudate nucleus and nearby white matter in humans are associated with a dysarthric form of aphasia that resembles Wernicke aphasia. DISEASES OF THE BASAL GANGLIA IN HUMANS Three distinct biochemical pathways in the basal ganglia nor-mally operate in a balanced fashion: (1) the nigrostriatal dopa-minergic system, (2) the intrastriatal cholinergic system, and (3) the GABAergic system, which projects from the striatum to the globus pallidus and substantia nigra. When one or more of these pathways become dysfunctional, characteristic motor ab-normalities occur. Diseases of the basal ganglia lead to two gen-eral types of disorders: hyperkinetic and hypokinetic. The hyperkinetic conditions are those in which movement is exces-sive and abnormal, including chorea, athetosis, and ballism. Hypokinetic abnormalities include akinesia and bradykinesia. Chorea is characterized by rapid, involuntary “dancing” movements. Athetosis is characterized by continuous, slow writhing movements. Choreiform and athetotic movements have been likened to the start of voluntary movements occurring in an involuntary, disorganized way. In ballism, involuntary CLINICAL BOX 16–4 Basal Ganglia Diseases The initial detectable damage in Huntington disease is to medium spiny neurons in the striatum. This loss of this GABAergic pathway to the globus pallidus external segment releases inhibition, permitting the hyperkinetic features of the disease to develop. An early sign is a jerky trajectory of the hand when reaching to touch a spot, especially toward the end of the reach. Later, hyperkinetic choreiform movements appear and gradually increase until they incapacitate the pa-tient. Speech becomes slurred and then incomprehensible, and a progressive dementia is followed by death, usually within 10–15 years after the onset of symptoms. Huntington disease affects 5 out of 100,000 people worldwide. It is inher-ited as an autosomal dominant disorder, and its onset is usu-ally between the ages of 30 and 50. The abnormal gene re-sponsible for the disease is located near the end of the short arm of chromosome 4. It normally contains 11–34 cytosine-adenine-guanine (CAG) repeats, each coding for glutamine. In patients with Huntington disease, this number is increased to 42–86 or more copies, and the greater the number of repeats, the earlier the age of onset and the more rapid the progres-sion of the disease. The gene codes for huntingtin, a protein of unknown function. Poorly soluble protein aggregates, which are toxic, form in cell nuclei and elsewhere. However, the correlation between aggregates and symptoms is less than perfect. It appears that a loss of the function of hunting-tin occurs that is proportionate to the size of the CAG insert. At present, no effective treatment is available, and the disease is uniformly fatal. However, there are a few glimmers of hope. In animal models of the disease, intrastriatal grafting of fetal striatal tissue improves cognitive performance. In addition, tissue caspase-1 activity is increased in the brains of humans and animals with the disease and in mice in which the gene for this apoptosis-regulating enzyme has been knocked out, progression of the disease is slowed. Another basal ganglia disorder is Wilson disease (or hepatolenticular degeneration), which is a rare disorder of copper metabolism which has an onset between 6 to 25 years of age, affecting about four times as many females as males. Wilson disease affects about 30,000 people world-wide. It is a genetic autosomal recessive disorder due to a mutation on the long arm of chromosome 13q. It affects the copper-transporting ATPase gene (ATP7B) in the liver, lead-ing to an accumulation of copper in the liver and resultant progressive liver damage. About 1% of the population carries a single abnormal copy of this gene but do not develop any symptoms. A child who inherits the gene from both parents may develop the disease. In affected individuals, copper ac-cumulates in the periphery of the cornea in the eye account-ing for the characteristic yellow Kayser–Fleischer rings. The dominant neuronal pathology is degeneration of the puta-men, a part of the lenticular nucleus. Motor disturbances in-clude “wing-beating” tremor or asterixis, dysarthria, un-steady gait, and rigidity. Treatment is to reduce the copper in the body. Another disease commonly referred to as a disease of the basal ganglia is tardive dyskinesia. This disease indeed in-volves the basal ganglia, but it is caused by medical treat-ment of another disorder with neuroleptic drugs such as phenothiazides or haloperidol. Therefore, tardive dyskinesia is iatrogenic in origin. Long-term use of these drugs may pro-duce biochemical abnormalities in the striatum. The motor disturbances include either temporary or permanent uncon-trolled involuntary movements of the face and tongue and cogwheel rigidity. The neuroleptic drugs act via blockade of dopaminergic transmission. Prolonged drug use leads to hy-persensitivity of D3 dopaminergic receptors and an imbal-ance in nigrostriatal influences on motor control. CHAPTER 16 Control of Posture & Movement 253 flailing, intense, and violent movements occur. Akinesia is difficulty in initiating movement and decreased spontaneous movement. Bradykinesia is slowness of movement. In addition to Parkinson disease, which is described below, there are several other disorders known to involve a malfunc-tion within the basal ganglia. A few of these are described in Clinical Box 16–4. Huntington disease is one of an increasing number of human genetic diseases affecting the nervous sys-tem that are characterized by trinucleotide repeat expansion. Most of these involve cytosine-adenine-guanine (CAG) repeats (Table 16–1), but one involves CGG repeats and another involves CTG repeats. All of these are in exons; however, a GAA repeat in an intron is associated with Friedreich’s ataxia. There is also preliminary evidence that increased numbers of a 12-nucleotide repeat are associated with a rare form of epilepsy. PARKINSON DISEASE (PARALYSIS AGITANS) Parkinson disease has both hypokinetic and hyperkinetic fea-tures. It was originally described by James Parkinson and is named for him. Parkinson disease is the first disease identified as being due to a deficiency in a specific neurotransmitter. In the 1960s, Parkinson disease was shown to result from the de-generation of dopaminergic neurons in the substantia nigra pars compacta. The fibers to the putamen are most severely affected. Par-kinsonism now occurs in sporadic idiopathic form in many middle-aged and elderly individuals and is one of the most common neurodegenerative diseases. It is estimated to occur in 1–2% of individuals over age 65. Dopaminergic neurons and dopamine receptors are steadily lost with age in the basal ganglia in normal individuals, and an acceleration of these losses apparently precipitates parkinsonism. Symptoms appear when 60–80% of the nigrostriatal dopaminergic neu-rons degenerate. Parkinsonism is also seen as a complication of treatment with the phenothiazine group of tranquilizer drugs and other drugs that block D2 receptors. It can be produced in rapid and dramatic form by injection of 1-methyl-4-phenyl-1,2,5,6-tet-rahydropyridine (MPTP). This effect was discovered by chance when a drug dealer in northern California supplied some of his clients with a homemade preparation of synthetic heroin that contained MPTP. MPTP is a prodrug that is metabolized in astrocytes by the enzyme MOA-B to produce a potent oxidant, 1-methyl-4-phenylpyridinium (MPP+). In rodents, MPP+ is rapidly removed from the brain, but in pri-mates it is removed more slowly and is taken up by the dopa-mine transporter into dopaminergic neurons in the substantia nigra, which it destroys without affecting other dopaminergic neurons to any appreciable degree. Consequently, MPTP can be used to produce parkinsonism in monkeys, and its avail-ability has accelerated research on the function of the basal ganglia. The hypokinetic features of Parkinson disease are akinesia and bradykinesia, and the hyperkinetic features are cogwheel rigidity and tremor at rest. The absence of motor activity and the difficulty in initiating voluntary movements are striking. There is a decrease in the normal, unconscious movements such as swinging of the arms during walking, the panorama of facial expressions related to the emotional con-tent of thought and speech, and the multiple “fidgety” actions and gestures that occur in all of us. The rigidity is different from spasticity because motor neuron discharge increases to both the agonist and antagonist muscles. Passive motion of an extremity meets with a plastic, dead-feeling resistance that has been likened to bending a lead pipe and is therefore called lead pipe rigidity. Sometimes a series of “catches” takes place during passive motion (cogwheel rigidity), but the sudden loss of resistance seen in a spastic extremity is absent. The tremor, which is present at rest and disappears with activity, is due to regular, alternating 8-Hz contractions of antagonistic muscles. A current view of the pathogenesis of the movement disor-ders in Parkinson disease is shown in Figure 16–11. In normal individuals, basal ganglia output is inhibitory via GABAergic nerve fibers. The dopaminergic neurons that project from the substantia nigra to the putamen normally have two effects: they stimulate the D1 dopamine receptors, which inhibit GPi via direct GABAergic receptors, and they inhibit D2 receptors, which also inhibit the GPi. In addition, the inhibition reduces the excitatory discharge from the subthalamic nucleus to the GPi. This balance between inhibition and excitation somehow maintains normal motor function. In Parkinson disease, the dopaminergic input to the putamen is lost. This results in decreased inhibition and increased excitation from the STN to the GPi. The overall increase in inhibitory output to the thalamus and brain stem disorganizes movement. TABLE 16–1 Examples of trinucleotide repeat diseases. Disease Expanded Trinucleotide Repeat Affected Protein Huntington disease CAG Huntingtin Spinocerebellar ataxia, types 1, 2, 3, 7 CAG Ataxin 1, 2, 3, 7 Spinocerebellar ataxia, type 6 CAG α1A subunit of Ca2+ channel Dentatorubral-pallidoluy-sian atrophy CAG Atrophin Spinobulbar muscular atrophy CAG Androgen receptor Fragile X syndrome CGG FMR-1 Myotonic dystrophy CTG DM protein kinase Friedreich ataxia GAA Frataxin 254 SECTION III Central & Peripheral Neurophysiology Treatment An important consideration in Parkinson disease is the bal-ance between the excitatory discharge of cholinergic interneu-rons and the inhibitory dopaminergic input in the striatum. Some improvement is produced by decreasing the cholinergic influence with anticholinergic drugs. More dramatic improve-ment is produced by administration of L-dopa (levodopa). Unlike dopamine, this dopamine precursor crosses the blood– brain barrier and helps repair the dopamine deficiency. How-ever, the degeneration of these neurons continues and in 5 to 7 y the beneficial effects of L-dopa disappear. Surgical treatment by making lesions in GPi (pallidotomy) or in the subthalamic nucleus helps to restore the output bal-ance toward normal (Figure 16–11). Surgical outcomes have been further improved by implanting electrodes attached to subcutaneous stimulators and administering high-frequency current. This produces temporary disruption of circuits at the electrode tip on demand. Another surgical approach is to implant dopamine-secret-ing tissue in or near the basal ganglia. Transplants of the patient’s own adrenal medullary tissue or carotid body works for a while, apparently by functioning as a sort of dopamine minipump, but long-term results have been disappointing. Results with transplantation of fetal striatal tissue have been better, and there is evidence that the transplanted cells not only survive but make appropriate connections in the host’s basal ganglia. However, some patients with transplants develop severe involuntary movements (dyskinesias). In monkeys with experimental parkinsonism, neurotrophic factors benefit the nigrostriatal neurons, and local injection of GDNF attached to a lentivirus vector so that it penetrates cells has produced promising results. Familial cases of Parkinson disease occur, but these are uncommon. The genes for at least five proteins can be mutated. These proteins appear to be involved in ubiquitina-tion. Two of the proteins, α-synuclein and barkin, interact and are found in Lewy bodies. The Lewy bodies are inclusion bodies in neurons that occur in all forms of Parkinson disease. However, the significance of these findings is still unsettled. CEREBELLUM ANATOMIC DIVISIONS The cerebellum sits astride the main sensory and motor sys-tems in the brain stem (Figure 16–12). It is connected to the brain stem on each side by a superior peduncle (brachium conjunctivum), middle peduncle (brachium pontis), and infe-rior peduncle (restiform body). The medial vermis and lateral cerebellar hemispheres are more extensively folded and fis-sured than the cerebral cortex (Figure 16–13). The cerebellum weighs only 10% as much as the cerebral cortex, but its surface area is about 75% of that of the cerebral cortex. Anatomically, the cerebellum is divided into three parts by two transverse fis-sures. The posterolateral fissure separates the medial nodulus and the lateral flocculus on either side from the rest of the cer-ebellum, and the primary fissure divides the remainder into an anterior and a posterior lobe. Lesser fissures divide the vermis into smaller sections, so that it contains 10 primary lobules numbered I–X from superior to inferior. These lobules are identified by name and number in Figure 16–13. ORGANIZATION The cerebellum has an external cerebellar cortex separated by white matter from the deep cerebellar nuclei. Its primary FIGURE 16–11 Probable basal ganglia-thalamocortical circuitry in Parkinson disease. Solid arrows indicate excitatory outputs and dashed arrows inhibitory outputs. The strength of each output is in-dicated by the width of the arrow. GPe, external segment of the globus pallidus; GPi, internal segment of the globus pallidus; SNC, pars compac-ta of the substantia nigra; STN, subthalamic nucleus; PPN, peduncu-lopontine nuclei; Thal, thalamus. See text for details. (Modified from Grafton SC, DeLong M: Tracing the brain circuitry with functional imaging. Nat Med 1997;3:602.) Normal CORTEX PUTAMEN SNC D2 GPe STN D1 PPN Brain stem, spinal cord Brain stem, spinal cord Thal Parkinsonism CORTEX SNC GPe STN PPN Thal PUTAMEN D2 D1 GPi GPi FIGURE 16–12 Diagrammatic representation of the principal parts of the brain. The parts are distorted to show the cerebellar pedun-cles and the way the cerebellum, pons, and middle peduncle form a “nap-kin ring” around the brain stem. (Reproduced with permission, from Goss CM [editor]: Gray’s Anatomy of the Human Body, 27th ed. Lea & Febiger, 1959.) CEREBRUM Cerebellar peduncles Cerebral peduncle Superior peduncle Middle peduncle Inferior peduncle Medulla oblongata PONS CEREBELLUM CHAPTER 16 Control of Posture & Movement 255 afferent inputs, the mossy and climbing fibers, send collaterals to the deep nuclei and pass to the cortex. There are four deep nuclei: the dentate, the globose, the emboliform, and the fas-tigial nuclei. The globose and the emboliform nuclei are sometimes lumped together as the interpositus nucleus. The cerebellar cortex contains five types of neurons: Purkinje, granule, basket, stellate, and Golgi cells. It has three layers (Figure 16–14): an external molecular layer, a Purkinje cell layer that is only one cell thick, and an internal granular layer. The Purkinje cells are among the biggest neurons in the body. They have very extensive dendritic arbors that extend throughout the molecular layer. Their axons, which are the only output from the cerebellar cortex, generally pass to the deep nuclei. The cerebellar cortex also contains granule cells, which receive input from the mossy fibers and innervate the Purkinje cells. The granule cells have their cell bodies in the granular layer. Each sends an axon to the molecular layer, where the axon bifurcates to form a T. The branches of the T are straight and run long distances. Consequently, they are called parallel fibers. The dendritic trees of the Purkinje cells are markedly flattened (Figure 16–14) and oriented at right angles to the parallel fibers. The parallel fibers thus make syn-aptic contact with the dendrites of many Purkinje cells, and the parallel fibers and Purkinje dendritic trees form a grid of remarkably regular proportions. The other three types of neurons in the cerebellar cortex are in effect inhibitory interneurons. Basket cells (Figure 16–14) are located in the molecular layer. They receive input from the parallel fibers and each projects to many Purkinje cells (Fig-ure 16–15). Their axons form a basket around the cell body FIGURE 16–13 Superior and inferior views and sagittal section of the human cerebellum. The 10 principal lobules are identified by name and by number (I–X). Superior surface Inferior surface MIDSAGITTAL SECTION Centralis Culmen Culmen Primary fissure Declive Fourth ventricle Superior cerebellar peduncle Middle cerebellar peduncle Nodulus of vermis Uvula Flocculus Anterior medullary velum Inferior cerebellar peduncle II III IV V Primary fissure Lobulus simplex VI Folium Tuber VII Pyramis VIII Prepyramidal sulcus Uvula IX Posterolateral fissure Nodulus X Fourth ventricle Pons Lingula I 256 SECTION III Central & Peripheral Neurophysiology and axon hillock of each Purkinje cell they innervate. Stellate cells are similar to the basket cells but more superficial in location. Golgi cells are located in the granular layer (Figure 16–14). Their dendrites, which project into the molecular layer, receive input from the parallel fibers (Figure 16–15). Their cell bodies receive input via collaterals from the incom-ing mossy fibers and the Purkinje cells. Their axons project to the dendrites of the granule cells. The two main inputs to the cerebellar cortex are climbing fibers and mossy fibers. Both are excitatory (Figure 16–15). The climbing fibers come from a single source, the inferior olivary nuclei. Each projects to the primary dendrites of a Purkinje cell, around which it entwines like a climbing plant. Proprioceptive input to the inferior olivary nuclei comes from all over the body. On the other hand, the mossy fibers provide direct proprioceptive input from all parts of the body plus input from the cerebral cortex via the pontine nuclei to the cerebellar cortex. They end on the dendrites of granule cells in complex synaptic groupings called glomeruli. The glomeruli also contain the inhibitory endings of the Golgi cells men-tioned above. The fundamental circuits of the cerebellar cortex are thus rel-atively simple (Figure 16–15). Climbing fiber inputs exert a strong excitatory effect on single Purkinje cells, whereas mossy FIGURE 16–14 Location and structure of five neuronal types in the cerebellar cortex. Drawings are based on Golgi-stained prepara-tions. Purkinje cells (1) have processes aligned in one plane; their axons are the only output from the cerebellum. Axons of granule cells (4) traverse and make connections with Purkinje cell processes in molecular layer. Golgi (2), basket (3), and stellate (5) cells have characteristic positions, shapes, branching patterns, and synaptic connections. (Reproduced with permission from Kuffler SW, Nicholls JG, Martin AR: From Neuron to Brain, 2nd ed. Sinauer, 1984.) 1 2 3 4 5 5 Stellate cell 1 Purkinje cell 2 Golgi cell 3 Basket cell 4 Granule cell Axon Axons Axon Molecular layer Purkinje layer Granular layer Parallel fibers: Axons of granule cells FIGURE 16–15 Diagram of neural connections in the cerebellum. Plus (+) and minus (–) signs indicate whether endings are excitatory or inhibitory. BC, basket cell; GC, Golgi cell; GR, granule cell; NC, cell in deep nucleus; PC, Purkinje cell. Note that PCs and BCs are in-hibitory. The connections of the stellate cells, which are not shown, are similar to those of the basket cells, except that they end for the most part on Purkinje cell dendrites. Other inputs Mossy fiber + + + + + + + + + + Climbing fiber − − − Parallel fiber PC BC GR GC NC Cerebellar cortex Deep nuclei CHAPTER 16 Control of Posture & Movement 257 fiber inputs exert a weak excitatory effect on many Purkinje cells via the granule cells. The basket and stellate cells are also excited by granule cells via the parallel fibers, and their output inhibits Purkinje cell discharge (feed-forward inhibition). Golgi cells are excited by the mossy fiber collaterals, Purkinje cell col-laterals, and parallel fibers, and they inhibit transmission from mossy fibers to granule cells. The transmitter secreted by the stellate, basket, Golgi, and Purkinje cells is GABA, whereas the granule cells secrete glutamate. GABA acts via GABAA recep-tors, but the combinations of subunits in these receptors vary from one cell type to the next. The granule cell is unique in that it appears to be the only type of neuron in the CNS that has a GABAA receptor containing the α6 subunit. The output of the Purkinje cells is in turn inhibitory to the deep cerebellar nuclei. As noted above, these nuclei also receive excitatory inputs via collaterals from the mossy and climbing fibers. It is interesting, in view of their inhibitory Purkinje cell input, that the output of the deep cerebellar nuclei to the brain stem and thalamus is always excitatory. Thus, almost all the cer-ebellar circuitry seems to be concerned solely with modulating or timing the excitatory output of the deep cerebellar nuclei to the brain stem and thalamus. The primary afferent systems that converge to form the mossy fiber or climbing fiber input to the cerebellum are summarized in Table 16–2. FUNCTIONAL DIVISIONS From a functional point of view, the cerebellum is divided into three parts (Figure 16–16). The nodulus in the vermis and the flanking flocculus in the hemisphere on each side form the ves-tibulocerebellum (or flocculonodular lobe). This lobe, which is phylogenetically the oldest part of the cerebellum, has vestibular connections and is concerned with equilibrium and eye move-ments. The rest of the vermis and the adjacent medial portions of the hemispheres form the spinocerebellum, the region that receives proprioceptive input from the body as well as a copy of the “motor plan” from the motor cortex. By comparing plan with performance, it smoothes and coordinates movements that are ongoing. The vermis projects to the brain stem area con-cerned with control of axial and proximal limb muscles (medial brain stem pathways), whereas the hemispheres project the brain stem areas concerned with control of distal limb muscles (lateral brain stem pathways). The lateral portions of the cerebellar hemispheres are called the cerebrocerebellum. They are the newest from a phylogenetic point of view, reaching their greatest development in humans. They interact with the motor cortex in planning and programming movements. Most of the vestibulocerebellar output passes directly to the brain stem, but the rest of the cerebellar cortex projects to the deep nuclei, which in turn project to the brain stem. The deep nuclei provide the only output for the spinocerebellum and the cerebrocerebellum. The medial portion of the spinocer-ebellum projects to the fastigial nuclei and from there to the brain stem. The adjacent hemispheric portions of the spino-cerebellum project to the emboliform and globose nuclei and from there to the brain stem. The cerebrocerebellum projects to the dentate nucleus and from there either directly or indi-rectly to the ventrolateral nucleus of the thalamus. MECHANISMS Although the functions of the flocculonodular lobe, spinocer-ebellum, and cerebrocerebellum are relatively clear and the cere-bellar circuits are simple, the exact ways their different parts carry out their functions are still unknown. The relation of the TABLE 16–2 Function of principal afferent systems to the cerebellum.a Afferent Tracts Transmits Vestibulocerebellar Vestibular impulses from labyrinths, di-rect and via vestibular nuclei Dorsal spinocerebellar Proprioceptive and exteroceptive impuls-es from body Ventral spinocerebellar Proprioceptive and exteroceptive impuls-es from body Cuneocerebellar Proprioceptive impulses, especially from head and neck Tectocerebellar Auditory and visual impulses via inferior and superior colliculi Pontocerebellar Impulses from motor and other parts of cerebral cortex via pontine nuclei Olivocerebellar Proprioceptive input from whole body via relay in inferior olive aThe olivocerebellar pathway projects to the cerebellar cortex via climbing fibers; the rest of the listed paths project via mossy fibers. Several other pathways transmit impulses from nuclei in the brain stem to the cerebellar cortex and to the deep nuclei, including a serotonergic input from the raphé nuclei to the granular and molecular layers and a noradrenergic input from the locus ceruleus to all three layers. FIGURE 16–16 Functional divisions of the cerebellum. (Modified from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Spinocerebellum To medial descending systems To lateral descending systems To motor and premotor cortices To vestibular nuclei Motor execution Motor planning Balance and eye movements Cerebrocerebellum Vestibulocerebellum 258 SECTION III Central & Peripheral Neurophysiology electrical events in the cerebellum to its function in motor control is another interesting problem. The cerebellar cortex has a basic, 150 to 300/s, 200-μV electrical rhythm and, super-imposed on this, a 1000 to 2000/s component of smaller am-plitude. The frequency of the basic rhythm is thus more than 10 times greater than that of the similarly recorded cerebral cortical alpha rhythm. Incoming stimuli generally alter the amplitude of the cerebellar rhythm like a broadcast signal modulating a carrier frequency in radio transmission. Howev-er, the significance of these electrical phenomena in terms of cerebellar function is unknown. CEREBELLAR DISEASE Damage to the cerebellum leads to several characteristic abnor-malities, including hypotonia, ataxia, and intention tremor. Figure 16–17 illustrates some of these abnormalities. Most abnor-malities are apparent during movement. The marked ataxia is characterized as incoordination due to errors in the rate, range, force, and direction of movement. Voluntary movements are also highly abnormal. For example, attempting to touch an object with a finger results in overshooting to one side or the other. This dysmetria, which is also called past-pointing, promptly initiates a gross corrective action, but the correction overshoots to the oth-er side. Consequently, the finger oscillates back and forth. This oscillation is the intention tremor of cerebellar disease. Another characteristic of cerebellar disease is inability to “put on the brakes,” that is, to stop movement promptly. Normally, for exam-ple, flexion of the forearm against resistance is quickly checked when the resistance force is suddenly broken off. The patient with cerebellar disease cannot brake the movement of the limb, and the forearm flies backward in a wide arc. This abnormal response is known as the rebound phenomenon, and similar impairment is detectable in other motor activities. This is one of the important reasons these patients show dysdiadochokinesia, the inability to perform rapidly alternating opposite movements such as repeat-ed pronation and supination of the hands. Finally, patients with cerebellar disease have difficulty performing actions that involve simultaneous motion at more than one joint. They dissect such movements and carry them out one joint at a time, a phenome-non known as decomposition of movement. Other signs of cerebellar deficit in humans provide addi-tional illustrations of the importance of the cerebellum in the control of movement. Ataxia is manifest not only in the wide-based, unsteady, “drunken” gait of patients, but also in defects of the skilled movements involved in the production of speech, so that slurred, scanning speech results. Motor abnormalities associated with cerebellar damage vary depending on the region involved. The major dysfunction seen after damage to the vestibulocerebellum is ataxia, dysequilib-rium, and nystagmus. Damage to the vermis and fastigial nucleus (part of the spinocerebellum) leads to disturbances in control of axial and trunk muscles during attempted antigrav-ity postures and scanning speech. Degeneration of this portion of the cerebellum can result from thiamine deficiency in alco-holics or malnourished individuals. The major dysfunction seen after damage to the cerebrocerebellum is delays in initiat-ing movements and decomposition of movement. THE CEREBELLUM & LEARNING The cerebellum is concerned with learned adjustments that make coordination easier when a given task is performed over and over. As a motor task is learned, activity in the brain shifts from the prefrontal areas to the parietal and motor cortex and the cerebellum. The basis of the learning in the cerebellum is probably the input via the olivary nuclei. It is worth noting that FIGURE 16–17 Typical defects associated with cerebellar disease. A) Lesion of the right cerebellar hemisphere delays initiation of movement. The patient is told to clench both hands simultaneously; right hand clenches later than left (shown by recordings from a pressure bulb transducer squeezed by the patient). B) Dysmetria and decomposition of movement shown by patient moving his arm from a raised position to his nose. Tremor increases on approaching the nose. C) Dysdiadochokinesia occurs in the abnormal position trace of hand and forearm as a cerebellar subject tries alternately to pronate and supinate forearm while flexing and extending elbow as rapidly as possible. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) A B C Go Normal Abnormal Normal Normal Abnormal Abnormal Delay Start Finish CHAPTER 16 Control of Posture & Movement 259 each Purkinje cell receives inputs from 250,000 to 1 million mossy fibers, but each has only a single climbing fiber from the inferior olive, and this fiber makes 2000–3000 synapses on the Purkinje cell. Climbing fiber activation produces a large, com-plex spike in the Purkinje cell and this spike in some way pro-duces long-term modification of the pattern of mossy fiber input to that particular Purkinje cell. Climbing fiber activity is increased when a new movement is being learned, and selective lesions of the olivary complex abolish the ability to produce long-term adjustments in certain motor responses. CHAPTER SUMMARY ■The ventral corticospinal tract and medial descending brain stem pathways (tectospinal, reticulospinal, and vestibulospinal tracts) regulate proximal muscles and posture. The lateral corti-cospinal and rubrospinal tracts control distal limb muscles and skilled voluntary movements. ■Spinal cord transection is followed by a period of spinal shock during which all spinal reflex responses are profoundly depressed. ■Decerebrate rigidity leads to hyperactivity in extensor muscles in all four extremities; it is actually spasticity due to facilitation of the myotatic stretch reflex. Decorticate posturing or decorti-cate rigidity is flexion of the upper extremities at the elbow and extensor hyperactivity in the lower extremities. ■The basal ganglia include the caudate nucleus, putamen, globus pallidus, subthalamic nucleus, and substantia nigra. The connec-tions between the parts of the basal ganglia include a dopaminergic nigrostriatal projection from the substantia nigra to the striatum and a GABAergic projection from the striatum to substantia nigra. ■Parkinson disease is due to degeneration of the nigrostriatal do-paminergic neurons and is characterized by akinesia, bradykin-esia, cogwheel rigidity, and tremor at rest. Huntington disease is characterized by choreiform movements due to the loss of the GABAergic pathway to the globus pallidus. ■The cerebellar cortex contains five types of neurons: Purkinje, granule, basket, stellate, and Golgi cells. The two main inputs to the cerebellar cortex are climbing fibers and mossy fibers. Purkinje cells are the only output from the cerebellar cortex and they generally project to the deep nuclei. ■Damage to the cerebellum leads to several characteristic abnor-malities, including hypotonia, ataxia, and intention tremor. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. A primary function of the basal ganglia is A) sensory integration. B) short-term memory. C) planning voluntary movement. D) neuroendocrine control. E) slow-wave sleep. 2. The therapeutic effect of L-dopa in patients with Parkinson dis-ease eventually wears off because A) antibodies to dopamine receptors develop. B) inhibitory pathways grow into the basal ganglia from the frontal lobe. C) there is an increase in circulating α-synuclein. D) the normal action of nerve growth factor (NGF) is dis-rupted. E) the dopaminergic neurons in the substantia nigra continue to degenerate. 3. Increased neural activity before a skilled voluntary movement is first seen in the A) spinal motor neurons. B) precentral motor cortex. C) midbrain. D) cerebellum. E) cortical association areas. 4. After falling down a flight of stairs, a young woman is found to have partial loss of voluntary movement on the right side of her body and loss of pain and temperature sensation on the left side below the midthoracic region. It is probable that she has a lesion A) transecting the left half of the spinal cord in the lumbar region. B) transecting the left half of the spinal cord in the upper tho-racic region. C) transecting sensory and motor pathways on the right side of the pons. D) transecting the right half of the spinal cord in the upper tho-racic region. E) transecting the dorsal half of the spinal cord in the upper thoracic region. 5. Patients with transected spinal cords frequently have a negative nitrogen balance because A) they develop hypercalcemia, and this causes dissolution of the protein in bone. B) they are paralyzed below the level of the transection. C) they lack the afferent input that normally maintains growth hormone secretion. D) they have difficulty voiding, and this causes nitrogen to accumulate in the urine in the bladder. E) their corticotropin response to stress is reduced. 6. Which of the following diseases is not known to be caused by overexpression of a trinucleotide repeat? A) Alzheimer disease B) Fragile X syndrome C) Spinocerebellar ataxia, type 3 D) Huntington disease E) Friedreich ataxia CHAPTER RESOURCES Alexi T, et al: Neuroprotective strategies for basal ganglia degeneration: Parkinson’s and Huntington’s diseases. Prog Neurbiol 2000;60:409. De Zeeuw CI, Strata P, Voogd J: The Cerebellum: From Structure to Control. Elsevier, 1997. Ditunno JF Jr, Formal CF: Chronic spinal cord injury. N Engl J Med 1994; 330:550. 260 SECTION III Central & Peripheral Neurophysiology Graybiel AM, Delong MR, Kitai ST: The Basal Ganglia VI. Springer, 2003. Haines DE (editor): Fundamental Neuroscience for Basic and Clinical Applications, 3rd ed. Elsevier, 2006. He SQ, Dum RP, Strick PL: Topographic organization of corticospinal projections from the frontal lobe: Motor areas on the lateral surface of the hemisphere. J Neurosci 1993; 13: 952. Holstege G, Kuypers HGJM: The anatomy of brain stem pathways to the spinal cord in cat. A labeled amino acid tracing study. Prog Brain Res 1982;57:145. Jueptner M, Weiller C: A review of differences between basal ganglia and cerebellar control of movements as revealed by functional imaging studies. Brain 1998;121:1437. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. Lemon RN: Descending pathways in motor control. Annu Rev Neurosci 2008;31:195. Manto MU, Pandolfo M: The Cerebellum and its Disorders. Cambridge University Press, 2001. McDonald JW, et al: Transplanted embryonic stem cells survive, differentiate and promote recovery in injured rat spinal cord. Nature Med 1999;5:1410. Nicholls JG, Martin AR, Wallace BG: From Neuron to Brain: A Cellular and Molecular Approach to the Function of the Nervous System, 4th ed. Sinauer, 2001. Nudo RJ: Postinfarct cortical plasticity and behavioral recovery. Stroke 2007;38:840. Ramer LM, Ramer MS, Steeves JD: Setting the stage for functional repair of spinal cord injuries: a cast of thousands. Spinal Cord 2005;43:134. 261 C H A P T E R 17 The Autonomic Nervous System O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the location of the cell bodies and axonal trajectories of preganglionic sympathetic and parasympathetic neurons. ■Describe the location and trajectories of postganglionic sympathetic and para-sympathetic neurons. ■Name the neurotransmitters that are released by preganglionic autonomic neu-rons, postganglionic sympathetic neurons, postganglionic parasympathetic neu-rons, and adrenal medullary cells. ■Outline the functions of the autonomic nervous system. ■List the ways that drugs act to increase or decrease the activity of the components of the autonomic nervous system. ■Describe the location of neurons that provide input to sympathetic preganglionic neurons. ■Describe the composition and functions of the enteric nervous system. INTRODUCTION The autonomic nervous system (ANS) is the part of the ner-vous system that is responsible for homeostasis. Except for skeletal muscle, which gets its innervation from the somato-motor nervous system, innervation to all other organs is sup-plied by the ANS. Nerve terminals are located in smooth muscle (eg, blood vessels, gut wall, urinary bladder), cardiac muscle, and glands (eg, sweat glands, salivary glands). Although survival is possible without an ANS, the ability to adapt to environmental stressors and other challenges is severely compromised (see Clinical Box 17–1). The ANS has two major divisions: the sympathetic and parasympathetic nervous systems. As will be described, some target organs are innervated by both divisions and others are controlled by only one. In addition, the ANS includes the enteric nervous sys-tem within the gastrointestinal tract. The classic definition of the ANS is the preganglionic and postganglionic neurons within the sympathetic and parasympathetic divisions. This would be equivalent to defining the somatomotor nervous sys-tem as the cranial and spinal motor neurons. A modern defini-tion of the ANS takes into account the descending pathways from several forebrain and brain stem regions as well as vis-ceral afferent pathways that set the level of activity in sympa-thetic and parasympathetic nerves. This is analogous to including the many descending and ascending pathways that influence the activity of somatic motor neurons as elements of the somatomotor nervous system. 262 SECTION III Central & Peripheral Neurophysiology ANATOMIC ORGANIZATION OF AUTONOMIC OUTFLOW GENERAL FEATURES Figure 17–1 compares some fundamental characteristics of the innervation to skeletal muscles and innervation to smooth muscle, cardiac muscle, and glands. As discussed in earlier chapters, the final common pathway linking the central ner-vous system (CNS) to skeletal muscles is the α-motor neuron. Similarly, sympathetic and parasympathetic neurons serve as the final common pathway from the CNS to visceral targets. However, in marked contrast to the somatomotor nervous system, the peripheral motor portions of the ANS are made up of two neurons: preganglionic and postganglionic neurons. CLINICAL BOX 17–1 Multiple System Atrophy & Shy–Drager Syndrome Multiple system atrophy (MSA) is a neurodegenerative dis-order associated with autonomic failure due to loss of preganglionic autonomic neurons in the spinal cord and brain stem. In the absence of an autonomic nervous system, it is difficult to regulate body temperature, fluid and electro-lyte balance, and blood pressure. In addition to these auto-nomic abnormalities, MSA presents with cerebellar, basal ganglia, locus coeruleus, inferior olivary nucleus, and pyrami-dal tract deficits. MSA is defined as “a sporadic, progressive, adult onset disorder characterized by autonomic dysfunc-tion, parkinsonism, and cerebellar ataxia in any combina-tion.” Shy–Drager syndrome is a subtype of MSA in which autonomic failure dominates. The pathological hallmark of MSA is cytoplasmic and nuclear inclusions in oligodendro-cytes and neurons in central motor and autonomic areas. There is also depletion of monoaminergic, cholinergic, and peptidergic markers in several brain regions and in the cere-brospinal fluid. Basal levels of sympathetic activity and plasma norepinephrine levels are normal in MSA patients, but they fail to increase in response to standing or other stimuli and leads to severe orthostatic hypotension. In addition to the fall in blood pressure, orthostatic hypotension leads to dizziness, dimness of vision, and even fainting. MSA is also ac-companied by parasympathetic dysfunction, including uri-nary and sexual dysfunction. MSA is most often diagnosed in individuals between 50 and 70 years of age; it affects more men than women. Erectile dysfunction is often the first symp-tom of the disease. There are also abnormalities in barorecep-tor reflex and respiratory control mechanisms. Although auto-nomic abnormalities are often the first symptoms, 75% of patients with MSA also experience motor disturbances. FIGURE 17–1 Comparison of peripheral organization and transmitters released by somatomotor and autonomic nervous systems (NS). ACh, acetylcholine; DA, dopamine; NE, norepinephrine; Epi, epinephrine. (From Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology. McGraw-Hill, 2008.) Autonomic nervous system: Parasympathetic division Somatic nervous system CNS CNS Effector organ Effector organ Effector organ Effector organ ACh ACh Ganglion NE ACh Adrenal medulla Epi (also NE, DA, peptides) (via bloodstream) ACh Ganglion CNS Autonomic nervous system: Sympathetic division CHAPTER 17 The Autonomic Nervous System 263 The cell bodies of the preganglionic neurons are located in the intermediolateral column (IML) of the spinal cord and in mo-tor nuclei of some cranial nerves. In contrast to the large diam-eter and rapidly conducting α-motor neurons, preganglionic axons are small-diameter, myelinated, relatively slowly con-ducting B fibers. A preganglionic axon diverges to an average of eight or nine postganglionic neurons. In this way, auto-nomic output is diffused. The axons of the postganglionic neurons are mostly unmyelinated C fibers and terminate on the visceral effectors. SYMPATHETIC DIVISION In contrast to α-motor neurons, which are located at all spinal segments, sympathetic preganglionic neurons are located in the IML of only the first thoracic to the third or fourth lumbar segments. This is why the sympathetic nervous system is sometimes called the thoracolumbar division of the ANS. The axons of the sympathetic preganglionic neurons leave the spi-nal cord at the level at which their cell bodies are located and exit via the ventral root along with axons of α- and γ-motor neurons (Figure 17–2). They then separate from the ventral root via the white rami communicans and project to the ad-jacent sympathetic paravertebral ganglion, where some of them end on the cell bodies of the postganglionic neurons. Paravertebral ganglia are located adjacent to each thoracic and upper lumbar spinal segments; in addition, there are a few ganglia adjacent to the cervical and sacral spinal segments. These ganglia form the sympathetic chain bilaterally. The ganglia are connected to each other via the axons of pregangli-onic neurons that travel rostrally or caudally to terminate on postganglionic neurons located at some distance. This ar-rangement is seen in Figures 17–2 and 17–3. Some preganglionic neurons pass through the paravertebral ganglion chain and end on postganglionic neurons located in prevertebral (or collateral) ganglia close to the viscera, including the celiac, superior mesenteric, and inferior mesen-teric ganglia (Figure 17–3). There are also preganglionic neu-rons whose axons terminate directly on the effector organ, the adrenal gland. The axons of some of the postganglionic neurons leave the chain ganglia and reenter the spinal nerves via the gray rami FIGURE 17–2 Projection of sympathetic preganglionic and postganglionic fibers. The drawing shows the thoracic spinal cord, paraver-tebral, and prevertebral ganglia. Preganglionic neurons are shown in red, postganglionic neurons in dark blue, afferent sensory pathways in blue, and interneurons in black. (Reproduced with permission from Boron WF, Boulpaep EL: Medical Physiology. Elsevier, 2005.) 264 SECTION III Central & Peripheral Neurophysiology FIGURE 17–3 Organization of sympathetic (left) and parasympathetic (right) nervous systems. Preganglionic sympathetic and para-sympathetic neurons are shown in red and orange, respectively; postganglionic sympathetic and parasympathetic neurons in blue and green, re-spectively. (Reproduced with permission from Boron WF, Boulpaep EL: Medical Physiology. Elsevier, 2005.) CHAPTER 17 The Autonomic Nervous System 265 communicans and are distributed to autonomic effectors in the areas supplied by these spinal nerves (Figure 17–2). These postganglionic sympathetic nerves terminate mainly on smooth muscle (eg, blood vessels, hair follicles, airways) and on sweat glands in the limbs. Other postganglionic fibers leave the chain ganglia to enter the thoracic cavity to termi-nate in visceral organs. Postganglionic fibers from preverte-bral ganglia also terminate in visceral targets. PARASYMPATHETIC DIVISION The parasympathetic nervous system is sometimes called the craniosacral division of the ANS because of the location of its preganglionic neurons (Figure 17–3). The parasympathetic nerves supply the visceral structures in the head via the oculo-motor, facial, and glossopharyngeal nerves, and those in the thorax and upper abdomen via the vagus nerves. The sacral outflow supplies the pelvic viscera via branches of the second to fourth sacral spinal nerves. Parasympathetic preganglionic fibers synapse on ganglia cells clustered within the walls of vis-ceral organs; thus these parasympathetic postganglionic fibers are very short. CHEMICAL TRANSMISSION AT AUTONOMIC JUNCTIONS ACETYLCHOLINE & NOREPINEPHRINE The first evidence for chemical neurotransmission was provid-ed by a simple yet dramatic study of heart rate control by the parasympathetic nervous system performed by Otto Loewi in 1920 (Clinical Box 17–2). Transmission at the synaptic junc-tions between pre- and postganglionic neurons and between the postganglionic neurons and the autonomic effectors is chemi-cally mediated. The principal transmitter agents involved are acetylcholine and norepinephrine (Figures 17–1 and 17–4). CLINICAL BOX 17–2 Pharmacological Control of Heart Rate Using drugs to control heart rate and other physiological pro-cesses is a very common therapy. It holds its roots in an obser-vation made by Otto Loewi in 1920 that served as the founda-tion for chemical transmission of nerve impulses. He provided the first decisive evidence that a chemical messenger was re-leased by cardiac nerves to affect heart rate. The experimental design came to him in a dream on Easter Sunday of that year. He awoke from the dream, jotted down notes, but the next morning they were indecipherable. The next night, the dream recurred and he went to his laboratory at 3:00 AM to conduct a simple experiment on a frog heart. He isolated the hearts from two frogs, one with and one without its innervation. Both hearts were attached to cannulas filled with Ringer solution. The vagus nerve of the first heart was stimulated, and then the Ringer solution from that heart was transferred to the nonin-nervated heart. The rate of its contractions slowed as if its vagus nerve had been stimulated. Loewi also showed that when the sympathetic nerve of the first heart was stimulated and its effluent was passed to the second heart, the rate of contractions of the “donor” heart increased as if its sympa-thetic fibers had been stimulated. These results proved that nerve terminals release chemicals which cause the well-known modifications of cardiac function that occur in re-sponse to stimulation of its nerve supply. Loewi called the chemical release by the vagus nerve Vagusstoff. Not long after, it was identified chemically to be acetylcholine. FIGURE 17–4 Chemical coding of sympathetic preganglionic and postganglionic neurons. CNS, central nervous system; PNS, periph-eral nervous system. (Reproduced with permission from Haines DE [editor]: Fundamental Neuroscience for Basic and Clinical Applications, 3rd ed. Elsevier, 2006.) 266 SECTION III Central & Peripheral Neurophysiology The neurons that are cholinergic (ie, release acetylcholine) are (1) all preganglionic neurons, (2) all parasympathetic postgan-glionic neurons, (3) sympathetic postganglionic neurons that innervate sweat glands, and (4) sympathetic postganglionic neurons that end on blood vessels in some skeletal muscles and produce vasodilation when stimulated (sympathetic vasodilator nerves). The remaining sympathetic postganglionic neurons are noradrenergic (ie, release norepinephrine). The adrenal me-dulla is essentially a sympathetic ganglion in which the postgan-glionic cells have lost their axons and secrete norepinephrine and epinephrine directly into the bloodstream. The cholinergic preganglionic neurons to these cells have consequently become the secretomotor nerve supply of this gland. Transmission in autonomic ganglia is mediated primarily by N2 nicotinic cholinergic receptors that are blocked by hexame-thonium. This is in contrast to the N1 nicotinic cholinergic receptors at the neuromuscular junction, which are blocked by D-tubocurare. The release of acetylcholine from postganglionic fibers acts on muscarinic receptors, which are blocked by atro-pine. The release of norepinephrine from sympathetic postgan-glionic fibers acts on α1, β1, or β2 adrenoreceptors, depending on the target organ. Table 17–1 shows the types of receptors at various junctions within the autonomic nervous system. In addition to these “classical neurotransmitters, ” some auto-nomic fibers also release neuropeptides. Figure 17–4 shows some examples for sympathetic postganglionic fibers. The small granulated vesicles in postganglionic noradrenergic neurons contain ATP and norepinephrine, and the large granulated vesi-cles contain neuropeptide Y. There is evidence that low-fre-quency stimulation promotes release of ATP , whereas high-frequency stimulation causes release of neuropeptide Y. The vis-cera contains purinergic receptors, and evidence is accumulating that ATP is a mediator in the autonomic nervous system along with norepinephrine. However, its exact role is unsettled. Acetylcholine does not usually circulate in the blood, and the effects of localized cholinergic discharge are generally discrete and of short duration because of the high concentration of acetylcho-linesterase at cholinergic nerve endings. Norepinephrine spreads farther and has a more prolonged action than acetylcholine. Nor-epinephrine, epinephrine, and dopamine are all found in plasma. The epinephrine and some of the dopamine come from the adre-nal medulla, but most of the norepinephrine diffuses into the bloodstream from noradrenergic nerve endings. Metabolites of norepinephrine and dopamine also enter the circulation, some from the sympathetic nerve endings and some from smooth mus-cle cells (Figure 17–5). It is worth noting that even when monoamine oxidase (MAO) and catechol-O-methyltransferase (COMT) are both inhibited, the metabolism of norepinephrine is still rapid. However, inhibition of reuptake prolongs its half-life. TRANSMISSION IN SYMPATHETIC GANGLIA At least in experimental animals, the responses produced in postganglionic neurons by stimulation of their preganglionic innervation include both a rapid depolarization (fast excitatory postsynaptic potential [EPSP]) that generates action potentials and a prolonged excitatory postsynaptic potential (slow EPSP). The slow response apparently modulates and regulates trans-mission through the sympathetic ganglia. As just described, the initial depolarization is produced by acetylcholine via the N2 nicotinic receptor. The slow EPSP is produced by acetylcholine acting on a muscarinic receptor on the membrane of the post-ganglionic neuron. The junctions in the peripheral autonomic motor pathways are a logical site for pharmacologic manipulation of visceral func-tion. The transmitter agents are synthesized, stored in the nerve endings, and released near the neurons, muscle cells, or gland cells on which they act. They bind to receptors on these cells, thus initiating their characteristic actions, and they are then removed from the area by reuptake or metabolism. Each of these steps can be stimulated or inhibited, with predictable consequences. Some of the drugs and toxins that affect the activity of the autonomic nervous system and the mechanisms by which they produce their effects are listed in Table 17–2. Compounds with muscarinic actions include congeners of acetylcholine and drugs that inhibit acetylcholinesterase. Among the latter are the insecticide parathion and diisopropyl fluorophosphate (DFP), a component of the so-called nerve gases, which kill by produc-ing massive inhibition of acetylcholinesterase. RESPONSES OF EFFECTOR ORGANS TO AUTONOMIC NERVE IMPULSES GENERAL PRINCIPLES The effects of stimulation of the noradrenergic and choliner-gic postganglionic nerve fibers are indicated in Figure 17–3 and Table 17–1. These findings point out another difference between the ANS and the somatomotor nervous system. The release of acetylcholine by α-motor neurons only leads to con-traction of skeletal muscles. In contrast, release of acetylcho-line onto smooth muscle of some organs leads to contraction (eg, walls of the gastrointestinal tract) while release onto other organs leads to relaxation (eg, sphincters in the gastrointesti-nal tract). The only way to relax a skeletal muscle is to inhibit the discharges of the α-motor neurons; but for some targets innervated by the ANS, one can shift from contraction to re-laxation by switching from activation of the parasympathetic nervous system to activation of the sympathetic nervous sys-tem. This is the case for the many organs which receive dual innervation with antagonistic effects, including the digestive tract, airways, and urinary bladder. The heart is another exam-ple of an organ with dual antagonistic control. Stimulation of sympathetic nerves increases heart rate, and stimulation of parasympathetic nerves decreases heart rate. In other cases, the effects of sympathetic and parasympa-thetic activation can be considered complementary. An example CHAPTER 17 The Autonomic Nervous System 267 TABLE 17–1 Responses of some effector organs to autonomic nerve activity. Sympathetic Nervous System Effector Organs Parasympathetic Nervous System Receptor Type Response Eyes Radial muscle of iris —a α1 Contraction (mydriasis) Sphincter muscle of iris Contraction (miosis) — Ciliary muscle Contraction for near vision — Heart S–A node Decreases heart rate β1 Increases heart rate Atria & ventricle Decreases contractility β1, β2 Increases contractility AV node & Purkinje Decreases conduction velocity β1, β2 Increases conduction velocity Arterioles Coronary — α1, α2 Constriction β2 Dilation Skin — α1, α2 Constriction Skeletal muscle — α1 Constriction β2, M Dilation Abdominal viscera — α1 Constriction Salivary glands Dilation α1, α2 Constriction Renal — α1 Constriction Systemic veins — α1, α2 Constriction β2 Dilation Lungs Bronchial muscle Contraction β2 Relaxation Stomach Motility and tone Increases α1, α2, β2 Decreases Sphincters Relaxation α1 Contraction Secretion Stimulation ? Inhibition Intestine Motility and tone Increases α1, α2, β1, β2 Decreases Sphincters Relaxation α1 Contraction (usually) Secretion Stimulation α2 Inhibition Gall bladder Contraction β2 Relaxation Urinary bladder Detrusor Contraction β2 Relaxation Sphincter Relaxation α1 Contraction (continued) 268 SECTION III Central & Peripheral Neurophysiology is the innervation of salivary glands. Parasympathetic activa-tion causes release of watery saliva, while sympathetic activa-tion causes the production of thick, viscous saliva. The two divisions of the ANS can also act in a synergistic or cooperative manner in the control of some functions. One example is the control of pupil diameter in the eye. Both sym-pathetic and parasympathetic innervations are excitatory, but the former contracts the radial muscle to cause mydriasis and the latter contracts the sphincter (or constrictor) muscle to cause meiosis. Another example is the synergistic actions of these nerves on sexual function. Activation of parasympa-thetic nerves to the penis increases blood flow and leads to erection while activation of sympathetic nerves to the penis causes ejaculation. There are also several organs that are innervated by only one division of the ANS. In addition to the adrenal gland, most blood vessels, the pilomotor muscles in the skin (hair follicles), and sweat glands are innervated exclusively by sympathetic nerves. The lacrimal muscle (tear gland), ciliary muscle (for accommodation for near vision), and the sublingual salivary gland are innervated exclusively by parasympathetic nerves. PARASYMPATHETIC CHOLINERGIC & SYMPATHETIC NORADRENERGIC DISCHARGE In a general way, the functions promoted by activity in the cholinergic division of the autonomic nervous system are those concerned with the vegetative aspects of day-to-day liv-ing. For example, parasympathetic action favors digestion and absorption of food by increasing the activity of the intestinal musculature, increasing gastric secretion, and relaxing the py-loric sphincter. For this reason, the cholinergic division is sometimes called the anabolic nervous system. The sympathetic (noradrenergic) division discharges as a unit in emergency situations and can be called the catabolic nervous system. The effects of this discharge prepares the indi-vidual to cope with an emergency. Sympathetic activity dilates the pupils (letting more light into the eyes), accelerates the heartbeat and raises the blood pressure (providing better per-fusion of the vital organs and muscles), and constricts the blood vessels of the skin (which limits bleeding from wounds). Uterus Variable α1 Contraction (pregnant) β2 Relaxation Male sex organs Erection α1 Ejaculation Skin Pilomotor muscles — α1 Contraction Sweat glands — α1 Slight, localized secretionb M Generalized abundant, dilute secretion Liver — α1, β2 Glycogenolysis Pancreas Exocrine glands Increases secretion α Decreases secretion Endocrine glands — α2 Inhibits secretion Salivary glands Profuse, watery secretion α1 Thick, viscous secretion β Amylase secretion Lacrimal glands Secretion — Adipose tissue — α2, β3 Lipolysis aA dash means these cells are not innervated by this division of the autonomic nervous system. bOn palms of hands and in some other locations (“adrenergic sweating”). Modified from Hardman JG, Limbird LE, Gilman AG (editors): Goodman and Gilman’s The Pharmacological Basis of Therapeutics, 10th ed. McGraw-Hill, 2001. TABLE 17–1 Responses of some effector organs to autonomic nerve activity. (Continued) Sympathetic Nervous System Effector Organs Parasympathetic Nervous System Receptor Type Response CHAPTER 17 The Autonomic Nervous System 269 Noradrenergic discharge also leads to elevated plasma glucose and free fatty acid levels (supplying more energy). On the basis of effects like these, Walter Cannon called the emergency-induced discharge of the noradrenergic nervous system the “preparation for flight or fight.” The emphasis on mass discharge in stressful situations should not obscure the fact that the sympathetic fibers also subserve other functions. For example, tonic sympathetic discharge to the arterioles maintains arterial pressure, and variations in this tonic discharge are the mechanism by which carotid sinus feed-back regulation of blood pressure is effected. In addition, sym-pathetic discharge is decreased in fasting animals and increased when fasted animals are refed. These changes may explain the decrease in blood pressure and metabolic rate produced by fast-ing and the opposite changes produced by feeding. DESCENDING INPUT TO AUTONOMIC PREGANGLIONIC NEURONS As is the case for α-motor neurons, the activity of autonomic nerves is dependent on both reflexes (eg, baroreceptor and chemoreceptor reflexes) and descending excitatory and inhib-itory input from several brain regions. Figure 17–6 shows the source of some forebrain and brain stem descending inputs to autonomic preganglionic neurons. For example, a major source of excitatory drive to sympathetic preganglionic neu-rons comes from the rostral ventrolateral medulla. Although not shown, medullary raphé neurons project to the spinal cord to inhibit or excite sympathetic activity. In addition to these direct pathways to preganglionic neurons, there are many brain stem nuclei that feed into these pathways. This is analo-gous to the control of somatomotor function by areas such as the basal ganglia. ENTERIC NERVOUS SYSTEM The enteric nervous system, which can be considered as the third division of the ANS, is located within the wall of the di-gestive tract, all the way from the esophagus to the anus. It is comprised of two well-organized neural plexuses. The myen-teric plexus is located between longitudinal and circular layers of muscle; it is involved in control of digestive tract motility. The submucosal plexus is located between the circular muscle and the luminal mucosa; it senses the environment of the lu-men and regulates gastrointestinal blood flow and epithelial cell function. The enteric nervous system contains as many neurons as the entire spinal cord. It is sometimes referred to as a “mini-brain” as it contains all the elements of a nervous system including sensory neurons, interneurons, and motor neurons. FIGURE 17–5 Catecholamine metabolism in the sympathetic nervous system. COMT, catechol-O-methyltransferase; DA, dopamine; DHPG, dihydroxyphenylglycol; DOPA, dihydroxyphenylalanine; DOPAC, dihydroxyphenylacetic acid; HVA, homovanillic acid; MHPG, 3-methoxy-4-hydroxyphenylglycol; MOA, monoamine oxidase; NE, norepinephrine; NMN, normetanephrine; TH, tyrosine hydroxylase; TYR, tyrosine; VMA, va-nillylmandelic acid. (Courtesy of DS Goldstein.) NE NE NE DA DOPA DOPAC DOPAC DHPG DHPG MHPG COMT COMT NMN VMA HVA TYR [TYR] [NE] [NMN] [VMA] [MHPG] [HVA] [DHPG] [DOPAC] TH Uptake–1 Sympathetic nerve ending Smooth muscle cell MAO MAO MAO Bloodstream 270 SECTION III Central & Peripheral Neurophysiology It contains sensory neurons innervating receptors in the mucosa that respond to mechanical, thermal, osmotic, and chemical stimuli. Motor neurons control motility, secretion, and absorption by acting on smooth muscle and secretory cells. Interneurons integrate information from sensory neu-rons and feedback to the enteric motor neurons. Parasympathetic and sympathetic nerves connect the cen-tral nervous system to the enteric nervous system or directly to the digestive tract. Although the enteric nervous system can function autonomously, normal digestive function often requires communication between the central nervous system and the enteric nervous system. TABLE 17–2 Some drugs and toxins that affect autonomic activity.a Site of Action Compounds That Augment Autonomic Activity Compounds That Depress Autonomic Activity Autonomic ganglia Stimulate postganglionic neurons Block conduction Nicotine Hexamethonium (C-6) Low concentration of acetylcholine Mecamylamine (Inversine) Inhibit acetylcholinesterase Pentolinium DFP (diisopropyl fluorophosphate) Trimethaphan (Arfonad) Physostigmine (Eserine) High concentration of acetylcholine Neostigmine (Prostigmin) Parathion Postganglionic sympathetic terminals Release norepinephrine Block norepinephrine synthesis Tyramine Metyrosine (Demser) Ephedrine Interfere with norepinephrine storage Amphetamine Reserpine Guanethidineb (Ismelin) Prevent norepinephrine release Bretylium (Bretylol) Guanethidineb (Ismelin) Form false transmitters Methyldopa (Aldomet) Muscarinic receptors Atropine, scopolamine α adrenergic receptors Stimulate α1 receptors Block α receptors Methoxamine (Vasoxyl) Phenoxybenzamine (Dibenzyline) Phenylephrine (Neosynephrine) Phentolamine (Regitine) Prazosin (Minipress) blocks α1 Yohimbine blocks α2 β adrenergic receptors Stimulate β receptors Block β receptors Isoproterenol (Isuprel) Propranolol (Inderal) blocks β1 and β2 Atenolol (Tenormin) blocks β1 Butoxamine blocks β2 aOnly the principal actions are listed. bGuanethidine is believed to have two principal actions. CHAPTER 17 The Autonomic Nervous System 271 CHAPTER SUMMARY ■Preganglionic sympathetic neurons are located in the IML of the thoracolumbar spinal cord and project to postganglionic neurons in the paravertebral or prevertebral ganglia or the adrenal medul-la. Preganglionic parasympathetic neurons are located in motor nuclei of cranial nerves III, VII, IX, and X and the sacral IML. ■Nerve terminals of postganglionic neurons are located in smooth muscle (eg, blood vessels, gut wall, urinary bladder), cardiac muscle, and glands (eg, sweat gland, salivary glands). ■Acetylcholine is released at nerve terminals of all preganglionic neurons, postganglionic parasympathetic neurons, and a few postganglionic sympathetic neurons (sweat glands, sympathetic vasodilator fibers). The remaining sympathetic postganglionic neurons release norepinephrine. ■Sympathetic activity prepares the individual to cope with an emergency by accelerating the heartbeat, raising blood pressure (perfusion of the vital organs), and constricting the blood ves-sels of the skin (limits bleeding from wounds). Parasympathetic activity is concerned with the vegetative aspects of day-to-day living and favors digestion and absorption of food by increasing the activity of the intestinal musculature, increasing gastric se-cretion, and relaxing the pyloric sphincter. ■Ganglionic transmission is blocked by N2 nicotinic antagonists. Postganglionic cholinergic transmission is blocked by musca-rinic antagonists. Postganglionic adrenergic transmission is blocked by antagonists of α1, β1, or β2 adrenoreceptors, depend-ing on the target organ. ■The enteric nervous system is located within the wall of the di-gestive tract and is composed of the myenteric plexus (control of digestive tract motility) and the submucosal plexus (regulates gastrointestinal blood flow and epithelial cell function). MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Which of the following drugs would not be expected to increase sympathetic discharge or mimic the effects of increased sympa-thetic discharge? A) Prazosin B) Neostigmine C) Amphetamine D) Isoproterenol E) Methoxamine FIGURE 17–6 Pathways that control autonomic responses. Direct projections (solid lines) to autonomic preganglionic neurons include the hypothalamic paraventricular nucleus, parabrachial nucleus, nucleus of the solitary tract, ventrolateral medulla, and medullary raphé (not shown). Indirect projections (dashed lines) include the cerebral cortex, amygdala, and periaqueductal grey matter. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Periaqueductal gray matter Parabrachial nucleus Dorsal motor vagal nucleus Nucleus of the solitary tract Cortex Parasympathetic input Sympathetic input Amygdala Hypothalamus Nucleus ambiguus Ventrolateral medulla Intermediolateral cell column Heart 272 SECTION III Central & Peripheral Neurophysiology 2. Sympathetic nerve activity A) is essential for survival. B) causes contraction of some smooth muscles and relaxation of others. C) causes relaxation of the radial muscle of the eye to dilate the pupil. D) relaxes smooth muscle of the gastrointestinal wall and gas-trointestinal sphincter. E) all of the above 3. Parasympathetic nerve activity A) is essential for survival. B) affects only smooth muscles and glands. C) causes contraction of the radial muscle of the eye to allow accommodation for near vision. D) contracts smooth muscle of the gastrointestinal wall and relaxes the gastrointestinal sphincter. E) all of the above 4. Which of the following is correctly paired? A) sinoatrial node : nicotinic cholinergic receptors B) autonomic ganglia : muscarinic cholinergic receptors C) pilomotor smooth muscle : β2-adrenergic receptors D) vasculature of some skeletal muscles : muscarinic choliner-gic receptors E) sweat glands : α2-adrenergic receptors CHAPTER RESOURCES Benarroch EE: Central Autonomic Network. Functional Organization and Clinical Correlations. Futura Publishing, 1997. Boron WF, Boulpaep EL: Medical Physiology. Elsevier, 2005. Brodal P: The Central Nervous System. Structure and Function. Oxford University Press, 1998. Elvin LG, Lindh B, Hokfelt T: The chemical neuroanatomy of sympathetic ganglia. Annu Rev Neurosci 1993;16:471. Jänig W: The Integrative Action of the Autonomic Nervous System. Neurobiology of Homeostasis. Cambridge University Press, 2006. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. Loewy AD, Spyer KM (editors): Central Regulation of Autonomic Function. Oxford University Press, 1990. Pick J: The Autonomic Nervous System. Lippincott, 1970. Squire LR, et al (editors): Fundamental Neuroscience, 3rd ed. Academic Press, 2008. 273 C H A P T E R 18 Hypothalamic Regulation of Hormonal Functions O B J E C T I V E S After reading this chapter you should be able to: ■Describe the anatomic connections between the hypothalamus and the pituitary gland and the functional significance of each connection. ■List the factors that control water intake, and outline the way they exert their effects. ■Describe the synthesis, processing, storage, and secretion of the hormones of the posterior pituitary. ■Discuss the effects of vasopressin, the receptors on which it acts, and how its secre-tion is regulated. ■Discuss the effects of oxytocin, the receptors on which it acts, and how its secretion is regulated. ■Name the hypophysiotropic hormones, and outline the effects that each has on anterior pituitary function. ■List the mechanisms by which heat is produced in and lost from the body, and comment on the differences in temperature in the hypothalamus, rectum, oral cavity, and skin. ■List the temperature-regulating mechanisms, and describe the way in which they are integrated under hypothalamic control to maintain normal body temperature. ■Discuss the pathophysiology of fever. INTRODUCTION Many of the complex autonomic mechanisms that maintain the chemical constancy and temperature of the internal environ-ment are integrated in the hypothalamus. The hypothalamus also functions with the limbic system as a unit that regulates emotional and instinctual behavior. HYPOTHALAMUS: ANATOMIC CONSIDERATIONS The hypothalamus (Figure 18–1) is the portion of the anterior end of the diencephalon that lies below the hypothalamic sul-cus and in front of the interpeduncular nuclei. It is divided into a variety of nuclei and nuclear areas. AFFERENT & EFFERENT CONNECTIONS OF THE HYPOTHALAMUS The principal afferent and efferent neural pathways to and from the hypothalamus are mostly unmyelinated. Many con-nect the hypothalamus to the limbic system. Important con-nections also exist between the hypothalamus and nuclei in the midbrain tegmentum, pons, and hindbrain. Norepinephrine-secreting neurons with their cell bodies in the hindbrain end in many different parts of the hypothalamus 274 SECTION III Central & Peripheral Neurophysiology (see Figure 15–5). Paraventricular neurons that probably secrete oxytocin and vasopressin project in turn to the hindbrain and the spinal cord. Neurons that secrete epinephrine have their cell bodies in the hindbrain and end in the ventral hypothalamus. An intrahypothalamic system of dopamine-secreting neu-rons have their cell bodies in the arcuate nucleus and end on or near the capillaries that form the portal vessels in the median eminence. Serotonin-secreting neurons project to the hypothalamus from the raphe nuclei. RELATION TO THE PITUITARY GLAND There are neural connections between the hypothalamus and the posterior lobe of the pituitary gland and vascular connec-tions between the hypothalamus and the anterior lobe. Embry-ologically, the posterior pituitary arises as an evagination of the floor of the third ventricle. It is made up in large part of the end-ings of axons that arise from cell bodies in the supraoptic and paraventricular nuclei and pass to the posterior pituitary (Fig-ure 18–2) via the hypothalamohypophysial tract. Most of the supraoptic fibers end in the posterior lobe itself, whereas some of the paraventricular fibers end in the median eminence. The anterior and intermediate lobes of the pituitary arise in the em-bryo from the Rathke pouch, an evagination from the roof of the pharynx (see Figure 24–1). Sympathetic nerve fibers reach the anterior lobe from its capsule, and parasympathetic fibers reach it from the petrosal nerves, but few if any nerve fibers pass to it from the hypothalamus. However, the portal hypophysial FIGURE 18–1 Human hypothalamus, with a superimposed diagrammatic representation of the portal hypophysial vessels. Dorsal hypothalamic area Paraventricular nucleus Anterior hypothalamic area Preoptic area Supraoptic nucleus Suprachiasmatic nucleus Arcuate nucleus Optic chiasm Median eminence Superior hypophysial artery Portal hypophysial vessel Anterior lobe Pituitary gland Posterior lobe Primary plexus Posterior hypothalamic nucleus Dorsomedial nucleus Ventromedial nucleus Premamillary nucleus Medial mamillary nucleus Lateral mamillary nucleus Mamillary body FIGURE 18–2 Secretion of hypothalamic hormones. The hor-mones of the posterior lobe (PL) are released into the general circulation from the endings of supraoptic and paraventricular neurons, whereas hypophysiotropic hormones are secreted into the portal hypophysial circulation from the endings of arcuate and other hypothalamic neu-rons. AL, anterior lobe; MB, mamillary bodies; OC, optic chiasm. MB OC PL Supraoptic and paraventricular nuclei Arcuate and other nuclei AL Anterior pituitary hormones Posterior pituitary hormones CHAPTER 18 Hypothalamic Regulation of Hormonal Functions 275 vessels form a direct vascular link between the hypothalamus and the anterior pituitary. Arterial twigs from the carotid arter-ies and circle of Willis form a network of fenestrated capillaries called the primary plexus on the ventral surface of the hypo-thalamus (Figure 18–1). Capillary loops also penetrate the me-dian eminence. The capillaries drain into the sinusoidal portal hypophysial vessels that carry blood down the pituitary stalk to the capillaries of the anterior pituitary. This system begins and ends in capillaries without going through the heart and is there-fore a true portal system. In birds and some mammals, includ-ing humans, there is no other anterior hypophysial arterial supply except capsular vessels and anastomotic connections from the capillaries of the posterior pituitary. The median emi-nence is generally defined as the portion of the ventral hypo-thalamus from which the portal vessels arise. This region is outside the blood–brain barrier (see Chapter 34). HYPOTHALAMIC FUNCTION The major functions of the hypothalamus are summarized in Ta-ble 18–1. Some are fairly clear-cut visceral reflexes, and others in-clude complex behavioral and emotional reactions; however, all involve a particular response to a particular stimulus. It is impor-tant to keep this in mind in considering hypothalamic function. RELATION TO AUTONOMIC FUNCTION Many years ago, Sherrington called the hypothalamus “the head ganglion of the autonomic system.” Stimulation of the hypothal-amus produces autonomic responses, but the hypothalamus does not seem to be concerned with the regulation of visceral func-tion per se. Rather, the autonomic responses triggered in the TABLE 18–1 Summary of principal hypothalamic regulatory mechanisms. Function Afferents from Integrating Areas Temperature regulation Temperature receptors in the skin, deep tissues, spinal cord, hypothalamus, and other parts of the brain Anterior hypothalamus, response to heat; posterior hypothalamus, response to cold Neuroendocrine control of: Catecholamines Limbic areas concerned with emotion Dorsal and posterior hypothalamus Vasopressin Osmoreceptors, “volume receptors,” others Supraoptic and paraventricular nuclei Oxytocin Touch receptors in breast, uterus, genitalia Supraoptic and paraventricular nuclei Thyroid-stimulating hormone (thyrotropin, TSH) via TRH Temperature receptors in infants, perhaps others Paraventricular nuclei and neighboring areas Adrenocorticotropic hormone (ACTH) and β-lipotropin (β-LPH) via CRH Limbic system (emotional stimuli); reticular formation (“systemic” stimuli); hypothalamic and anterior pituitary cells sensitive to circulating blood cortisol level; suprachi-asmatic nuclei (diurnal rhythm) Paraventricular nuclei Follicle-stimulating hormone (FSH) and lu-teinizing hormone (LH) via GnRH Hypothalamic cells sensitive to estrogens, eyes, touch re-ceptors in skin and genitalia of reflex ovulating species Preoptic area; other areas Prolactin via PIH and PRH Touch receptors in breasts, other unknown receptors Arcuate nucleus; other areas (hypothala-mus inhibits secretion) Growth hormone via somatostatin and GRH Unknown receptors Periventricular nucleus, arcuate nucleus “Appetitive” behavior Thirst Osmoreceptors, probably located in the organum vasculo-sum of the lamina terminalis; angiotensin II uptake in the subfornical organ Lateral superior hypothalamus Hunger Glucostat cells sensitive to rate of glucose utilization; leptin receptors; receptors for other polypeptides Ventromedial, arcuate, and paraventricu-lar nuclei; lateral hypothalamus Sexual behavior Cells sensitive to circulating estrogen and androgen, others Anterior ventral hypothalamus plus, in the male, piriform cortex Defensive reactions (fear, rage) Sense organs and neocortex, paths unknown Diffuse, in limbic system and hypothala-mus Control of body rhythms Retina via retinohypothalamic fibers Suprachiasmatic nuclei 276 SECTION III Central & Peripheral Neurophysiology hypothalamus are part of more complex phenomena such as eat-ing, and emotions such as rage. For example, stimulation of vari-ous parts of the hypothalamus, especially the lateral areas, produces diffuse sympathetic discharge and increased adrenal medullary secretion, the mass sympathetic discharge seen in ani-mals exposed to stress (the flight or fight reaction; see Chapter 17). It has been claimed that separate hypothalamic areas con-trol epinephrine and norepinephrine secretion. Differential secretion of one or the other of these adrenal medullary cate-cholamines does occur in certain situations (see Chapter 22), but the selective increases are small. Body weight depends on the balance between caloric intake and utilization of calories. Obesity results when the former exceeds the latter. The hypothalamus and related parts of the brain play a key role in the regulation of food intake. Obesity is considered in detail in Chapter 27, and the relation of obe-sity to diabetes mellitus is discussed in Chapter 21. Hypothalamic regulation of sleep and circadian rhythms are discussed in Chapter 15. THIRST Another appetitive mechanism under hypothalamic control is thirst. Drinking is regulated by plasma osmolality and extra-cellular fluid (ECF) volume in much the same fashion as vaso-pressin secretion. Water intake is increased by increased effective osmotic pressure of the plasma (Figure 18–3), by de-creases in ECF volume, and by psychologic and other factors. Osmolality acts via osmoreceptors, receptors that sense the osmolality of the body fluids. These osmoreceptors are located in the anterior hypothalamus. Decreases in ECF volume also stimulate thirst by a pathway independent of that mediating thirst in response to increased plasma osmolality (Figure 18–4). Thus, hemorrhage causes increased drinking even if there is no change in the osmolality of the plasma. The effect of ECF volume depletion on thirst is mediated in part via the renin–angiotensin system (see Chap-ter 39). Renin secretion is increased by hypovolemia and results in an increase in circulating angiotensin II. The angio-tensin II acts on the subfornical organ, a specialized receptor area in the diencephalon (see Figure 34–7), to stimulate the neural areas concerned with thirst. Some evidence suggests that it acts on the organum vasculosum of the lamina termi-nalis (OVLT) as well. These areas are highly permeable and are two of the circumventricular organs located outside the blood–brain barrier (see Chapter 34). However, drugs that block the action of angiotensin II do not completely block the thirst response to hypovolemia, and it appears that the barore-ceptors in the heart and blood vessels are also involved. The intake of liquids is increased during eating (prandial drinking). The increase has been called a learned or habit response, but it has not been investigated in detail. One factor is an increase in plasma osmolality that occurs as food is absorbed. Another may be an action of one or more gas-trointestinal hormones on the hypothalamus. When the sensation of thirst is obtunded, either by direct dam-age to the diencephalon or by depressed or altered states of con-sciousness, patients stop drinking adequate amounts of fluid. Dehydration results if appropriate measures are not instituted to maintain water balance. If the protein intake is high, the products of protein metabolism cause an osmotic diuresis (see Chapter 38), and the amounts of water required to maintain hydration are large. Most cases of hypernatremia are actually due to simple dehydration in patients with psychoses or hypothalamic disease who do not or cannot increase their water intake when their thirst mechanism is stimulated. Lesions of the anterior commu-nicating artery can also obtund thirst because branches of this artery supply the hypothalamic areas concerned with thirst. FIGURE 18–3 Relation of plasma osmolality to thirst in healthy adult humans during infusion of hypertonic saline. The intensity of thirst is measured on a special analog scale. (Reproduced with permission from Thompson CJ et al: The osmotic thresholds for thirst and vasopressin release are similar in healthy humans. Clin Sci Lond 1986;71:651.) 10 8 6 4 2 0 280 300 320 Plasma osmolality (mosm/kg) Intensity of thirst FIGURE 18–4 Diagrammatic representation of the way in which changes in plasma osmolality and changes in ECF volume affect thirst by separate pathways. Hypertonicity Osmoreceptors Hypovolemia Hypothalamus Thirst Baroreceptors Angiotensin II CHAPTER 18 Hypothalamic Regulation of Hormonal Functions 277 OTHER FACTORS REGULATING WATER INTAKE A number of other well-established factors contribute to the regulation of water intake. Psychologic and social factors are important. Dryness of the pharyngeal mucous membrane causes a sensation of thirst. Patients in whom fluid intake must be restricted sometimes get appreciable relief of thirst by sucking ice chips or a wet cloth. Dehydrated dogs, cats, camels, and some other animals rapidly drink just enough water to make up their water deficit. They stop drinking before the water is absorbed (while their plasma is still hypertonic), so some kind of pharyngeal gastrointestinal “meter-ing” must be involved. Some evidence suggests that humans have a similar metering ability, though it is not well developed. CONTROL OF POSTERIOR PITUITARY SECRETION VASOPRESSIN & OXYTOCIN In most mammals, the hormones secreted by the posterior pi-tuitary gland are arginine vasopressin (AVP) and oxytocin. In hippopotami and most pigs, arginine in the vasopressin molecule is replaced by lysine to form lysine vasopressin. The posterior pituitaries of some species of pigs and marsupials contain a mixture of arginine and lysine vasopressin. The pos-terior lobe hormones are nonapeptides with a disulfide ring at one end (Figure 18–5). BIOSYNTHESIS, INTRANEURONAL TRANSPORT, & SECRETION The hormones of the posterior pituitary gland are synthesized in the cell bodies of the magnocellular neurons in the supraoptic and paraventricular nuclei and transported down the axons of these neurons to their endings in the posterior lobe, where they are secreted in response to electrical activity in the endings. Some of the neurons make oxytocin and others make vaso-pressin, and oxytocin-containing and vasopressin-containing cells are found in both nuclei. Oxytocin and vasopressin are typical neural hormones, that is, hormones secreted into the circulation by nerve cells. This type of neural regulation is compared with other types in Figure 18–6. The term neurosecretion was originally coined to describe the secretion of hormones by neurons, but the term is somewhat misleading because it appears that all neu-rons secrete chemical messengers (see Chapter 7). Like other peptide hormones, the posterior lobe hormones are synthesized as part of larger precursor molecules. Vaso-pressin and oxytocin each have a characteristic neurophysin associated with them in the granules in the neurons that secrete them—neurophysin I in the case of oxytocin and neu-rophysin II in the case of vasopressin. The neurophysins were originally thought to be binding polypeptides, but it now appears that they are simply parts of the precursor molecules. The precursor for arginine vasopressin, prepropressophysin, contains a 19-amino-acid residue leader sequence followed by FIGURE 18–5 Arginine vasopressin and oxytocin. Cys-Tyr-Phe-Gln-Asn-Cys-Pro-Arg-Gly-NH2 1 2 3 4 5 6 7 8 9 S S Cys-Tyr-Ile-Gln-Asn-Cys-Pro-Leu-Gly-NH2 1 2 3 4 5 6 7 8 9 S S Arginine vasopressin Oxytocin FIGURE 18–6 Neural control mechanisms. In the two situations on the left, neurotransmitters act at nerve endings on muscle; in the two in the middle, neurotransmitters regulate the secretion of endocrine glands; and in the two on the right, neurons secrete hormones into the hy-pophysial portal or general circulation. Acetylcholine Acetylcholine Acetylcholine Acetylcholine Vasopressin General circulation Norepinephrine Epinephrine, norepinephrine ACTH, TSH, GH, FSH, LH, prolactin Norepinephrine or acetylcholine Releasing and inhibiting hormones Portal vessels Motor nerves to skeletal muscle Motor nerves to smooth and cardiac muscle Juxta-glomerular cells Adrenal medulla Anterior pituitary Posterior pituitary Renin 278 SECTION III Central & Peripheral Neurophysiology arginine vasopressin, neurophysin II, and a glycopeptide (Fig-ure 18–7). Prepro-oxyphysin, the precursor for oxytocin, is a similar but smaller molecule that lacks the glycopeptide. The precursor molecules are synthesized in the ribosomes of the cell bodies of the neurons. They have their leader sequences removed in the endoplasmic reticulum, are pack-aged into secretory granules in the Golgi apparatus, and are transported down the axons by axoplasmic flow to the end-ings in the posterior pituitary. The secretory granules, called Herring bodies, are easy to stain in tissue sections, and they have been extensively studied. Cleavage of the precursor mol-ecules occurs as they are being transported, and the storage granules in the endings contain free vasopressin or oxytocin and the corresponding neurophysin. In the case of vaso-pressin, the glycopeptide is also present. All these products are secreted, but the functions of the components other than the established posterior pituitary hormones are unknown. ELECTRICAL ACTIVITY OF MAGNOCELLULAR NEURONS The oxytocin-secreting and vasopressin-secreting neurons also generate and conduct action potentials, and action poten-tials reaching their endings trigger release of hormone from them by Ca2+-dependent exocytosis. At least in anesthetized rats, these neurons are silent at rest or discharge at low, irreg-ular rates (0.1–3 spikes/s). However, their response to stimu-lation varies (Figure 18–8). Stimulation of the nipples causes a synchronous, high-frequency discharge of the oxytocin neu-rons after an appreciable latency. This discharge causes release of a pulse of oxytocin and consequent milk ejection in post-partum females. On the other hand, stimulation of the vaso-pressin-secreting neurons by a stimulus such as hemorrhage causes an initial steady increase in firing rate followed by a prolonged pattern of phasic discharge in which periods of high-frequency discharge alternate with periods of electrical quiescence (phasic bursting). These phasic bursts are gener-ally not synchronous in different vasopressin-secreting neu-rons. They are well suited to maintain a prolonged increase in the output of vasopressin, as opposed to the synchronous, rel-atively short, high-frequency discharge of oxytocin-secreting neurons in response to stimulation of the nipples. FIGURE 18–7 Structure of bovine prepropressophysin (left) and prepro-oxyphysin (right). Gly in the 10 position of both peptides is necessary for amidation of the Gly residue in position 9. aa, amino acid residues. (Reproduced with permission from Richter D: Molecular events in expression of vasopressin and oxytocin and their cognate receptors. Am J Physiol 1988;255:F207.) 1 1 2 3 4 2 3 4 Signal peptide Vasopressin Neurophysin II Glycopeptide 19 aa 9 aa 95 aa 39 aa -Gly-Lys-Arg--Arg-1 1 2 3 2 3 Signal peptide Oxytocin Neurophysin I 19 aa 9 aa 93 aa -Gly-Lys-Arg--Arg/His FIGURE 18–8 Responses of magnocellular neurons to stimulation. The tracings show individual extracellularly recorded ac-tion potentials, discharge rates, and intramammary duct pressure. A) Response of an oxytocin-secreting neuron. HFD, high-frequency dis-charge; ME, milk ejection. Stimulation of nipples started before the on-set of recording. B) Responses of a vasopressin-secreting neuron, showing no change in the slow firing rate in response to stimulation of nipples and a prompt increase in the firing rate when 5 mL of blood was drawn, followed by typical phasic discharge. (Modified from Wakerly JB: Hypothalamic neurosecretory function: Insights from electrophysiological studies of the magno-cellular nuclei. IBRO News 1985;4:15.) Unit Rate Control 5 mL blood removed 5 mL blood removed (+ 20 min) 1 min 10/s ME ME 50/s HFD A B Intramammary pressure CHAPTER 18 Hypothalamic Regulation of Hormonal Functions 279 VASOPRESSIN & OXYTOCIN IN OTHER LOCATIONS Vasopressin-secreting neurons are found in the suprachias-matic nuclei, and vasopressin and oxytocin are also found in the endings of neurons that project from the paraventricular nuclei to the brain stem and spinal cord. These neurons ap-pear to be involved in cardiovascular control. In addition, va-sopressin and oxytocin are synthesized in the gonads and the adrenal cortex, and oxytocin is present in the thymus. The functions of the peptides in these organs are unsettled. Vasopressin Receptors There are at least three kinds of vasopressin receptors: V1A, V1B, and V2. All are G protein-coupled. The V1A and V1B re-ceptors act through phosphatidylinositol hydrolysis to in-crease the intracellular Ca2+ concentration. The V2 receptors act through Gs to increase cAMP levels. Effects of Vasopressin Because one of its principal physiologic effects is the retention of water by the kidney, vasopressin is often called the antidi-uretic hormone (ADH). It increases the permeability of the collecting ducts of the kidney so that water enters the hyper-tonic interstitium of the renal pyramids (see Chapter 38). The urine becomes concentrated and its volume decreases. The overall effect is therefore retention of water in excess of solute; consequently, the effective osmotic pressure of the body fluids is decreased. In the absence of vasopressin, the urine is hypo-tonic to plasma, urine volume is increased, and there is a net water loss. Consequently, the osmolality of the body fluid rises. Effects of Oxytocin In humans, oxytocin acts primarily on the breasts and uterus, though it appears to be involved in luteolysis as well (see Chapter 25). A G protein-coupled serpentine oxytocin recep-tor has been identified in human myometrium, and a similar or identical receptor is found in mammary tissue and the ova-ry. It triggers increases in intracellular Ca2+ levels. The Milk Ejection Reflex Oxytocin causes contraction of the myoepithelial cells, smooth-muscle-like cells that line the ducts of the breast. This squeezes the milk out of the alveoli of the lactating breast into the large ducts (sinuses) and thence out of the nipple (milk ejection). Many hormones acting in concert are responsible for breast growth and the secretion of milk into the ducts (see Chapter 25), but milk ejection in most species requires oxytocin. Milk ejection is normally initiated by a neuroendocrine reflex. The receptors involved are the touch receptors, which are plentiful in the breast—especially around the nipple. Impulses generated in these receptors are relayed from the somatic touch pathways to the supraoptic and paraventricular nuclei. Dis-charge of the oxytocin-containing neurons causes secretion of oxytocin from the posterior pituitary (Figure 18–8). The infant suckling at the breast stimulates the touch receptors, the nuclei are stimulated, oxytocin is released, and the milk is expressed into the sinuses, ready to flow into the mouth of the waiting infant. In lactating women, genital stimulation and emotional stimuli also produce oxytocin secretion, sometimes causing milk to spurt from the breasts. Other Actions of Oxytocin Oxytocin causes contraction of the smooth muscle of the uter-us. The sensitivity of the uterine musculature to oxytocin is enhanced by estrogen and inhibited by progesterone. The in-hibitory effect of progesterone is due to a direct action of the steroid on uterine oxytocin receptors. In late pregnancy, the uterus becomes very sensitive to oxytocin coincident with a marked increase in the number of oxytocin receptors and ox-ytocin receptor mRNA (see Chapter 25). Oxytocin secretion is increased during labor. After dilation of the cervix, descent of the fetus down the birth canal initiates impulses in the afferent nerves that are relayed to the supraoptic and paraventricular nuclei, causing secretion of sufficient oxytocin to enhance la-bor (Figure 25-32). The amount of oxytocin in plasma is nor-mal at the onset of labor. It is possible that the marked increase in oxytocin receptors at this time causes normal oxytocin le-vels to initiate contractions, setting up a positive feedback. However, the amount of oxytocin in the uterus is also in-creased, and locally produced oxytocin may also play a role. Oxytocin may also act on the nonpregnant uterus to facilitate sperm transport. The passage of sperm up the female genital tract to the uterine tubes, where fertilization normally takes place, depends not only on the motile powers of the sperm but also, at least in some species, on uterine contractions. The geni-tal stimulation involved in coitus releases oxytocin, but it has not been proved that it is oxytocin which initiates the rather specialized uterine contractions that transport the sperm. The secretion of oxytocin is increased by stressful stimuli and, like that of vasopressin, is inhibited by alcohol. Circulating oxytocin increases at the time of ejaculation in males, and it is possible that this increase causes increased contraction of the smooth muscle of the vas deferens, propel-ling sperm toward the urethra. CONTROL OF ANTERIOR PITUITARY SECRETION ANTERIOR PITUITARY HORMONES The anterior pituitary secretes six hormones: adrenocortico-tropic hormone (corticotropin, ACTH), thyroid-stimulat-ing hormone (thyrotropin, TSH), growth hormone, follicle-stimulating hormone (FSH), luteinizing hormone (LH), 280 SECTION III Central & Peripheral Neurophysiology and prolactin (PRL). An additional polypeptide, β-lipotropin (β-LPH), is secreted with ACTH, but its physiologic role is un-known. The actions of the anterior pituitary hormones are summarized in Figure 18–9. The hormones are discussed in detail in the chapters on the endocrine system. The hypothal-amus plays an important stimulatory role in regulating the se-cretion of ACTH, β-LPH, TSH, growth hormone, FSH, and LH. It also regulates prolactin secretion, but its effect is pre-dominantly inhibitory rather than stimulatory. NATURE OF HYPOTHALAMIC CONTROL Anterior pituitary secretion is controlled by chemical agents carried in the portal hypophysial vessels from the hypothala-mus to the pituitary. These substances used to be called releas-ing and inhibiting factors, but now they are commonly called hypophysiotropic hormones. The latter term seems appro-priate since they are secreted into the bloodstream and act at a distance from their site of origin. Small amounts escape into the general circulation, but they are in high concentration in portal hypophysial blood. HYPOPHYSIOTROPIC HORMONES There are six established hypothalamic releasing and inhibiting hormones (Figure 18–10): corticotropin-releasing hormone (CRH); thyrotropin-releasing hormone (TRH); growth hor-mone-releasing hormone (GRH); growth hormone-inhibit-ing hormone (GIH), now generally called somatostatin; luteinizing hormone-releasing hormone (LHRH), now gener-ally known as gonadotropin-releasing hormone (GnRH); and prolactin-inhibiting hormone (PIH). In addition, hypotha-lamic extracts contain prolactin-releasing activity, and a prolac-tin-releasing hormone (PRH) has been postulated to exist. TRH, VIP, and several other polypeptides found in the hypothal-amus stimulate prolactin secretion, but it is uncertain whether one or more of these peptides is the physiologic PRH. Recently, an orphan receptor was isolated from the anterior pituitary, and FIGURE 18–9 Anterior pituitary hormones. In women, FSH and LH act in sequence on the ovary to produce growth of the ovarian follicle, ovulation, and formation and maintenance of the corpus luteum. Prolactin stimulates lactation. In men, FSH and LH control the functions of the testes. Anterior pituitary ACTH TSH FSH LH Prolactin Growth hormone β-LPH ? Breast 17-Hydroxy-corticoids Aldosterone, sex hormones Somato-medins Thyroxine Estrogen Progesterone FIGURE 18–10 Effects of hypophysiotropic hormones on the secretion of anterior pituitary hormones. Hypothalamus CRH TRH Anterior pituitary GnRH GRH GIH PRH PIH β-LPH ACTH TSH LH FSH Prolactin Growth hormone CHAPTER 18 Hypothalamic Regulation of Hormonal Functions 281 the search for its ligand led to the isolation of a 31-amino-acid polypeptide from the human hypothalamus. This polypeptide stimulated prolactin secretion by an action on the anterior pitu-itary receptor, but additional research is needed to determine if it is the physiologic PRH. GnRH stimulates the secretion of FSH as well as that of LH, and it seems unlikely that a separate follicle-stimulating hormone-releasing hormone exists. The structures of the six established hypophysiotropic hor-mones are shown in Figure 18–11. The structures of the genes and preprohormones for TRH, GnRH, somatostatin, CRH, and GRH are known. PreproTRH contains six copies of TRH. Several other preprohormones may contain other hormonally active peptides in addition to the hypophysiotro-pic hormones. The area from which the hypothalamic releasing and inhib-iting hormones are secreted is the median eminence of the hypothalamus. This region contains few nerve cell bodies, but many nerve endings are in close proximity to the capillary loops from which the portal vessels originate. The locations of the cell bodies of the neurons that project to the external layer of the median eminence and secrete the hypophysiotropic hormones are shown in Figure 18–12, which also shows the location of the neurons secreting oxyto-cin and vasopressin. The GnRH-secreting neurons are pri-marily in the medial preoptic area, the somatostatin-secreting neurons are in the periventricular nuclei, the TRH-secreting and CRH-secreting neurons are in the medial parts of the paraventricular nuclei, and the GRH-secreting and dopa-mine-secreting neurons are in the arcuate nuclei. Most, if not all, of the hypophysiotropic hormones affect the secretion of more than one anterior pituitary hormone (Figure 18–10). The FSH-stimulating activity of GnRH has been mentioned previously. TRH stimulates the secretion of prolactin as well as TSH. Somatostatin inhibits the secretion of TSH as well as growth hormone. It does not normally inhibit the secretion of the other anterior pituitary hormones, but it inhibits the abnormally elevated secretion of ACTH in patients with Nelson’s syndrome. CRH stimulates the secre-tion of ACTH and β-LPH. FIGURE 18–11 Structure of hypophysiotropic hormones in humans. Preprosomatostatin is processed to a tetradecapeptide (soma-tostatin 14, [SS14], shown above) and also to a polypeptide containing 28 amino acid residues (SS28). TRH (pyro)Glu-His-Pro-NH2 GnRH (pyro)Glu-His-Trp-Ser-Tyr-Gly-Leu-Arg-Pro-Gly-NH2 Somatostatin Ala-Gly-Cys-Lys-Asn-Phe-Phe-Trp-Lys-Thr-Phe-Thr-Ser-Cys Ser-Glu-Glu-Pro-Pro-Ile-Ser-Leu-Asp-Leu-Thr-Phe-His-Leu-Leu-Arg-Glu-Val-Leu-Glu-Met-Ala-Arg-Ala-Glu-Gln-Leu-Ala-Gln-Gln-Ala-His-Ser-Asn-Arg-Lys-Leu-Met-Glu-Ile-Ile-NH2 CRH GRH Tyr-Ala-Asp-Ala-Ile-Phe-Thr-Asn-Ser-Tyr-Arg-Lys-Val-Leu-Gly-Gln-Leu-Ser-Ala-Arg-Lys-Leu-Leu-Gln-Asp-Ile-Met-Ser-Arg-Gln-Gln-Gly-Glu-Ser-Asn-Gln-Glu-Arg-Gly-Ala-Arg-Ala-Arg-Leu-NH2 PIH Dopamine S S FIGURE 18–12 Location of cell bodies of hypophysiotropic hormone-secreting neurons projected on a ventral view of the hypothalamus and pituitary of the rat. AL, anterior lobe; ARC, arcu-ate nucleus; BA, basilar artery; DA, dopamine; IC, internal carotid artery; IL, intermediate lobe; MC, middle cerebral artery; ME, median emi-nence; PC, posterior cerebral artery; Peri, periventricular nucleus; PL, posterior lobe; PV, paraventricular nucleus; SO, supraoptic nucleus. The names of the hormones are enclosed in boxes. (Courtesy of LW Swanson and ET Cunningham Jr.) 0.5 mm Oxytocin Vasopressin SS DA CRH GnRH TRH GRH Peri SO PV ME ARC IC PL IL AL BA PC MC TRH GRH DA 282 SECTION III Central & Peripheral Neurophysiology Hypophysiotropic hormones function as neurotransmit-ters in other parts of the brain, the retina, and the autonomic nervous system (see Chapter 7). In addition, somatostatin is found in the pancreatic islets (see Chapter 21), GRH is secreted by pancreatic tumors, and somatostatin and TRH are found in the gastrointestinal tract (see Chapter 26). Receptors for most of the hypophysiotropic hormones are serpentine and coupled to G proteins. There are two human CRH receptors: hCRH-RI, and hCRHRII. The latter differs from the former in having a 29-amino-acid insert in its first cytoplasmic loop. The physiologic role of hCRH-RII is unset-tled, though it is found in many parts of the brain. In addition, a CRH-binding protein in the peripheral circulation inacti-vates CRH. It is also found in the cytoplasm of corticotropes in the anterior pituitary, and in this location it might play a role in receptor internalization. However, the exact physio-logic role of this protein is unknown. Other hypophysiotropic hormones do not have known binding proteins. SIGNIFICANCE & CLINICAL IMPLICATIONS Research delineating the multiple neuroendocrine regulatory functions of the hypothalamus is important because it helps explain how endocrine secretion is made appropriate to the demands of a changing environment. The nervous system re-ceives information about changes in the internal and external environment from the sense organs. It brings about adjust-ments to these changes through effector mechanisms that in-clude not only somatic movement but also changes in the rate at which hormones are secreted. The manifestations of hypothalamic disease are neurologic defects, endocrine changes, and metabolic abnormalities such as hyperphagia and hyperthermia. The relative frequencies of the signs and symptoms of hypothalamic disease in one large series of cases are shown in Table 18–2. The possibility of hypothalamic pathology should be kept in mind in evaluating all patients with pituitary dysfunction, especially those with isolated deficiencies of single pituitary tropic hormones. A condition of considerable interest in this context is Kall-mann syndrome, the combination of hypogonadism due to low levels of circulating gonadotropins (hypogonadotropic hypo-gonadism) with partial or complete loss of the sense of smell (hyposmia or anosmia). Embryologically, GnRH neurons develop in the nose and migrate up the olfactory nerves and then through the brain to the hypothalamus. If this migration is prevented by congenital abnormalities in the olfactory path-ways, the GnRH neurons do not reach the hypothalamus and pubertal maturation of the gonads fails to occur. The syndrome is most common in men, and the cause in many cases is muta-tion of the KALIG1 gene, a gene on the X chromosome that codes for what is apparently an adhesion molecule necessary for normal development of the olfactory nerve on which the GnRH neurons migrate into the brain. However, the condition also occurs in women and can be due to other genetic abnormalities. TEMPERATURE REGULATION In the body, heat is produced by muscular exercise, assimilation of food, and all the vital processes that contribute to the basal metabolic rate (see Chapter 27). It is lost from the body by radi-ation, conduction, and vaporization of water in the respiratory passages and on the skin. Small amounts of heat are also re-moved in the urine and feces. The balance between heat produc-tion and heat loss determines the body temperature. Because the speed of chemical reactions varies with the temperature and be-cause the enzyme systems of the body have narrow temperature ranges in which their function is optimal, normal body function depends on a relatively constant body temperature. Invertebrates generally cannot adjust their body temperatures and so are at the mercy of the environment. In vertebrates, mechanisms for maintaining body temperature by adjusting heat production and heat loss have evolved. In reptiles, amphibians, and fish, the adjusting mechanisms are relatively rudimentary, and these species are called “cold-blooded” (poikilothermic) because their body temperature fluctuates over a considerable TABLE 18–2 Symptoms and signs in 60 autopsied pa-tients with hypothalamic disease. Symptoms and Signs Percentage of Cases Endocrine and metabolic findings Precocious puberty 40 Hypogonadism 32 Diabetes insipidus 35 Obesity 25 Abnormalities of temperature regulation 22 Emaciation 18 Bulimia 8 Anorexia 7 Neurologic findings Eye signs 78 Pyramidal and sensory deficits 75 Headache 65 Extrapyramidal signs 62 Vomiting 40 Psychic disturbances, rage attacks, etc 35 Somnolence 30 Convulsions 15 Data from Bauer HG: Endocrine and other clinical manifestations of hypothalamic disease. J Clin Endocrinol 1954;14:13. See also Kahana L, et al: Endocrine manifesta-tions of intracranial extrasellar lesions. J Clin Endocrinol 1962;22:304. CHAPTER 18 Hypothalamic Regulation of Hormonal Functions 283 range. In birds and mammals, the “warm-blooded” (homeo-thermic) animals, a group of reflex responses that are primarily integrated in the hypothalamus, operate to maintain body tem-perature within a narrow range in spite of wide fluctuations in environmental temperature. The hibernating mammals are a partial exception. While awake they are homeothermic, but dur-ing hibernation their body temperature falls. NORMAL BODY TEMPERATURE In homeothermic animals, the actual temperature at which the body is maintained varies from species to species and, to a lesser degree, from individual to individual. In humans, the traditional normal value for the oral temperature is 37 °C (98.6 °F), but in one large series of normal young adults, the morning oral tem-perature averaged 36.7 °C, with a standard deviation of 0.2 °C. Therefore, 95% of all young adults would be expected to have a morning oral temperature of 36.3–37.1 °C (97.3–98.8 °F; mean ± 1.96 standard deviations; see Appendix). Various parts of the body are at different temperatures, and the magnitude of the temperature difference between the parts varies with the envi-ronmental temperature (Figure 18–13). The extremities are generally cooler than the rest of the body. The temperature of the scrotum is carefully regulated at 32 °C. The rectal temperature is representative of the temperature at the core of the body and var-ies least with changes in environmental temperature. The oral temperature is normally 0.5 °C lower than the rectal tempera-ture, but it is affected by many factors, including ingestion of hot or cold fluids, gum chewing, smoking, and mouth breathing. The normal human core temperature undergoes a regular circadian fluctuation of 0.5–0.7 °C. In individuals who sleep at night and are awake during the day (even when hospitalized at bed rest), it is lowest at about 6:00 AM and highest in the eve-nings (Figure 18–14). It is lowest during sleep, is slightly higher in the awake but relaxed state, and rises with activity. In women, an additional monthly cycle of temperature variation is characterized by a rise in basal temperature at the time of ovu-lation (Figure 25–38). Temperature regulation is less precise in young children and they may normally have a temperature that is 0.5 ° or so above the established norm for adults. During exercise, the heat produced by muscular contraction accumulates in the body and the rectal temperature normally rises as high as 40 °C (104 °F). This rise is due in part to the inability of the heat-dissipating mechanisms to handle the greatly increased amount of heat produced, but evidence sug-gests that in addition there is an elevation of the body tempera-ture at which the heat-dissipating mechanisms are activated during exercise. Body temperature also rises slightly during emotional excitement, probably owing to unconscious tensing of the muscles. It is chronically elevated by as much as 0.5 °C when the metabolic rate is high, as in hyperthyroidism, and lowered when the metabolic rate is low, as in hypothyroidism (Figure 18–14). Some apparently normal adults chronically have a tem-perature above the normal range (constitutional hyperthermia). HEAT PRODUCTION Heat production and energy balance are discussed in Chapter 27. A variety of basic chemical reactions contribute to body heat production at all times. Ingestion of food increases heat production because of the specific dynamic action of the food (see Chapter 27), but the major source of heat is the contrac-tion of skeletal muscle (Table 18–3). Heat production can be varied by endocrine mechanisms in the absence of food intake or muscular exertion. Epinephrine and norepinephrine pro-duce a rapid but short-lived increase in heat production; thy-roid hormones produce a slowly developing but prolonged increase. Furthermore, sympathetic discharge decreases dur-ing fasting and is increased by feeding. A source of considerable heat, particularly in infants, is brown fat. This fat has a high rate of metabolism and its thermogenic function has been likened to that of an electric blanket. FIGURE 18–13 Temperatures of various parts of the body of a naked subject at various ambient temperatures in a calorimeter. (Redrawn and reproduced, with permission, from Hardy JD, DuBois EF: Basal metabolism, radiation, convection and vaporization at temperatures of 22–35 °C. J Nutr 1938;15:477.) 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 23 24 25 26 Feet Hands 27 28 29 30 31 32 33 34 Calorimetric temperature (°C) Temperature (°C) of subject Average skin Trunk Head Rectum FIGURE 18–14 Typical temperature chart of a hospitalized patient who does not have a febrile disease. Note the slight rise in temperature, due to excitement and apprehension, at the time of ad-mission to the hospital, and the regular circadian temperature cycle. 38 37 36 1 2 3 4 5 Admitted to hospital Days Oral temp (°C) Hyper-thyroidism Hypo-thyroidism Normal 284 SECTION III Central & Peripheral Neurophysiology HEAT LOSS The processes by which heat is lost from the body when the environmental temperature is below body temperature are listed in Table 18–3. Conduction is heat exchange between objects or substances at different temperatures that are in contact with one another. A basic characteristic of matter is that its molecules are in motion, with the amount of motion proportionate to the temperature. These molecules collide with the molecules in cooler objects, transferring thermal en-ergy to them. The amount of heat transferred is proportionate to the temperature difference between the objects in contact (thermal gradient). Conduction is aided by convection, the movement of molecules away from the area of contact. Thus, for example, an object in contact with air at a different tem-perature changes the specific gravity of the air, and because warm air rises and cool air falls, a new supply of air is brought into contact with the object. Of course, convection is greatly aided if the object moves about in the medium or the medium moves past the object, for example, if a subject swims through water or a fan blows air through a room. Radiation is the transfer of heat by infrared electromagnetic radiation from one object to another at a different temperature with which it is not in contact. When an individual is in a cold environ-ment, heat is lost by conduction to the surrounding air and by radiation to cool objects in the vicinity. Conversely, of course, heat is transferred to an individual and the heat load is in-creased by these processes when the environmental tempera-ture is above body temperature. Note that because of radiation, an individual can feel chilly in a room with cold walls even though the room is relatively warm. On a cold but sunny day, the heat of the sun reflected off bright objects ex-erts an appreciable warming effect. It is the heat reflected from the snow, for example, that makes it possible to ski in fairly light clothes even though the air temperature is below freezing. Because conduction occurs from the surface of one object to the surface of another, the temperature of the skin deter-mines to a large extent the degree to which body heat is lost or gained. The amount of heat reaching the skin from the deep tissues can be varied by changing the blood flow to the skin. When the cutaneous vessels are dilated, warm blood wells into the skin, whereas in the maximally vasoconstricted state, heat is held centrally in the body. The rate at which heat is transferred from the deep tissues to the skin is called the tis-sue conductance. Birds have a layer of feathers next to the skin, and most mammals have a significant layer of hair or fur. Heat is conducted from the skin to the air trapped in this layer and from the trapped air to the exterior. When the thickness of the trapped layer is increased by fluffing the feathers or erection of the hairs (horripilation), heat transfer across the layer is reduced and heat losses (or, in a hot environment, heat gains) are decreased. “Goose pimples” are the result of horrip-ilation in humans; they are the visible manifestation of cold-induced contraction of the piloerector muscles attached to the rather meager hair supply. Humans usually supplement this layer of hair with one or more layers of clothes. Heat is con-ducted from the skin to the layer of air trapped by the clothes, from the inside of the clothes to the outside, and from the out-side of the clothes to the exterior. The magnitude of the heat transfer across the clothing, a function of its texture and thickness, is the most important determinant of how warm or cool the clothes feel, but other factors, especially the size of the trapped layer of warm air, are important also. Dark clothes absorb radiated heat and light-colored clothes reflect it back to the exterior. The other major process transferring heat from the body in humans and other animals that sweat is vaporization of water on the skin and mucous membranes of the mouth and respira-tory passages. Vaporization of 1 g of water removes about 0.6 kcal of heat. A certain amount of water is vaporized at all times. This insensible water loss amounts to 50 mL/h in humans. When sweat secretion is increased, the degree to which the sweat vaporizes depends on the humidity of the environment. It is common knowledge that one feels hotter on a humid day. This is due in part to the decreased vaporization of sweat, but even under conditions in which vaporization of sweat is complete, an individual in a humid environment feels warmer than an individual in a dry environment. The reason for this difference is unknown, but it seems related to the fact that in the humid environment sweat spreads over a greater area of skin before it evaporates. During muscular exertion in a hot environment, sweat secretion reaches values as high as 1600 mL/h, and in a dry atmosphere, most of this sweat is vaporized. Heat loss by vaporization of water therefore varies from 30 to over 900 kcal/h. Some mammals lose heat by panting. This rapid, shallow breathing greatly increases the amount of water vaporization in the mouth and respiratory passages and therefore the amount of heat lost. Because the breathing is shallow, it pro-duces relatively little change in the composition of alveolar air (see Chapter 35). The relative contribution of each of the processes that transfer heat away from the body (Table 18–3) varies with the TABLE 18–3 Body heat production and heat loss. Body heat is produced by: Basic metabolic processes Food intake (specific dynamic action) Muscular activity Body heat is lost by: Percentage of heat lost at 21 °C Radiation and conduction 70 Vaporization of sweat 27 Respiration 2 Urination and defecation 1 CHAPTER 18 Hypothalamic Regulation of Hormonal Functions 285 environmental temperature. At 21 °C, vaporization is a minor component in humans at rest. As the environmental tempera-ture approaches body temperature, radiation losses decline and vaporization losses increase. TEMPERATURE-REGULATING MECHANISMS The reflex and semireflex thermoregulatory responses in hu-mans are listed in Table 18–4. They include autonomic, so-matic, endocrine, and behavioral changes. One group of responses increases heat loss and decreases heat production; the other decreases heat loss and increases heat production. In general, exposure to heat stimulates the former group of re-sponses and inhibits the latter, whereas exposure to cold does the opposite. Curling up “in a ball” is a common reaction to cold in ani-mals and has a counterpart in the position some people assume on climbing into a cold bed. Curling up decreases the body surface exposed to the environment. Shivering is an involuntary response of the skeletal muscles, but cold also causes a semiconscious general increase in motor activity. Examples include foot stamping and dancing up and down on a cold day. Increased catecholamine secretion is an important endocrine response to cold. Mice unable to make norepineph-rine and epinephrine because their dopamine β-hydroxylase gene is knocked out do not tolerate cold; they have deficient vasoconstriction and are unable to increase thermogenesis in brown adipose tissue through UCP 1. TSH secretion is increased by cold and decreased by heat in laboratory ani-mals, but the change in TSH secretion produced by cold in adult humans is small and of questionable significance. It is common knowledge that activity is decreased in hot weather—the “it’s too hot to move” reaction. Thermoregulatory adjustments involve local responses as well as more general reflex responses. When cutaneous blood vessels are cooled they become more sensitive to catechol-amines and the arterioles and venules constrict. This local effect of cold directs blood away from the skin. Another heat-conserving mechanism that is important in animals living in cold water is heat transfer from arterial to venous blood in the limbs. The deep veins (venae comitantes) run alongside the arteries supplying the limbs and heat is transferred from the warm arterial blood going to the limbs to the cold venous blood coming from the extremities (countercurrent exchange; see Chapter 38). This keeps the tips of the extremities cold but con-serves body heat. The reflex responses activated by cold are controlled from the posterior hypothalamus. Those activated by warmth are controlled primarily from the anterior hypothalamus, although some thermoregulation against heat still occurs after decerebration at the level of the rostral midbrain. Stimulation of the anterior hypothalamus causes cutaneous vasodilation and sweating, and lesions in this region cause hyperthermia, with rectal temperatures sometimes reaching 43 °C (109.4 °F). Posterior hypothalamic stimulation causes shivering, and the body temperature of animals with posterior hypothalamic lesions falls toward that of the environment. AFFERENTS The hypothalamus is said to integrate body temperature infor-mation from sensory receptors (primarily cold receptors) in the skin, deep tissues, spinal cord, extrahypothalamic portions of the brain, and the hypothalamus itself. Each of these five in-puts contributes about 20% of the information that is integrat-ed. There are threshold core temperatures for each of the main temperature-regulating responses and when the threshold is reached the response begins. The threshold is 37 °C for sweat-ing and vasodilation, 36.8 °C for vasoconstriction, 36 °C for nonshivering thermogenesis, and 35.5 °C for shivering. FEVER Fever is perhaps the oldest and most universally known hall-mark of disease. It occurs not only in mammals but also in birds, reptiles, amphibia, and fish. When it occurs in homeo-thermic animals, the thermoregulatory mechanisms behave as if they were adjusted to maintain body temperature at a higher than normal level, that is, “as if the thermostat had been reset” to a new point above 37 °C. The temperature receptors then TABLE 18–4 Temperature-regulating mechanisms. Mechanisms activated by cold Shivering Hunger Increased voluntary activity Increased secretion of norepinephrine and epinephrine Decreased heat loss Cutaneous vasoconstriction Curling up Horripilation Mechanisms activated by heat Increased heat loss Cutaneous vasodilation Sweating Increased respiration Decreased heat production Anorexia Apathy and inertia 286 SECTION III Central & Peripheral Neurophysiology signal that the actual temperature is below the new set point, and the temperature-raising mechanisms are activated. This usually produces chilly sensations due to cutaneous vasocon-striction and occasionally enough shivering to produce a shaking chill. However, the nature of the response depends on the ambient temperature. The temperature rise in experimen-tal animals injected with a pyrogen is due mostly to increased heat production if they are in a cold environment and mostly to decreased heat loss if they are in a warm environment. The pathogenesis of fever is summarized in Figure 18–15. Toxins from bacteria such as endotoxin act on monocytes, macrophages, and Kupffer cells to produce cytokines that act as endogenous pyrogens (EPs). There is good evidence that IL-1β, IL-6, β-IFN, γ-IFN, and TNF-α (see Chapter 3) can act independently to produce fever. These cytokines are polypep-tides and it is unlikely that circulating cytokines penetrate the brain. Instead, evidence suggests that they act on the OVLT, one of the circumventricular organs (see Chapter 34). This in turn activates the preoptic area of the hypothalamus. Cyto-kines are also produced by cells in the central nervous system (CNS) when these are stimulated by infection, and these may act directly on the thermoregulatory centers. The fever produced by cytokines is probably due to local release of prostaglandins in the hypothalamus. Intrahypotha-lamic injection of prostaglandins produces fever. In addition, the antipyretic effect of aspirin is exerted directly on the hypo-thalamus, and aspirin inhibits prostaglandin synthesis. PGE2 is one of the prostaglandins that causes fever. It acts on four sub-types of prostaglandin receptors—EP1, EP2, EP3, and EP4—and knockout of the EP3 receptor impairs the febrile response to PGE2, IL-1β, and bacterial lipopolysaccharide (LPS). The benefit of fever to the organism is uncertain. It is presum-ably beneficial because it has evolved and persisted as a response to infections and other diseases. Many microorganisms grow best within a relatively narrow temperature range and a rise in temperature inhibits their growth. In addition, antibody pro-duction is increased when body temperature is elevated. Before the advent of antibiotics, fevers were artificially induced for the treatment of neurosyphilis and proved to be beneficial. Hyper-thermia benefits individuals infected with anthrax, pneumococ-cal pneumonia, leprosy, and various fungal, rickettsial, and viral diseases. Hyperthermia also slows the growth of some tumors. However, very high temperatures are harmful. A rectal tempera-ture over 41 °C (106 °F) for prolonged periods results in some permanent brain damage. When the temperature is over 43 °C, heat stroke develops and death is common. In malignant hyperthermia, various mutations of the gene coding for the ryanodine receptor (see Chapter 5) lead to excess Ca2+ release during muscle contraction triggered by stress. This in turn leads to contractures of the muscles, increased muscle metabolism, and a great increase in heat production in muscle. The increased heat production causes a marked rise in body temperature that is fatal if not treated. Periodic fevers also occur in humans with mutations in the gene for pyrin, a protein found in neutrophils; the gene for mevalonate kinase, an enzyme involved in cholesterol synthe-sis; and the gene for the type 1 TNF receptor, which is involved in inflammatory responses. However, how any of these three mutant gene products cause fever is unknown. HYPOTHERMIA In hibernating mammals, body temperature drops to low le-vels without causing any demonstrable ill effects on subse-quent arousal. This observation led to experiments on induced hypothermia. When the skin or the blood is cooled enough to lower the body temperature in nonhibernating animals and in humans, metabolic and physiologic processes slow down. Respiration and heart rate are very slow, blood pressure is low, and consciousness is lost. At rectal temperatures of about 28 °C, the ability to spontaneously return the temperature to normal is lost, but the individual continues to survive and, if rewarmed with external heat, returns to a normal state. If care is taken to prevent the formation of ice crystals in the tissues, the body temperature of experimental animals can be lowered to subfreezing levels without producing any detectable dam-age after subsequent rewarming. Humans tolerate body temperatures of 21–24 °C (70–75 °F) without permanent ill effects, and induced hypothermia has been used in surgery. On the other hand, accidental hypother-mia due to prolonged exposure to cold air or cold water is a serious condition and requires careful monitoring and prompt rewarming. CHAPTER SUMMARY ■Neural connections run between the hypothalamus and the pos-terior lobe of the pituitary gland, and vascular connections be-tween the hypothalamus and the anterior lobe of the pituitary. ■In most mammals, the hormones secreted by the posterior pitu-itary gland are vasopressin and oxytocin. Vasopressin increases FIGURE 18–15 Pathogenesis of fever. Endotoxin Inflammation Other pyrogenic stimuli Monocytes Macrophages Kupffer cells Cytokines Preoptic area of hypothalamus Prostaglandins Raise temperature set point Fever CHAPTER 18 Hypothalamic Regulation of Hormonal Functions 287 the permeability of the collecting ducts of the kidney to water, thus concentrating the urine. Oxytocin acts on the breasts (lac-tation) and the uterus (contraction). ■The anterior pituitary secretes six hormones: adrenocortico-tropic hormone (corticotropin, ACTH), thyroid-stimulating hormone (thyrotropin, TSH), growth hormone, follicle-stimu-lating hormone (FSH), luteinizing hormone (LH), and prolactin (PRL). ■Other complex autonomic mechanisms that maintain the chemical constancy and temperature of the internal environ-ment are integrated in the hypothalamus. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Thirst is stimulated by A) increases in plasma osmolality and volume. B) an increase in plasma osmolality and a decrease in plasma volume. C) a decrease in plasma osmolality and an increase in plasma volume. D) decreases in plasma osmolality and volume. E) injection of vasopressin into the hypothalamus. 2. When an individual is naked in a room in which the air temper-ature is 21 °C (69.8 °F) and the humidity 80%, the greatest amount of heat is lost from the body by A) elevated metabolism. B) respiration. C) urination. D) vaporization of sweat. E) radiation and conduction. In questions 3–8, select A if the item is associated with (a) below, B if the item is associated with (b) below, C if the item is associated with both (a) and (b), and D if the item is associated with neither (a) nor (b). (a) V1A vasopressin receptors (b) V2 vasopressin receptors 3. Activation of Gs 4. Vasoconstriction 5. Increase in intracellular inositol triphosphate 6. Movement of aquaporin 7. Proteinuria 8. Milk ejection CHAPTER RESOURCES Brunton PJ, Russell JA, Douglas AJ: Adaptive responses of the maternal hypothalamic-pituitary-adrenal axis during pregnancy and lactation. J Neuroendocrinol. 2008;20:764. Lamberts SWJ, Hofland LJ, Nobels FRE: Neuroendocrine tumor markers. Front Neuroendocrinol 2001;22:309. Loh JA, Verbalis JG: Disorders of water and salt metabolism associated with pituitary disease. Endocrinol Metab Clin 2008;37:213. McKinley MS, Johnson AK: The physiologic regulation of thirst and fluid intake. News Physiol Sci 2004;19:1. This page intentionally left blank 289 C H A P T E R 19 Learning, Memory, Language, & Speech O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the various types of long-term memory. ■Define synaptic plasticity, long-term potentiation (LTP), long-term depression (LTD), habituation, and sensitization, and their roles in learning and memory. ■List the parts of the brain that appear to be involved in memory in mammals and summarize the proposed role of each in memory processing and storage. ■Describe the abnormalities of brain structure and function found in Alzheimer disease. ■Define the terms categorical hemisphere and representational hemisphere and summarize the difference between these hemispheres. ■Summarize the differences between fluent and nonfluent aphasia, and explain each type on the basis of its pathophysiology. INTRODUCTION A revolution in our understanding of brain function in humans has been brought about by the development and widespread availability of positron emission tomographic (PET) scanning, functional magnetic resonance imaging (fMRI), and related techniques. PET is often used to measure local glucose metabolism, which is proportionate to neural activity, and fMRI is used to measure local amounts of oxy-genated blood. These techniques make it possible to deter-mine the activity of the various parts of the brain in completely intact normal humans and in humans with many different diseases. They have been used to study not only sim-ple responses but complex aspects of learning, memory, and perception. An example of the use of PET scans to study the functions of the cerebral cortex in processing words is shown in Figure 19–1. Different portions of the cortex are activated when hearing, seeing, speaking, or generating words. Other techniques that have provided information on corti-cal function include stimulation of the exposed cerebral cortex in conscious humans undergoing neurosurgical procedures and, in a few instances, studies with chronically implanted electrodes. Valuable information has also been obtained from investigations in laboratory primates, but it is worth remem-bering that in addition to the difficulties in communicating with them, the brain of the rhesus monkey is only one-fourth the size of the brain of the chimpanzee, our nearest primate relative, and the chimpanzee brain is in turn one-fourth the size of the human brain. LEARNING & MEMORY A characteristic of animals and particularly of humans is the ability to alter behavior on the basis of experience. Learning is acquisition of the information that makes this possible and memory is the retention and storage of that information. The two are obviously closely related and should be considered together. 290 SECTION III Central & Peripheral Neurophysiology FORMS OF MEMORY From a physiologic point of view, memory is appropriately di-vided into explicit and implicit forms (Figure 19–2). Explicit or declarative memory is associated with consciousness—or at least awareness—and is dependent on the hippocampus and other parts of the medial temporal lobes of the brain for its re-tention. Clinical Box 19–1 describes how tracking a patient with brain damage has led to an awareness of the role of the temporal lobe in declarative memory. Implicit or nondeclar-ative memory does not involve awareness, and its retention does not usually involve processing in the hippocampus. Explicit memory is divided into episodic memory for events and semantic memory for facts (eg, words, rules, and language). Explicit memories initially required for activities such as riding a bicycle can become implicit once the task is thoroughly learned. Implicit memory is subdivided into four types. Procedural memory includes skills and habits, which, once acquired, become unconscious and automatic. Priming is facilitation of recognition of words or objects by prior exposure to them. An example is improved recall of a word when presented with the first few letters of it. In nonassociative learning, the organism learns about a single stimulus. In associative learning, the organism learns about the relation of one stimulus to another. Explicit memory and many forms of implicit memory involve (1) short-term memory, which lasts seconds to hours, during which processing in the hippocampus and elsewhere lays down long-term changes in synaptic strength; and (2) long-term memory, which stores memories for years and sometimes for life. During short-term memory, the memory traces are sub-ject to disruption by trauma and various drugs, whereas long-term memory traces are remarkably resistant to disruption. Working memory is a form of short-term memory that keeps information available, usually for very short periods, while the individual plans action based on it. NEURAL BASIS OF MEMORY The key to memory is alteration in the strength of selected synaptic connections. In all but the simplest of cases, the alter-ation involves protein synthesis and activation of genes. This occurs during the change from short-term working memory to long-term memory. In animals, acquisition of long-term learned responses is prevented if, within 5 min after each training session, the animals are anesthetized, given elec-troshock, subjected to hypothermia, or given drugs, antibod-ies, or oligonucleotides that block the synthesis of proteins. If these interventions are performed 4 h after the training ses-sions, there is no effect on acquisition. The human counterpart of this phenomenon is the loss of memory for the events immediately preceding brain concus-sion or electroshock therapy (retrograde amnesia). This amnesia encompasses longer periods than it does in experi-mental animals—sometimes many days—but remote memo-ries remain intact. FIGURE 19–1 Drawings of PET scans of the left cerebral hemisphere showing areas of greatest neuronal activation when subjects performed various language-based activities. (From Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology. McGraw-Hill, 2008.) FIGURE 19–2 Forms of long-term memory. (Modified from Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Two forms of long term memory Implicit (nondeclarative) Explicit (declarative) Facts (Semantic) Medial temporal lobe Hippocampus Neocortex Striatum Amygdala Cerebellum Emotional responses Skeletal musculature Reflex pathways Events (Episodic) Priming Procedural (skills and habits) Associative learning: classical and operant conditioning Nonassociative learni habituation and sensitization CHAPTER 19 Learning, Memory, Language, & Speech 291 SYNAPTIC PLASTICITY & LEARNING Short- and long-term changes in synaptic function can occur as a result of the history of discharge at a synapse; that is, syn-aptic conduction can be strengthened or weakened on the ba-sis of past experience. These changes are of great interest because they represent forms of learning and memory. They can be presynaptic or postsynaptic in location. One form of plastic change is posttetanic potentiation, the production of enhanced postsynaptic potentials in response to stimulation. This enhancement lasts up to 60 seconds and occurs after a brief (tetanizing) train of stimuli in the presyn-aptic neuron. The tetanizing stimulation causes Ca2+ to accu-mulate in the presynaptic neuron to such a degree that the intracellular binding sites that keep cytoplasmic Ca2+ low are overwhelmed. Habituation is a simple form of learning in which a neutral stimulus is repeated many times. The first time it is applied it is novel and evokes a reaction (the orienting reflex or “what is it?” response). However, it evokes less and less electrical response as it is repeated. Eventually, the subject becomes habituated to the stimulus and ignores it. This is associated with decreased release of neurotransmitter from the presynaptic terminal because of decreased intracellular Ca2+. The decrease in intra-cellular Ca2+ is due to a gradual inactivation of Ca2+channels. It can be short term, or it can be prolonged if exposure to the benign stimulus is repeated many times. Habituation is a classic example of nonassociative learning. Sensitization is in a sense the opposite of habituation. Sensi-tization is the prolonged occurrence of augmented postsynaptic responses after a stimulus to which one has become habituated is paired once or several times with a noxious stimulus. At least in the sea snail Aplysia, the noxious stimulus causes discharge of serotonergic neurons that end on the presynaptic endings of sensory neurons. Thus, sensitization is due to presynaptic facil-itation. Sensitization may occur as a transient response, or if it is reinforced by additional pairings of the noxious stimulus and the initial stimulus, it can exhibit features of short-term or long-term memory. The short-term prolongation of sensitization is due to a Ca2+-mediated change in adenylyl cyclase that leads to a greater production of cAMP. The long-term potentiation also involves protein synthesis and growth of the presynaptic and postsynaptic neurons and their connections. Long-term potentiation (LTP) is a rapidly developing per-sistent enhancement of the postsynaptic potential response to presynaptic stimulation after a brief period of rapidly repeated stimulation of the presynaptic neuron. It resembles posttetanic potentiation but is much more prolonged and can last for days. Unlike posttetanic potentiation, it is initiated by an increase in intracellular Ca2+ in the postsynaptic rather than the presynap-tic neuron. It occurs in many parts of the nervous system but has been studied in greatest detail in the hippocampus. There are two forms in the hippocampus: mossy fiber LTP, which is presynaptic and independent of N-methyl-D-aspartate (NMDA) receptors; and Schaffer collateral LTP, which is postsynaptic and NMDA receptor-dependent. The hypotheti-cal basis of the latter form is summarized in Figure 19–3. The basis of mossy fiber LTP is unsettled, although it appears to include cAMP and Ih, a hyperpolarization-activated cation channel. Other parts of the nervous system have not been as well studied, but it is interesting that NMDA-independent LTP can be produced in GABAergic neurons in the amygdala. Long-term depression (LTD) was first noted in the hippo-campus but was subsequently shown to be present through-out the brain in the same fibers as LTP. LTD is the opposite of LTP. It resembles LTP in many ways, but it is characterized by a decrease in synaptic strength. It is produced by slower stim-ulation of presynaptic neurons and is associated with a smaller rise in intracellular Ca2+ than occurs in LTP. In the cerebellum, its occurrence appears to require the phosphory-lation of the GluR2 subunit of the α-amino-3-hydroxy-5-methylisoxazole-4 propionic acid (AMPA) receptors. It may be involved in the mechanism by which learning occurs in the cerebellum. CLINICAL BOX 19–1 The Case of HM: Defining a Link between Brain Function & Memory HM is an anonymous patient who suffered from bilateral temporal lobe seizures that began following a bicycle acci-dent at age 9. His case has been studied by many scientists and has led to a greater understanding of the link between the temporal lobe and declarative memory. HM had par-tial seizures for many years, and then several tonic-clonic sei-zures by age 16. In 1953, at the age of 27, HM underwent bi-lateral surgical removal of the amygdala, large portions of the hippocampal formation, and portions of the association area of the temporal cortex. HM’s seizures were better con-trolled after surgery, but removal of the temporal lobes led to devastating memory deficits. He maintained long-term memory for events that occurred prior to surgery, but he suffered from anterograde amnesia. His short-term mem-ory was intact, but he could not commit new events to long-term memory. He had normal procedural memory, and he could learn new puzzles and motor tasks. His case is the first to bring attention to the critical role of temporal lobes in for-mation of long-term declarative memories and to implicate this region in the conversion of short-term to long-term memories. Later work showed that the hippocampus is the primary structure within the temporal lobe involved in this conversion. Because HM retained memories from before sur-gery, his case also shows that the hippocampus is not in-volved in the storage of declarative memory. An audio-record-ing from the 1990s of HM talking to scientists was released in 2007 and is available at story.php?storyId=7584970. 292 SECTION III Central & Peripheral Neurophysiology CONDITIONED REFLEXES A classic example of associative learning is a conditioned re-flex. A conditioned reflex is a reflex response to a stimulus that previously elicited little or no response, acquired by repeatedly pairing the stimulus with another stimulus that normally does produce the response. In Pavlov’s classic experiments, the sali-vation normally induced by placing meat in the mouth of a dog was studied. A bell was rung just before the meat was placed in the dog’s mouth, and this was repeated a number of times until the animal would salivate when the bell was rung even though no meat was placed in its mouth. In this experiment, the meat placed in the mouth was the unconditioned stimulus (US), the stimulus that normally produces a particular innate response. The conditioned stimulus (CS) was the bell ringing. After the CS and US had been paired a sufficient number of times, the CS produced the response originally evoked only by the US. The CS had to precede the US. An immense number of somatic, viscer-al, and other neural changes can be made to occur as condi-tioned reflex responses. Conditioning of visceral responses is often called biofeed-back. The changes that can be produced include alterations in heart rate and blood pressure. Conditioned decreases in blood pressure have been advocated for the treatment of hyperten-sion; however, the depressor response produced in this fashion is small. INTERCORTICAL TRANSFER OF MEMORY If a cat or monkey is conditioned to respond to a visual stim-ulus with one eye covered and then tested with the blindfold transferred to the other eye, it performs the conditioned re-sponse. This is true even if the optic chiasm has been cut, mak-ing the visual input from each eye go only to the ipsilateral cortex. If, in addition to the optic chiasm, the anterior and posterior commissures and the corpus callosum are sectioned (“split-brain animal”), no memory transfer occurs. Partial cal-losal section experiments indicate that the memory transfer occurs in the anterior portion of the corpus callosum. Similar results have been obtained in humans in whom the corpus cal-losum is congenitally absent or in whom it has been sectioned surgically in an effort to control epileptic seizures. This dem-onstrates that the neural coding necessary for “remembering with one eye what has been learned with the other” has been transferred to the opposite cortex via the commissures. Evi-dence suggests that similar transfer of information is acquired through other sensory pathways. WORKING MEMORY As noted above, working memory keeps incoming informa-tion available for a short time while deciding what to do with it. It is that form of memory which permits us, for example, to look up a telephone number, then remember the number while we pick up the telephone and dial the number. It con-sists of what has been called a central executive located in the prefrontal cortex, and two “rehearsal systems:” a verbal sys-tem for retaining verbal memories and a parallel visuospatial system for retaining visual and spatial aspects of objects. The executive steers information into these rehearsal systems. HIPPOCAMPUS & MEDIAL TEMPORAL LOBE Working memory areas are connected to the hippocampus and the adjacent parahippocampal portions of the medial temporal cortex (Figure 19–4). In humans, bilateral destruc-tion of the ventral hippocampus, or Alzheimer disease and similar disease processes that destroy its CA1 neurons, cause striking defects in short-term memory, as do bilateral lesions of the same area in monkeys. Humans with such destruction have intact working memory and remote memory. Their FIGURE 19–3 Production of LTP in Schaffer collaterals in the hippocampus. Glutamate (Glu) released from the presynaptic neuron binds to AMPA and NMDA receptors in the membrane of the postsyn-aptic neuron. The depolarization triggered by activation of the AMPA receptors relieves the Mg2+ block in the NMDA receptor channel, and Ca2+ enters the neuron with Na+. The increase in cytoplasmic Ca2+ ac-tivates calmodulin (CaM), which in turn activates Ca2+/calmodulin ki-nase II (CaM kII). The kinase phosphorylates the AMPA receptors (P), increasing their conductance, and moves more AMPA receptors into the synaptic cell membrane from cytoplasmic storage sites. In addi-tion, a chemical signal (PS) may pass to the presynaptic neuron, pro-ducing a long-term increase in the quantal release of glutamate. (Courtesy of R Nicoll.) Na+ Ca2+ Ca2+ Mg2+ AMPA NMDA Glu PS AMPA CaM CaM kII P CHAPTER 19 Learning, Memory, Language, & Speech 293 implicit memory processes are generally intact. They perform adequately in terms of conscious memory as long as they con-centrate on what they are doing. However, if they are distract-ed for even a very short period, all memory of what they were doing and what they proposed to do is lost. They are thus ca-pable of new learning and retain old prelesion memories, but they cannot form new long-term memories. The hippocampus is closely associated with the overlying parahippocampal cortex in the medial frontal lobe (Figure 19–4). Memory processes have now been studied not only with fMRI but with measurement of evoked potentials (event-related potentials; ERPs) in epileptic patients with implanted electrodes. When subjects recall words, activity in their left frontal lobe and their left parahippocampal cortex increases, but when they recall pictures or scenes, activity takes place in their right frontal lobe and the parahippocampal cortex on both sides. The connections of the hippocampus to the diencephalon are also involved in memory. Some people with alcoholism-related brain damage develop impairment of recent memory, and the memory loss correlates well with the presence of pathologic changes in the mamillary bodies, which have extensive efferent connections to the hippocampus via the fornix. The mamillary bodies project to the anterior thalamus via the mamillothalamic tract, and in monkeys, lesions of the thalamus cause loss of recent memory. From the thalamus, the fibers concerned with memory project to the prefrontal cortex and from there to the basal forebrain. From the basal forebrain, a diffuse cholinergic projection goes to all of the neocortex, the amygdala, and the hippocampus from the nucleus basalis of Meynert. Severe loss of these fibers occurs in Alzheimer disease. The amygdala is closely associated with the hippocampus and is concerned with encoding and recalling emotionally charged memories. During retrieval of fearful memories, the theta rhythms of the amygdala and the hippocampus become synchronized. In normal humans, events associated with strong emotions are remembered better than events without an emotional charge, but in patients with bilateral lesions of the amygdala, this difference is absent. Confabulation is an interesting though poorly understood condition that sometimes occurs in individuals with lesions of the ventromedial portions of the frontal lobes. These individ-uals perform poorly on memory tests, but they spontaneously describe events that never occurred. This has been called “honest lying.” NEW BRAIN CELLS? It is now established that the traditional view that brain cells are not added after birth is wrong; new neurons form from stem cells throughout life in two areas: the olfactory bulb and the hippocampus. This is a process called neurogenesis. There is evidence implicating a role of neurogenesis in the hippocam-pus with learning and memory. A reduction in the number of new neurons formed reduces at least one form of hippocampal memory production. However, a great deal more is still to be done before the relation of new cells to memory processing can be considered established. LONG-TERM MEMORY While the encoding process for short-term explicit memory involves the hippocampus, long-term memories are stored in various parts of the neocortex. Apparently, the various parts of the memories—visual, olfactory, auditory, etc—are located in the cortical regions concerned with these functions, and the pieces are tied together by long-term changes in the strength of transmission at relevant synaptic junctions so that all the components are brought to consciousness when the memory is recalled. Once long-term memories have been established, they can be recalled or accessed by a large number of different associa-tions. For example, the memory of a vivid scene can be evoked not only by a similar scene but also by a sound or smell associ-ated with the scene and by words such as “scene,” “vivid,” and “view.” Thus, each stored memory must have multiple routes or keys. Furthermore, many memories have an emotional component or “color,” that is, in simplest terms, memories can be pleasant or unpleasant. STRANGENESS & FAMILIARITY It is interesting that stimulation of some parts of the temporal lobes in humans causes a change in interpretation of one’s sur-roundings. For example, when the stimulus is applied, the subject may feel strange in a familiar place or may feel that what is happening now has happened before. The occurrence of a sense of familiarity or a sense of strangeness in appropri-ate situations probably helps the normal individual adjust to FIGURE 19–4 Areas concerned with encoding explicit memories. The prefrontal cortex and the parahippocampal cortex of the brain are active during the encoding of memories. (Modified from Russ MD: Memories are made of this. Science 1998;281:1151.) Prefrontal cortex Hippocampus Parahippocampal cortex 294 SECTION III Central & Peripheral Neurophysiology the environment. In strange surroundings, one is alert and on guard, whereas in familiar surroundings, vigilance is relaxed. An inappropriate feeling of familiarity with new events or in new surroundings is known clinically as the déjà vu phenom-enon, from the French words meaning “already seen.” The phenomenon occurs from time to time in normal individuals, but it also may occur as an aura (a sensation immediately pre-ceding a seizure) in patients with temporal lobe epilepsy. SUMMARY In summary, much is still to be learned about the encoding of explicit memory. However, according to current views, infor-mation from the senses is temporarily stored in various areas of the prefrontal cortex as working memory. It is also passed to the medial temporal lobe, and specifically to the parahippoc-ampal gyrus. From there, it enters the hippocampus and is pro-cessed in a way that is not yet fully understood. At this time, the activity is vulnerable, as described above. Output from the hip-pocampus leaves via the subiculum and the entorhinal cortex and somehow binds together and strengthens circuits in many different neocortical areas, forming over time the stable remote memories that can now be triggered by many different cues. ALZHEIMER DISEASE & SENILE DEMENTIA Alzheimer disease is the most common age-related neurode-generative disorder. Memory decline initially manifests as a loss of episodic memory, which impedes recollection of recent events. Loss of short-term memory is followed by general loss of cognitive and other brain functions, the need for constant care, and, eventually, death. It was originally characterized in middle-aged people, and similar deterioration in elderly individuals is technically senile dementia of the alzheimer type, though it is frequently just called Alzheimer disease. Most cases are sporadic, but some are familial. Senile dementia can be caused by vascular disease and other disorders, but Alzheimer disease is the most com-mon cause, accounting for 50–60% of the cases. It is present in about 17% of the population aged 65–69, but its incidence increases steadily with age, and in those who are 95 and older, the incidence is 40–50%. Thus, Alzheimer disease plus the other forms of senile dementia are a major medical problem. Figure 19–5 summarizes some of the risk factors, patho-genic processes, and clinical signs linked to cellular abnor-malities that occur in Alzheimer disease. The cytopathologic hallmarks of Alzheimer disease are intracellular neurofibril-lary tangles, made up in part of hyperphosphorylated forms of the tau protein that normally binds to microtubules, and extracellular senile plaques, which have a core of β-amyloid peptides (Aβ) surrounded by altered nerve fibers and reactive glial cells. Figure 19–6 compares a normal nerve cell to one showing abnormalities associated with Alzheimer disease. The Aβ peptides are products of a normal protein, amyloid precursor protein (APP), a transmembrane protein that projects into the extracellular fluid (ECF) from all nerve cells. This protein is hydrolyzed at three different sites by α-secre-tase, β-secretase, and γ-secretase, respectively. When APP is hydrolyzed by α-secretase, nontoxic peptide products are pro-duced. However, when it is hydrolyzed by β-secretase and γ-secretase, polypeptides with 40 to 42 amino acids are produced; the actual length varies because of variation in the site at which γ-secretase cuts the protein chain. These polypeptides are toxic, the most toxic being Aβσ1–42. The polypeptides form extracellular aggregates, which can stick to AMPA receptors and Ca2+ ion channels, increasing Ca2+ influx. The polypep-tides also initiate an inflammatory response, with production of intracellular tangles. The damaged cells eventually die. An interesting finding that may well have broad physiologic implications is the observation—now confirmed in a rigorous prospective study—that frequent effortful mental activities, such as doing difficult crossword puzzles and playing board games, slow the onset of cognitive dementia due to Alzheimer disease and vascular disease. The explanation for this “use it or FIGURE 19–5 Relationships of risk factors, pathogenic processes, and clinical signs to cellular abnormalities in the brain during Alzheimer disease. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) Risk factors - Age - Presenilin 1 mutations (chromosome 14) - Presenilin 2 mutations (chromosome 1) - Amyloid precursor protein gene mutations (chromosome 21) - apoE alleles (chromosome 19) - Trisomy 21 Recently a mutation in the α-2 macroglobulin gene has been implicated in the late-onset disease Vulnerable neurons Monoaminergic systems, basal forebrain cholinergic system, hippocampus, entorhinal cortex, and neocortex Cytopathology Neurofibrillary tangles, neurites, Aβ peptide deposition, other cellular abnormalities End-stage disease Senile plaques, death of neurons, gliosis Pathogenic mechanisms Clinical signs Memory loss, cognitive deficits CHAPTER 19 Learning, Memory, Language, & Speech 295 lose it” phenomenon is as yet unknown, but it certainly sug-gests that the hippocampus and its connections have plasticity like other parts of the brain and skeletal and cardiac muscles. LANGUAGE & SPEECH Memory and learning are functions of large parts of the brain, but the centers controlling some of the other “higher functions of the nervous system,” particularly the mechanisms related to language, are more or less localized to the neocortex. Speech and other intellectual functions are especially well developed in humans—the animal species in which the neocortical man-tle is most highly developed. COMPLEMENTARY SPECIALIZATION OF THE HEMISPHERES VERSUS “CEREBRAL DOMINANCE” One group of functions more or less localized to the neocortex in humans consists of those related to language, that is, under-standing the spoken and printed word and expressing ideas in speech and writing. It is a well-established fact that human lan-guage functions depend more on one cerebral hemisphere than on the other. This hemisphere is concerned with categorization and symbolization and has often been called the dominant hemisphere. However, it is clear that the other hemisphere is not simply less developed or “nondominant;” instead, it is spe-cialized in the area of spatiotemporal relations. It is this hemi-sphere that is concerned, for example, with the identification of objects by their form and the recognition of musical themes. It also plays a primary role in the recognition of faces. Conse-quently, the concept of “cerebral dominance” and a dominant and nondominant hemisphere has been replaced by a concept of complementary specialization of the hemispheres, one for se-quential-analytic processes (the categorical hemisphere) and one for visuospatial relations (the representational hemi-sphere). The categorical hemisphere is concerned with lan-guage functions, but hemispheric specialization is also present in monkeys, so it antedates the evolution of language. Clinical Box 19–2 describes deficits that occur in subjects with represen-tational or categorical hemisphere lesions. Hemispheric specialization is related to handedness. Hand-edness appears to be genetically determined. In 96% of right-handed individuals, who constitute 91% of the human popu-lation, the left hemisphere is the dominant or categorical hemisphere, and in the remaining 4%, the right hemisphere is dominant. In approximately 15% of left-handed individuals, the right hemisphere is the categorical hemisphere and in 15%, there is no clear lateralization. However, in the remain-ing 70% of left-handers, the left hemisphere is the categorical hemisphere. It is interesting that learning disabilities such as dyslexia (see Clinical Box 19–3), an impaired ability to learn to read, are 12 times as common in left-handers as they are in right-handers, possibly because some fundamental abnormal-ity in the left hemisphere led to a switch in handedness early in development. However, the spatial talents of left-handers may be well above average; a disproportionately large number FIGURE 19–6 Comparison of a normal neuron and one with abnormalities associated with Alzheimer disease. (From Kandel ER, Schwartz JH, Jessell TM [editors]: Principles of Neural Science, 4th ed. McGraw-Hill, 2000.) A Normal Neuropil threads Neurofibrillary tangles B Alzheimer disease Neurites Senile plaque Abnormal membranous organelles Aβ (fibrillar) Paired helical filaments Nerve terminals 296 SECTION III Central & Peripheral Neurophysiology of artists, musicians, and mathematicians are left-handed. For unknown reasons, left-handers have slightly but significantly shorter life spans than right-handers. Some anatomic differences between the two hemispheres may correlate with the functional differences. The planum temporale, an area of the superior temporal gyrus that is involved in language-related auditory processing, is regularly larger on the left side than the right. It is also larger on the left in the brain of chimpanzees, even though language is almost exclusively a human trait. Imaging studies show that other por-tions of the upper surface of the left temporal lobe are larger in right-handed individuals, the right frontal lobe is normally thicker than the left, and the left occipital lobe is wider and pro-trudes across the midline. Chemical differences also exist between the two sides of the brain. For example, the concentra-tion of dopamine is higher in the nigrostriatal pathway on the left side in right-handed humans but higher on the right in left-handers. The physiologic significance of these differences is unknown. In patients with schizophrenia, MRI studies have demon-strated reduced volumes of gray matter on the left side in the anterior hippocampus, amygdala, parahippocampal gyrus, and posterior superior temporal gyrus. The degree of reduc-tion in the left superior temporal gyrus correlates with the degree of disordered thinking in the disease. There are also apparent abnormalities of dopaminergic systems and cerebral blood flow in this disease. PHYSIOLOGY OF LANGUAGE Language is one of the fundamental bases of human intelli-gence and a key part of human culture. The primary brain ar-eas concerned with language are arrayed along and near the CLINICAL BOX 19–2 Lesions of Representational & Categorical Hemispheres Lesions in the categorical hemisphere produce language disorders, whereas extensive lesions in the representational hemisphere do not. Instead, lesions in the representational hemisphere produce astereognosis—the inability to iden-tify objects by feeling them—and other agnosias. Agnosia is the general term used for the inability to recognize ob-jects by a particular sensory modality even though the sen-sory modality itself is intact. Lesions producing these de-fects are generally in the parietal lobe. Especially when they are in the representational hemisphere, lesions of the infe-rior parietal lobule, a region in the posterior part of the pa-rietal lobe that is close to the occipital lobe, cause unilat-eral inattention and neglect. Individuals with such lesions do not have any apparent primary visual, auditory, or so-matesthetic defects, but they ignore stimuli from the con-tralateral portion of their bodies or the space around these portions. This leads to failure to care for half their bodies and, in extreme cases, to situations in which individuals shave half their faces, dress half their bodies, or read half of each page. This inability to put together a picture of visual space on one side is due to a shift in visual attention to the side of the brain lesion and can be improved, if not totally corrected, by wearing eyeglasses that contain prisms. Hemispheric specialization extends to other parts of the cortex as well. Patients with lesions in the categorical hemi-sphere are disturbed about their disability and often de-pressed, whereas patients with lesions in the representa-tional hemisphere are sometimes unconcerned and even euphoric. Lesions of different parts of the categorical hemi-sphere produce fluent, nonfluent, and anomic aphasias (see text for more details). Although aphasias are produced by lesions of the categorical hemisphere, lesions in the rep-resentational hemisphere also have effects. For example, they may impair the ability to tell a story or make a joke. They may also impair a subject’s ability to get the point of a joke and, more broadly, to comprehend the meaning of dif-ferences in inflection and the “color” of speech. This is one more example of the way the hemispheres are specialized rather than simply being dominant and nondominant. CLINICAL BOX 19–3 Dyslexia Dyslexia, which is a broad term applied to impaired ability to read, is characterized by difficulties with learning how to de-code at the word level, to spell, and to read accurately and fluently. It is frequently due to an inherited abnormality that affects 5% of the population. Many individuals with dyslexic symptoms also have problems with short-term memory skills and problems processing spoken language. Although its pre-cise cause is unknown, there is evidence that dyslexia is of neurological origin. Acquired dyslexias occur due to brain damage in the left hemisphere’s key language areas. Also, in many cases, there is a decreased blood flow in the angular gyrus in the categorical hemisphere. There are numerous the-ories to explain the causes of dyslexia. The phonological hy-pothesis is that dyslexics have a specific impairment in the representation, storage, and/or retrieval of speech sounds. The rapid auditory processing theory proposes that the pri-mary deficit is the perception of short or rapidly varying sounds. The visual theory is that a defect in the magnocellu-lar portion of the visual system slows processing and also leads to phonemic deficit. More selective speech defects have also been described. For example, lesions limited to the left temporal pole (area 38) cause inability to retrieve names of places and persons but preserves the ability to retrieve com-mon nouns, that is, the names of nonunique objects. The abil-ity to retrieve verbs and adjectives is also intact. CHAPTER 19 Learning, Memory, Language, & Speech 297 sylvian fissure (lateral cerebral sulcus) of the categorical hemi-sphere. A region at the posterior end of the superior temporal gyrus called Wernicke’s area (Figure 19–7) is concerned with comprehension of auditory and visual information. It projects via the arcuate fasciculus to Broca’s area (area 44) in the frontal lobe immediately in front of the inferior end of the mo-tor cortex. Broca’s area processes the information received from Wernicke’s area into a detailed and coordinated pattern for vocalization and then projects the pattern via a speech ar-ticulation area in the insula to the motor cortex, which ini-tiates the appropriate movements of the lips, tongue, and larynx to produce speech. The probable sequence of events that occurs when a subject names a visual object is shown in Figure 19–8. The angular gyrus behind Wernicke’s area ap-pears to process information from words that are read in such a way that they can be converted into the auditory forms of the words in Wernicke’s area. It is interesting that in individuals who learn a second lan-guage in adulthood, fMRI reveals that the portion of Broca’s area concerned with it is adjacent to but separate from the area concerned with the native language. However, in chil-dren who learn two languages early in life, only a single area is involved with both. It is well known, of course, that chil-dren acquire fluency in a second language more easily than adults. LANGUAGE DISORDERS Aphasias are abnormalities of language functions that are not due to defects of vision or hearing or to motor paralysis. They are caused by lesions in the categorical hemisphere (see Clinical Box 19–2). The most common cause is embolism or thrombosis of a cerebral blood vessel. Many different classifications of the aphasias have been published, but a convenient classification divides them into fluent, nonfluent, and anomic aphasias. In nonfluent aphasia, the lesion is in Broca’s area (Table 19–1). Speech is slow, and words are hard to come by. Patients with se-vere damage to this area are limited to two or three words with which to express the whole range of meaning and emotion. Sometimes the words retained are those that were being spoken at the time of the injury or vascular accident that caused the aphasia. In one form of fluent aphasia, the lesion is in Wernicke’s area. In this condition, speech itself is normal and sometimes the patients talk excessively. However, what they say is full of jargon and neologisms that make little sense. The patient also fails to comprehend the meaning of spoken or written words, so other aspects of the use of language are compromised. Another form of fluent aphasia is a condition in which patients can speak relatively well and have good auditory comprehension but cannot put parts of words together or conjure up words. This is called conduction aphasia because it was thought to be due to lesions of the arcuate fasciculus connecting Wernicke’s and Broca’s areas. However, it now FIGURE 19–7 Location of some of the areas in the categorical hemisphere that are concerned with language functions. Arcuate fasciculus Angular gyrus Wernicke’s area Broca’s area FIGURE 19–8 Path taken by impulses when a subject names a visual object projected on a horizontal section of the human brain. TABLE 19–1 Aphasias. Characteristic responses of patients with lesions in various areas when shown a picture of a chair. Type of Aphasia and Site of Lesion Characteristic Naming Errors Nonfluent (Broca’s area) “Tssair” Fluent (Wernicke’s area) “Stool” or “choss” (neologism) Fluent (areas 40, 41, and 42; conduc-tion aphasia) “Flair . . . no, swair . . . tair.” Anomic (angular gyrus) “I know what it is . . . I have a lot of them.” Modified from Goodglass H: Disorders of naming following brain injury. Am Sci 1980;68:647. Wernicke’s area (area 22) Angular gyrus (area 39) Higher order visual cortical areas (area 18) Primary visual cortex (area 17) From lateral geniculate nucleus Arcuate fasciculus Facial area of motor cortex (area 4) Broca’s area Left Right 6 5 4 3 2 1 298 SECTION III Central & Peripheral Neurophysiology appears that it is due to lesions in and around the auditory cortex (areas 40, 41, and 42). When a lesion damages the angular gyrus in the categorical hemisphere without affecting Wernicke’s or Broca’s areas, there is no difficulty with speech or the understanding of auditory information; instead there is trouble understanding written language or pictures, because visual information is not processed and transmitted to Wernicke’s area. The result is a condition called anomic aphasia. The isolated lesions that cause the selective defects described above occur in some patients, but brain destruction is often more general. Consequently, more than one form of aphasia is often present. Frequently, the aphasia is general (global), involving both receptive and expressive functions. In this situation, speech is scant as well as nonfluent. Writing is abnormal in all aphasias in which speech is abnormal, but the neural circuits involved are unknown. In addition, deaf subjects who develop a lesion in the categorical hemisphere lose their ability to communicate in sign language. Stuttering has been found to be associated with right cere-bral dominance and widespread overactivity in the cerebral cortex and cerebellum. This includes increased activity of the supplementary motor area. Stimulation of part of this area has been reported to produce laughter, with the duration and intensity of the laughter proportionate to the intensity of the stimulus. RECOGNITION OF FACES An important part of the visual input goes to the inferior tem-poral lobe, where representations of objects, particularly faces, are stored (Figure 19–9). Faces are particularly important in distinguishing friends from foes and the emotional state of those seen. In humans, storage and recognition of faces is more strongly represented in the right inferior temporal lobe in right-handed individuals, though the left lobe is also active. Lesions in this area cause prosopagnosia, the inability to rec-ognize faces. Patients with this abnormality can recognize forms and reproduce them. They can recognize people by their voices, and many of them show autonomic responses when they see familiar as opposed to unfamiliar faces. Howev-er, they cannot identify the familiar faces they see. The left hemisphere is also involved, but the role of the right hemi-sphere is primary. The presence of an autonomic response to a familiar face in the absence of recognition has been ex-plained by postulating the existence of a separate dorsal path-way for processing information about faces that leads to recognition at only a subconscious level. LOCALIZATION OF OTHER FUNCTIONS Use of fMRI and PET scanning combined with study of pa-tients with strokes and head injuries has provided further in-sights—or at least glimpses—into the ways serial processing of sensory information produce cognition, reasoning, compre-hension, and language. Analysis of the brain regions involved in arithmetic calculations has highlighted two areas. In the in-ferior portion of the left frontal lobe is an area concerned with number facts and exact calculations. Frontal lobe lesions can cause acalculia, a selective impairment of mathematical abili-ty. There are areas around the intraparietal sulci of both pari-etal lobes that are concerned with visuospatial representations of numbers and, presumably, finger counting. Two right-sided subcortical structures play a role in accurate navigation in humans. One is the right hippocampus, which is concerned with learning where places are located, and the other is the right caudate nucleus, which facilitates movement to the places. Men have larger brains than women and are said to have superior spatial skills and ability to navigate. Other defects seen in patients with localized cortical lesions include, for example, the inability to name animals, though the ability to name other living things and objects is intact. One patient with a left parietal lesion had difficulty with the second half but not the first half of words. Some patients with pari-etooccipital lesions write only with consonants and omit vow-els. The pattern that emerges from studies of this type is one of precise sequential processing of information in localized brain areas. Additional research of this type should greatly expand our understanding of the functions of the neocortex. CHAPTER SUMMARY ■Long-term memory is divided into explicit (declarative) and implicit (nondeclarative). Explicit is further subdivided into semantic and episodic. Implicit is further subdivided into priming, procedural, associative learning, and nonassociative learning. ■Synaptic plasticity is the ability of neural tissue to change as re-flected by LTP (an increased effectiveness of synaptic activity) or LTD (a reduced effectiveness of synaptic activity) after contin-ued use. ■Hippocampal and other temporal lobe structures and associa-tion cortex are involved in declarative memory. ■Alzheimer disease is characterized by progressive loss of short-term memory followed by general loss of cognitive function. FIGURE 19–9 Areas in the right cerebral hemisphere, in right-handed individuals, that are concerned with recognition of faces. (Modified from Szpir M: Accustomed to your face. Am Sci 1992;80:539.) Stores biographical information Extracts facial features Connects facial features to biographical information CHAPTER 19 Learning, Memory, Language, & Speech 299 The cytopathologic hallmarks of Alzheimer disease are intra-cellular neurofibrillary tangles and extracellular senile plaques. ■Categorical and representational hemispheres are for sequen-tial-analytic processes and visuospatial relations, respectively. Lesions in the categorical hemisphere produce language disor-ders, whereas lesions in the representational hemisphere pro-duce astereognosis. ■Aphasias are abnormalities of language functions and are caused by lesions in the categorical hemisphere. They are classi-fied as fluent (Wernicke’s area; areas 40, 41, 42), nonfluent (Bro-ca’s area), and anomic (angular gyrus) based on the location of brain lesions. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. The representational hemisphere A) is the right cerebral hemisphere in most right-handed individuals. B) is the left cerebral hemisphere in most left-handed individuals. C) includes the part of the brain concerned with language functions. D) is the site of lesions in most patients with aphasia. E) is morphologically identical to the opposite nonrepresenta-tional hemisphere. 2. The optic chiasm and corpus callosum are sectioned in a dog, and with the right eye covered, the animal is trained to bark when it sees a red square. The right eye is then uncovered and the left eye covered. The animal will now A) fail to respond to the red square because the square does not produce impulses that reach the right occipital cortex. B) fail to respond to the red square because the animal has bitemporal hemianopia. C) fail to respond to the red square if the posterior commissure is also sectioned. D) respond to the red square only after retraining. E) respond promptly to the red square in spite of the lack of input to the left occipital cortex. 3. The effects of bilateral loss of hippocampal function include A) disappearance of remote memories. B) loss of working memory. C) loss of the ability to encode events of the recent past in long-term memory. D) loss of the ability to recall faces and forms but not the ability to recall printed or spoken words. E) production of inappropriate emotional responses when recalling events of the recent past. 4. Which of the following are incorrectly paired? A) lesion of the parietal lobe of the representational hemisphere : unilateral inattention and neglect B) loss of cholinergic neurons in the nucleus basalis of Meynert and related areas of the forebrain : loss of recent memory C) lesions of mamillary bodies : loss of recent memory D) lesion of the angular gyrus in the categorical hemisphere : nonfluent aphasia E) lesion of Broca’s area in the categorical hemisphere : slow speech 5. The representational hemisphere is better than the categorical hemisphere at A) language functions. B) recognition of objects by their form. C) understanding printed words. D) understanding spoken words. E) mathematical calculations. 6. A lesion of Wernicke’s area (the posterior end of the superior temporal gyrus) in the categorical hemisphere causes patients to A) lose short-term memory. B) speak in a slow, halting voice. C) experience déjà vu. D) talk rapidly but make little sense. E) lose the ability to recognize faces. 7. Which of the following is most likely not to be involved in pro-duction of LTP? A) NO B) Ca2+ C) NMDA receptors D) membrane hyperpolarization E) membrane depolarization CHAPTER RESOURCES Andersen P, Morris R, Amaral D, Bliss T, O’Keefe J: The Hippocampus Book. Oxford University Press, New York, 2007. Bird CM, Burgess N: The hippocampus and memory: Insights from spatial processing. Nature Rev Neurosci 2008;9:182. Charlton MH: Apasia: A review. Am J Psychiatry 1963;119:872. Goodglass H: Understanding Aphasia. Academic Press, 1993. Ingram VM: Alzheimer’s disease. Am Scientist 2003;91:312. Kandel ER: The molecular biology of memory: A dialogue between genes and synapses. Science 2001;294:1028. Kandel ER, Schwartz JH, Jessell TM (editors): Principles of Neural Science, 4th ed. McGraw-Hill, 2000. LaFerla FM, Green KN, Oddo S: Intracellular amyloid-β in Alzheimer’s disease. Nature Rev Neurosci 2007;8:499. Ramus F: Developmental dyslexia: Specific phonological defect or general sensorimotor dysfunction. Curr Opin Neurobiol 2003;13:212. Russ MD: Memories are made of this. Science 1998;281:1151. Selkoe DJ: Translating cell biology into therapeutic advances in Alzheimer’s disease. Nature 1999;399 (Suppl): A23. Shaywitz S: Dyslexia. N Engl J Med 1998;338:307. Squire LR, Stark CE, Clark RE: The medial temporal lobe. Annu Rev Neurosci 2004; 27:279. Squire LR, Zola SM: Structure and function of declarative and nondeclarative memory systems. Proc Natl Acad Sci 1996;93:13515. This page intentionally left blank 301 C H A P T E R SECTION IV ENDOCRINE & REPRODUCTIVE PHYSIOLOGY 20 The Thyroid Gland O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the structure of the thyroid gland and how it relates to its function. ■Define the chemical nature of the thyroid hormones and how they are synthesized. ■Understand the critical role of iodine in the thyroid gland and how its transport is controlled. ■Describe the role of protein binding in the transport of thyroid hormones and peripheral metabolism. ■Identify the role of the hypothalamus and pituitary in regulating thyroid function. ■Define the effects of the thyroid hormones in homeostasis and development. ■Understand the basis of conditions where thyroid function is abnormal and how they can be treated. INTRODUCTION The thyroid gland is one of the larger endocrine glands of the body. The gland has two primary functions. The first is to secrete the thyroid hormones, which maintain the level of metabolism in the tissues that is optimal for their normal function. Thyroid hor-mones stimulate O2 consumption by most of the cells in the body, help regulate lipid and carbohydrate metabolism, and thereby influence body mass and mentation. Consequences of thyroid gland dysfunction depend on the life stage at which they occur. The thyroid is not essential for life, but its absence or hypofunc-tion during fetal and neonatal life results in severe mental retarda-tion and dwarfism. In adults, hypothyroidism is accompanied by mental and physical slowing and poor resistance to cold. Con-versely, excess thyroid secretion leads to body wasting, nervous-ness, tachycardia, tremor, and excess heat production. Thyroid function is controlled by the thyroid-stimulating hormone (TSH, thyrotropin) of the anterior pituitary. The secretion of this hor-mone is in turn increased by thyrotropin-releasing hormone (TRH) from the hypothalamus and is also subject to negative feedback control by high circulating levels of thyroid hormones acting on the anterior pituitary and the hypothalamus. The second function of the thyroid gland is to secrete calcito-nin, a hormone that regulates circulating levels of calcium. This function of the thyroid gland is discussed in Chapter 23 in the broader context of whole body calcium homeostasis. 302 SECTION IV Endocrine & Reproductive Physiology ANATOMIC CONSIDERATIONS The thyroid is a butterfly-shaped gland that straddles the trachea in the front of the neck. It develops from an evagination of the floor of the pharynx, and a thyroglossal duct marking the path of the thyroid from the tongue to the neck sometimes persists in the adult. The two lobes of the human thyroid are connected by a bridge of tissue, the thyroid isthmus, and there is sometimes a pyramidal lobe arising from the isthmus in front of the larynx (Figure 20–1). The gland is well vascularized, and the thyroid has one of the highest rates of blood flow per gram of tissue of any or-gan in the body. The portion of the thyroid concerned with the production of thyroid hormone consists of multiple acini (follicles). Each spherical follicle is surrounded by a single layer of polarized epi-thelial cells and filled with pink-staining proteinaceous material called colloid. Colloid consists predominantly of the glycopro-tein, thyroglobulin. When the gland is inactive, the colloid is abundant, the follicles are large, and the cells lining them are flat. When the gland is active, the follicles are small, the cells are cuboid or columnar, and areas where the colloid is being actively reabsorbed into the thyrocytes are visible as “reabsorption lacu-nae” (Figure 20–2). Microvilli project into the colloid from the apexes of the thyroid cells and canaliculi extend into them. The endoplas-mic reticulum is prominent, a feature common to most glan-dular cells, and secretory granules containing thyroglobulin are seen (Figure 20–3). The individual thyroid cells rest on a basal lamina that separates them from the adjacent capillaries. The capillaries are fenestrated, like those of other endocrine glands (see Chapter 32). FORMATION & SECRETION OF THYROID HORMONES CHEMISTRY The primary hormone secreted by the thyroid is thyroxine (T4), along with much lesser amounts of triiodothyronine FIGURE 20–1 The human thyroid. Left lobe Right lobe Larynx Pyramidal lobe Hyoid bone Left lobe Right lobe Larynx Pyramidal lobe Hyoid bone FIGURE 20–2 Thyroid histology. Note the small, punched-out “reabsorption lacunae” in the colloid next to the cells in the active gland. FIGURE 20–3 Thyroid cell. Left: Normal pattern. Right: After TSH stimulation. The arrows on the right show the secretion of thyro-globulin into the colloid. On the right, endocytosis of the colloid and merging of a colloid-containing vacuole with a lysosome are also shown. The cell rests on a capillary with gaps (fenestrations) in the endothelial wall. Inactive Active Colloid Reabsorption lacunae Parafollicular cells Secretory droplets Lumen of follicle Uptake of colloid by endocytosis Lysosome coalescing with endo-cytotic vacuole Golgi apparatus Cell basal lamina Capillary basal lamina Capillary endothelium Normal TSH-stimulated CHAPTER 20 The Thyroid Gland 303 (T3). T3 has much greater biological activity than T4 and is specifically generated at its site of action in peripheral tissues by deiodination of T4 (see below). Both hormones are iodine-containing amino acids (Figure 20–4). Small amounts of re-verse triiodothyronine (3,3',5'-triiodothyronine, RT3) and other compounds are also found in thyroid venous blood. RT3 is not biologically active. IODINE HOMEOSTASIS Iodine is an essential raw material for thyroid hormone syn-thesis. Dietary iodide is absorbed by the intestine and enters the circulation; its subsequent fate is summarized in Figure 20–5. The minimum daily iodine intake that will maintain normal thyroid function is 150 μg in adults. In most developed countries, supplementation of table salt means that the aver-age dietary intake is approximately 500 μg/d. The principal or-gans that take up circulating I– are the thyroid, which uses it to make thyroid hormones, and the kidneys, which excrete it in the urine. About 120 μg/d enter the thyroid at normal rates of thyroid hormone synthesis and secretion. The thyroid secretes 80 μg/d in the form of T3 and T4, while 40 μg/d diffuses back into the extracellular fluid (ECF). Circulating T3 and T4 are metabolized in the liver and other tissues, with the release of a further 60 μg of I– per day into the ECF. Some thyroid hor-mone derivatives are excreted in the bile, and some of the io-dine in them is reabsorbed (enterohepatic circulation), but there is a net loss of I– in the stool of approximately 20 μg/d. The total amount of I– entering the ECF is thus 500 + 40 + 60, or 600 μg/d; 20% of this I– enters the thyroid, whereas 80% is excreted in the urine. IODIDE TRANSPORT ACROSS THYROCYTES The basolateral membranes of thyrocytes facing the capillaries contain a symporter that transports two Na+ ions and one I– ion into the cell with each cycle, against the electrochemical gradient for I–. This Na+/I– symporter (NIS) is capable of pro-ducing intracellular I– concentrations that are 20 to 40 times as great as the concentration in plasma. The process involved is secondary active transport (see Chapter 2), with the energy provided by active transport of Na+ out of thyroid cells by Na, K ATPase. NIS is regulated both by transcriptional means and by active trafficking into and out of the thyrocyte basolat-eral membrane; in particular, thyroid stimulating hormone (TSH; see below) induces both NIS expression and the reten-tion of NIS in the basolateral membrane where it can mediate sustained iodide uptake. Iodide must also exit the thyrocyte across the apical mem-brane to access the colloid, where the initial steps of thyroid hormone synthesis occur. This transport step is believed to be mediated, at least in part, by a Cl–/I– exchanger known as pendrin. This protein was first identified as the product of the gene responsible for the Pendred syndrome, whose patients suffer from thyroid dysfunction and deafness. Pendrin (SLC26A4) is one member of the larger family of SLC26 anion exchangers. The relation of thyroid function to iodide is unique. As dis-cussed in more detail below, iodide is essential for normal thyroid function, but iodide deficiency and iodide excess both inhibit thyroid function. The salivary glands, the gastric mucosa, the placenta, the ciliary body of the eye, the choroid plexus, the mammary glands, and certain cancers derived from these tissues also express NIS and can transport iodide against a concentration gradient, but the transporter in these tissues is not affected by TSH. The physiologic significance of all these extrathyroidal iodide-concentrating mechanisms is obscure, but they may provide pathways for radioablation of NIS-expressing cancer cells using iodide radioisotopes. This approach is also useful for the ablation of thyroid cancers. FIGURE 20–4 Thyroid hormones. The numbers in the rings in the T4 formula indicate the number of positions in the molecule. RT3 is 3,3',5'-triiodothyronine. HO O CH2 CH NH2 OH C O Ι Ι Ι Ι 3' 5' 3 5 3,5,3',5',-Tetraiodothyronine (thyroxine, T4) HO O CH2 CH NH2 OH C O Ι Ι Ι 3,5,3',-Triiodothyronine (T3) FIGURE 20–5 Iodine metabolism. Liver and other tissues Thyroid 120 μg I− 40 μg I− 60 μg I− Extracellular fluid 500 μg I− in diet 480 μg I− in urine 20 μg I− in stool 80 μg in T3, T4 Bile 304 SECTION IV Endocrine & Reproductive Physiology THYROID HORMONE SYNTHESIS & SECRETION At the interface between the thyrocyte and the colloid, iodide undergoes a process referred to as organification. First, it is oxidized to iodine, and then incorporated into the carbon 3 position of tyrosine residues that are part of the thyroglobulin molecule in the colloid (Figure 20–6). Thyroglobulin is a gly-coprotein made up of two subunits and has a molecular weight of 660 kDa. It contains 10% carbohydrate by weight. It also contains 123 tyrosine residues, but only 4 to 8 of these are nor-mally incorporated into thyroid hormones. Thyroglobulin is synthesized in the thyroid cells and secreted into the colloid by exocytosis of granules. The oxidation and reaction of iodide with the secreted thyroglobulin is mediated by thyroid perox-idase, a membrane-bound enzyme found in the thyrocyte api-cal membrane. The thyroid hormones so produced remain part of the thyroglobulin molecule until needed. As such, col-loid represents a reservoir of thyroid hormones, and humans can ingest a diet completely devoid of iodide for up to 2 months before a decline in circulating thyroid hormone levels is seen. When there is a need for thyroid hormone secretion, colloid is internalized by the thyrocytes by endocytosis, and directed toward lysosomal degradation. Thus, the peptide bonds of thyroglobulin are hydrolyzed, and free T4 and T3 are discharged into cytosol and thence to the capillaries (see be-low). Thyrocytes thus have four functions: They collect and transport iodine, they synthesize thyroglobulin and secrete it into the colloid, they fix iodine to the thyroglobulin to gener-ate thyroid hormones, and they remove the thyroid hormones from thyroglobulin and secrete them into the circulation. Thyroid hormone synthesis is a multistep process. Thyroid peroxidase generates reactive iodine species that can attack thyroglobulin. The first product is monoiodotyrosine (MIT). MIT is next iodinated on the carbon 5 position to form diiodotyrosine (DIT). Two DIT molecules then undergo an oxidative condensation to form T4 with the elimination of the alanine side chain from the molecule that forms the outer ring. There are two theories of how this coupling reaction occurs. One holds that the coupling occurs with both DIT molecules attached to thyroglobulin (intramolecular cou-pling). The other holds that the DIT that forms the outer ring is first detached from thyroglobulin (intermolecular cou-pling). In either case, thyroid peroxidase is involved in cou-pling as well as iodination. T3 is formed by condensation of MIT with DIT. A small amount of RT3 is also formed, proba-bly by condensation of DIT with MIT. In the normal human thyroid, the average distribution of iodinated compounds is FIGURE 20–6 Outline of thyroid hormone biosynthesis. Iodination of tyrosine takes place at the apical border of the thyroid cells while the molecules are bound in peptide linkage in thyroglobulin. HO CH2−CH CH CH HO CH2− Tyrosine Ι Ι− Ι− (Iodide) Ι (Iodine) HO CH2− Ι Ι CH O CH2− Ι Ι DIT + DIT Alanine + HO Ι Ι 3-Monoiodotyrosine (MIT) 3,5-Diiodotyrosine (DIT) Ι− Ι− Ι MIT + DIT Alanine + 3,5,3'-Triiodothyronine (T3) DIT + MIT Alanine + 3,3',5'-Triiodothyronine (reverse T3) Thyroxine (T4) THYROID CELL Active transport PLASMA COLLOID THYRO-GLOBULIN MOLECULE CHAPTER 20 The Thyroid Gland 305 23% MIT, 33% DIT, 35% T4, and 7% T3. Only traces of RT3 and other components are present. The human thyroid secretes about 80 μg (103 nmol) of T4, 4 μg (7 nmol) of T3, and 2 μg (3.5 nmol) of RT3 per day (Fig-ure 20–7). MIT and DIT are not secreted. These iodinated tyrosines are deiodinated by a microsomal iodotyrosine deiodinase. This represents a mechanism to recover iodine and bound tyrosines and recycle them for additional rounds of hormone synthesis. The iodine liberated by deiodination of MIT and DIT is reutilized in the gland and normally provides about twice as much iodide for hormone synthesis as NIS does. In patients with congenital absence of the iodotyrosine deiodinase, MIT and DIT appear in the urine and there are symptoms of iodine deficiency (see below). Iodinated thy-ronines are resistant to the activity of iodotyrosine deiodinase, thus allowing T4 and T3 to pass into the circulation. TRANSPORT & METABOLISM OF THYROID HORMONES PROTEIN BINDING The normal total plasma T4 level in adults is approximately 8 μg/dL (103 nmol/L), and the plasma T3 level is approximately 0.15 μg/dL (2.3 nmol/L). T4 and T3 are relatively lipophilic; thus, their free forms in plasma are in equilibrium with a much larger pool of protein-bound thyroid hormones in plas-ma and in tissues. Free thyroid hormones are added to the cir-culating pool by the thyroid. It is the free thyroid hormones in plasma that are physiologically active and that feed back to in-hibit pituitary secretion of TSH (Figure 20–8). The function of protein-binding appears to be maintenance of a large pool of hormone that can readily be mobilized as needed. In addition, at least for T3, hormone binding prevents excess uptake by the first cells encountered and promotes uniform tissue distribu-tion. Both total T4 and T3 can be measured by radioimmu-noassay. There are also direct assays that specifically measure only the free forms of the hormones. The latter are the more clinically relevant measures given that these are the active forms, and also due to both acquired and congenital variations in the concentrations of binding proteins between individuals. The plasma proteins that bind thyroid hormones are albu-min, a prealbumin called transthyretin (formerly called thy-roxine-binding prealbumin), and a globulin known as thyroxine-binding globulin (TBG). Of the three proteins, albumin has the largest capacity to bind T4 (ie, it can bind the most T4 before becoming saturated) and TBG has the smallest capacity. However, the affinities of the proteins for T4 (ie, the avidity with which they bind T4 under physiologic condi-tions) are such that most of the circulating T4 is bound to TBG (Table 20–1), with over a third of the binding sites on the protein occupied. Smaller amounts of T4 are bound to transthyretin and albumin. The half-life of transthyretin is 2 d, that of TBG is 5 d, and that of albumin is 13 d. Normally, 99.98% of the T4 in plasma is bound; the free T4 level is only about 2 ng/dL. There is very little T4 in the urine. Its biologic half-life is long (about 6–7 d), and its volume of distribution is less than that of ECF (10 L, or about 15% of body weight). All of these properties are characteristic of a substance that is strongly bound to protein. T3 is not bound to quite as great an extent; of the 0.15 μg/dL normally found in plasma, 0.2% (0.3 ng/dL) is free. The remaining 99.8% is protein-bound, 46% to TBG and most of the remainder to albumin, with very little binding to tran-sthyretin (Table 20–1). The lesser binding of T3 correlates with the facts that T3 has a shorter half-life than T4 and that FIGURE 20–7 Secretion and interconversion of thyroid hormones in normal adult humans. Figures are in micrograms per day. Note that most of the T3 and RT3 are formed from T4 deiodination in the tissues and only small amounts are secreted by the thyroid. Conjugates, etc Thyroid 80 4 27 36 2 17 RT3 38 μg T4 80 μg T3 31 μg FIGURE 20–8 Regulation of thyroid hormone synthesis. TABLE 20–1 Binding of thyroid hormones to plasma proteins in normal adult humans. Protein Plasma Concentration (mg/dL) Amount of Circulating Hormone Bound (%) T4 T3 Thyroxine-binding globulin (TBG) 2 67 46 Transthyretin (thyrox-ine-binding prealbu-min, TBPA) 15 20 1 Albumin 3500 13 53 Thyroid T4 Pituitary TSH Free T4 (0.002 μg/dL) Plasma protein-bound T4 (8 μg/dL) Tissue pro-tein-bound T4 306 SECTION IV Endocrine & Reproductive Physiology its action on the tissues is much more rapid. RT3 also binds to TBG. FLUCTUATIONS IN BINDING When a sudden, sustained increase in the concentration of thy-roid-binding proteins in the plasma takes place, the concentra-tion of free thyroid hormones falls. This change is temporary, however, because the decrease in the concentration of free thy-roid hormones in the circulation stimulates TSH secretion, which in turn causes an increase in the production of free thy-roid hormones. A new equilibrium is eventually reached at which the total quantity of thyroid hormones in the blood is elevated but the concentration of free hormones, the rate of their metabolism, and the rate of TSH secretion are normal. Corresponding changes in the opposite direction occur when the concentration of thyroid-binding protein is reduced. Con-sequently, patients with elevated or decreased concentrations of binding proteins, particularly TBG, are typically neither hy-per- nor hypothyroid; that is, they are euthyroid. TBG levels are elevated in estrogen-treated patients and dur-ing pregnancy, as well as after treatment with various drugs (Table 20–2). They are depressed by glucocorticoids, andro-gens, the weak androgen danazol, and the cancer chemo-therapeutic agent L-asparaginase. A number of other drugs, including salicylates, the anti-convulsant phenytoin, and the cancer chemotherapeutic agents mitotane (o, p'-DDD) and 5-fluorouracil inhibit binding of T4 and T3 to TBG and conse-quently produce changes similar to those produced by a decrease in TBG concentration. Changes in total plasma T4 and T3 can also be produced by changes in plasma concentra-tions of albumin and prealbumin. METABOLISM OF THYROID HORMONES T4 and T3 are deiodinated in the liver, the kidneys, and many other tissues. These deiodination reactions serve not only to catabolize the hormones, but also to provide a local supply specifically of T3, which is believed to be the primary mediator of the physiological effects of thyroid secretion. One third of the circulating T4 is normally converted to T3 in adult hu-mans, and 45% is converted to RT3. As shown in Figure 20–7, only about 13% of the circulating T3 is secreted by the thyroid while 87% is formed by deiodination of T4; similarly, only 5% of the circulating RT3 is secreted by the thyroid and 95% is formed by deiodination of T4. It should be noted as well that marked differences in the ratio of T3 to T4 occur in various tis-sues. Two tissues that have very high T3/T4 ratios are the pitu-itary and the cerebral cortex, due to the expression of specific deiodinases, as discussed below. Three different deiodinases act on thyroid hormones: D1, D2, and D3. All are unique in that they contain the rare amino acid selenocysteine, with selenium in place of sulfur, which is essential for their enzymatic activity. D1 is present in high concentrations in the liver, kidneys, thyroid, and pituitary. It appears primarily to be responsible for monitor-ing the formation of T3 from T4 in the periphery. D2 is present in the brain, pituitary, and brown fat. It also contrib-utes to the formation of T3. In the brain, it is located in astro-glia and produces a supply of T3 to neurons. D3 is also present in the brain and in reproductive tissues. It acts only on the 5 position of T4 and T3 and is probably the main source of RT3 in the blood and tissues. Overall, the deiodi-nases appear to be responsible for maintaining differences in T3/T4 ratios in the various tissues in the body. In the brain, in particular, high levels of deiodinase activity ensure an ample supply of active T3. Some of the T4 and T3 is further converted to deiodoty-rosines by deiodinases. T4 and T3 are also conjugated in the liver to form sulfates and glucuronides. These conjugates enter the bile and pass into the intestine. The thyroid conju-gates are hydrolyzed, and some are reabsorbed (enterohepatic circulation), but some are excreted in the stool. In addition, some T4 and T3 passes directly from the circulation to the intestinal lumen. The iodide lost by these routes amounts to about 4% of the total daily iodide loss. FLUCTUATIONS IN DEIODINATION Much more RT3 and much less T3 are formed during fetal life, and the ratio shifts to that of adults about 6 wk after birth. Var-ious drugs inhibit deiodinases, producing a fall in plasma T3 TABLE 20–2 Effect of variations in the concentrations of thyroid hormone-binding proteins in the plasma on various parameters of thyroid function after equilibrium has been reached. Condition Concentrations of Binding Proteins Total Plasma T4, T3, RT3 Free Plasma T4, T3, RT3 Plasma TSH Clinical State Hyperthyroidism Normal High High Low Hyperthyroid Hypothyroidism Normal Low Low High Hypothyroid Estrogens, methadone, heroin, major tranquilizers, clofibrate High High Normal Normal Euthyroid Glucocorticoids, androgens, danazol, asparaginase Low Low Normal Normal Euthyroid CHAPTER 20 The Thyroid Gland 307 levels and a reciprocal rise in RT3. Selenium deficiency has the same effect. A wide variety of nonthyroidal illnesses also sup-press deiodinases. These include burns, trauma, advanced cancer, cirrhosis, renal failure, myocardial infarction, and feb-rile states. The low-T3 state produced by these conditions dis-appears with recovery. It is difficult to decide whether individuals with the low-T3 state produced by drugs and ill-ness have mild hypothyroidism. Diet also has a clear-cut effect on conversion of T4 to T3. In fasted individuals, plasma T3 is reduced by 10–20% within 24 h and by about 50% in 3 to 7 d, with a corresponding rise in RT3 (Figure 20–9). Free and bound T4 levels remain essen-tially normal. During more prolonged starvation, RT3 returns to normal but T3 remains depressed. At the same time, the basal metabolic rate (BMR) falls and urinary nitrogen excre-tion, an index of protein breakdown, is decreased. Thus, the decline in T3 conserves calories and protein. Conversely, overfeeding increases T3 and reduces RT3. REGULATION OF THYROID SECRETION Thyroid function is regulated primarily by variations in the circulating level of pituitary TSH (Figure 20–8). TSH secretion is increased by the hypothalamic hormone thyrotropin-releas-ing hormone (TRH; see Chapter 18) and inhibited in a nega-tive feedback fashion by circulating free T4 and T3. The effect of T4 is enhanced by production of T3 in the cytoplasm of the pituitary cells by the 5'-D2 they contain. TSH secretion is also inhibited by stress, and in experimental animals it is increased by cold and decreased by warmth. CHEMISTRY & METABOLISM OF TSH Human TSH is a glycoprotein that contains 211 amino acid res-idues. It is made up of two subunits, designated α and β. The α subunit is encoded by a gene on chromosome 6 and the β sub-unit by a gene on chromosome 1. The α and β subunits become noncovalently linked in the pituitary thyrotropes. TSH-α is identical to the α subunit of LH, FSH, and hCG-α (see Chapters 24 and 25). The functional specificity of TSH is conferred by the β subunit. The structure of TSH varies from species to species, but other mammalian TSHs are biologically active in humans. The biologic half-life of human TSH is about 60 min. TSH is degraded for the most part in the kidneys and to a lesser extent in the liver. Secretion is pulsatile, and mean output starts to rise at about 9:00 PM, peaks at midnight, and then declines during the day. The normal secretion rate is about 110 μg/d. The aver-age plasma level is about 2 μg/mL (Figure 20–10). Because the α subunit in hCG is the same as that in TSH, large amounts of hCG can activate thyroid receptors nonspe-cifically. In some patients with benign or malignant tumors of placental origin, plasma hCG levels can rise so high that they produce mild hyperthyroidism. EFFECTS OF TSH ON THE THYROID When the pituitary is removed, thyroid function is depressed and the gland atrophies; when TSH is administered, thyroid FIGURE 20–9 Effect of starvation on plasma levels of T4, T3, and RT3 in humans. Similar changes occur in wasting diseases. The scale for T3 and RT3 is on the left and the scale for T4 is on the right. (Reproduced with permission from Burger AG: New aspects of the peripheral action of thyroid hormones. Triangle, Sandoz J Med Sci 1983;22:175. Copyright © 1983 Sandoz Ltd., Basel, Switzerland.) Starvation Days 240 200 160 120 80 40 0 −4 −2 0 2 4 6 8 10 +2 +4 12 10 8 6 4 2 0 T4 T3 RT3 ng/dL μg/dL FIGURE 20–10 Relation between plasma TSH, measured by a highly sensitive radioimmunoassay, and plasma free T4, measured by dialysis (FT4). Note that the TSH scale is a log scale. Levothyroxine suppressed (euthyroid) (n = 20) Hypothyroid (primary) (n = 49) Hyperthyroid (nonpituitary) (n = 56) Euthyroid (n = 194) 1000 100 10 1 0.1 0.01 0 1 2 3 4 5 6 30 TSH (μU/mL) FT4 (ng/dL) 308 SECTION IV Endocrine & Reproductive Physiology function is stimulated. Within a few minutes after the injec-tion of TSH, there are increases in iodide binding; synthesis of T3, T4, and iodotyrosines; secretion of thyroglobulin into the colloid; and endocytosis of colloid. Iodide trapping is in-creased in a few hours; blood flow increases; and, with chronic TSH treatment, the cells hypertrophy and the weight of the gland increases. Whenever TSH stimulation is prolonged, the thyroid becomes detectably enlarged. Enlargement of the thyroid is called a goiter. TSH RECEPTORS The TSH receptor is a typical G protein-coupled, seven-trans-membrane segment receptor that activates adenylyl cyclase through Gs. It also activates phospholipase C (PLC). Like oth-er glycoprotein hormone receptors, it has an extended, glyco-sylated extracellular domain. OTHER FACTORS AFFECTING THYROID GROWTH In addition to TSH receptors, thyrocytes express receptors for insulin-like growth factor I (IGF-I), EGF, and other growth factors. IGF-I and EGF promote growth, whereas interferon γ and tumor necrosis factor α inhibit growth. The exact physio-logic role of these factors in the thyroid has not been estab-lished, but the effect of the cytokines implies that thyroid function might be inhibited in the setting of chronic inflam-mation, which could contribute to cachexia, or weight loss. CONTROL MECHANISMS The mechanisms regulating thyroid secretion are summarized in Figure 20–8. The negative feedback effect of thyroid hor-mones on TSH secretion is exerted in part at the hypothalamic level, but it is also due in large part to an action on the pituitary, since T4 and T3 block the increase in TSH secretion produced by TRH. Infusion of either T4 or T3 reduces the circulating level of TSH, which declines measurably within 1 hour. In experi-mental animals, there is an initial rise in pituitary TSH content before the decline, indicating that thyroid hormones inhibit se-cretion before they inhibit synthesis. The effects on secretion and synthesis of TSH both appear to depend on protein synthe-sis, even though the former is relatively rapid. The day-to-day maintenance of thyroid secretion depends on the feedback interplay of thyroid hormones with TSH and TRH (Figure 20–8). The adjustments that appear to be medi-ated via TRH include the increased secretion of thyroid hor-mones produced by cold and, presumably, the decrease produced by heat. It is worth noting that although cold pro-duces clear-cut increases in circulating TSH in experimental animals and human infants, the rise produced by cold in adult humans is negligible. Consequently, in adults, increased heat production due to increased thyroid hormone secretion (thy-roid hormone thermogenesis) plays little if any role in the response to cold. Stress has an inhibitory effect on TRH secre-tion. Dopamine and somatostatin act at the pituitary level to inhibit TSH secretion, but it is not known whether they play a physiologic role in the regulation of TSH secretion. Glucocor-ticoids also inhibit TSH secretion. The amount of thyroid hormone necessary to maintain nor-mal cellular function in thyroidectomized individuals used to be defined as the amount necessary to normalize the BMR, but it is now defined as the amount necessary to return plasma TSH to normal. Indeed, with the accuracy and sensitivity of modern assays for TSH and the marked inverse correlation between plasma free thyroid hormone levels and plasma TSH, measurement of TSH is now widely regarded as one of the best tests of thyroid function. The amount of T4 that normalizes plasma TSH in athyreotic individuals averages 112 μg of T4 by mouth per day in adults. About 80% of this dose is absorbed from the gastrointestinal tract. It produces a slightly greater than normal FT4I but a normal FT3I, indicating that in humans, unlike some experimental animals, it is circulating T3 rather than T4 that is the principal feedback regulator of TSH secretion (see Clinical Boxes 20–1 and 20–2). EFFECTS OF THYROID HORMONES Some of the widespread effects of thyroid hormones in the body are secondary to stimulation of O2 consumption (calori-genic action), although the hormones also affect growth and development in mammals, help regulate lipid metabolism, and increase the absorption of carbohydrates from the intestine (Table 20–5 on page 311). They also increase the dissociation of oxygen from hemoglobin by increasing red cell 2,3-diphos-phoglycerate (DPG) (see Chapter 36). MECHANISM OF ACTION Thyroid hormones enter cells and T3 binds to thyroid recep-tors (TR) in the nuclei. T4 can also bind, but not as avidly. The hormone-receptor complex then binds to DNA via zinc fin-gers and increases (or in some cases, decreases) the expression of a variety of different genes that code for proteins that regu-late cell function (see Chapter 1). Thus, the nuclear receptors for thyroid hormones are members of the superfamily of hor-mone-sensitive nuclear transcription factors. There are two human TR genes: an α receptor gene on chromosome 17 and a β receptor gene on chromosome 3. By alternative splicing, each forms at least two different mRNAs and therefore two different receptor proteins. TRβ2 is found only in the brain, but TRα1, TRα2, and TRβ1 are widely dis-tributed. TRα2 differs from the other three in that it does not bind T3 and its function is not yet fully established. TRs bind to DNA as monomers, homodimers, and heterodimers with other nuclear receptors, particularly the retinoid X receptor CHAPTER 20 The Thyroid Gland 309 (RXR). The TR/RXR heterodimer does not bind 9-cis retinoic acid, the usual ligand for RXR, but TR binding to DNA is greatly enhanced in response to thyroid hormones when the receptor is in the form of this heterodimer. There are also coactivator and corepressor proteins that affect the actions of TRs. Presumably, this complexity underlies the ability of thy-roid hormones to produce many different effects in the body. In most of its actions, T3 acts more rapidly and is three to five times more potent than T4 (Figure 20–13). This is because T3 is less tightly bound to plasma proteins than is T4, but binds more avidly to thyroid hormone receptors. RT3 is inert (see Clinical Box 20–3). CLINICAL BOX 20–1 Reduced Thyroid Function The syndrome of adult hypothyroidism is generally called myx-edema, although this term is also used to refer specifically to the skin changes in the syndrome. Hypothyroidism may be the end result of a number of diseases of the thyroid gland, or it may be secondary to pituitary or hypothalamic failure. In the latter two conditions, the thyroid remains able to respond to TSH. Thyroid function may be reduced by a number of conditions (Table 20– 3). For example, when the dietary iodine intake falls below 50 μg/d, thyroid hormone synthesis is inadequate and secretion de-clines. As a result of increased TSH secretion, the thyroid hyper-trophies, producing an iodine deficiency goiter that may be-come very large. Such “endemic goiters” have been substantially reduced by the practice of adding iodide to table salt. Drugs may also inhibit thyroid function. Most do so either by interfering with the iodide-trapping mechanism or by blocking the organic binding of iodine. In either case, TSH secretion is stimulated by the decline in circulating thyroid hormones, and a goiter is pro-duced. The thioureylenes, a group of compounds related to thiourea, inhibit the iodination of monoiodotyrosine and block the coupling reaction. The two used clinically are propylthioura-cil and methimazole (Figure 20–11). Iodination of tyrosine is in-hibited because propylthiouracil and methimazole compete with tyrosine residues for iodine and become iodinated. In addi-tion, propylthiouracil but not methimazole inhibits D2 deiodi-nase, reducing the conversion of T4 to T3 in many extrathyroidal tissues. Paradoxically, another substance that inhibits thyroid function under certain conditions is iodide itself. In normal indi-viduals, large doses of iodide act directly on the thyroid to pro-duce a mild and transient inhibition of organic binding of iodide and hence of hormone synthesis. This inhibition is known as the Wolff–Chaikoff effect. In completely athyreotic adults, the BMR falls to about 40%. The hair is coarse and sparse, the skin is dry and yellowish (carotenemia), and cold is poorly tolerated. Mentation is slow, memory is poor, and in some patients there are severe mental symptoms (“myxedema madness”). Plasma cholesterol is ele-vated. Children who are hypothyroid from birth or before are called cretins. They are dwarfed and mentally retarded. Worldwide, congenital hypothyroidism is one of the most common causes of preventable mental retardation. The main causes are included in Table 20–3. They include not only ma-ternal iodine deficiency and various congenital abnormalities of the fetal hypothalamo–pituitary–thyroid axis, but also ma-ternal antithyroid antibodies that cross the placenta and damage the fetal thyroid. T4 crosses the placenta, and unless the mother is hypothyroid, growth and development are nor-mal until birth. If treatment is started at birth, the prognosis for normal growth and development is good, and mental retarda-tion can generally be avoided; for this reason, screening tests for congenital hypothyroidism are becoming routine. When the mother is hypothyroid as well, as in the case of iodine defi-ciency, the mental deficiency is more severe and less respon-sive to treatment after birth. It has been estimated that 20 mil-lion people in the world now have various degrees of brain damage caused by iodine deficiency in utero. Uptake of tracer doses of radioactive iodine can be used to assess thyroid function (contrast this with the use of large doses to ablate thyroid tissue in cases of hyperthyroidism (Clinical Box 20–2). An analysis of the kinetics of iodine han-dling also provides insights into the basic physiology of the gland (Figure 20–12). TABLE 20–3 Causes of congenital hypothyroidism. Maternal iodine deficiency Fetal thyroid dysgenesis Inborn errors of thyroid hormone synthesis Maternal antithyroid antibodies that cross the placenta Fetal hypopituitary hypothyroidism FIGURE 20–11 Structure of commonly used thioureylenes. C C C3H7 N H O C C H N H S Propylthiouracil C C H N CH3 H C N HS Methimazole 310 SECTION IV Endocrine & Reproductive Physiology CALORIGENIC ACTION T4 and T3 increase the O2 consumption of almost all metabol-ically active tissues. The exceptions are the adult brain, testes, uterus, lymph nodes, spleen, and anterior pituitary. T4 actual-ly depresses the O2 consumption of the anterior pituitary, pre-sumably because it inhibits TSH secretion. The increase in metabolic rate produced by a single dose of T4 becomes mea-surable after a latent period of several hours and lasts 6 days or more. Some of the calorigenic effect of thyroid hormones is due to metabolism of the fatty acids they mobilize. In addition, thy-roid hormones increase the activity of the membrane-bound Na, K ATPase in many tissues. EFFECTS SECONDARY TO CALORIGENESIS When the metabolic rate is increased by T4 and T3 in adults, nitrogen excretion is increased; if food intake is not increased, endogenous protein and fat stores are catabolized and weight is lost. In hypothyroid children, small doses of thyroid hor-mones cause a positive nitrogen balance because they stimu-late growth, but large doses cause protein catabolism similar to that produced in the adult. The potassium liberated during protein catabolism appears in the urine, and there is also an increase in urinary hexosamine and uric acid excretion. When the metabolic rate is increased, the need for all vita-mins is increased and vitamin deficiency syndromes may be precipitated. Thyroid hormones are necessary for hepatic con-version of carotene to vitamin A, and the accumulation of car-otene in the bloodstream (carotenemia) in hypothyroidism is responsible for the yellowish tint of the skin. Carotenemia can CLINICAL BOX 20–2 Hyperthyroidism The symptoms of an overactive thyroid gland follow logi-cally from the actions of thyroid hormone discussed in this chapter. Thus, hyperthyroidism is characterized by ner-vousness; weight loss; hyperphagia; heat intolerance; in-creased pulse pressure; a fine tremor of the outstretched fingers; warm, soft skin; sweating; and a BMR from +10 to as high as +100. It has various causes (Table 20–4); however, the most common cause is Graves disease (Graves hyper-thyroidism), which accounts for 60–80% of the cases. This is an autoimmune disease, more common in women, in which antibodies to the TSH receptor stimulate the recep-tor. This produces marked T4 and T3 secretion and enlarge-ment of the thyroid gland (goiter). However, due to the feedback effects of T4 and T3, plasma TSH is low, not high. Another hallmark of Graves disease is the occurrence of swelling of tissues in the orbits, producing protrusion of the eyeballs (exophthalmos). This occurs in 50% of patients and often precedes the development of obvious hyperthy-roidism. Other antithyroid antibodies are present in Graves disease, including antibodies to thyroglobulin and thyroid peroxidase. In Hashimoto thyroiditis, autoimmune antibod-ies ultimately destroy the thyroid, but during the early stage the inflammation of the gland causes excess thyroid hormone secretion and thyrotoxicosis similar to that seen in Graves disease. In general, some of the symptoms of hy-perthyroidism can be controlled by the thioureylene drugs discussed above, or by the administration of radioactive io-dine that destroys part of the gland. FIGURE 20–12 Distribution of radioactive iodine in individuals on a relatively low-iodine diet. Percentages are plot-ted against time after an oral dose of radioactive iodine. In hyperthy-roidism, plasma radioactivity falls rapidly and then rises again as a result of release of labeled T4 and T3 from the thyroid. 75 50 25 0 75 50 25 0 75 50 25 0 Euthyroid Urine Thyroid Plasma Hyperthyroid Hypothyroid Time (h) Percent of ingested dose 2 4 6 8 12 24 CHAPTER 20 The Thyroid Gland 311 be distinguished from jaundice because in the former condi-tion the scleras are not yellow. The skin normally contains a variety of proteins combined with polysaccharides, hyaluronic acid, and chondroitin sulfu-ric acid. In hypothyroidism, these complexes accumulate, promoting water retention and the characteristic puffiness of the skin (myxedema). When thyroid hormones are adminis-tered, the proteins are metabolized, and diuresis continues until the myxedema is cleared. Milk secretion is decreased in hypothyroidism and stimu-lated by thyroid hormones, a fact sometimes put to practical use in the dairy industry. Thyroid hormones do not stimulate TABLE 20–4 Causes of hyperthyroidism. Thyroid overactivity Solitary toxic adenoma Toxic multinodular goiter Hashimoto thyroiditis TSH-secreting pituitary tumor Mutations causing constitutive activation of TSH receptor Other rare causes Extrathyroidal Administration of T3 or T4 (factitious or iatrogenic hyperthyroidism) Ectopic thyroid tissue TABLE 20–5 Physiologic effects of thyroid hormones. Target Tissue Effect Mechanism Heart Chronotropic Inotropic Increased number of β-adrenergic receptors Enhanced responses to circulating catecholamines Increased proportion of α-myosin heavy chain (with higher ATPase activity) Adipose tissue Catabolic Stimulated lipolysis Muscle Catabolic Increased protein breakdown Bone Developmental Promote normal growth and skel-etal development Nervous system Developmental Promote normal brain develop-ment Gut Metabolic Increased rate of carbohydrate absorption Lipoprotein Metabolic Formation of LDL receptors Other Calorigenic Stimulated oxygen consumption by metabolically active tissues (ex-ceptions: testes, uterus, lymph nodes, spleen, anterior pituitary) Increased metabolic rate Modified and reproduced with permission from McPhee SJ, Lingarra VR, Ganong WF (editors): Pathophysiology of Disease, 4th ed. McGraw-Hill, 2003. CLINICAL BOX 20–3 Thyroid Hormone Resistance Some mutations in the gene that codes for TRβ are associ-ated with resistance to the effects of T3 and T4. Most com-monly, there is resistance to thyroid hormones in the pe-ripheral tissues and the anterior pituitary gland. Patients with this abnormality are usually not clinically hypothyroid, because they maintain plasma levels of T3 and T4 that are high enough to overcome the resistance, and hTRα is unaf-fected. However, plasma TSH is inappropriately high given the high circulating T3 and T4 levels and is difficult to sup-press with exogenous thyroid hormone. Some patients have thyroid hormone resistance only in the pituitary. They have hypermetabolism and elevated plasma T3 and T4 le-vels with normal, nonsuppressible levels of TSH. A few pa-tients apparently have peripheral resistance with normal pituitary sensitivity. They have hypometabolism despite normal plasma levels of T3, T4, and TSH, and they require large doses of thyroid hormones to increase their meta-bolic rate. An interesting finding is that attention deficit hyperactivity disorder, a condition frequently diagnosed in children who are overactive and impulsive, is much more common in individuals with thyroid hormone resistance than in the general population. This suggests that hTRβ may play a special role in brain development. FIGURE 20–13 Calorigenic responses of thyroidectomized rats to subcutaneous injections of T4 and T3. (Redrawn and reproduced with permission from Barker SB: Peripheral actions of thyroid hormones. Fed Proc 1962;21:635.) 80 60 40 20 20 40 60 Increased metabolism (mL O2/100 g/h) 80 100 Dose (μg/kg/d) T3 T4 312 SECTION IV Endocrine & Reproductive Physiology the metabolism of the uterus but are essential for normal menstrual cycles and fertility. EFFECTS ON THE CARDIOVASCULAR SYSTEM Large doses of thyroid hormones cause enough extra heat pro-duction to lead to a slight rise in body temperatures (Chapter 18), which in turn activates heat-dissipating mechanisms. Pe-ripheral resistance decreases because of cutaneous vasodila-tion, and this increases levels of renal Na+ and water absorption, expanding blood volume. Cardiac output is in-creased by the direct action of thyroid hormones, as well as that of catecholamines, on the heart, so that pulse pressure and cardiac rate are increased and circulation time is shortened. T3 is not formed from T4 in myocytes to any degree, but circulatory T3 enters the myocytes, combines with its recep-tors, and enters the nucleus, where it promotes the expression of some genes and inhibits the expression of others. Those that are enhanced include the genes for α-myosin heavy chain, sarcoplasmic reticulum Ca2+ ATPase, β-adrenergic receptors, G proteins, Na, K ATPase, and certain K+ channels. Those that are inhibited include the genes for β-myosin heavy chain, phospholamban, two types of adenylyl cyclase, T3 nuclear receptors, and NCX, the Na+–Ca2+ exchanger. The net result is increased heart rate and force of contraction. The heart contains two myosin heavy chain (MHC) iso-forms, α-MHC and β-MHC. They are encoded by two highly homologous genes located on the short arm of chromosome 17. Each myosin molecule consists of two heavy chains and two pairs of light chains (see Chapter 5). The myosin contain-ing β-MHC has less ATPase activity than the myosin contain-ing α-MHC. α-MHC predominates in the atria in adults, and its level is increased by treatment with thyroid hormone. This increases the speed of cardiac contraction. Conversely, expression of the α-MHC gene is depressed and that of the β-MHC gene is enhanced in hypothyroidism. EFFECTS ON THE NERVOUS SYSTEM In hypothyroidism, mentation is slow and the cerebrospinal fluid (CSF) protein level elevated. Thyroid hormones reverse these changes, and large doses cause rapid mentation, irrita-bility, and restlessness. Overall, cerebral blood flow and glu-cose and O2 consumption by the brain are normal in adult hypo- and hyperthyroidism. However, thyroid hormones en-ter the brain in adults and are found in gray matter in numer-ous different locations. In addition, astrocytes in the brain convert T4 to T3, and there is a sharp increase in brain D2 ac-tivity after thyroidectomy that is reversed within 4 h by a sin-gle intravenous dose of T3. Some of the effects of thyroid hormones on the brain are probably secondary to increased responsiveness to catecholamines, with consequent increased activation of the reticular activating system (see Chapter 15). In addition, thyroid hormones have marked effects on brain development. The parts of the central nervous system (CNS) most affected are the cerebral cortex and the basal ganglia. In addition, the cochlea is also affected. Consequently, thyroid hormone deficiency during development causes mental retar-dation, motor rigidity, and deaf–mutism. Deficiencies in thy-roid hormone synthesis secondary to a failure of thyrocytes to transport iodide presumably also contribute to deafness in Pendred syndrome, discussed above. Thyroid hormones also exert effects on reflexes. The reac-tion time of stretch reflexes (see Chapter 9) is shortened in hyperthyroidism and prolonged in hypothyroidism. Measure-ment of the reaction time of the ankle jerk (Achilles reflex) has attracted attention as a clinical test for evaluating thyroid function, but this reaction time is also affected by other dis-eases and thus is not a specific assessment of thyroid activity. RELATION TO CATECHOLAMINES The actions of thyroid hormones and the catecholamines nor-epinephrine and epinephrine are intimately interrelated. Epi-nephrine increases the metabolic rate, stimulates the nervous system, and produces cardiovascular effects similar to those of thyroid hormones, although the duration of these actions is brief. Norepinephrine has generally similar actions. The toxic-ity of the catecholamines is markedly increased in rats treated with T4. Although plasma catecholamine levels are normal in hyperthyroidism, the cardiovascular effects, tremulousness, and sweating produced by thyroid hormones can be reduced or abolished by sympathectomy. They can also be reduced by drugs such as propranolol that block β-adrenergic receptors. Indeed, propranolol and other β blockers are used extensively in the treatment of thyrotoxicosis and in the treatment of the severe exacerbations of hyperthyroidism called thyroid storms. However, even though β blockers are weak inhibitors of extrathyroidal conversion of T4 to T3, and consequently may produce a small fall in plasma T3, they have little effect on the other actions of thyroid hormones. Presumably, the functional synergism observed between catecholamines and thyroid hor-mones, particularly in pathological settings, arises from their overlapping biological functions as well as the ability of thyroid hormones to increase expression of catecholamine receptors and the signaling effectors to which they are linked. EFFECTS ON SKELETAL MUSCLE Muscle weakness occurs in most patients with hyperthyroid-ism (thyrotoxic myopathy), and when the hyperthyroidism is severe and prolonged, the myopathy may be severe. The mus-cle weakness may be due in part to increased protein catabo-lism. Thyroid hormones affect the expression of the MHC genes in skeletal as well as cardiac muscle (see Chapter 5). However, the effects produced are complex and their relation to the myopathy is not established. Hypothyroidism is also as-sociated with muscle weakness, cramps, and stiffness. CHAPTER 20 The Thyroid Gland 313 EFFECTS ON CARBOHYDRATE METABOLISM Thyroid hormones increase the rate of absorption of carbohy-drates from the gastrointestinal tract, an action that is probably independent of their calorigenic action. In hyperthyroidism, therefore, the plasma glucose level rises rapidly after a carbohy-drate meal, sometimes exceeding the renal threshold. However, it falls again at a rapid rate. EFFECTS ON CHOLESTEROL METABOLISM Thyroid hormones lower circulating cholesterol levels. The plasma cholesterol level drops before the metabolic rate rises, which indicates that this action is independent of the stimu-lation of O2 consumption. The decrease in plasma cholesterol concentration is due to increased formation of low-density lipoprotein (LDL) receptors in the liver, resulting in in-creased hepatic removal of cholesterol from the circulation. Despite considerable effort, however, it has not been possible to produce a clinically useful thyroid hormone analog that lowers plasma cholesterol without increasing metabolism. EFFECTS ON GROWTH Thyroid hormones are essential for normal growth and skele-tal maturation (see Chapter 23). In hypothyroid children, bone growth is slowed and epiphysial closure delayed. In the absence of thyroid hormones, growth hormone secretion is also depressed. This further impairs growth and development, since thyroid hormones normally potentiate the effect of growth hormone on tissues. CHAPTER SUMMARY ■The thyroid gland transports and fixes iodide to amino acids present in thyroglobulin to generate the thyroid hormones thyroxine (T4) and triiodothyronine (T3). ■Synthesis and secretion of thyroid hormones is stimulated by thyroid-stimulating hormone (TSH) from the pituitary, which in turn is released in response to thyrotropin-releasing hormone (TRH) from the hypothalamus. These releasing factors are con-trolled by changes in whole body status (eg, exposure to cold or stress). ■Thyroid hormones circulate in the plasma predominantly in protein-bound forms. Only the free hormones are biologically active, and both feed back to reduce secretion of TSH. ■Thyroid hormones exert their effects by entering cells and bind-ing to thyroid receptors. The liganded forms of thyroid recep-tors are nuclear transcription factors that alter gene expression. ■Thyroid hormones stimulate metabolic rate, calorigenesis, cardi-ac function, and normal mentation, and interact synergistically with catecholamines. Thyroid hormones also play critical roles in development, particularly of the nervous system, and growth. ■Disease results with both under- and overactivity of the thyroid gland. Hypothyroidism is accompanied by mental and physical slowing in adults, and by mental retardation and dwarfism if it occurs in neonatal life. Overactivity of the thyroid gland, which most commonly is caused by autoantibodies that trigger secre-tion (Graves disease) results in body wasting, nervousness, and tachycardia. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. In which of the following conditions is it most likely that the TSH response to TRH will be reduced? A) hypothyroidism due to tissue resistance to thyroid hormone B) hypothyroidism due to disease destroying the thyroid gland C) hyperthyroidism due to circulating antithyroid antibodies with TSH activity D) hyperthyroidism due to diffuse hyperplasia of thyrotropes of the anterior pituitary E) iodine deficiency 2. A young woman has puffy skin and a hoarse voice. Her plasma TSH concentration is low but increases markedly when she is given TRH. She probably has A) hyperthyroidism due to a thyroid tumor. B) hypothyroidism due to a primary abnormality in the thyroid gland. C) hypothyroidism due to a primary abnormality in the pitu-itary gland. D) hypothyroidism due to a primary abnormality in the hypo-thalamus. E) hyperthyroidism due to a primary abnormality in the hypo-thalamus. 3. The enzyme primarily responsible for the conversion of T4 to T3 in the periphery is A) D1 thyroid deiodinase. B) D2 thyroid deiodinase. C) D3 thyroid deiodinase. D) thyroid peroxidase. E) none of the above 4. The metabolic rate is least affected by an increase in the plasma level of A) TSH. B) TRH. C) TBG. D) free T4. E) free T3. 5. Which of the following is not essential for normal biosynthesis of thyroid hormones? A) iodine B) ferritin C) thyroglobulin D) protein synthesis E) TSH 314 SECTION IV Endocrine & Reproductive Physiology 6. Which of the following would be least affected by injections of TSH? A) thyroidal uptake of iodine B) synthesis of thyroglobulin C) cyclic adenosine monophosphate (AMP) in thyroid cells D) cyclic guanosine monophosphate (GMP) in thyroid cells E) size of the thyroid 7. Hypothyroidism due to disease of the thyroid gland is associated with increased plasma levels of A) cholesterol. B) albumin. C) RT3. D) iodide. E) TBG. 8. Thyroid hormone receptors bind to DNA in which of the follow-ing forms? A) a heterodimer with the prolactin receptor B) a heterodimer with the growth hormone receptor C) a heterodimer with the retinoid X receptor D) a heterodimer with the insulin receptor E) a heterodimer with the progesterone receptor 9. Increasing intracellular I– due to the action of NIS is an example of A) endocytosis. B) passive diffusion. C) Na+ and K+ cotransport. D) primary active transport. E) secondary active transport. CHAPTER RESOURCES Brent GA: Graves’ disease. N Engl J Med 2008;358:2594. Dohan O, Carrasco N: Advances in Na+/I- symporter (NIS) research in the thyroid and beyond. Mol Cell Endocrinol 2003;213:59. Glaser B: Pendred syndrome. Pediatr Endocrinol Rev 2003;1(Suppl 2):199. Peeters RP, van der Deure WM, Visser TJ: Genetic variation in thyroid hormone pathway genes: Polymorphisms in the TSH receptor and the iodothyronine deiodinases. Eur J Endocrinol 2006;155:655. 315 C H A P T E R 21 Endocrine Functions of the Pancreas & Regulation of Carbohydrate Metabolism O B J E C T I V E S After reading this chapter, you should be able to: ■List the hormones that affect the plasma glucose concentration and briefly de-scribe the action of each. ■Describe the structure of the pancreatic islets and name the hormones secreted by each of the cell types in the islets. ■Describe the structure of insulin and outline the steps involved in its biosynthesis and release into the bloodstream. ■List the consequences of insulin deficiency and explain how each of these abnor-malities is produced. ■Describe insulin receptors, the way they mediate the effects of insulin, and the way they are regulated. ■Describe the types of glucose transporters found in the body and the function of each. ■List the major factors that affect the secretion of insulin. ■Describe the structure of glucagon and other physiologically active peptides pro-duced from its precursor. ■List the physiologically significant effects of glucagon and the factors that regulate glucagon secretion. ■Describe the physiologic effects of somatostatin in the pancreas. ■Outline the mechanisms by which thyroid hormones, adrenal glucocorticoids, catecholamines, and growth hormone affect carbohydrate metabolism. ■Understand the major differences between type 1 and type 2 diabetes. INTRODUCTION At least four polypeptides with regulatory activity are secreted by the islets of Langerhans in the pancreas. Two of these are hormones insulin and glucagon, and have important functions in the regulation of the intermediary metabolism of carbohy-drates, proteins, and fats. The third polypeptide, somatostatin, plays a role in the regulation of islet cell secretion, and the fourth, pancreatic polypeptide, is probably concerned primar-ily with the regulation of HCO– 3 secretion to the intestine. Glu-cagon, somatostatin, and possibly pancreatic polypeptide are also secreted by cells in the mucosa of the gastrointestinal tract. Insulin is anabolic, increasing the storage of glucose, fatty acids, and amino acids. Glucagon is catabolic, mobilizing glu-cose, fatty acids, and the amino acids from stores into the bloodstream. The two hormones are thus reciprocal in their overall action and are reciprocally secreted in most circum-stances. Insulin excess causes hypoglycemia, which leads to convulsions and coma. Insulin deficiency, either absolute or relative, causes diabetes mellitus (chronic elevated blood glu-cose), a complex and debilitating disease that if untreated is eventually fatal. Glucagon deficiency can cause hypoglycemia, 316 SECTION IV Endocrine & Reproductive Physiology and glucagon excess makes diabetes worse. Excess pancreatic production of somatostatin causes hyperglycemia and other manifestations of diabetes. A variety of other hormones also have important roles in the regulation of carbohydrate metabolism. ISLET CELL STRUCTURE The islets of Langerhans (Figure 21–1) are ovoid, 76- × 175-μm collections of cells. The islets are scattered throughout the pancreas, although they are more plentiful in the tail than in the body and head. β-islets make up about 2% of the volume of the gland, whereas the exocrine portion of the pancreas (see Chapter 26) makes up 80%, and ducts and blood vessels make up the remainder. Humans have 1 to 2 million islets. Each has a copious blood supply; blood from the islets, like that from the gastrointestinal tract (but unlike that from any other endo-crine organs) drains into the hepatic portal vein. The cells in the islets can be divided into types on the basis of their staining properties and morphology. Humans have at least four distinct cell types: A, B, D, and F cells. A, B, and D cells are also called α, β, and δ cells. However, this leads to confusion in view of the use of Greek letters to refer to other structures in the body, particularly adrenergic receptors (see Chapter 7). The A cells secrete glucagon, the B cells secrete insulin, the D cells secrete somatostatin, and the F cells secrete pancreatic polypeptide. The B cells, which are the most common and account for 60–75% of the cells in the islets, are generally located in the center of each islet. They tend to be surrounded by the A cells, which make up 20% of the total, and the less common D and F cells. The islets in the tail, the body, and the anterior and superior part of the head of the human pancreas have many A cells and few if any F cells in the outer rim, whereas in rats and probably in humans, the islets in the posterior part of the head of the pan-creas have a relatively large number of F cells and few A cells. The A-cell-rich (glucagon-rich) islets arise embryologically from the dorsal pancreatic bud, and the F-cell-rich (pancre-atic polypeptide-rich) islets arise from the ventral pancreatic bud. These buds arise separately from the duodenum. The B cell granules are packets of insulin in the cell cyto-plasm. The shape of the packets varies from species to species; in humans, some are round whereas others are rectangular (Figure 21–2). In the B cells, the insulin molecule forms poly-mers and also complexes with zinc. The differences in the shape of the packets are probably due to differences in the size of polymers or zinc aggregates of insulin. The A granules, which contain glucagon, are relatively uniform from species to species (Figure 21–3). The D cells also contain large num-bers of relatively homogeneous granules. STRUCTURE, BIOSYNTHESIS, & SECRETION OF INSULIN STRUCTURE & SPECIES SPECIFICITY Insulin is a polypeptide containing two chains of amino ac-ids linked by disulfide bridges (Table 21–1). Minor differ-ences occur in the amino acid composition of the molecule from species to species. The differences are generally not sufficient to affect the biologic activity of a particular insulin FIGURE 21–1 Islet of Langerhans in the rat pancreas. Darkly stained cells are B cells. Surrounding pancreatic acinar tissue is light-colored (× 400). (Courtesy of LL Bennett.) FIGURE 21–2 Electron micrograph of two adjoining B cells in a human pancreatic islet. The B granules are the crystals in the membrane-lined vesicles. They vary in shape from rhombic to round (× 26,000). (Courtesy of A Like. Reproduced, with permission, from Fawcett DW: Bloom and Fawcett, A Textbook of Histology, 11th ed. Saunders, 1986.) CHAPTER 21 Endocrine Functions of the Pancreas & Regulation of Carbohydrate Metabolism 317 in heterologous species but are sufficient to make the insulin antigenic. If insulin of one species is injected for a prolonged period into another species, the anti-insulin antibodies formed inhibit the injected insulin. Almost all humans who have received commercial bovine insulin for more than 2 mo have antibodies against bovine insulin, but the titer is usually low. Porcine insulin differs from human insulin by only one amino acid residue and has low antigenicity. Human insulin produced in bacteria by recombinant DNA technology is now widely used to avoid antibody formation. BIOSYNTHESIS & SECRETION Insulin is synthesized in the rough endoplasmic reticulum of the B cells (Figure 21–3). It is then transported to the Golgi ap-paratus, where it is packaged into membrane-bound granules. These granules move to the plasma membrane by a process in-volving microtubules, and their contents are expelled by exo-cytosis (see Chapter 2). The insulin then crosses the basal lamina of the B cell and a neighboring capillary and the fenes-trated endothelium of the capillary to reach the bloodstream. The fenestrations are discussed in detail in Chapter 32. Like other polypeptide hormones and related proteins that enter the endoplasmic reticulum, insulin is synthesized as part of a larger preprohormone (see Chapter 1). The gene for insulin is located on the short arm of chromosome 11 in humans. It has two introns and three exons. Preproinsulin FIGURE 21–3 A and B cells, showing their relation to blood vessels. RER, rough endoplasmic reticulum. Insulin from the B cell and glucagon from the A cell are secreted by exocytosis and cross the basal lamina of the cell and the basal lamina of the capillary before entering the lumen of the fenestrated capillary. (Reproduced with permission from Junqueira IC, Carneiro J: Basic Histology: Text and Atlas, 10th ed. McGraw-Hill, 2003.) Basal laminae Fenestrations Capillary A cell B cell RER Desmosome Golgi TABLE 21–1 Structure of human insulin (molecular weight 5808) and (below) variations in this structure in other mammalian species.a Variations from Human Amino Acid Sequence Species A Chain Position 8 9 10 B Chain Position 30 Pig, dog, sperm whale Thr-Ser-Ile Ala Rabbit Thr-Ser-Ile Ser Cattle, goat Ala-Ser-Val Ala Sheep Ala-Gly-Val Ala Horse Thr-Gly-Ile Ala Sei whale Ala-Ser-Thr Ala a In the rat, the islet cells secrete two slightly different insulins, and in certain fish four different chains are found. A chain Gly-Ile-Val-Glu-Gin-Cys-Cys-Thr-Ser-Ile-Cys-Ser-Leu-Tyr-Gin-Leu-Glu-Asn-Tyr-Cys-Asn 1 2 3 4 5 6 S S S S S S 8 9 10 11 12 13 14 15 16 17 18 19 21 B chain Phe-Val-Asn-Gln-His-Leu-Cys-Gly-Ser-His-Leu-Val-Glu-Ala-Leu-Tyr-Leu-Val-Cys-Gly-Glu-Arg-Gly-Phe-Phe-Tyr-Thr-Pro-Lys-Thr 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 21 20 22 23 24 25 26 27 28 29 30 318 SECTION IV Endocrine & Reproductive Physiology has a 23-amino-acid signal peptide removed as it enters the endoplasmic reticulum. The remainder of the molecule is then folded, and the disulfide bonds are formed to make pro-insulin. The peptide segment connecting the A and B chains, the connecting peptide (C peptide), facilitates the folding and then is detached in the granules before secretion. Two proteases are involved in processing the proinsulin; to date it has no other established physiologic activity. Normally, 90– 97% of the product released from the B cells is insulin along with equimolar amounts of C peptide. The rest is mostly pro-insulin. C peptide can be measured by radioimmunoassay, and its level in blood provides an index of B cell function in patients receiving exogenous insulin. FATE OF SECRETED INSULIN INSULIN & INSULINLIKE ACTIVITY IN BLOOD Plasma contains a number of substances with insulin-like ac-tivity in addition to insulin (Table 21–2). The activity that is not suppressed by anti-insulin antibodies has been called non-suppressible insulin-like activity (NSILA). Most, if not all, of this activity persists after pancreatectomy and is due to the in-sulinlike growth factors IGF-I and IGF-II (see Chapter 24). These IGFs are polypeptides. Small amounts are free in the plasma (low-molecular-weight fraction), but large amounts are bound to proteins (high-molecular-weight fraction). One may well ask why pancreatectomy causes diabetes mel-litus when NSILA persists in the plasma. However, the insulinlike activities of IGF-I and IGF-II are weak compared to that of insulin and likely play other specific functions. METABOLISM The half-life of insulin in the circulation in humans is about 5 min. Insulin binds to insulin receptors, and some is internal-ized. It is destroyed by proteases in the endosomes formed by the endocytotic process. EFFECTS OF INSULIN The physiologic effects of insulin are far-reaching and com-plex. They are conveniently divided into rapid, intermediate, and delayed actions, as listed in Table 21–3. The best known is the hypoglycemic effect, but there are additional effects on amino acid and electrolyte transport, many enzymes, and growth. The net effect of the hormone is storage of carbohy-drate, protein, and fat. Therefore, insulin is appropriately called the “hormone of abundance.” The actions of insulin on adipose tissue; skeletal, cardiac, and smooth muscle; and the liver are summarized in Table 21–4. GLUCOSE TRANSPORTERS Glucose enters cells by facilitated diffusion (see Chapter 1) or, in the intestine and kidneys, by secondary active transport with Na+. In muscle, adipose, and some other tissues, insulin stimulates glucose entry into cells by increasing the number of glucose transporters in the cell membranes. The glucose transporters (GLUTs) that are responsible for facilitated diffusion of glucose across cell membranes are a family of closely related proteins that span the cell membrane 12 times and have their amino and carboxyl terminals inside the cell. They differ from and have no homology with the sodium-dependent glucose transporters, SGLT 1 and SGLT 2, responsible for the secondary active transport of glucose in the intestine (see Chapter 27) and renal tubules (see Chapter 38), although the SGLTs also have 12 transmembrane domains. Seven different glucose transporters, named GLUT 1–7 in order of discovery, have been characterized (Table 21–5). They TABLE 21–2 Substances with insulin-like activity in human plasma. Insulin Proinsulin Nonsuppressible insulin-like activity (NSILA) Low-molecular-weight fraction IGF-I IGF-II High-molecular-weight fraction (mostly IGF bound to protein) TABLE 21–3 Principal actions of insulin. Rapid (seconds) Increased transport of glucose, amino acids, and K+ into insulin-sen-sitive cells Intermediate (minutes) Stimulation of protein synthesis Inhibition of protein degradation Activation of glycolytic enzymes and glycogen synthase Inhibition of phosphorylase and gluconeogenic enzymes Delayed (hours) Increase in mRNAs for lipogenic and other enzymes Courtesy of ID Goldfine. CHAPTER 21 Endocrine Functions of the Pancreas & Regulation of Carbohydrate Metabolism 319 contain 492 to 524 amino acid residues and their affinity for glucose varies. Each transporter appears to have evolved for special tasks. GLUT 4 is the transporter in muscle and adipose tissue that is stimulated by insulin. A pool of GLUT 4 mole-cules is maintained within vesicles in the cytoplasm of insulin-sensitive cells. When the insulin receptors of these cells are activated, the vesicles move rapidly to the cell membrane and fuse with it, inserting the transporters into the cell membrane (Figure 21–4). When insulin action ceases, the transporter-containing patches of membrane are endocytosed and the ves-icles are ready for the next exposure to insulin. Activation of the insulin receptor brings about the movement of the vesicles to the cell membrane by activating phosphatidylinositol 3-kinase (Figure 21–4), but how this activation triggers vesicle movement is still unsettled. Most of the other GLUT trans-porters that are not insulin-sensitive appear to be constitu-tively expressed in the cell membrane. In the tissues in which insulin increases the number of glu-cose transporters in the cell membranes, the rate of phosphor-ylation of the glucose, once it has entered the cells, is regulated by other hormones. Growth hormone and cortisol both inhibit phosphorylation in certain tissues. Transport is normally so rapid that it is not a rate-limiting step in glucose metabolism. However, it is rate-limiting in the B cells. Insulin also increases the entry of glucose into liver cells, but it does not exert this effect by increasing the number of GLUT 4 transporters in the cell membranes. Instead, it induces glucokinase, and this increases the phosphorylation of glucose, so that the intracellular free glucose concentration stays low, facilitating the entry of glucose into the cell. Insulin-sensitive tissues also contain a population of GLUT 4 vesicles that move into the cell membrane in response to exercise, a process that occurs independent of the action of insulin. This is why exercise lowers blood sugar. A 5'-AMP-activated kinase may be responsible for the insertion of these vesicles into the cell membrane. INSULIN PREPARATIONS The maximal decline in plasma glucose occurs 30 min after in-travenous injection of insulin. After subcutaneous adminis-tration, the maximal fall occurs in 2 to 3 h. A wide variety of insulin preparations are now available commercially. These include insulins that have been complexed with protamine and other polypeptides to delay absorption and degradation, and synthetic insulins in which there have been changes in amino acid residues. In general, they fall into three categories: rapid, intermediate-acting, and long-acting (24–36 h). RELATION TO POTASSIUM Insulin causes K+ to enter cells, with a resultant lowering of the extracellular K+ concentration. Infusions of insulin and glucose significantly lower the plasma K+ level in normal individuals and are very effective for the temporary relief of hyperkalemia in patients with renal failure. Hypokalemia often develops when patients with diabetic acidosis are treated with insulin. The reason for the intracellular migration of K+ is still uncer-tain. However, insulin increases the activity of Na+–K+ ATPase in cell membranes, so that more K+ is pumped into cells. OTHER ACTIONS The hypoglycemic and other effects of insulin are summarized in temporal terms in Table 21–3, and the net effects on various tissues are summarized in Table 21–4. The action on glycogen synthase fosters glycogen storage, and the actions on glycolytic enzymes favor glucose metabolism to two carbon fragments (see Chapter 1), with resulting promotion of lipogenesis. Stim-TABLE 21–4 Effects of insulin on various tissues. Adipose tissue Increased glucose entry Increased fatty acid synthesis Increased glycerol phosphate synthesis Increased triglyceride deposition Activation of lipoprotein lipase Inhibition of hormone-sensitive lipase Increased K+ uptake Muscle Increased glucose entry Increased glycogen synthesis Increased amino acid uptake Increased protein synthesis in ribosomes Decreased protein catabolism Decreased release of gluconeogenic amino acids Increased ketone uptake Increased K+ uptake Liver Decreased ketogenesis Increased protein synthesis Increased lipid synthesis Decreased glucose output due to decreased gluconeogenesis, increased glycogen synthesis, and increased glycolysis General Increased cell growth 320 SECTION IV Endocrine & Reproductive Physiology ulation of protein synthesis from amino acids entering the cells and inhibition of protein degradation foster growth. The anabolic effect of insulin is aided by the protein-sparing action of adequate intracellular glucose supplies. Fail-ure to grow is a symptom of diabetes in children, and insulin stimulates the growth of immature hypophysectomized rats to almost the same degree as growth hormone. MECHANISM OF ACTION INSULIN RECEPTORS Insulin receptors are found on many different cells in the body, including cells in which insulin does not increase glu-cose uptake. TABLE 21–5 Glucose transporters in mammals. Function Km (mM)a Major Sites of Expression Secondary active transport (Na1-glucose cotransport) SGLT 1 Absorption of glucose 0.1–1.0 Small intestine, renal tubules SGLT 2 Absorption of glucose 1.6 Renal tubules Facilitated diffusion GLUT 1 Basal glucose uptake 1–2 Placenta, blood-brain barrier, brain, red cells, kidneys, colon, many other organs GLUT 2 B-cell glucose sensor; transport out of intestinal and renal epithelial cells 12–20 B cells of islets, liver, epithelial cells of small in-testine, kidneys GLUT 3 Basal glucose uptake <1 Brain, placenta, kidneys, many other organs GLUT 4 Insulin-stimulated glucose uptake 5 Skeletal and cardiac muscle, adipose tissue, other tissues GLUT 5 Fructose transport 1–2 Jejunum, sperm GLUT 6 None — Pseudogene GLUT 7 Glucose 6-phosphate ransporter in endoplasmic reticulum — Liver, ? other tissues aThe Km is the glucose concentration at which transport is half-maximal. Modified from Stephens JM, Pilch PF: The metabolic regulation and vesicular transport of GLUT 4, the major insulin-responsive glucose transporter. Endocr Rev 1995;16:529. FIGURE 21–4 Cycling of GLUT 4 transporters through endosomes in insulin-sensitive tissues. Activation of the insulin receptor causes activation of phosphatidylinositol 3-kinase, which speeds translocation of the GLUT 4-containing endosomes into the cell membrane. The GLUT 4 transporters then mediate glucose transport into the cell. Insulin receptor Fusion Glucose transport Phosphatidylinositol 3-kinase Internalization CHAPTER 21 Endocrine Functions of the Pancreas & Regulation of Carbohydrate Metabolism 321 The insulin receptor, which has a molecular weight of approximately 340,000, is a tetramer made up of two α and two β glycoprotein subunits (Figure 21–5). All these are syn-thesized on a single mRNA and then proteolytically separated and bound to each other by disulfide bonds. The gene for the insulin receptor has 22 exons and in humans is located on chromosome 19. The α subunits bind insulin and are extra-cellular, whereas the β subunits span the membrane. The intracellular portions of the β subunits have tyrosine kinase activity. The α and β subunits are both glycosylated, with sugar residues extending into the interstitial fluid. Binding of insulin triggers the tyrosine kinase activity of the β subunits, producing autophosphorylation of the β subunits on tyrosine residues. The autophosphorylation, which is neces-sary for insulin to exert its biologic effects, triggers phosphory-lation of some cytoplasmic proteins and dephosphorylation of others, mostly on serine and threonine residues. Insulin recep-tor substrate (IRS-1) mediates some of the effects in humans but there are other effector systems as well (Figure 21–6). For example, mice in which the insulin receptor gene is knocked out show marked growth retardation in utero, have abnormali-ties of the central nervous system (CNS) and skin, and die at birth of respiratory failure, whereas IRS-1 knockouts show only moderate growth retardation in utero, survive, and are insulin-resistant but otherwise nearly normal. The growth-promoting protein anabolic effects of insulin are mediated via phosphatidylinositol 3-kinase (PI3K), and evi-dence indicates that in invertebrates, this pathway is involved in the growth of nerve cells and axon guidance in the visual system. It is interesting to compare the insulin receptor with other related receptors. The insulin receptor is very similar to the receptor for IGF-I but different from the receptor for IGF-II (Figure 21–5). Other receptors for growth factors and receptors for various oncogenes also are tyrosine kinases. However, the amino acid composition of these receptors is quite different. When insulin binds to its receptors, they aggregate in patches and are taken into the cell by receptor-mediated endocytosis (see Chapter 2). Eventually, the insulin–receptor complexes enter lysosomes, where the receptors are broken down or recy-cled. The half-life of the insulin receptor is about 7 h. CONSEQUENCES OF INSULIN DEFICIENCY The far-reaching physiologic effects of insulin are highlighted by a consideration of the extensive and serious consequences of insulin deficiency (Clinical Box 21–1). In humans, insulin deficiency is a common pathologic con-dition. In animals, it can be produced by pancreatectomy; by administration of alloxan, streptozocin, or other toxins that in appropriate doses cause selective destruction of the B cells of the pancreatic islets; by administration of drugs that inhibit insulin secretion; and by administration of anti-insulin anti-bodies. Strains of mice, rats, hamsters, guinea pigs, miniature swine, and monkeys that have a high incidence of spontane-ous diabetes mellitus have also been described. GLUCOSE TOLERANCE In diabetes, glucose piles up in the bloodstream, especially af-ter meals. If a glucose load is given to a diabetic, the plasma FIGURE 21–5 Insulin, IGF-I, and IGF-II receptors. Each hor-mone binds primarily to its own receptor, but insulin also binds to the IGF-I receptor, and IGF-I and IGF-II bind to all three. The purple boxes are intracellular tyrosine kinase domains. Note the marked similarity between the insulin receptor and the IGF-I receptor; also note the 15 repeat sequences in the extracellular portion of the IGF-II receptor. ISF, interstitial fluid. α β β β β α α α Insulin IGF-I IGF-II Insulin receptor IGF-I receptor IGF-II receptor ISF Cytoplasm 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 FIGURE 21–6 Intracellular responses triggered by insulin binding to the insulin receptor. Red balls and balls labeled P repre-sent phosphate groups. IRS-1, insulin receptor substrate-1. Insulin Insulin receptor Other effector systems and secondary mediators ATP P P P P IRS-1 Various effects 322 SECTION IV Endocrine & Reproductive Physiology glucose rises higher and returns to the baseline more slowly than it does in normal individuals. The response to a standard oral test dose of glucose, the oral glucose tolerance test, is used in the clinical diagnosis of diabetes (Figure 21–7). Impaired glucose tolerance in diabetes is due in part to reduced entry of glucose into cells (decreased peripheral uti-lization). In the absence of insulin, the entry of glucose into skeletal, cardiac, and smooth muscle and other tissues is decreased (Figure 21–8). Glucose uptake by the liver is also reduced, but the effect is indirect. Intestinal absorption of glu-cose is unaffected, as is its reabsorption from the urine by the cells of the proximal tubules of the kidneys. Glucose uptake by most of the brain and the red blood cells is also normal. The second and the major cause of hyperglycemia in diabe-tes is derangement of the glucostatic function of the liver (see Chapter 29). The liver takes up glucose from the bloodstream and stores it as glycogen, but because the liver contains glu-cose 6-phosphatase it also discharges glucose into the blood-stream. Insulin facilitates glycogen synthesis and inhibits hepatic glucose output. When the plasma glucose is high, insulin secretion is normally increased and hepatic glucogen-esis is decreased. This response does not occur in type 1 dia-betes (as insulin is absent) and in type 2 diabetes (as tissues are insulin resistant). Glucagon can contribute to hyperglyce-mia as it stimulates gluconeogenesis. Glucose output by the CLINICAL BOX 21–1 Diabetes Mellitus The constellation of abnormalities caused by insulin defi-ciency is called diabetes mellitus. Greek and Roman physi-cians used the term “diabetes” to refer to conditions in which the cardinal finding was a large urine volume, and two types were distinguished: “diabetes mellitus,” in which the urine tasted sweet; and “diabetes insipidus,” in which the urine had little taste. Today, the term “diabetes insipi-dus” is reserved for conditions in which there is a deficiency of the production or action of vasopressin (see Chapter 39), and the unmodified word “diabetes” is generally used as a synonym for diabetes mellitus. The cause of clinical diabetes is always a deficiency of the effects of insulin at the tissue level. Type 1 diabetes, or in-sulin-dependent diabetes mellitus (IDDM), is due to in-sulin deficiency caused by autoimmune destruction of the B cells in the pancreatic islets, and it accounts for 3–5% of cases and usually presents in children. Type 2 diabetes, or non-insulin-dependent diabetes mellitus (NIDDM), is characterized by the dysregulation of insulin release from the B cells, along with insulin resistance in peripheral tis-sues such as skeletal muscle, brain, and liver. Type 2 diabe-tes usually presents in overweight or obese adults. Diabetes is characterized by polyuria (passage of large volumes of urine), polydipsia (excessive drinking), weight loss in spite of polyphagia (increased appetite), hyperglyce-mia, glycosuria, ketosis, acidosis, and coma. Widespread biochemical abnormalities are present, but the fundamen-tal defects to which most of the abnormalities can be traced are (1) reduced entry of glucose into various “pe-ripheral” tissues and (2) increased liberation of glucose into the circulation from the liver. Therefore there is an extracel-lular glucose excess and, in many cells, an intracellular glu-cose deficiency—a situation that has been called “starva-tion in the midst of plenty.” Also, the entry of amino acids into muscle is decreased and lipolysis is increased. FIGURE 21–7 Oral glucose tolerance test. Adults are given 75 g of glucose in 300 mL of water. In normal individuals, the fasting venous plasma glucose is less than 115 mg/dL, the 2-hour value is less than 140 mg/dL, and no value is greater than 200 mg/dL. Diabetes mellitus is present if the 2-hour value and one other value are greater than 200 mg/dL. Impaired glucose tolerance is diagnosed when the values are above the upper limits of normal but below the values diag-nostic of diabetes. FIGURE 21–8 Disordered plasma glucose homeostasis in insulin deficiency. The heavy arrows indicate reactions that are ac-centuated. The rectangles across arrows indicate reactions that are blocked. 250 200 150 100 50 0 1 Time after oral glucose (h) 2 Normal Diabetes Plasma glucose (mg/dL) Plasma glucose 300 mg/dL Diet Intestine Kidney Urine (glycosuria) Brain Fat Muscle and some other tissues Lactic acid Liver Amino acids Glycerol CHAPTER 21 Endocrine Functions of the Pancreas & Regulation of Carbohydrate Metabolism 323 liver can be stimulated by catecholamines, cortisol, and growth hormone (ie, during a stress response). EFFECTS OF HYPERGLYCEMIA Hyperglycemia by itself can cause symptoms resulting from the hyperosmolality of the blood. In addition, there is glycos-uria because the renal capacity for glucose reabsorption is ex-ceeded. Excretion of the osmotically active glucose molecules entails the loss of large amounts of water (osmotic diuresis; see Chapter 38). The resultant dehydration activates the mecha-nisms regulating water intake, leading to polydipsia. There is an appreciable urinary loss of Na+ and K+ as well. For every gram of glucose excreted, 4.1 kcal is lost from the body. In-creasing the oral caloric intake to cover this loss simply raises the plasma glucose further and increases the glycosuria, so mobilization of endogenous protein and fat stores and weight loss are not prevented. When plasma glucose is episodically elevated over time, small amounts of hemoglobin A are nonenzymatically gly-cated to form HbAIc (see Chapter 32). Careful control of the diabetes with insulin reduces the amount formed and conse-quently HbAIc concentration is measured clinically as an inte-grated index of diabetic control for the 4- to 6-wk period before the measurement. The role of chronic hyperglycemia in production of the long-term complications of diabetes is discussed below. EFFECTS OF INTRACELLULAR GLUCOSE DEFICIENCY The plethora of glucose outside the cells in diabetes contrasts with the intracellular deficit. Glucose catabolism is normally a major source of energy for cellular processes, and in diabetes energy requirements can be met only by drawing on protein and fat reserves. Mechanisms are activated that greatly in-crease the catabolism of protein and fat, and one of the conse-quences of increased fat catabolism is ketosis. Deficient glucose utilization and deficient hormone sensing (insulin, leptin, CCK) in the cells of the hypothalamus that regulate satiety are the probable causes of hyperphagia in dia-betes. The feeding area of the hypothalamus is not inhibited and thus satiety is not sensed so food intake is increased. Glycogen depletion is a common consequence of intracellu-lar glucose deficit, and the glycogen content of liver and skele-tal muscle in diabetic animals is usually reduced. CHANGES IN PROTEIN METABOLISM In diabetes, the rate at which amino acids are catabolized to CO2 and H2O is increased. In addition, more amino acids are converted to glucose in the liver. The increased gluconeogen-esis has many causes. Glucagon stimulates gluconeogenesis, and hyperglucagonemia is generally present in diabetes. Adre-nal glucocorticoids also contribute to increased gluconeogen-esis when they are elevated in severely ill diabetics. The supply of amino acids is increased for gluconeogenesis because, in the absence of insulin, less protein synthesis occurs in muscle and hence blood amino acid levels rise. Alanine is particularly eas-ily converted to glucose. In addition, the activity of the en-zymes that catalyze the conversion of pyruvate and other two-carbon metabolic fragments to glucose is increased. These in-clude phosphoenolpyruvate carboxykinase, which facilitates the conversion of oxaloacetate to phosphoenolpyruvate (see Chapter 1). They also include fructose 1,6-diphosphatase, which catalyzes the conversion of fructose diphosphate to fructose 6-phosphate, and glucose 6-phosphatase, which con-trols the entry of glucose into the circulation from the liver. In-creased acetyl-CoA increases pyruvate carboxylase activity, and insulin deficiency increases the supply of acetyl-CoA be-cause lipogenesis is decreased. Pyruvate carboxylase catalyzes the conversion of pyruvate to oxaloacetate (see Figure 1–22). In diabetes, the net effect of accelerated protein conversion to CO2, H2O, and glucose, plus diminished protein synthesis, is protein depletion and wasting. Protein depletion from any cause is associated with poor “resistance” to infections. FAT METABOLISM IN DIABETES The principal abnormalities of fat metabolism in diabetes are acceleration of lipid catabolism, with increased formation of ketone bodies, and decreased synthesis of fatty acids and tri-glycerides. The manifestations of the disordered lipid metab-olism are so prominent that diabetes has been called “more a disease of lipid than of carbohydrate metabolism.” Fifty percent of an ingested glucose load is normally burned to CO2 and H2O; 5% is converted to glycogen; and 30–40% is converted to fat in the fat depots. In diabetes, less than 5% of ingested glucose is converted to fat, despite a decrease in the amount burned to CO2 and H2O, and no change in the amount converted to glycogen. Therefore, glucose accumu-lates in the bloodstream and spills over into the urine. The role of lipoprotein lipase and hormone-sensitive lipase in the regulation of the metabolism of fat depots is discussed in Chapter 1. In diabetes, conversion of glucose to fatty acids in the depots is decreased because of the intracellular glucose deficiency. Insulin inhibits the hormone-sensitive lipase in adipose tissue, and, in the absence of this hormone, the plasma level of free fatty acids (NEFA, UFA, FFA) is more than dou-bled. The increased glucagon also contributes to the mobiliza-tion of FFA. Thus, the FFA level parallels the plasma glucose level in diabetes and in some ways is a better indicator of the severity of the diabetic state. In the liver and other tissues, the fatty acids are catabolized to acetyl-CoA. Some of the acetyl-CoA is burned along with amino acid residues to yield CO2 and H2O in the citric acid cycle. However, the supply exceeds the capacity of the tissues to catabolize the acetyl-CoA. In addition to the previously mentioned increase in gluco-neogenesis and marked outpouring of glucose into the circu-324 SECTION IV Endocrine & Reproductive Physiology lation, the conversion of acetyl-CoA to malonyl-CoA and thence to fatty acids is markedly impaired. This is due to a defi-ciency of acetyl-CoA carboxylase, the enzyme that catalyzes the conversion. The excess acetyl-CoA is converted to ketone bodies. In uncontrolled diabetes, the plasma concentration of tri-glycerides and chylomicrons as well as FFA is increased, and the plasma is often lipemic. The rise in these constituents is due mainly to decreased removal of triglycerides into the fat depots. The decreased activity of lipoprotein lipase contrib-utes to this decreased removal (Clinical Box 21–2). ACIDOSIS As noted in Chapter 1, acetoacetate and β-hydroxybutyrate are anions of the fairly strong acids acetoacetic acid and β-hy-droxybutyric acids. The hydrogen ions from these acids are buffered, but the buffering capacity is soon exceeded if pro-duction is increased. The resulting acidosis stimulates respira-tion, producing the rapid, deep respiration described by Kussmaul as “air hunger” and named (for him) Kussmaul breathing. The urine becomes acidic. However, when the ability of the kidneys to replace the plasma cations accompa-nying the organic anions with H+ and NH4 + is exceeded, Na+ and K+ are lost in the urine. The electrolyte and water losses lead to dehydration, hypovolemia, and hypotension. Finally, the acidosis and dehydration depress consciousness to the point of coma. Diabetic acidosis is a medical emergency. Now that the infections that used to complicate the disease can be controlled with antibiotics, acidosis is the most common cause of early death in clinical diabetes. In severe acidosis, total body Na+ is markedly depleted, and when Na+ loss exceeds water loss, plasma Na+ may also be low. Total body K+ is also low, but the plasma K+ is usually normal, partly because extracellular fluid (ECF) volume is reduced and partly because K+ moves from cells to ECF when the ECF H+ concentration is high. Another factor tending to maintain the plasma K+ is the lack of insulin-induced entry of K+ into cells. COMA Coma in diabetes can be due to acidosis and dehydration. However, the plasma glucose can be elevated to such a degree that independent of plasma pH, the hyperosmolarity of the plasma causes unconsciousness (hyperosmolar coma). Accu-mulation of lactate in the blood (lactic acidosis) may also complicate diabetic ketoacidosis if the tissues become hy-poxic, and lactic acidosis may itself cause coma. Brain edema occurs in about 1% of children with ketoacidosis, and it can cause coma. Its cause is unsettled, but it is a serious complica-tion, with a mortality rate of about 25%. CHOLESTEROL METABOLISM In diabetes, the plasma cholesterol level is usually elevated and this plays a role in the accelerated development of the athero-sclerotic vascular disease that is a major long-term complica-tion of diabetes in humans. The rise in plasma cholesterol level is due to an increase in the plasma concentration of very low-density lipoprotein (VLDL) and low-density lipoprotein (LDL) (see Chapter 1). These in turn may be due to increased hepatic production of VLDL or decreased removal of VLDL and LDL from the circulation. SUMMARY Because of the complexities of the metabolic abnormalities in diabetes, a summary is in order. One of the key features of in-sulin deficiency (Figure 21–9) is decreased entry of glucose into many tissues (decreased peripheral utilization). Also, the net release of glucose from the liver is increased (increased production), due in part to glucagon excess. The resultant hyperglycemia leads to glycosuria and a dehydrating osmotic diuresis. Dehydration leads to polydipsia. In the face of in-tracellular glucose deficiency, appetite is stimulated, glucose is formed from protein (gluconeogenesis), and energy sup-plies are maintained by metabolism of proteins and fats. Weight loss, debilitating protein deficiency, and inanition are the result. Fat catabolism is increased and the system is flooded with triglycerides and FFA. Fat synthesis is inhibited and the over-loaded catabolic pathways cannot handle the excess acetyl-CoA that is formed. In the liver, the acetyl-CoA is converted to ketone bodies. Two of these are organic acids, and meta-bolic acidosis develops as ketones accumulate. Na+ and K+ depletion is added to the acidosis because these plasma cat-ions are excreted with the organic anions not covered by the H+ and NH4 + secreted by the kidneys. Finally, the acidotic, hypovolemic, hypotensive, depleted animal or patient becomes comatose because of the toxic effects of acidosis, dehydra-CLINICAL BOX 21–2 Ketosis When excess acetyl-CoA is present in the body, some of it is converted to acetoacetyl-CoA and then, in the liver, to ace-toacetate. Acetoacetate and its derivatives, acetone and β-hydroxybutyrate, enter the circulation in large quantities (see Chapter 1). These circulating ketone bodies are an important source of energy in fasting. Half of the metabolic rate in fasted nor-mal dogs is said to be due to metabolism of ketones. The rate of ketone utilization in diabetics is also appreciable. It has been calculated that the maximal rate at which fat can be catabolized without significant ketosis is 2.5 g/kg body weight/d in diabetic humans. In untreated diabetes, pro-duction is much greater than this, and ketone bodies pile up in the bloodstream. CHAPTER 21 Endocrine Functions of the Pancreas & Regulation of Carbohydrate Metabolism 325 tion, and hyperosmolarity on the nervous system and dies if treatment is not instituted. All of these abnormalities are corrected by administration of insulin. Although emergency treatment of acidosis also includes administration of alkali to combat the acidosis and parenteral water, Na+, and K+ to replenish body stores, only insulin repairs the fundamental defects in a way that permits a return to normal. INSULIN EXCESS SYMPTOMS All the known consequences of insulin excess are manifesta-tions, directly or indirectly, of the effects of hypoglycemia on the nervous system. Except in individuals who have been fast-ing for some time, glucose is the only fuel used in appreciable quantities by the brain. The carbohydrate reserves in neural tissue are very limited and normal function depends on a con-tinuous glucose supply. As the plasma glucose level falls, the first symptoms are palpitations, sweating, and nervousness due to autonomic discharge. These appear at plasma glucose values slightly lower than the value at which autonomic acti-vation first begins, because the threshold for symptoms is slightly above the threshold for initial activation. At lower plasma glucose levels, so-called neuroglycopenic symptoms begin to appear. These include hunger as well as confusion and the other cognitive abnormalities. At even lower plasma glucose levels, lethargy, coma, convulsions, and eventually death occur. Obviously, the onset of hypoglycemic symptoms calls for prompt treatment with glucose or glucose-containing drinks such as orange juice. Although a dramatic disappear-ance of symptoms is the usual response, abnormalities ranging from intellectual dulling to coma may persist if the hypoglyce-mia was severe or prolonged. COMPENSATORY MECHANISMS One important compensation for hypoglycemia is cessation of the secretion of endogenous insulin. Inhibition of insulin se-cretion is complete at a plasma glucose level of about 80 mg/ dL (Figures 21–10 and 21–11). In addition, hypoglycemia trig-gers increased secretion of at least four counter-regulatory hormones: glucagon, epinephrine, growth hormone, and cor-tisol. The epinephrine response is reduced during sleep. Glu-cagon and epinephrine increase the hepatic output of glucose by increasing glycogenolysis. Growth hormone decreases the FIGURE 21–9 Effects of insulin deficiency. (Courtesy of RJ Havel.) Insulin deficiency (and glucagon excess) Increased protein catabolism Dehydration, acidosis Decreased glucose uptake Increased lipolysis Increased plasma amino acids, nitrogen loss in urine Coma, death Hyperglycemia, glycosuria, osmotic diuresis, electrolyte depletion Increased plasma FFA, ketogenesis, ketonuria, ketonemia FIGURE 21–10 Plasma glucose levels at which various effects of hypoglycemia appear. FIGURE 21–11 Mean rates of insulin and glucagon delivery from an artificial pancreas at various plasma glucose levels. The device was programmed to establish and maintain various plasma glu-cose levels in insulin-requiring diabetic humans, and the values for hormone output approximate the output of the normal human pan-creas. The shape of the insulin curve also resembles the insulin re-sponse of incubated B cells to graded concentrations of glucose. (Reproduced with permission from Marliss EB, et al: Normalization of glycemia in diabetics during meals with insulin and glucagon delivery by the artificial pancreas. Diabetes 1977;26:663.) Plasma glucose mmol/L mg/dL 90 75 60 45 30 15 0 4.6 Inhibition of insulin secretion Glucagon, epinephrine, growth hormone secretion Cortisol secretion Cognitive dysfunction Lethargy Coma Convulsions Permanent brain damage, death 0 1.1 0.6 1.7 2.8 2.2 3.2 3.8 4.50 3.00 1.50 0 40 80 120 160 200 240 500 400 300 200 100 Glucagon Insulin Plasma glucose (mg/dL) Glucagon secretion rate (μg/min) Insulin secretion rate (mU/min) 326 SECTION IV Endocrine & Reproductive Physiology utilization of glucose in various peripheral tissues, and cortisol has a similar action. The keys to counter-regulation appear to be epinephrine and glucagon: if the plasma concentration of either increases, the decline in the plasma glucose level is re-versed; but if both fail to increase, there is little if any compen-satory rise in the plasma glucose level. The actions of the other hormones are supplementary. Note that the autonomic discharge and release of counter-regulatory hormones normally occurs at a higher plasma glu-cose level than the cognitive deficits and other more serious CNS changes (Figure 21–10). For diabetics treated with insulin, the symptoms caused by the autonomic discharge serve as a warning to seek glucose replacement. However, particularly in long-term diabetics who have been tightly regulated, the auto-nomic symptoms may not occur, and the resulting hypoglyce-mia unawareness can be a clinical problem of some magnitude. REGULATION OF INSULIN SECRETION The normal concentration of insulin measured by radioim-munoassay in the peripheral venous plasma of fasting normal humans is 0–70 μU/mL (0–502 pmol/L). The amount of insu-lin secreted in the basal state is about 1 U/h, with a fivefold to tenfold increase following ingestion of food. Therefore, the av-erage amount secreted per day in a normal human is about 40 U (287 nmol). Factors that stimulate and inhibit insulin secretion are sum-marized in Table 21–6. EFFECTS OF THE PLASMA GLUCOSE LEVEL It has been known for many years that glucose acts directly on pancreatic B cells to increase insulin secretion. The response to glucose is biphasic; there is a rapid but short-lived increase in secretion followed by a more slowly developing prolonged increase (Figure 21–12). Glucose enters the B cells via GLUT 2 transporters and is phosphorylated by glucokinase then metabolized to pyruvate in the cytoplasm (Figure 21–13). The pyruvate enters the mito-chondria and is metabolized to CO2 and H2O via the citric acid cycle with the formation of ATP by oxidative phosphorylation. The ATP enters the cytoplasm, where it inhibits ATP-sensitive K+ channels, reducing K+ efflux. This depolarizes the B cell, and Ca2+ enters the cell via voltage-gated Ca2+ channels. The Ca2+ influx causes exocytosis of a readily releasable pool of insulin-containing secretory granules, producing the initial spike of insulin secretion. Metabolism of pyruvate via the citric acid cycle also causes an increase in intracellular glutamate. The glutamate appears to act on a second pool of secretory granules, committing them to the releasable form. The action of glutamate may be to decrease the pH in the secretory granules, a necessary step in their maturation. The release of these granules then pro-duces the prolonged second phase of the insulin response to glucose. Thus, glutamate appears to act as an intracellular sec-ond messenger that primes secretory granules for secretion. TABLE 21–6 Factors affecting insulin secretion. Stimulators Inhibitors Glucose Somatostatin Mannose 2-Deoxyglucose Amino acids (leucine, arginine, others) Mannoheptulose Intestinal hormones (GIP, GLP-1 [7– 36], gastrin, secretin, CCK; others?) α-Adrenergic stimulators (nor-epinephrine, epinephrine) β-Keto acids β-Adrenergic blockers (propranolol) Acetylcholine Glucagon Galanin Cyclic AMP and various cAMP-generating substances Diazoxide Thiazide diuretics β-Adrenergic stimulators K+ depletion Theophylline Phenytoin Sulfonylureas Alloxan Microtubule inhibitors Insulin FIGURE 21–12 Insulin secretion from perfused rat pancreas in response to sustained glucose infusion. Values are means of three preparations. The top record shows the glucose concentration in the effluent perfusion mixture. (Reproduced with permission, from Curry DL Bennett LL, Grodsky GM: Dynamics of insulin secretion by the perfused rat pancreas. Endocrinology 1968;83:572.) 300 250 200 150 100 50 0 5 10 15 20 Time (min) 25 30 35 40 45 Insulin release (ng/30 s) Glucose (mg/dL) 300 200 100 0 CHAPTER 21 Endocrine Functions of the Pancreas & Regulation of Carbohydrate Metabolism 327 The feedback control of plasma glucose on insulin secre-tion normally operates with great precision so that plasma glucose and insulin levels parallel each other with remarkable consistency. PROTEIN & FAT DERIVATIVES Insulin stimulates the incorporation of amino acids into pro-teins and combats the fat catabolism that produces the β-keto acids. Therefore, it is not surprising that arginine, leucine, and certain other amino acids stimulate insulin secretion, as do β-keto acids such as acetoacetate. Like glucose, these com-pounds generate ATP when metabolized, and this closes ATP-sensitive K+ channels in the B cells. In addition, L-arginine is the precursor of NO, and NO stimulates insulin secretion. ORAL HYPOGLYCEMIC AGENTS Tolbutamide and other sulfonylurea derivatives such as aceto-hexamide, tolazamide, glipizide, and glyburide are orally active hypoglycemic agents that lower blood glucose by increasing the secretion of insulin. They only work in patients with some remaining B cells and are ineffective after pancreatectomy or in type 1 diabetes. They bind to the ATP-inhibited K+ channels in the B cell membranes and inhibit channel activity, depolarizing the B cell membrane and increasing Ca2+ influx and hence in-sulin release, independent of increases in plasma glucose. Persistent hyperinsulinemic hypoglycemia of infancy is a condition in which plasma insulin is elevated despite the hypoglycemia. The condition is caused by mutations in the genes for various enzymes in B cells that decrease K+ efflux via the ATP-sensible K+ channels. Treatment consists of adminis-tration of diazoxide, a drug that increases the activity of the K+ channels or, in more severe cases, subtotal pancreatectomy. The biguanide metformin is an oral hypoglycemic agent that acts in the absence of insulin. Metformin acts primarily by reducing gluconeogenesis and therefore decreasing hepatic glucose output. It is sometimes combined with a sulfonylurea in the treatment of type 2 diabetes. Metformin can cause lac-tic acidosis, but the incidence is usually low. Troglitazone (Rezulin) and related thiazolidinediones are also used in the treatment of diabetes because they increase insulin-mediated peripheral glucose disposal, thus reducing insulin resistance. They bind to and activate peroxisome prolif-erator-activated receptor γ (PPARγ) in the nucleus of cells. Acti-vation of this receptor, which is a member of the superfamily of hormone-sensitive nuclear transcription factors, has a unique ability to normalize a variety of metabolic functions. CYCLIC AMP & INSULIN SECRETION Stimuli that increase cAMP levels in B cells increase insulin se-cretion, including β-adrenergic agonists, glucagon, and phos-phodiesterase inhibitors such as theophylline. Catecholamines have a dual effect on insulin secretion; they inhibit insulin secretion via α2-adrenergic receptors and stim-ulate insulin secretion via β-adrenergic receptors. The net effect of epinephrine and norepinephrine is usually inhibi-tion. However, if catecholamines are infused after administra-tion of α-adrenergic blocking drugs, the inhibition is converted to stimulation. EFFECT OF AUTONOMIC NERVES Branches of the right vagus nerve innervate the pancreatic is-lets, and stimulation of this parasympathetic pathway causes increased insulin secretion via M4 receptors (see Table 7–2). Atropine blocks the response and acetylcholine stimulates in-sulin secretion. The effect of acetylcholine, like that of glucose, is due to increased cytoplasmic Ca2+, but acetylcholine acti-vates phospholipase C, with the released IP3 releasing the Ca2+ from the endoplasmic reticulum. Stimulation of the sympathetic nerves to the pancreas inhib-its insulin secretion. The inhibition is produced by released norepinephrine acting on α2-adrenergic receptors. However, if α-adrenergic receptors are blocked, stimulation of the sympa-thetic nerves causes increased insulin secretion mediated by β2-adrenergic receptors. The polypeptide galanin is found in some of the autonomic nerves innervating the islets, and galanin inhibits insulin secretion by activating the K+ channels that are inhibited by ATP. Thus, although the denervated pancreas FIGURE 21–13 Insulin secretion. Glucose enters B cells by GLUT 2 transporters. It is phosphorylated and metabolized to pyruvate (Pyr) in the cytoplasm. The Pyr enters the mitochondria and is metab-olized via the citric acid cycle. The ATP formed by oxidative phosphory-lation inhibits ATP-sensitive K+ channels, reducing K+ efflux. This depolarizes the B cell, and Ca2+ influx is increased. The Ca2+ stimulates release of insulin by exocytosis. Glutamate (Glu) is also formed, and this primes secretory granules, preparing them for exocytosis. Glucokinase Glucose-P Citric acid cycle ATP ATP Pyr Glu Insulin Ca2+ K+ K+ Glucose GLUT 2 328 SECTION IV Endocrine & Reproductive Physiology responds to glucose, the autonomic innervation of the pancreas is involved in the overall regulation of insulin secretion. INTESTINAL HORMONES Orally administered glucose exerts a greater insulin-stimulating effect than intravenously administered glucose, and orally ad-ministered amino acids also produce a greater insulin response than intravenous amino acids. These observations led to explo-ration of the possibility that a substance secreted by the gas-trointestinal mucosa stimulated insulin secretion. Glucagon, glucagon derivatives, secretin, cholecystokinin (CCK), gastrin, and gastric inhibitory peptide (GIP) all have such an action (see Chapter 26), and CCK potentiates the insulin-stimulating ef-fects of amino acids. However, GIP is the only one of these pep-tides that produces stimulation when administered in doses that reflect blood GIP levels produced by an oral glucose load. Recently, attention has focused on glucagon-like polypep-tide 1 (7–36) (GLP-1 [7–36]) as an additional gut factor that stimulates insulin secretion. This polypeptide is a product of preproglucagon. B cells have GLP-1 (7–36) receptors as well as GIP recep-tors, and GLP-1 (7–36) is a more potent insulinotropic hor-mone than GIP. GIP and GLP-1 (7–36) both appear to act by increasing Ca2+ influx through voltage-gated Ca2+ channels. The possible roles of pancreatic somatostatin and glucagon in the regulation of insulin secretion are discussed below (Clinical Box 21–3). LONG-TERM CHANGES IN B CELL RESPONSES The magnitude of the insulin response to a given stimulus is determined in part by the secretory history of the B cells. Indi-viduals fed a high-carbohydrate diet for several weeks not only have higher fasting plasma insulin levels but also show a great-er secretory response to a glucose load than individuals fed an isocaloric low-carbohydrate diet. Although the B cells respond to stimulation with hypertro-phy like other endocrine cells, they become exhausted and stop secreting (B cell exhaustion) when the stimulation is marked or prolonged. The pancreatic reserve is large and it is difficult to produce B cell exhaustion in normal animals, but if the pancreatic reserve is reduced by partial pancreatectomy, exhaustion of the remaining B cells can be initiated by any pro-cedure that chronically raises the plasma glucose level. For example, diabetes can be produced in animals with limited pancreatic reserves by anterior pituitary extracts, growth hor-mone, thyroid hormones, or the prolonged continuous infu-sion of glucose alone. The diabetes precipitated by hormones in animals is at first reversible, but with prolonged treatment it becomes permanent. The transient diabetes is usually named for the agent producing it, for example, “hypophysial diabetes” or “thyroid diabetes.” Permanent diabetes persisting after treat-ment has been discontinued is indicated by the prefix meta-, for example, “metahypophysial diabetes” or “metathyroid diabetes.” When insulin is administered along with the diabe-togenic hormones, the B cells are protected, probably because the plasma glucose is lowered, and diabetes does not develop. It is interesting in this regard that genetic factors may be involved in the control of B cell reserve. In mice in which the gene for IRS-1 has been knocked out (see above), a robust com-pensatory B cell response occurs. However, in IRS-2 knockouts, the compensation is reduced and a more severe diabetic pheno-type is produced. GLUCAGON CHEMISTRY Human glucagon, a linear polypeptide with a molecular weight of 3485, is produced by the A cells of the pancreatic islets and the upper gastrointestinal tract. It contains 29 amino acid residues. All mammalian glucagons appear to have the same structure. Human preproglucagon (Figure 21–14) is a 179-amino-acid protein that is found in pancreatic A cells, in L cells in the lower gastrointestinal tract, and in the brain. It is the product of a single mRNA, but it is processed differently in different tissues. In A cells, it is processed primarily to glucagon and major progluca-gon fragment (MPGF). In L cells, it is processed primarily to glicentin, a polypeptide that consists of glucagon extended by additional amino acid residues at either end, plus glucagon-like polypeptides 1 and 2 (GLP-1 and GLP-2). Some oxyntomodu-lin is also formed, and in both A and L cells, residual glicentin-related polypeptide (GRPP) is left. Glicentin has some glucagon activity. GLP-1 and GLP-2 have no definite biologic activity by themselves. However, GLP-1 is processed further by removal of its amino-terminal amino acid residues and the product, GLP-1 (7–36), is a potent stimulator of insulin secretion that also in-creases glucose utilization (see above). GLP-1 and GLP-2 are also produced in the brain. The function of GLP-1 in this location is uncertain, but GLP-2 appears to be the mediator in a pathway from the nucleus tractus solitarius (NTS) to the dorsomedial nu-clei of the hypothalamus, and injection of GLP-2 lowers food in-take. Oxyntomodulin inhibits gastric acid secretion, though its CLINICAL BOX 21–3 Effects of K+ Depletion K+ depletion decreases insulin secretion, and K+-depleted patients, for example, patients with primary hyperaldoster-onism (see Chapter 22), develop diabetic glucose tolerance curves. These curves are restored to normal by K+ repletion. The thiazide diuretics, which cause loss of K+ as well as Na+ in the urine (see Chapter 38), decrease glucose tolerance and make diabetes worse. They apparently exert this effect primarily because of their K+-depleting effects, although some of them also cause pancreatic islet cell damage. CHAPTER 21 Endocrine Functions of the Pancreas & Regulation of Carbohydrate Metabolism 329 physiologic role is unsettled, and GRPP does not have any estab-lished physiologic effects. ACTION Glucagon is glycogenolytic, gluconeogenic, lipolytic, and keto-genic. It acts on G-protein coupled receptors with a molecular weight of about 190,000. In the liver, it acts via Gs to activate adenylyl cyclase and increase intracellular cAMP. This leads via protein kinase A to activation of phosphorylase and there-fore to increased breakdown of glycogen and an increase in plasma glucose. However, glucagon acts on different glucagon receptors located on the same hepatic cells to activate phospho-lipase C, and the resulting increase in cytoplasmic Ca2+ also stimulates glycogenolysis. Protein kinase A also decreases the metabolism of glucose 6-phosphate (Figure 21–15) by inhibit-ing the conversion of phosphoenolpyruvate to pyruvate. It also decreases the concentration of fructose 2,6-diphosphate and this in turn inhibits the conversion of fructose 6-phosphate to fructose 1,6-diphosphate. The resultant buildup of glucose 6-phosphate leads to increased glucose synthesis and release. Glucagon does not cause glycogenolysis in muscle. It increases gluconeogenesis from available amino acids in the liver and ele-vates the metabolic rate. It increases ketone body formation by decreasing malonyl-CoA levels in the liver. Its lipolytic activity, which leads in turn to increased ketogenesis, is discussed in Chapter 1. The calorigenic action of glucagon is not due to the hyperglycemia per se but probably to the increased hepatic deamination of amino acids. Large doses of exogenous glucagon exert a positively ino-tropic effect on the heart (see Chapter 31) without producing increased myocardial excitability, presumably because they increase myocardial cAMP. Use of this hormone in the treat-ment of heart disease has been advocated, but there is no evi-dence for a physiologic role of glucagon in the regulation of cardiac function. Glucagon also stimulates the secretion of growth hormone, insulin, and pancreatic somatostatin. METABOLISM Glucagon has a half-life in the circulation of 5 to 10 min. It is degraded by many tissues but particularly by the liver. Because glucagon is secreted into the portal vein and reaches the liver before it reaches the peripheral circulation, peripheral blood levels are relatively low. The rise in peripheral blood glucagon levels produced by excitatory stimuli is exaggerated in patients with cirrhosis, presumably because of decreased hepatic deg-radation of the hormone. REGULATION OF SECRETION The principal factors known to affect glucagon secretion are summarized in Table 21–7. Secretion is increased by hypogly-cemia and decreased by a rise in plasma glucose. Pancreatic B cells contain GABA, and evidence suggests that coincident with the increased insulin secretion produced by hyperglyce-mia, GABA is released and acts on the A cells to inhibit gluca-gon secretion by activating GABAA receptors. The GABAA receptors are Cl– channels, and the resulting Cl– influx hyper-polarizes the A cells. Secretion is also increased by stimulation of the sympa-thetic nerves to the pancreas, and this sympathetic effect is mediated via β-adrenergic receptors and cAMP. It appears that the A cells are like the B cells in that stimulation of β-adrenergic receptors increases secretion and stimulation of α-adrenergic receptors inhibits secretion. However, the pancre-atic response to sympathetic stimulation in the absence of blocking drugs is increased secretion of glucagon, so the effect of β-receptors predominates in the glucagon-secreting cells. The stimulatory effects of various stresses and possibly of exercise and infection are mediated at least in part via the FIGURE 21–14 Posttranslational processing of preproglucagon in A and L cells. S, signal peptide; GRPP, glicentin-related polypeptide; GLP, glucagon-like polypeptide; Oxy, oxynto-modulin; MPGF, major proglucagon fragment. (Modified from Drucker, DJ: Glucagon and glucagon-like peptides. Pancreas 1990;5:484.) Glucagon GLP-1 GLP-2 GRPP S A cells Glucagon MPGF GRPP L cells Glicentin GLP-1 GLP-2 Oxyntomodulin GRPP Oxy Glicentin MPGF FIGURE 21–15 Mechanisms by which glucagon increases glucose output from the liver. Solid arrows indicate facilitation; dashed arrows indicate inhibition. Pyruvate Phosphoenolpyruvate Fructose 1, 6-biPO4 Fructose 2, 6-biPO4 Protein kinase A cAMP Fructose 6-PO4 Glucose 6-PO4 Glucose Glycogen Glucagon 330 SECTION IV Endocrine & Reproductive Physiology sympathetic nervous system. Vagal stimulation also increases glucagon secretion. A protein meal and infusion of various amino acids increase glucagon secretion. It seems appropriate that the glucogenic amino acids are particularly potent in this regard, since these are the amino acids that are converted to glucose in the liver under the influence of glucagon. The increase in glucagon secretion following a protein meal is also valuable, since the amino acids stimulate insulin secretion and the secreted gluca-gon prevents the development of hypoglycemia while the insu-lin promotes storage of the absorbed carbohydrates and lipids. Glucagon secretion increases during starvation. It reaches a peak on the third day of a fast, at the time of maximal gluco-neogenesis. Thereafter, the plasma glucagon level declines as fatty acids and ketones become the major sources of energy. During exercise, there is an increase in glucose utilization that is balanced by an increase in glucose production caused by an increase in circulating glucagon levels. The glucagon response to oral administration of amino acids is greater than the response to intravenous infusion of amino acids, suggesting that a glucagon-stimulating factor is secreted from the gastrointestinal mucosa. CCK and gastrin increase glucagon secretion, whereas secretin inhibits it. Because CCK and gastrin secretion are both increased by a protein meal, either hormone could be the gastrointestinal mediator of the glucagon response. The inhibition produced by somatostatin is discussed below. Glucagon secretion is also inhibited by FFA and ketones. However, this inhibition can be overridden, since plasma glu-cagon levels are high in diabetic ketoacidosis. INSULIN–GLUCAGON MOLAR RATIOS As noted previously, insulin is glycogenic, antigluconeogenet-ic, antilipolytic, and antiketotic in its actions. It thus favors storage of absorbed nutrients and is a “hormone of energy storage.” Glucagon, on the other hand, is glycogenolytic, glu-coneogenetic, lipolytic, and ketogenic. It mobilizes energy stores and is a “hormone of energy release.” Because of their opposite effects, the blood levels of both hormones must be considered in any given situation. It is convenient to think in terms of the molar ratios of these hormones. The insulin–glucagon molar ratios fluctuate markedly because the secretion of glucagon and insulin are both modi-fied by the conditions that preceded the application of any given stimulus (Table 21–8). Thus, for example, the insulin– glucagon molar ratio on a balanced diet is approximately 2.3. An infusion of arginine increases the secretion of both hor-mones and raises the ratio to 3.0. After 3 days of starvation, the ratio falls to 0.4, and an infusion of arginine in this state lowers the ratio to 0.3. Conversely, the ratio is 25 in individuals receiv-ing a constant infusion of glucose and rises to 170 on ingestion of a protein meal during the infusion. The rise occurs because insulin secretion rises sharply, while the usual glucagon response to a protein meal is abolished. Thus, when energy is needed during starvation, the insulin–glucagon molar ratio is low, favoring glycogen breakdown and gluconeogenesis; con-versely, when the need for energy mobilization is low, the ratio is high, favoring the deposition of glycogen, protein, and fat. OTHER ISLET CELL HORMONES In addition to insulin and glucagon, the pancreatic islets secrete somatostatin and pancreatic polypeptide into the bloodstream. In addition, somatostatin may be involved in regulatory pro-cesses within the islets that adjust the pattern of hormones se-creted in response to various stimuli. SOMATOSTATIN Somatostatin and its receptors are discussed in Chapter 7. So-matostatin 14 (SS 14) and its amino terminal-extended form somatostatin 28 (SS 28) are found in the D cells of pancreatic islets. Both forms inhibit the secretion of insulin, glucagon, and pancreatic polypeptide and act locally within the pancreatic is-lets in a paracrine fashion. SS 28 is more active than SS 14 in in-hibiting insulin secretion, and it apparently acts via the SSTR5 receptor (see Chapter 7). Patients with somatostatin-secreting TABLE 21–7 Adipokines. Agent Effect on Insulin Resistance Leptin Decreases TNFα Increases Adiponectin Decreases Resistin Increases TABLE 21–8 Factors affecting glucagon secretion. Stimulators Inhibitors Amino acids (particularly the glucogenic amino acids: alanine, serine, glycine, cys-teine, and threonine) Glucose CCK, gastrin Somatostatin Cortisol Secretin Exercise FFA Infections Ketones Other stresses Insulin β-Adrenergic stimulators Phenytoin Theophylline α-Adrenergic stimulators Acetylcholine GABA CHAPTER 21 Endocrine Functions of the Pancreas & Regulation of Carbohydrate Metabolism 331 pancreatic tumors (somatostatinomas) develop hyperglyce-mia and other manifestations of diabetes that disappear when the tumor is removed. They also develop dyspepsia due to slow gastric emptying and decreased gastric acid secretion, and gall-stones, which are precipitated by decreased gallbladder con-traction due to inhibition of CCK secretion. The secretion of pancreatic somatostatin is increased by several of the same stimuli that increase insulin secretion, that is, glucose and ami-no acids, particularly arginine and leucine. It is also increased by CCK. Somatostatin is released from the pancreas and the gastrointestinal tract into the peripheral blood. PANCREATIC POLYPEPTIDE Human pancreatic polypeptide is a linear polypeptide that contains 36 amino acid residues and is produced by F cells in the islets. It is closely related to two other 36-amino acid polypeptides, polypeptide YY, a gastrointestinal peptide (see Chapter 26), and neuropeptide Y, which is found in the brain and the autonomic nervous system (see Chapter 7). All end in tyrosine and are amidated at their carboxyl terminal. At least in part, pancreatic polypeptide secretion is under cholinergic control; plasma levels fall after administration of atropine. Its secretion is increased by a meal containing protein and by fasting, exercise, and acute hypoglycemia. Secretion is de-creased by somatostatin and intravenous glucose. Infusions of leucine, arginine, and alanine do not affect it, so the stimula-tory effect of a protein meal may be mediated indirectly. Pan-creatic polypeptide slows the absorption of food in humans, and it may smooth out the peaks and valleys of absorption. However, its exact physiologic function is still uncertain. ORGANIZATION OF THE PANCREATIC ISLETS The presence in the pancreatic islets of hormones that affect the secretion of other islet hormones suggests that the islets function as secretory units in the regulation of nutrient ho-meostasis. Somatostatin inhibits the secretion of insulin, glu-cagon, and pancreatic polypeptide (Figure 21–16); insulin inhibits the secretion of glucagon; and glucagon stimulates the secretion of insulin and somatostatin. As noted above, A and D cells and pancreatic polypeptide-secreting cells are general-ly located around the periphery of the islets, with the B cells in the center. There are clearly two types of islets, glucagon-rich islets and pancreatic polypeptide-rich islets, but the functional significance of this separation is not known. The islet cell hor-mones released into the ECF probably diffuse to other islet cells and influence their function (paracrine communication; see Chapter 26). It has been demonstrated that gap junctions are present between A, B, and D cells and that these permit the passage of ions and other small molecules from one cell to an-other, which could coordinate their secretory functions. EFFECTS OF OTHER HORMONES & EXERCISE ON CARBOHYDRATE METABOLISM Exercise has direct effects on carbohydrate metabolism. Many hormones in addition to insulin, IGF-I, IGF-II, glucagon, and somatostatin also have important roles in the regulation of carbohydrate metabolism. They include epinephrine, thyroid hormones, glucocorticoids, and growth hormone. The other functions of these hormones are considered elsewhere, but it seems wise to summarize their effects on carbohydrate metab-olism in the context of the present chapter. EXERCISE The entry of glucose into skeletal muscle is increased during exercise in the absence of insulin by causing an insulin-inde-pendent increase in the number of GLUT 4 transporters in muscle cell membranes (see above). This increase in glucose entry persists for several hours after exercise, and regular ex-ercise training can also produce prolonged increases in insulin sensitivity. Exercise can precipitate hypoglycemia in diabetics not only because of the increase in muscle uptake of glucose but also because absorption of injected insulin is more rapid during exercise. Patients with diabetes should take in extra calories or reduce their insulin dosage when they exercise. CATECHOLAMINES The activation of phosphorylase in liver by catecholamines is discussed in Chapter 1. Activation occurs via β-adrenergic re-ceptors, which increase intracellular cAMP, and α-adrenergic receptors, which increase intracellular Ca2+. Hepatic glucose output is increased, producing hyperglycemia. In muscle, the phosphorylase is also activated via cAMP and presumably via Ca2+, but the glucose 6-phosphate formed can be catabolized only to pyruvate because of the absence of glucose 6-phospha-tase. For reasons that are not entirely clear, large amounts of pyruvate are converted to lactate, which diffuses from the muscle into the circulation (Figure 21–17). The lactate is oxi-dized in the liver to pyruvate and converted to glycogen. FIGURE 21–16 Effects of islet cell hormones on the secretion of other islet cell hormones. Solid arrows indicate stimu-lation; dashed arrows indicate inhibition. Glucagon Pancreatic polypeptide Insulin Somatostatin 332 SECTION IV Endocrine & Reproductive Physiology Therefore, the response to an injection of epinephrine is an initial glycogenolysis followed by a rise in hepatic glycogen content. Lactate oxidation may be responsible for the calori-genic effect of epinephrine (see Chapter 22). Epinephrine and norepinephrine also liberate FFA into the circulation, and epi-nephrine decreases peripheral utilization of glucose. THYROID HORMONES Thyroid hormones make experimental diabetes worse; thyro-toxicosis aggravates clinical diabetes; and metathyroid diabe-tes can be produced in animals with decreased pancreatic reserve. The principal diabetogenic effect of thyroid hormones is to increase absorption of glucose from the intestine, but the hormones also cause (probably by potentiating the effects of catecholamines) some degree of hepatic glycogen depletion. Glycogen-depleted liver cells are easily damaged. When the liver is damaged, the glucose tolerance curve is diabetic be-cause the liver takes up less of the absorbed glucose. Thyroid hormones may also accelerate the degradation of insulin. All these actions have a hyperglycemic effect and, if the pancreatic reserve is low, may lead to B cell exhaustion. ADRENAL GLUCOCORTICOIDS Glucocorticoids from the adrenal cortex (see Chapter 22) ele-vate blood glucose and produce a diabetic type of glucose toler-ance curve. In humans, this effect may occur only in individuals with a genetic predisposition to diabetes. Glucose tolerance is reduced in 80% of patients with Cushing syndrome (see Chap-ter 22), and 20% of these patients have frank diabetes. The glu-cocorticoids are necessary for glucagon to exert its gluconeogenic action during fasting. They are gluconeogenic themselves, but their role is mainly permissive. In adrenal insuf-ficiency, the blood glucose is normal as long as food intake is maintained, but fasting precipitates hypoglycemia and collapse. The plasma-glucose-lowering effect of insulin is greatly en-hanced in patients with adrenal insufficiency. In animals with experimental diabetes, adrenalectomy markedly ameliorates the diabetes. The major diabetogenic effects are an increase in protein catabolism with increased gluconeogenesis in the liver; increased hepatic glycogenesis and ketogenesis; and a decrease in peripheral glucose utilization relative to the blood insulin lev-el that may be due to inhibition of glucose phosphorylation. GROWTH HORMONE Human growth hormone makes clinical diabetes worse, and 25% of patients with growth hormone-secreting tumors of the anterior pituitary have diabetes. Hypophysectomy amelio-rates diabetes and decreases insulin resistance even more than adrenalectomy, whereas growth hormone treatment increases insulin resistance. The effects of growth hormone are partly direct and partly mediated via IGF-I (see Chapter 24). Growth hormone mobi-lizes FFA from adipose tissue, thus favoring ketogenesis. It decreases glucose uptake into some tissues (“anti-insulin action”), increases hepatic glucose output, and may decrease tissue binding of insulin. Indeed, it has been suggested that the ketosis and decreased glucose tolerance produced by star-vation are due to hypersecretion of growth hormone. Growth hormone does not stimulate insulin secretion directly, but the hyperglycemia it produces secondarily stimulates the pan-creas and may eventually exhaust the B cells. HYPOGLYCEMIA & DIABETES MELLITUS IN HUMANS HYPOGLYCEMIA “Insulin reactions” are common in type 1 diabetics and occa-sional hypoglycemic episodes are the price of good diabetic control in most diabetics. Glucose uptake by skeletal muscle and absorption of injected insulin both increase during exer-cise (see above). Symptomatic hypoglycemia also occurs in nondiabetics, and a review of some of the more important causes serves to emphasize the variables affecting plasma glucose homeostasis. Chronic mild hypoglycemia can cause incoordination and slurred speech, and the condition can be mistaken for drunk-enness. Mental aberrations and convulsions in the absence of frank coma also occur. When the level of insulin secretion is chronically elevated by an insulinoma, a rare, insulin-secret-ing tumor of the pancreas, symptoms are most common in the morning. This is because a night of fasting has depleted hepatic glycogen reserves. However, symptoms can develop at any time, and in such patients, the diagnosis may be missed. Some cases of insulinoma have been erroneously diagnosed as epilepsy or psychosis. Hypoglycemia also occurs in some patients with large malignant tumors that do not involve the pancreatic islets, and the hypoglycemia in these cases is apparently due to excess secretion of IGF-II. FIGURE 21–17 Effect of epinephrine on tissue glycogen, plasma glucose, and blood lactate levels in fed rats. (Reproduced with permission from Ruch TC, Patton HD [editors]: Physiology and Biophysics, 20th ed. Vol. 3. Saunders, 1973.) +100 +50 0 −50 0 1 2 3 % change from initial level Time after injection of epinephrine (h) Liver glycogen Plasma glucose Blood lactate Muscle glycogen CHAPTER 21 Endocrine Functions of the Pancreas & Regulation of Carbohydrate Metabolism 333 As noted above, the autonomic discharge caused by lowered blood glucose that produces shakiness, sweating, anxiety, and hunger normally occurs at plasma glucose levels that are higher than the glucose levels that cause cognitive dysfunction, thereby serving as a warning to ingest sugar. However, in some individuals, these warning symptoms fail to occur before the cognitive symptoms, due to cerebral dysfunction (desensitiza-tion), and this hypoglycemia unawareness is potentially dan-gerous. The condition is prone to develop in patients with insulinomas and in diabetics receiving intensive insulin ther-apy, so it appears that repeated bouts of hypoglycemia cause the eventual development of hypoglycemia unawareness. If blood sugar rises again for some time, the warning symptoms again appear at a higher plasma glucose level than cognitive abnormalities and coma. The reason why prolonged hypogly-cemia causes loss of the warning symptoms is unsettled. In liver disease, the glucose tolerance curve is diabetic but the fasting plasma glucose level is low (Figure 21–18). In functional hypoglycemia, the plasma glucose rise is normal after a test dose of glucose, but the subsequent fall overshoots to hypoglycemic levels, producing symptoms 3 to 4 h after meals. This pattern is sometimes seen in individuals who later develop diabetes. Patients with this syndrome should be dis-tinguished from the more numerous patients with similar symptoms due to psychologic or other problems who do not have hypoglycemia when blood is drawn during the sympto-matic episode. It has been postulated that the overshoot of the plasma glucose is due to insulin secretion stimulated by impulses in the right vagus, but cholinergic blocking agents do not routinely correct the abnormality. In some thyrotoxic patients and in patients who have had gastrectomies or other operations that speed the passage of food into the intestine, glucose absorption is abnormally rapid. The plasma glucose rises to a high, early peak, but it then falls rapidly to hypogly-cemic levels because the wave of hyperglycemia evokes a greater than normal rise in insulin secretion. Symptoms char-acteristically occur about 2 h after meals. DIABETES MELLITUS The incidence of diabetes mellitus in the human population has reached epidemic proportions worldwide and it is increas-ing at a rapid rate. In 2000, there were an estimated 150 mil-lion cases in the world; this number is projected to increase to 221 million by 2010. Ninety percent of the present cases are type 2 diabetes, and most of the increase will be in type 2, par-alleling the increase in the incidence of obesity. Diabetes is sometimes complicated by acidosis and coma, and in long-standing diabetes additional complications occur. These include microvascular, macrovascular, and neuropathic disease. The microvascular abnormalities are proliferative scar-ring of the retina (diabetic retinopathy) leading to blindness; and renal disease (diabetic nephropathy) leading to renal fail-ure. The macrovascular abnormalities are due to accelerated atherosclerosis, which is secondary to increased plasma LDL. The result is an increased incidence of stroke and myocardial infarction. The neuropathic abnormalities (diabetic neuropa-thy) involve the autonomic nervous system and peripheral nerves. The neuropathy plus the atherosclerotic circulatory insufficiency in the extremities and reduced resistance to infec-tion can lead to chronic ulceration and gangrene, particularly in the feet. The ultimate cause of the microvascular and neuropathic complications is chronic hyperglycemia, and tight control of the diabetes reduces their incidence. Intracellular hyperglyce-mia activates the enzyme aldose reductase. This increases the formation of sorbitol in cells, which in turn reduces cellular Na+–K+ ATPase. In addition, intracellular glucose can be con-verted to so-called Amadori products, and these in turn can FIGURE 21–18 Typical glucose tolerance curves after an oral glucose load in liver disease and in conditions causing excessively rapid absorption of glucose from the intestine. The horizontal line is the approximate plasma glucose level at which hy-poglycemic symptoms may appear. 150 125 100 75 50 25 0 1 2 3 4 Time (h) Normal Liver disease Excessively rapid carbohydrate absorption Plasma glucose (mg/dL) CLINICAL BOX 21–4 Macrosomia & GLUT 1 Deficiency Infants born to diabetic mothers often have high birth weights and large organs (macrosomia). This condition is caused by excess circulating insulin in the fetus, which in turn is caused in part by stimulation of the fetal pancreas by glucose and amino acids from the blood of the mother. Free insulin in maternal blood is destroyed by proteases in the placenta, but antibody-bound insulin is protected, so it reaches the fetus. Therefore, fetal macrosomia also occurs in women who develop antibodies against various animal insulin and then continue to receive the animal insulin dur-ing pregnancy. Infants with GLUT 1 deficiency have defective transport of glucose across the blood–brain barrier. They have low cerebrospinal fluid glucose in the presence of normal plasma glucose, seizures, and developmental delay. 334 SECTION IV Endocrine & Reproductive Physiology form advanced glycosylation end products (AGEs), which cross-link matrix proteins. This damages blood vessels. The AGEs also interfere with leukocyte responses to infection. TYPES OF DIABETES The cause of clinical diabetes is always a deficiency of the ef-fects of insulin at the tissue level, but the deficiency may be rel-ative. One of the common forms, type 1, or insulin-dependent diabetes mellitus (IDDM), is due to insulin defi-ciency caused by autoimmune destruction of the B cells in the pancreatic islets; the A, D, and F cells remain intact. The sec-ond common form, type 2, or non-insulin-dependent diabe-tes mellitus (NIDDM), is characterized by insulin resistance. In addition, some cases of diabetes are due to other diseases or conditions such as chronic pancreatitis, total pancreatec-tomy, Cushing syndrome (see Chapter 22), and acromegaly (see Chapter 24). These make up 5% of the total cases and are sometimes classified as secondary diabetes. Type 1 diabetes usually develops before the age of 40 and hence is called juvenile diabetes. Patients with this disease are not obese and they have a high incidence of ketosis and acidosis. Various anti-B cell antibodies are present in plasma, but the current thinking is that type 1 diabetes is primarily a T lymphocyte-mediated disease. Definite genetic susceptibility is present as well; if one identical twin develops the disease, the chances are 1 in 3 that the other twin will also do so. In other words, the concordance rate is about 33%. The main genetic abnormality is in the major histocompatibility com-plex on chromosome 6, making individuals with certain types of histocompatibility antigens (see Chapter 3) much more prone to develop the disease. Other genes are also involved. Immunosuppression with drugs such as cyclosporine ame-liorate type 1 diabetes if given early in the disease before all B cells are lost. Attempts have been made to treat type 1 diabetes by transplanting pancreatic tissue or isolated islet cells, but results to date have been poor, largely because B cells are eas-ily damaged and it is difficult to transplant enough of them to normalize glucose responses. As mentioned above, type 2 is the most common type of diabetes and is usually associated with obesity. It usually develops after age 40 and is not associated with total loss of the ability to secrete insulin. It has an insidious onset, is rarely associated with ketosis, and is usually associated with normal B cell morphology and insulin content if the B cells have not become exhausted. The genetic component in type 2 diabetes is actually stronger than the genetic component in type 1 dia-betes; in identical twins, the concordance rate is higher, rang-ing in some studies to nearly 100%. In some patients, type 2 diabetes is due to defects in identi-fied genes. Over 60 of these defects have been described. They include defects in glucokinase (about 1% of the cases), the insulin molecule itself (about 0.5% of the cases), the insulin receptor (about 1% of the cases), GLUT 4 (about 1% of the cases), or IRS-1 (about 15% of the cases). In maturity-onset diabetes occurring in young individuals (MODY), which accounts for about 1% of the cases of type 2 diabetes, loss-of-function mutations have been described in six different genes. Five code for transcription factors affecting the production of enzymes involved in glucose metabolism. The sixth is the gene for glucokinase (Figure 21–13), the enzyme that controls the rate of glucose phosphorylation and hence its metabolism in the B cells. However, the vast majority of cases of type 2 diabe-tes are almost certainly polygenic in origin, and the actual genes involved are still unknown. OBESITY, THE METABOLIC SYNDROME, & TYPE 2 DIABETES Obesity is increasing in incidence, and relates to the regulation of food intake and energy balance and overall nutrition. It de-serves additional consideration in this chapter because of its special relation to disordered carbohydrate metabolism and di-abetes. As body weight increases, insulin resistance increases, that is, there is a decreased ability of insulin to move glucose into fat and muscle and to shut off glucose release from the liver. Weight reduction decreases insulin resistance. Associated with obesity there is hyperinsulinemia, dyslipidemia (characterized by high circulating triglycerides and low high-density lipopro-tein [HDL]), and accelerated development of atherosclerosis. This combination of findings is commonly called the metabolic syndrome, or syndrome X. Some of the patients with the syn-drome are prediabetic, whereas others have type 2 diabetes. It has not been proved but it is logical to assume that the hyperin-sulinemia is a compensatory response to the increased insulin resistance and that frank diabetes develops in individuals with reduced B cell reserves. These observations and other data strongly suggest that fat produces a chemical signal or signals that act on muscles and the liver to increase insulin resistance. Evidence for this includes the recent observation that when glucose transport-ers are selectively knocked out in adipose tissue, there is an associated decrease in glucose transport in muscle in vivo, but when the muscles of those animals are tested in vitro their transport is normal. One possible signal is the circulating free fatty acid level, which is elevated in many insulin-resistant states. Other possi-bilities are peptides and proteins secreted by fat cells. It is now clear that white fat depots are not inert lumps but are actually endocrine tissues that secrete not only leptin but also other hormones that affect fat metabolism. The most intensively studied of these adipokines are listed in Table 21–9. Some of the adipokines decrease, rather than increase, insulin resis-tance. Leptin and adiponectin, for example, decrease insulin resistance, whereas resistin increases insulin resistance. Fur-ther complicating the situation, marked insulin resistance is present in the rare metabolic disease congenital lipodystro-phy, in which fat depots fail to develop. This resistance is reduced by leptin and adiponectin. Finally, a variety of knock-outs of intracellular second messengers have been reported to CHAPTER 21 Endocrine Functions of the Pancreas & Regulation of Carbohydrate Metabolism 335 increase insulin resistance. It is unclear how, or indeed if, these findings fit together to provide an explanation of the relation of obesity to insulin tolerance, but the topic is obviously an important one and it is under intensive investigation. CHAPTER SUMMARY ■Four polypeptides with hormonal activity are secreted by the pancreas: insulin, glucagon, somatostatin, and pancreatic polypeptide. ■Insulin increases the entry of glucose into cells. In skeletal mus-cle cell it increases the number of GLUT 4 transporters in the cell membranes. In liver it induces glucokinase, which increases the phosphorylation of glucose, facilitating the entry of glucose into the cell. ■Insulin causes K+ to enter cells, with a resultant lowering of the extracellular K+ concentration. Insulin increases the activity of Na+–K+ ATPase in cell membranes, so that more K+ is pumped into cells. Hypokalemia often develops when patients with dia-betic acidosis are treated with insulin. ■Insulin receptors are found on many different cells in the body and have two subunits, α and β. Binding of insulin to its recep-tor triggers a signaling pathway that involves autophosphoryla-tion of the β subunits on tyrosine residues. This triggers phosphorylation of some cytoplasmic proteins and dephos-phorylation of others, mostly on serine and threonine residues. ■The constellation of abnormalities caused by insulin deficiency is called diabetes mellitus. Type 1 diabetes is due to insulin defi-ciency caused by autoimmune destruction of the B cells in the pancreatic islets; Type 2 diabetes is characterized by the dysreg-ulation of insulin release from the B cells, along with insulin re-sistance in peripheral tissues such as skeletal muscle, brain, and liver. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Which of the following are incorrectly paired? A) B cells : insulin B) D cells : somatostatin C) A cells : glucagon D) pancreatic exocrine cells : chymotrypsinogen E) F cells : gastrin 2. Which of the following are incorrectly paired? A) epinephrine : increased glycogenolysis in skeletal muscle B) insulin : increased protein synthesis C) glucagon : increased gluconeogenesis D) progesterone : increased plasma glucose level E) growth hormone : increased plasma glucose level 3. Which of the following would be least likely to be seen 14 days after a rat is injected with a drug that kills all of its pancreatic B cells? A) a rise in the plasma H+ concentration B) a rise in the plasma glucagon concentration C) a fall in the plasma HCO3 + concentration D) a fall in the plasma amino acid concentration E) a rise in plasma osmolality 4. When the plasma glucose concentration falls to low levels, a number of different hormones help combat the hypoglycemia. After intravenous administration of a large dose of insulin, the return of a low blood sugar level to normal is delayed in A) adrenal medullary insufficiency. B) glucagon deficiency. C) combined adrenal medullary insufficiency and glucagon deficiency. D) thyrotoxicosis. E) acromegaly. 5. Insulin increases the entry of glucose into A) all tissues. B) renal tubular cells. C) the mucosa of the small intestine. D) most neurons in the cerebral cortex. E) skeletal muscle. 6. Glucagon increases glycogenolysis in liver cells but ACTH does not because A) cortisol increases the plasma glucose level. B) liver cells have an adenylyl cyclase different from that in adrenocortical cells. C) ACTH cannot enter the nucleus of liver cells. D) the membranes of liver cells contain receptors different from those in adrenocortical cells. E) liver cells contain a protein that inhibits the action of ACTH. 7. A meal rich in proteins containing the amino acids that stimu-late insulin secretion but low in carbohydrates does not cause hypoglycemia because A) the meal causes a compensatory increase in T4 secretion. B) cortisol in the circulation prevents glucose from entering muscle. C) glucagon secretion is also stimulated by the meal. D) the amino acids in the meal are promptly converted to glucose. E) insulin does not bind to insulin receptors if the plasma con-centration of amino acids is elevated. TABLE 21–9 Insulin-glucagon molar ratios (I/G) in blood in various conditions. Condition Hepatic Glucose Storage (S) or Production (P)a I/G Glucose availability Large carbohydrate meal 4+ (S) 70 Intravenous glucose 2+ (S) 25 Small meal 1+ (S) 7 Glucose need Overnight fast 1+ (P) 2.3 Low-carbohydrate diet 2+ (P) 1.8 Starvation 4+ (P) 0.4 a1+ to 4+ indicate relative magnitude. Courtesy of RH Unger. 336 SECTION IV Endocrine & Reproductive Physiology CHAPTER RESOURCES Bannerjee RK, et al: Regulation of fasted blood glucose by resistin. Science 2004;303:1195. Gehlert DR: Multiple receptors for the pancreatic polypeptide (PP-fold) family: Physiological implications. Proc Soc Exper Biol Med 1998;218:7. Harmel AP, Mothur R: Davidson’s Diabetes Mellitus, 5th ed. Elsvier, 2004. Kjos SL, Buchanan TA: Gestational diabetes mellitus. N Engl J Med 1999;341:1749. Kulkarni RN, Kahn CR: HNFs-linking the liver and pancreatic islets in diabetes. Science 2004;303:1311. Larsen PR, et al (editors): Williams Textbook of Endocrinology, 9th ed. Saunders, 2003. Lechner D, Habner JF: Stem cells for the treatment of diabetes mellitus. Endocrinology Rounds 2003;2:issue 2. LeRoith D: Insulin-like growth factors. N Engl J Med 1997;336:633. Meigs JB, Avruch J: The metabolic syndrome. Endocrinology Rounds 2003;2:issue 5. Sealey RJ (basic research), Rolls BJ (clinical research), Hensrud DD (clinical practice): Three perspectives on obesity. Endocrine News 2004;29:7. 337 C H A P T E R 22 The Adrenal Medulla & Adrenal Cortex O B J E C T I V E S After reading this chapter, you should be able to: ■Name the three catecholamines secreted by the adrenal medulla and summarize their biosynthesis, metabolism, and function. ■List the stimuli that increase adrenal medullary secretion. ■Differentiate between C18, C19, and C21 steroids and give examples of each. ■Outline the steps involved in steroid biosynthesis in the adrenal cortex. ■Name the plasma proteins that bind adrenocortical steroids and discuss their physiologic role. ■Name the major site of adrenocortical hormone metabolism and the principal metabolites produced from glucocorticoids, adrenal androgens, and aldosterone. ■Describe the mechanisms by which glucocorticoids and aldosterone produce changes in cellular function. ■List and briefly describe the physiologic and pharmacologic effects of glucocorticoids. ■Contrast the physiologic and pathologic effects of adrenal androgens. ■Describe the mechanisms that regulate secretion of glucocorticoids and adrenal sex hormones. ■List the actions of aldosterone and describe the mechanisms that regulate aldos-terone secretion. ■Describe the main features of the diseases caused by excess or deficiency of each of the hormones of the adrenal gland. INTRODUCTION There are two endocrine organs in the adrenal gland, one sur-rounding the other. The main secretions of the inner adrenal medulla (Figure 22–1) are the catecholamines epinephrine, norepinephrine, and dopamine; the outer adrenal cortex secretes steroid hormones. The adrenal medulla is in effect a sympathetic ganglion in which the postganglionic neurons have lost their axons and become secretory cells. The cells secrete when stimulated by the preganglionic nerve fibers that reach the gland via the splanchnic nerves. Adrenal medullary hormones work mostly to prepare the body for emergencies, the “fight-or-flight” responses. The adrenal cortex secretes glucocorticoids, steroids with widespread effects on the metabolism of carbohydrate and protein; a mineralocorticoid essential to the maintenance of Na+ balance and extracellular fluid (ECF) volume; and sex hormones that exert effects on reproductive function. Of these, the mineralocorticoids and the glucocorticoids are nec-essary for survival. Adrenocortical secretion is controlled pri-marily by adrenocorticotropic hormone (ACTH) from the anterior pituitary, but mineralocorticoid secretion is also sub-ject to independent control by circulating factors, of which the most important is angiotensin II, a peptide formed in the bloodstream by the action of renin. 338 SECTION IV Endocrine & Reproductive Physiology ADRENAL MORPHOLOGY The adrenal medulla, which constitutes 28% of the mass of the adrenal gland, is made up of interlacing cords of densely inner-vated granule-containing cells that abut on venous sinuses. Two cell types can be distinguished morphologically: an epineph-rine-secreting type that has larger, less dense granules; and a norepinephrine-secreting type in which smaller, very dense granules fail to fill the vesicles in which they are contained. In humans, 90% of the cells are the epinephrine-secreting type and 10% are the norepinephrine-secreting type. The type of cell that secretes dopamine is unknown. Paraganglia, small groups of cells resembling those in the adrenal medulla, are found near the thoracic and abdominal sympathetic ganglia (Figure 22–1). In adult mammals, the adrenal cortex is divided into three zones (Figure 22–2). The outer zona glomerulosa is made up of whorls of cells that are continuous with the columns of cells that form the zona fasciculata. These columns are separated by venous sinuses. The inner portion of the zona fasciculata merges into the zona reticularis, where the cell columns become inter-laced in a network. The zona glomerulosa makes up 15% of the mass of the adrenal gland; the zona fasciculata, 50%; and the zona reticularis, 7%. The adrenocortical cells contain abundant lipid, especially in the outer portion of the zona fasciculata. All three cortical zones secrete corticosterone, but the active enzy-matic mechanism for aldosterone biosynthesis is limited to the zona glomerulosa, whereas the enzymatic mechanisms for form-ing cortisol and sex hormones are found in the two inner zones. Furthermore, subspecialization occurs within the inner two zones, the zona fasciculata, secreting mostly glucocorticoids and the zona reticularis, secreting mainly sex hormones. Arterial blood reaches the adrenal from many small branches of the phrenic and renal arteries and the aorta. From a plexus in the capsule, blood flows through the cortex to the sinusoids of the medulla. The medulla is also supplied by a few arterioles that pass directly to it from the capsule. In most species, including humans, blood from the medulla flows into a central adrenal vein. The blood flow through the adrenal is large, as it is in most endocrine glands. During fetal life, the human adrenal is large and under pitu-itary control, but the three zones of the permanent cortex repre-sent only 20% of the gland. The remaining 80% is the large fetal adrenal cortex, which undergoes rapid degeneration at the time of birth. A major function of this fetal adrenal is synthesis and secretion of sulfate conjugates of androgens that are converted in the placenta to estrogens (see Chapter 25). No structure is com-parable to the human fetal adrenal in laboratory animals. An important function of the zona glomerulosa, in addition to aldosterone synthesis, is the formation of new cortical cells. The adrenal medulla does not regenerate, but when the inner two zones of the cortex are removed, a new zona fasciculata and zona reticularis regenerate from glomerular cells attached to the capsule. Small capsular remnants regrow large pieces of adreno-cortical tissue. Immediately after hypophysectomy, the zona fasciculata and zona reticularis begin to atrophy, whereas the zona glomerulosa is unchanged because of the action of angiotensin II on this zone. The ability to secrete aldosterone and conserve Na+ is normal for some time after hypophysectomy, but in long-standing hypopituitarism, aldos-terone deficiency may develop, apparently because of the absence of a pituitary factor that maintains the responsiveness of the zona glomerulosa. Injections of ACTH and stimuli that cause endogenous ACTH secretion produce hypertrophy of the zona fasciculata and zona reticularis but actually decrease, rather than increase, the size of the zona glomerulosa. The cells of the adrenal cortex contain large amounts of smooth endoplasmic reticulum, which is involved in the ster-oid-forming process. Other steps in steroid biosynthesis occur in the mitochondria. The structure of steroid-secreting cells is very similar throughout the body. The typical features of such cells are shown in Figure 22–3. ADRENAL MEDULLA: STRUCTURE & FUNCTION OF MEDULLARY HORMONES CATECHOLAMINES Norepinephrine, epinephrine, and dopamine are secreted by the adrenal medulla. Cats and some other species secrete mainly nor-epinephrine, but in dogs and humans, most of the catecholamine output in the adrenal vein is epinephrine. Norepinephrine also enters the circulation from noradrenergic nerve endings. FIGURE 22–1 Human adrenal glands. Adrenocortical tissue is yellow; adrenal medullary tissue is orange. Note the location of the adrenals at the superior pole of each kidney. Also shown are extra-adrenal sites (grey) at which cortical and medullary tissue is sometimes found. (Reproduced with permission from Williams RH: Textbook of Endocrinology, 4th ed. Williams RH [editor]: Saunders, 1968.) CHAPTER 22 The Adrenal Medulla & Adrenal Cortex 339 The structures of norepinephrine, epinephrine, and dopamine and the pathways for their biosynthesis and metabolism are dis-cussed in Chapter 7. Norepinephrine is formed by hydroxylation and decarboxylation of tyrosine, and epinephrine by methylation of norepinephrine. Phenylethanolamine-N-methyltransferase (PNMT), the enzyme that catalyzes the formation of epinephrine from norepinephrine, is found in appreciable quantities only in the brain and the adrenal medulla. Adrenal medullary PNMT is induced by glucocorticoids. Although relatively large amounts are required, the glucocorticoid concentration is high in the blood draining from the cortex to the medulla. After hypophys-ectomy, the glucocorticoid concentration of this blood falls and epinephrine synthesis is decreased. In addition, glucocorticoids are apparently necessary for the normal development of the adre-nal medulla; in 21β-hydroxylase deficiency, glucocorticoid secre-tion is reduced during fetal life and the adrenal medulla is dysplastic. In untreated 21β-hydroxylase deficiency, circulating catecholamines are low after birth. In plasma, about 95% of the dopamine and 70% of the norepi-nephrine and epinephrine are conjugated to sulfate. Sulfate con-jugates are inactive and their function is unsettled. In recumbent humans, the normal plasma level of free norepinephrine is about 300 pg/mL (1.8 nmol/L). On standing, the level increases 50– 100% (Figure 22–4). The plasma norepinephrine level is gener-ally unchanged after adrenalectomy, but the free epinephrine level, which is normally about 30 pg/mL (0.16 nmol/L), falls to essentially zero. The epinephrine found in tissues other than the adrenal medulla and the brain is for the most part absorbed from the bloodstream rather than synthesized in situ. Interestingly, low levels of epinephrine reappear in the blood some time after bilat-eral adrenalectomy, and these levels are regulated like those secreted by the adrenal medulla. They may come from cells such as the intrinsic cardiac adrenergic (ICA) cells (see Chapter 17), but their exact source is unknown. The plasma free dopamine level is about 35 pg/mL (0.23 nmol/L), and appreciable quantities of dopamine are present in the urine. Half the plasma dopamine comes from the adre-nal medulla, whereas the remaining half presumably comes from the sympathetic ganglia or other components of the autonomic nervous system. The catecholamines have a half-life of about 2 min in the circulation. For the most part, they are methoxylated and then oxidized to 3-methoxy-4-hydroxymandelic acid (vanillyl-mandelic acid [VMA]; see Chapter 7). About 50% of the secreted catecholamines appear in the urine as free or conju-gated metanephrine and normetanephrine, and 35% as VMA. Only small amounts of free norepinephrine and epinephrine are excreted. In normal humans, about 30 μg of norepinephrine, FIGURE 22–2 Section through an adrenal gland showing both the medulla and the zones of the cortex, as well as the hormones they secrete. (Reproduced with permission from Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology: The Mechanisms of Body Function, 11th ed. McGraw-Hill, 2008.) Zona glomerulosa Zona fasciculata Cortex Medulla Zona reticularis Aldosterone Cortisol and androgens Epinephrine and norepinephrine Cortex Medulla 340 SECTION IV Endocrine & Reproductive Physiology 6 μg of epinephrine, and 700 μg of VMA are excreted per day. OTHER SUBSTANCES SECRETED BY THE ADRENAL MEDULLA In the medulla, norepinephrine and epinephrine are stored in granules with ATP. The granules also contain chromogranin A (see Chapter 7). Secretion is initiated by acetylcholine re-leased from the preganglionic neurons that innervate the secretory cells. Acetylcholine activates cation channels allow-ing Ca2+ to enter the cells from the extracellular fluid (ECF) and trigger the exocytosis of the granules. In this fashion, cate-cholamines, (adenosine triphosphate) ATP, and proteins from the granules are all released into the blood together. Epinephrine-containing cells of the medulla also contain and secrete opioid peptides (see Chapter 7). The precursor molecule is preproenkephalin. Most of the circulating meten-kephalin comes from the adrenal medulla. The circulating opioid peptides do not cross the blood–brain barrier. Adrenomedullin, a vasodepressor polypeptide found in the adrenal medulla, is discussed in Chapter 33. EFFECTS OF EPINEPHRINE & NOREPINEPHRINE In addition to mimicking the effects of noradrenergic nervous discharge, norepinephrine and epinephrine exert metabolic effects that include glycogenolysis in liver and skeletal muscle, mobilization of free fatty acids (FFA), increased plasma lac-tate, and stimulation of the metabolic rate. The effects of nor-epinephrine and epinephrine are brought about by actions on two classes of receptors: α- and β-adrenergic receptors. Alpha receptors are subdivided into two groups, α1 and α2 receptors, and β receptors into β1, β2, and β3 receptors, as outlined in Chapter 4. There are three subtypes of α1 receptors and three subtypes of α2 receptors (see Table 7–2). Norepinephrine and epinephrine both increase the force and rate of contraction of the isolated heart. These responses are mediated by β1 receptors. The catecholamines also increase myocardial excitability, causing extrasystoles and, occasionally, more serious cardiac arrhythmias. Norepinephrine produces vasoconstriction in most if not all organs via α1 receptors, but epinephrine dilates the blood vessels in skeletal muscle and the liver via β2 receptors. This usually overbalances the vasocon-striction produced by epinephrine elsewhere, and the total peripheral resistance drops. When norepinephrine is infused slowly in normal animals or humans, the systolic and diastolic blood pressures rise. The hypertension stimulates the carotid and aortic baroreceptors, producing reflex bradycardia that overrides the direct cardioacceleratory effect of norepineph-rine. Consequently, cardiac output per minute falls. Epineph-rine causes a widening of the pulse pressure, but because baroreceptor stimulation is insufficient to obscure the direct effect of the hormone on the heart, cardiac rate and output increase. These changes are summarized in Figure 22–5. Catecholamines increase alertness (see Chapter 15). Epi-nephrine and norepinephrine are equally potent in this FIGURE 22–3 Schematic overview of the structures of steroid-secreting cells and the intracellular pathway of steroid synthesis. PKA: protein kinase A; LDL: low-density lipoprotein. (Re-produced with permission from Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology: The Mechanisms of Body Function, 11th ed. McGraw-Hill, 2008.) 7 6 5 4 3 1 Receptor 2 Smooth endoplasmic reticulum Mitochondrion Shuttling of intermediates P450 enzymes located on inner membrane Free cholesterol Lipid droplet (from LDL) Phosphoproteins (cholesterol esterase) Proteins PKA active cAMP PKA inactive Adenylyl cyclase G-protein Nucleus ATP H Diffu sion of ste roid horm one into bloo d CHAPTER 22 The Adrenal Medulla & Adrenal Cortex 341 regard, although in humans epinephrine usually evokes more anxiety and fear. The catecholamines have several different actions that affect blood glucose. Epinephrine and norepinephrine both cause glycogenolysis. They produce this effect via β-adrener-gic receptors that increase cyclic adenosine monophosphate (cAMP), with activation of phosphorylase, and via α-adrener-gic receptors that increase intracellular Ca2+ (see Chapter 7). In addition, the catecholamines increase the secretion of insu-lin and glucagon via β-adrenergic mechanisms and inhibit the secretion of these hormones via α-adrenergic mechanisms. Norepinephrine and epinephrine also produce a prompt rise in the metabolic rate that is independent of the liver and a smaller, delayed rise that is abolished by hepatectomy and coincides with the rise in blood lactate concentration. The ini-tial rise in metabolic rate may be due to cutaneous vasocon-striction, which decreases heat loss and leads to a rise in body temperature, or to increased muscular activity, or both. The second rise is probably due to oxidation of lactate in the liver. Mice unable to make norepinephrine or epinephrine because their dopamine β-hydroxylase gene is knocked out are intol-erant to cold, but surprisingly, their basal metabolic rate is ele-vated. The cause of this elevation is unknown. When injected, epinephrine and norepinephrine cause an initial rise in plasma K+ because of release of K+ from the liver and then a prolonged fall in plasma K+ because of an increased entry of K+ into skeletal muscle that is mediated by β2-adrenergic receptors. Some evidence suggests that activa-tion of α receptors opposes this effect. The increases in plasma norepinephrine and epinephrine that are needed to produce the various effects listed above have been determined by infusion of catecholamines in resting humans. In general, the threshold for the cardiovascular and the metabolic effects of norepinephrine is about 1500 pg/mL, that is, about five times the resting value (Figure 22–4). Epi-nephrine, on the other hand, produces tachycardia when the plasma level is about 50 pg/mL, that is, about twice the resting value. The threshold for increased systolic blood pressure and lipolysis is about 75 pg/mL; the threshold for hyperglycemia, increased plasma lactate, and decreased diastolic blood pres-sure is about 150 pg/mL; and the threshold for the α-mediated decrease in insulin secretion is about 400 pg/mL. Plasma epi-nephrine often exceeds these thresholds. On the other hand, FIGURE 22–4 Norepinephrine and epinephrine levels in human venous blood in various physiologic and pathologic states. Note that the horizontal scales are different. The numbers to the left in parentheses are the numbers of subjects tested. In each case, the vertical dashed line identifies the threshold plasma concentration at which detectable physiologic changes are observed. (Modified and reproduced with permission from Cryer PE: Physiology and pathophysiology of the human sympathoadrenal neuroendocrine system. N Engl J Med 1980;303:436.) Pheochromocytoma (16) Cigarette smoking (10) To < 40 mg/dL (6) 95 → 60 mg/dL (10) Mild (8) Moderate (8) Heavy (8) During (11) After (11) Ketoacidosis (10) Myocardial infarction (11) Quiet standing (40) Resting supine (60) Hypoglycemia Exercise Surgery 0 500 1000 1500 2000 2500 0 500 1000 1500 2000 2500 (5310) Plasma norepinephrine (pg/mL) 0 0 100 Plasma epinephrine (pg/mL) 200 300 400 500 1000 5000 100 200 300 400 500 1000 5000 FIGURE 22–5 Circulatory changes produced in humans by the slow intravenous infusion of epinephrine and norepinephrine. 150 100 50 30 20 10 15 20 35 40 Epi Nor 4 6 8 50 100 Time (min) Epi = Epinephrine Nor = Norepinephrine Heart rate Cardiac output (L/min) Total peripheral resistance Arterial BP (mm Hg) 342 SECTION IV Endocrine & Reproductive Physiology plasma norepinephrine rarely exceeds the threshold for its car-diovascular and metabolic effects, and most of its effects are due to its local release from postganglionic sympathetic neurons. Most adrenal medullary tumors (pheochromocytomas) secrete norepinephrine, or epinephrine, or both, and produce sustained hypertension. However, 15% of epinephrine-secreting tumors secrete this catecholamine episodically, producing intermittent bouts of palpitations, headache, glycosuria, and extreme systolic hypertension. These same symptoms are produced by intrave-nous injection of a large dose of epinephrine. EFFECTS OF DOPAMINE The physiologic function of the dopamine in the circulation is unknown. However, injected dopamine produces renal vasodila-tion, probably by acting on a specific dopaminergic receptor. It also produces vasodilation in the mesentery. Elsewhere, it pro-duces vasoconstriction, probably by releasing norepinephrine, and it has a positively inotropic effect on the heart by an action on β1-adrenergic receptors. The net effect of moderate doses of do-pamine is an increase in systolic pressure and no change in dia-stolic pressure. Because of these actions, dopamine is useful in the treatment of traumatic and cardiogenic shock (see Chapter 33). Dopamine is made in the renal cortex. It causes natriuresis and may exert this effect by inhibiting renal Na+–K+ ATPase. REGULATION OF ADRENAL MEDULLARY SECRETION NEURAL CONTROL Certain drugs act directly on the adrenal medulla, but physio-logic stimuli affect medullary secretion through the nervous system. Catecholamine secretion is low in basal states, but the secretion of epinephrine and, to a lesser extent, that of norepi-nephrine is reduced even further during sleep. Increased adrenal medullary secretion is part of the diffuse sympathetic discharge provoked in emergency situations, which Cannon called the “emergency function of the sympathoadrenal system. ” The ways in which this discharge prepares the individual for flight or fight are described in Chapter 17, and the increases in plasma catecholamines under various conditions are shown in Figure 22–4. The metabolic effects of circulating catecholamines are probably important, especially in certain situations. The calo-rigenic action of catecholamines in animals exposed to cold is an example, and so is the glycogenolytic effect (see Chapter 21) in combating hypoglycemia. SELECTIVE SECRETION When adrenal medullary secretion is increased, the ratio of nor-epinephrine to epinephrine in the adrenal effluent is generally unchanged. However, norepinephrine secretion tends to be se-lectively increased by emotional stresses with which the individ-ual is familiar, whereas epinephrine secretion rises selectively in situations in which the individual does not know what to expect. ADRENAL CORTEX: STRUCTURE & BIOSYNTHESIS OF ADRENOCORTICAL HORMONES CLASSIFICATION & STRUCTURE The hormones of the adrenal cortex are derivatives of cholesterol. Like cholesterol, bile acids, vitamin D, and ovarian and testicular steroids, they contain the cyclopentanoperhydrophenanthrene nucleus (Figure 22–6). Gonadal and adrenocortical steroids are of three types: C21 steroids, which have a two-carbon side chain at position 17; C19 steroids, which have a keto or hydroxyl group at position 17; and C18 steroids, which, in addition to a 17-keto or hydroxyl group, have no angular methyl group attached to posi-tion 10. The adrenal cortex secretes primarily C21 and C19 ster-oids. Most of the C19 steroids have a keto group at position 17 and are therefore called 17-ketosteroids. The C21 steroids that have a FIGURE 22–6 Basic structure of adrenocortical and gonadal steroids. The letters in the formula for cholesterol identify the four ba-sic rings, and the numbers identify the positions in the molecule. As shown here, the angular methyl groups (positions 18 and 19) are usu-ally indicated simply by straight lines. HO 3 4 5 6 1 A 2 19 10 14 15 18 13 17 D 16 12 9 8 7 B C 11 20 22 23 24 26 25 27 21 Cyclopentanoperhydrophenanthrene nucleus CH3 C O Progesterone Corticoids Androgens Estrogens Cholesterol (27 carbons) Pregnane derivatives (21 carbons) Androstane derivatives (19 carbons) Estrane derivatives (18 carbons) CHAPTER 22 The Adrenal Medulla & Adrenal Cortex 343 hydroxyl group at the 17 position in addition to the side chain are often called 17-hydroxycorticoids or 17-hydroxycorticosteroids. The C19 steroids have androgenic activity. The C21 steroids are classified, using Selye’s terminology, as mineralocorticoids or glucocorticoids. All secreted C21 steroids have both miner-alocorticoid and glucocorticoid activity; mineralocorticoids are those in which effects on Na+ and K+ excretion predomi-nate and glucocorticoids are those in which effects on glu-cose and protein metabolism predominate. The details of steroid nomenclature and isomerism can be found elsewhere. However, it is pertinent to mention that the Greek letter Δ indicates a double bond and that the groups that lie above the plane of each of the steroid rings are indicated by the Greek letter α and a solid line (—OH), whereas those that lie below the plane are indicated by α and a dashed line (- - -OH). Thus, the C21 steroids secreted by the adrenal have a Δ4-3-keto configuration in the A ring. In most naturally occurring adrenal steroids, 17-hydroxy groups are in the α configuration, whereas 3-, 11-, and 21-hydroxy groups are in the β configuration. The 18-aldehyde configuration on naturally occurring aldosterone is the D form. L-Aldosterone is physiologically inactive. SECRETED STEROIDS Innumerable steroids have been isolated from adrenal tissue, but the only steroids normally secreted in physiologically sig-nificant amounts are the mineralocorticoid aldosterone, the glucocorticoids cortisol and corticosterone, and the andro-gens dehydroepiandrosterone (DHEA) and androstenedi-one. The structures of these steroids are shown in Figures 22–7 and 22–8. Deoxycorticosterone is a mineralocorticoid that is normally secreted in about the same amount as aldosterone (Table 22–1) but has only 3% of the mineralocorticoid activity of aldosterone. Its effect on mineral metabolism is usually neg-ligible, but in diseases in which its secretion is increased, its ef-fect can be appreciable. Most of the estrogens that are not formed in the ovaries are produced in the circulation from ad-renal androstenedione. Almost all the dehydroepiandrosterone is secreted conjugated with sulfate, although most if not all of the other steroids are secreted in the free, unconjugated form. The secretion rate for individual steroids can be determined by injecting a very small dose of isotopically labeled steroid and determining the degree to which the radioactive steroid excreted in the urine is diluted by unlabeled secreted hor-mone. This technique is used to measure the output of many different hormones (see Clinical Box 22–1). SPECIES DIFFERENCES In all species from amphibia to humans, the major C21 steroid hormones secreted by adrenocortical tissue appear to be aldos-terone, cortisol, and corticosterone, although the ratio of cortisol to corticosterone varies. Birds, mice, and rats secrete corticoster-one almost exclusively; dogs secrete approximately equal amounts of the two glucocorticoids; and cats, sheep, monkeys, and humans secrete predominantly cortisol. In humans, the ra-tio of secreted cortisol to corticosterone is approximately 7:1. FIGURE 22–7 Outline of hormone biosynthesis in the zona fasciculata and zona reticularis of the adrenal cortex. The major secreto-ry products are underlined. The enzymes for the reactions are shown on the left and at the top of the chart. When a particular enzyme is deficient, hormone production is blocked at the points indicated by the shaded bars. C O O CH2OH C O HO O CH2OH Corticosterone 11-Deoxy-corticosterone C O O CH3 Progesterone C O HO CH3 Pregnenolone ACTH C O O CH2OH C O HO O CH2OH C O O CH3 C O OH HO CH3 17-Hydroxy-pregnenolone 17-Hydroxy-progesterone 11-Deoxycortisol Cortisol Cholesterol Cholesterol desmolase 3β-Hydroxysteroid dehydrogenase 17α-Hydroxylase 17,20-Lyase 21β-Hydroxylase 11β-Hydroxylase OH O O HO O Dehydroepiandros-terone Androstenedione Testosterone Estradiol OH OH Sulfo-kinase DHEA sulfate 344 SECTION IV Endocrine & Reproductive Physiology STEROID BIOSYNTHESIS The major paths by which the naturally occurring adrenocortical hormones are synthesized in the body are summarized in Fig-ures 22–7 and 22–8. The precursor of all steroids is cholesterol. Some of the cholesterol is synthesized from acetate, but most of it is taken up from LDL in the circulation. LDL receptors are es-pecially abundant in adrenocortical cells. The cholesterol is es-terified and stored in lipid droplets. Cholesterol ester hydrolase catalyzes the formation of free cholesterol in the lipid droplets (Figure 22–9). The cholesterol is transported to mitochondria by a sterol carrier protein. In the mitochondria, it is converted to pregnenolone in a reaction catalyzed by an enzyme known as cholesterol desmolase or side-chain cleavage enzyme. This en-zyme, like most of the enzymes involved in steroid biosynthesis, is a member of the cytochrome P450 superfamily and is also known as P450scc or CYP11A1. For convenience, the various FIGURE 22–8 Hormone synthesis in the zona glomerulosa. The zona glomerulosa lacks 17α-hydroxylase activity, and only the zona glomerulosa can convert corticosterone to aldosterone because it is the only zone that normally contains aldosterone synthase. ANG II, angiotensin II. TABLE 22–1 Principal adrenocortical hormones in adult humans.a Name Synonyms Average Plasma Concentration (Free and Bound)a (μg/dL) Average Amount Secreted (mg/24 h) Cortisol Compound F, hydrocortisone 13.9 10 Corticosterone Compound B 0.4 3 Aldosterone 0.0006 0.15 Deoxycorticos-terone DOC 0.0006 0.20 Dehydroepi-androsterone sulfate DHEAS 175.0 20 aAll plasma concentration values except DHEAS are fasting morning values after overnight recumbency. C O HO HO O CH2OH CH2 C O O HC HO O CH2OH Cortisol and sex steroids Cholesterol Pregnenolone Progesterone Deoxycorticosterone Corticosterone 18-Hydroxycorticosterone Aldosterone Aldosterone synthase Aldosterone synthase ANG II Aldosterone synthase ACTH ANG II CLINICAL BOX 22–1 Synthetic Steroids As with many other naturally occurring substances, the ac-tivity of adrenocortical steroids can be increased by alter-ing their structure. A number of synthetic steroids are avail-able that have many times the activity of cortisol. The relative glucocorticoid and mineralocorticoid potencies of the natural steroids are compared with those of the syn-thetic steroids 9α-fluorocortisol, prednisolone, and dexa-methasone in Table 22–2. The potency of dexamethasone is due to its high affinity for glucocorticoid receptors and its long half-life. Prednisolone also has a long half-life. TABLE 22–2 Relative potencies of corticosteroids compared with cortisol.a Steroid Glucocorticoid Activity Mineralocorticoid Activity Cortisol 1.0 1.0 Corticosterone 0.3 15 Aldosterone 0.3 3000 Deoxycorticosterone 0.2 100 Costisone 0.7 0.8 Prednisolone 4 0.8 9α-Fluorocortisol 10 125 Dexamethasone 25 –0 aValues are approximations based on liver glycogen deposition or anti-inflamma-tory assays for glucocorticoid activity, and effect on urinary Na+/K+ or maintenance of adrena-lectomized animals for mineralocorticoid activity. The last three steroids listed are synthetic compounds that do not occur naturally. CHAPTER 22 The Adrenal Medulla & Adrenal Cortex 345 names of the enzymes involved in adrenocortical steroid biosyn-thesis are summarized in Table 22–3. Pregnenolone moves to the smooth endoplasmic reticulum, where some of it is dehydrogenated to form progesterone in a reaction catalyzed by 3β-hydroxysteroid dehydrogenase. This enzyme has a molecular weight of 46,000 and is not a cytochrome P450. It also catalyzes the conversion of 17α-hydroxypregnenolone to 17α-hydroxyprogesterone, and dehy-droepiandrosterone to androstenedione (Figure 22–7) in the smooth endoplasmic reticulum. The 17α-hydroxypreg-nenolone and the 17α-hydroxyprogesterone are formed from pregnenolone and progesterone, respectively (Figure 22–7) by the action of 17α-hydroxylase. This is another mitochon-drial P450, and it is also known as P450c17 or CYP17. Located in another part of the same enzyme is 17,20-lyase activity that breaks the 17,20 bond, converting 17α-preg-nenolone and 17α-progesterone to the C19 steroids dehy-droepiandrosterone and androstenedione. Hydroxylation of progesterone to 11-deoxycorticosterone and of 17α-hydroxyprogesterone to 11-deoxycortisol occurs in the smooth endoplasmic reticulum. These reactions are catalyzed by 21β-hydroxylase, a cytochrome P450 that is also known as P450c21 or CYP21A2. 11-deoxycorticosterone and the 11-deoxycortisol move back to the mitochondria, where they are 11-hydroxylated to form corticosterone and cortisol. These reactions occur in the zona fasciculata and zona reticularis and are catalyzed by 11β-hydrox-ylase, a cytochrome P450 also known as P450c11 or CYP11B1. In the zona glomerulosa there is no 11β-hydroxylase but a closely related enzyme called aldosterone synthase is present. This cytochrome P450 is 95% identical to 11β-hydroxylase and is also known as P450c11AS or CYP11B2. The genes that code CYP11B1 and CYP11B2 are both located on chromo-some 8. However, aldosterone synthase is normally found only in the zona glomerulosa. The zona glomerulosa also lacks 17α-hydroxylase. This is why the zona glomerulosa makes aldosterone but fails to make cortisol or sex hormones. Furthermore, subspecialization occurs within the inner two zones. The zona fasciculata has more 3β-hydroxysteroid dehydrogenase activity than the zona reticularis, and the zona reticularis has more of the cofactors required for the expres-sion of the 17,20-lyase activity of 17α-hydroxylase. Therefore, the zona fasciculata makes more cortisol and corticosterone, and the zona reticularis makes more androgens. Most of the dehydroepiandrosterone that is formed is converted to dehy-droepiandrosterone sulfate by adrenal sulfokinase, and this enzyme is localized in the zona reticularis as well. ACTION OF ACTH ACTH binds to high-affinity receptors on the plasma membrane of adrenocortical cells. This activates adenylyl cyclase via Gs. The resulting reactions (Figure 22–9) lead to a prompt increase in the formation of pregnenolone and its derivatives, with secretion of the latter. Over longer periods, ACTH also increases the syn-thesis of the P450s involved in the synthesis of glucocorticoids. ACTIONS OF ANGIOTENSIN II Angiotensin II binds to AT1 receptors (see Chapter 39) in the zona glomerulosa which act via a G protein to activate phos-pholipase C. The resulting increase in protein kinase C fosters the conversion of cholesterol to pregnenolone (Figure 22–8) and facilitates the action of aldosterone synthase, resulting in increased secretion of aldosterone. FIGURE 22–9 Mechanism of action of ACTH on cortisol-secreting cells in the inner two zones of the adrenal cortex. When ACTH binds to its receptor (R), adenylyl cyclase (AC) is activated via Gs. The resulting increase in cAMP activates protein kinase A, and the kinase phosphorylates cholesteryl ester hydrolase (CEH), increasing its activity. Consequently, more free cholesterol is formed and converted to preg-nenolone. Note that in the subsequent steps in steroid biosynthesis, products are shuttled between the mitochondria and the smooth endo-plasmic reticulum (SER). Corticosterone is also synthesized and secreted. TABLE 22–3 Nomenclature for adrenal steroidogenic enzymes and their location in adrenal cells. Trivial Name P450 CYP Location Cholesterol desmolase; side-chain cleavage enzyme P450SCC CYP11A1 Mitochondria 3β-Hydroxysteroid dehydrogenase . . . . . . SER 17α-Hydroxylase, 17,20-lyase P450C17 CYP17 Mitochondria 21β-Hydroxylase P450C21 CYP21A2 SER 11β-Hydroxylase P450C11 CYP11B1 Mitochondria Aldosterone synthase P450C11AS CYP11B2 Mitochondria SER = smooth endoplasmic reticulum. LDL Cholesteryl esters Cholesterol Cortisol Preg 11-Deoxy-cortisol 17-OH preg cAMP Protein kinase A ATP CEH SER Lipid droplet Mitochondrion ACTH R GS AC 346 SECTION IV Endocrine & Reproductive Physiology ENZYME DEFICIENCIES The consequences of inhibiting any of the enzyme systems in-volved in steroid biosynthesis can be predicted from Figures 22–7 and 22–8. Congenital defects in the enzymes lead to defi-cient cortisol secretion and the syndrome of congenital adrenal hyperplasia. The hyperplasia is due to increased ACTH secre-tion. Cholesterol desmolase deficiency is fatal in utero because it prevents the placenta from making the progesterone necessary for pregnancy to continue. A cause of severe congenital adrenal hyperplasia in newborns is a loss of function mutation of the gene for the steroidogenic acute regulatory (StAR) protein. This protein is essential in the adrenals and gonads but not in the pla-centa for the normal movement of cholesterol into the mitochon-dria to reach cholesterol desmolase, which is located on the matrix space side of the internal mitochondrial membrane. In its absence, only small amounts of steroids are formed. The degree of ACTH stimulation is marked, resulting eventually in accumu-lation of large numbers of lipoid droplets in the adrenal. For this reason, the condition is called congenital lipoid adrenal hyper-plasia. Because androgens are not formed, female genitalia develop regardless of genetic sex (see Chapter 25). In 3β hydroxy-steroid dehydrogenase deficiency, another rare condition, DHEA secretion is increased. This steroid is a weak androgen that can cause some masculinization in females with the disease, but it is not adequate to produce full masculinization of the genitalia in genetic males. Consequently, hypospadias is common. In fully developed 17α-hydroxylase deficiency, a third rare condition due to a mutated gene for CYP17, no sex hormones are produced, so female external genitalia are present. However, the pathway lead-ing to corticosterone and aldosterone is intact, and elevated levels of 11-deoxycorticosterone and other mineralocorticoids pro-duce hypertension and hypokalemia. Cortisol is deficient, but this is partially compensated by the glucocorticoid activity of corticosterone. Unlike the defects discussed in the preceding paragraph, 21β-hydroxylase deficiency is common, accounting for 90% or more of the enzyme deficiency cases. The 21β-hydroxylase gene, which is in the human leukocyte antigen (HLA) com-plex of genes on the short arm of chromosome 6 (see Chapter 3) is one of the most polymorphic in the human genome. Mutations occur at many different sites in the gene, and the abnormalities that are produced therefore range from mild to severe. Production of cortisol and aldosterone are generally reduced, so ACTH secretion and consequently production of precurose steroids are increased. These steroids are converted to androgens, producing virilization. The characteristic pat-tern that develops in females in the absence of treatment is the adrenogenital syndrome. Masculization may not be marked until later in life and mild cases can be detected only by labo-ratory tests. In 75% of the cases, aldosterone deficiency causes appreciable loss of Na+ (salt-losing form of adrenal hyperpla-sia). The resulting hypovolemia can be severe. In 11β-hydroxylase deficiency, virilization plus excess secretion of 11-deoxycortisol and 11-deoxycorticosterone take place. Because the former is an active mineralocorticoid, patients with this condition also have salt and water retention and, in two-thirds of the cases, hypertension (hypertensive form of congenital adrenal hyperplasia). Glucocorticoid treatment is indicated in all of the virilizing forms of congenital adrenal hyperplasia because it repairs the glucocorticoid deficit and inhibits ACTH secretion, reducing the abnormal secretion of androgens and other steroids. Expression of the cytochrome P450 enzymes responsible for steroid hormone biosynthesis depends on steroid factor-1 (SF-1), an orphan nuclear receptor. If Ft2-F1, the gene for SF-1, is knocked out, gonads as well as adrenals fail to develop and additional abnormalities are present at the pituitary and hypothalamic level. TRANSPORT, METABOLISM, & EXCRETION OF ADRENOCORTICAL HORMONES GLUCOCORTICOID BINDING Cortisol is bound in the circulation to an α globulin called transcortin or corticosteroid-binding globulin (CBG). A mi-nor degree of binding to albumin also takes place (see Table 25– 5). Corticosterone is similarly bound, but to a lesser degree. The half-life of cortisol in the circulation is therefore longer (about 60–90 min) than that of corticosterone (50 min). Bound ster-oids are physiologically inactive. In addition, relatively little free cortisol and corticosterone are found in the urine because of protein binding. The equilibrium between cortisol and its binding protein and the implications of binding in terms of tissue supplies and ACTH secretion are summarized in Figure 22–10. The bound cortisol functions as a circulating reservoir of hormone that keeps a sup-ply of free cortisol available to the tissues. The relationship is sim-ilar to that of T4 and its binding protein (see Chapter 20). At normal levels of total plasma cortisol (13.5 μg/dL or 375 nmol/L), very little free cortisol is present in the plasma, but the binding sites on CBG become saturated when the total plasma cortisol FIGURE 22–10 The interrelationships of free and bound cortisol. The dashed arrow indicates that cortisol inhibits ACTH secre-tion. The value for free cortisol is an approximation; in most studies, it is calculated by subtracting the protein-bound cortisol from the total plasma cortisol. ACTH Adrenal cortex Anterior pituitary Tissue cortisol Protein-bound cortisol in plasma (13 μg/dL) Free cortisol in plasma (~0.5 μg/dL) CHAPTER 22 The Adrenal Medulla & Adrenal Cortex 347 exceeds 20 μg/dL. At higher plasma levels, binding to albumin increases, but the main increase is in the unbound fraction. CBG is synthesized in the liver and its production is increased by estrogen. CBG levels are elevated during pregnancy and depressed in cirrhosis, nephrosis, and multiple myeloma. When the CBG level rises, more cortisol is bound, and initially the free cortisol level drops. This stimulates ACTH secretion, and more cortisol is secreted until a new equilibrium is reached at which the bound cortisol is elevated but the free cortisol is normal. Changes in the opposite direction occur when the CBG level falls. This explains why pregnant women have high total plasma cortisol levels without symptoms of glucocorticoid excess and, conversely, why some patients with nephrosis have low total plasma cortisol without symptoms of glucocorticoid deficiency. METABOLISM & EXCRETION OF GLUCOCORTICOIDS Cortisol is metabolized in the liver, which is the principal site of glucocorticoid catabolism. Most of the cortisol is reduced to dihydrocortisol and then to tetrahydrocortisol, which is conju-gated to glucuronic acid (Figure 22–11). The glucuronyl trans-ferase system responsible for this conversion also catalyzes the formation of the glucuronides of bilirubin (see Chapter 29) and a number of hormones and drugs. Competitive inhibition takes place between these substrates for the enzyme system. The liver and other tissues contain the enzyme 11β hydroxy-steroid dehydrogenase. There are at least two forms of this enzyme. Type 1 catalyzes the conversion of cortisol to cortisone and the reverse reaction, though it functions primarily as a reductase, forming cortisol from corticosterone. Type 2 cata-lyzes almost exclusively the one-way conversion of cortisol to cortisone. Cortisone is an active glucocorticoid because it is converted to cortisol, and it is well known because of its exten-sive use in medicine. It is not secreted in appreciable quantities by the adrenal glands. Little, if any, of the cortisone formed in the liver enters the circulation, because it is promptly reduced and conjugated to form tetrahydrocortisone glucuronide. The tetrahydroglucuronide derivatives (“conjugates”) of cortisol and corticosterone are freely soluble. They enter the circulation, where they do not become bound to protein. They are rapidly excreted in the urine. About 10% of the secreted cortisol is converted in the liver to the 17-ketosteroid derivatives of cortisol and cortisone. The ketosteroids are conjugated for the most part to sulfate and then excreted in the urine. Other metabolites, including 20-hydroxy derivatives, are formed. There is an enterohepatic circulation of glucocorticoids and about 15% of the secreted cortisol is excreted in the stool. The metabolism of corticos-terone is similar to that of cortisol, except that it does not form a 17-ketosteroid derivative (see Clinical Box 22–2). ALDOSTERONE Aldosterone is bound to protein to only a slight extent, and its half-life is short (about 20 min). The amount secreted is small FIGURE 22–11 Outline of hepatic metabolism of cortisol. O HO C O OH Δ4-Hydrogenase; NADPH Cortisol Cortisone Tetrahydrocortisol glucuronide Tetrahydrocortisol Dihydrocortisol 17-Ketosteroids Tetrahydrocortisone glucuronide 17-Ketosteroids 11β-Hydroxysteroid dehydrogenase Glucuronyl transferase; uridine-diphospho-glucuronic acid 3α-Hydroxysteroid dehydrogenase; NADPH or NADH CH2OH C O OH CH2OH C O OH CH2OH C O OH CH2OH C O OH CH2OH O O O HO H H HO HO H O HC HO HCOH HOCH HCOH HC O COO– 348 SECTION IV Endocrine & Reproductive Physiology (Table 22–1), and the total plasma aldosterone level in humans is normally about 0.006 μg/dL (0.17 nmol/L), compared with a cortisol level (bound and free) of about 13.5 μg/dL (375 nmol/ L). Much of the aldosterone is converted in the liver to the tet-rahydroglucuronide derivative, but some is changed in the liver and in the kidneys to an 18-glucuronide. This glucuronide, which is unlike the breakdown products of other steroids, is converted to free aldosterone by hydrolysis at pH 1.0, and it is therefore often referred to as the “acid-labile conjugate.” Less than 1% of the secreted aldosterone appears in the urine in the free form. Another 5% is in the form of the acid-labile conju-gate, and up to 40% is in the form of the tetrahydroglucuronide. 17-KETOSTEROIDS The major adrenal androgen is the 17-ketosteroid dehydro-epiandrosterone, although androstenedione is also secreted. The 11-hydroxy derivative of androstenedione and the 17-ketosteroids formed from cortisol and cortisone by side chain cleavage in the liver are the only 17-ketosteroids that have an =O or an —OH group in the 11 position (“11-oxy-17-keto-steroids”). Testosterone is also converted to 17-ketosteroids. Because the daily 17-ketosteroid excretion in normal adults is 15 mg in men and 10 mg in women, about two thirds of the urinary ketosteroids in men are secreted by the adrenal or formed from cortisol in the liver and about one third are of testicular origin. Etiocholanolone, one of the metabolites of the adrenal androgens and testosterone, can cause fever when it is uncon-jugated (see Chapter 18). Certain individuals have episodic bouts of fever due to periodic accumulation in the blood of unconjugated etiocholanolone (“etiocholanolone fever”). EFFECTS OF ADRENAL ANDROGENS & ESTROGENS ANDROGENS Androgens are the hormones that exert masculinizing effects and they promote protein anabolism and growth (see Chapter 25). Testosterone from the testes is the most active androgen and the adrenal androgens have less than 20% of its activity. Secretion of the adrenal androgens is controlled acutely by ACTH and not by gonadotropins. However, the concentra-tion of dehydroepiandrosterone sulfate (DHEAS) increases until it peaks at about 225 mg/dL in the early 20s, then falls to very low values in old age (Figure 22–12). These long-term changes are not due to changes in ACTH secretion and appear to be due instead to a rise and then a gradual fall in the lyase activity of 17α-hydroxylase. All but about 0.3% of the circulating DHEA is conjugated to sulfate (DHEAS). The secretion of adrenal androgens is nearly as great in castrated males and females as it is in nor-mal males, so it is clear that these hormones exert very little masculinizing effect when secreted in normal amounts. How-ever, they can produce appreciable masculinization when secreted in excessive amounts. In adult males, excess adrenal androgens merely accentuate existing characteristics, but in prepubertal boys they can cause precocious development of the secondary sex characteristics without testicular growth (precocious pseudopuberty). In females they cause female pseudo-hermaphroditism and the adrenogenital syndrome. Some health practitioners recommend injections of dehy-droepiandrosterone to combat the effects of aging (see Chap-ter 1), but results to date are controversial at best. ESTROGENS The adrenal androgen androstenedione is converted to testos-terone and to estrogens (aromatized) in fat and other periph-eral tissues. This is an important source of estrogens in men and postmenopausal women (see Chapter 25). CLINICAL BOX 22–2 Variations in the Rate of Hepatic Metabolism The rate of hepatic inactivation of glucocorticoids is de-pressed in liver disease and, interestingly, during surgery and other stresses. Thus, in stressed humans, the plasma-free cortisol level rises higher than it does with maximal ACTH stimulation in the absence of stress. FIGURE 22–12 Change in serum dehydroepiandrosterone sulfate (DHEAS) with age. The middle line is the mean, and the dashed lines identify ±1.96 standard deviations. (Reproduced, with permission, from Smith MR, et al: A radioimmunoassay for the estimation of serum dehydroepiandrosterone sulfate in normal and pathological sera. Clin Chim Acta 1975;65:5.) 600 500 400 300 200 100 0 0 10 20 30 40 60 70 80 50 Age (years) DHEAS (μg/dL) Males Females CHAPTER 22 The Adrenal Medulla & Adrenal Cortex 349 PHYSIOLOGIC EFFECTS OF GLUCOCORTICOIDS ADRENAL INSUFFICIENCY In untreated adrenal insufficiency, Na+ loss and shock occurs due to the lack of mineralocorticoid activity, as well as abnor-malities of water, carbohydrate, protein, and fat metabolism due to the lack of glucocorticoids. These metabolic abnormal-ities are eventually fatal despite mineralocorticoid treatment. Small amounts of glucocorticoids correct the metabolic ab-normalities, in part directly and in part by permitting other re-actions to occur. It is important to separate these physiologic actions of glucocorticoids from the quite different effects pro-duced by large amounts of the hormones. MECHANISM OF ACTION The multiple effects of glucocorticoids are triggered by bind-ing to glucocorticoid receptors, and the steroid–receptor complexes act as transcription factors that promote the trans-cription of certain segments of DNA (see Chapter 1). This, in turn, leads via the appropriate mRNAs to synthesis of en-zymes that alter cell function. In addition, it seems likely that glucocorticoids have nongenomic actions. EFFECTS ON INTERMEDIARY METABOLISM The actions of glucocorticoids on the intermediary metabolism of carbohydrate, protein, and fat are discussed in Chapter 21. They include increased protein catabolism and increased he-patic glycogenesis and gluconeogenesis. Glucose 6-phospha-tase activity is increased, and the plasma glucose level rises. Glucocorticoids exert an anti-insulin action in peripheral tis-sues and make diabetes worse. However, the brain and the heart are spared, so the increase in plasma glucose provides ex-tra glucose to these vital organs. In diabetics, glucocorticoids raise plasma lipid levels and increase ketone body formation, but in normal individuals, the increase in insulin secretion pro-voked by the rise in plasma glucose obscures these actions. In adrenal insufficiency, the plasma glucose level is normal as long as an adequate caloric intake is maintained, but fasting causes hypoglycemia that can be fatal. The adrenal cortex is not essential for the ketogenic response to fasting. PERMISSIVE ACTION Small amounts of glucocorticoids must be present for a num-ber of metabolic reactions to occur, although the glucocorti-coids do not produce the reactions by themselves. This effect is called their permissive action. Permissive effects include the requirement for glucocorticoids to be present for glucagon and catecholamines to exert their calorigenic effects (see above and Chapter 21), for catecholamines to exert their lip-olytic effects, and for catecholamines to produce pressor re-sponses and bronchodilation. EFFECTS ON ACTH SECRETION Glucocorticoids inhibit ACTH secretion, and ACTH secre-tion is increased in adrenalectomized animals. The conse-quences of the feedback action of cortisol on ACTH secretion are discussed below in the section on regulation of glucocorti-coid secretion. VASCULAR REACTIVITY In adrenally insufficient animals, vascular smooth muscle be-comes unresponsive to norepinephrine and epinephrine. The capillaries dilate and, terminally, become permeable to colloi-dal dyes. Failure to respond to the norepinephrine liberated at noradrenergic nerve endings probably impairs vascular com-pensation for the hypovolemia of adrenal insufficiency and promotes vascular collapse. Glucocorticoids restore vascular reactivity. EFFECTS ON THE NERVOUS SYSTEM Changes in the nervous system in adrenal insufficiency that are reversed only by glucocorticoids include the appearance of electroencephalographic waves slower than the normal α rhythm and personality changes. The latter, which are mild, include irritability, apprehension, and inability to concentrate. EFFECTS ON WATER METABOLISM Adrenal insufficiency is characterized by an inability to ex-crete a water load, causing the possibility of water intoxica-tion. Only glucocorticoids repair this deficit. In patients with adrenal insufficiency who have not received glucocorticoids, glucose infusion may cause high fever (“glucose fever”) fol-lowed by collapse and death. Presumably, the glucose is me-tabolized, the water dilutes the plasma, and the resultant osmotic gradient between the plasma and the cells causes the cells of the thermoregulatory centers in the hypothalamus to swell to such an extent that their function is disrupted. The cause of defective water excretion in adrenal insuffi-ciency is unsettled. Plasma vasopressin levels are elevated in adrenal insufficiency and reduced by glucocorticoid treat-ment. The glomerular filtration rate is low, and this probably contributes to the reduction in water excretion. The selective effect of glucocorticoids on the abnormal water excretion is consistent with this possibility, because even though the min-eralocorticoids improve filtration by restoring plasma vol-ume, the glucocorticoids raise the glomerular filtration rate to a much greater degree. 350 SECTION IV Endocrine & Reproductive Physiology EFFECTS ON THE BLOOD CELLS & LYMPHATIC ORGANS Glucocorticoids decrease the number of circulating eosino-phils by increasing their sequestration in the spleen and lungs. Glucocorticoids also lower the number of basophils in the cir-culation and increase the number of neutrophils, platelets, and red blood cells (Table 22–4). Glucocorticoids decrease the circulating lymphocyte count and the size of the lymph nodes and thymus by inhibiting lymphocyte mitotic activity. They reduce secretion of cyto-kines by inhibiting the effect of NF-κB on the nucleus. The reduced secretion of the cytokine IL-2 leads to reduced prolif-eration of lymphocytes (see Chapter 3), and these cells undergo apoptosis. RESISTANCE TO STRESS The term stress as used in biology has been defined as any change in the environment that changes or threatens to change an existing optimal steady state. Most, if not all, of these stresses activate processes at the molecular, cellular, or systemic level that tend to restore the previous state, that is, they are homeostatic reactions. Some, but not all, of the stress-es stimulate ACTH secretion. The increase in ACTH secretion is essential for survival when the stress is severe. If animals are then hypophysectomized, or adrenalectomized but treated with maintenance doses of glucocorticoids, they die when ex-posed to the same stress. The reason an elevated circulating ACTH, and hence gluco-corticoid level, is essential for resisting stress remains for the most part unknown. Most of the stressful stimuli that increase ACTH secretion also activate the sympathetic nervous sys-tem, and part of the function of circulating glucocorticoids may be maintenance of vascular reactivity to catecholamines. Glucocorticoids are also necessary for the catecholamines to exert their full FFA-mobilizing action, and the FFAs are an important emergency energy supply. However, sympathecto-mized animals tolerate a variety of stresses with relative impu-nity. Another theory holds that glucocorticoids prevent other stress-induced changes from becoming excessive. At present, all that can be said is that stress causes increases in plasma glucocorticoids to high “pharmacologic” levels that in the short run are life-saving. It should also be noted that the increase in ACTH, which is beneficial in the short term, becomes harmful and disruptive in the long term, causing among other things, the abnormali-ties of Cushing syndrome. PHARMACOLOGIC & PATHOLOGIC EFFECTS OF GLUCOCORTICOIDS CUSHING SYNDROME The clinical picture produced by prolonged increases in plas-ma glucocorticoids was described by Harvey Cushing and is called Cushing syndrome (Figure 22–13). It may be ACTH-independent or ACTH-dependent. The causes of ACTH-in-dependent Cushing syndrome include glucocorticoid-secret-ing adrenal tumors, adrenal hyperplasia, and prolonged administration of exogenous glucocorticoids for diseases such as rheumatoid arthritis. Rare but interesting ACTH-indepen-dent cases have been reported in which adrenocortical cells abnormally express receptors for gastric inhibitory polypep-tide (GIP) (see Chapter 26), vasopressin (see Chapter 39), β-adrenergic agonists, IL-1, or gonadotropin-releasing hormone (GnRH; see Chapter 25), causing these peptides to increase glucocorticoid secretion. The causes of ACTH-dependent Cushing syndrome include ACTH-secreting tumors of the an-terior pituitary gland and tumors of other organs, usually the TABLE 22–4 Typical effects of cortisol on the white and red blood cell counts in humans (cells/μL). Cell Normal Cortisol-Treated White blood cells Total 9000 10,000 PMNs 5760 8330 Lymphocytes 2370 1080 Eosinophils 270 20 Basophils 60 30 Monocytes 450 540 Red blood cells 5 million 5.2 million FIGURE 22–13 Typical findings in Cushing syndrome. (Reproduced with permission from Forsham PH, Di Raimondo VC: Traumatic Medicine and Surgery for the Attorney. Butterworth, 1960.) Moon face Red cheeks Bruisability with ecchymoses Thin skin Poor muscle development Poor wound healing Striae Pendulous abdomen Fat pads CHAPTER 22 The Adrenal Medulla & Adrenal Cortex 351 lungs, that secrete ACTH (ectopic ACTH syndrome) or corti-cotropin releasing hormone (CRH). Cushing syndrome due to anterior pituitary tumors is often called Cushing disease be-cause these tumors were the cause of the cases described by Cushing. However, it is confusing to speak of Cushing disease as a subtype of Cushing syndrome, and the distinction seems to be of little more than historical value. Patients with Cushing syndrome are protein-depleted as a result of excess protein catabolism. The skin and subcutane-ous tissues are therefore thin and the muscles are poorly developed. Wounds heal poorly, and minor injuries cause bruises and ecchymoses. The hair is thin and scraggly. Many patients with the disease have some increase in facial hair and acne, but this is caused by the increased secretion of adrenal androgens and often accompanies the increase in glucocorti-coid secretion. Body fat is redistributed in a characteristic way. The extrem-ities are thin, but fat collects in the abdominal wall, face, and upper back, where it produces a “buffalo hump.” As the thin skin of the abdomen is stretched by the increased subcutane-ous fat depots, the subdermal tissues rupture to form promi-nent reddish purple striae. These scars are seen normally whenever a rapid stretching of skin occurs, but in normal indi-viduals the striae are usually inconspicuous and lack the intense purplish color. Many of the amino acids liberated from catabolized pro-teins are converted into glucose in the liver and the resultant hyperglycemia and decreased peripheral utilization of glucose may be sufficient to precipitate insulin-resistant diabetes mel-litus, especially in patients genetically predisposed to diabetes. Hyperlipemia and ketosis are associated with the diabetes, but acidosis is usually not severe. The glucocorticoids are present in such large amounts in Cushing syndrome that they may exert a significant mineralo-corticoid action. Deoxycorticosterone secretion is also ele-vated in cases due to ACTH hypersecretion. The salt and water retention plus the facial obesity cause the characteristic plethoric, rounded “moon-faced” appearance, and there may be significant K+ depletion and weakness. About 85% of patients with Cushing syndrome are hypertensive. The hypertension may be due to increased deoxycorticosterone secretion, increased angiotensinogen secretion, or a direct glucocorti-coid effect on blood vessels (see Chapter 33). Glucocorticoid excess leads to bone dissolution by decreas-ing bone formation and increasing bone resorption. This leads to osteoporosis, a loss of bone mass that leads eventu-ally to collapse of vertebral bodies and other fractures. The mechanisms by which glucocorticoids produce their effects on bone are discussed in Chapter 23. Glucocorticoids in excess accelerate the basic electroen-cephalographic rhythms and produce mental aberrations ranging from increased appetite, insomnia, and euphoria to frank toxic psychoses. As noted above, glucocorticoid defi-ciency is also associated with mental symptoms, but the symptoms produced by glucocorticoid excess are more severe. ANTI-INFLAMMATORY & ANTI-ALLERGIC EFFECTS OF GLUCOCORTICOIDS Glucocorticoids inhibit the inflammatory response to tissue injury. The glucocorticoids also suppress manifestations of al-lergic disease that are due to the release of histamine from tis-sues. Both of these effects require high levels of circulating glucocorticoids and cannot be produced by administering steroids without producing the other manifestations of gluco-corticoid excess. Furthermore, large doses of exogenous glu-cocorticoids inhibit ACTH secretion to the point that severe adrenal insufficiency can be a dangerous problem when ther-apy is stopped. However, local administration of glucocorti-coids, for example, by injection into an inflamed joint or near an irritated nerve, produces a high local concentration of the steroid, often without enough systemic absorption to cause se-rious side effects. The actions of glucocorticoids in patients with bacterial infections are dramatic but dangerous. For example, in pneu-mococcal pneumonia or active tuberculosis, the febrile reac-tion, the toxicity, and the lung symptoms disappear, but unless antibiotics are given at the same time, the bacteria spread throughout the body. It is important to remember that the symptoms are the warning that disease is present; when these symptoms are masked by treatment with glucocorticoids, there may be serious and even fatal delays in diagnosis and the institution of treatment with antimicrobial drugs. The role of NF-κB in the anti-inflammatory and anti-aller-gic effects of glucocorticoids has been mentioned above and is discussed in Chapter 3. An additional action that combats local inflammation is inhibition of phospholipase A2. This reduces the release of arachidonic acid from tissue phospho-lipids and consequently reduces the formation of leuko-trienes, thromboxanes, prostaglandins, and prostacyclin (see Chapter 33). OTHER EFFECTS Large doses of glucocorticoids inhibit growth, decrease growth hormone secretion (see Chapter 24), induce PNMT, and decrease thyroid-stimulating hormone (TSH) secretion. During fetal life, glucocorticoids accelerate the maturation of surfactant in the lungs (see Chapter 35). REGULATION OF GLUCOCORTICOID SECRETION ROLE OF ACTH Both basal secretion of glucocorticoids and the increased se-cretion provoked by stress are dependent upon ACTH from the anterior pituitary. Angiotensin II also stimulates the adre-nal cortex, but its effect is mainly on aldosterone secretion. Large doses of a number of other naturally occurring 352 SECTION IV Endocrine & Reproductive Physiology substances, including vasopressin, serotonin, and vasoactive intestinal polypeptide (VIP), are capable of stimulating the ad-renal directly, but there is no evidence that these agents play any role in the physiologic regulation of glucocorticoid secretion. CHEMISTRY & METABOLISM OF ACTH ACTH is a single-chain polypeptide containing 39 amino ac-ids. Its origin from proopiomelanocortin (POMC) in the pitu-itary is discussed in Chapter 24. The first 23 amino acids in the chain generally constitute the active “core” of the molecule. Amino acids 24–39 constitute a “tail” that stabilizes the mole-cule and varies slightly in composition from species to species (Figure 22–14). The ACTHs that have been isolated are gener-ally active in all species but antigenic in heterologous species. ACTH is inactivated in blood in vitro more slowly than in vivo; its half-life in the circulation in humans is about 10 min. A large part of an injected dose of ACTH is found in the kid-neys, but neither nephrectomy nor evisceration appreciably enhances its in vivo activity, and the site of its inactivation is not known. EFFECT OF ACTH ON THE ADRENAL After hypophysectomy, glucocorticoid synthesis and output de-cline within 1 h to very low levels, although some hormone is still secreted. Within a short time after an injection of ACTH (in dogs, less than 2 min), glucocorticoid output is increased (Fig-ure 22–15). With low doses of ACTH, the relationship between the log of the dose and the increase in glucocorticoid secretion is linear. However, the maximal rate at which glucocorticoids can be secreted is rapidly reached, and in dogs, doses larger than 10 mU only prolong the period of maximal secretion. A similar “ceiling on output” exists in humans. The effects of ACTH on adrenal morphology and the mechanism by which it increases steroid secretion have been discussed above. ADRENAL RESPONSIVENESS ACTH not only produces prompt increases in glucocorticoid secretion but also increases the sensitivity of the adrenal to sub-sequent doses of ACTH. Conversely, single doses of ACTH do not increase glucocorticoid secretion in chronically hypophy-sectomized animals and patients with hypopituitarism, and re-peated injections or prolonged infusions of ACTH are necessary to restore normal adrenal responses to ACTH. Decreased re-sponsiveness is also produced by doses of glucocorticoids that inhibit ACTH secretion. The decreased adrenal responsiveness to ACTH is detectable within 24 h after hypophysectomy and increases progressively with time (Figure 22–16). It is marked when the adrenal is atrophic but develops before visible changes occur in adrenal size or morphology. CIRCADIAN RHYTHM ACTH is secreted in irregular bursts throughout the day and plasma cortisol tends to rise and fall in response to these bursts (Figure 22–17). In humans, the bursts are most frequent in the early morning, and about 75% of the daily production of cor-tisol occurs between 4:00 AM and 10:00 AM. The bursts are least frequent in the evening. This diurnal (circadian) rhythm in ACTH secretion is present in patients with adrenal insufficiency receiving constant doses of glucocorticoids. It is not due to the stress of getting up in the morning, traumatic as that may be, because the increased ACTH secretion occurs be-fore waking up. If the “day” is lengthened experimentally to more than 24 h, that is, if the individual is isolated and the day’s activities are spread over more than 24 h, the adrenal cy-cle also lengthens, but the increase in ACTH secretion still oc-curs during the period of sleep. The biologic clock responsible FIGURE 22–14 Structure of ACTH. In the species shown, the amino acid composition varies only at positions 25, 31, and 33. (Reproduced with permission from Li CH: Adrenocorticotropin 45: Revised amino acid sequences for sheep and bovine hormones. Biochem Biophys Res Commun 1972;49:835.) Tyr Ser Ser Met Glu His Phe Arg Trp Gly Lys Pro Val Gly Lys Lys Arg Arg Pro Val Lys Val Tyr Pro Asn Asn Asp Asn Gly Ala Glu Asp Glu Ala Ser Ser Ser Leu Gln Glu Ala Phe Pro Leu Glu Phe Gln Glu 1 5 10 15 20 25 30 35 Human Bovine Ovine Porcine FIGURE 22–15 Changes in glucocorticoid output from the adrenal in hypophysectomized dogs following the intravenous (IV) administration of various doses of ACTH. 5 0 0 4 8 12 16 20 24 28 32 2 mU 5 mU 10 mU 50 mU Maximum output Time after ACTH IV (min) Glucocorticoid secretion rate (μg/min) CHAPTER 22 The Adrenal Medulla & Adrenal Cortex 353 for the diurnal ACTH rhythm is located in the suprachiasmat-ic nuclei of the hypothalamus (see Chapter 15). THE RESPONSE TO STRESS The morning plasma ACTH concentration in a healthy rest-ing human is about 25 pg/mL (5.5 pmol/L). ACTH and corti-sol values in various abnormal conditions are summarized in Figure 22–18. During severe stress, the amount of ACTH se-creted exceeds the amount necessary to produce maximal glu-cocorticoid output. However, prolonged exposure to ACTH in conditions such as the ectopic ACTH syndrome increases the adrenal maximum. Increases in ACTH secretion to meet emergency situations are mediated almost exclusively through the hypothalamus via release of CRH. This polypeptide is produced by neurons in the paraventricular nuclei. It is secreted in the median emi-nence and transported in the portal-hypophysial vessels to the anterior pituitary, where it stimulates ACTH secretion (see Chapter 18). If the median eminence is destroyed, increased secretion in response to many different stresses is blocked. Afferent nerve pathways from many parts of the brain con-verge on the paraventricular nuclei. Fibers from the amygdal-oid nuclei mediate responses to emotional stresses, and fear, FIGURE 22–16 Loss of ACTH responsiveness when ACTH secretion is decreased in humans. The 1- to 24-amino-acid se-quence of ACTH was infused intravenously (IV) in a dose of 250 μg over 8 hours. N, normal subjects; DX, dexamethasone 0.75 mg every 8 h for 3 days; CST, long-term corticosteroid therapy; HI, anterior pituitary in-sufficiency. (Reproduced with permission from Kolanowski J, et al: Adrenocortical response upon repeated stimulation with corticotropin in patients lacking endogenous corticotropin secretion. Acta Endocrinol [Kbh] 1977;85:595.) 24 8 4 Time (h) 2 0 10 20 30 40 50 ACTH IV N DX CST HI Plasma cortisol (μg/dL) FIGURE 22–17 Fluctuations in plasma ACTH and glucocorticoids throughout the day in a normal girl (age 16). The ACTH was measured by immunoassay and the glucocorticoids as 11-oxysteroids (11-OHCS). Note the greater ACTH and glucocorticoid rises in the morning, before awakening. (Reproduced, with permission, from Krieger DT, et al: Characterization of the normal temporal pattern of plasma corticosteroid levels. J Clin Endocrinol Metab 1971;32:266.) 200 180 160 140 120 100 80 60 40 20 0 25 20 15 10 5 0 Noon Noon 4 PM 8 PM Mid-night 4 AM 8 AM Lunch Dinner Snack Sleep B'kfast Snack Plasma ACTH (pg/mL) Plasma 11-OHCS (μg/dL) FIGURE 22–18 Plasma concentrations of ACTH and cortisol in various clinical states. (Reproduced with permission from Textbook of Endocrinology, 5th ed. Williams RH [editor]. Saunders, 1974.) Normal, morning Normal, evening Normal, dexamethasone Normal, metyrapone Normal, stress Addison disease Hypopituitarism Congenital adrenal hyperplasia Cushing, hyperplasia Cushing, dexamethasone Cushing, postadrenalectomy Cushing, ectopic ACTH syndrome Cushing, adrenal tumor 0 5 50 500 5000 5 50 500 5000 0 0 0 12 25 50 100 12 25 50 100 Condition Plasma ACTH (pg/mL) Plasma cortisol (μg/dL) 354 SECTION IV Endocrine & Reproductive Physiology anxiety, and apprehension cause marked increases in ACTH secretion. Input from the suprachiasmatic nuclei provides the drive for the diurnal rhythm. Impulses ascending to the hypo-thalamus via the nociceptive pathways and the reticular for-mation trigger increased ACTH secretion in response to injury (Figure 22–18). The baroreceptors exert an inhibitory input via the nucleus of the tractus solitarius. GLUCOCORTICOID FEEDBACK Free glucocorticoids inhibit ACTH secretion, and the degree of pituitary inhibition is proportional to the circulating gluco-corticoid level. The inhibitory effect is exerted at both the pi-tuitary and the hypothalamic levels. The inhibition is due primarily to an action on DNA, and maximal inhibition takes several hours to develop, although more rapid “fast feedback” also occurs. The ACTH-inhibiting activity of the various ster-oids parallels their glucocorticoid potency. A drop in resting corticoid levels stimulates ACTH secretion, and in chronic ad-renal insufficiency the rate of ACTH synthesis and secretion is markedly increased. Thus, the rate of ACTH secretion is determined by two opposing forces: the sum of the neural and possibly other stimuli converging through the hypothalamus to increase ACTH secretion, and the magnitude of the braking action of glucocorticoids on ACTH secretion, which is proportional to their level in the circulating blood (Figure 22–19). The dangers involved when prolonged treatment with anti-inflammatory doses of glucocorticoids is stopped deserve emphasis. Not only is the adrenal atrophic and unresponsive after such treatment, but even if its responsiveness is restored by injecting ACTH, the pituitary may be unable to secrete normal amounts of ACTH for as long as a month. The cause of the deficiency is presumably diminished ACTH synthesis. Thereafter, ACTH secretion slowly increases to supranormal levels. These in turn stimulate the adrenal, and glucocorticoid output rises, with feedback inhibition gradually reducing the elevated ACTH levels to normal (Figure 22–20). The compli-cations of sudden cessation of steroid therapy can usually be avoided by slowly decreasing the steroid dose over a long period of time. EFFECTS OF MINERALOCORTICOIDS ACTIONS Aldosterone and other steroids with mineralocorticoid activi-ty increase the reabsorption of Na+ from the urine, sweat, sa-liva, and the contents of the colon. Thus, mineralocorticoids cause retention of Na+ in the ECF. This expands ECF volume. In the kidneys, they act primarily on the principal cells (P cells) of the collecting ducts (see Chapter 38). Under the influ-ence of aldosterone, increased amounts of Na+ are in effect exchanged for K+ and H+ in the renal tubules, producing a K+ diuresis (Figure 22–21) and an increase in urine acidity. MECHANISM OF ACTION Like many other steroids, aldosterone binds to a cytoplasmic receptor, and the receptor-hormone complex moves to the FIGURE 22–19 Feedback control of the secretion of cortisol and other glucocorticoids via the hypothalamic-pituitary-adrenal axis. The dashed arrows indicate inhibitory effects and the solid arrows indicate stimulating effects. NTS, nucleus tractus solitarius. FIGURE 22–20 Pattern of plasma ACTH and cortisol values in patients recovering from prior long-term daily treatment with large doses of glucocorticoids. (Courtesy of R Ney.) CRH Trauma via nociceptive pathways Systemic effects Cortisol Afferents from NTS Emotion via limbic system Drive for circadian rhythm CRH ACTH Hypothalamus Anterior pituitary Adrenal cortex 2 4 6 8 10 12 0 ACTH Cortisol Months after stopping glucocorticoid treatment High Normal Low Plasma concentration CHAPTER 22 The Adrenal Medulla & Adrenal Cortex 355 nucleus where it alters the transcription of mRNAs. This in turn increases the production of proteins that alter cell function. The aldosterone-stimulated proteins have two effects—a rapid ef-fect, to increase the activity of epithelial sodium channels (EN-aCs) by increasing the insertion of these channels into the cell membrane from a cytoplasmic pool; and a slower effect to in-crease the synthesis of ENaCs. Among the genes activated by aldosterone is the gene for serum- and glucocorticoid-regulated kinase (sgk), a serine-threonine protein kinase. The gene for sgk is an early response gene, and sgk increases ENaC activity. Aldosterone also increases the mRNAs for the three subunits that make up ENaCs. The fact that sgk is activated by glucocorticoids as well as aldosterone is not a problem because glucocorticoids are inactivated at mineralocorticoid receptor sites. However, aldosterone activates the genes for other pro-teins in addition to sgk and ENaCs and inhibits others. There-fore, the exact mechanism by which aldosterone-induced proteins increase Na+ reabsorption is still unsettled. Evidence is accumulating that aldosterone also binds to the cell membrane and by a rapid, nongenomic action increases the activity of membrane Na+–K+ exchangers. This produces an increase in intracellular Na+, and the second messenger involved is probably IP3. In any case, the principal effect of aldosterone on Na+ transport takes 10 to 30 min to develop and peaks even later (Figure 22–21), indicating that it depends on the synthesis of new proteins by genomic mechanism. RELATION OF MINERALOCORTICOID TO GLUCOCORTICOID RECEPTORS It is intriguing that in vitro, the mineralocorticoid receptor has an appreciably higher affinity for glucocorticoids than the glu-cocorticoid receptor does, and glucocorticoids are present in large amounts in vivo. This raises the question of why gluco-corticoids do not bind to the mineralocorticoid receptors in the kidneys and other locations and produce mineralocorti-coid effects. At least in part, the answer is that the kidneys and other mineralocorticoid-sensitive tissues also contain the en-zyme 11β-hydroxysteroid dehydrogenase type 2. This en-zyme leaves aldosterone untouched, but it converts cortisol to cortisone (Figure 22–11) and corticosterone to its 11-oxy de-rivative. These 11-oxy derivatives do not bind to the receptor (Clinical Box 22–3). OTHER STEROIDS THAT AFFECT Na+ EXCRETION Aldosterone is the principal mineralocorticoid secreted by the adrenal, although corticosterone is secreted in sufficient amounts to exert a minor mineralocorticoid effect (Tables 22–1 and 22–2). Deoxycorticosterone, which is secreted in apprecia-ble amounts only in abnormal situations, has about 3% of the activity of aldosterone. Large amounts of progesterone and some other steroids cause natriuresis, but there is little evidence that they play any normal role in the control of Na+ excretion. FIGURE 22–21 Effect of aldosterone (5 μg as a single dose injected into the aorta) on electrolyte excretion in an adrenalectomized dog. The scale for creatinine clearance is on the right. 45 40 35 98 96 94 92 90 80 70 400 300 200 60 40 20 0 0 30 60 80 110 140 170 200 230 260 Time (min) Aldosterone Creatinine clearance K+ excretion Percent filtered K+ reabsorbed Percent filtered Na+ reabsorbed Na+ excretion Percent mL/min μeq/min CLINICAL BOX 22–3 Apparent Mineralocorticoid Excess If 11β-hydroxysteroid dehydrogenase type 2 is inhibited or absent, cortisol has marked mineralocorticoid effects. The resulting syndrome is called apparent mineralocorticoid excess (AME). Patients with this condition have the clini-cal picture of hyperaldosteronism because cortisol is act-ing on their mineralocorticoid receptors, and their plasma aldosterone level as well as their plasma renin activity is low. The condition can be due to congenital absence of the enzyme or to prolonged ingestion of licorice. Outside of the United States, licorice contains glycyrrhetinic acid, which inhibits 11β-hydroxysteroid dehydrogenase type 2. Individuals who eat large amounts of licorice have an in-crease in MR-activated sodium absorption via the epithe-lial sodium channel ENaC in the renal collecting duct, and blood pressure can rise. 356 SECTION IV Endocrine & Reproductive Physiology EFFECT OF ADRENALECTOMY In adrenal insufficiency, Na+ is lost in the urine; K+ is retained, and the plasma K+ rises. When adrenal insufficiency develops rapidly, the amount of Na+ lost from the ECF exceeds the amount excreted in the urine, indicating that Na+ also must be entering cells. When the posterior pituitary is intact, salt loss exceeds water loss, and the plasma Na+ falls (Table 22–5). However, the plasma volume also is reduced, resulting in hy-potension, circulatory insufficiency, and, eventually, fatal shock. These changes can be prevented to a degree by increas-ing the dietary NaCl intake. Rats survive indefinitely on extra salt alone, but in dogs and most humans, the amount of sup-plementary salt needed is so large that it is almost impossible to prevent eventual collapse and death unless mineralocorti-coid treatment is also instituted (see Clinical Box 22–4). REGULATION OF ALDOSTERONE SECRETION STIMULI The principal conditions that increase aldosterone secretion are summarized in Table 22–6. Some of them also increase glucocorticoid secretion; others selectively affect the output of aldosterone. The primary regulatory factors involved are ACTH from the pituitary, renin from the kidney via angioten-sin II, and a direct stimulatory effect of a rise in plasma K+ concentration on the adrenal cortex. EFFECT OF ACTH When first administered, ACTH stimulates the output of aldos-terone as well as that of glucocorticoids and sex hormones. Al-though the amount of ACTH required to increase aldosterone output is somewhat greater than the amount that stimulates maximal glucocorticoid secretion (Figure 22–23), it is well within the range of endogenous ACTH secretion. The effect is transient, and even if ACTH secretion remains elevated, aldos-terone output declines in 1 or 2 days. On the other hand, the output of the mineralocorticoid deoxycorticosterone remains elevated. The decline in aldosterone output is partly due to de-creased renin secretion secondary to hypervolemia, but it is pos-sible that some other factor also decreases the conversion of corticosterone to aldosterone. After hypophysectomy, the basal rate of aldosterone secretion is normal. The increase normally produced by surgical and other stresses is absent, but the in-crease produced by dietary salt restriction is unaffected for some time. Later on, atrophy of the zona glomerulosa compli-cates the picture in long-standing hypopituitarism, and this may lead to salt loss and hypoaldosteronism. Normally, glucocorticoid treatment does not suppress aldos-terone secretion. However, an interesting recently described syn-drome is glucocorticoid-remediable aldosteronism (GRA). This is an autosomal dominant disorder in which the increase in aldosterone secretion produced by ACTH is no longer transient. The hypersecretion of aldosterone and the accompanying hyper-tension are remedied when ACTH secretion is suppressed by administering glucocorticoids. The genes encoding aldosterone synthase and 11β-hydroxylase are 95% identical and are close together on chromosome 8. In individuals with GRA, there is unequal crossing over so that the 5 regulatory region of the 11β-hydroxylase gene is fused to the coding region of the aldosterone synthase. The product of this hybrid gene is an ACTH-sensitive aldosterone synthase. EFFECTS OF ANGIOTENSIN II & RENIN The octapeptide angiotensin II is formed in the body from angiotensin I, which is liberated by the action of renin on TABLE 22–5 Typical plasma electrolyte levels in normal humans and patients with adrenocortical diseases. Plasma Electolytes (mEq/L) State Na+ K+ Cl– HCO3 – Normal 142 4.5 105 25 Adrenal insufficiency 120 6.7 85 25 Primary hyperaldosteronism 145 2.4 96 41 CLINICAL BOX 22–4 Secondary Effects of Excess Mineralocorticoids A prominent feature of prolonged mineralocorticoid excess (Table 22–5) is K+ depletion due to prolonged K+ diuresis. H+ is also lost in the urine. Na+ is retained initially, but the plasma Na+ is elevated only slightly if at all, because water is retained with the osmotically active sodium ions. Consequently, ECF volume is expanded and the blood pressure rises. When the ECF expansion passes a certain point, Na+ excretion is usually increased in spite of the continued action of mineralocorti-coids on the renal tubules. This escape phenomenon (Figure 22–22) is probably due to increased secretion of ANP (see Chapter 39). Because of increased excretion of Na+ when the ECF volume is expanded, mineralocorticoids do not produce edema in normal individuals and patients with hyperaldos-teronism. However, escape may not occur in certain disease states, and in these situations, continued expansion of ECF volume leads to edema (see Chapters 38 and 39). CHAPTER 22 The Adrenal Medulla & Adrenal Cortex 357 circulating angiotensinogen (see Chapter 39). Injections of an-giotensin II stimulate adrenocortical secretion and, in small doses, affect primarily the secretion of aldosterone (Figure 22–24). The sites of action of angiotensin II are both early and late in the steroid biosynthetic pathway. The early action is on the conversion of cholesterol to pregnenolone, and the late ac-tion is on the conversion of corticosterone to aldosterone (Figure 22–8). Angiotensin II does not increase the secretion of deoxycorticosterone, which is controlled by ACTH. Renin is secreted from the juxtaglomerular cells that sur-round the renal afferent arterioles as they enter the glomeruli (see Chapter 39). Aldosterone secretion is regulated via the FIGURE 22–22 “Escape” from the sodium-retaining effect of desoxycorticosterone acetate (DOCA) in an adrenalectomized patient. ECF, extracellular fluid volume; TBV, total blood volume; RCV, red cell volume; PV, plasma volume. (Courtesy of EG Biglieri.) 150 100 50 0 6 Serum K+ (meq/L) Serum Na+ (meq/L) 4 DOCA 10 mg IM every 12 h 300 200 100 0 150 130 Intake Intake 60 58 56 1 Days Body weight (kg) 3 5 7 9 11 13 15 17 11.40 L 4.26 L 1.14 L 3.12 L 9.14 L 3.55 L 1.48 L 2.07 L ECF TBV RCV PV Urinary Na+ (meq/24 h) Urinary K+ (meq/24 h) Male, age 29 Bilateral adrenalectomy Dexamethasone, 0.25 mg/6 h TABLE 22–6 Conditions that increase aldosterone secretion. Glucocorticoid secretion also increased Surgery Anxiety Physical trauma Hemorrhage Glucocorticoid secretion unaffected High potassium intake Low sodium intake Constriction of inferior vena cava in thorax Standing Secondary hyperaldosteronism (in some cases of congestive heart fail-ure, cirrhosis, and nephrosis) FIGURE 22–23 Changes in adrenal venous output of steroids produced by ACTH in nephrectomized hypophysectomized dogs. 12 40 10 8 Change in 17-hydroxy-corticoid output (μg/min) Change in aldosterone output (ng/min) 6 4 2 0 30 25 20 15 10 5 0 2 5 10 100 1000 Dose of ACTH (mU) No. of dogs (4) (8) (6) (3) (10) 358 SECTION IV Endocrine & Reproductive Physiology renin–angiotensin system in a feedback fashion (Figure 22–25). A drop in ECF volume or intra-arterial vascular volume leads to a reflex increase in renal nerve discharge and decreases renal arterial pressure. Both changes increase renin secretion, and the angiotensin II formed by the action of the renin increases the rate of secretion of aldosterone. The aldosterone causes Na+ and, secondarily, water retention, expanding ECF volume and shutting off the stimulus that initiated increased renin secretion. Hemorrhage stimulates ACTH and renin secretion. Like hemorrhage, standing and constriction of the thoracic inferior vena cava decrease intrarenal arterial pressure. Dietary sodium restriction also increases aldosterone secretion via the renin– angiotensin system (Figure 22–26). Such restriction reduces ECF volume, but aldosterone and renin secretion are increased before any consistent decrease in blood pressure takes place. Consequently, the initial increase in renin secretion produced by dietary sodium restriction is probably due to a reflex increase in the activity of the renal nerves. The increase in cir-culating angiotensin II produced by salt depletion upregulates the angiotensin II receptors in the adrenal cortex and hence increases the response to angiotensin II, whereas it down-regu-lates the angiotensin II receptors in the blood vessels. ELECTROLYTES & OTHER FACTORS An acute decline in plasma Na+ of about 20 mEq/L stimulates aldosterone secretion, but changes of this magnitude are rare. However, the plasma K+ level need increase only 1 mEq/L to stimulate aldosterone secretion, and transient increases of this magnitude may occur after a meal, particularly if it is rich in K+. Like angiotensin II, K+ stimulates the conversion of cho-lesterol to pregnenolone and the conversion of deoxycorticos-terone to aldosterone. It appears to act by depolarizing the cell, which opens voltage-gated Ca2+ channels, increasing intra-cellular Ca2+. The sensitivity of the zona glomerulosa to an-giotensin II and consequently to a low-sodium diet is decreased by a low-potassium diet. In normal individuals, plasma aldosterone concentrations increase during the portion of the day that the individual is carrying on activities in the upright position. This increase is due to a decrease in the rate of removal of aldosterone from the circulation by the liver and an increase in aldosterone secretion due to a postural increase in renin secretion. Indi-viduals who are confined to bed show a circadian rhythm of aldosterone and renin secretion, with the highest values in the early morning before awakening. Atrial natriuretic peptide (ANP) inhibits renin secretion and decreases the responsiveness of the zona glomerulosa to angiotensin II (see Chapter 39). The mechanisms by which ACTH, angiotensin II, and K+ stimulate aldosterone secretion are summarized in Table 22–7. FIGURE 22–24 Changes in adrenal venous output of steroids produced by angiotensin II in nephrectomized hypophysectomized dogs. 8 Change in 17-hydroxy-corticoid (μg/min) Change in aldosterone output (ng/min) 6 4 2 0 25 20 15 10 5 0 0.042 0.083 0.167 0.42 1.67 Dose of angiotensin II (μg/min) No. of dogs (5) (2) (8) (7) (7) Aldosterone values in 3 dogs FIGURE 22–25 Feedback mechanism regulating aldosterone secretion. The dashed arrow indicates inhibiition. Angiotensin-converting enzyme Renin Angiotensinogen Angiotensin I Angiotensin II Aldosterone Decreased Na+ (and water) excretion Increased extracellular fluid volume Increased renal arterial mean pressure, decreased discharge of renal nerves Juxtaglomerular apparatus Adrenal cortex CHAPTER 22 The Adrenal Medulla & Adrenal Cortex 359 ROLE OF MINERALOCORTICOIDS IN THE REGULATION OF SALT BALANCE Variation in aldosterone secretion is only one of many factors affecting Na+ excretion. Other major factors include the glo-merular filtration rate, ANP, the presence or absence of os-motic diuresis, and changes in tubular reabsorption of Na+ independent of aldosterone. It takes some time for aldoster-one to act. When one rises from the supine to the standing po-sition, aldosterone secretion increases and Na+ is retained from the urine. However, the decrease in Na+ excretion devel-ops too rapidly to be explained solely by increased aldosterone secretion. The primary function of the aldosterone-secreting mechanism is the defense of intravascular volume, but it is only one of the homeostatic mechanisms involved. SUMMARY OF THE EFFECTS OF ADRENOCORTICAL HYPER- & HYPOFUNCTION IN HUMANS Recapitulating the manifestations of excess and deficiency of the adrenocortical hormones in humans is a convenient way to summarize the multiple and complex actions of these ster-oids. A characteristic clinical syndrome is associated with ex-cess secretion of each of the types of hormones. Excess androgen secretion causes masculinization (adreno-genital syndrome) and precocious pseudopuberty or female pseudohermaphroditism. Excess glucocorticoid secretion produces a moon-faced, plethoric appearance, with trunk obesity, purple abdominal striae, hypertension, osteoporosis, protein depletion, mental abnormalities, and, frequently, diabetes mellitus (Cushing syndrome). The causes of Cushing syndrome have been dis-cussed previously. Excess mineralocorticoid secretion leads to K+ depletion and Na+ retention, usually without edema but with weakness, hypertension, tetany, polyuria, and hypokalemic alkalosis (hyperaldosteronism). This condition may be due to primary adrenal disease (primary hyperaldosteronism; Conn syn-drome) such as an adenoma of the zona glomerulosa, unilateral or bilateral adrenal hyperplasia, adrenal carcinoma, or GRA. In patients with primary hyperaldosteronism, renin secretion is depressed. Secondary hyperaldosteronism with high plasma renin activity is caused by cirrhosis, heart failure, and nephro-sis. Increased renin secretion is also found in individuals with the salt-losing form of the adrenogenital syndrome (see above), because their ECF volume is low. In patients with elevated renin secretion due to renal artery constriction, aldosterone secretion is increased; in those in whom renin secretion is not elevated, aldosterone secretion is normal. The relationship of aldoster-one to hypertension is discussed in Chapter 33. FIGURE 22–26 Effect of low-, normal-, and high-sodium diets on sodium metabolism and plasma renin activity, aldosterone, vasopressin, and ANP in normal humans. (Data from Sagnella GA, et al: Plasma atrial natriuretic peptide: Its relationship to changes in sodium in-take, plasma renin activity, and aldosterone in man. Clin Sci 1987;72:25.) 150 140 130 120 15 10 400 300 200 100 2000 0 1500 1000 500 0 15 20 10 5 0 5 0 2.0 2.5 1.5 1.0 0.5 0 Low Normal High Sodium intake Plasma vasopressin (pg/mL) Plasma renin activity (ng AI/mL/h) Plasma Na+ (mmol/L) Plasma aldosterone (pmol/L) Urinary Na+ excretion (mmol/day) Plasma ANP (pg/mL) TABLE 22–7 Second messengers involved in the regulation of aldosterone secretion. Secretagogue Intracellular Mediator ACTH Cyclic AMP, protein kinase A Angiotensin II Diacylglycerol, protein kinase C K+ Ca2+ via voltage-gated Ca2+ channels 360 SECTION IV Endocrine & Reproductive Physiology Primary adrenal insufficiency due to disease processes that destroy the adrenal cortex is called Addison disease. The condi-tion used to be a relatively common complication of tuberculo-sis, and now it is usually due to autoimmune inflammation of the adrenal. Patients lose weight, are tired, and become chroni-cally hypotensive. They have small hearts, probably because the hypotension decreases the work of the heart. Eventually they develop severe hypotension and shock (addisonian crisis). This is due not only to mineralocorticoid deficiency but to glucocor-ticoid deficiency as well. Fasting causes fatal hypoglycemia, and any stress causes collapse. Water is retained, and there is always the danger of water intoxication. Circulating ACTH levels are elevated. The diffuse tanning of the skin and the spotty pigmen-tation characteristic of chronic glucocorticoid deficiency (Fig-ure 22–27) are due, at least in part, to the melanocyte-stimulating hormone (MSH) activity of the ACTH in the blood. Minor menstrual abnormalities occur in women, but the defi-ciency of adrenal sex hormones usually has little effect in the presence of normal testes or ovaries. Secondary adrenal insufficiency is caused by pituitary dis-eases that decrease ACTH secretion, and tertiary adrenal insufficiency is caused by hypothalamic disorders disrupting CRH secretion. Both are usually milder than primary adrenal insufficiency because electrolyte metabolism is affected to a lesser degree. In addition, there is no pigmentation because in both of these conditions, plasma ACTH is low, not high. Cases of isolated aldosterone deficiency have also been reported in patients with renal disease and a low circulating renin level (hyporeninemic hypoaldosteronism). In addi-tion, pseudohypoaldosteronism is produced when there is resistance to the action of aldosterone. Patients with these syndromes have marked hyperkalemia, salt wasting, and hypotension, and they may develop metabolic acidosis. CHAPTER SUMMARY ■The adrenal gland consists of the adrenal medulla which se-cretes dopamine and the catecholamines epinephrine and FIGURE 22–27 Pigmentation in Addison disease. A) Tan and vitiligo. B) Pigmentation of scars from lesions that occurred after the devel-opment of the disease. C) Pigmentation of skin creases. D) Darkening of areolas. E) Pigmentation of pressure points. F) Pigmentation of the gums. (Reproduced with permission from Forsham PH, Di Raimondo V: Traumatic Medicine and Surgery for the Attorney. Butterworth, 1960.) A B C D E F CHAPTER 22 The Adrenal Medulla & Adrenal Cortex 361 norepinephrine, and the adrenal cortex which secretes steroid hormones. ■Norepinephrine and epinephrine act on two classes of receptors, α- and β-adrenergic receptors, and exert metabolic effects that include glycogenolysis in liver and skeletal muscle, mobilization of FFA, increased plasma lactate, and stimulation of the meta-bolic rate. ■The hormones of the adrenal cortex are derivatives of cholester-ol and include the mineralocorticoid aldosterone, the glucocor-ticoids cortisol and corticosterone, and the androgens dehydroepiandrosterone (DHEA) and androstenedione. ■Androgens are the hormones that exert masculinizing effects, and they promote protein anabolism and growth. The adrenal androgen androstenedione is converted to testosterone and to estrogens (aromatized) in fat and other peripheral tissues. This is an important source of estrogens in men and postmenopausal women. ■The mineralocorticoid aldosterone has effects on Na+ and K+ excretion and glucocorticoids affect glucose and protein metabolism. ■Glucocorticoid secretion is dependent upon ACTH from the anterior pituitary and is increased by stress. Angiotensin II increases the secretion of aldosterone. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Which of the following is produced only by large amounts of glu-cocorticoids? A) normal responsiveness of fat depots to norepinephrine B) maintenance of normal vascular reactivity C) increased excretion of a water load D) inhibition of the inflammatory response E) inhibition of ACTH secretion 2. Which of the following are incorrectly paired? A) gluconeogenesis : cortisol B) free fatty acid mobilization : dehydroepiandrosterone C) muscle glycogenolysis : epinephrine D) kaliuresis : aldosterone E) hepatic glycogenesis : insulin 3. Which of the following hormones has the shortest plasma half-life? A) corticosterone B) renin C) dehydroepiandrosterone D) aldosterone E) norepinephrine 4. Mole for mole, which of the following has the greatest effect on Na+ excretion? A) progesterone B) cortisol C) vasopressin D) aldosterone E) dehydroepiandrosterone 5. Mole for mole, which of the following has the greatest effect on plasma osmolality? A) progesterone B) cortisol C) vasopressin D) aldosterone E) dehydroepiandrosterone 6. The secretion of which of the following would be least affected by a decrease in extracellular fluid volume? A) CRH B) arginine vasopressin C) dehydroepiandrosterone D) estrogens E) aldosterone 7. A young man presents with a blood pressure of 175/110 mm Hg. He is found to have a high circulating aldosterone but a low cir-culating cortisol. Glucocorticoid treatment lowers his circulating aldosterone and lowers his blood pressure to 140/85 mm Hg. He probably has an abnormal A) 17α-hydroxylase. B) 21β-hydroxylase. C) 3β-hydroxysteroid dehydrogenase. D) aldosterone synthase. E) cholesterol desmolase. 8. A 32-year-old woman presents with a blood pressure of 155/96 mm Hg. In response to questioning, she admits that she loves lic-orice and eats some at least three times a week. She probably has a low level of A) type 2 11β-hydroxysteroid dehydrogenase activity. B) ACTH. C) 11β-hydroxylase activity. D) glucuronyl transferase. E) norepinephrine. 9. In its action in cells, aldosterone A) increases transport of ENaCs from the cytoplasm to the cell membrane. B) does not act on the cell membrane. C) binds to a receptor in the nucleus. D) may activate a heat shock protein. E) also binds to glucocorticoid receptors. CHAPTER RESOURCES Goldstein JL, Brown MS: The cholesterol quartet. Science 2001;292:1510. Goodman HM (editor): Handbook of Physiology, Section 7: The Endocrine System. Oxford University Press, 2000. Larsen PR et al (editors): Williams Textbook of Endocrinology, 9th ed. Saunders, 2003. Stocco DM: A review of the characteristics of the protein required for the acute regulation of steroid hormone biosynthesis: The case for the steroidogenic acute regulatory (StAR) protein. Proc Soc Exp Biol Med 1998;217:123. White PC: Disorders of aldosterone biosynthesis and action. N Engl J Med 1994;331:250. This page intentionally left blank 363 C H A P T E R 23 Hormonal Control of Calcium & Phosphate Metabolism & the Physiology of Bone O B J E C T I V E S After studying this chapter, you should be able to: ■Understand the importance of maintaining homeostasis of bodily calcium and phosphate concentrations, and how this is accomplished. ■Describe the bodily pools of calcium, their rates of turnover, and the organs that play central roles in regulating movement of calcium between stores. ■Delineate the mechanisms of calcium and phosphate absorption and excretion. ■Identify the major hormones and other factors that regulate calcium and phos-phate homeostasis and their sites of synthesis as well as targets of their action. ■Define the basic anatomy of bone. ■Delineate cells and their functions in bone formation and resorption. INTRODUCTION Calcium is an essential intracellular-signaling molecule and also plays a variety of extracellular functions, thus the control of bodily calcium concentrations is vitally important. The components of the system that maintain calcium homeostasis include cell types that sense changes in extracellular calcium and release calcium-regulating hormones, and the targets of these hormones, including the kidneys, bones, and intestine, that respond with changes in calcium mobilization, excretion, or uptake. Three hormones are primarily concerned with the regulation of calcium metabolism. 1,25-Dihydroxycholecal-ciferol is a steroid hormone formed from vitamin D by suc-cessive hydroxylations in the liver and kidneys. Its primary action is to increase calcium absorption from the intestine. Parathyroid hormone (PTH) is secreted by the parathyroid glands. Its main action is to mobilize calcium from bone and increase urinary phosphate excretion. Calcitonin, a calcium-lowering hormone that in mammals is secreted primarily by cells in the thyroid gland, inhibits bone resorption. Although the role of calcitonin seems to be relatively minor, all three hormones probably operate in concert to maintain the con-stancy of the Ca2+ level in the body fluids. Phosphate homeo-stasis is likewise critical to normal body function, particularly given its inclusion in adenosine triphosphate (ATP), its role as a biological buffer, and its role as a modifier of proteins, thereby altering their functions. Many of the systems that reg-ulate calcium homeostasis also contribute to that of phos-phate, albeit sometimes in a reciprocal fashion, and thus will also be discussed in this chapter. 364 SECTION IV Endocrine & Reproductive Physiology CALCIUM & PHOSPHORUS METABOLISM CALCIUM The body of a young adult human contains about 1100 g (27.5 mol) of calcium. Ninety-nine percent of the calcium is in the skeleton. Plasma calcium, normally at a concentration of around 10 mg/dL (5 mEq/L, 2.5 mmol/L), is partly bound to protein and partly diffusible (Table 23–1). The distribution of calcium inside cells is discussed in Chapter 2. It is the free, ionized calcium in the body fluids that is a vital second messenger (see Chapter 2) and is necessary for blood coagulation, muscle contraction, and nerve function. A decrease in extracellular Ca2+ exerts a net excitatory effect on nerve and muscle cells in vivo (see Chapters 4 and 5). The result is hypocalcemic tetany, which is characterized by extensive spasms of skeletal muscle, involving especially the muscles of the extremities and the larynx. Laryngospasm can become so severe that the airway is obstructed and fatal asphyxia is produced. Ca2+ also plays an important role in blood clotting (see Chapter 32), but in vivo, fatal tetany would occur before compromising the clotting reaction. Because the extent of Ca2+ binding by plasma proteins is proportional to the plasma protein level, it is important to know the plasma protein level when evaluating the total plasma calcium. Other electrolytes and pH also affect the free Ca2+ level. Thus, for example, symptoms of tetany appear at higher total calcium levels if the patient hyperventilates, thereby increasing plasma pH. Plasma proteins are more ion-ized when the pH is high, providing more protein anion to bind with Ca2+. The calcium in bone is of two types: a readily exchangeable reservoir and a much larger pool of stable calcium that is only slowly exchangeable. Two independent but interacting homeostatic systems affect the calcium in bone. One is the system that regulates plasma Ca2+, providing for the move-ment of about 500 mmol of Ca2+ per day into and out of the readily exchangeable pool in the bone (Figure 23–1). The other system involves bone remodeling by the constant inter-play of bone resorption and deposition (see following text). However, the Ca2+ interchange between plasma and this sta-ble pool of bone calcium is only about 7.5 mmol/d. Ca2+ is transported across the brush border of intestinal epi-thelial cells via channels known as transient receptor potential vanilloid type 6 (TRPV6) and binds to an intracellular protein known as calbindin-D9k. Calbindin sequesters the absorbed calcium so that it does not disturb epithelial signaling pro-cesses that involve calcium. The absorbed Ca2+ is thereby delivered to the basolateral membrane of the epithelial cell, from where it can be transported into the bloodstream by either a sodium/calcium exchanger (NCX1) or a calcium-dependent ATPase. Nevertheless, it should be noted that TABLE 23–1 Distribution (mmol/L) of calcium in normal human plasma. Total diffusible 1.34 Ionized (Ca2+) 1.18 Complexed to HCO3 –, citrate, etc 0.16 Total nondiffusible (protein-bound) 1.16 Bound to albumin 0.92 Bound to globulin 0.24 Total plasma calcium 2.50 FIGURE 23–1 Calcium metabolism in an adult human. A typical daily intake of 25 mmol Ca2+ (1000 mg) moves through many body compartments. Diet 25 mmol GI tract Feces 22.5 mmol Absorption 15 mmol Secretion 12.5 mmol Reabsorption 247.5 mmol ECF 35 mmol Glomerular filtrate 250 mmol Urine 2.5 mmol Rapid exchange Accretion 500 mmol 7.5 mmol Reabsorption 7.5 mmol Bone Exchangeable 100 mmol Stable 27,200 mmol CHAPTER 23 Hormonal Control of Calcium & Phosphate Metabolism & the Physiology of Bone 365 recent studies indicate that some intestinal calcium uptake persists even in the absence of TRPV6 and calbindin-D9k, sug-gesting that additional pathways are likely also involved in this critical process. The overall transport process is regulated by 1,25-dihydroxycholecalciferol (see below). As Ca2+ uptake rises, moreover, 1,25-dihydroxycholecalciferol levels fall in response to increased plasma Ca2+. Plasma Ca2+ is filtered in the kidneys, but 98–99% of the fil-tered Ca2+ is reabsorbed. About 60% of the reabsorption occurs in the proximal tubules and the remainder in the ascending limb of the loop of Henle and the distal tubule. Distal tubular reabsorption depends on the TRPV5 channel, which is related to TRPV6 discussed previously, and whose expression is regu-lated by parathyroid hormone. PHOSPHORUS Phosphate is found in ATP, cyclic adenosine monophosphate (cAMP), 2,3-diphosphoglycerate, many proteins, and other vital compounds in the body. Phosphorylation and dephos-phorylation of proteins are involved in the regulation of cell function (see Chapter 2). Therefore, it is not surprising that, like calcium, phosphate metabolism is closely regulated. Total body phosphorus is 500 to 800 g (16.1–25.8 mol), 85–90% of which is in the skeleton. Total plasma phosphorus is about 12 mg/dL, with two-thirds of this total in organic compounds and the remaining inorganic phosphorus (Pi) mostly in PO4 3–, HPO4 2–, and H2PO4 –. The amount of phosphorus normally entering bone is about 3 mg (97 +mol)/kg/d, with an equal amount leaving via reabsorption. Pi in the plasma is filtered in the glomeruli, and 85–90% of the filtered Pi is reabsorbed. Active transport in the proximal tubule accounts for most of the reabsorption and involves two related sodium-dependent Pi cotransporters, NaPi-IIa and NaPi-IIc. NaPi-IIa is powerfully inhibited by parathyroid hor-mone, which causes its internalization and degradation and thus a reduction in renal Pi reabsorption (see below). Pi is absorbed in the duodenum and small intestine. Uptake occurs by a transporter related to those in the kidney, NaPi-IIb, that takes advantage of the low intracellular sodium concentra-tion established by the Na, K ATPase on the basolateral mem-brane of intestinal epithelial cells to load Pi against its concentration gradient. However, the pathway by which Pi exits into the bloodstream is not known. Many stimuli that increase Ca2+ absorption, including 1,25-dihydroxycholecalciferol, also increase Pi absorption via increased NaPi-IIb expression. VITAMIN D & THE HYDROXYCHOLECALCIFEROLS CHEMISTRY The active transport of Ca2+ and PO4 3– from the intestine is increased by a metabolite of vitamin D. The term “vitamin D” is used to refer to a group of closely related sterols produced by the action of ultraviolet light on certain provitamins (Fig-ure 23–2). Vitamin D3, which is also called cholecalciferol, is produced in the skin of mammals from 7-dehydrocholesterol by the action of sunlight. The reaction involves the rapid for-mation of previtamin D3, which is then converted more slowly to vitamin D3. Vitamin D3 and its hydroxylated derivatives are transported in the plasma bound to a globulin vitamin D-binding protein (DBP). Vitamin D3 is also ingested in the diet. Vitamin D3 is metabolized by enzymes that are members of the cytochrome P450 (CYP) superfamily (see Chapters 1 and 29). In the liver, vitamin D3 is converted to 25-hydroxychole-calciferol (calcidiol, 25-OHD3). The 25-hydroxycholecalciferol is converted in the cells of the proximal tubules of the kidneys to the more active metabolite 1,25-dihydroxycholecalciferol, which is also called calcitriol or 1,25-(OH)2D3. 1,25-Dihydrox-ycholecalciferol is also made in the placenta, in keratinocytes in the skin, and in macrophages. The normal plasma level of 25-hydroxycholecalciferol is about 30 ng/mL, and that of 1,25-dihydroxycholecalciferol is about 0.03 ng/mL (approximately 100 pmol/L). The less active metabolite 24,25-dihydroxychole-calciferol is also formed in the kidneys (Figure 23–2). MECHANISM OF ACTION 1,25 dihydroxycholecalciferol stimulates the expression of a number of gene products involved in calcium transport and handling via its receptor, which acts as a transcriptional regu-lator in its ligand-bound form. One group is the family of cal-bindin-D proteins. These are members of the troponin C superfamily of Ca2+-binding proteins that also includes cal-modulin (see Chapter 2). Calbindin-Ds are found in human intestine, brain, and kidneys. In the intestinal epithelium and many other tissues, two calbindins are induced: calbindin-D9K and calbindin-D28K, with molecular weights of 9,000 and 28,000, respectively. 1,25-dihydroxycholecalciferol also in-creases the number of Ca2+–ATPase and TRPV6 molecules in the intestinal cells, thus, the overall capacity for absorption of dietary calcium is enhanced. In addition to increasing Ca2+ absorption from the intestine, 1,25-dihydroxycholecalciferol facilitates Ca2+ reabsorption in the kidneys via increased TRPV5 expression in the proximal tubules, increases the synthetic activity of osteoblasts, and is necessary for normal calcification of matrix (see Clinical Box 23–1). The stimulation of osteoblasts brings about a secondary increase in the activity of osteoclasts (see below). REGULATION OF SYNTHESIS The formation of 25-hydroxycholecalciferol does not appear to be stringently regulated. However, the formation of 1,25-di-hydroxycholecalciferol in the kidneys, which is catalyzed by the renal 1_-hydroxylase, is regulated in a feedback fashion by plasma Ca2+ and PO4 3+ (Figure 23–3). When the plasma Ca2+ level is high, little 1,25-dihydroxycholecalciferol is produced, 366 SECTION IV Endocrine & Reproductive Physiology and the kidneys produce the relatively inactive metabolite 24,25-dihydroxycholecalciferol instead. This effect of Ca2+ on production of 1,25-dihydroxycholecalciferol is the mecha-nism that brings about adaptation of Ca2+ absorption from the intestine (see previous text). Conversely, expression of 1_-hydroxylase is stimulated by PTH, and when the plasma Ca2+ level is low, PTH secretion is increased. The production of FIGURE 23–2 Formation and hydroxylation of vitamin D3. 25-hydroxylation takes place in the liver, and the other hydroxylations occur primarily in the kidneys. The formulas of 7-dehydrocholesterol, vitamin D3, and 1,25-dihydroxycholecalciferol are also shown below. HO 7-Dehydrocholesterol 7-Dehydrocholesterol 1,25-Dihydroxycholecalciferol 1,25-Dihydroxycholecalciferol 24,25-Dihydroxycholecalciferol 25-Hydroxycholecalciferol 24 25 Vitamin D3 CH2 26 27 HO CH2 HO OH OH KIDNEY Other metabolites LIVER SKIN 25-Hydroxylase 24-Hydroxylase 1α-Hydroxylase Vitamin D3 (cholecalciferol) Previtamin D3 Sunlight FIGURE 23–3 Effects of PTH and 1,25-dihydroxycholecalciferol on whole body calcium homeostasis. Note that these hormones are also involved in the regulation of circulating phosphate levels. (Repro-duced with permission from Widmaier EP, Raff H, Strang KT: Vander’s Human Physiol-ogy,10th ed., McGraw-Hill, 2006.) Begin Release of calcium into plasma Bone Resorption Intestine Calcium absorption Urinary excretion of calcium Plasma 1,25–(OH)2D Kidneys Calcium reabsorption 1,25–(OH)2D formation Plasma parathyroid hormone Parathyroid glands Parathyroid hormone secretion Plasma calcium Restoration of plasma calcium toward normal FIGURE 23–4 The human parathyroid glands, viewed from behind. Common carotid artery Right para-thyroids Pharynx Recurrent laryngeal nerve Inferior thyroid artery CHAPTER 23 Hormonal Control of Calcium & Phosphate Metabolism & the Physiology of Bone 367 1,25-dihydroxycholecalciferol is also increased by low and in-hibited by high plasma PO4 3– levels, by a direct inhibitory ef-fect of PO4 3– on the 1_-hydroxylase. Additional control of 1,25-dihydroxycholecalciferol formation is exerted by a direct negative feedback effect of the metabolite on 1_-hydroxylase, a positive feedback action on the formation of 24,25-dihy-droxycholecalciferol, and a direct action on the parathyroid gland to inhibit PTH expression. An “anti-aging” protein called -Klotho (named after Klotho, a daughter of Zeus in Greek mythology who spins the thread of life) has also recently been discovered to play impor-tant roles in calcium and phosphate homeostasis, in part by reciprocal effects on 1,25-dihydroxycholecalciferol levels. Mice deficient in -Klotho displayed accelerated aging, decreased bone mineral density, calcifications, and hypercalcemia and hyperphosphatemia. -Klotho plays an important role in sta-bilizing the membrane localization of proteins important in calcium and phosphate (re)absorption, such as TRPV5 and Na, K ATPase. Likewise, it enhances the activity of another fac-tor, fibroblast growth factor 23 (FGF23), at its receptor. FGF23 thereby decreases renal NaPi-IIa and NaPi-IIc expression and inhibits the production of 1-hydroxylase, reducing levels of 1,25-dihydroxycholecalciferol (Clinical Box 23–1). THE PARATHYROID GLANDS ANATOMY Humans usually have four parathyroid glands: two embedded in the superior poles of the thyroid and two in its inferior poles (Figure 23–4). Each parathyroid gland is a richly vascularized disk, about 3 × 6 × 2 mm, containing two distinct types of cells (Figure 23–5). The abundant chief cells, which contain a prominent Golgi apparatus plus endoplasmic reticulum and secretory granules, synthesize and secrete parathyroid hor-mone (PTH). The less abundant and larger oxyphil cells con-tain oxyphil granules and large numbers of mitochondria in their cytoplasm. In humans, few are seen before puberty, and thereafter they increase in number with age. Their function is unknown. Consequences of loss of parathyroid gland are dis-cussed in Clinical Box 23–2. SYNTHESIS & METABOLISM OF PTH Human PTH is a linear polypeptide with a molecular weight of 9500 that contains 84 amino acid residues (Figure 23–6). It is synthesized as part of a larger molecule containing 115 amino acid residues (preproPTH). On entry of preproPTH into the endoplasmic reticulum, a leader sequence is removed from the amino terminal to form the 90-amino-acid polypeptide proPTH. Six additional amino acid residues are removed from the amino terminal of proPTH in the Golgi apparatus, and the 84-amino-acid polypeptide PTH is packaged in secretory gran-ules and released as the main secretory product of the chief cells. The normal plasma level of intact PTH is 10 to 55 pg/mL. The half-life of PTH is approximately 10 min, and the secreted polypeptide is rapidly cleaved by the Kupffer cells in the liver into fragments that are probably biologically inactive. PTH and these fragments are then cleared by the kidneys. Modern immunoassays for PTH are designed only to measure mature PTH (1–84) and not these fragments to obtain an accurate measure of “active” PTH. CLINICAL BOX 23–1 Rickets & Osteomalacia Vitamin D deficiency causes defective calcification of bone matrix and the disease called rickets in children and osteo-malacia in adults. Even though 1,25-dihydroxycholecalcif-erol is necessary for normal mineralization of bone matrix, the main defect in this condition is failure to deliver ade-quate amounts of Ca2+ and PO4 3– to the sites of mineraliza-tion. The full-blown condition in children is characterized by weakness and bowing of weight-bearing bones, dental defects, and hypocalcemia. In adults, the condition is less obvious. It used to be most commonly due to inadequate exposure to the sun in smoggy cities, but now it is more commonly due to inadequate intake of the provitamins on which the sun acts in the skin. These cases respond to ad-ministration of vitamin D. The condition can also be caused by inactivating mutations of the gene for renal 1_-hydroxy-lase, in which case there is no response to vitamin D but a normal response to 1,25-dihydroxycholecalciferol (type I vitamin D-resistant rickets). In rare instances, it can be due to inactivating mutations of the gene for the 1,25-dihy-droxycholecalciferol receptor (type II vitamin D-resistant rickets), in which case there is a deficient response to both vitamin D and 1,25-dihydroxycholecalciferol. FIGURE 23–5 Section of human parathyroid. (Reduced 50% from × 960.) Small cells are chief cells; large stippled cells (especially prominent in the lower left of picture) are oxyphil cells. (Reproduced with permission from Fawcett DW: Bloom and Fawcett, A Textbook of Histology, 11th ed. Saunders, 1986.) 368 SECTION IV Endocrine & Reproductive Physiology ACTIONS PTH acts directly on bone to increase bone resorption and mo-bilize Ca2+. In addition to increasing the plasma Ca2+, PTH in-creases phosphate excretion in the urine and thereby depresses plasma phosphate levels. This phosphaturic action is due to a decrease in reabsorption of phosphate via effects on NaPi-IIa in the proximal tubules, as discussed previously. PTH also increas-es reabsorption of Ca2+ in the distal tubules, although Ca2+ ex-cretion in the urine is often increased in hyperparathyroidism because the increase in the load of filtered calcium overwhelms the effect on reabsorption (Clinical Box 23-3). PTH also in-creases the formation of 1,25-dihydroxycholecalciferol, and this increases Ca2+ absorption from the intestine. On a longer time scale, PTH stimulates both osteoblasts and osteoclasts. MECHANISM OF ACTION It now appears that there are at least three different PTH recep-tors. One also binds parathyroid hormone-related protein (PTHrP; see below) and is known as the hPTH/PTHrP receptor. CLINICAL BOX 23–2 Effects of Parathyroidectomy Occasionally, inadvertent parathyroidectomy occurs in hu-mans during thyroid surgery. This can have serious conse-quences as PTH is essential for life. After parathyroidectomy, there is a steady decline in the plasma Ca2+ level. Signs of neuromuscular hyperexcitability appear, followed by full-blown hypocalcemic tetany (see above). Plasma phosphate levels usually rise as the plasma calcium level falls. Symp-toms usually develop 2 to 3 d postoperatively but may not appear for several weeks or more. Injections of PTH can be given to correct the chemical abnormalities, and the symp-toms then disappear. Injections of Ca2+ salts can also give temporary relief. The signs of tetany in humans include Chvostek’s sign, a quick contraction of the ipsilateral facial muscles elicited by tapping over the facial nerve at the angle of the jaw, and Trousseau’s sign, a spasm of the mus-cles of the upper extremity that causes flexion of the wrist and thumb with extension of the fingers. In individuals with mild tetany in whom spasm is not yet evident, Trousseau sign can sometimes be produced by occluding the circula-tion for a few minutes with a blood pressure cuff. FIGURE 23–6 Parathyroid hormone. The symbols above and below the human structure show where amino acid residues are differ-ent in bovine and porcine PTH. (Reproduced with permission from Keutmann HT, et al: Complete amino acid sequence of human parathyroid hormone. Biochemistry 1978;17:5723. Copyright © 1978 by the American Chemical Society.) V S E I Q S A M H N L G K H L E V R R L W L K K V D N F V A L G A P L A P R D H Q N S S I V H L S M E L F S S I Y E H S L S E A D K A D V D V L G K Q A Q T K A I I K S Q P P A G G G S Q V N D E V L R P R K K E S 5 10 15 20 25 30 35 45 60 65 70 55 75 80 Human Porcine Bovine 50 CLINICAL BOX 23–3 Diseases of Parathyroid Excess Hyperparathyroidism due to injections of parathyroid extract in animals or hypersecretion of a functioning parathyroid tumor in humans is characterized by hypercalcemia and hy-pophosphatemia. Humans with PTH-secreting adenomas are usually asymptomatic, with the condition detected when plasma Ca2+ is measured in conjunction with a routine physi-cal examination. However, there may be minor changes in personality, and calcium-containing kidney stones occasion-ally form. In conditions such as chronic renal disease and rick-ets, in which the plasma Ca2+ level is chronically low, stimulation of the parathyroid glands causes compensatory parathyroid hypertrophy and secondary hyperparathyroid-ism. The plasma Ca2+ level is low in chronic renal disease pri-marily because the diseased kidneys lose the ability to form 1,25-dihydroxycholecalciferol. Finally, mutations in the calcium receptor, CaR, gene cause predictable long-term changes in plasma Ca2+. Individuals heterozygous for inacti-vating mutations have familial benign hypocalciuric hyper-calcemia, a condition in which there is a chronic moderate el-evation in plasma Ca2+ because the feedback inhibition of PTH secretion by Ca2+ is reduced. Plasma PTH levels are nor-mal or even elevated. However, children who are homozy-gous for inactivating mutations develop neonatal severe pri-mary hyperparathyroidism. Conversely, individuals with gain-of-function mutations of the CaR gene develop familial hy-percalciuric hypocalcemia due to increased sensitivity of the parathyroid glands to plasma Ca2+. CHAPTER 23 Hormonal Control of Calcium & Phosphate Metabolism & the Physiology of Bone 369 A second receptor, PTH2 (hPTH2-R), does not bind PTHrP and is found in the brain, placenta, and pancreas. In addition, there is evidence for a third receptor, CPTH, which reacts with the carboxyl terminal rather than the amino terminal of PTH. The first two receptors are coupled to Gs, and via this hetero-trimeric G protein they activate adenylyl cyclase, increasing intracellular cAMP. The hPTH/PTHrP receptor also activates PLC via Gq, increasing intracellular Ca2+ and activating pro-tein kinase C (Figure 23–7). However, the way these second messengers affect Ca2+ in bone is unsettled. In the disease called pseudohypoparathyroidism, the signs and symptoms of hypoparathyroidism develop but the circu-lating level of PTH is normal or elevated. Because the tissues fail to respond to the hormone, this is a receptor disease. There are two forms. In the more common form, a congenital 50% reduction of the activity of Gs occurs and PTH fails to produce a normal increase in cAMP concentration. In a dif-ferent, less common form, the cAMP response is normal but the phosphaturic action of the hormone is defective. REGULATION OF SECRETION Circulating ionized calcium acts directly on the parathyroid glands in a negative feedback fashion to regulate the secretion of PTH (Figure 23–8). The key to this regulation is a cell mem-brane Ca2+ receptor, CaR. Activation of this G-protein cou-pled receptor leads to phosphoinositide turnover in many tissues. In the parathyroid, its activation inhibits PTH secre-tion. In this way, when the plasma Ca2+ level is high, PTH se-cretion is inhibited and the Ca2+ is deposited in the bones. When it is low, secretion is increased and Ca2+ is mobilized from the bones. 1,25-dihydroxycholecalciferol acts directly on the parathy-roid glands to decrease preproPTH mRNA. Increased plasma phosphate stimulates PTH secretion by lowering plasma levels of free Ca2+ and inhibiting the formation of 1,25-dihydroxycholecalciferol. Magnesium is required to maintain normal parathyroid secretory responses. Impaired PTH release along with diminished target organ responses to PTH account for the hypocalcemia that occasionally occurs in magnesium deficiency (Clinical Box 23–2 and Clinical Box 23–3). PTHrP Another protein with PTH activity, parathyroid hormone-related protein (PTHrP), is produced by many different tis-sues in the body. It has 140 amino acid residues, compared with 84 in PTH, and is encoded by a gene on human chromo-some 12, whereas PTH is encoded by a gene on chromosome 11. PTHrP and PTH have marked homology at their amino terminal ends and they both bind to the hPTH/ PTHrP recep-tor, yet their physiologic effects are very different. How is this possible when they bind to the same receptor? For one thing, PTHrP is primarily a paracrine factor, acting close to where it is produced. It may be that circulating PTH cannot reach at least some of these sites. Second, subtle conformational differ-ences may be produced by binding of PTH versus PTHrP to their receptor, despite their structural similarities. Another possibility is action of one or the other hormone on other, more selective receptors. PTHrP has a marked effect on the growth and development of cartilage in utero. Mice in which both alleles of the PTHrP gene are knocked out have severe skeletal deformities and die soon after birth. In normal animals, on the other hand, PTHrP-stimulated cartilage cells proliferate and their termi-nal differentiation is inhibited. PTHrP is also expressed in the brain, where evidence indicates that it inhibits excitotoxic damage to developing neurons. In addition, there is evidence that it is involved in Ca2+ transport in the placenta. PTHrP is also found in keratinocytes in the skin, in smooth muscle, and FIGURE 23–7 Signal transduction pathways activated by PTH or PTHrP binding to the hPTH/hPTHrP receptor. Intracellular cAMP is increased via Gs and adenylyl cyclase (AC). Diacylglycerol and IP3 (1,4,5-InsP3) are increased via Gq and phospholipase C (PLC). (Modi-fied and reproduced with permission from McPhee SJ, Lingappa VR, Ganong WF [editors]: Pathophysiology of Disease, 4th ed. McGraw-Hill, 2003.) Gs Gq AC PLC PTHrP PTH ATP cAMP PIP2 PTH-R Diacylglycerol Protein kinase C activation Intracellular calcium mobilization 1,4,5-InsP3 FIGURE 23–8 Relation between plasma Ca2+ concentration and PTH response in humans.The set point is the plasma Ca2+ at which half the maximal response occurred (ie, 1.2 mmol/L). (Modified and reproduced with permission from Brown E: Extracellular Ca2+ sensing, regulation of parathyroid cell functions, and role of Ca2+ and other ions as extracellular (first) messengers. Physiol Rev 1991;71:371.) 1.5 1.25 1.0 0 20 40 60 80 100 Ionized calcium (mmol/L) Set point Maximal PTH response (%) 370 SECTION IV Endocrine & Reproductive Physiology in the teeth, where it is present in the enamel epithelium that caps each tooth. In the absence of PTHrP, teeth cannot erupt. HYPERCALCEMIA OF MALIGNANCY Hypercalcemia is a common metabolic complication of can-cer. About 20% of hypercalcemic patients have bone metasta-ses that produce the hypercalcemia by eroding bone (local osteolytic hypercalcemia). Evidence suggests that this ero-sion is produced by prostaglandins such as prostaglandin E2 from the tumor. The hypercalcemia in the remaining 80% of the patients is due to elevated circulating levels of PTHrP (hu-moral hypercalcemia of malignancy). The tumors responsi-ble for the hypersecretion include cancers of the breast, kidney, ovary, and skin. CALCITONIN ORIGIN In dogs, perfusion of the thyroparathyroid region with solutions containing high concentrations of Ca2+ leads to a fall in periph-eral plasma calcium, and after damage to this region, Ca2+ infu-sions cause a greater increase in plasma Ca2+ than they do in control animals. These and other observations led to the discov-ery that a Ca2+-lowering as well as a Ca2+-elevating hormone was secreted by structures in the neck. The Ca2+-lowering hor-mone has been named calcitonin. In mammals, calcitonin is produced by the parafollicular cells of the thyroid gland, which are also known as the clear or C cells. STRUCTURE Human calcitonin has a molecular weight of 3500 and con-tains 32 amino acid residues (Figure 23–9). Much of the mRNA transcribed from the calcitonin gene is processed to a different mRNA in the nervous system, so that calcitonin gene-related peptide (CGRP) is formed rather than calcito-nin (see Chapter 4). SECRETION & METABOLISM Secretion of calcitonin is increased when the thyroid gland is exposed to plasma calcium level of approximately 9.5 mg/dL. Above this level, plasma calcitonin is directly proportionate to plasma calcium. -adrenergic agonists, dopamine, and estro-gens also stimulate calcitonin secretion. Gastrin, cholecystoki-nin (CCK), glucagon, and secretin have all been reported to stimulate calcitonin secretion, with gastrin being the most po-tent stimulus (see Chapter 26). Thus, the plasma calcitonin lev-el is elevated in Zollinger–Ellison syndrome and in pernicious anemia (see Chapter 26). However, the dose of gastrin needed to stimulate calcitonin secretion is supraphysiological and not seen after eating in normal individuals, so dietary calcium in the intestine probably does not induce secretion of a calcium-lowering hormone prior to the calcium being absorbed. In any event, the actions of calcitonin are short-lived because it has a half-life of less than 10 min in humans. ACTIONS Receptors for calcitonin are found in bones and the kidneys. Calcitonin lowers circulating calcium and phosphate levels. It exerts its calcium-lowering effect by inhibiting bone resorption. This action is direct, and calcitonin inhibits the activity of osteo-clasts in vitro. It also increases Ca2+ excretion in the urine. The exact physiologic role of calcitonin is uncertain. The cal-citonin content of the human thyroid is low, and after thyroidec-tomy, bone density and plasma Ca2+ level are normal as long as the parathyroid glands are intact. In addition, there are only transient abnormalities of Ca2+ metabolism when a Ca2+ load is injected after thyroidectomy. This may be explained in part by secretion of calcitonin from tissues other than the thyroid. How-ever, there is general agreement that the hormone has little long-term effect on the plasma Ca2+ level in adult animals and humans. Further, unlike PTH and 1,25-dihydroxycholecalcif-erol, calcitonin does not appear to be involved in phosphate homeostasis. Moreover, patients with medullary carcinoma of the thyroid have a very high circulating calcitonin level but no symptoms directly attributable to the hormone, and their bones are essentially normal. No syndrome due to calcitonin defi-ciency has been described. More hormone is secreted in young individuals, and it may play a role in skeletal development. In addition, it may protect the bones of the mother from excess cal-cium loss during pregnancy. Bone formation in the infant and lactation are major drains on Ca2+ stores, and plasma concen-trations of 1,25-dihydroxycholecalciferol are elevated in preg-nancy. They would cause bone loss in the mother if bone resorption were not simultaneously inhibited by an increase in the plasma calcitonin level. SUMMARY The actions of the three principal hormones that regulate the plasma concentration of Ca2+ can now be summarized. PTH increases plasma Ca2+ by mobilizing this ion from bone. It in-creases Ca2+ reabsorption in the kidney, but this may be offset by the increase in filtered Ca2+. It also increases the formation of 1,25-dihydroxycholecalciferol. 1,25-Dihydroxycholecalciferol FIGURE 23–9 Human calcitonin. The sequence is shown using the three letter abbreviations for constituent amino acids. Cys-Gly-Asn-Leu-Ser-Thr-Cys-Met-Leu-Gly-Thr-Tyr-Thr-Gln-Asp-Phe-Asn-1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Lys-Phe-His-Thr-Phe-Pro-Gln-Thr-Ala-Ile-Gly-Val-Gly-Ala-Pro-NH2 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 S S CHAPTER 23 Hormonal Control of Calcium & Phosphate Metabolism & the Physiology of Bone 371 increases Ca2+ absorption from the intestine and increases Ca2+ reabsorption in the kidneys. Calcitonin inhibits bone re-sorption and increases the amount of Ca2+ in the urine. EFFECTS OF OTHER HORMONES & HUMORAL AGENTS ON CALCIUM METABOLISM Calcium metabolism is affected by various hormones in addi-tion to 1,25-dihydroxycholecalciferol, PTH, and calcitonin. Glucocorticoids lower plasma Ca2+ levels by inhibiting osteo-clast formation and activity, but over long periods they cause osteoporosis by decreasing bone formation and increasing bone resorption. They decrease bone formation by inhibiting protein synthesis in osteoblasts. They also decrease the absorp-tion of Ca2+ and PO4 3– from the intestine and increase the re-nal excretion of these ions. The decrease in plasma Ca2+ concentration also increases the secretion of PTH, and bone resorption is facilitated. Growth hormone increases calcium excretion in the urine, but it also increases intestinal absorp-tion of Ca2+, and this effect may be greater than the effect on excretion, with a resultant positive calcium balance. Insulin-like growth factor I (IGF-I) generated by the action of growth hormone stimulates protein synthesis in bone. As noted previ-ously, thyroid hormones may cause hypercalcemia, hypercal-ciuria, and, in some instances, osteoporosis. Estrogens prevent osteoporosis by inhibiting the stimulatory effects of certain cytokines on osteoclasts. Insulin increases bone for-mation, and there is significant bone loss in untreated diabetes. BONE PHYSIOLOGY Bone is a special form of connective tissue with a collagen framework impregnated with Ca2+ and PO4 3– salts, particu-larly hydroxyapatites, which have the general formula Ca10(PO4)6(OH)2. Bone is also involved in overall Ca2+ and PO4 3– homeostasis. It protects vital organs, and the rigidity it provides permits locomotion and the support of loads against gravity. Old bone is constantly being resorbed and new bone formed, permitting remodeling that allows it to respond to the stresses and strains that are put upon it. It is a living tissue that is well vascularized and has a total blood flow of 200 to 400 mL/min in adult humans. STRUCTURE Bone in children and adults is of two types: compact or cortical bone, which makes up the outer layer of most bones (Figure 23–10) and accounts for 80% of the bone in the body; and tra-becular or spongy bones inside the cortical bone, which makes up the remaining 20% of bone in the body. In compact bone, the surface-to-volume ratio is low, and bone cells lie in lacunae. They receive nutrients by way of canaliculi that ramify through-out the compact bone (Figure 23–10). Trabecular bone is made up of spicules or plates, with a high surface to volume ratio and many cells sitting on the surface of the plates. Nutrients diffuse from bone extracellular fluid (ECF) into the trabeculae, but in compact bone, nutrients are provided via haversian canals (Figure 23–10), which contain blood vessels. Around each Haversian canal, collagen is arranged in concentric layers, forming cylinders called osteons or haversian systems. The protein in bone matrix is over 90% type I collagen, which is also the major structural protein in tendons and skin. This collagen, which weight for weight is as strong as steel, is made up of a triple helix of three polypeptides bound tightly together. Two of these are identical _1 polypeptides encoded by one gene, and one is an _2 polypeptide encoded by a differ-ent gene. Collagens make up a family of structurally related proteins that maintain the integrity of many different organs. Fifteen different types of collagens encoded by more than 20 different genes have so far been identified. BONE GROWTH During fetal development, most bones are modeled in cartilage and then transformed into bone by ossification (enchondral bone formation). The exceptions are the clavicles, the mandi-bles, and certain bones of the skull in which mesenchymal cells form bone directly (intramembranous bone formation). During growth, specialized areas at the ends of each long bone (epiphyses) are separated from the shaft of the bone by a plate of actively proliferating cartilage, the epiphysial plate (Figure 23–11). The bone increases in length as this plate lays down new bone on the end of the shaft. The width of the epi-physial plate is proportionate to the rate of growth. The width is affected by a number of hormones, but most markedly by the pituitary growth hormone and IGF-I (see Chapter 24). Linear bone growth can occur as long as the epiphyses are separated from the shaft of the bone, but such growth ceases after the epiphyses unite with the shaft (epiphysial closure). The cartilage cells stop proliferating, become hypertrophic, and secrete vascular endothelial growth factor (VEGF), lead-ing to vascularization and ossification. The epiphyses of the various bones close in an orderly temporal sequence, the last epiphyses closing after puberty. The normal age at which each of the epiphyses closes is known, and the “bone age” of a young individual can be determined by x-raying the skeleton and noting which epiphyses are open and which are closed. BONE FORMATION & RESORPTION The cells responsible for bone formation are osteoblasts and the cells responsible for bone resorption are osteoclasts. Osteoblasts are modified fibroblasts. Their early develop-ment from the mesenchyme is the same as that of fibroblasts, with extensive growth factor regulation. Later, ossification-specific transcription factors, such as Cbfa1/Runx2, contribute to their differentiation. The importance of this transcription 372 SECTION IV Endocrine & Reproductive Physiology factor in bone development is underscored in knockout mice deficient for the Cbfa1/Runx gene. These mice develop to term with their skeletons made exclusively of cartilage; no ossification occurs. Normal osteoblasts are able to lay down type 1 collagen and form new bone. Osteoclasts, on the other hand, are members of the mono-cyte family. Stromal cells in the bone marrow, osteoblasts, and T lymphocytes all express receptor activator for nuclear factor kappa beta ligand (RANKL) on their surface. When these cells come in contact with appropriate monocytes expressing RANK (ie, the RANKL receptor) two distinct signaling path-ways are initiated: (1) there is a RANKL/RANK interaction between the cell pairs, (2) mononuclear phagocyte colony stimulating factor (M-CSF) is secreted by the nonmonocytic cells and it binds to its corresponding receptor on the mono-cytes (c-fins). The combination of these two signaling events leads to differentiation of the monocytes into osteoclasts. The precursor cells also secrete osteoprotegerin (OPG), which controls for differentiation of the monocytes by competing with RANK for binding of RANKL. Osteoclasts erode and absorb previously formed bone. They become attached to bone via integrins in a membrane exten-sion called the sealing zone. This creates an isolated area between the bone and a portion of the osteoclast. Proton pumps (ie, H+-dependent ATPases) then move from endo-somes into the cell membrane apposed to the isolated area, and they acidify the area to approximately pH 4.0. Similar proton pumps are found in the endosomes and lysosomes of all eukaryotic cells, but in only a few other instances do they move into the cell membrane. Note in this regard that the sealed-off space formed by the osteoclast resembles a large lysosome. The acidic pH dissolves hydroxyapatite, and acid FIGURE 23–10 Structure of compact and trabecular bone. The compact bone is shown in horizontal section (top) and vertical section (bottom). (Reproduced with permission from Williams PL et al (editors): Gray’s Anatomy, 37th ed. Churchill Livingstone, 1989.) Cortical (compact) bone Lacunae Trabecular (cancellous) bone Osteons Canaliculi Haversian canal Resorption spaces CHAPTER 23 Hormonal Control of Calcium & Phosphate Metabolism & the Physiology of Bone 373 proteases secreted by the cell break down collagen, forming a shallow depression in the bone (Figure 23–12). The products of digestion are then endocytosed and move across the osteo-clast by transcytosis (see Chapter 2), with release into the interstitial fluid. The collagen breakdown products have pyri-dinoline structures, and pyridinolines can be measured in the urine as an index of the rate of bone resorption. Throughout life, bone is being constantly resorbed and new bone is being formed. The calcium in bone turns over at a rate of 100% per year in infants and 18% per year in adults. Bone remodeling is mainly a local process carried out in small areas by populations of cells called bone-remodeling units. First, osteoclasts resorb bone, and then osteoblasts lay down new bone in the same general area. This cycle takes about 100 days. Modeling drifts also occur in which the shapes of bones change as bone is resorbed in one location and added in another. Osteoclasts tunnel into cortical bone followed by osteoblasts, whereas trabecular bone remodeling occurs on the surface of the trabeculae. About 5% of the bone mass is being remodeled by about 2 million bone-remodeling units in the human skeleton at any one time. The renewal rate for bone is about 4% per year for compact bone and 20% per year for trabecular bone. The remodeling is related in part to the stresses and strains imposed on the skeleton by gravity. At the cell–cell level, there is some regulation of osteoclast formation by osteoblasts via the RANKL–RANK and the M-CSF–OPG mechanism; however, specific feedback mecha-nisms of osteoclasts on osteoblasts are not well defined. In a broader sense, the bone remodeling process is primarily under endocrine control. Parathyroid hormone accelerates bone resorption, and estrogens slow bone resorption by inhibiting the production of bone-eroding cytokines. An interesting new observation is that intracerebroventricular but not intravenous leptin decreases bone formation. This finding is consistent with the observations that obesity protects against bone loss and that most obese humans are resistant to the effects of lep-tin on appetite. Thus, there may be neuroendocrine regulation of bone mass via leptin. BONE DISEASE The diseases produced by selective abnormalities of the cells and processes discussed above illustrate the interplay of fac-tors that maintain normal bone function. In osteopetrosis, a rare and often severe disease, the osteo-clasts are defective and are unable to resorb bone in their usual fashion so the osteoblasts operate unopposed. The result is a steady increase in bone density, neurologic defects due to narrowing and distortion of foramina through which nerves normally pass, and hematologic abnormalities due to crowding out of the marrow cavities. Mice lacking the protein encoded by the immediate-early gene c-fos develop osteo-petrosis, and osteopetrosis also occurs in mice lacking the PU.1 transcription factor. This suggests that all these factors are involved in normal osteoclast development and function. On the other hand, osteoporosis is caused by a relative excess of osteoclastic function. Loss of bone matrix in this con-dition (Figure 23–13) is marked, and the incidence of fractures is increased. Fractures are particularly common in the distal forearm (Colles fracture), vertebral body, and hip. All of these areas have a high content of trabecular bone, and because tra-becular bone is more active metabolically, it is lost more rap-idly. Fractures of the vertebrae with compression cause kyphosis, with the production of a typical “widow’s hump” that is common in elderly women with osteoporosis. Fractures of FIGURE 23–11 Structure of a typical long bone before (left) and after (right) epiphysial closure. Note the rearrangement of cells and growth of the bone as the epiphysial plate closes (see text for details). FIGURE 23–12 Osteoclast resorbing bone. The edges of the cell are tightly sealed to bone, permitting secretion of acid from the ruffled apical membrane and consequent erosion of the bone under-neath the cell. Note the multiple nuclei (n) and mitochondria (mi). (Courtesy of R Baron.) Epiphysis Epiphysis Diaphysis Epiphysial plate Marrow cavity Compact bone Periosteum Trabecular bone Bone-resorbing compartment Ruffled apical membrane Sealing zone Integrins Basolateral membrane Bone matrix n n n mi n 374 SECTION IV Endocrine & Reproductive Physiology the hip in elderly individuals are associated with a mortality rate of 12–20%, and half of those who survive require pro-longed expensive care. Osteoporosis has multiple causes, but by far the most com-mon form is involutional osteoporosis. All normal humans gain bone early in life, during growth. After a plateau, they begin to lose bone as they grow older (Figure 23–14). When this loss is accelerated or exaggerated, it leads to osteoporosis (see Clinical Box 23–4). Increased intake of calcium, particu-larly from natural sources such as milk, and moderate exercise may help prevent or slow the progress of osteoporosis, although their effects are not great. Bisphosphonates such as etidronate, which inhibit osteoclastic activity, increase the mineral content of bone when administered in a cyclic fash-ion and decrease the rate of new vertebral fractures. Fluoride stimulates osteoblasts, making bone more dense, but it has proved to be of little value in the treatment of the disease. CHAPTER SUMMARY ■Circulating levels of calcium and phosphate ions are controlled by cells that sense the levels of these electrolytes in the blood and release hormones, and effects of these hormones are evident in mobilization of the minerals from the bones, intestinal absorp-tion, and/or renal wasting. ■The majority of the calcium in the body is stored in the bones but it is the free, ionized calcium in the cells and extracellular fluids that fulfills physiological roles in cell signaling, nerve function, muscle contraction, and blood coagulation, among others. ■Phosphate is likewise predominantly stored in the bones and reg-ulated by many of the same factors that influence calcium levels. ■The two major hormones regulating calcium and phosphate ho-meostasis are 1,25-dihydroxycholecalciferol (a derivative of vi-tamin D) and parathyroid hormone; calcitonin is also capable of regulating levels of these ions, but its full physiologic contribu-tion is unclear. ■1,25-dihydroxycholecalciferol acts to elevate plasma calcium and phosphate by predominantly transcriptional mechanisms, whereas parathyroid hormone elevates calcium but decreases FIGURE 23–13 Normal trabecular bone (left) compared with trabecular bone from a patient with osteoporosis (right). The loss of mass in osteoporosis leaves bones more susceptible to breakage. FIGURE 23–14 Total body calcium, an index of bone mass, at various ages in men and women. Note the rapid increase to young adult levels (phase I) followed by the steady loss of bone with advancing age in both sexes (phase III) and the superimposed rapid loss in women after menopause (phase II). (Reproduced by permission of Oxford University Press from Riggs BL, Melton LJ III: Involutional osteoporosis. In Evans TG, Williams TF (editors): Oxford Textbook of Geriatric Medicine. Oxford University Press, 1992.) 1500 1000 500 0 0 20 40 60 80 100 Age (years) Total body calcium (grams) I I II III III CLINICAL BOX 23–4 Osteoporosis Adult women have less bone mass than adult men, and after menopause they initially lose it more rapidly than men of comparable age do. Consequently, they are more prone to development of serious osteoporosis. The cause of the bone loss after menopause is primarily estrogen deficiency, and estrogen treatment arrests the progress of the disease. Estrogens inhibit secretion of cytokines such as interleukin-1 (IL-1), IL-6, and tumor necrosis factor (TNF-_), and these cyto-kines foster the development of osteoclasts. Estrogen also stimulates production of transforming growth factor (TGF-), and this cytokine increases apoptosis of osteoclasts. How-ever, it now appears that even small doses of estrogens may increase the incidence of uterine and breast cancer, and in carefully controlled studies, estrogens do not protect against cardiovascular disease. Therefore, the decision to treat a postmenopausal woman with estrogens depends on a care-ful weighing of the risk–benefit ratio. Bone loss can also occur in both men and women as a result of inactivity. In pa-tients who are immobilized for any reason, and during space flight, bone resorption exceeds bone formation and disuse osteoporosis develops. The plasma calcium level is not mark-edly elevated, but plasma concentrations of parathyroid hormone and 1,25-dihydroxycholecalciferol fall and large amounts of calcium are lost in the urine. CHAPTER 23 Hormonal Control of Calcium & Phosphate Metabolism & the Physiology of Bone 375 phosphate by increasing the latter’s renal excretion. Calcitonin lowers both calcium and phosphate levels. ■Deficiencies of 1,25-dihydroxycholecalciferol or mutations in its receptor, lead to decreases in circulating calcium, defective calcification of the bones, and bone weakness. Disease states also result from either deficiencies or overproduction of parathyroid hormone, with reciprocal effects on calcium and phosphate. ■Bone is a highly structured mass with outer cortical and inner trabecular layers. The larger cortical layer has a high surface to volume layer with haversian canals that provide nutrients and gaps (lacunae) inhabited by bone cells that are connected by a canaliculi network. The smaller trabecular layer has a much higher surface to volume layer that relies on diffusion for nutri-ents supply. ■Regulated bone growth through puberty occurs through epiphy-sial plates. These plates are located near the end of the bone shaft and fuse with the shaft of the bone to cease linear bone growth. ■Bone is constantly remodeled by osteoclasts, which erode and absorb bone, and osteoblasts, which lay down new bone. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. A patient with parathyroid deficiency 10 days after inadvertent damage to the parathyroid glands during thyroid surgery would probably have A) low plasma phosphate and Ca2+ levels and tetany. B) low plasma phosphate and Ca2+ levels and tetanus. C) a low plasma Ca2+ level, increased muscular excitability, and a characteristic spasm of the muscles of the upper extremity (Trousseau sign). D) high plasma phosphate and Ca2+ levels and bone demineral-ization. E) increased muscular excitability, a high plasma Ca2+ level, and bone demineralization. 2. A high plasma Ca2+ level causes A) bone demineralization. B) increased formation of 1,25-dihydroxycholecalciferol. C) decreased secretion of calcitonin. D) decreased blood coagulability. E) increased formation of 24,25-dihydroxycholecalciferol. 3. Which of the following is not involved in regulating plasma Ca2+ levels? A) kidneys B) skin C) liver D) lungs E) intestine 4. 1,25-dihydroxycholecalciferol affects intestinal Ca2+ absorption through a mechanism that A) includes alterations in the activity of genes. B) activates adenylyl cyclase. C) decreases cell turnover. D) changes gastric acid secretion. E) is comparable to the action of polypeptide hormones. 5. Which of the following would you expect to find in a patient whose diet has been low in calcium for 2 mo? A) increased formation of 24,25-dihydroxycholecalciferol B) decreased amounts of calcium-binding protein in intestinal epithelial cells C) increased parathyroid hormone secretion D) a high plasma calcitonin concentration E) increased plasma phosphates 6. In osteopetrosis, which of the following is defective? A) phosphate deposition in trabecular bone B) structure of parathyroid hormone related protein (PTHrP) C) osteoblasts D) osteoclasts E) bone collagen 7. At epiphysial closure A) cortical bone and trabecular bone merge. B) osteoclasts and osteoblasts undergo differentiation. C) there is an extended amount of proliferating cartilage that contributes to bone elongation. D) lacunae meet the trabecular bone. E) ephyses unite with the shaft to end normal linear bone growth. CHAPTER RESOURCES Brown EM: The calcium-sensing receptor: Physiology, pathophysiology and CaR-based therapeutics. Subcell Biochem 2007;45:139. Murer H, Hernanado N, Forster L, Biber J: Molecular mechanisms in proximal tubular and small intestinal phosphate reabsorption. Mol Membr Biol 2001;18:3. Nijenhuis T, Hoenderop JGJ, Bindels RJM: TRPV5 and TRPV6 in Ca2+ (re)absorption: Regulating Ca2+ entry at the gate. Pflugers Arch Eur J Physiol 2005;451:181. Renkema KY, Alexander RT, Bindels FJ, Hoenderop JF: Calcium and phosphate homeostasis: Concerted interplay of new regulators. Ann Med 2008;40:82. This page intentionally left blank 377 C H A P T E R 24 The Pituitary Gland O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the structure of the pituitary gland and how it relates to its function. ■Define the cell types present in the anterior pituitary and understand how their numbers are controlled in response to physiologic demands. ■Understand the function of hormones derived from proopiomelanocortin in hu-mans, and how they are involved in regulating pigmentation in humans, other mammals, and lower vertebrates. ■Define the effects of the growth hormone in growth and metabolic function, and how insulin-like growth factor I (IGF-I) may mediate some of its actions in the periphery. ■List the stimuli that regulate growth hormone secretion and define their underly-ing mechanisms. ■Understand the basis of conditions where pituitary function and growth hormone secretion and function are abnormal, and how they can be treated. INTRODUCTION The pituitary gland, or hypophysis, lies in a pocket of the sphe-noid bone at the base of the brain. It is a coordinating center for control of many downstream endocrine glands, some of which are discussed in other chapters. In many ways, it can be considered to consist of at least two (and in some species, three) separate endocrine organs that contain a plethora of hormonally active substances. The anterior pituitary secretes thyroid-stimulating hormone (TSH, thyrotropin), adreno-corticotropic hormone (ACTH), luteinizing hormone (LH), follicle-stimulating hormone (FSH), prolactin, and growth hormone (see Figure 18–9), and receives almost all of its blood supply from the portal hypophysial vessels that pass ini-tially through the median eminence, a structure immediately below the hypothalamus. This vascular arrangement positions the cells of the anterior pituitary to respond efficiently to regu-latory factors released from the hypothalamus. Of the listed hormones, prolactin acts on the breast. The remaining five are, at least in part, tropic hormones; that is, they stimulate secre-tion of hormonally active substances by other endocrine glands or, in the case of growth hormone, the liver and other tissues (see below). The hormones tropic for a particular endo-crine gland are discussed in the chapter on that gland: TSH in Chapter 20; ACTH in Chapter 22; and the gonadotropins FSH and LH in Chapter 25, along with prolactin. The posterior pituitary in mammals consists predominantly of nerves that have their cell bodies in the hypothalamus, and stores oxytocin and vasopressin in the termini of these neurons, to be released into the bloodstream. The secretion of these hor-mones, as well as a discussion of the overall role of the hypothal-amus and median eminence in regulating both the anterior and posterior pituitary, were covered in Chapter 18. Finally, in some species there is a well-developed intermediate lobe of the pitu-itary, whereas in humans it is rudimentary. Nevertheless, the intermediate lobe, as well as the anterior pituitary, contain hor-monally active derivatives of the proopiomelanocortin molecule that regulate skin pigmentation, among other functions (see 378 SECTION IV Endocrine & Reproductive Physiology below). To avoid redundancy, this chapter will focus particularly on growth hormone and its role in growth and facilitating the activity of other hormones, along with a number of general con-siderations about the pituitary. The melanocyte-stimulating hor-mones (MSHs) of the intermediate lobe of the pituitary, α-MSH and β-MSH, will also be touched upon. MORPHOLOGY GROSS ANATOMY The anatomy of the pituitary gland is summarized in Figure 24–1 and discussed in detail in Chapter 18. The posterior pi-tuitary is made up largely of the endings of axons from the su-praoptic and paraventricular nuclei of the hypothalamus and arises initially as an extension of this structure. The anterior pituitary, on the other hand, contains endocrine cells that store its characteristic hormones and arises embryologically as an invagination of the pharynx (Rathke’s pouch). In species where it is well developed, the intermediate lobe is formed in the embryo from the dorsal half of Rathke’s pouch, but is closely adherent to the posterior lobe in the adult. It is separat-ed from the anterior lobe by the remains of the cavity in Rath-ke’s pouch, the residual cleft. HISTOLOGY In the posterior lobe, the endings of the supraoptic and paraventricular axons can be observed in close relation to blood vessels. Pituicytes, stellate cells that are modified astro-cytes, are also present. As noted above, the intermediate lobe is rudimentary in humans and a few other mammalian species. In these species, most of its cells are incorporated in the anterior lobe. Along the residual cleft are small thyroid-like follicles, some containing a little colloid. The function of the colloid, if any, is unknown. The anterior pituitary is made up of interlacing cell cords and an extensive network of sinusoidal capillaries. The endothelium of the capillaries is fenestrated, like that in other endocrine organs. The cells contain granules of stored hormone that are extruded from the cells by exocytosis. Their constituents then enter the capillaries to be conveyed to target tissues. CELL TYPES IN THE ANTERIOR PITUITARY Five types of secretory cells have been identified in the anterior pituitary by immunocytochemistry and electron microscopy. Traditionally, they were also characterized by their affinity for either acidic or basic histological stains. The cell types are the somatotropes, which secrete growth hormone; the lactotropes (also called mammotropes), which secrete prolactin; the corti-cotropes, which secrete ACTH; the thyrotropes, which secrete TSH; and the gonadotropes, which secrete FSH and LH. The characteristics of these cells are summarized in Table 24–1. Some cells may contain two or more hormones. It is also nota-ble that the three pituitary glycoprotein hormones, FSH, LH, and TSH, while being made up of two subunits, all share a com-mon α subunit that is the product of a single gene and has the same amino acid composition in each hormone, although their carbohydrate residues vary. The α subunit must be combined with a β subunit characteristic of each hormone for maximal physiologic activity. The β subunits, which are produced by sep-arate genes and differ in structure, confer hormonal specificity. The α subunits are remarkably interchangeable and hybrid molecules can be created. In addition, the placental glycopro-tein gonadotropin human chorionic gonadotropin (hCG) has α and β subunits (see Chapter 25). The anterior pituitary also contains folliculostellate cells that send processes between the granulated secretory cells. These cells produce paracrine factors that regulate the growth and function of the secretory cells discussed above. Indeed, the anterior pituitary can adjust the relative proportion of secretory cell types to meet varying requirements for different hormones at different life stages. This plasticity has recently been ascribed to the presence of a small number of pluripo-tent stem cells that persist in the adult gland. PROOPIOMELANOCORTIN & DERIVATIVES BIOSYNTHESIS Intermediate-lobe cells and corticotropes of the anterior lobe both synthesize a large precursor protein that is cleaved to form a family of hormones. After removal of the signal pep-tide, this prohormone is known as proopiomelanocortin (POMC). This molecule is also synthesized in the hypothala-mus, the lungs, the gastrointestinal tract, and the placenta. The structure of POMC, as well as its derivatives, is shown in Fig-ure 24–2. In corticotropes, it is hydrolyzed to ACTH and a polypeptide of unknown function called β-lipotropin (LPH), FIGURE 24–1 Diagrammatic outline of the formation of the pituitary (left) and the various parts of the organ in the adult (right). Pars tuberalis Anterior lobe Posterior lobe Intermediate lobe Rathke’s pouch Third ventricle Third ventricle CHAPTER 24 The Pituitary Gland 379 plus a small amount of β-endorphin, and these substances are secreted. In the intermediate lobe cells, POMC is hydrolyzed to corticotropin-like intermediate-lobe peptide (CLIP), γ-LPH, and appreciable quantities of β-endorphin. The func-tions, if any, of CLIP and γ-LPH are unknown, whereas β-en-dorphin is an opioid peptide (see Chapter 7) that has the five amino acid residues of met-enkephalin at its amino terminal end. The melanotropins α- and β-MSH are also formed. However, the intermediate lobe in humans is rudimentary, and it appears that neither α-MSH nor β-MSH is secreted in adults. In some species, however, the melanotropins have im-portant physiological functions, as discussed below. CONTROL OF SKIN COLORATION & PIGMENT ABNORMALITIES Fish, reptiles, and amphibia change the color of their skin for thermoregulation, camouflage, and behavioral displays. They do this in part by moving black or brown granules into or out of the periphery of pigment cells called melanophores. The granules are made up of melanins, which are synthesized from dopamine (see Chapter 7) and dopaquinone. The move-ment of these granules is controlled by a variety of hormones and neurotransmitters, including α- and β-MSH, melanin-concentrating hormone, melatonin, and catecholamines. Mammals have no melanophores containing pigment gran-ules that disperse and aggregate, but they do have melanocytes, which have multiple processes containing melanin granules. Melanocytes express melanotropin-1 receptors. Treatment with MSHs accelerates melanin synthesis, causing readily detectable darkening of the skin in humans in 24 h. As noted TABLE 24–1 Hormone-secreting cells of the human anterior pituitary gland. Cell Type Hormones Secreted % of Total Secretory Cells Stain Affinity Diameter of Secretory Granules (nm) Somatotrope Growth hormone 50 Acidophilic 300–400 Lactotrope Prolactin 10–30 Acidophilic 200 Corticotrope ACTH 10 Basophilic 400–550 Thyrotrope TSH 5 Basophilic 120–200 Gonadotrope FSH, LH 20 Basophilic 250–400 FIGURE 24–2 Schematic representation of the preproopiomelanocortin molecule formed in pituitary cells, neurons, and other tissues. The numbers in parentheses identify the amino acid sequences in each of the polypeptide fragments. For convenience, the amino acid sequences are numbered from the amino terminal of ACTH and read toward the carboxyl terminal portion of the parent molecule, whereas the amino acid sequences in the other portion of the molecule read to the left to –131, the amino terminal of the parent molecule. The locations of Lys–Arg and other pairs of basic amino acids residues are also indicated; these are the sites of proteolytic cleavage in the formation of the smaller fragments of the parent molecule. AL, anterior lobe; IL, intermediate lobe. Signal peptide (–131) γ−MSH (–55 to –44) β-LPH (42–134) β-LPH ACTH (1–39) AL and IL γ−MSH α-MSH (1–13) β-MSH (84–101) Met-enkephalin (104–108) γ-LPH (42–101) β-Endorphin (104–134) CLIP (18–39) Arg-Lys Arg-Arg Lys-Arg Lys-Arg Lys-Arg Lys-Arg Lys-Lys Lys-Lys Lys-Lys Arg-Arg Fast in IL Slow in AL ACTH IL only AL and IL Amino terminal fragment 380 SECTION IV Endocrine & Reproductive Physiology above, α- and β-MSH do not circulate in adult humans, and their function is unknown. However, ACTH binds to melan-otropin-1 receptors. Indeed, the pigmentary changes in sev-eral human endocrine diseases are due to changes in circulating ACTH. For example, abnormal pallor is a hall-mark of hypopituitarism. Hyperpigmentation occurs in patients with adrenal insufficiency due to primary adrenal disease. Indeed, the presence of hyperpigmentation in associ-ation with adrenal insufficiency rules out the possibility that the insufficiency is secondary to pituitary or hypothalamic disease because in these conditions, plasma ACTH is not increased (see Chapter 22). Other disorders of pigmentation result from peripheral mechanisms. Thus, albinos have a con-genital inability to synthesize melanin. This can result from a variety of different genetic defects in the pathways for mela-nin synthesis. Piebaldism is characterized by patches of skin that lack melanin as a result of congenital defects in the migration of pigment cell precursors from the neural crest during embryonic development. Not only the condition but also the precise pattern of the loss is passed from one genera-tion to the next. Vitiligo involves a similar patchy loss of mel-anin, but the loss develops progressively after birth secondary to an autoimmune process that targets melanocytes. GROWTH HORMONE BIOSYNTHESIS & CHEMISTRY The long arm of human chromosome 17 contains the growth hormone-hCS cluster that contains five genes: one, hGH-N, codes for the most abundant (“normal”) form of growth hor-mone; a second, hGH-V, codes for the variant form of growth hormone (see below); two code for human chorionic so-matomammotropin (hCS) (see Chapter 25); and the fifth is probably an hCS pseudogene. The structure of hGH-N is shown in Figure 24–3, where it is also compared with that of hCS. Growth hormone that is secreted into the circulation by the pituitary gland consists of a complex mixture of hGH-N, peptides derived from this molecule with varying degrees of post-translational modifica-tions, such as glycosylation, and a splice variant of hGH-N that lacks amino acids 32–46. The physiologic significance of this complex array of hormones has yet to be fully under-stood, particularly since their structural similarities make it difficult to assay for each species separately. Nevertheless, there is emerging evidence that while the various peptides share a broad range of functions, they may occasionally exert actions in opposition to one another. hGH-V and hCS, on the other hand, are primarily products of the placenta, and as a FIGURE 24–3 Structure of the principal human growth hormone (continuous chain). The red bars indicate disulfide bridges. The 29 residues alongside the chain identify residues that differ in human chorionic somatomammotropin (hCS; see Chapter 23). All the other residues in hCS are the same, and hCS also has 191 amino acid residues. S S L V Y N V M Q R F A N F L L W S Gin I L L L S I R L L Q L N S K Q Q T E E R M N S P T P I S E S D E E E P R 95 V 90 85 80 75 70 65 60 55 100 G A S N D D S D V D Y D L L K D L E E H T 105 110 35 30 25 20 15 10 5 1 115 G I Q T L 120 M G R L E D G S P R R T G Q I F K Q T Y S K F D T N S H N D H D A L L K N Y G L L Y C C L S T Q P N S D H Q L F S Y K Q E K P I Y A T E E F E Q Y T D F A L Q H L A R H A R L M A D F L R S L P I T P F V Q V Q N H I D F F F R K D M D K V E T F L R C Q V I M C G F S G E V S R L 125 130 135 140 145 155 160 165 170 50 45 40 175 180 185 150 191 COOH NH2 CHAPTER 24 The Pituitary Gland 381 consequence are only found in appreciable quantities in the circulation during pregnancy (see Chapter 25). SPECIES SPECIFICITY The structure of growth hormone varies considerably from one species to another. Porcine and simian growth hormones have only a transient effect in the guinea pig. In monkeys and humans, bovine and porcine growth hormones do not even have a transient effect on growth, although monkey and hu-man growth hormones are fully active in both monkeys and humans. These facts are relevant to public health discussions surrounding the presence of bovine growth hormones (used to increase milk production) in dairy products, as well as the popularity of growth hormone supplements, marketed via the Internet, with body builders. Controversially, recombinant human growth hormone has also been given to children who are short in stature, but otherwise healthy (ie, without growth hormone deficiency), with apparently limited results. PLASMA LEVELS, BINDING, & METABOLISM A portion of circulating growth hormone is bound to a plasma protein that is a large fragment of the extracellular domain of the growth hormone receptor (see below). It appears to be produced by cleavage of receptors in humans, and its concen-tration is an index of the number of growth hormone recep-tors in the tissues. Approximately 50% of the circulating pool of growth hormone activity is in the bound form, providing a reservoir of the hormone to compensate for the wide fluctua-tions that occur in secretion (see below). The basal plasma growth hormone level measured by radioimmunoassay in adult humans is normally less than 3 ng/mL. This represents both the protein-bound and free forms. Growth hormone is metabolized rapidly, probably at least in part in the liver. The half-life of circulating growth hormone in humans is 6–20 min, and the daily growth hor-mone output has been calculated to be 0.2–1.0 mg/d in adults. GROWTH HORMONE RECEPTORS The growth hormone receptor is a 620-amino-acid protein with a large extracellular portion, a transmembrane domain, and a large cytoplasmic portion. It is a member of the cytokine receptor superfamily, which is discussed in Chapter 3. Growth hormone has two domains that can bind to its receptor, and when it binds to one receptor, the second binding site attracts another, producing a homodimer (Figure 24–4). Dimeriza-tion is essential for receptor activation. FIGURE 24–4 Some of the principal signaling pathways activated by the dimerized growth hormone receptor (GHR). Solid arrows in-dicate established pathways; dashed arrows indicate probable pathways. The details of the PLC pathway and the pathway from Grb2 to MAP K are shown in Chapter 2. GLE-1 and GLE-2, interferon γ-activated response elements; IRS, insulin receptor substrate; p90RSK, an S6 kinase; PLA2, phospho-lipase A2; SIE, Sis-induced element; SRE, serum response element; SRF, serum response factor; TCF, ternary complex factor. The variants of IGR-II are also shown: a 21-amino-acid extension of the carboxyl terminal, a tetrapeptide substitution at Ser-29, and a tripeptide substitution of Ser-33. P P P P P P P P GH GHR JAK2 IRS P P SHC DAG PKC P MAP K Grb2 STATs STAT5 STAT1 STAT3 SRF SRF TCF PLC p90RSK PLA2 GLE-2 P450-3A10 GLE-1 Spi 2.1 SIE SRE c-fos STAT5 Ca2+ 382 SECTION IV Endocrine & Reproductive Physiology Growth hormone has widespread effects in the body (see below), so even though it is not yet possible precisely to corre-late intracellular and whole body effects, it is not surprising that, like insulin, growth hormone activates many different intracellular signaling cascades (Figure 24–4). Of particular note is its activation of the JAK2–STAT pathway. JAK2 is a member of the Janus family of cytoplasmic tyrosine kinases. STATs (for signal transducers and activators of transcription) are a family of inactive cytoplasmic transcription factors that, upon phosphorylation by JAK kinases, migrate to the nucleus and activate various genes. JAK–STAT pathways are known also to mediate the effects of prolactin and various other growth factors. EFFECTS ON GROWTH In young animals in which the epiphyses have not yet fused to the long bones (see Chapter 23), growth is inhibited by hypo-physectomy and stimulated by growth hormone. Chondrogen-esis is accelerated, and as the cartilaginous epiphysial plates widen, they lay down more bone matrix at the ends of long bones. In this way, stature is increased. Prolonged treatment of animals with growth hormone leads to gigantism. When the epiphyses are closed, linear growth is no longer possible. In this case, an overabundance of growth hormone produces the pattern of bone and soft tissue deformities known in humans as acromegaly. The sizes of most of the viscera are increased. The protein content of the body is increased, and the fat content is decreased (see Clinical Box 24–1). EFFECTS ON PROTEIN & ELECTROLYTE METABOLISM Growth hormone is a protein anabolic hormone and produces a positive nitrogen and phosphorus balance, a rise in plasma phosphorus, and a fall in blood urea nitrogen and amino acid levels. In adults with growth hormone deficiency, recombi-nant human growth hormone produces an increase in lean body mass and a decrease in body fat, along with an increase in metabolic rate and a fall in plasma cholesterol. Gastrointes-tinal absorption of Ca2+ is increased. Na+ and K+ excretion is reduced by an action independent of the adrenal glands, prob-ably because these electrolytes are diverted from the kidneys to the growing tissues. On the other hand, excretion of the amino acid 4-hydroxyproline is increased during this growth, reflec-tive of the ability of growth hormone to stimulate the synthesis of soluble collagen. EFFECTS ON CARBOHYDRATE & FAT METABOLISM The actions of growth hormone on carbohydrate metabolism are discussed in Chapter 21. At least some forms of growth hormone are diabetogenic because they increase hepatic glu-cose output and exert an anti-insulin effect in muscle. Growth hormone is also ketogenic and increases circulating free fatty acid (FFA) levels. The increase in plasma FFA, which takes several hours to develop, provides a ready source of energy for the tissues during hypoglycemia, fasting, and stressful stimuli. Growth hormone does not stimulate beta cells of the pancreas directly, but it increases the ability of the pancreas to respond to insulinogenic stimuli such as arginine and glucose. This is an additional way growth hormone promotes growth, since insulin has a protein anabolic effect (see Chapter 21). SOMATOMEDINS The effects of growth hormone on growth, cartilage, and protein metabolism depend on an interaction between growth hormone and somatomedins, which are polypeptide growth factors secret-ed by the liver and other tissues. The first of these factors isolated was called sulfation factor because it stimulated the incorporation of sulfate into cartilage. However, it also stimulated collagen for-mation, and its name was changed to somatomedin. It then be-came clear that there are a variety of different somatomedins and that they are members of an increasingly large family of growth factors that affect many different tissues and organs. The principal (and in humans probably the only) circulat-ing somatomedins are insulin-like growth factor I (IGF-I, somatomedin C) and insulin-like growth factor II (IGF-II). These factors are closely related to insulin, except that their C CLINICAL BOX 24–1 Gigantism & Acromegaly Tumors of the somatotropes of the anterior pituitary (pitu-itary adenoma) secrete large amounts of growth hormone, leading in children to gigantism and in adults to acromeg-aly. If the tumor arises before puberty, the individual may grow to an extraordinary height. After linear growth is no longer possible, on the other hand, the characteristic fea-tures of acromegaly arise, including greatly enlarged hands and feet, vertebral changes attributable to osteoarthritis, soft tissue swelling, hirsutism, and protrusion of the brow and jaw. Abnormal growth of internal organs may eventually im-pair their function such that the condition, which has an in-sidious onset, can prove fatal if left untreated. Hypersecre-tion of growth hormone is accompanied by hypersecretion of prolactin in 20–40% of patients with acromegaly. About 25% of patients have abnormal glucose tolerance tests, and 4% develop lactation in the absence of pregnancy. Acromeg-aly can be caused by extra-pituitary as well as intrapituitary growth hormone-secreting tumors and by hypothalamic tu-mors that secrete GHRH, but the latter are rare. Treatment in-volves surgical removal of the tumor where possible, the use of long-acting analogues of somatostatin, or both. CHAPTER 24 The Pituitary Gland 383 chains are not separated (Figure 24–5) and they have an extension of the A chain called the D domain. The hormone relaxin (see Chapter 25) is also a member of this family. Humans have two related relaxin isoforms, and both resemble IGF-II. In humans a variant form of IGF-I lacking three amino terminal amino acid residues has been found in the brain, and there are several variant forms of human IGF-II (Figure 24–5). The mRNAs for IGF-I and IGF-II are found in the liver, in cartilage, and in many other tissues, indicating that they are synthesized in these tissues. The properties of IGF-I, IGF-II, and insulin are compared in Table 24–2. Both are tightly bound to proteins in the plasma, and, at least for IGF-I, this prolongs their half-life in the circu-lation. Six different IGF-binding proteins, with different pat-terns of distribution in various tissues, have been identified. All are present in plasma, with IGF-binding protein-3 (IGFBP-3) accounting for 95% of the binding in the circulation. The contribution of the IGFs to the insulin-like activity in blood is discussed in Chapter 21. The IGF-I receptor is very similar to the insulin receptor and probably uses similar or identical intracellular signaling pathways. The IGF-II receptor has a FIGURE 24-5 Structure of human IGF-I, IGF-II, and insulin (ins) (top). The lower panel shows the structure of human IGF-II with its disul-fide bonds, as well as three variant structures shown: a 21-aa extension of the c-terminus, a tetrapeptide substitution at Ser-29, and a tripeptide substitution of Ser-33. GPETLCGAELVDALQFVCGDRGFYFNKPTGYGSSSRRAPQTGIVDECCFRSCDLRRLEMYCAPLKPAKSA AYRPSETLCGGELVDTLQFVCGDRGFYFSRPA--SRVSRRSR--GIVEECCFRSCDLALLETYCAT--PAKSE FVNQHLCGSHLVEALYLVCGERGFFYTPKT GIVEQCCTSICSLYQLENYCN hlGF-I hlGF-II h ins I II II II I B C A D C C C G C D C E E E L L P P P P L A A A A A K S E 1 21 T T V I T Y Y C Q C G G G G G G E E S S S S R R R R R R R R S Y L L L V V V D D D T L F F F F R S TABLE 24–2 Comparison of insulin and the insulin-like growth factors. Insulin IGF-I IGF-II Other names … Somatomedin C Multiplication-stimulating activity (MSA) Number of amino acids 51 70 67 Source Pancreatic B cells Liver and other tissues Diverse tissues Level regulated by Glucose Growth hormone after birth, nutritional status Unknown Plasma levels 0.3–2 ng/mL 10–700 ng/mL; peaks at puberty 300–800 ng/mL Plasma-binding proteins No Yes Yes Major physiologic role Control of metabolism Skeletal and cartilage growth Growth during fetal development 384 SECTION IV Endocrine & Reproductive Physiology distinct structure (see Figure 21–5) and is involved in the intracellular targeting of acid hydrolases and other proteins to intracellular organelles. Secretion of IGF-I is independent of growth hormone before birth but is stimulated by growth hor-mone after birth, and it has pronounced growth-stimulating activity. Its concentration in plasma rises during childhood and peaks at the time of puberty, then declines to low levels in old age. IGF-II is largely independent of growth hormone and plays a role in the growth of the fetus before birth. In human fetuses in which it is overexpressed, growth of organs, espe-cially the tongue, other muscles, kidneys, heart, and liver, is disproportionate. In adults, the gene for IGF-II is expressed only in the choroid plexus and meninges. DIRECT & INDIRECT ACTIONS OF GROWTH HORMONE Our understanding of the mechanism of action of growth hor-mone has evolved recently as new information has become available. Growth hormone was originally thought to produce growth by a direct action on tissues, then later was believed to act solely through its ability to induce somatomedins. Howev-er, if growth hormone is injected into one proximal tibial epi-physis, a unilateral increase in cartilage width is produced, and cartilage, like other tissues, makes IGF-I. A current hypothesis to explain these results holds that growth hormone acts on car-tilage to convert stem cells into cells that respond to IGF-I and then locally produced and circulating IGF-I makes the cartilage grow. However, the independent role of circulating IGF-I re-mains important, since infusion of IGF-I to hypophysecto-mized rats restores bone and body growth. Overall, it seems that growth hormone and somatomedins can act both in coop-eration and independently to stimulate pathways that lead to growth. The situation is almost certainly complicated further by the existence of multiple forms of growth hormone in the circulation that can, in some situations, have opposing actions. Figure 24–6 is a summary of current views of the other actions of growth hormone and IGF-I. However, growth hor-mone probably combines with circulating and locally pro-duced IGF-I in various proportions to produce at least some of these effects. Indeed, while the mainstay of therapy for acromegaly remains somatostatin analogues that inhibit the secretion of growth hormone, a growth hormone receptor antagonist has recently become available and has been found to reduce plasma IGF-I and produce clinical improvement in cases of acromegaly that fail to respond to other treatments. HYPOTHALAMIC & PERIPHERAL CONTROL OF GROWTH HORMONE SECRETION The secretion of growth hormone is not stable over time. Ado-lescents have the highest circulating levels of growth hormone, followed by children and finally adults. Levels decline in old age, and there has been considerable interest in injecting growth hormone to counterbalance the effects of aging. The hormone increases lean body mass and decreases body fat, but it does not produce statistically significant increases in muscle strength or mental status. There are also diurnal variations in growth hor-mone secretion superimposed on these developmental stages. Growth hormone is found at relatively low levels during the day, unless specific triggers for its release are present (see be-low). During sleep, on the other hand, large pulsatile bursts of growth hormone secretion occur. Therefore, it is not surprising that the secretion of growth hormone is under hypothalamic control. The hypothalamus controls growth hormone produc-tion by secreting growth hormone-releasing hormone (GHRH) as well as somatostatin, which inhibits growth hormone release (see Chapter 18). Thus, the balance between the effects of these hypothalamic factors on the pituitary will determine the level of growth hormone release. The stimuli of growth hormone se-cretion discussed as follows can therefore act by increasing hypothalamic secretion of GHRH, decreasing secretion of so-matostatin, or both. A third regulator of growth hormone secre-tion is ghrelin. The main site of ghrelin synthesis and secretion is the stomach, but it is also produced in the hypothalamus and has marked growth hormone-stimulating activity. In addition, it appears to be involved in the regulation of food intake. Growth hormone secretion is under feedback control, like the secretion of other anterior pituitary hormones. It acts on the hypothalamus to antagonize GHRH release. Growth hormone also increases circulating IGF-I, and IGF-I in turn exerts a direct inhibitory action on growth hormone secretion from the pitu-itary. It also stimulates somatostatin secretion (Figure 24–7). STIMULI AFFECTING GROWTH HORMONE SECRETION The basal plasma growth hormone concentration ranges from 0–3 ng/mL in normal adults. However, secretory rates cannot be estimated from single values because of their irregular na-ture. Thus, average values over 24 h (see below) and peak val-ues may be more meaningful, albeit difficult to assess in the clinical setting. The stimuli that increase growth hormone FIGURE 24–6 Actions believed to be mediated by growth hormone (GH) and IGF-I. (Courtesy of R Clark and N Gesundheit.) GH Na+ retention Decreased insulin sensitivity Lipolysis Protein synthesis Epiphysial growth IGF-I Insulin-like activity Antilipolytic activity Protein synthesis Epiphysial growth CHAPTER 24 The Pituitary Gland 385 secretion are summarized in Table 24–3. Most of them fall into three general categories: (1) conditions such as hypoglycemia and fasting in which there is an actual or threatened decrease in the substrate for energy production in cells, (2) conditions in which the amounts of certain amino acids are increased in the plasma, and (3) stressful stimuli. The response to glucagon has been used as a test of growth hormone reserve. Growth hor-mone secretion is also increased in subjects deprived of rapid eye movement (REM) sleep (see Chapter 15) and inhibited during normal REM sleep. Glucose infusions lower plasma growth hormone levels and inhibit the response to exercise. The increase produced by 2-deoxyglucose is presumably due to intracellular glucose defi-ciency, since this compound blocks the catabolism of glucose 6-phosphate. Sex hormones induce growth hormone secre-tion, increase growth hormone responses to provocative stim-uli such as arginine and insulin, and also serve as permissive factors for the action of growth hormone in the periphery. This likely contributes to the relatively high levels of circulat-ing growth hormone and associated growth spurt in puberty. Growth hormone secretion is also induced by thyroid hor-mones. Growth hormone secretion is inhibited, on the other hand, by cortisol, FFA, and medroxyprogesterone. Growth hormone secretion is increased by L-dopa, which increases the release of dopamine and norepinephrine in the brain, and by the dopamine receptor agonist apomorphine. PHYSIOLOGY OF GROWTH Growth hormone, while being essentially unimportant for fetal development, is the most important hormone for postnatal growth. However, growth overall is a complex phenomenon that is affected not only by growth hormone and somatome-dins, but also, as would be predicted by the previous discussion, by thyroid hormones, androgens, estrogens, glucocorticoids, and insulin. It is also affected, of course, by genetic factors, and it depends on adequate nutrition. It is normally accompanied by an orderly sequence of maturational changes, and it involves accretion of protein and increase in length and size, not just an increase in weight (which could reflect the formation of fat or retention of salt and water rather than growth per se). ROLE OF NUTRITION The food supply is the most important extrinsic factor affect-ing growth. The diet must be adequate not only in protein content but also in essential vitamins and minerals (see Chap-ter 27) and in calories, so that ingested protein is not burned for energy. However, the age at which a dietary deficiency FIGURE 24–7 Feedback control of growth hormone secretion. Solid arrows represent positive effects and dashed arrows represent inhibition. SS Anterior pituitary Liver (and other organs) GH Tissues IGF-I GHRH Hypothalamus Ghrelin TABLE 24–3 Stimuli that affect growth hormone secretion in humans. Stimuli that increase secretion Hypoglycemia 2-Deoxyglucose Exercise Fasting Increase in circulating levels of certain amino acids Protein meal Infusion of arginine and some other amino acids Glucagon Stressful stimuli Pyrogen Lysine vasopressin Various psychologic stresses Going to sleep L-Dopa and α-adrenergic agonists that penetrate the brain Apomorphine and other dopamine receptor agonists Estrogens and androgens Stimuli that decrease secretion REM sleep Glucose Cortisol FFA Medroxyprogesterone Growth hormone and IGF-I 386 SECTION IV Endocrine & Reproductive Physiology occurs appears to be an important consideration. For exam-ple, once the pubertal growth spurt has commenced, consid-erable linear growth continues even if caloric intake is reduced. Injury and disease likewise stunt growth because they increase protein catabolism. GROWTH PERIODS Patterns of growth vary somewhat from species to species. Rats continue to grow, although at a declining rate, through-out life. In humans, two periods of rapid growth occur (Figure 24–8): the first in infancy and the second in late puberty just before growth stops. The first period of accelerated growth is partly a continuation of the fetal growth period. The second growth spurt, at the time of puberty, is due to growth hor-mone, androgens, and estrogens, and the subsequent cessa-tion of growth is due in large part to closure of the epiphyses in the long bones by estrogens (see Chapter 25). After this time, further increases in height are not possible. Because girls mature earlier than boys, this growth spurt appears earlier in girls. Of course, in both sexes the rate of growth of individual tissues varies (Figure 24–9). It is interesting that at least during infancy, growth is not a con-tinuous process but is episodic or saltatory. Increases in length of human infants of 0.5 to 2.5 cm in a few days are separated by periods of 2 to 63 d during which no measurable growth can be detected. The cause of the episodic growth is unknown. HORMONAL EFFECTS The contributions of hormones to growth after birth are shown diagrammatically in Figure 24–10. Plasma growth hor-mone is elevated in newborns. Subsequently, average resting levels fall but the spikes of growth hormone secretion are larg-er, especially during puberty, so the mean plasma level over 24 h is increased; it is 2 to 4 ng/mL in normal adults, but 5 to 8 ng/mL in children. One of the factors stimulating IGF-I se-cretion is growth hormone, and plasma IGF-I levels rise during childhood, reaching a peak at 13 to 17 years of age. In contrast, IGF-II levels are constant throughout postnatal growth. The growth spurt that occurs at the time of puberty (Figure 24–8) is due in part to the protein anabolic effect of andro-gens, and the secretion of adrenal androgens increases at this time in both sexes; however, it is also due to an interaction among sex steroids, growth hormone, and IGF-I. Treatment FIGURE 24–8 Rate of growth in boys and girls from birth to age 20. 20 18 16 14 12 10 Girls Height gain (cm/yr) Boys Age in years 8 6 4 2 0 5.1 10.2 15.2 20.3 25.4 FIGURE 24–9 Growth of different tissues at various ages as a percentage of size at age 20. The curves are composites that include data for both boys and girls. FIGURE 24–10 Relative importance of hormones in human growth at various ages. (Courtesy of DA Fisher.) 20 18 16 10 12 14 Brain and head Percent of size at age 20 Body and most visceral organs Age in years 8 6 4 2 0 20 40 60 80 100 120 140 160 180 200 Lymphoid tissue Reproductive organs Birth 2 4 8 10 12 14 16 18 20 6 Thyroid hormones Growth hormone Androgens and estrogens Age (years) CHAPTER 24 The Pituitary Gland 387 with estrogens and androgens increases the secretion of growth hormone in response to various stimuli and increases plasma IGF-I secondary to this increase in circulating growth hormone. This, in turn, causes growth. Although androgens and estrogens initially stimulate growth, estrogens ultimately terminate growth by causing the epiphyses to fuse to the long bones (epiphysial closure). Once the epiphyses have closed, linear growth ceases (see Chapter 23). This is why patients with sexual precocity are apt to be dwarfed. On the other hand, men who were castrated before puberty tend to be tall because their estrogen production is decreased and their epiphyses remain open, allowing some growth to continue past the normal age of puberty. When growth hormone is administered to hypophysecto-mized animals, the animals do not grow as rapidly as they do when treated with growth hormone plus thyroid hormones. Thyroid hormones alone have no effect on growth in this situ-ation. Their action is therefore permissive to that of growth hormone, possibly via potentiation of the actions of somatomedins. Thyroid hormones also appear to be necessary for a completely normal rate of growth hormone secretion; basal growth hormone levels are normal in hypothyroidism, but the response to hypoglycemia is frequently blunted. Thy-roid hormones have widespread effects on the ossification of cartilage, the growth of teeth, the contours of the face, and the proportions of the body. Hypothyroid dwarfs (also known as cretins) therefore have infantile features (Figure 24–11). Patients who are dwarfed because of panhypopituitarism have features consistent with their chronologic age until puberty, but since they do not mature sexually, they have juvenile fea-tures in adulthood (Clinical Box 24–2). The effect of insulin on growth is discussed in Chapter 21. Diabetic animals fail to grow, and insulin causes growth in hypophysectomized animals. However, the growth is appre-ciable only when large amounts of carbohydrate and protein are supplied with the insulin. Adrenocortical hormones other than androgens exert a permissive action on growth in the sense that adrenalecto-mized animals fail to grow unless their blood pressures and circulations are maintained by replacement therapy. On the other hand, glucocorticoids are potent inhibitors of growth because of their direct action on cells, and treatment of chil-dren with pharmacologic doses of steroids slows or stops growth for as long as the treatment is continued. CATCH-UP GROWTH Following illness or starvation in children, a period of catch-up growth (Figure 24–12) takes place during which the growth rate is greater than normal. The accelerated growth FIGURE 24–11 Normal and abnormal growth. Hypothyroid dwarfs (cretins) retain their infantile proportions, whereas dwarfs of the con-stitutional type and, to a lesser extent, of the hypopituitary type have proportions characteristic of their chronologic age. (Reproduced, with permission, from Wilkins L: The Diagnosis and Treatment of Endocrine Disorders in Childhood and Adolescence, 3rd ed. Thomas, 1966.) Level of symphysis Normal 2 years Normal 8 years Hypothyroid 8 years Dwarf–not hypothyroid 8 years Inches Centimeters 60 55 50 45 40 35 30 25 20 15 10 5 0 0 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 388 SECTION IV Endocrine & Reproductive Physiology usually continues until the previous growth curve is reached, then slows to normal. The mechanisms that bring about and control catch-up growth are unknown. EFFECTS OF PITUITARY INSUFFICIENCY CHANGES IN OTHER ENDOCRINE GLANDS The widespread changes that develop when the pituitary is re-moved surgically or destroyed by disease in humans or ani-mals are predictable in terms of the known hormonal functions of the gland. In hypopituitarism, the adrenal cortex atrophies, and the secretion of adrenal glucocorticoids and sex hormones falls to low levels. Stress induced increases in aldos-terone secretion are absent, but basal aldosterone secretion and increases induced by salt depletion are normal, at least for some time. Since no mineralocorticoid deficiency is present, salt loss and hypovolemic shock do not develop, but the inabil-ity to increase glucocorticoid secretion makes patients with pi-tuitary insufficiency sensitive to stress. The development of salt loss in long-standing hypopituitarism is discussed in Chapter 22. Growth is inhibited (see above). Thyroid function is depressed to low levels, and cold is tolerated poorly. The go-nads atrophy, sexual cycles stop, and some of the secondary sex characteristics disappear. CLINICAL BOX 24–2 Dwarfism The accompanying discussion of growth control should suggest several possible etiologies of short stature. It can be due to GHRH deficiency, growth hormone deficiency, or deficient secretion of IGF-I. Isolated growth hormone defi-ciency is often due to GHRH deficiency, and in these in-stances, the growth hormone response to GHRH is normal. However, some patients with isolated growth hormone de-ficiency have abnormalities of their growth hormone se-creting cells. In another group of dwarfed children, the plasma growth hormone concentration is normal or ele-vated but their growth hormone receptors are unrespon-sive as a result of loss-of-function mutations. The resulting condition is known as growth hormone insensitivity or Laron dwarfism. Plasma IGF-I is markedly reduced, along with IGFBP 3, which is also growth hormone-dependent. African pygmies have normal plasma growth hormone le-vels and a modest reduction in the plasma level of growth hormone-binding protein. However, their plasma IGF-I con-centration fails to increase at the time of puberty and they experience less growth than non-pygmy controls through-out the prepubertal period. Short stature may also be caused by mechanisms inde-pendent of specific defects in the growth hormone axis. It is characteristic of childhood hypothyroidism (cretinism) and occurs in patients with precocious puberty. It is also part of the syndrome of gonadal dysgenesis seen in patients who have an XO chromosomal pattern instead of an XX or XY pattern (see Chapter 25). Various bone and metabolic dis-eases also cause stunted growth, and in many cases there is no known cause (“constitutional delayed growth”). Chronic abuse and neglect can also cause dwarfism in children, in-dependent of malnutrition. This condition is known as psy-chosocial dwarfism or the Kaspar Hauser syndrome, named for the patient with the first reported case. Finally, achondroplasia, the most common form of dwarfism in humans, is characterized by short limbs with a normal trunk. It is an autosomal dominant condition caused by a mutation in the gene that codes for fibroblast growth fac-tor receptor 3 (FGFR3). This member of the fibroblast growth receptor family is normally expressed in cartilage and the brain. The treatment of dwarfism is dictated by its underlying cause. If treatment is commenced promptly in childhood, almost normal stature can often be attained. The availabil-ity of recombinant forms of growth hormone and IGF-I has greatly improved treatment in cases where these hor-mones are deficient. FIGURE 24–12 Growth curve for a normal boy who had an illness beginning at age 5 and ending at age 7. Catch-up growth eventually returned his height to his previous normal growth curve. (Modified from Boersma B, Wit JM: Catch-up growth. Endocr Rev 1997;18:646.) 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 1 2 3 4 5 6 7 8 9 10111213141516171819 Age (y) Height (cm) Illness Catch-up growth CHAPTER 24 The Pituitary Gland 389 INSULIN SENSITIVITY Hypophysectomized animals have a tendency to become hy-poglycemic, especially when fasted. Hypophysectomy amelio-rates diabetes mellitus (see Chapter 21) and markedly increases the hypoglycemic effect of insulin. This is due in part to the deficiency of adrenocortical hormones, but hypophy-sectomized animals are more sensitive to insulin than adrenal-ectomized animals because they also lack the anti-insulin effect of growth hormone. WATER METABOLISM Although selective destruction of the supraoptic–posterior pi-tuitary causes diabetes insipidus (see Chapter 18), removal of both the anterior and posterior pituitary usually causes no more than a transient polyuria. In the past, there was specula-tion that the anterior pituitary secreted a “diuretic hormone,” but the amelioration of the diabetes insipidus is actually ex-plained by a decrease in the osmotic load presented for excre-tion. Osmotically active particles hold water in the renal tubules (see Chapter 38). Because of the ACTH deficiency, the rate of protein catabolism is decreased in hypophysectomized animals. Because of the TSH deficiency, the metabolic rate is low. Consequently, fewer osmotically active products of catab-olism are filtered and urine volume declines, even in the ab-sence of vasopressin. Growth hormone deficiency contributes to the depression of the glomerular filtration rate in hypophy-sectomized animals, and growth hormone increases the glo-merular filtration rate and renal plasma flow in humans. Finally, because of the glucocorticoid deficiency, there is the same defective excretion of a water load that is seen in adrenal-ectomized animals. The “diuretic” activity of the anterior pitu-itary can thus be explained in terms of the actions of ACTH, TSH, and growth hormone. OTHER DEFECTS When growth hormone deficiency develops in adulthood, it is usually accompanied by deficiencies in other anterior pitu-itary hormones. The deficiency of ACTH and other pituitary hormones with MSH activity may be responsible for the pallor of the skin in patients with hypopituitarism. There may be some loss of protein in adults, but wasting is not a feature of hypopituitarism in humans, and most patients with pituitary insufficiency are well nourished. CAUSES OF PITUITARY INSUFFICIENCY IN HUMANS Tumors of the anterior pituitary cause pituitary insufficiency. Suprasellar cysts, remnants of Rathke’s pouch that enlarge and compress the pituitary, are another cause of hypopituitarism. In women who have an episode of shock due to postpartum hemorrhage, the pituitary may become infarcted, with the subsequent development of postpartum necrosis (Sheehan syndrome). The blood supply to the anterior lobe is vulnera-ble because it descends on the pituitary stalk through the rigid diaphragma sellae, and during pregnancy the pituitary is en-larged. Pituitary infarction is usually extremely rare in men. CHAPTER SUMMARY ■The pituitary gland plays a critical role in regulating the func-tion of downstream glands, and also exerts independent endo-crine actions on a wide variety of peripheral organs and tissues. It consists of two functional sections in humans: the anterior pi-tuitary, which secretes mainly tropic hormones; and the poster-ior pituitary, which contains nerve endings that release oxytocin and vasopressin. The intermediate lobe is prominent in lower vertebrates but not in humans or other mammals. ■Corticotropes of the anterior lobe synthesize proopiomelano-cortin, the precursor of ACTH, endorphins, and melanocortins. The latter have a critical role in the control of skin coloration, whereas ACTH is a primary regulator of skin pigmentation in mammals. ■Growth hormone is synthesized by somatotropes and is highly species-specific. It is secreted in an episodic fashion in response to hypothalamic factors, and secretion is subject to feedback inhibition. A portion of the circulating pool is protein-bound. ■Growth hormone activates growth and influences protein, car-bohydrate, and fat metabolism to react to stressful conditions. Many, but not all, of the peripheral actions of growth hormone can be attributed to its ability to stimulate production of IGF-I. ■Growth reflects a complex interplay of growth hormone, IGF-I, and many other hormones as well as extrinsic influences and ge-netic factors. The consequences of over- or underproduction of such influences depends on whether this occurs before or after puberty. Deficiencies in components of the growth hormone pathway in childhood lead to dwarfism; overproduction results in gigantism, acromegaly, or both. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Which of the following hormones exerts the least effect on growth? A) growth hormone B) testosterone C) T4 D) insulin E) vasopressin 2. Which of the following pituitary hormones is an opioid peptide? A) α-melanocyte-stimulating hormone (α-MSH) B) β-MSH C) ACTH D) growth hormone E) β-endorphin 390 SECTION IV Endocrine & Reproductive Physiology 3. Which of the following is not characteristic of hypopituitarism? A) cachexia B) infertility C) pallor D) low basal metabolic rate E) intolerance to stress 4. A scientist finds that infusion of growth hormone into the median eminence of the hypothalamus in experimental animals inhibits the secretion of growth hormone and concludes that this proves that growth hormone feeds back to inhibit GHRH secre-tion. Do you accept this conclusion? A) No, because growth hormone does not cross the blood– brain barrier. B) No, because the infused growth hormone could be stimulat-ing dopamine secretion. C) No, because substances placed in the median eminence could be transported to the anterior pituitary. D) Yes, because systemically administered growth hormone inhibits growth hormone secretion. E) Yes, because growth hormone binds GHRH, inactivating it. 5. The growth hormone receptor A) activates Gs. B) requires dimerization to exert its effects. C) must be internalized to exert its effects. D) resembles the IGF-I receptor. E) resembles the ACTH receptor. CHAPTER RESOURCES Ayuk J, Sheppard MC: Growth hormone and its disorders. Postgrad Med J 2006;82:24. Boissy RE, Nordlund JJ: Molecular basis of congenital hypopigmentary disorders in humans: A review. Pigment Cell Res 1997;10:12. Buzi F, Mella P, Pilotta A, Prandi E, Lanfranchi F, Carapella T: Growth hormone receptor polymorphisms. Endocr Dev 2007;11:28. Fauquier T, Rizzoti K, Dattani M, Lovell-Badge R, Robinson ICAF: SOX2-expressing progenitor cells generate all of the major cell types in the adult mouse pituitary gland. Proc Natl Acad Sci USA 2008;105:2907. Hindmarsh PC, Dattani MT: Use of growth hormone in children. Nat Clin Pract Endocrinol Metab 2006;2:260. 391 C H A P T E R 25 The Gonads: Development & Function of the Reproductive System O B J E C T I V E S After studying this chapter, you should be able to: ■Name the key hormones secreted by Leydig cells and Sertoli cells of the testes and by graafian follicles and corpora lutea of the ovaries. ■Outline the role of chromosomes, hormones, and related factors in sex determina-tion and development. ■Summarize the hormonal changes that occur at puberty in males and females. ■Outline the hormonal changes and their physiologic effects during perimeno-pause and menopause. ■List the physiologic stimuli and the drugs that affect prolactin secretion. ■Outline the steps involved in spermatogenesis and the mechanisms that produce erection and ejaculation. ■Know the general structure of testosterone, and describe its biosynthesis, trans-port, metabolism, and actions. ■Describe the processes involved in regulation of testosterone secretion. ■Describe the physiologic changes that occur in the female reproductive organs during the menstrual cycle. ■Know the general structures of 17β-estradiol and progesterone, and describe their biosynthesis, transport, metabolism, and actions. ■Describe the roles of the pituitary and the hypothalamus in the regulation of ovar-ian function, and the role of feedback loops in this process. ■Describe the hormonal changes that accompany pregnancy and parturition. ■Outline the processes involved in lactation. INTRODUCTION Modern genetics and experimental embryology make it clear that, in most species of mammals, the multiple differences between the male and the female depend primarily on a sin-gle chromosome (the Y chromosome) and a single pair of endocrine structures, the testes in the male and the ovaries in the female. The differentiation of the primitive gonads into testes or ovaries in utero is genetically determined in humans, but the formation of male genitalia depends on the presence of a functional, secreting testis; in the absence of testicular tissue, development is female. Evidence indicates that male sexual behavior and, in some species, the male pat-tern of gonadotropin secretion are due to the action of male hormones on the brain in early development. After birth, the gonads remain quiescent until adolescence, when they are activated by gonadotropins from the anterior pituitary. Hor-mones secreted by the gonads at this time cause the appear-ance of features typical of the adult male or female and the onset of the sexual cycle in the female. In human females, ovarian function regresses after a number of years and sexual cycles cease (the menopause). In males, gonadal function 392 SECTION IV Endocrine & Reproductive Physiology slowly declines with advancing age, but the ability to produce viable gametes persists. In both sexes, the gonads have a dual function: the produc-tion of germ cells (gametogenesis) and the secretion of sex hor-mones. The androgens are the steroid sex hormones that are masculinizing in their action; the estrogens are those that are feminizing. Both types of hormones are normally secreted in both sexes. The testes secrete large amounts of androgens, prin-cipally testosterone, but they also secrete small amounts of estrogens. The ovaries secrete large amounts of estrogens and small amounts of androgens. Androgens are secreted from the adrenal cortex in both sexes, and some of the androgens are converted to estrogens in fat and other extragonadal and extra-adrenal tissues. The ovaries also secrete progesterone, a steroid that has special functions in preparing the uterus for pregnancy. Particularly during pregnancy, the ovaries secrete the polypep-tide hormone relaxin, which loosens the ligaments of the pubic symphysis and softens the cervix, facilitating delivery of the fetus. In both sexes, the gonads secrete other polypeptides, including inhibin B, a polypeptide that inhibits follicle-stimulating hormone (FSH) secretion. The secretory and gametogenic functions of the gonads are both dependent on the secretion of the anterior pituitary gonadotropins, FSH, and luteinizing hormone (LH). The sex hormones and inhibin B feed back to inhibit gonadotropin secretion. In males, gonadotropin secretion is noncyclic; but in postpubertal females an orderly, sequential secretion of gonadotropins is necessary for the occurrence of menstrua-tion, pregnancy, and lactation. SEX DIFFERENTIATION & DEVELOPMENT CHROMOSOMAL SEX The Sex Chromosomes Sex is determined genetically by two chromosomes, called the sex chromosomes, to distinguish them from the somatic chromo-somes (autosomes). In humans and many other mammals, the sex chromosomes are called X and Y chromosomes. The Y chro-mosome is necessary and sufficient for the production of testes, and the testis-determining gene product is called SRY (for sex-determining region of the Y chromosome). SRY is a DNA-binding regulatory protein. It bends the DNA and acts as a trans-cription factor that initiates transcription of a cascade of genes necessary for testicular differentiation, including the gene for müllerian inhibiting substance (MIS; see below). The gene for SRY is located near the tip of the short arm of the human Y chro-mosome. Male cells with the diploid number of chromosomes contain an X and a Y chromosome (XY pattern), whereas female cells contain two X chromosomes (XX pattern). As a conse-quence of meiosis during gametogenesis, each normal ovum con-tains a single X chromosome, but half of the normal sperm contain an X chromosome and half contain a Y chromosome (Figure 25–1). When a sperm containing a Y chromosome fertil-izes an ovum, an XY pattern results and the zygote develops into a genetic male. When fertilization occurs with an X-containing sperm, an XX pattern and a genetic female result. Cell division and the chemical nature of chromosomes are discussed in Chapter 1. Human Chromosomes Human chromosomes can be studied in detail. Human cells are grown in tissue culture; treated with the drug colchicine, which arrests mitosis at the metaphase; exposed to a hypotonic solution that makes the chromosomes swell and disperse; and then “squashed” onto slides. Staining techniques make it possible to identify the individual chromosomes and study them in detail (Figure 25–2). There are 46 chromosomes: in males, 22 pairs of autosomes plus an X chromosome and a Y chromosome; in fe-males, 22 pairs of autosomes plus two X chromosomes. The indi-vidual chromosomes are usually arranged in an arbitrary pattern (karyotype). The individual autosome pairs are identified by the numbers 1–22 on the basis of their morphologic characteristics. Sex Chromatin Soon after cell division has started during embryonic develop-ment, one of the two X chromosomes of the somatic cells in normal females becomes functionally inactive. In abnormal indi-viduals with more than two X chromosomes, only one remains ac-tive. The process that is normally responsible for inactivation is initiated in an X-inactivation center in the chromosome, probably via the transactivating factor CTCF (for CCCTC-binding factor), which is also induced in gene imprinting. However, the details of the inactivation process are still incompletely understood. The choice of which X chromosome remains active is random, so nor-mally one X chromosome remains active in approximately half of the cells and the other X chromosome is active in the other half. The selection persists through subsequent divisions of these cells, and consequently some of the somatic cells in adult females con-tain an active X chromosome of paternal origin and some contain an active X chromosome of maternal origin. In normal cells, the inactive X chromosome condenses and can be seen in various types of cells, usually near the nuclear membrane, as the Barr body, also called sex chromatin (Figure 25–3). Thus, there is a Barr body for each X chromosome in excess of one in the cell. The inactive X chromosome is also visi-ble as a small “drumstick” of chromatin projecting from the nuclei of 1–15% of the polymorphonuclear leukocytes in females but not in males (Figure 25–3). CHAPTER 25 The Gonads: Development & Function of the Reproductive System 393 FIGURE 25–1 Basis of genetic sex determination. In the two-stage meiotic division in the female, only one cell survives as the mature ovum. In the male, the meiotic division results in the formation of four sperms, two containing the X and two the Y chromosome. Fertilization thus produces a male zygote with 22 pairs of autosomes plus an X and a Y or a female zygote with 22 pairs of autosomes and two X chromosomes. Note that for clarity, this figure and Figures 25–6 and 25–7 differ from the current international nomenclature for karyotypes, which lists the total num-ber of chromosomes followed by the sex chromosome pattern. Thus, XO is 45, X; XY is 46, XY; XXY is 47, XXY, and so on. FIGURE 25–2 Karyotype of chromosomes from a normal male. The chromosomes have been stained with Giemsa’s stain, which produc-es a characteristic banding pattern. (Reproduced with permission, from Lingappa VJ, Farey K: Physiological Medicine. McGraw-Hill, 2000.) FIGURE 25–3 Left: Barr body (arrows) in the epidermal spinous cell layer. Right: Nuclear appendage (“drumstick”) identified by arrow in white blood cells. (Reproduced with permission from Grumbach MM, Barr ML: Cytologic tests of chromosomal sex in relation to sex anomalies in man. Recent Prog Horm Res 1958;14:255.) 44 XY Fertilization Meiosis 44 XX Zygotes 44 XY 44 XX 44 XX 22 Y 22 X 22 X 22 X 22 X 22 X 22Y 22Y 22X 22X 22X 22X Mature sperms Mature ova Immature ova Primary spermatocyte Meiosis 22X (First polar body) 22X (Second polar body) 1 6 13 19 20 21 22 14 15 16 17 18 7 8 9 10 11 12 2 3 4 5 X Y a b a b 394 SECTION IV Endocrine & Reproductive Physiology EMBRYOLOGY OF THE HUMAN REPRODUCTIVE SYSTEM Development of the Gonads On each side of the embryo, a primitive gonad arises from the genital ridge, a condensation of tissue near the adrenal gland. The gonad develops a cortex and a medulla. Until the sixth week of development, these structures are identical in both sex-es. In genetic males, the medulla develops during the seventh and eighth weeks into a testis, and the cortex regresses. Leydig and Sertoli cells appear, and testosterone and MIS are secreted. In genetic females, the cortex develops into an ovary and the medulla regresses. The embryonic ovary does not secrete hor-mones. Hormonal treatment of the mother has no effect on go-nadal (as opposed to ductal and genital) differentiation in humans, although it does in some experimental animals. Embryology of the Genitalia The embryology of the gonads is summarized in Figures 25–4 and 25–5. In the seventh week of gestation, the embryo has FIGURE 25–4 Embryonic differentiation of male and female internal genitalia (genital ducts) from wolffian (male) and müllerian (female) primordia. (After Corning HK, Wilkins L. Redrawn and reproduced with permission from Williams Textbook of Endocrinology, 7th ed. Wilson JD, Foster DW [editors]. Saunders,1985.) Gubernaculum Testis Hydatid Vaginal rudiment Urethra Prostate Seminal vesicle Vas deferens Epididymis MALE Fimbria Gonad Gonadal ligament Mullerian ligament : Uterovaginal canal Bladder Mesonephros Wolffian duct Urogenital sinus INDIFFERENT Epoophoron Ovary Uterine tube Ovarian ligament Uterus Gartner’s duct FEMALE CHAPTER 25 The Gonads: Development & Function of the Reproductive System 395 both male and female primordial genital ducts (Figure 25–4). In a normal female fetus, the müllerian duct system then devel-ops into uterine tubes (oviducts) and a uterus. In the normal male fetus, the wolffian duct system on each side develops into the epididymis and vas deferens. The external genitalia are sim-ilarly bipotential until the eighth week (Figure 25–5). Thereaf-ter, the urogenital slit disappears and male genitalia form, or, alternatively, it remains open and female genitalia form. When the embryo has functional testes, male internal and external genitalia develop. The Leydig cells of the fetal testis secrete testosterone, and the Sertoli cells secrete müllerian inhibiting substance (MIS; also called müllerian regression factor, or MRF). MIS is a 536-amino-acid homodimer that is a member of the transforming growth factor β (TGF-β) super-family of growth factors, which includes inhibins and activins. In their effects on the internal as opposed to the external geni-talia, MIS and testosterone act unilaterally. MIS causes regres-sion of the müllerian ducts by apoptosis on the side on which it is secreted, and testosterone fosters the development of the vas deferens and related structures from the wolffian ducts. The testosterone metabolite dihydrotestosterone induces the forma-tion of male external genitalia and male secondary sex charac-teristics (Figure 25–6). MIS continues to be secreted by the Sertoli cells, and it reaches mean values of 48 ng/mL in plasma in 1- to 2-year-old boys. Thereafter, it declines to low levels by the time of puberty and persists at low but detectable levels throughout life. In girls, MIS is produced by granulosa cells in small folli-cles in the ovaries, but plasma levels are very low or undetect-able until puberty. Thereafter, plasma MIS is about the same FIGURE 25–5 Differentiation of male and female external genitalia from indifferent primordial structures in the embryo. INDIFFERENT STAGE MALE FEMALE Glans penis Clitoris Urethral meatus Urethral meatus Labia minora Vaginal orifice Labia majora Scrotum Raphe Anus Phallus Genital swelling Urethral slit Tail Glans Genital tubercle Urogenital slit Urethral fold Labioscrotal swelling Anus SEVENTH TO EIGHTH WEEK TWELFTH WEEK 396 SECTION IV Endocrine & Reproductive Physiology as in adult men, that is, about 2 ng/mL. The functions of MIS after early embryonic life are unsettled, but it is probably involved in germ cell maturation in both sexes and in control of testicular descent in boys. Development of the Brain At least in some species, the development of the brain as well as the external genitalia is affected by androgens early in life. In rats, a brief exposure to androgens during the first few days of life causes the male pattern of sexual behavior and the male pattern of hypothalamic control of gonadotropin secretion to develop after puberty. In the absence of androgens, female patterns develop (see Chapter 18). In monkeys, similar effects on sexual behavior are produced by exposure to androgens in utero, but the pattern of gonadotropin secretion remains cyclical. Early exposure of female human fetuses to androgens also appears to cause subtle but significant masculinizing ef-fects on behavior. However, women with adrenogenital syn-drome due to congenital adrenocortical enzyme deficiency (see Chapter 22) develop normal menstrual cycles when treat-ed with cortisol. Thus, the human, like the monkey, appears to retain the cyclical pattern of gonadotropin secretion despite exposure to androgens in utero. ABERRANT SEXUAL DIFFERENTIATION Chromosomal Abnormalities From the preceding discussion, it might be expected that ab-normalities of sexual development could be caused by genetic or hormonal abnormalities as well as by other nonspecific ter-atogenic influences, and this is indeed the case. The major classes of abnormalities are listed in Table 25–1. Nondisjunction of sex chromosomes during the first divi-sion in meiosis results in distinct defects (see Clinical Box 25–1). Meiosis is a two-stage process, and although nondis-junction usually occurs during the first meiotic division, it can occur in the second, producing more complex chromosomal abnormalities. In addition, nondisjunction or simple loss of a sex chromosome can occur during the early mitotic divisions after fertilization. The result of faulty mitoses in the early zygote is the production of mosaicism, in which two or more populations of cells have different chromosome complements. True hermaphroditism, the condition in which the individual has both ovaries and testes, is probably due to XX/XY mosa-icism and related mosaic patterns, although other genetic aberrations are possible. Chromosomal abnormalities also include transposition of parts of chromosomes to other chromosomes. Rarely, genetic males are found to have the XX karyotype because the short FIGURE 25–6 Diagrammatic summary of normal sex determination, differentiation, and development in humans. MIS, müllerian inhibiting substance; T, testosterone; DHT, dihydrotestosterone. Bipotential gonad Embryonic testis Adult testis SRY No female internal genitalia Male internal genitalia Male external genitalia "Male brain" Male secondary sex characteristics Bipotential primordia DHT DHT T MIS 44 XY Y + 22 X 22 X 22 MALE Bipotential gonad Embryonic ovary Adult ovary Female internal genitalia Female external genitalia "Female brain" Female secondary sex characteristics Bipotential primordia Estrogens 44 XX X + 22 FEMALE T CHAPTER 25 The Gonads: Development & Function of the Reproductive System 397 arm of their father’s Y chromosome was transposed to their father’s X chromosome during meiosis and they received that X chromosome along with their mother’s. Similarly, deletion of the small portion of the Y chromosome containing SRY produces females with the XY karyotype. Hormonal Abnormalities Development of the male external genitalia occurs normally in genetic males in response to androgen secreted by the embry-onic testes, but male genital development may also occur in ge-netic females exposed to androgens from some other source during the 8th to the 13th weeks of gestation. The syndrome that results is female pseudohermaphroditism. A pseudoher-maphrodite is an individual with the genetic constitution and gonads of one sex and the genitalia of the other. After the 13th week, the genitalia are fully formed, but exposure to androgens can cause hypertrophy of the clitoris. Female pseudohermaph-roditism may be due to congenital virilizing adrenal hyperpla-sia (see Chapter 22), or it may be caused by androgens administered to the mother. Conversely, one cause of the de-velopment of female external genitalia in genetic males (male pseudohermaphroditism) is defective testicular development. Because the testes also secrete MIS, genetic males with defec-tive testes have female internal genitalia. Another cause of male pseudohermaphroditism is andro-gen resistance, in which, as a result of various congenital abnormalities, male hormones cannot exert their full effects on the tissues. One form of androgen resistance is a 5α-reductase deficiency, in which the enzyme responsible for the formation of dihydrotestosterone, the active form of testoster-one, is decreased. The consequences of thisdeficiency are dis-cussed in the section on the male reproductive system. Other forms of androgen resistance are due to various mutations in the androgen receptor gene, and the resulting defects in recep-tor function range from minor to severe. Mild defects cause TABLE 25–1 Classification of the major disorders of sex differentiation in humans. Chromosomal disorders Gonadal dysgenesis (XO and variants) “Superfemales” (XXX) Seminiferous tubule dysgenesis (XXY and variants) True hermaphroditism Developmental disorders Female pseudohermaphroditism Congenital virilizing adrenal hyperplasia of fetus Maternal androgen excess Virilizing ovarian tumor Iatrogenic: Treatment with androgens or certain synthetic progestational drugs Male pseudohermaphroditism Androgen resistance Defective testicular development Congenital 17α-hydroxylase deficiency Congenital adrenal hyperplasia due to blockade of pregnenolone formation Various nonhormonal anomalies Many of these syndromes can have great variation in degree and, consequently, in manifestations. CLINICAL BOX 25–1 Chromosomal Abnormalities An established defect in gametogenesis is nondisjunction, a phenomenon in which a pair of chromosomes fail to sepa-rate, so that both go to one of the daughter cells during mei-osis. Four of the abnormal zygotes that can form as a result of nondisjunction of one of the X chromosomes during oo-genesis are shown in Figure 25–7. In individuals with the XO chromosomal pattern, the gonads are rudimentary or ab-sent, so that female external genitalia develop, stature is short, other congenital abnormalities are often present, and no sexual maturation occurs at puberty. This syndrome is called gonadal dysgenesis or, alternatively, ovarian agene-sis or Turner syndrome. Individuals with the XXY pattern, the most common sex chromosome disorder, have the geni-talia of a normal male. Testosterone secretion at puberty is often great enough for the development of male characteris-tics, however, the seminiferous tubules are abnormal, and the incidence of mental retardation is higher than normal. This syndrome is known as seminiferous tubule dysgene-sis or Klinefelter syndrome. The XXX (“superfemale”) pat-tern is second in frequency only to the XXY pattern and may be even more common in the general population, since it does not seem to be associated with any characteristic ab-normalities. The YO combination is probably lethal. Nondisjunction of chromosome 21 produces trisomy 21, the chromosomal abnormality associated with Down syndrome (mongolism). The additional chromosome 21 is normal, so Down syndrome is a pure case of gene excess causing abnormalities. Many other chromosomal abnormalities occur as well as numerous diseases due to defects in single genes. These conditions are generally diagnosed in utero by analysis of fetal cells in a sample of amniotic fluid collected by inserting a needle through the abdominal wall (amniocentesis) or, earlier in pregnancy, by examining fetal cells obtained by a needle biopsy of chorionic villi (chorionic villus sampling). 398 SECTION IV Endocrine & Reproductive Physiology infertility with or without gynecomastia. When the loss of receptor function is complete, the testicular feminizing syndrome, now known as complete androgen resistance syndrome, results. In this condition, MIS is present and testos-terone is secreted at normal or even elevated rates. The exter-nal genitalia are female, but the vagina ends blindly because there are no female internal genitalia. Individuals with this syndrome develop enlarged breasts at puberty and usually are considered to be normal women until they are diagnosed when they seek medical advice because of lack of menstruation. It is worth noting that genetic males with congenital block-age of the formation of pregnenolone are pseudohermaphro-dites because testicular as well as adrenal androgens are normally formed from pregnenolone. Male pseudohermaph-roditism also occurs when there is a congenital deficiency of 17α-hydroxylase (see Chapter 22). PUBERTY As noted above, a burst of testosterone secretion occurs in male fetuses before birth (Figure 25–8). In the neonatal period there is another burst, with unknown function, but thereafter the Leydig cells become quiescent. There follows in all mam-mals a period in which the gonads of both sexes are quiescent until they are activated by gonadotropins from the pituitary to bring about the final maturation of the reproductive system. This period of final maturation is known as adolescence. It is often also called puberty, although puberty, strictly defined, is the period when the endocrine and gametogenic functions of the gonads have first developed to the point where reproduc-tion is possible. In girls, the first event is thelarche, the devel-opment of breasts, followed by pubarche, the development of axillary and pubic hair, and then by menarche, the first men-strual period. Initial menstrual periods are generally anovula-tory, and regular ovulation appears about a year later. In contrast to the situation in adulthood, removal of the gonads during the period from soon after birth to puberty causes only a small increase in gonadotropin secretion, so gonadotropin secretion is not being held in check by the gonadal hormones. In children between the ages of 7 and 10, a slow increase in es-trogen and androgen secretion precedes the more rapid rise in the early teens (Figure 25–9). The age at the time of puberty is variable. In Europe and the United States, it has been declining at the rate of 1 to 3 mo per decade for more than 175 y. In the United States in recent years, puberty generally occurs between the ages of 8 and 13 in girls and 9 and 14 in boys. Another event that occurs in humans at the time of puberty is an increase in the secretion of adrenal androgens (see Figure 22–12). The onset of this increase is called adrenarche. It occurs at age 8 to 10 y in girls and age 10 to 12 y in boys. Dehydroepiandrosterone (DHEA) values peak at about age 25 in females and slightly later than that in males. They then decline slowly to low values in old age. The rise appears to be due to an increase in the activity of 17 α-hydroxylase. Control of the Onset of Puberty The gonads of children can be stimulated by gonadotropins; their pituitaries contain gonadotropins and their hypothalami contain gonadotropin-releasing hormone (GnRH) (see Chap-ter 18). However, their gonadotropins are not secreted. In im-mature monkeys, normal menstrual cycles can be brought on by pulsatile injection of GnRH, and they persist as long as the pulsatile injection is continued. Thus, it seems clear that pulsa-tile secretion of GnRH brings on puberty. During the period FIGURE 25–7 Summary of four possible defects produced by maternal nondisjunction of the sex chromosomes at the time of meiosis. The YO combination is believed to be lethal, and the fetus dies in utero. 44 XO 44 XX 22 O Gonadal dysgenesis 22X 44 XXX 44 XX 22 XX Superfemale Zygote Ovum Sperm 22X 44 YO 44 XX 22 O Lethal 22Y 44 XXY 44 XX 22 XX Seminiferous tubule dysgenesis 22Y Abnormal meiosis FIGURE 25–8 Plasma testosterone levels at various ages in human males. Fetal Neo-natal Pre-pubertal Pubertal Adult Senes-cence 600 500 400 300 200 100 0 Plasma testosterone (ng/dL) CHAPTER 25 The Gonads: Development & Function of the Reproductive System 399 from birth to puberty, a neural mechanism is operating to pre-vent the normal pulsatile release of GnRH. The nature of the mechanism inhibiting the GnRH pulse generator is unknown. However, one or more genes produce products that stimulate secretion of GnRH, and inhibition of these genes before puberty is an interesting possibility (see Clinical Box 25–2). PRECOCIOUS & DELAYED PUBERTY Sexual Precocity The major causes of precocious sexual development in hu-mans are listed in Table 25–2. Early development of secondary sexual characteristics without gametogenesis is caused by ab-normal exposure of immature males to androgen or females to estrogen. This syndrome should be called precocious pseudo-puberty to distinguish it from true precocious puberty due to an early but otherwise normal pubertal pattern of gonadotro-pin secretion from the pituitary. Constitutional precocious puberty; that is, precocious puberty in which no cause can be determined, is more common in girls than in boys. In both sexes, tumors or infections involving the hypothalamus cause precocious puberty. Indeed, in one large series of cases, precocious puberty was the most common endo-crine symptom of hypothalamic disease. In experimental FIGURE 25–9 Changes in plasma hormone concentrations during puberty in boys (top) and girls (bottom). Stage 1 of puberty is preadolescence in both sexes. In boys, stage 2 is characterized by be-ginning enlargement of the testes, stage 3 by penile enlargement, stage 4 by growth of the glans penis, and stage 5 by adult genitalia. In girls, stage 2 is characterized by breast buds, stage 3 by elevation and enlargement of the breasts, stage 4 by projection of the areolas, and stage 5 by adult breasts. (Modified and reproduced with permission from Berenberg SR [editor]: Puberty: Biologic and Psychosocial Components. HE Stenfoert Kroese BV, 1975.) 3.0 2.0 1.0 0.5 6 4 2 FSH LH (ng/mL) 1 7.7 12 13.7 15.7 Bone age 2 3 Stage of puberty 4–5 Adult Testosterone (ng/mL) 3.0 2.0 1.0 0.5 60 40 20 FSH LH (ng/mL) 1 7.0 10.5 11.6 13.0 14.0 Bone age 2 3 Stage of puberty 4 5 17β-Estradiol (pg/mL) FSH LH FSH LH CLINICAL BOX 25–2 Leptin It has been argued for some time that a critical body weight must normally be reached for puberty to occur. Thus, for example, young women who engage in strenuous athletics lose weight and stop menstruating, as do girls with an-orexia nervosa. If these girls start to eat and gain weight, they menstruate again, that is, they “go back through pu-berty.” It now appears that leptin, the satiety-producing hormone secreted by fat cells, may be the link between body weight and puberty. Obese ob/ob mice that cannot make leptin are infertile, and their fertility is restored by in-jections of leptin. Leptin treatment also induces precocious puberty in immature female mice. However, the way that leptin fits into the overall control of puberty remains to be determined. TABLE 25–2 Classification of the causes of precocious sexual development in humans. True precocious puberty Constitutional Cerebral: Disorders involving posterior hypothalamus Tumors Infections Developmental abnormalities Gonadotropin-independent precocity Precocious pseudopuberty (no spermatogenesis or ovarian development) Adrenal Congenital virilizing adrenal hyperplasia Androgen-secreting tumors (in males) Estrogen-secreting tumors (in females) Gonadal Leydig cell tumors of testis Granulosa cell tumors of ovary Miscellaneous 400 SECTION IV Endocrine & Reproductive Physiology animals, precocious puberty can be produced by hypothalamic lesions. Apparently the lesions interrupt a pathway that nor-mally holds pulsatile GnRH secretion in check. Pineal tumors are sometimes associated with precocious puberty, but evidence indicates that these tumors are associated with precocity only when there is secondary damage to the hypothalamus. Precocious gametogenesis and steroidogenesis can occur without the pubertal pattern of gonadotropin secretion (gonad-otropin-independent precocity). At least in some cases of this condition, the sensitivity of LH receptors to gonadotropins is increased because of an activating mutation in the G protein that couples the receptors to adenylyl cyclase. Delayed or Absent Puberty The normal variation in the age at which adolescent changes oc-cur is so wide that puberty cannot be considered to be patholog-ically delayed until the menarche has failed to occur by the age of 17 or testicular development by the age of 20. Failure of mat-uration due to panhypopituitarism is associated with dwarfing and evidence of other endocrine abnormalities. Patients with the XO chromosomal pattern and gonadal dysgenesis are also dwarfed. In some individuals, puberty is delayed even though the gonads are present and other endocrine functions are nor-mal. In males, this clinical picture is called eunuchoidism. In females, it is called primary amenorrhea. MENOPAUSE The human ovaries become unresponsive to gonadotropins with advancing age, and their function declines, so that sexual cycles disappear (menopause). This unresponsiveness is asso-ciated with and probably caused by a decline in the number of primordial follicles, which becomes precipitous at the time of menopause (Figure 25–10). The ovaries no longer secrete progesterone and 17β-estradiol in appreciable quantities, and estrogen is formed only in small amounts by aromatization of androstenedione in peripheral tissues (see Chapter 22). The uterus and the vagina gradually become atrophic. As the neg-ative feedback effect of estrogens and progesterone is reduced, secretion of FSH is increased, and plasma FSH increases to high levels, LH levels are moderately high. Old female mice and rats have long periods of diestrus and increased levels of gonadotropin secretion. In women, a period called perimeno-pause precedes menopause, and can last up to 10 y. During perimenopause the menses become irregular and the level of inhibins decrease, usually between the ages of 45 and 55. The average age at onset of the menopause has been increasing since the end of the 19th century and is currently 52 y. The loss of ovarian function causes many symptoms such as sensations of warmth spreading from the trunk to the face (hot flushes; also called hot flashes) and night sweats. In addi-tion, the onset of menopause increases the risk of many dis-eases such as osteoporosis, ischemic heart disease, and renal disease. Hot flushes are said to occur in 75% of menopausal women and may continue intermittently for as long as 40 y. They also occur when early menopause is produced by bilateral ovariec-tomy, and they are prevented by estrogen treatment. In addi-tion, they occur after castration in men. Their cause is unknown. However, they coincide with surges of LH secre-tion. LH is secreted in episodic bursts at intervals of 30 to 60 min or more (circhoral secretion), and in the absence of gonadal hormones these bursts are large. Each hot flush begins with the start of a burst. However, LH itself is not responsible for the symptoms, because they can continue after removal of the pituitary. Instead, it appears that some estro-gen-sensitive event in the hypothalamus initiates both the release of LH and the episode of flushing. Although the function of the testes tends to decline slowly with advancing age, the evidence is unclear whether there is a “male menopause” (andropause) similar to that occurring in women. PITUITARY GONADOTROPINS & PROLACTIN CHEMISTRY FSH and LH are each made up of an α and a β subunit whose nature is discussed in Chapter 24. They are glycoproteins that contain the hexoses mannose and galactose, the hexosamines N-acetylgalactosamine and N-acetylglycosamine, and the methylpentose fucose. They also contain sialic acid. The car-bohydrate in the gonadotropin molecules increases their FIGURE 25–10 Number of primordial follicles per ovary in women at various ages. Blue squares, premenopausal women (regu-lar menses); red squares, perimenopausal women (irregular menses for at least 1 y); red triangles, postmenopausal women (no menses for at least 1 y). Note that the vertical scale is a log scale and that the values are from one rather than two ovaries. (Redrawn by PM Wise and reproduced with permission from Richardson SJ, Senikas V, Nelson JF: Follicular depletion during the menopausal transition: Evidence for accelerated loss and ultimate exhaustion. J Clin Endocrinol Metab 1987;65:1231.) 100,000 10,000 1000 100 10 1 0 10 20 30 40 50 60 Age in years Primordial follicles/ovary CHAPTER 25 The Gonads: Development & Function of the Reproductive System 401 potency by markedly slowing their metabolism. The half-life of human FSH is about 170 min; the half-life of LH is about 60 min. Loss-of-function mutations in the FSH receptor cause hypogonadism. Gain-of-function mutations cause a sponta-neous form of ovarian hyperstimulation syndrome, a condi-tion in which many follicles are stimulated and cytokines are released from the ovary, causing increased vascular perme-ability and shock. Human pituitary prolactin contains 199 amino acid resi-dues and three disulfide bridges and has considerable struc-tural similarity to human growth hormone and human chorionic somatomammotropin (hCS). The half-life of pro-lactin, like that of growth hormone, is about 20 min. Structur-ally similar prolactins are secreted by the endometrium and by the placenta. RECEPTORS The receptors for FSH and LH are G-protein coupled recep-tors coupled to adenylyl cyclase through a stimulatory G pro-tein (Gs; see Chapter 2). In addition, each has an extended, glycosylated extracellular domain. The human prolactin receptor resembles the growth hor-mone receptor and is one of the superfamily of receptors that includes the growth hormone receptor and receptors for many cytokines and hematopoietic growth factors (see Chap-ters 2 and 3). It dimerizes and activates the Janus kinase/signal transducers and activators of transcription (JAK–STAT) path-way and other intracellular enzyme cascades. ACTIONS The testes and ovaries become atrophic when the pituitary is removed or destroyed. The actions of prolactin and the gona-dotropins FSH and LH, as well as those of the gonadotropin secreted by the placenta, are described in detail in succeeding sections of this chapter. In brief, FSH helps maintain the sper-matogenic epithelium by stimulating Sertoli cells in the male and is responsible for the early growth of ovarian follicles in the female. LH is tropic to the Leydig cells and, in females, is responsible for the final maturation of the ovarian follicles and estrogen secretion from them. It is also responsible for ovula-tion, the initial formation of the corpus luteum, and secretion of progesterone. Prolactin causes milk secretion from the breast after estro-gen and progesterone priming. Its effect on the breast involves increased action of mRNA and increased production of casein and lactalbumin. However, the action of the hormone is not exerted on the cell nucleus and is prevented by inhibitors of microtubules. Prolactin also inhibits the effects of gonadotro-pins, possibly by an action at the level of the ovary. Its role in preventing ovulation in lactating women is discussed below. The function of prolactin in normal males is unsettled, but excess prolactin secreted by tumors causes impotence. REGULATION OF PROLACTIN SECRETION The normal plasma prolactin concentration is approximately 5 ng/mL in men and 8 ng/mL in women. Secretion is tonically inhibited by the hypothalamus, and section of the pituitary stalk leads to an increase in circulating prolactin. Thus, the effect of the hypothalamic prolactin-inhibiting hormone (PIH) dopa-mine is normally greater than the effects of the various hypotha-lamic peptides with prolactin-releasing activity. In humans, prolactin secretion is increased by exercise, surgical and psy-chologic stresses, and stimulation of the nipple (Table 25–3). The plasma prolactin level rises during sleep, the rise starting af-ter the onset of sleep and persisting throughout the sleep period. Secretion is increased during pregnancy, reaching a peak at the time of parturition. After delivery, the plasma concentration falls to nonpregnant levels in about 8 days. Suckling produces a prompt increase in secretion, but the magnitude of this rise gradually declines after a woman has been nursing for more than 3 months. With prolonged lactation, milk secretion occurs with prolactin levels that are in the normal range. TABLE 25–3 Factors affecting the secretion of human prolactin and growth hormone. Factor Prolactina Growth Hormonea Sleep I+ I+ Nursing I++ N Breast stimulation in nonlactating women I N Stress I+ I+ Hypoglycemia I I+ Strenuous exercise I I Sexual intercourse in women I N Pregnancy I++ N Estrogens I I Hypothyroidism I N TRH I+ N Phenothiazines, butyrophenones I+ N Opioids I I Glucose N D Somatostatin N D+ L-Dopa D+ I+ Apomorphine D+ I+ Bromocriptine and related ergot derivatives D+ I aI, moderate increase; I+, marked increase; I++, very marked increase; N, no change; D, moderate decrease; D+, marked decrease; TRH, thyrotropin-releasing hormone. 402 SECTION IV Endocrine & Reproductive Physiology L-Dopa decreases prolactin secretion by increasing the forma-tion of dopamine; bromocriptine and other dopamine agonists inhibit secretion because they stimulate dopamine receptors. Chlorpromazine and related drugs that block dopamine recep-tors increase prolactin secretion. Thyrotropin-releasing hormone (TRH) stimulates the secretion of prolactin in addition to thyroid-stimulating hormone (TSH), and additional polypep-tides with prolactin-releasing activity are present in hypotha-lamic tissue. Estrogens produce a slowly developing increase in prolactin secretion as a result of a direct action on the lactotropes. It has now been established that prolactin facilitates the secretion of dopamine in the median eminence. Thus, prolac-tin acts in the hypothalamus in a negative feedback fashion to inhibit its own secretion (see Clinical Box 25–3). THE MALE REPRODUCTIVE SYSTEM STRUCTURE The testes are made up of loops of convoluted seminiferous tubules, in the walls of which the spermatozoa are formed from the primitive germ cells (spermatogenesis). Both ends of each loop drain into a network of ducts in the head of the epididymis. From there, spermatozoa pass through the tail of the epididymis into the vas deferens. They enter through the ejaculatory ducts into the urethra in the body of the prostate at the time of ejaculation (Figure 25–11). Between the tubules in the testes are nests of cells containing lipid granules, the in-terstitial cells of Leydig (Figures 25–12 and 25–13), which se-crete testosterone into the bloodstream. The spermatic arteries to the testes are tortuous, and blood in them runs par-allel but in the opposite direction to blood in the pampiniform plexus of spermatic veins. This anatomic arrangement may permit countercurrent exchange of heat and testosterone. The principles of countercurrent exchange are considered in detail in relation to the kidney in Chapter 38. GAMETOGENESIS & EJACULATION Blood–Testis Barrier The walls of the seminiferous tubules are lined by primitive germ cells and Sertoli cells, large, complex glycogen-contain-ing cells that stretch from the basal lamina of the tubule to the lumen (Figure 25–13). Germ cells must stay in contact with Sertoli cells to survive, and this contact is maintained by CLINICAL BOX 25–3 Hyperprolactinemia Up to 70% of the patients with chromophobe adenomas of the anterior pituitary have elevated plasma prolactin levels. In some instances, the elevation may be due to damage to the pituitary stalk, but in most cases, the tumor cells are ac-tually secreting the hormone. The hyperprolactinemia may cause galactorrhea, but in many individuals no demonstra-ble endocrine abnormalities are present. Conversely, most women with galactorrhea have normal prolactin levels; definite elevations are found in less than a third of patients with this condition. Another interesting observation is that 15–20% of women with secondary amenorrhea have elevated prolac-tin levels, and when prolactin secretion is reduced, normal menstrual cycles and fertility return. It appears that the pro-lactin may produce amenorrhea by blocking the action of gonadotropins on the ovaries, but definitive proof of this hypothesis must await further research. The hypogonadism produced by prolactinomas is associated with osteoporosis due to estrogen deficiency. As noted previously, hyperprolactinemia in men is asso-ciated with impotence and hypogonadism that disappear when prolactin secretion is reduced. FIGURE 25–11 Anatomical features of the male reproductive system. Left: Male reproductive system. Right: Duct system of the testis. Bladder Ureter Vas deferens Vas deferens Head of epididymis Tail of epididymis Seminiferous tubules Tunica albuginea Rete testis Septa Symphysis Urethra Epididymis Testis Scrotum Cowper’s (bulbourethral) gland Ejaculatory duct Prostate Seminal vesicle CHAPTER 25 The Gonads: Development & Function of the Reproductive System 403 cytoplasmic bridges. Tight junctions between adjacent Sertoli cells near the basal lamina form a blood–testis barrier that prevents many large molecules from passing from the intersti-tial tissue and the part of the tubule near the basal lamina (basal compartment) to the region near the tubular lumen (adluminal compartment) and the lumen. However, steroids penetrate this barrier with ease, and evidence suggests that some proteins pass from the Sertoli cells to the Leydig cells and vice versa in a paracrine fashion. In addition, maturing germ cells must pass through the barrier as they move to the lumen. This appears to occur without disruption of the barrier by progressive break-down of the tight junctions above the germ cells, with concom-itant formation of new tight junctions below them. The fluid in the lumen of the seminiferous tubules is quite different from plasma; it contains very little protein and glu-cose but is rich in androgens, estrogens, K+, inositol, and glutamic and aspartic acids. Maintenance of its composition presumably depends on the blood–testis barrier. The barrier also protects the germ cells from bloodborne noxious agents, prevents antigenic products of germ cell division and matura-tion from entering the circulation and generating an autoim-mune response, and may help establish an osmotic gradient that facilitates movement of fluid into the tubular lumen. Spermatogenesis Spermatogonia, the primitive germ cells next to the basal lami-na of the seminiferous tubules, mature into primary spermato-cytes (Figure 25–13). This process begins during adolescence. FIGURE 25–12 Section of human testis. Seminiferous tubules Primary spermatocyte Spermato-gonium Interstitial cell FIGURE 25–13 Seminiferous epithelium. Note that maturing germ cells remain connected by cytoplasmic bridges through the early sper-matid stage and that these cells are closely invested by Sertoli cell cytoplasm as they move from the basal lamina to the lumen. (Reproduced with permission from Junqueira LC, Carneiro J: Basic Histology: Text & Atlas, 10th ed. McGraw-Hill, 2003.) Late spermatids Spermiogenesis Meiosis Basal lamina Fibroblast Capillary Capillary Myoid cells Interstitial cells Spermatogonium Primary spermatocyte Secondary spermatocytes Early spermatids Cytoplasmic bridges Sertoli cell Sertoli cell [Low resolution, for proofing only] 404 SECTION IV Endocrine & Reproductive Physiology The primary spermatocytes undergo meiotic division, reducing the number of chromosomes. In this two-stage process, they di-vide into secondary spermatocytes and then into spermatids, which contain the haploid number of 23 chromosomes. The spermatids mature into spermatozoa (sperms). As a single spermatogonium divides and matures, its descendants remain tied together by cytoplasmic bridges until the late spermatid stage. This apparently ensures synchrony of the differentiation of each clone of germ cells. The estimated number of spermatids formed from a single spermatogonium is 512. In humans, it takes an average of 74 d to form a mature sperm from a primi-tive germ cell by this orderly process of spermatogenesis. Each sperm is an intricate motile cell, rich in DNA, with a head that is made up mostly of chromosomal material (Figure 25–14). Covering the head like a cap is the acrosome, a lyso-some-like organelle rich in enzymes involved in sperm pene-tration of the ovum and other events involved in fertilization. The motile tail of the sperm is wrapped in its proximal por-tion by a sheath holding numerous mitochondria. The mem-branes of late spermatids and spermatozoa contain a special small form of angiotensin-converting enzyme called germi-nal angiotensin-converting enzyme. The function of this enzyme in the sperms is unknown, although male mice in which the function of the angiotensin-converting enzyme gene has been disrupted have reduced fertility. The spermatids mature into spermatozoa in deep folds of the cytoplasm of the Sertoli cells (Figure 25–13). Mature sper-matozoa are released from the Sertoli cells and become free in the lumens of the tubules. The Sertoli cells secrete androgen-binding protein (ABP), inhibin, and MIS. They do not syn-thesize androgens, but they contain aromatase (CYP19), the enzyme responsible for conversion of androgens to estrogens, and they can produce estrogens. ABP probably functions to maintain a high, stable supply of androgen in the tubular fluid. Inhibin inhibits FSH secretion. FSH and androgens maintain the gametogenic function of the testis. After hypophysectomy, injection of LH produces a high local concentration of androgen in the testes, and this maintains spermatogenesis. The stages from spermatogonia to spermatids appear to be androgen-independent. However, the maturation from spermatids to spermatozoa depends on androgen acting on the Sertoli cells in which the developing spermatozoa are embedded. FSH acts on the Sertoli cells to facilitate the last stages of spermatid maturation. In addition, it promotes the production of ABP. An interesting observation is that the estrogen content of the fluid in the rete testis (Figure 25–11) is high, and the walls of the rete contain numerous ERα estrogen receptors. In this region, fluid is reabsorbed and the spermatozoa are concen-trated. If this does not occur, the sperm entering the epididymis are diluted in a large volume of fluid, and infertility results. Further Development of Spermatozoa Spermatozoa leaving the testes are not fully mobile. They con-tinue their maturation and acquire motility during their passage through the epididymis. Motility is obviously important in vivo, but fertilization occurs in vitro if an immotile spermatozoon from the head of the epididymis is microinjected directly into an ovum. The ability to move forward (progressive motility), which is acquired in the epididymis, involves activation of a unique protein called CatSper, which is localized to the princi-pal piece of the sperm tail. This protein appears to be a Ca2+ ion channel that permits cAMP-generalized Ca2+ influx. In addi-tion, spermatozoa express olfactory receptors, and ovaries pro-duce odorant-like molecules. Recent evidence indicates that these molecules and their receptors interact, fostering move-ment of the spermatozoa toward the ovary (chemotaxis). Ejaculation of the spermatozoon involves contractions of the vas deferens mediated in part by P2X receptors, ligand-gated cation channels that respond to ATP (see Chapter 7), and fertility is reduced in mice in which these receptors are knocked out. Once ejaculated into the female, the spermatozoa move up the uterus to the isthmus of the uterine tubes, where they slow down and undergo capacitation. This further maturation process involves two components: increasing the motility of the spermatozoa and facilitating their preparation for the acrosome reaction. However, the role of capacitation appears to be facilitatory rather than obligatory, because fertilization is readily produced in vitro. From the isthmuses the capacitated spermatozoa move rapidly to the tubal ampullas, where fertil-ization takes place. FIGURE 25–14 Human spermatozoon, profile view. Note the acrosome, an organelle that covers half the sperm head inside the plasma membrane of the sperm. (Reproduced with permission from Junqueira LC, Carneiro J: Basic Histology: Text & Atlas, 10th ed. McGraw-Hill, 2003.) 5 μm End piece 5 μm Middle piece 5 μm Head 50 μm Principal piece Acrosome Mitochondria CHAPTER 25 The Gonads: Development & Function of the Reproductive System 405 Effect of Temperature Spermatogenesis requires a temperature considerably lower than that of the interior of the body. The testes are normally maintained at a temperature of about 32 °C. They are kept cool by air circulating around the scrotum and probably by heat ex-change in a countercurrent fashion between the spermatic ar-teries and veins. When the testes are retained in the abdomen or when, in experimental animals, they are held close to the body by tight cloth binders, degeneration of the tubular walls and sterility result. Hot baths (43–45 °C for 30 min/d) and in-sulated athletic supporters reduce the sperm count in humans, in some cases by 90%. However, the reductions produced in this manner are not consistent enough to make the procedures reliable forms of male contraception. In addition, evidence suggests a seasonal effect in men, with sperm counts being greater in the winter regardless of the temperature to which the scrotum is exposed. Semen The fluid that is ejaculated at the time of orgasm, the semen, contains sperms and the secretions of the seminal vesicles, prostate, Cowper’s glands, and, probably, the urethral glands (Table 25–4). An average volume per ejaculate is 2.5 to 3.5 mL after several days of abstinence. The volume of semen and the sperm count decrease rapidly with repeated ejaculation. Even though it takes only one sperm to fertilize the ovum, each mil-liliter of semen normally contains about 100 million sperms. Fifty percent of men with counts of 20 to 40 million/mL and essentially all of those with counts under 20 million/mL are sterile. The presence of many morphologically abnormal or immotile spermatozoa also correlates with infertility. The prostaglandins in semen, which actually come from the sem-inal vesicles, are in high concentration, but the function of these fatty acid derivatives in semen is unknown. Human sperms move at a speed of about 3 mm/min through the female genital tract. Sperms reach the uterine tubes 30 to 60 min after copulation. In some species, contrac-tions of the female organs facilitate the transport of the sperms to the uterine tubes, but it is unknown if such contrac-tions are important in humans. Erection Erection is initiated by dilation of the arterioles of the penis. As the erectile tissue of the penis fills with blood, the veins are compressed, blocking outflow and adding to the turgor of the organ. The integrating centers in the lumbar segments of the spinal cord are activated by impulses in afferents from the genitalia and descending tracts that mediate erection in re-sponse to erotic psychologic stimuli. The efferent parasympa-thetic fibers are in the pelvic splanchnic nerves (nervi erigentes). The fibers presumably release acetylcholine and the vasodilator vasoactive intestinal polypeptide (VIP) as cotransmitters (see Chapter 7). Nonadrenergic noncholinergic fibers are also present in the nervi erigentes, and these contain large amounts of NO syn-thase, the enzyme that catalyzes the formation of nitric oxide (NO; see Chapter 33). NO activates guanylyl cyclase, resulting in increased production of cyclic GMP (cGMP), and cGMP is a potent vasodilator. Injection of inhibitors of NO synthase prevents the erection normally produced by stimulation of the pelvic nerve in experimental animals. Thus, it seems clear that NO plays a prominent role in the production of erection. The drugs sildenafil, tadalafil, and vardenafil all inhibit the break-down of cGMP by phosphodiesterases and have gained world-wide fame for the treatment of impotence. The multiple phosphodiesterases (PDEs) in the body have been divided into seven isoenzyme families, and these drugs are all most active against PDE V, the type of phosphodiesterase found in the cor-pora cavernosa. It is worth noting, however, that these drugs can also produces significant inhibition of PDE VI (and others, if taken at high doses). Phosphodiesterase VI is found in the retina, and one of the side effects of these drugs is a transient loss of the ability to discriminate between blue and green (see Chapter 12). TABLE 25–4 Composition of human semen. Color: White, opalescent Specific gravity: 1.028 pH: 7.35–7.50 Sperm count: Average about 100 million/mL, with fewer than 20% abnormal forms Other components: From seminal vesicles (contributes 60% of total volume) Fructose (1.5-6.5 mg/mL) Phosphorylcholine Ergothioneine Ascorbic acid Flavins Prostaglandins Spermine From prostate (contributes 20% of total volume) Citric acid Cholesterol, phospholipids Fibrinolysin, fibrinogenase Zinc Acid phosphatase Phosphate Buffers Bicarbonate Hyaluronidase } } } 406 SECTION IV Endocrine & Reproductive Physiology Normally, erection is terminated by sympathetic vasocon-strictor impulses to the penile arterioles. Ejaculation Ejaculation is a two-part spinal reflex that involves emission, the movement of the semen into the urethra; and ejaculation proper, the propulsion of the semen out of the urethra at the time of orgasm. The afferent pathways are mostly fibers from touch receptors in the glans penis that reach the spinal cord through the internal pudendal nerves. Emission is a sympa-thetic response, integrated in the upper lumbar segments of the spinal cord and effected by contraction of the smooth muscle of the vasa deferentia and seminal vesicles in response to stimuli in the hypogastric nerves. The semen is propelled out of the urethra by contraction of the bulbocavernosus mus-cle, a skeletal muscle. The spinal reflex centers for this part of the reflex are in the upper sacral and lowest lumbar segments of the spinal cord, and the motor pathways traverse the first to third sacral roots and the internal pudendal nerves. PSA The prostate produces and secretes into the semen and the bloodstream a 30 kDa serine protease generally called pros-tate-specific antigen (PSA). The gene for PSA has two andro-gen response elements. PSA hydrolyzes the sperm motility inhibitor semenogelin in semen, and it has several substrates in plasma, but its precise function in the circulation is un-known. An elevated plasma PSA occurs in prostate cancer and is widely used as a screening test for this disease, though PSA is also elevated in benign prostatic hyperplasia and prostatitis. Vasectomy Bilateral ligation of the vas deferens (vasectomy) has proved to be a relatively safe and convenient contraceptive procedure. However, it has proven difficult to restore the patency of the vas in those wishing to restore fertility, and the current success rate for such operations, as measured by the subsequent pro-duction of pregnancy, is about 50%. Half of the men who have been vasectomized develop antibodies against spermatozoa, and in monkeys, the presence of such antibodies is associated with a higher incidence of infertility after restoration of the patency of the vas. However, the anti-sperm antibodies do not appear to have any other adverse effects. ENDOCRINE FUNCTION OF THE TESTES Chemistry & Biosynthesis of Testosterone Testosterone, the principal hormone of the testes, is a C19 ster-oid with an –OH group in the 17 position (Figure 25–15). It is synthesized from cholesterol in the Leydig cells and is also formed from androstenedione secreted by the adrenal cortex. The biosynthetic pathways in all endocrine organs that form steroid hormones are similar, the organs differing only in the enzyme systems they contain. In the Leydig cells, the 11- and 21-hydroxylases found in the adrenal cortex (see Figure 22–7) are absent, but 17α-hydroxylase is present. Pregnenolone is therefore hydroxylated in the 17 position and then subjected to side chain cleavage to form dehydroepiandrosterone. An-drostenedione is also formed via progesterone and 17-hydrox-yprogesterone, but this pathway is less prominent in humans. Dehydroepiandrosterone and androstenedione are then con-verted to testosterone. The secretion of testosterone is under the control of LH, and the mechanism by which LH stimulates the Leydig cells involves FIGURE 25–15 Biosynthesis of testosterone. The formulas of the precursor steroids are shown in Figure 22–7. Although the main secre-tory product of the Leydig cells is testosterone, some of the precursors also enter the circulation. Cholesterol Pregnenolone 17α-Hydroxypregnenolone Dehydroepiandrosterone Progesterone 17α-Hydroxyprogesterone Androstenedione O O OH Testosterone OH Dihydrotestosterone H In some target tissues 5α-reductase, type 1 or type 2 CHAPTER 25 The Gonads: Development & Function of the Reproductive System 407 increased formation of cAMP via the G protein-coupled LH receptor and Gs. Cyclic AMP increases the formation of choles-terol from cholesteryl esters and the conversion of cholesterol to pregnenolone via the activation of protein kinase A. Secretion The testosterone secretion rate is 4 to 9 mg/d (13.9–31.33 μmol/d) in normal adult males. Small amounts of testosterone are also secreted in females, with the major source being the ovary, but possibly from the adrenal as well. Transport & Metabolism Ninety-eight percent of the testosterone in plasma is bound to protein: 65% is bound to a β-globulin called gonadal steroid-binding globulin (GBG) or sex steroid-binding globulin, and 33% to albumin (Table 25–5). GBG also binds estradiol. The plasma testosterone level (free and bound) is 300 to 1000 ng/dL (10.4–34.7 nmol/L) in adult men (Figure 25–8), com-pared with 30 to 70 ng/dL (1.04–2.43 nmol/L) in adult women. It declines somewhat with age in males. A small amount of circulating testosterone is converted to estradiol, but most of the testosterone is converted to 17-keto-steroids, principally androsterone and its isomer etiocholano-lone (Figure 25–16), and excreted in the urine. About two thirds of the urinary 17-ketosteroids are of adrenal origin, and one third are of testicular origin. Although most of the 17-ketosteroids are weak androgens (they have 20% or less the potency of testosterone), it is worth emphasizing that not all 17-ketosteroids are androgens and not all androgens are 17-ketosteroids. Etiocholanolone, for example, has no andro-genic activity, and testosterone itself is not a 17-ketosteroid. Actions In addition to their actions during development, testosterone and other androgens exert an inhibitory feedback effect on pituitary LH secretion; develop and maintain the male secon-dary sex characteristics; exert an important protein-anabolic, growth-promoting effect; and, along with FSH, maintain spermatogenesis. Secondary Sex Characteristics The widespread changes in hair distribution, body configura-tion, and genital size that develop in boys at puberty—the male secondary sex characteristics—are summarized in Table 25–6. The prostate and seminal vesicles enlarge, and the seminal vesicles begin to secrete fructose. This sugar appears to function as the main nutritional supply for the spermato-zoa. The psychic effects of testosterone are difficult to define in humans, but in experimental animals, androgens provoke boisterous and aggressive play. The effects of androgens and estrogens on sexual behavior are considered in detail in TABLE 25–5 Distribution of gonadal steroids and cortisol in plasma. % Bound to Steroid % Free CBG GBG Albumin Testosterone 2 0 65 33 Androstenedione 7 0 8 85 Estradiol 2 0 38 60 Progesterone 2 18 0 80 Cortisol 4 90 0 6 CBG, corticosteroid-binding globulin; GBG, gonadal steroid-binding globulin. (Courtesy of S Munroe.) FIGURE 25–16 Two 17-ketosteroid metabolites of testosterone. TABLE 25–6 Changes at puberty in boys (male secondary sex characteristics). External genitalia: Penis increases in length and width. Scrotum becomes pigmented and rugose. Internal genitalia: Seminal vesicles enlarge and secrete and begin to form fructose. Prostate and bulbourethral glands enlarge and secrete. Voice: Larynx enlarges, vocal cords increase in length and thickness, and voice becomes deeper. Hair growth: Beard appears. Hairline on scalp recedes anterolaterally. Pubic hair grows with male (triangle with apex up) pattern. Hair ap-pears in axillas, on chest, and around anus; general body hair increases. Mental: More aggressive, active attitude. Interest in opposite sex develops. Body conformation: Shoulders broaden, muscles enlarge. Skin: Sebaceous gland secretion thickens and increases (predisposing to acne). H Etiocholanolone O O Androsterone H HO HO 408 SECTION IV Endocrine & Reproductive Physiology Chapter 15. Although body hair is increased by androgens, scalp hair is decreased (Figure 25–17). Hereditary baldness of-ten fails to develop unless dihydrotestosterone is present. Anabolic Effects Androgens increase the synthesis and decrease the breakdown of protein, leading to an increase in the rate of growth. It used to be argued that they cause the epiphyses to fuse to the long bones, thus eventually stopping growth, but it now appears that epiphysial closure is due to estrogens (see Chapter 23). Second-ary to their anabolic effects, androgens cause moderate Na+, K+, H2O, Ca2+, SO4 –, and PO4 – retention; and they also increase the size of the kidneys. Doses of exogenous testosterone that exert significant anabolic effects are also masculinizing and increase libido, which limits the usefulness of the hormone as an anabol-ic agent in patients with wasting diseases. Attempts to develop synthetic steroids in which the anabolic action is divorced from the androgenic action have not been successful. Mechanism of Action Like other steroids, testosterone binds to an intracellular re-ceptor, and the receptor–steroid complex then binds to DNA in the nucleus, facilitating transcription of various genes. In addition, testosterone is converted to dihydrotestosterone (DHT) by 5α-reductase in some target cells (Figures 25–15 and 25–18), and DHT binds to the same intracellular receptor as testosterone. DHT also circulates, with a plasma level that is about 10% of the testosterone level. Testosterone–receptor complexes are less stable than DHT–receptor complexes in target cells, and they conform less well to the DNA-binding state. Thus, DHT formation is a way of amplifying the action of testosterone in target tissues. Humans have two 5α-reductases, encoded by different genes. Type 1 5α-reductase is present in skin throughout the body and is the dominant enzyme in the scalp. Type 2 5α-reductase is present in genital skin, the pros-tate, and other genital tissues. Testosterone–receptor complexes are responsible for the maturation of wolffian duct structures and consequently for the formation of male internal genitalia during development, but DHT–receptor complexes are needed to form male exter-nal genitalia (Figure 25–18). DHT–receptor complexes are FIGURE 25–17 Hairline in children and adults. The hairline of the woman is like that of the child, whereas that of the man is indented in the lateral frontal region. FIGURE 25–18 Schematic diagram of the actions of testosterone (solid arrows) and dihydrotestosterone (dashed arrows). Note that they both bind to the same receptor, but DHT binds more effectively. (Reproduced with permission from Wilson JD, Griffin JE, Russell W: Steroid 5α-reductase 2 deficiency. Endocr Rev 1993;14:577. Copyright © 1993 by The Endocrine Society.) Testosterone Testis 5α-Reductase Dihydrotestosterone Receptor Gonadotropin regulation Spermato-genesis Sexual maturation at puberty Sexual differentiation Wolffian stimulation External virilization Luteinizing hormone Target cell CHAPTER 25 The Gonads: Development & Function of the Reproductive System 409 also primarily responsible for enlargement of the prostate and probably of the penis at the time of puberty, as well as for the facial hair, the acne, and the temporal recession of the hair-line. On the other hand, the increase in muscle mass and the development of male sex drive and libido depend primarily on testosterone rather than DHT (see Clinical Box 25–4). Testicular Production of Estrogens Over 80% of the estradiol and 95% of the estrone in the plasma of adult men is formed by extragonadal and extraadrenal aro-matization of circulating testosterone and androstenedione. The remainder comes from the testes. Some of the estradiol in testicular venous blood comes from the Leydig cells, but some is also produced by aromatization of androgens in Sertoli cells. In men, the plasma estradiol level is 20 to 50 pg/mL (73–184 pmol/L) and the total production rate is approximately 50 μg/ d (184 nmol/d). In contrast to the situation in women, estrogen production moderately increases with advancing age in men. CONTROL OF TESTICULAR FUNCTION FSH is tropic for Sertoli cells, and FSH and androgens main-tain the gametogenic function of the testes. FSH also stimu-lates the secretion of ABP and inhibin. Inhibin feeds back to inhibit FSH secretion. LH is tropic for Leydig cells and stimu-lates the secretion of testosterone, which in turn feeds back to inhibit LH secretion. Hypothalamic lesions in animals and hy-pothalamic disease in humans lead to atrophy of the testes and loss of their function. Inhibins Testosterone reduces plasma LH but, except in large doses, it has no effect on plasma FSH. Plasma FSH is elevated in patients who have atrophy of the seminiferous tubules but normal levels of testosterone and LH secretion. These observations led to the search for inhibin, a factor of testicular origin that inhibits FSH secretion. There are two inhibins in extracts of testes in men and in antral fluid from ovarian follicles in women. They are formed from three polypeptide subunits: a glycosylated α subunit with a molecular weight of 18,000; and two nonglycosylated β sub-units, βA and βB, each with a molecular weight of 14,000. The subunits are formed from precursor proteins (Figure 25–19). The α subunit combines with βA to form a heterodimer and with βB to form another heterodimer, with the subunits linked by disulfide bonds. Both αβA (inhibin A) and αβB (inhibin B) inhibit FSH secretion by a direct action on the pituitary, though it now appears that it is inhibin B that is the FSH-regulating in-hibin in adult men and women. Inhibins are produced by Ser-toli cells in males and granulosa cells in females. The heterodimer βAβB and the homodimers βAβA and βBβB are also formed. They stimulate rather than inhibit FSH secre-tion and consequently are called activins. Their function in reproduction is unsettled. However, the inhibins and activins are members of the TGFβ superfamily of dimeric growth fac-tors that also includes MIS. Activin receptors have been iden-tified and belong to the serine/threonine kinase receptor family. Inhibins and activins are found not only in the gonads but also in the brain and many other tissues. In the bone mar-row, activins are involved in the development of white blood cells. In embryonic life, activins are involved in the formation of mesoderm. All mice with a targeted deletion of the α-inhibin subunit gene initially exhibit normal growth but then develop gonadal stromal tumors, so the gene is a tumor sup-pressor gene. In plasma, α2-macroglobulin binds activins and inhibins. In tissues, activins bind to a family of four glycoproteins called follistatins. Binding of the activins inactivates their CLINICAL BOX 25–4 Congenital 5α-Reductase Deficiency Congenital 5α-reductase deficiency, in which the gene for type 2 5α-reductase is mutated, is common in certain parts of the Dominican Republic. It produces an interesting form of male pseudohermaphroditism. Individuals with this syn-drome are born with male internal genitalia including testes, but they have female external genitalia and are usually raised as girls. However, when they reach puberty, LH secre-tion and circulating testosterone levels are increased. Conse-quently, they develop male body contours and male libido. At this point, they usually change their gender identities and “become boys.” The clitoris enlarges (“penis-at-12 syn-drome”) to the point that some of the individuals can have intercourse with women. This enlargement probably occurs because with the high LH, enough testosterone is produced to overcome the need for DHT amplification in the genitalia. 5α-Reductase-inhibiting drugs are now being used clini-cally to treat benign prostatic hyperplasia, and finasteride, the most extensively used drug, has its greatest effect on type 2 5α-reductase. FIGURE 25–19 Inhibin precursor proteins and the various inhibins and activins that are formed from the carboxyl terminal regions of these precursors. SS, disulfide bonds. α precursor βA precursor βB precursor Inhibin A S S S S Activins α α βA βB S S βA βA S S βB βB S S βA βB Inhibin B 410 SECTION IV Endocrine & Reproductive Physiology biologic activity, but the relation of follistatins to inhibin and their physiologic function remain unsettled. Steroid Feedback A current “working hypothesis” of the way the functions of the testes are regulated by steroids is shown in Figure 25–20. Cas-tration is followed by a rise in the pituitary content and secre-tion of FSH and LH, and hypothalamic lesions prevent this rise. Testosterone inhibits LH secretion by acting directly on the anterior pituitary and by inhibiting the secretion of GnRH from the hypothalamus. Inhibin acts directly on the anterior pituitary to inhibit FSH secretion. In response to LH, some of the testosterone secreted from the Leydig cells bathes the seminiferous epithelium and provides the high local concentration of androgen to the Sertoli cells that is necessary for normal spermatogenesis. Systemically adminis-tered testosterone does not raise the androgen level in the testes to as great a degree, and it inhibits LH secretion. Consequently, the net effect of systemically administered testosterone is gen-erally a decrease in sperm count. Testosterone therapy has been suggested as a means of male contraception. However, the dose of testosterone needed to suppress spermatogenesis causes sodium and water retention. The possible use of inhibins as male contraceptives is now being explored. ABNORMALITIES OF TESTICULAR FUNCTION Cryptorchidism The testes develop in the abdominal cavity and normally mi-grate to the scrotum during fetal development. Testicular de-scent to the inguinal region depends on MIS, and descent from the inguinal region to the scrotum depends on other fac-tors. Descent is incomplete on one or, less commonly, both sides in 10% of newborn males, with the testes remaining in the abdominal cavity or inguinal canal. Gonadotropic hor-mone treatment speeds descent in some cases, or the defect can be corrected surgically. Spontaneous descent of the testes is the rule, and the proportion of boys with undescended testes (cryptorchidism) falls to 2% at age 1 y and 0.3% after puberty. However, early treatment is now recommended despite these figures because the incidence of malignant tumors is higher in undescended than in scrotal testes and because after puberty the higher temperature in the abdomen eventually causes irre-versible damage to the spermatogenic epithelium. Male Hypogonadism The clinical picture of male hypogonadism depends on whether testicular deficiency develops before or after puberty. In adults, if it is due to testicular disease, circulating gonado-tropin levels are elevated (hypergonadotropic hypogonad-ism); if it is secondary to disorders of the pituitary or the hypothalamus (eg, Kallmann syndrome), circulating gonado-tropin levels are depressed (hypogonadotropic hypogonad-ism). If the endocrine function of the testes is lost in adulthood, the secondary sex characteristics regress slowly be-cause it takes very little androgen to maintain them once they are established. The growth of the larynx during adolescence is permanent, and the voice remains deep. Men castrated in adulthood suffer some loss of libido, although the ability to copulate persists for some time. They occasionally have hot flushes and are generally more irritable, passive, and de-pressed than men with intact testes. When the Leydig cell de-ficiency dates from childhood, the clinical picture is that of eunuchoidism. Eunuchoid individuals over the age of 20 are characteristically tall, although not as tall as hyperpituitary gi-ants, because their epiphyses remain open and some growth continues past the normal age of puberty. They have narrow shoulders and small muscles, a body configuration resembling that of the adult female. The genitalia are small and the voice high-pitched. Pubic hair and axillary hair are present because of adrenocortical androgen secretion. However, the hair is sparse, and the pubic hair has the female “triangle with the base up” distribution rather than the “triangle with the base down” pattern (male escutcheon) seen in normal males. Androgen-Secreting Tumors “Hyperfunction” of the testes in the absence of tumor forma-tion is not a recognized entity. Androgen-secreting Leydig cell tumors are rare and cause detectable endocrine symptoms only in prepubertal boys, who develop precocious pseudopu-berty (Table 25–2). FIGURE 25–20 Postulated interrelationships between the hypothalamus, anterior pituitary, and testes. Solid arrows indicate excitatory effects; dashed arrows indicate inhibitory effects. GnRH LH Androgenic and anabolic effects Testosterone FSH Leydig cells Sertoli cells Inhibin B Hypothalamus Anterior pituitary Testis CHAPTER 25 The Gonads: Development & Function of the Reproductive System 411 THE FEMALE REPRODUCTIVE SYSTEM THE MENSTRUAL CYCLE The reproductive system of women (Figure 25–21), unlike that of men, shows regular cyclic changes that teleologically may be regarded as periodic preparations for fertilization and pregnancy. In humans and other primates, the cycle is a men-strual cycle, and its most conspicuous feature is the periodic vaginal bleeding that occurs with the shedding of the uterine mucosa (menstruation). The length of the cycle is notoriously variable in women, but an average figure is 28 days from the start of one menstrual period to the start of the next. By com-mon usage, the days of the cycle are identified by number, starting with the first day of menstruation. Ovarian Cycle From the time of birth, there are many primordial follicles under the ovarian capsule. Each contains an immature ovum (Figure 25–21). At the start of each cycle, several of these fol-licles enlarge, and a cavity forms around the ovum (antrum formation). This cavity is filled with follicular fluid. In hu-mans, usually one of the follicles in one ovary starts to grow FIGURE 25–21 Functional anatomy of the female reproductive tract. The female reproductive organs include the ovaries, the uterus and the fallopian tubes, and the breast/mammary glands. The sequential development of a follicle, the formation of a corpus luteum and follicular atresia are shown. Vagina Urethra Bladder Rectum Ovary Uterine tube Uterus Vagina Uterine cavity Cervix Broad ligament Ovary Uterine tube Ligament of the ovary Fundus Ovarian artery Interstitial cell mass Antral follicle Theca externa Theca interna Granulosa Follicular fluid Mature follicle Antrum Corpus hemorrhagicum Ovulated oocyte Primordial follicle Germinal epithelium Ovarian stroma Atretic follicle Young corpus luteum Mature corpus luteum Regressing corpus luteum Corpus albicans Blood vessels 412 SECTION IV Endocrine & Reproductive Physiology rapidly on about the sixth day and becomes the dominant fol-licle, while the others regress, forming atretic follicles. The atretic process involves apoptosis. It is uncertain how one fol-licle is selected to be the dominant follicle in this follicular phase of the menstrual cycle, but it seems to be related to the ability of the follicle to secrete the estrogen inside it that is needed for final maturation. When women are given highly purified human pituitary gonadotropin preparations by injec-tion, many follicles develop simultaneously. The structure of a maturing ovarian (graafian) follicle is shown in Figure 25–21. The primary source of circulating estrogen is the granulosa cells of the ovaries; however, the cells of the theca interna of the follicle are necessary for the production of estrogen as they secrete androgens that are aro-matized to estrogen by the granulosa cells. At about the 14th day of the cycle, the distended follicle rup-tures, and the ovum is extruded into the abdominal cavity. This is the process of ovulation. The ovum is picked up by the fim-briated ends of the uterine tubes (oviducts). It is transported to the uterus and, unless fertilization occurs, out through the vagina. The follicle that ruptures at the time of ovulation promptly fills with blood, forming what is sometimes called a corpus hemorrhagicum. Minor bleeding from the follicle into the abdominal cavity may cause peritoneal irritation and fleeting lower abdominal pain (“mittelschmerz”). The granulosa and theca cells of the follicle lining promptly begin to proliferate, and the clotted blood is rapidly replaced with yellowish, lipid-rich luteal cells, forming the corpus luteum. This initiates the luteal phase of the menstrual cycle, during which the luteal cells secrete estrogen and progesterone. Growth of the corpus luteum depends on its developing an adequate blood supply, and there is evidence that vascular endothelial growth factor (VEGF) (see Chapter 32) is essential for this process. If pregnancy occurs, the corpus luteum persists and usually there are no more periods until after delivery. If pregnancy does not occur, the corpus luteum begins to degenerate about 4 d before the next menses (24th day of the cycle) and is even-tually replaced by scar tissue, forming a corpus albicans. The ovarian cycle in other mammals is similar, except that in many species more than one follicle ovulates and multiple births are the rule. Corpora lutea form in some submamma-lian species but not in others. In humans, no new ova are formed after birth. During fetal development, the ovaries contain over 7 million primordial fol-licles. However, many undergo atresia (involution) before birth and others are lost after birth. At the time of birth, there are 2 million ova, but 50% of these are atretic. The million that are normal undergo the first part of the first meiotic division at about this time and enter a stage of arrest in prophase in which those that survive persist until adulthood. Atresia continues during development, and the number of ova in both of the ova-ries at the time of puberty is less than 300,000 (Figure 25–10). Only one of these ova per cycle (or about 500 in the course of a normal reproductive life) normally reaches maturity; the remainder degenerate. Just before ovulation, the first meiotic division is completed. One of the daughter cells, the secondary oocyte, receives most of the cytoplasm, while the other, the first polar body, fragments and disappears. The secondary oocyte immediately begins the second meiotic division, but this division stops at metaphase and is completed only when a sperm penetrates the oocyte. At that time, the second polar body is cast off and the fertilized ovum proceeds to form a new individual. The arrest in metaphase is due, at least in some spe-cies, to formation in the ovum of the protein pp39mos, which is encoded by the c-mos protooncogene. When fertilization occurs, the pp39mos is destroyed within 30 min by calpain, a calcium-dependent cysteine protease. Uterine Cycle At the end of menstruation, all but the deep layers of the en-dometrium have sloughed. A new endometrium then regrows under the influence of estrogens from the developing follicle. The endometrium increases rapidly in thickness from the 5th to the 14th days of the menstrual cycle. As the thickness in-creases, the uterine glands are drawn out so that they lengthen (Figure 25–22), but they do not become convoluted or secrete to any degree. These endometrial changes are called prolifera-tive, and this part of the menstrual cycle is sometimes called the proliferative phase. It is also called the preovulatory or follic-ular phase of the cycle. After ovulation, the endometrium be-comes more highly vascularized and slightly edematous under the influence of estrogen and progesterone from the corpus lu-teum. The glands become coiled and tortuous and they begin to secrete a clear fluid. Consequently, this phase of the cycle is called the secretory or luteal phase. Late in the luteal phase, the endometrium, like the anterior pituitary, produces prolactin, but the function of this endometrial prolactin is unknown. The endometrium is supplied by two types of arteries. The superficial two thirds of the endometrium that is shed during menstruation, the stratum functionale, is supplied by long, coiled spiral arteries (Figure 25–23), whereas the deep layer that is not shed, the stratum basale, is supplied by short, straight basilar arteries. When the corpus luteum regresses, hormonal support for the endometrium is withdrawn. The endometrium becomes thinner, which adds to the coiling of the spiral arteries. Foci of necrosis appear in the endometrium, and these coalesce. In addition, spasm and degeneration of the walls of the spiral arteries take place, leading to spotty hemorrhages that become confluent and produce the menstrual flow. The vasospasm is probably produced by locally released prostaglandins. Large quantities of prostaglandins are present in the secretory endometrium and in menstrual blood, and infusions of prostagladin F2α (PGF2α) produce endometrial necrosis and bleeding. From the point of view of endometrial function, the prolif-erative phase of the menstrual cycle represents restoration of the epithelium from the preceding menstruation, and the secretory phase represents preparation of the uterus for implantation of the fertilized ovum. The length of the secretory CHAPTER 25 The Gonads: Development & Function of the Reproductive System 413 phase is remarkably constant at about 14 d, and the variations seen in the length of the menstrual cycle are due for the most part to variations in the length of the proliferative phase. When fertilization fails to occur during the secretory phase, the endometrium is shed and a new cycle starts. Normal Menstruation Menstrual blood is predominantly arterial, with only 25% of the blood being of venous origin. It contains tissue debris, prostaglandins, and relatively large amounts of fibrinolysin from endometrial tissue. The fibrinolysin lyses clots, so that menstrual blood does not normally contain clots unless the flow is excessive. The usual duration of the menstrual flow is 3 to 5 d, but flows as short as 1 d and as long as 8 d can occur in normal women. The amount of blood lost may range normally from slight spotting to 80 mL; the average amount lost is 30 mL. Loss of more than 80 mL is abnormal. Obviously, the amount of flow can be affected by various factors, including the thick-ness of the endometrium, medication, and diseases that affect the clotting mechanism. Anovulatory Cycles In some instances, ovulation fails to occur during the men-strual cycle. Such anovulatory cycles are common for the first 12 to 18 mo after menarche and again before the onset of the menopause. When ovulation does not occur, no corpus lu-teum is formed and the effects of progesterone on the en-dometrium are absent. Estrogens continue to cause growth, FIGURE 25–22 Relationship between ovarian and uterine changes during the menstrual cycle. (Reproduced with permission from Windmaier EP, Raff H, Strang KT: Vander’s Human Physiology: The Mechanisms of Body Function, 11th ed. McGraw-Hill, 2008.) Follicle Ovum Corpus luteum Follicular Follicular Luteal Menstrual Day Ovarian phase Uterine phase Endometrial thickness Ovarian event Menstrual Secretory Proliferative Ovulation Estrogen Progesterone Estrogen 1 5 10 15 20 25 28 5 FIGURE 25–23 Spiral artery of endometrium. Drawing of a spiral artery (left) and two uterine glands (right) from the endometri-um of a rhesus monkey; early secretory phase. (Reproduced with permission from Daron GH: The arterial pattern of the tunica mucosa of the uterus in the Macacus rhesus. Am J Anat 1936;58:349.) Myometrium Uterine lumen 414 SECTION IV Endocrine & Reproductive Physiology however, and the proliferative endometrium becomes thick enough to break down and begins to slough. The time it takes for bleeding to occur is variable, but it usually occurs in less than 28 d from the last menstrual period. The flow is also vari-able and ranges from scanty to relatively profuse. Cyclical Changes in the Uterine Cervix Although it is continuous with the body of the uterus, the cer-vix of the uterus is different in a number of ways. The mucosa of the uterine cervix does not undergo cyclical desquamation, but there are regular changes in the cervical mucus. Estrogen makes the mucus thinner and more alkaline, changes that pro-mote the survival and transport of sperms. Progesterone makes it thick, tenacious, and cellular. The mucus is thinnest at the time of ovulation, and its elasticity, or spinnbarkeit, increases so that by midcycle, a drop can be stretched into a long, thin thread that may be 8 to 12 cm or more in length. In addition, it dries in an arborizing, fern-like pattern (Figure 25–24) when a thin layer is spread on a slide. After ovulation and during preg-nancy, it becomes thick and fails to form the fern pattern. Vaginal Cycle Under the influence of estrogens, the vaginal epithelium be-comes cornified, and cornified epithelial cells can be identified in the vaginal smear. Under the influence of progesterone, a thick mucus is secreted, and the epithelium proliferates and becomes infiltrated with leukocytes. The cyclical changes in the vaginal smear in rats are relatively marked. The changes in humans and other species are similar but not so clear-cut. Cyclical Changes in the Breasts Although lactation normally does not occur until the end of pregnancy, cyclical changes take place in the breasts during the menstrual cycle. Estrogens cause proliferation of mamma-ry ducts, whereas progesterone causes growth of lobules and alveoli. The breast swelling, tenderness, and pain experienced by many women during the 10 d preceding menstruation are probably due to distention of the ducts, hyperemia, and edema of the interstitial tissue of the breast. All these changes regress, along with the symptoms, during menstruation. Changes During Intercourse During sexual excitement in women, fluid is secreted onto the vaginal walls, probably because of release of VIP from vaginal nerves. A lubricating mucus is also secreted by the vestibular glands. The upper part of the vagina is sensitive to stretch, while tactile stimulation from the labia minora and clitoris adds to the sexual excitement. These stimuli are reinforced by tactile stimuli from the breasts and, as in men, by visual, audi-tory, and olfactory stimuli, which may build to the crescendo known as orgasm. During orgasm, autonomically mediated rhythmic contractions occur in the vaginal walls. Impulses also travel via the pudendal nerves and produce rhythmic con-traction of the bulbocavernosus and ischiocavernosus mus-cles. The vaginal contractions may aid sperm transport but are not essential for it, since fertilization of the ovum is not depen-dent on orgasm. Indicators of Ovulation Knowing when during the menstrual cycle ovulation occurs is important in increasing fertility or, conversely, in family plan-ning. A convenient and reasonably reliable indicator of the time of ovulation is a change—usually a rise—in the basal body temperature (Figure 25–25). The rise starts 1 to 2 d after ovulation. Women interested in obtaining an accurate tem-perature chart should use a digital thermometer and take their temperatures (oral or rectal) in the morning before getting out of bed. The cause of the temperature change at the time of ovulation is probably the increase in progesterone secretion, since progesterone is thermogenic. A surge in LH secretion triggers ovulation, and ovulation normally occurs about 9 h after the peak of the LH surge at midcycle (Figure 25–25). The ovum lives for approximately 72 h after it is extruded from the follicle, but it is fertilizable for a much shorter time than this. In a study of the relation of isolated intercourse to pregnancy, 36% of women had a detected pregnancy following intercourse on the day of ovula-tion, but with intercourse on days after ovulation, the percent-age was zero. Isolated intercourse on the first and second day before ovulation also led to pregnancy in about 36% of FIGURE 25–24 Patterns formed when cervical mucus is smeared on a slide, permitted to dry, and examined under the microscope. Progesterone makes the mucus thick and cellular. In the smear from a patient who failed to ovulate (bottom), no progesterone is present to inhibit the estrogen-induced fern pattern. Normal cycle, 14th day Midluteal phase, normal cycle Anovulatory cycle with estrogen present CHAPTER 25 The Gonads: Development & Function of the Reproductive System 415 women. A few pregnancies resulted from isolated intercourse on day 3, 4, or 5 before ovulation, although the percentage was much lower, for example, 8% on day 5 before ovulation. Thus, some sperms can survive in the female genital tract and fertilize the ovum for up to 120 h before ovulation, but the most fertile period is clearly the 48 h before ovulation. How-ever, for those interested in the “rhythm method” of contra-ception, it should be noted that there are rare but documented cases in the literature of pregnancy resulting from isolated coitus on every day of the cycle. The Estrous Cycle Mammals other than primates do not menstruate, and their sexual cycle is called an estrous cycle. It is named for the con-spicuous period of “heat” (estrus) at the time of ovulation, normally the only time during which the sexual interest of the female is aroused. In spontaneously ovulating species with es-trous cycles, such as the rat, no episodic vaginal bleeding oc-curs but the underlying endocrine events are essentially the same as those in the menstrual cycle. In other species, ovula-tion is induced by copulation (reflex ovulation). FIGURE 25–25 Basal body temperature and plasma hormone concentrations (mean ± standard error) during the normal human menstrual cycle. Values are aligned with respect to the day of the midcycle LH peak. FSH, follicle-stimulating hormone; LH, luteinizing hormone; M, menses. ˚C 36.6 36.8 36.4 20 80 0 40 60 50 0 100 150 200 Follicular phase Luteal phase Basal body temperature Inhibin A (pg/mL) Inhibin B (pg/mL) 20 50 0 10 30 40 500 0 1000 1500 Progesterone (nmol/L) Estradiol (pmol/L) 20 40 0 10 30 5 0 10 15 LH (U/L) FSH (U/L) -14 -7 0 7 14 Days relative to midcycle LH peak M 416 SECTION IV Endocrine & Reproductive Physiology OVARIAN HORMONES Chemistry, Biosynthesis, & Metabolism of Estrogens The naturally occurring estrogens are 17β-estradiol, estrone, and estriol (Figure 25–26). They are C18 steroids which do not have an angular methyl group attached to the 10 position or a Δ4-3-keto configuration in the A ring. They are secreted primarily by the granulosa cells of the ovarian follicles, the cor-pus luteum, and the placenta. Their biosynthesis depends on the enzyme aromatase (CYP19), which converts testosterone to estradiol and androstenedione to estrone (Figure 25–26). The latter reaction also occurs in fat, liver, muscle, and the brain. Theca interna cells have many LH receptors, and LH acts via cAMP to increase conversion of cholesterol to androstenedi-one. The theca interna cells supply androstenedione to the granulosa cells. The granulosa cells make estradiol when pro-vided with androgens (Figure 25–27), and it appears that the estradiol they form in primates is secreted into the follicular fluid. Granulosa cells have many FSH receptors, and FSH facili-tates their secretion of estradiol by acting via cAMP to increase their aromatase activity. Mature granulosa cells also acquire LH receptors, and LH also stimulates estradiol production. Two percent of the circulating estradiol is free, and the remainder is bound to protein: 60% to albumin and 38% to the same gonadal steroid-binding globulin (GBG) that binds testosterone (Table 25–5). In the liver, estradiol, estrone, and estriol are converted to glucuronide and sulfate conjugates. All these compounds, along with other metabolites, are excreted in the urine. Appreciable amounts are secreted in the bile and reabsorbed into the blood-stream (enterohepatic circulation). FIGURE 25–26 Biosynthesis and metabolism of estrogens. The formulas of the precursor steroids are shown in Figure 22–7. HO O HO OH HO OH Estrone (E1) 17β-Estradiol (E2) Estriol (E3) 16-Ketoestrone 16α-Hydroxyestrone OH Aromatase Aromatase Testosterone Androstenedione 17α-Hydroxyprogesterone 17α-Hydroxypregnenolone Progesterone Pregnenolone Cholesterol Dehydroepiandrosterone FIGURE 25–27 Interactions between theca and granulosa cells in estradiol synthesis and secretion. Theca Interna Cells Cholesterol Cholesterol Androstenedione Estrone Granulosa Cells Circulation Antrum Androstenedione Estrone Estradiol Estradiol CHAPTER 25 The Gonads: Development & Function of the Reproductive System 417 Secretion The concentration of estradiol in the plasma during the men-strual cycle is shown in Figure 25–25. Almost all of this estro-gen comes from the ovary, and two peaks of secretion occur: one just before ovulation and one during the midluteal phase. The estradiol secretion rate is 36 μg/d (133 nmol/d) in the ear-ly follicular phase, 380 μg/d just before ovulation, and 250 μg/d during the midluteal phase (Table 25–7). After menopause, estrogen secretion declines to low levels. As noted previously, the estradiol production rate in men is about 50 μg/d (184 nmol/d). Effects on the Female Genitalia Estrogens facilitate the growth of the ovarian follicles and in-crease the motility of the uterine tubes. Their role in the cyclic changes in the endometrium, cervix, and vagina has been dis-cussed previously. They increase uterine blood flow and have important effects on the smooth muscle of the uterus. In im-mature and castrated females, the uterus is small and the my-ometrium atrophic and inactive. Estrogens increase the amount of uterine muscle and its content of contractile pro-teins. Under the influence of estrogens, the muscle becomes more active and excitable, and action potentials in the individ-ual fibers become more frequent. The “estrogen-dominated” uterus is also more sensitive to oxytocin. Chronic treatment with estrogens causes the endometrium to hypertrophy. When estrogen therapy is discontinued, sloughing takes place with withdrawal bleeding. Some “break-through” bleeding may occur during treatment when estrogens are given for long periods. Effects on Endocrine Organs Estrogens decrease FSH secretion. Under some circumstanc-es, they inhibit LH secretion (negative feedback); in other cir-cumstances, they increase LH secretion (positive feedback). Women are sometimes given large doses of estrogens for 4 to 6 d to prevent conception after coitus during the fertile period (postcoital or “morning-after” contraception). However, in this instance, pregnancy is probably prevented by interference with implantation of the ovum rather than changes in gonad-otropin secretion. Estrogens cause increased secretion of angiotensinogen and thyroid-binding globulin. They exert an important protein anabolic effect in chickens and cattle, possibly by stimulating the secretion of androgens from the adrenal, and estrogen treatment has been used commercially to increase the weight of domestic animals. They cause epiphysial closure in humans (see Chapter 23). Effects on the Central Nervous System The estrogens are responsible for estrous behavior in animals, and they increase libido in humans. They apparently exert this action by a direct effect on certain neurons in the hypo-thalamus (Figure 25–28). Estrogens also increase the prolifer-ation of dendrites on neurons and the number of synaptic knobs in rats. Effects on the Breasts Estrogens produce duct growth in the breasts and are largely re-sponsible for breast enlargement at puberty in girls; they have been called the growth hormones of the breast. They are re-sponsible for the pigmentation of the areolas, although pigmen-tation usually becomes more intense during the first pregnancy than it does at puberty. The role of the estrogens in the overall control of breast growth and lactation is discussed below. TABLE 25–7 Twenty-four-hour production rates of sex steroids in women at different stages of the menstrual cycle. Sex Steroids Early Follicular Preovulatory Midluteal Progesterone (mg) 1.0 4.0 25.0 17-hydroxyprogesterone (mg) 0.5 4.0 4.0 Dehydroepiandrosterone (mg) 7.0 7.0 7.0 Androstenedione (mg) 2.6 4.7 3.4 Testosterone (μg) 144.0 171.0 126.0 Estrone (μg) 50.0 350.0 250.0 Estradiol (μg) 36.0 380.0 250.0 Modified and reproduced, with permission, from Yen SSC, Jaffe RB, Barbieri RL: Reproductive Endocrinology, 4th ed. Saunders, 1999. FIGURE 25–28 Loci where implantations of estrogen in the hypothalamus affect ovarian weight and sexual behavior in rats, projected on a sagittal section of the hypothalamus. The implants that stimulate sex behavior are located in the suprachiasmatic area above the optic chiasm (blue area), whereas ovarian atrophy is pro-duced by implants in the arcuate nucleus and surrounding ventral hy-pothalamus (red). MB, mamillary body. Third ventricle Massa intermedia Midbrain MB Pituitary Optic chiasm 418 SECTION IV Endocrine & Reproductive Physiology Female Secondary Sex Characteristics The body changes that develop in girls at puberty—in addition to enlargement of breasts, uterus, and vagina—are due in part to estrogens, which are the “feminizing hormones,” and in part simply to the absence of testicular androgens. Women have nar-row shoulders and broad hips, thighs that converge, and arms that diverge (wide carrying angle). This body configuration, plus the female distribution of fat in the breasts and buttocks, is seen also in castrate males. In women, the larynx retains its pre-pubertal proportions and the voice stays high-pitched. Women have less body hair and more scalp hair, and the pubic hair gen-erally has a characteristic flat-topped pattern (female escutch-eon). However, growth of pubic and axillary hair in both sexes is due primarily to androgens rather than estrogens. Other Actions Normal women retain salt and water and gain weight just be-fore menstruation. Estrogens cause some degree of salt and water retention. However, aldosterone secretion is slightly ele-vated in the luteal phase, and this also contributes to the pre-menstrual fluid retention. Estrogens are said to make sebaceous gland secretions more fluid and thus to counter the effect of testosterone and inhibit formation of comedones (“black-heads”) and acne. The liver palms, spider angiomas, and slight breast enlargement seen in advanced liver disease are due to increased circulating estro-gens. The increase appears to be due to decreased hepatic metabolism of androstenedione, making more of this andro-gen available for conversion to estrogens. Estrogens have a significant plasma cholesterol-lowering action, and they rapidly produce vasodilation by increasing the local production of NO. Their action on bone is discussed in Chapter 23. Mechanism of Action There are two principal types of nuclear estrogen receptors: estrogen receptor α (ERα) encoded by a gene on chromosome 6; and estrogen receptor β (ERβ), encoded by a gene on chro-mosome 14. Both are members of the nuclear receptor super-family (see Chapter 2). After binding estrogen, they form homodimers and bind to DNA, altering its transcription. Some tissues contain one type or the other, but overlap also occurs, with some tissues containing both ERα and ERβ. ERα is found primarily in the uterus, kidneys, liver, and heart, whereas ERβ is found primarily in the ovaries, prostate, lungs, gastrointestinal tract, hemopoietic system, and central ner-vous system (CNS). They also form heterodimers with ERα binding to ERβ. Male and female mice in which the gene for ERα has been knocked out are sterile, develop osteoporosis, and continue to grow because their epiphyses do not close. ERβ female knockouts are infertile, but ERβ male knockouts are fertile even though they have hyperplastic prostates and loss of fat. Both receptors exist in isoforms and, like thyroid re-ceptors, can bind to various activating and stimulating factors. In some situations, ERβ can inhibit ERα transcription. Thus, their actions are complex, multiple, and varied. Most of the effects of estrogens are genomic, that is, due to actions on the nucleus, but some are so rapid that it is difficult to believe they are mediated via production of mRNAs. These include effects on neuronal discharge in the brain and, possi-bly, feedback effects on gonadotropin secretion. Evidence is accumulating that these effects are mediated by cell mem-brane receptors that appear to be structurally related to the nuclear receptors and produce their effects by intracellular mitogen-activated protein kinase pathways. Similar rapid effects of progesterone, testosterone, glucocorticoids, aldos-terone, and 1,25-dihydroxycholecalciferol may also be pro-duced by membrane receptors. Synthetic and Environmental Estrogens The ethinyl derivative of estradiol is a potent estrogen and, unlike the naturally occurring estrogens, is relatively active when given by mouth because it is resistant to hepatic metab-olism. The activity of the naturally occurring hormones is low when they are administered by mouth because the portal venous drainage of the intestine carries them to the liver, where they are inactivated before they can reach the general circulation. Some nonsteroidal substances and a few com-pounds found in plants have estrogenic activity. The plant es-trogens are rarely a problem in human nutrition, but they may cause undesirable effects in farm animals. Dioxins, which are found in the environment and are produced by a variety of industrial processes, can activate estrogen response elements on genes. However, they have been reported to have antiestrogenic as well as estrogenic effects, and their role, if any, in the production of human disease remains a matter of disagreement and debate. Because natural estrogens have undesirable as well as desir-able effects (for example, they preserve bone in osteoporosis but can cause uterine and breast cancer), there has been an active search for “tailor-made” estrogens that have selective effects in humans. Two compounds, tamoxifen and ralox-ifene, show promise in this regard. Neither combats the symptoms of menopause, but both have the bone-preserving effects of estradiol. In addition, tamoxifen does not stimulate the breast, and raloxifene does not stimulate the breast or uterus. The way the effects of these selective estrogen receptor modulators (SERMs) are brought about is related to the com-plexity of the estrogen receptors and hence to differences in the way receptor–ligand complexes they form bind to DNA. Chemistry, Biosynthesis, & Metabolism of Progesterone Progesterone is a C21 steroid (Figure 25–29) secreted by the corpus luteum, the placenta, and (in small amounts) the folli-cle. It is an important intermediate in steroid biosynthesis in all tissues that secrete steroid hormones, and small amounts apparently enter the circulation from the testes and adrenal CHAPTER 25 The Gonads: Development & Function of the Reproductive System 419 cortex. About 2% of the circulating progesterone is free (Table 25–5), whereas 80% is bound to albumin and 18% is bound to corticosteroid-binding globulin. Progesterone has a short half-life and is converted in the liver to pregnanediol, which is conjugated to glucuronic acid and excreted in the urine. Secretion In men, the plasma progesterone level is approximately 0.3 ng/mL (1 nmol/L). In women, the level is approximately 0.9 ng/mL (3 nmol/L) during the follicular phase of the menstrual cycle (Figure 25–25). The difference is due to secretion of small amounts of progesterone by cells in the ovarian follicles; theca cells provide pregnenolone to the granulosa cells, which con-vert it to progesterone. Late in the follicular phase, progester-one secretion begins to increase. During the luteal phase, the corpus luteum produces large quantities of progesterone (Table 25–7) and plasma progesterone is markedly increased to a peak value of approximately 18 ng/mL (60 nmol/L). The stimulating effect of LH on progesterone secretion by the corpus luteum is due to activation of adenylyl cyclase and involves a subsequent step that is dependent on protein synthesis. Actions The principal target organs of progesterone are the uterus, the breasts, and the brain. Progesterone is responsible for the progestational changes in the endometrium and the cyclic changes in the cervix and vagina described above. It has an antiestrogenic effect on the myometrial cells, decreasing their excitability, their sensitivity to oxytocin, and their spontaneous electrical activity while increasing their membrane poten-tial. It also decreases the number of estrogen receptors in the endometrium and increases the rate of conversion of 17β-estradiol to less active estrogens. In the breast, progesterone stimulates the development of lobules and alveoli. It induces differentiation of estrogen-pre-pared ductal tissue and supports the secretory function of the breast during lactation. The feedback effects of progesterone are complex and are exerted at both the hypothalamic and pituitary levels. Large doses of progesterone inhibit LH secretion and potentiate the inhibitory effect of estrogens, preventing ovulation. Progesterone is thermogenic and is probably responsible for the rise in basal body temperature at the time of ovulation. It stimulates respiration, and the alveolar PCO2 (see Chapter 35) in women during the luteal phase of the menstrual cycle is lower than that in men. In pregnancy, the PCO2 falls as progesterone secretion rises. However, the physiologic signifi-cance of this respiratory response is unknown. Large doses of progesterone produce natriuresis, probably by blocking the action of aldosterone on the kidney. The hor-mone does not have a significant anabolic effect. Mechanism of Action The effects of progesterone, like those of other steroids, are brought about by an action on DNA to initiate synthesis of new mRNA. The progesterone receptor is bound to a heat shock protein in the absence of the steroid, and progesterone binding releases the heat shock protein, exposing the DNA-binding do-main of the receptor. The synthetic steroid mifepristone (RU 486) binds to the receptor but does not release the heat shock protein, and it blocks the binding of progesterone. Because the maintenance of early pregnancy depends on the stimulatory ef-fect of progesterone on endometrial growth and its inhibition of uterine contractility, mifepristone combined with a prostaglan-din can be used to produce elective abortions. There are two isoforms of the progesterone receptor—PRA and PRB—that are produced by differential processing from a single gene. PRA is a truncated form, but it is likely that both isoforms mediate unique subsets of progesterone action. Substances that mimic the action of progesterone are some-times called progestational agents, gestagens, or progestins. They are used along with synthetic estrogens as oral contra-ceptive agents. FIGURE 25–29 Biosynthesis of progesterone and major pathway for its metabolism. Other metabolites are also formed. HO HO CH3 C O Cholesterol Sodium pregnanediol-20-glucuronide Pregnenolone CH3 CHOH Pregnanediol O CH3 C O Progesterone 420 SECTION IV Endocrine & Reproductive Physiology Relaxin Relaxin is a polypeptide hormone that is produced in the cor-pus luteum, uterus, placenta, and mammary glands in women and in the prostate gland in men. During pregnancy, it relaxes the pubic symphysis and other pelvic joints and softens and dilates the uterine cervix. Thus, it facilitates delivery. It also in-hibits uterine contractions and may play a role in the develop-ment of the mammary glands. In nonpregnant women, relaxin is found in the corpus luteum and the endometrium during the secretory but not the proliferative phase of the menstrual cycle. Its function in nonpregnant women is un-known. In men, it is found in semen, where it may help main-tain sperm motility and aid in sperm penetration of the ovum. In most species there is only one relaxin gene, but in humans there are two genes on chromosome 9 that code for two struc-turally different polypeptides that both have relaxin activity. However, only one of these genes is active in the ovary and the prostate. The structure of the polypeptide produced in these two tissues is shown in Figure 25–30. CONTROL OF OVARIAN FUNCTION FSH from the pituitary is responsible for the early maturation of the ovarian follicles, and FSH and LH together are responsi-ble for their final maturation. A burst of LH secretion (Figure 25–25) is responsible for ovulation and the initial formation of the corpus luteum. A smaller midcycle burst of FSH secretion also occurs, the significance of which is uncertain. LH stimu-lates the secretion of estrogen and progesterone from the cor-pus luteum. Hypothalamic Components The hypothalamus occupies a key position in the control of gonadotropin secretion. Hypothalamic control is exerted by GnRH secreted into the portal hypophysial vessels. GnRH stimulates the secretion of FSH as well as LH. GnRH is normally secreted in episodic bursts, and these bursts produce the circhoral peaks of LH secretion. They are essential for normal secretion of gonadotropins. If GnRH is administered by constant infusion, the GnRH receptors in the anterior pituitary down-regulate and LH secretion declines to zero. However, if GnRH is administered episodically at a rate of one pulse per hour, LH secretion is stimulated. This is true even when endogenous GnRH secretion has been prevented by a lesion of the ventral hypothalamus. It is now clear not only that episodic secretion of GnRH is a general phenomenon but also that fluctuations in the frequency and amplitude of the GnRH bursts are important in generating the other hormonal changes that are responsible for the men-strual cycle. Frequency is increased by estrogens and decreased by progesterone and testosterone. The frequency increases late in the follicular phase of the cycle, culminating in the LH surge. During the secretory phase, the frequency decreases as a result of the action of progesterone (Figure 25–31), but when estro-gen and progesterone secretion decrease at the end of the cycle, the frequency once again increases. At the time of the midcycle LH surge, the sensitivity of the gonadotropes to GnRH is greatly increased because of their FIGURE 25–30 Structure of human luteal and seminal relaxin. Pca, pyroglutamic acid. (Modified and reproduced with permission from Winslow JW et al: Human seminal relaxin is a product of the same gene as human luteal relaxin. Endocrinology 1992;130:2660. Copyright © 1992 by The Endocrine Society.) Pca Leu Leu Leu Leu Ser Ser Ser Ser Ser Ala Ala Ala Asn Tyr Leu Lys Lys Lys Arg Arg Arg Arg Ala Gly Ala Gln Ile Ile Ile Phe Cys Cys Cys His Val Val Val Gly Glu Glu Glu Gly Met Met Cys Cys Cys Thr Thr Trp Trp Asp S S COO− COO− A1 A5 A10 A15 A20 A24 B25 B29 B20 B15 B10 B5 B1 NH3 S S S S FIGURE 25–31 Episodic secretion of LH (s) and FSH (d) during the follicular stage (top) and the luteal stage (bottom) of the menstrual cycle. The numbers above each graph indicate the numbers of LH pulses per 12 hours and the plasma estradiol (E2) and progesterone (P) concentrations at these two times of the cycle. (Reproduced with permission from Marshall JC, Kelch RO: Gonadotropin-releasing hormone: Role of pulsatile secretion in the regulation of reproduction. N Engl J Med 1986;315:1459.) 0 4 8 12 16 (7.5/12h) P 0 4 8 12 16 20 8 PM 12 AM 4 PM 8 AM Time LH ( ), FSH ( ) mlU/mL 98 pg/mL 0.4 ng/mL E2 LH pulse (4/12h) P 192 pg/mL 20 ng/mL E2 LH pulse CHAPTER 25 The Gonads: Development & Function of the Reproductive System 421 exposure to GnRH pulses of the frequency that exist at this time. This self-priming effect of GnRH is important in pro-ducing a maximum LH response. The nature and the exact location of the GnRH pulse gener-ator in the hypothalamus are still unsettled. However, it is known in a general way that norepinephrine and possibly epi-nephrine in the hypothalamus increase GnRH pulse frequen-cies. Conversely, opioid peptides such as the enkephalins and β-endorphin reduce the frequency of GnRH pulses. The down-regulation of pituitary receptors and the conse-quent decrease in LH secretion produced by constantly ele-vated levels of GnRH has led to the use of long-acting GnRH analogs to inhibit LH secretion in precocious puberty and in cancer of the prostate. Feedback Effects Changes in plasma LH, FSH, sex steroids, and inhibin during the menstrual cycle are shown in Figure 25–25, and their feed-back relations are diagrammed in Figure 25–32. During the early part of the follicular phase, inhibin B is low and FSH is modestly elevated, fostering follicular growth. LH secretion is held in check by the negative feedback effect of the rising plas-ma estrogen level. At 36 to 48 h before ovulation, the estrogen feedback effect becomes positive, and this initiates the burst of LH secretion (LH surge) that produces ovulation. Ovulation occurs about 9 h after the LH peak. FSH secretion also peaks, despite a small rise in inhibin, probably because of the strong stimulation of gonadotropes by GnRH. During the luteal phase, the secretion of LH and FSH is low because of the ele-vated levels of estrogen, progesterone, and inhibin. It should be emphasized that a moderate, constant level of circulating estrogen exerts a negative feedback effect on LH secretion, whereas during the cycle, an elevated estrogen level exerts a positive feedback effect and stimulates LH secretion. It has been demonstrated that in monkeys estrogens must also be elevated for a minimum time to produce positive feedback. When circulating estrogen was increased about 300% for 24 h, only negative feedback was seen; but when it was increased about 300% for 36 h or more, a brief decline in secretion was followed by a burst of LH secretion that resembled the midcy-cle surge. When circulating levels of progesterone were high, the positive feedback effect of estrogen was inhibited. There is evidence that in primates, both the negative and the positive feedback effects of estrogen are exerted in the mediobasal hypothalamus, but exactly how negative feedback is switched to positive feedback and then back to negative feedback in the luteal phase remains unknown. Control of the Cycle In an important sense, regression of the corpus luteum (luteol-ysis) starting 3 to 4 d before menses is the key to the menstrual cycle. PGF2α appears to be a physiologic luteolysin, but this prostaglandin is only active when endothelial cells producing ET-1 (see Chapter 33) are present. Therefore, it appears that at least in some species luteolysis is produced by the combined action of PGF2α and ET-1. In some domestic animals, oxytocin secreted by the corpus luteum appears to exert a local luteolytic effect, possibly by causing the release of prostaglandins. Once luteolysis begins, the estrogen and progesterone levels fall and the secretion of FSH and LH increases. A new crop of follicles develops, and then a single dominant follicle matures as a result of the action of FSH and LH. Near midcycle, estrogen secretion from the follicle rises. This rise augments the responsiveness of the pituitary to GnRH and triggers a burst of LH secretion. The resulting ovulation is followed by formation of a corpus lu-teum. Estrogen secretion drops, but progesterone and estrogen levels then rise together, along with inhibin B. The elevated le-vels inhibit FSH and LH secretion for a while, but luteolysis again occurs and a new cycle starts. Reflex Ovulation Female cats, rabbits, mink, and some other animals have long periods of estrus, during which they ovulate only after copula-tion. Such reflex ovulation is brought about by afferent im-pulses from the genitalia and the eyes, ears, and nose that FIGURE 25–32 Feedback regulation of ovarian function. The cells of the theca interna provide androgens to the granulosa cells, and theca cells also produce the circulating estrogens that inhibit the secretion of GnRH, LH, and FSH. Inhibin from the granulosa cells inhib-its FSH secretion. LH regulates the thecal cells, whereas the granulosa cells are regulated by both LH and FSH. The dashed arrows indicate in-hibitory effects and the solid arrows stimulatory effects. Androgens GnRH LH Estrogenic effects Estrogen FSH Theca interna Granu-losa Inhibin B Hypothalamus Anterior pituitary Ovary 422 SECTION IV Endocrine & Reproductive Physiology converge on the ventral hypothalamus and provoke an ovula-tion-inducing release of LH from the pituitary. In species such as rats, monkeys, and humans, ovulation is a spontaneous pe-riodic phenomenon, but neural mechanisms are also involved. Ovulation can be delayed 24 h in rats by administering pento-barbital or various other neurally active drugs 12 h before the expected time of follicle rupture. Contraception Methods commonly used to prevent conception are listed in Table 25–8, along with their failure rates. Once conception has occurred, abortion can be produced by progesterone antago-nists such as mifepristone. Implantation of foreign bodies in the uterus causes changes in the duration of the sexual cycle in a number of mammalian species. In humans, such foreign bodies do not alter the men-strual cycle, but they act as effective contraceptive devices. Intrauterine implantation of pieces of metal or plastic (intra-uterine devices, IUDs) has been used in programs aimed at controlling population growth. Although the mechanism of action of IUDs is still unsettled, they seem in general to pre-vent sperms from fertilizing ova. Those containing copper appear to exert a spermatocidal effect. IUDs that slowly release progesterone or synthetic progestins have the additional effect of thickening cervical mucus so that entry of sperms into the uterus is impeded. IUDs can cause intrauterine infections, but these usually occur in the first month after insertion and in women exposed to sexually transmitted diseases. Women undergoing long-term treatment with relatively large doses of estrogen do not ovulate, probably because they have depressed FSH levels and multiple irregular bursts of LH secretion rather than a single midcycle peak. Women treated with similar doses of estrogen plus a progestational agent do not ovulate because the secretion of both gonadotropins is suppressed. In addition, the progestin makes the cervical mucus thick and unfavorable to sperm migration, and it may also interfere with implantation. For contraception, an orally active estrogen such as ethinyl estradiol is often combined with a synthetic progestin such as norethindrone. The pills are administered for 21 d, then withdrawn for 5 to 7 d to permit menstrual flow, and started again. Like ethinyl estradiol, nor-ethindrone has an ethinyl group on position 17 of the steroid nucleus, so it is resistant to hepatic metabolism and conse-quently is effective by mouth. In addition to being a progestin, it is partly metabolized to ethinyl estradiol, and for this reason it also has estrogenic activity. Small as well as large doses of estrogen are effective (Table 25–8). Implants made up primarily of progestins such as levonor-gestrel are now seeing increased use in some parts of the world. These are inserted under the skin and can prevent pregnancy for up to 5 y. They often produce amenorrhea, but otherwise they appear to be effective and well tolerated. ABNORMALITIES OF OVARIAN FUNCTION Menstrual Abnormalities Some women who are infertile have anovulatory cycles; they fail to ovulate but have menstrual periods at fairly regular in-tervals. As noted above, anovulatory cycles are the rule for the first 1 to 2 y after menarche and again before the menopause. Amenorrhea is the absence of menstrual periods. If menstrual bleeding has never occurred, the condition is called primary amenorrhea. Some women with primary amenorrhea have small breasts and other signs of failure to mature sexually. Cessation of cycles in a woman with previously normal peri-ods is called secondary amenorrhea. The most common cause of secondary amenorrhea is pregnancy, and the old clin-ical maxim that “secondary amenorrhea should be considered to be due to pregnancy until proved otherwise” has consider-able merit. Other causes of amenorrhea include emotional stimuli and changes in the environment, hypothalamic diseas-es, pituitary disorders, primary ovarian disorders, and various systemic diseases. Evidence suggests that in some women with hypothalamic amenorrhea, the frequency of GnRH pulses is slowed as a result of excess opioid activity in the hypothala-mus. In encouraging preliminary studies, the frequency of GnRH pulses has been increased by administration of the orally active opioid blocker naltrexone. TABLE 25–8 Relative effectiveness of frequently used contraceptive methods. Method Failures per 100 Woman-Years Vasectomy 0.02 Tubal ligation and similar procedures 0.13 Oral contraceptives > 50 mg estrogen and progestin 0.32 < 50 mg estrogen and progestin 0.27 Progestin only 1.2 IUD Copper 7 1.5 Loop D 1.3 Diaphragm 1.9 Condom 3.6 Withdrawal 6.7 Spermicide 11.9 Rhythm 15.5 Data from Vessey M, Lawless M, Yeates D: Efficacy of different contraceptive meth-ods. Lancet 1982;1:841. Reproduced with permission. CHAPTER 25 The Gonads: Development & Function of the Reproductive System 423 The terms hypomenorrhea and menorrhagia refer to scanty and abnormally profuse flow, respectively, during regu-lar periods. Metrorrhagia is bleeding from the uterus between periods, and oligomenorrhea is reduced frequency of periods. Dysmenorrhea is painful menstruation. The severe menstrual cramps that are common in young women quite often disap-pear after the first pregnancy. Most of the symptoms of dys-menorrhea are due to accumulation of prostaglandins in the uterus, and symptomatic relief has been obtained by treatment with inhibitors of prostaglandin synthesis. Some women develop symptoms such as irritability, bloat-ing, edema, decreased ability to concentrate, depression, headache, and constipation during the last 7 to 10 d of their menstrual cycles. These symptoms of the premenstrual syn-drome (PMS) have been attributed to salt and water reten-tion. However, it seems unlikely that this or any of the other hormonal alterations that occur in the late luteal phase are responsible because the time course and severity of the symp-toms are not modified if the luteal phase is terminated early and menstruation produced by administration of mifepris-tone. The antidepressant fluoxetine (Prozac), which is a sero-tonin reuptake inhibitor, and the benzodiazepine alprazolam (Xanax) produce symptomatic relief, and so do GnRH-releas-ing agonists in doses that suppress the pituitary–ovarian axis. How these diverse clinical observations fit together to pro-duce a picture of the pathophysiology of PMS is still unknown (see Clinical Box 25–5). PREGNANCY Fertilization & Implantation In humans, fertilization of the ovum by the sperm usually oc-curs in the ampulla of the uterine tube. Fertilization involves (1) chemoattraction of the sperm to the ovum by substances produced by the ovum; (2) adherence to the zona pellucida, the membranous structure surrounding the ovum; (3) pene-tration of the zona pellucida and the acrosome reaction; and (4) adherence of the sperm head to the cell membrane of the ovum, with breakdown of the area of fusion and release of the sperm nucleus into the cytoplasm of the ovum (Figure 25–33). Millions of sperms are deposited in the vagina during inter-course. Eventually, 50 to 100 sperms reach the ovum, and many of them contact the zona pellucida. Sperms bind to a sperm receptor in the zona, and this is followed by the acroso-mal reaction, that is, the breakdown of the acrosome, the lyso-some-like organelle on the head of the sperm (Figure 25–14). Various enzymes are released, including the trypsin-like pro-tease acrosin. Acrosin facilitates but is not required for the penetration of the sperm through the zona pellucida. When one sperm reaches the membrane of the ovum, fusion to the ovum membrane is mediated by fertilin, a protein on the sur-face of the sperm head that resembles the viral fusion proteins that permit some viruses to attack cells. The fusion provides the signal that initiates development. In addition, the fusion sets off a reduction in the membrane potential of the ovum that prevents polyspermy, the fertilization of the ovum by more than one sperm. This transient potential change is followed by a structural change in the zona pellucida that provides protec-tion against polyspermy on a more long-term basis. The developing embryo, now called a blastocyst, moves down the tube into the uterus. This journey takes about 3 d, during which the blastocyst reaches the 8- or 16-cell stage. Once in contact with the endometrium, the blastocyst becomes surrounded by an outer layer of syncytiotrophoblast, a multi-nucleate mass with no discernible cell boundaries, and an inner layer of cytotrophoblast made up of individual cells. The syn-cytiotrophoblast erodes the endometrium, and the blastocyst CLINICAL BOX 25–5 Genetic Defects Causing Reproductive Abnormalities A number of single-gene mutations cause reproductive ab-normalities when they occur in women. Examples include (1) Kallmann syndrome, which causes hypogonadotropic hy-pogonadism; (2) GnRH resistance, FSH resistance, and LH re-sistance, which are due to defects in the GnRH, FSH, or LH receptors, respectively; and (3) aromatase deficiency, which prevents the formation of estrogens. These are all caused by loss-of-function mutations. An interesting gain-of-function mutation causes the McCune–Albright syndrome, in which Gsα becomes constitutively active in certain cells but not others (mosaicism) because a somatic mutation after initial cell division has occurred in the embryo. It is associated with multiple endocrine abnormalities, including precocious pu-berty and amenorrhea with galactorrhea. FIGURE 25–33 Sequential events in fertilization in mammals. Sperm are attracted to the ovum, bind to the zona pellucida, release acrosomal enzymes, penetrate the zona pellucida, and fuse with the membrane of the ovum, releasing the sperm nucleus into its cyto-plasm. Current evidence indicates that the side, rather than the tip, of the sperm head fuses with the egg cell membrane. (Modified from Vacquier VD: Evolution of gamete recognition proteins. Science 1999;281:1995.) Zona pellucida Egg cytoplasm Egg cell membrane Sperm tail Nucleus Acrosome 424 SECTION IV Endocrine & Reproductive Physiology burrows into it (implantation). The implantation site is usually on the dorsal wall of the uterus. A placenta then develops, and the trophoblast remains associated with it. Failure to Reject the “Fetal Graft” It should be noted that the fetus and the mother are two genet-ically distinct individuals, and the fetus is in effect a transplant of foreign tissue in the mother. However, the transplant is tol-erated, and the rejection reaction that is characteristically pro-duced when other foreign tissues are transplanted (see Chapter 3) fails to occur. The way the “fetal graft” is protected is unknown. However, one explanation may be that the pla-cental trophoblast, which separates maternal and fetal tissues, does not express the polymorphic class I and class II MHC genes and instead expresses HLA-G, a nonpolymorphic gene. Therefore, antibodies against the fetal proteins do not devel-op. In addition, there is a Fas ligand on the surface of the pla-centa, and this bonds to T cells, causing them to undergo apoptosis (see Chapter 3). Infertility The vexing clinical problem of infertility often requires exten-sive investigation before a cause is found. In 30% of cases the problem is in the man; in 45%, the problem is in the woman; in 20%, both partners have a problem; and in 5% no cause can be found. In vitro fertilization, that is, removing mature ova, fertilizing them with sperm, and implanting one or more of them in the uterus at the four-cell stage is of some value in these cases. It has a 5–10% chance of producing a live birth. Endocrine Changes In all mammals, the corpus luteum in the ovary at the time of fertilization fails to regress and instead enlarges in response to stimulation by gonadotropic hormones secreted by the placen-ta. The placental gonadotropin in humans is called human chorionic gonadotropin (hCG). The enlarged corpus luteum of pregnancy secretes estrogens, progesterone, and relaxin. The relaxin helps maintain pregnancy by inhibiting myometrial contractions. In most species, removal of the ovaries at any time during pregnancy precipitates abortion. In humans, however, the placenta produces sufficient estrogen and progesterone from maternal and fetal precursors to take over the function of the corpus luteum after the sixth week of pregnancy. Ovariecto-my before the sixth week leads to abortion, but ovariectomy thereafter has no effect on the pregnancy. The function of the corpus luteum begins to decline after 8 wk of pregnancy, but it persists throughout pregnancy. hCG secretion decreases after an initial marked rise, but estrogen and progesterone secretion increase until just before parturition (Table 25–9). Human Chorionic Gonadotropin hCG is a glycoprotein that contains galactose and hexosamine. It is produced by the syncytiotrophoblast. Like the pituitary glycoprotein hormones, it is made up of α and β subunits. hCG-α is identical to the α subunit of LH, FSH, and TSH. The molecular weight of hCG-α is 18,000, and that of hCG-β is 28,000. hCG is primarily luteinizing and luteotropic and has little FSH activity. It can be measured by radioimmunoassay and detected in the blood as early as 6 d after conception. Its presence in the urine in early pregnancy is the basis of the var-ious laboratory tests for pregnancy, and it can sometimes be detected in the urine as early as 14 d after conception. It ap-pears to act on the same receptor as LH. hCG is not absolutely specific for pregnancy. Small amounts are secreted by a variety of gastrointestinal and other tumors in both sexes, and hCG has been measured in individuals with suspected tumors as a “tumor marker.” It also appears that the fetal liver and kidney normally produce small amounts of hCG. Human Chorionic Somatomammotropin The syncytiotrophoblast also secretes large amounts of a pro-tein hormone that is lactogenic and has a small amount of growth-stimulating activity. This hormone has been called chorionic growth hormone-prolactin (CGP) and human placental lactogen (hPL), but it is now generally called hu-man chorionic somatomammotropin (hCS). The structure of hCS is very similar to that of human growth hormone (see Figure 24–3), and it appears that these two hormones and pro-lactin evolved from a common progenitor hormone. Large quantities of hCS are found in maternal blood, but very little reaches the fetus. Secretion of growth hormone from the ma-ternal pituitary is not increased during pregnancy and may ac-tually be decreased by hCS. However, hCS has most of the actions of growth hormone and apparently functions as a “maternal growth hormone of pregnancy” to bring about the nitrogen, potassium, and calcium retention, lipolysis, and de-creased glucose utilization seen in this state. These latter two actions divert glucose to the fetus. The amount of hCS secreted is proportionate to the size of the placenta, which normally weighs about one-sixth as much as the fetus, and low hCS le-vels are a sign of placental insufficiency. TABLE 25–9 Hormone levels in human maternal blood during normal pregnancy. Hormone Approximate Peak Value Time of Peak Secretion hCG 5 mg/mL First trimester Relaxin 1 ng/mL First trimester hCS 15 mg/mL Term Estradiol 16 ng/mL Term Estriol 14 ng/mL Term Progesterone 190 ng/mL Term Prolactin 200 ng/mL Term CHAPTER 25 The Gonads: Development & Function of the Reproductive System 425 Other Placental Hormones In addition to hCG, hCS, progesterone, and estrogens, the pla-centa secretes other hormones. Human placental fragments probably produce proopiomelanocortin (POMC). In culture, they release corticotropin-releasing hormone (CRH), β-endorphin, α-melanocyte-stimulating hormone (MSH), and dynorphin A, all of which appear to be identical to their hypo-thalamic counterparts. They also secrete GnRH and inhibin, and since GnRH stimulates and inhibin inhibits hCG secre-tion, locally produced GnRH and inhibin may act in a paracrine fashion to regulate hCG secretion. The trophoblast cells and amnion cells also secrete leptin, and moderate amounts of this satiety hormone enter the maternal circulation. Some also en-ters the amniotic fluid. Its function in pregnancy is unknown. The placenta also secretes prolactin in a number of forms. Finally, the placenta secretes the α subunits of hCG, and the plasma concentration of free α subunits rises throughout pregnancy. These α subunits acquire a carbohydrate composi-tion that makes them unable to combine with β subunits, and their prominence suggests that they have a function of their own. It is interesting in this regard that the secretion of the prolactin produced by the endometrium also appears to increase throughout pregnancy, and it may be that the circu-lating α subunits stimulate endometrial prolactin secretion. The cytotrophoblast of the human chorion contains prore-nin (see Chapter 39). A large amount of prorenin is also present in amniotic fluid, but its function in this location is unknown. Fetoplacental Unit The fetus and the placenta interact in the formation of steroid hormones. The placenta synthesizes pregnenolone and proges-terone from cholesterol. Some of the progesterone enters the fe-tal circulation and provides the substrate for the formation of cortisol and corticosterone in the fetal adrenal glands (Figure 25–34). Some of the pregnenolone enters the fetus and, along with pregnenolone synthesized in the fetal liver, is the substrate for the formation of dehydroepiandrosterone sulfate (DHEAS) and 16-hydroxydehydroepiandrosterone sulfate (16-OHDH-EAS) in the fetal adrenal. Some 16-hydroxylation also occurs in the fetal liver. DHEAS and 16-OHDHEAS are transported back to the placenta, where DHEAS forms estradiol and 16-OHDH-EAS forms estriol. The principal estrogen formed is estriol, and since fetal 16-OHDHEAS is the principal substrate for the es-trogens, the urinary estriol excretion of the mother can be mon-itored as an index of the state of the fetus. Parturition The duration of pregnancy in humans averages 270 d from fertilization (284 d from the first day of the menstrual period preceding conception). Irregular uterine contractions increase in frequency in the last month of pregnancy. The difference between the body of the uterus and the cer-vix becomes evident at the time of delivery. The cervix, which is firm in the nonpregnant state and throughout pregnancy until near the time of delivery, softens and dilates, while the body of the uterus contracts and expels the fetus. There is still considerable uncertainty about the mecha-nisms responsible for the onset of labor. One factor is the increase in circulating estrogens produced by increased circu-lating DHEAS. This makes the uterus more excitable, increases the number of gap junctions between myometrial cells, and causes production of more prostaglandins, which in turn cause uterine contractions. In humans, CRH secretion by the fetal hypothalamus increases and is supplemented by increased pla-cental production of CRH. This increases circulating adreno-corticotropic hormone (ACTH) in the fetus, and the resulting increase in cortisol hastens the maturation of the respiratory system. Thus, in a sense, the fetus picks the time to be born by increasing CRH secretion. The number of oxytocin receptors in the myometrium and the decidua (the endometrium of pregnancy) increases more than 100-fold during pregnancy and reaches a peak during early labor. Estrogens increase the number of oxytocin recep-tors, and uterine distention late in pregnancy may also increase their formation. In early labor, the oxytocin concen-tration in maternal plasma is not elevated from the prelabor value of about 25 pg/mL. It is possible that the marked increase in oxytocin receptors causes the uterus to respond to normal plasma oxytocin concentrations. However, at least in rats, the amount of oxytocin mRNA in the uterus increases, reaching a peak at term; this suggests that locally produced oxytocin also participates in the process. Premature onset of labor is a problem because premature infants have a high mortality rate and often require intensive, expensive care. Intramuscular 17α-hydroxyprogesterone causes a significant decrease in the incidence of premature labor. The mechanism by which it exerts its effect is uncertain, but it may be that the steroid provides a stable level of circulating progesterone. Progesterone relaxes uterine smooth muscle, inhibits the action of oxytocin on the muscle, and reduces FIGURE 25–34 Interactions between the placenta and the fetal adrenal cortex in the production of steroids. Placenta Cholesterol Pregnenolone DHEAS Fetal Adrenal 16-OHDHEAS Cortisol, corticosterone Progesterone Estradiol DHEAS Estriol 16-OHDHEAS 426 SECTION IV Endocrine & Reproductive Physiology the formation of gap junctions between the muscle fibers. All these actions would be expected to inhibit the onset of labor. Once labor is started, the uterine contractions dilate the cervix, and this dilation in turn sets up signals in afferent nerves that increase oxytocin secretion (Figure 25–35). The plasma oxytocin level rises and more oxytocin becomes avail-able to act on the uterus. Thus, a positive feedback loop is established that aids delivery and terminates on expulsion of the products of conception. Oxytocin increases uterine con-tractions in two ways: (1) It acts directly on uterine smooth muscle cells to make them contract, and (2) it stimulates the formation of prostaglandins in the decidua. The prostaglan-dins enhance the oxytocin-induced contractions. During labor, spinal reflexes and voluntary contractions of the abdominal muscles (“bearing down”) also aid in delivery. However, delivery can occur without bearing down and with-out a reflex increase in secretion of oxytocin from the poster-ior pituitary gland, since paraplegic women can go into labor and deliver. LACTATION Development of the Breasts Many hormones are necessary for full mammary develop-ment. In general, estrogens are primarily responsible for pro-liferation of the mammary ducts and progesterone for the development of the lobules. In rats, some prolactin is also needed for development of the glands at puberty, but it is not known if prolactin is necessary in humans. During pregnancy, prolactin levels increase steadily until term, and levels of estro-gens and progesterone are elevated as well, producing full lob-uloalveolar development. Secretion & Ejection of Milk The composition of human and cows’ milk is shown in Table 25–10. In estrogen- and progesterone-primed rodents, injec-tions of prolactin cause the formation of milk droplets and their secretion into the ducts. Oxytocin causes contraction of the myoepithelial cells lining the duct walls, with consequent ejection of the milk through the nipple. FIGURE 25–35 Role of oxytocin in parturition. Increase in oxytocin receptors Prostaglandins Uterine contractions Dilation of cervix and distention of vagina Stimuli from cervix and vagina Increased secretion of oxytocin TABLE 25–10 Composition of colostrum and milk. Component Human Colostrum Human Milk Cows’ Milk Water, g . . . 88 88 Lactose, g 5.3 6.8 5.0 Protein, g 2.7 1.2 3.3 Casein: lactalbumin ratio . . . 1:2 3:1 Fat, g 2.9 3.8 3.7 Linoleic acid . . . 8.3% of fat 1.6% of fat Sodium, mg 92 15 58 Potassium, mg 55 55 138 Chloride, mg 117 43 103 Calcium, mg 31 33 125 Magnesium, mg 4 4 12 Phosphorus, mg 14 15 100 Iron, mg 0.092 0.15a 0.10a Vitamin A, μg 89 53 34 Vitamin D, μg . . . 0.03a 0.06a Thiamine, μg 15 16 42 Riboflavin, μg 30 43 157 Nicotinic acid, μg 75 172 85 Ascorbic acid, mg 4.4a 4.3a 1.6a Weights per deciliter. aPoor source. Reproduced with permission from Findlay ALR: Lactation. Res Reprod (Nov) 1974;6(6). CHAPTER 25 The Gonads: Development & Function of the Reproductive System 427 Initiation of Lactation after Delivery The breasts enlarge during pregnancy in response to high cir-culating levels of estrogens, progesterone, prolactin, and pos-sibly hCG. Some milk is secreted into the ducts as early as the fifth month, but the amounts are small compared with the surge of milk secretion that follows delivery. In most animals, milk is secreted within an hour after delivery, but in women it takes 1 to 3 d for the milk to “come in.” After expulsion of the placenta at parturition, the levels of circulating estrogens and progesterone abruptly decline. The drop in circulating estrogen initiates lactation. Prolactin and estrogen synergize in producing breast growth, but estrogen antagonizes the milk-producing effect of prolactin on the breast. Indeed, in women who do not wish to nurse their babies, estrogens may be administered to stop lactation. Suckling not only evokes reflex oxytocin release and milk ejection, it also maintains and augments the secretion of milk because of the stimulation of prolactin secretion produced by suckling. Effect of Lactation on Menstrual Cycles Women who do not nurse their infants usually have their first menstrual period 6 wk after delivery. However, women who nurse regularly have amenorrhea for 25 to 30 wk. Nursing stimulates prolactin secretion, and evidence suggests that pro-lactin inhibits GnRH secretion, inhibits the action of GnRH on the pituitary, and antagonizes the action of gonadotropins on the ovaries. Ovulation is inhibited, and the ovaries are in-active, so estrogen and progesterone output falls to low levels. Consequently, only 5–10% of women become pregnant again during the suckling period, and nursing has long been known to be an important, if only partly effective, method of birth control. Furthermore, almost 50% of the cycles in the first 6 mo after resumption of menses are anovulatory (see Clinical Box 25–6). Gynecomastia Breast development in the male is called gynecomastia. It may be unilateral but is more commonly bilateral. It is common, occurring in about 75% of newborns because of transplacental passage of maternal estrogens. It also occurs in mild, transient form in 70% of normal boys at the time of puberty and in many men over the age of 50. It occurs in androgen resistance. It is a complication of estrogen therapy and is seen in patients with estrogen-secreting tumors. It is found in a wide variety of seemingly unrelated conditions, including eunuchoidism, hy-perthyroidism, and cirrhosis of the liver. Digitalis can produce it, apparently because cardiac glycosides are weakly estrogen-ic. It can also be caused by many other drugs. It has been seen in malnourished prisoners of war, but only after they were lib-erated and eating an adequate diet. A feature common to many and perhaps all cases of gynecomastia is an increase in the plasma estrogen:androgen ratio due to either increased circulating estrogens or decreased circulating androgens. HORMONES & CANCER About 35% of carcinomas of the breast in women of childbear-ing age are estrogen-dependent; their continued growth de-pends on the presence of estrogens in the circulation. The tumors are not cured by decreasing estrogen secretion, but symptoms are dramatically relieved, and the tumor regresses for months or years before recurring. Women with estrogen-dependent tumors often have a remission when their ovaries are removed. Inhibition of the action of estrogens with tamox-ifen also produces remissions, and inhibition of estrogen for-mation with drugs that inhibit aromatase (Figure 25–26) is even more effective. Some carcinomas of the prostate are androgen-dependent and regress temporarily after the removal of the testes or treatment with GnRH agonists in doses that are sufficient to produce down-regulation of the GnRH receptors on gonadot-ropes and decrease LH secretion. CHAPTER SUMMARY ■Differences between males and females depend primarily on a single chromosome (the Y chromosome) and a single pair of endocrine structures (the gonads); testes in the male and ovaries in the female. ■The gonads have a dual function: the production of germ cells (gametogenesis) and the secretion of sex hormones. The testes secrete large amounts of androgens, principally testosterone, but they also secrete small amounts of estrogens. The ovaries secrete large amounts of estrogens and small amounts of androgens. ■Spermatogonia develop into mature spermatozoa that start in the seminiferous tubules in a process called spermatogenesis. This is a multistep process that includes maturation of spermatogonia into primary spermatocytes, which undergo meiotic division, resulting in haploid secondary spermato-cytes and several further divisions result in spermatids. Each cell division from a spermatogonium to a spermatid is incom-plete with cells remaining connected via cytoplasmic bridges. Spermatids eventually mature into motile spermatozoa to CLINICAL BOX 25–6 Chiari–Frommel Syndrome An interesting, although rare, condition is persistence of lac-tation (galactorrhea) and amenorrhea in women who do not nurse after delivery. This condition, called the Chiari– Frommel syndrome, may be associated with some genital atrophy and is due to persistent prolactin secretion without the secretion of the FSH and LH necessary to produce matu-ration of new follicles and ovulation. A similar pattern of ga-lactorrhea and amenorrhea with high circulating prolactin levels is seen in nonpregnant women with chromophobe pituitary tumors and in women in whom the pituitary stalk has been sectioned during treatment of cancer. 428 SECTION IV Endocrine & Reproductive Physiology complete spermatogenesis; this last part of maturation is called spermiogenesis. ■Testosterone is the principal hormone of the testis. It is synthe-sized from cholesterol in Leydig cells. The secretion of testoster-one from Leydig cells is under control of luteinizing hormone at a rate of 4 to 9 mg/day in adult males. Most testosterone is bound to albumin or to gonadal steroid-binding globulin in the plasma. Testosterone plays an important role in the develop-ment and maintenance of male secondary sex characteristics, as well as other defined functions. ■Ovaries also secrete progesterone, a steroid that has special functions in preparing the uterus for pregnancy. During preg-nancy the ovaries secrete relaxin, which facilitates the delivery of the fetus. In both sexes, the gonads secrete other polypeptides, including inhibin B, a polypeptide that inhibits FSH secretion. ■In women, a period called perimenopause precedes menopause, and can last up to ten years; during this time the menstrual cycles become irregular and the level of inhibins decrease. ■Once in menopause, the ovaries no longer secrete progesterone and 17β-estradiol and estrogen is formed only in small amounts by aromatization of androstenedione in peripheral tissues. ■The naturally occurring estrogens are 17β-estradiol, estrone, and estriol. They are secreted primarily by the granulosa cells of the ovarian follicles, the corpus luteum, and the placenta. Their bio-synthesis depends on the enzyme aromatase (CYP19), which con-verts testosterone to estradiol and androstenedione to estrone. The latter reaction also occurs in fat, liver, muscle, and the brain. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. If a young woman has high plasma levels of T3, cortisol, and renin activity but her blood pressure is only slightly elevated and she has no symptoms or signs of thyrotoxicosis or Cushing syn-drome, the most likely explanation is that A) she has been treated with TSH and ACTH. B) she has been treated with T3 and cortisol. C) she is in the third trimester of pregnancy. D) she has an adrenocortical tumor. E) she has been subjected to chronic stress. 2. Full development and function of the seminiferous tubules require A) somatostatin. B) LH. C) oxytocin. D) FSH. E) androgens and FSH. 3. In humans, fertilization usually occurs in the A) vagina. B) cervix. C) uterine cavity. D) uterine tubes. E) abdominal cavity. 4. In human males, testosterone is produced mainly by the A) Leydig cells. B) Sertoli cells. C) seminiferous tubules. D) epididymis. E) vas deferens. 5. Home-use kits for determining a woman’s fertile period depend on the detection of one hormone in the urine. This hormone is A) FSH. B) progesterone. C) estradiol. D) hCG. E) LH. 6. Which of the following is not a steroid? A) 17α-hydroxyprogesterone B) estrone C) relaxin D) pregnenolone E) etiocholanolone 7. Which of the following probably triggers the onset of labor? A) ACTH in the fetus B) ACTH in the mother C) prostaglandins D) oxytocin E) placental renin CHAPTER RESOURCES Bole-Feysot C et al: Prolactin (PRL) and its receptor: Actions, signal transduction pathways, and phenotypes observed in PRL receptor knockout mice. Endocrinol Rev 1998;19:225. Mather JP, Moore A, Li R-H: Activins, inhibins, and follistatins: Further thoughts on a growing family of regulators. Proc Soc Exper Biol Med 1997;215:209. Matthews J, Gustafson J-A: Estrogen signaling: A subtle balance between ERα and ERβ. Mol Interv 2003;3:281. McLaughlin DT, Donahoe PR: Sex determination and differentiation. N Engl J Med 2004;350:367. Naz RK (editor): Endocrine Disruptors. CRC Press, 1998. Norwitz ER, Robinson JN, Challis JRG: The control of labor. N Engl J Med 1999;341:660. Primakoff P, Nyles DG: Penetration, adhesion, and fusion in mammalian sperm–egg interaction. Science 2002;296:2183. Simpson ER, et al: Aromatose—A brief overview. Annu Rev Physiol 2002;64:93. Yen SSC, Jaffe RB, Barbieri RL: Reproductive Endocrinology: Physiology, Pathophysiology, and Clinical Management, 4th ed. Saunders, 1999. 429 C H A P T E R SECTION V GASTROINTESTINAL PHYSIOLOGY 26 Overview of Gastrointestinal Function & Regulation O B J E C T I V E S After studying this chapter, you should be able to: ■Understand the functional significance of the gastrointestinal system, and in par-ticular, its roles in nutrient assimilation, excretion, and immunity. ■Describe the structure of the gastrointestinal tract, the glands that drain into it, and its subdivision into functional segments. ■List the major gastrointestinal secretions, their components, and the stimuli that regulate their production. ■Describe water balance in the gastrointestinal tract and explain how the level of luminal fluidity is adjusted to allow for digestion and absorption. ■Identify the major hormones, other peptides, and key neurotransmitters of the gastrointestinal system. ■Describe the special features of the enteric nervous system and the splanchnic circulation. INTRODUCTION The gastrointestinal tract is a continuous tube that stretches from the mouth to the anus. Its primary function is to serve as a portal whereby nutrients and water can be absorbed into the body. In fulfilling this function, the meal is mixed with a vari-ety of secretions that arise from both the gastrointestinal tract itself and organs that drain into it, such as the pancreas, gall-bladder, and salivary glands. Likewise, the intestine displays a variety of motility patterns that serve to mix the meal with digestive secretions and move it along the length of the gas-trointestinal tract. Ultimately, residues of the meal that cannot be absorbed, along with cellular debris and lipid-soluble met-abolic end products that are excreted in the bile rather than the urine, are expelled from the body. All of these functions are tightly regulated in concert with the ingestion of meals. Thus, the gastrointestinal system has evolved a large number of regulatory mechanisms that act both locally and to coordi-nate the function of the gut, and the organs that drain into it, over long distances. The lumen of the gastrointestinal tract is functionally contig-uous with the outside of the body. The intestine also has a very substantial surface area, which is important for its absorptive function. Finally, the gut is an unusual organ in that it becomes colonized, almost from birth, with a large number of commen-sal bacteria (particularly in the colon, or large intestine). This 430 SECTION V Gastrointestinal Physiology relationship is mutually beneficial, because the bacteria perform several metabolic functions that cannot be accomplished with mammalian enzymes, and likely also provide some degree of protection against subsequent infection with pathogenic micro-organisms that might cause disease. Nevertheless, the constant presence of bacterial and other stimuli, as well as the large sur-face area that must be defended against potentially harmful sub-stances, doubtlessly accounts for the fact that the intestine has a very well-developed local immune system that comprises both innate and adaptive immune effectors (see Chapter 3). Indeed, there are more lymphocytes in the wall of the intestine than there are circulating in the blood. STRUCTURAL CONSIDERATIONS The parts of the gastrointestinal tract that are encountered by the meal or its residues include, in order, the mouth, esopha-gus, stomach, duodenum, jejunum, ileum, cecum, colon, rec-tum, and anus. Throughout the length of the intestine, glandular structures deliver secretions into the lumen, particu-larly in the stomach and mouth. Also important in the process of digestion are secretions from the pancreas and the biliary system of the liver. The intestinal tract is also functionally di-vided into segments that restrict the flow of intestinal contents to optimize digestion and absorption. These sphincters include the upper and lower esophageal sphincters, the pylorus that re-tards emptying of the stomach, the ileocecal valve that retains colonic contents (including large numbers of bacteria) in the large intestine, and the inner and outer anal sphincters. After toilet training, the latter permit delaying the elimination of wastes until a time when it is socially convenient. The intestine is composed of functional layers (Figure 26–1). Immediately adjacent to nutrients in the lumen is a single layer of columnar epithelial cells. This represents the barrier that nutrients must traverse to enter the body. Below the epithelium is a layer of loose connective tissue known as the lamina pro-pria, which in turn is surrounded by concentric layers of smooth muscle, oriented circumferentially and then longitudi-nally to the axis of the gut (the circular and longitudinal muscle layers, respectively). The intestine is also amply supplied with blood vessels, nerve endings, and lymphatics, which are all important in its function. The epithelium of the intestine is also further specialized in a way that maximizes the surface area available for nutrient absorption. Throughout the small intestine, it is folded up into fingerlike projections called villi (Figure 26–2). Between the villi are infoldings known as crypts. Stem cells that give rise to both crypt and villus epithelial cells reside toward the base of the crypts and are responsible for completely renewing the epithelium every few days or so. Indeed, the gastrointesti-nal epithelium is one of the most rapidly dividing tissues in the body. Daughter cells undergo several rounds of cell divi-sion in the crypts then migrate out onto the villi, where they are eventually shed and lost in the stool. The villus epithelial cells are also notable for the extensive microvilli that charac-terize their apical membranes. These microvilli are endowed with a dense glycocalyx (the brush border) that probably FIGURE 26–1 Organization of the wall of the intestine into functional layers. (Adapted from Barrett KE: Gastrointestinal Physiology. McGraw-Hill, 2006.) Lumen Epithelium Basement memdrane Lamina propria Muscularis mucosa Submucosa Circular muscle Myenteric plexus Longitudinal muscle Mesothelium (Serosa) Mucosa Muscularis propria CHAPTER 26 Overview of Gastrointestinal Function & Regulation 431 protects the cells to some extent from the effects of digestive enzymes. Some digestive enzymes are also actually part of the brush border, being membrane-bound proteins. These so-called “brush border hydrolases” perform the final steps of digestion for specific nutrients. GASTROINTESTINAL SECRETIONS SALIVARY SECRETION The first secretion encountered when food is ingested is saliva. Saliva is produced by three pairs of salivary glands that drain into the oral cavity. It has a number of organic constituents that serve to initiate digestion (particularly of starch, mediated by amylase) and which also protect the oral cavity from bacteria (such as immunoglobulin A and lysozyme). Saliva also serves to lubricate the food bolus (aided by mucins). Saliva is also hypotonic com-pared with plasma and alkaline; the latter feature is important to neutralize any gastric secretions that reflux into the esophagus. The salivary glands consist of blind end pieces (acini) that produce the primary secretion containing the organic constit-uents dissolved in a fluid that is essentially identical in its composition to plasma. The salivary glands are actually extremely active when maximally stimulated, secreting their own weight in saliva every minute. To accomplish this, they are richly endowed with surrounding blood vessels that dilate when salivary secretion is initiated. The composition of the saliva is then modified as it flows from the acini out into ducts that eventually coalesce and deliver the saliva into the mouth. Na+ and Cl– are extracted and K+ and bicarbonate are added. Because the ducts are relatively impermeable to water, the loss of NaCl renders the saliva hypotonic, particularly at low secretion rates. As the rate of secretion increases, there is less time for NaCl to be extracted and the tonicity of the saliva rises, but it always stays somewhat hypotonic with respect to plasma. Overall, the three pairs of salivary glands that drain into the mouth supply 1000 to 1500 mL of saliva per day. Salivary secretion is almost entirely controlled by neural influences, with the parasympathetic branch of the autonomic nervous system playing the most prominent role (Figure 26–3). Sympathetic input slightly modifies the composition of saliva (particularly by increasing proteinaceous content), but has little influence on volume. Secretion is triggered by reflexes that are stimulated by the physical act of chewing, but is actually initi-ated even before the meal is taken into the mouth as a result of central triggers that are prompted by thinking about, seeing, or smelling food. Indeed, salivary secretion can readily be condi-tioned, as in the classical experiments of Pavlov where dogs were conditioned to salivate in response to a ringing bell by associating this stimulus with a meal. Salivary secretion is also prompted by nausea, but inhibited by fear or during sleep. Saliva performs a number of important functions: it facili-tates swallowing, keeps the mouth moist, serves as a solvent for the molecules that stimulate the taste buds, aids speech by facilitating movements of the lips and tongue, and keeps the mouth and teeth clean. The saliva also has some antibacterial action, and patients with deficient salivation (xerostomia) have a higher than normal incidence of dental caries. The buffers in saliva help maintain the oral pH at about 7.0. They also help neutralize gastric acid and relieve heartburn when gastric juice is regurgitated into the esophagus. GASTRIC SECRETION Food is stored in the stomach; mixed with acid, mucus, and pepsin; and released at a controlled, steady rate into the duodenum (see Clinical Box 26–1). ANATOMIC CONSIDERATIONS The gross anatomy of the stomach is shown in Figure 26–4. The gastric mucosa contains many deep glands. In the cardia and the pyloric region, the glands secrete mucus. In the body of the stomach, including the fundus, the glands also contain parietal (oxyntic) cells, which secrete hydrochloric acid and intrinsic factor, and chief (zymogen, peptic) cells, which se-crete pepsinogens (Figure 26–5) These secretions mix with FIGURE 26–2 The structure of intestinal villi and crypts. (Reproduced with permission, from Fox SI: Human Physiology, 10th ed. McGraw-Hill, 2008.) Simple columnar epithelium Lacteal Capillary network Goblet cells Intestinal crypt Lymph vessel Arteriole Venule Villus 432 SECTION V Gastrointestinal Physiology mucus secreted by the cells in the necks of the glands. Several of the glands open on a common chamber (gastric pit) that opens in turn on the surface of the mucosa. Mucus is also se-creted along with HCO3 – by mucus cells on the surface of the epithelium between glands. The stomach has a very rich blood and lymphatic supply. Its parasympathetic nerve supply comes from the vagi and its sympathetic supply from the celiac plexus. ORIGIN & REGULATION OF GASTRIC SECRETION The stomach also adds a significant volume of digestive juices to the meal. Like salivary secretion, the stomach actually read-ies itself to receive the meal before it is actually taken in, dur-ing the so-called cephalic phase that can be influenced by food preferences. Subsequently, there is a gastric phase of secretion that is quantitatively the most significant, and finally an intes-tinal phase once the meal has left the stomach. Each phase is closely regulated by both local and distant triggers. The gastric secretions (Table 26–1) arise from glands in the wall of the stomach that drain into its lumen, and also from the surface cells that secrete primarily mucus and bicarbonate to protect the stomach from digesting itself, as well as sub-stances known as trefoil peptides that stabilize the mucus-bicarbonate layer. The glandular secretions of the stomach differ in different regions of the organ. The most characteris-tic secretions derive from the glands in the fundus or body of the stomach. These contain two distinctive cell types from which the gastric secretions arise: the parietal cells, which secrete hydrochloric acid and intrinsic factor; and the chief cells, which produce pepsinogens and gastric lipase (Figure 26–5). The acid secreted by parietal cells serves to sterilize the meal and also to begin the hydrolysis of dietary macromole-cules. Intrinsic factor is important for the later absorption of vitamin B12, or cobalamin (Figure 26–6). Pepsinogen is the precursor of pepsin, which initiates protein digestion. Lipase similarly begins the digestion of dietary fats. There are three primary stimuli of gastric secretion, each with a specific role to play in matching the rate of secretion to functional requirements (Figure 26–7). Gastrin is a hormone that is released by G cells in the antrum of the stomach both in response to a specific neurotransmitter released from enteric nerve endings, known as gastrin releasing peptide (GRP, or bombesin), and also in response to the presence of oligopeptides in the gastric lumen. Gastrin is then carried through the bloodstream to the fundic glands, where it binds to receptors not only on parietal (and likely, chief cells) to activate secretion, but also on so-called enterochromaffin-like cells (ECL cells) that are located in the gland, and release his-tamine. Histamine is also a trigger of parietal cell secretion, via binding to H2 histamine receptors. Finally, parietal and chief cells can also be stimulated by acetylcholine, released from enteric nerve endings in the fundus. During the cephalic phase of gastric secretion, secretion is predominantly activated by vagal input that originates from the brain region known as the dorsal vagal complex, which coordi-nates input from higher centers. Vagal outflow to the stomach then releases GRP and acetylcholine, thereby initiating secretory FIGURE 26–3 Regulation of salivary secretion by the parasympathetic nervous system. ACh, acetylcholine. (Adapted from Barrett KE: Gas-trointestinal Physiology. McGraw-Hill, 2006.) Smell Taste Sound Sight Pressure in mouth ACh ACh Parasympathetics Sleep Fatigue Fear Increased salivary secretion via effects on Salivatory nucleus of medulla Higher centers Otic ganglion • Acinar secretion • Vasodilatation Parotid gland Submandibular gland Submandibular ganglion − CHAPTER 26 Overview of Gastrointestinal Function & Regulation 433 function. However, before the meal enters the stomach, there are few additional triggers and thus the amount of secretion is lim-ited. Once the meal is swallowed, on the other hand, meal con-stituents trigger substantial release of gastrin and the physical presence of the meal also distends the stomach and activates stretch receptors, which provoke a “vago-vagal” as well as local reflexes that further amplify secretion. The presence of the meal also buffers gastric acidity that would otherwise serve as a feed-back inhibitory signal to shut off secretion secondary to the release of somatostatin, which inhibits both G and ECL cells as well as secretion by parietal cells themselves (Figure 26–7). This probably represents a key mechanism whereby gastric secretion CLINICAL BOX 26–1 Peptic Ulcer Disease Gastric and duodenal ulceration in humans is related primar-ily to a breakdown of the barrier that normally prevents irrita-tion and autodigestion of the mucosa by the gastric secre-tions. Infection with the bacterium Helicobacter pylori disrupts this barrier, as do aspirin and other nonsteroidal anti-inflammatory drugs (NSAIDs), which inhibit the produc-tion of prostaglandins and consequently decrease mucus and HCO3 – secretion. The NSAIDs are widely used to combat pain and treat arthritis. An additional cause of ulceration is prolonged excess secretion of acid. An example of this is the ulcers that occur in the Zollinger–Ellison syndrome. This syndrome is seen in patients with gastrinomas. These tumors can occur in the stomach and duodenum, but most of them are found in the pancreas. The gastrin causes prolonged hy-persecretion of acid, and severe ulcers are produced. Gastric and duodenal ulcers can be given a chance to heal by inhibi-tion of acid secretion with drugs such as cimetidine that block the H2 histamine receptors on parietal cells or omepra-zole and related drugs that inhibit H+–K+ ATPase. H. pylori can be eradicated with antibiotics, and NSAID-induced ulcers can be treated by stopping the NSAID or, when this is not ad-visable, by treatment with the prostaglandin agonist miso-prostol. Gastrinomas can sometimes be removed surgically. FIGURE 26–4 Anatomy of the stomach. The principal secre-tions of the body and antrum are listed in parentheses. (Reproduced with permission from Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology: The Mechanisms of Body Function, 11th ed. McGraw-Hill, 2008.) Esophagus Body (secretes mucus, pepsinogen, and HCI) Duodenum Pyloric sphincter Antrum (secretes mucus, pepsinogen, and gastrin) Fundus Lower esophageal sphincter FIGURE 26–5 Structure of a gastric gland from the fundus and body of the stomach. These acid- and pepsinogen-producing glands are referred to as “oxyntic” glands in some sources. (Adapted from Barrett KE: Gastrointestinal Physiology. McGraw-Hill, 2006.) TABLE 26–1 Contents of normal gastric juice (fasting state). Cations: Na+, K+, Mg2+, H+ (pH approximately 1.0) Anions: Cl–, HPO4 2–, SO4 2– Pepsins Lipase Mucus Intrinsic factor Cell migration Acid, intrinsic factor, pepsinogen Mucus layer Surface mucous cells (mucus, trefoil peptide, bicarbonate secretion) Mucous neck cells (stem cell compartment) Parietal cells (acid, intrinsic factor secretion) ECL cell (histamine secretion) Chief cells (pepsinogen secretion) 434 SECTION V Gastrointestinal Physiology is terminated after the meal moves from the stomach into the small intestine. Gastric parietal cells are highly specialized for their unusual task of secreting concentrated acid (Figure 26–8). The cells are packed with mitochondria that supply energy to drive the api-cal H,K-ATPase, or proton pump, that moves H+ ions out of the parietal cell against a concentration gradient of more than a million-fold. At rest, the proton pumps are sequestered within the parietal cell in a series of membrane compartments known as tubulovesicles. When the parietal cell begins to secrete, on the other hand, these vesicles fuse with invaginations of the api-cal membrane known as canaliculi, thereby substantially ampli-fying the apical membrane area and positioning the proton pumps to begin acid secretion (Figure 26–9). The apical mem-brane also contains potassium channels, which supply the K+ ions to be exchanged for H+, and Cl– channels that supply the counterion for HCl secretion (Figure 26–10). The secretion of protons is also accompanied by the release of equivalent num-bers of bicarbonate ions into the bloodstream, which as we will see, are later used to neutralize gastric acidity once its function is complete (Figure 26–10). The three agonists of the parietal cell—gastrin, histamine, and acetylcholine—each bind to distinct receptors on the basolateral membrane (Figure 26–9). Gastrin and acetylcho-line promote secretion by elevating cytosolic free calcium con-centrations, whereas histamine increases intracellular cyclic adenosine 3',5'-monophosphate (cAMP). The net effect of these second messengers are the transport and morphological changes described above. However, it is important to be aware that the two distinct pathways for activation are synergistic, with a greater than additive effect on secretion rates when his-tamine plus gastrin or acetylcholine, or all three, are present simultaneously. The physiologic significance of this synergism is that high rates of secretion can be stimulated with relatively small changes in availability of each of the stimuli. Synergism is also therapeutically significant because secretion can be markedly inhibited by blocking the action of only one of the triggers (most commonly that of histamine, via H2 histamine antagonists that are widely used therapies for adverse effects of excessive gastric secretion, such as reflux). FIGURE 26–6 Cyanocobalamin (vitamin B12). N N O H H H H HO CH2 CH3 CH3 HO O P O O O CHCH2NH CH3 C O CH2 CH2 CH3 CH3 CH3 CH3 CH3 CH3 − + C C Co CN CH2CH2CONH2 NH2COCH2 CH2CH2CONH2 N N CH N B C A D N CH2CONH2 CH2CH2CONH2 NH2COCH2 H3C H3C FIGURE 26–7 Regulation of gastric acid and pepsin secretion by soluble mediators and neural input. Gastrin is released from G cells in the antrum and travels through the circulation to influence the activity of ECL cells and parietal cells. The specific agonists of the chief cell are not well under-stood. Gastrin release is negatively regulated by luminal acidity via the release of somatostatin from antral D cells. (Adapted from Barrett KE: Gastrointestinal Physiology. McGraw-Hill, 2006.) − G cell GRP ANTRUM Peptides/amino acids D cell SST Gastrin H+ Nerve ending Circulation ECL cell Histamine ? ? ACh ACh P Chief cell Parietal cell ACh H+ FUNDUS CHAPTER 26 Overview of Gastrointestinal Function & Regulation 435 Gastric secretion adds about 2.5 L per day to the intestinal contents. However, despite their substantial volume and fine control, gastric secretions are dispensable for the full digestion and absorption of a meal, with the exception of cobalamin absorption. This illustrates an important facet of gastrointesti-nal physiology, that digestive and absorptive capacity are mark-edly in excess of normal requirements. On the other hand, if gastric secretion is chronically reduced, individuals may display increased susceptibility to infections acquired via the oral route. PANCREATIC SECRETION The pancreatic juice contains enzymes that are of major im-portance in digestion (see Table 26–2). Its secretion is con-trolled in part by a reflex mechanism and in part by the gastrointestinal hormones secretin and cholecystokinin (CCK). ANATOMIC CONSIDERATIONS The portion of the pancreas that secretes pancreatic juice is a compound alveolar gland resembling the salivary glands. Granules containing the digestive enzymes (zymogen gran-ules) are formed in the cell and discharged by exocytosis (see Chapter 2) from the apexes of the cells into the lumens of the pancreatic ducts (Figure 26–11). The small duct radicles coa-lesce into a single duct (pancreatic duct of Wirsung), which usually joins the common bile duct to form the ampulla of Vater (Figure 26–12). The ampulla opens through the duode-nal papilla, and its orifice is encircled by the sphincter of Oddi. Some individuals have an accessory pancreatic duct (duct of Santorini) that enters the duodenum more proximally. COMPOSITION OF PANCREATIC JUICE The pancreatic juice is alkaline (Table 26–3) and has a high HCO3 – content (approximately 113 mEq/L vs. 24 mEq/L in plasma). About 1500 mL of pancreatic juice is secreted per day. Bile and intestinal juices are also neutral or alkaline, and these three secretions neutralize the gastric acid, raising the pH of the duodenal contents to 6.0 to 7.0. By the time the chyme reaches the jejunum, its pH is nearly neutral, but the intestinal contents are rarely alkaline. FIGURE 26–8 Composite diagram of a parietal cell, showing the resting state (lower left) and the active state (upper right). The resting cell has intracellular canaliculi (IC), which open on the api-cal membrane of the cell, and many tubulovesicular structures (TV) in the cytoplasm. When the cell is activated, the TVs fuse with the cell membrane and microvilli (MV) project into the canaliculi, so the area of cell membrane in contact with gastric lumen is greatly increased. M, mitochondrion; G, Golgi apparatus. (Adapted from Junqueira LC, Carneiro J: Basic Histology: Text & Atlas, 10th ed. McGraw-Hill, 2003.) IC IC IC G TV M IC MV M M FIGURE 26–9 Parietal cell receptors and schematic representation of the morphological changes depicted in Figure 26–7. Amplifi-cation of the apical surface area is accompanied by an increased density of H+, K+–ATPase molecules at this site. Note that acetylcholine (ACh) and gastrin signal via calcium, whereas histamine signals via cAMP. (Adapted from Barrett KE: Gastrointestinal Physiology. McGraw-Hill, 2006.) Resting Canaliculus H+, K+ ATPase Tubulo-vesicle M3 H2 M3 H2 CCK−B CCK−B Secreting ACh Histamine cAMP Gastrin Ca++ Ca++ 436 SECTION V Gastrointestinal Physiology The potential danger of the release into the pancreas of a small amount of trypsin is apparent; the resulting chain reac-tion would produce active enzymes that could digest the pan-creas. It is therefore not surprising that the pancreas normally contains a trypsin inhibitor. Another enzyme activated by trypsin is phospholipase A2. This enzyme splits a fatty acid off phosphatidylcholine (PC), forming lyso-PC. Lyso-PC damages cell membranes. It has been hypothesized that in acute pancreatitis, a severe and sometimes fatal disease, phospholipase A2 is activated in the pancreatic ducts, with the formation of lyso-PC from the PC that is a normal constituent of bile. This causes disruption of pancreatic tissue and necrosis of surrounding fat. Small amounts of pancreatic digestive enzymes normally leak into the circulation, but in acute pancreatitis, the circulat-ing levels of the digestive enzymes rise markedly. Measure-ment of the plasma amylase or lipase concentration is therefore of value in diagnosing the disease. REGULATION OF THE SECRETION OF PANCREATIC JUICE Secretion of pancreatic juice is primarily under hormonal control. Secretin acts on the pancreatic ducts to cause copious secretion of a very alkaline pancreatic juice that is rich in HCO3 – and poor in enzymes. The effect on duct cells is due to an increase in intracellular cAMP. Secretin also stimulates bile secretion. CCK acts on the acinar cells to cause the release of zymogen granules and production of pancreatic juice rich in enzymes but low in volume. Its effect is mediated by phospho-lipase C (see Chapter 2). The response to intravenous secretin is shown in Figure 26–13. Note that as the volume of pancreatic secretion increases, its Cl– concentration falls and its HCO3 – concen-tration increases. Although HCO3 – is secreted in the small ducts, it is reabsorbed in the large ducts in exchange for Cl– (Figure 26–14). The magnitude of the exchange is inversely proportionate to the rate of flow. Like CCK, acetylcholine acts on acinar cells via phospholi-pase C to cause discharge of zymogen granules, and stimulation of the vagi causes secretion of a small amount of pancreatic juice rich in enzymes. There is evidence for vagally mediated conditioned reflex secretion of pancreatic juice in response to the sight or smell of food. BILIARY SECRETION An additional secretion important for gastrointestinal function, bile, arises from the liver. The bile acids contained therein are important in the digestion and absorption of fats. In addition, bile serves as a critical excretory fluid by which the body dispos-es of lipid soluble end products of metabolism as well as lipid soluble xenobiotics. Bile is also the only route by which the body can dispose of cholesterol—either in its native form, or follow-ing conversion to bile acids. In this chapter and the next, we will be concerned with the role of bile as a digestive fluid. In Chapter 29, a more general consideration of the transport and metabolic functions of the liver will be presented. FIGURE 26–10 Ion transport proteins of parietal cells. Protons are generated in the cytoplasm via the action of carbonic anhydrase II (C.A. II). Bicarbonate ions are exported from the basolateral pole of the cell either by vesicular fusion or via a chloride/bicarbonate exchanger. (Adapted from Barrett KE: Gastrointestinal Physiology. McGraw-Hill, 2006.) Potassium channel Chloride channel Lumen Blood Stream H+, K+ ATPase Na+, K+ ATPase 3Na+ 2K+ Na+ H+ NHE-1 K+ H+ Cl− Cl− ClC Apical Basolateral Cl−/HCO3 HCO3 HCO3 exchanger H2O + CO2 H+ + HCO3 C.A.II − − − − CHAPTER 26 Overview of Gastrointestinal Function & Regulation 437 TABLE 26–2 Principal digestive enzymes. Source Enzyme Activator Substrate Catalytic Function or Products Salivary glands Salivary α-amylase Cl– Starch Hydrolyzes 1:4α linkages, producing α-limit dextrins, maltotriose, and maltose Lingual glands Lingual lipase Triglycerides Fatty acids plus 1,2-diacylglycerols Stomach Pepsins (pepsinogens) HCl Proteins and polypeptides Cleave peptide bonds adjacent to aromatic amino acids Gastric lipase Triglycerides Fatty acids and glycerol Exocrine pancreas Trypsin (trypsinogen) Enteropepti-dase Proteins and polypeptides Cleave peptide bonds on carboxyl side of basic ami-no acids (arginine or lysine) Chymotrypsins (chymotrypsinogens) Trypsin Proteins and polypeptides Cleave peptide bonds on carboxyl side of aromatic amino acids Elastase (proelastase) Trypsin Elastin, some other proteins Cleaves bonds on carboxyl side of aliphatic amino acids Carboxypeptidase A (procarboxypeptidase A) Trypsin Proteins and polypeptides Cleave carboxyl terminal amino acids that have aromatic or branched aliphatic side chains Carboxypeptidase B (procarboxypeptidase B) Trypsin Proteins and polypeptides Cleave carboxyl terminal amino acids that have basic side chains Colipase (procolipase) Trypsin Fat droplets Facilitates exposure of active site of pancreaticlipase Pancreatic lipase . . . Triglycerides Monoglycerides and fatty acids Bile salt-acid lipase Cholesteryl esters Cholesterol Cholesteryl ester hydrolase . . . Cholesteryl esters Cholesterol Pancreatic α-amylase Cl– Starch Same as salivary α-amylase Ribonuclease . . . RNA Nucleotides Deoxyribonuclease . . . DNA Nucleotides Phospholipase A2 (pro-phospholipase A2) Trypsin Phospholipids Fatty acids, lysophospholipids Intestinal mucosa Enteropeptidase . . . Trypsinogen Trypsin Aminopeptidases . . . Polypeptides Cleave amino terminal amino acid from peptide Carboxypeptidases . . . Polypeptides Cleave carboxyl terminal amino acid from peptide Endopeptidases . . . Polypeptides Cleave between residues in midportion of peptide Dipeptidases . . . Dipeptides Two amino acids Maltase . . . Maltose, maltotriose, α-dextrins Glucose Lactase . . . Lactose Galactose and glucose Sucrasea . . . Sucrose; also maltotriose and maltose Fructose and glucose α-Dextrinasea . . . α-Dextrins, maltose maltotriose Glucose Trehalase . . . Trehalose Glucose Nuclease and related enzymes . . . Nucleic acids Pentoses and purine and pyrimidine bases Cytoplasm of mucosal cells Various peptidases . . . Di-, tri-, and tetrapeptides Amino acids Corresponding proenzymes, where relevant, are shown in parentheses aSucrase and a-dextrinase are separate subunits of a single protein. 438 SECTION V Gastrointestinal Physiology BILE Bile is made up of the bile acids, bile pigments, and other sub-stances dissolved in an alkaline electrolyte solution that re-sembles pancreatic juice (Table 26–4). About 500 mL is secreted per day. Some of the components of the bile are reab-sorbed in the intestine and then excreted again by the liver (enterohepatic circulation). The glucuronides of the bile pigments, bilirubin and biliverdin, are responsible for the golden yellow color of bile. The formation of these breakdown products of hemoglobin is discussed in detail in Chapter 29, and their excretion is dis-cussed below. The bile acids secreted into the bile are conjugated to gly-cine or taurine, a derivative of cysteine. The bile acids are syn-thesized from cholesterol. The four major bile acids found in humans are listed in Figure 26–15. In common with vitamin D, cholesterol, a variety of steroid hormones, and the digitalis glycosides, the bile acids contain the steroid nucleus (see Chapter 22). The two principal (primary) bile acids formed in the liver are cholic acid and chenodeoxycholic acid. In the colon, bacteria convert cholic acid to deoxycholic acid and chenodeoxycholic acid to lithocholic acid. In addition, small quantities of ursodeoxycholic acid are formed from chenode-oxycholic acid. Ursodeoxycholic acid is a tautomer of cheno-deoxycholic acid at the 7-position. Because they are formed FIGURE 26–11 Structure of the pancreas. (Reproduced with per-mission from Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology: The Mecha-nisms of Body Function, 11th ed. McGraw-Hill, 2008.) FIGURE 26–12 Connections of the ducts of the gallbladder, liver, and pancreas. (Adapted from Bell GH, Emslie-Smith D, Paterson CR: Textbook of Physiology and Biochemistry, 9th ed. Churchill Livingstone, 1976.) Common bile duct from gallbladder Duodenum Gallbladder Pancreatic duct Exocrine cells (secrete enzymes) Endocrine cells of pancreas Duct cells (secrete bicarbonate) Pancreas Right hepatic duct Left hepatic duct Common hepatic duct Bile duct Cystic duct Gall-bladder Accessory pancreatic duct Ampulla of bile duct Duodenum Pancreas Pancreatic duct TABLE 26–3 Composition of normal human pancreatic juice. Cations: Na+, K+, Ca2+, Mg2+ (pH approximately 8.0) Anions: HCO3 –, Cl–, SO4 2–, HPO4 2– Digestive enzymes (see Table 26–1; 95% of protein in juice) Other proteins FIGURE 26–13 Effect of a single dose of secretin on the composition and volume of the pancreatic juice in humans. Secretin 12.5 units/kg IV (K+) (HCO3 −) (CI−) (Amylase) 150 120 90 60 30 0 −20 −10 0 +10 +20 +30 +40 Time (min) Concentration of electrolytes (meq/L) and amylase (U/mL) Volume of secretion (mL) 0.3 0.2 17.7 15.2 5.1 0.6 CHAPTER 26 Overview of Gastrointestinal Function & Regulation 439 by bacterial action, deoxycholic, lithocholic, and ursodeoxy-cholic acids are called secondary bile acids. The bile salts have a number of important actions: they reduce surface tension and, in conjunction with phospholip-ids and monoglycerides, are responsible for the emulsification of fat preparatory to its digestion and absorption in the small intestine (see Chapter 27). They are amphipathic, that is, they have both hydrophilic and hydrophobic domains; one surface of the molecule is hydrophilic because the polar peptide bond and the carboxyl and hydroxyl groups are on that surface, whereas the other surface is hydrophobic. Therefore, the bile salts tend to form cylindrical disks called micelles. A top view of micelles is shown in Figure 26–16 and a side view of one in FIGURE 26–14 Ion transport pathways present in pancreatic duct cells. CA, carbonic anhydrase; NHE-1, sodium/hydrogen exchanger-1; NBC, sodium-bicarbonate cotransporter. (Adapted from Barrett KE: Gastrointestinal Physiology. McGraw-Hill, 2006.) cAMP CFTR Basolateral Duct lumen Cl− NHE-1 Na+ Na+ Na+, K+ 3Na+ 2K+ K+ H+ 2HCO3 ATPase CO2 + H2O HCO3 + H+ C.A NBC + − Cl−/HCO3 Exchanger − HCO3 − − channel TABLE 26–4 Composition of human hepatic duct bile. Water 97.0% Bile salts 0.7% Bile pigments 0.2% Cholesterol 0.06% Inorganic salts 0.7% Fatty acids 0.15% Phosphatidylcholine 0.2% Fat 0.1% Alkaline phosphatase … FIGURE 26–15 Human bile acids. The numbers in the formula for cholic acid refer to the positions in the steroid ring. HO OH COOH OH 3 7 CH3 CH3 Cholic acid 12 Cholic acid Chenodeoxycholic acid Deoxycholic acid Lithocholic acid 3 OH OH OH OH Percent in human bile 50 30 15 5 OH H OH H OH OH H H 7 12 Group at position FIGURE 26–16 Lipid digestion and passage to intestinal mucosa. Fatty acids (FA) are liberated by the action of pancreatic li-pase on dietary triglycerides and, in the presence of bile salts (BS), form micelles (the circular structures), which diffuse through the unstirred layer to the mucosal surface. (Adapted from Thomson ABR: Intestinal absorption of lipids: Influence of the unstirred water layer and bile acid micelle. In: Disturbances in Lipid and Lipoprotein Metabolism. Dietschy JM, Gotto AM Jr, Ontko JA [editors]: American Physiological Society, 1978.) Dietary triglyceride Pancreatic lipase Mucosa BULK SOLUTION OF INTESTINAL CONTENTS UNSTIRRED LAYER FA absorption in presence of BS FA absorption in absence of BS 440 SECTION V Gastrointestinal Physiology Figure 26–17. Their hydrophilic portions face out and their hydrophobic portions face in. Above a certain concentration, called the critical micelle concentration, all bile salts added to a solution form micelles. Lipids collect in the micelles, with cholesterol in the hydrophobic center and amphipathic phos-pholipids and monoglycerides lined up with their hydrophilic heads on the outside and their hydrophobic tails in the center. The micelles play an important role in keeping lipids in solu-tion and transporting them to the brush border of the intesti-nal epithelial cells, where they are absorbed (see Chapter 27). Ninety to 95% of the bile salts are absorbed from the small intestine. Once they are deconjugated, they can be absorbed by nonionic diffusion, but most are absorbed in their conju-gated forms from the terminal ileum (Figure 26–18) by an extremely efficient Na+–bile salt cotransport system powered by basolateral Na+–K+ ATPase. The remaining 5–10% of the bile salts enter the colon and are converted to the salts of deoxycholic acid and lithocholic acid. Lithocholate is rela-tively insoluble and is mostly excreted in the stools; only 1% is absorbed. However, deoxycholate is absorbed. The absorbed bile salts are transported back to the liver in the portal vein and reexcreted in the bile (enterohepatic circu-lation) (Figure 26–18). Those lost in the stool are replaced by synthesis in the liver; the normal rate of bile salt synthesis is 0.2 to 0.4 g/d. The total bile salt pool of approximately 3.5 g recycles repeatedly via the enterohepatic circulation; it has been calculated that the entire pool recycles twice per meal and six to eight times per day. When bile is excluded from the intestine, up to 50% of ingested fat appears in the feces. A severe malabsorption of fat-soluble vitamins also results. When bile salt reabsorption is prevented by resection of the terminal ileum or by disease in this portion of the small intes-tine, the amount of fat in the stools is also increased because when the enterohepatic circulation is interrupted, the liver cannot increase the rate of bile salt production to a sufficient degree to compensate for the loss. INTESTINAL FLUID & ELECTROLYTE TRANSPORT The intestine itself also supplies a fluid environment in which the processes of digestion and absorption can occur. Then, when the meal has been assimilated, fluid used during diges-tion and absorption is reclaimed by transport back across the epithelium to avoid dehydration. Water moves passively into and out of the gastrointestinal lumen, driven by electrochem-ical gradients established by the active transport of ions and other solutes. In the period after a meal, much of the fluid re-uptake is driven by the coupled transport of nutrients, such as glucose, with sodium ions. In the period between meals, ab-sorptive mechanisms center exclusively around electrolytes. In both cases, secretory fluxes of fluid are largely driven by the active transport of chloride ions into the lumen, although ab-sorption still predominates overall. FIGURE 26–17 Physical forms adopted by bile acids in solution. Micelles are shown in cross-section, and are actually thought to be cylindrical in shape. Mixed micelles of bile acids present in hepat-ic bile also incorporate cholesterol and phosphatidylcholine. (Adapted from Barrett KE: Gastrointestinal Physiology. McGraw-Hill, 2006.) Charged side chain OH group Simple micelle Bile acid monomers Mixed micelle Phosphatidylcholine Cholesterol FIGURE 26–18 Quantitative aspects of the circulation of bile acids. The majority of the bile acid pool circulates between the small intestine and liver. A minority of the bile acid pool is in the sys-temic circulation (due to incomplete hepatocyte uptake from the por-tal blood) or spills over into the colon and is lost to the stool. Fecal loss must be equivalent to hepatic synthesis of bile acids at steady state. (Adapted from Barrett KE: Gastrointestinal Physiology. McGraw-Hill, 2006.) Hepatic synthesis Sphincter of Oddi Small intestine Large intestine Spillover into colon Fecal loss ( = hepatic synthesis) Passive uptake of deconjugated bile acids from colon Return to liver Active ileal uptake Gallbladder Spillover from liver into systemic circulation CHAPTER 26 Overview of Gastrointestinal Function & Regulation 441 Overall water balance in the gastrointestinal tract is sum-marized in Table 26–5. The intestines are presented each day with about 2000 mL of ingested fluid plus 7000 mL of secre-tions from the mucosa of the gastrointestinal tract and associ-ated glands. Ninety-eight percent of this fluid is reabsorbed, with a daily fluid loss of only 200 mL in the stools. In the small intestine, secondary active transport of Na+ is important in bringing about absorption of glucose, some amino acids, and other substances such as bile acids (see above). Con-versely, the presence of glucose in the intestinal lumen facili-tates the reabsorption of Na+. In the period between meals, when nutrients are not present, sodium and chloride are absorbed together from the lumen by the coupled activity of a sodium/hydrogen exchanger (NHE) and chloride/bicarbonate exchanger in the apical membrane, in a so-called electroneutral mechanism (Figure 26–19). Water then follows to maintain an osmotic balance. In the colon, moreover, an additional electro-genic mechanism for sodium absorption is expressed, particu-larly in the distal colon. In this mechanism, sodium enters across the apical membrane via an ENaC (epithelial sodium) channel that is identical to that expressed in the distal tubule of the kidney (Figure 26–20). This underpins the ability of the colon to desiccate the stool and ensure that only a small portion of the fluid load used daily in the digestion and absorption of meals is lost from the body. Following a low-salt diet, increased expression of ENaC in response to aldosterone increases the ability to reclaim sodium from the stool. Despite the predominance of absorptive mechanisms, secretion also takes place continuously throughout the small intestine and colon to adjust the local fluidity of the intestinal contents as needed for mixing, diffusion, and movement of the meal and its residues along the length of the gastrointesti-nal tract. Cl– normally enters enterocytes from the interstitial fluid via Na+–K+–2Cl– cotransporters in their basolateral membranes (Figure 26–21), and the Cl– is then secreted into the intestinal lumen via channels that are regulated by various protein kinases. The cystic fibrosis transmembrane conduc-tance regulator (CFTR) channel that is defective in the disease of cystic fibrosis is quantitatively most important, and is acti-vated by protein kinase A and hence by cAMP (see Clinical Box 26–2). Water moves into or out of the intestine until the osmotic pressure of the intestinal contents equals that of the plasma. The osmolality of the duodenal contents may be hypertonic or hypotonic, depending on the meal ingested, but by the time the meal enters the jejunum, its osmolality is close to that of TABLE 26–5 Daily water turnover (mL) in the gastrointestinal tract. Ingested 2000 Endogenous secretions 7000 Salivary glands 1500 Stomach 2500 Bile 500 Pancreas 1500 Intestine +1000 7000 Total input 9000 Reabsorbed 8800 Jejunum 5500 Ileum 2000 Colon +1300 8800 Balance in stool 200 Data from Moore EW: Physiology of Intestinal Water and Electrolyte Absorption. American Gastroenterological Society, 1976. FIGURE 26–19 Electroneutral NaCl absorption in the small intestine and colon. NaCl enters across the apical membrane via the coupled activity of a sodium/hydrogen exchanger and a chloride/bi-carbonate exchanger. FIGURE 26–20 Electrogenic sodium absorption in the colon. Sodium enters the epithelial cell via epithelial sodium channels (ENaC). Na+,K+ -ATPase 3Na+ 2K+ Na+ H+ NHE-3? NHE-2? KCC1 ? K+ Cl− Cl− CLD HCO3– 2K+ Na+ Cl− ENaC K+ Na+,K+ -ATPase 3Na+ 442 SECTION V Gastrointestinal Physiology plasma. This osmolality is maintained throughout the rest of the small intestine; the osmotically active particles produced by digestion are removed by absorption, and water moves passively out of the gut along the osmotic gradient thus gener-ated. In the colon, Na+ is pumped out and water moves pas-sively with it, again along the osmotic gradient. Saline cathartics such as magnesium sulfate are poorly absorbed salts that retain their osmotic equivalent of water in the intes-tine, thus increasing intestinal volume and consequently exerting a laxative effect. Some K+ is secreted into the intestinal lumen, especially as a component of mucus. K+ channels are present in the luminal as well as the basolateral membrane of the enterocytes of the colon, so K+ is secreted into the colon. In addition, K+ moves passively down its electrochemical gradient. The accumulation of K+ in the colon is partially offset by H+–K+ ATPase in the luminal membrane of cells in the distal colon, with resulting active transport of K+ into the cells. Nevertheless, loss of ileal or colonic fluids in chronic diarrhea can lead to severe hypo-kalemia. When the dietary intake of K+ is high for a prolonged period, aldosterone secretion is increased and more K+ enters the colon. This is due in part to the appearance of more Na+– K+ ATPase pumps in the basolateral membranes of the cells, with a consequent increase in intracellular K+ and K+ diffu-sion across the luminal membranes of the cells. GASTROINTESTINAL REGULATION The various functions of the gastrointestinal tract, including secretion, digestion, and absorption (Chapter 27) and motility (Chapter 28) must be regulated in an integrated way to ensure efficient assimilation of nutrients after a meal. There are three main modalities for gastrointestinal regulation that operate in a complementary fashion to ensure that function is appropri-ate. First, endocrine regulation is mediated by the release of hormones by triggers associated with the meal. These hor-mones travel through the bloodstream to change the activity of a distant segment of the gastrointestinal tract, an organ draining into it (eg, the pancreas), or both. Second, some sim-ilar mediators are not sufficiently stable to persist in the blood-stream, but instead alter the function of cells in the local area where they are released, in a paracrine fashion. Finally, the in-testinal system is endowed with extensive neural connections. These include connections to the central nervous system (ex-trinsic innervation), but also the activity of a largely autono-mous enteric nervous system that comprises both sensory and secreto-motor neurons. The enteric nervous system inte-grates central input to the gut, but can also regulate gut function independently in response to changes in the luminal environment. In some cases, the same substance can mediate regulation by endocrine, paracrine, and neurocrine pathways (eg, cholecystokinin, see below). FIGURE 26–21 Chloride secretion in the small intestine and colon. Chloride uptake occurs via the sodium/potassium/2 chloride cotransporter, NKCC1. Chloride exit is via the cystic fibrosis transmem-brane conductance regulator (CFTR) as well as perhaps via other chlo-ride channels, not shown. 2K+ 2CI _ Na+ Na+ K+ Cl− CFTR K+ Na+, K+ -ATPase 3Na+ NKCC1 CLINICAL BOX 26–2 Cholera Cholera is a severe secretory diarrheal disease that often oc-curs in epidemics associated with natural disasters where normal sanitary practices break down. Along with other secretory diarrheal illnesses produced by bacteria and vi-ruses, cholera causes a significant amount of morbidity and mortality, particularly among the young and in developing countries. The cAMP concentration in intestinal epithelial cells is increased in cholera. The cholera bacillus stays in the intestinal lumen, but it produces a toxin that binds to GM-1 ganglioside receptors on the apical membrane of intestinal epithelial cells, and this permits part of the A subunit (A1 peptide) of the toxin to enter the cell. The A1 peptide binds adenosine diphosphate ribose to the α subunit of Gs, inhibit-ing its GTPase activity (see Chapter 2). Therefore, the consti-tutively activated G protein produces prolonged stimulation of adenylyl cyclase and a marked increase in the intracellular cAMP concentration. In addition to increased Cl– secretion, the function of the mucosal NHE carrier for Na+ is reduced, thus reducing NaCl absorption. The resultant increase in electrolyte and water content of the intestinal contents causes the diarrhea. However, Na+–K+ ATPase and the Na+/ glucose cotransporter are unaffected, so coupled reabsorp-tion of glucose and Na+ bypasses the defect. This is the phys-iologic basis for the treatment of Na+ and water loss in diar-rhea by oral administration of solutions containing NaCl and glucose. Cereals containing carbohydrates are also useful in the treatment of diarrhea. CHAPTER 26 Overview of Gastrointestinal Function & Regulation 443 HORMONES/PARACRINES Biologically active polypeptides that are secreted by nerve cells and gland cells in the mucosa act in a paracrine fashion, but they also enter the circulation. Measurement of their concen-trations in blood after a meal has shed light on the roles these gastrointestinal hormones play in the regulation of gas-trointestinal secretion and motility. When large doses of the hormones are given, their actions overlap. However, their physiologic effects appear to be relatively discrete. On the basis of structural similarity (Table 26–6) and, to a degree, similarity of function, the key hormones fall into one of two families: the gastrin family, the primary members of which are gastrin and CCK; and the secretin family, the primary members of which are secretin, glucagon, glicentin (GLI), vaso-active intestinal peptide (VIP; actually a neurotransmitter, or neurocrine), and gastric inhibitory polypeptide (also known as glucose-dependent insulinotropic peptide, or GIP). There are also other hormones that do not fall readily into these families. ENTEROENDOCRINE CELLS More than 15 types of hormone-secreting enteroendocrine cells have been identified in the mucosa of the stomach, small intestine, and colon. Many of these secrete only one hormone and are identified by letters (G cells, S cells, etc). Others man-ufacture serotonin or histamine and are called enterochromaf-fin or enterochromaffin-like (ECL) cells, respectively. GASTRIN Gastrin is produced by cells called G cells in the antral portion of the gastric mucosa (Figure 26–22). G cells are flask-shaped, with a broad base containing many gastrin granules and a nar-row apex that reaches the mucosal surface. Microvilli project from the apical end into the lumen. Receptors mediating gas-trin responses to changes in gastric contents are present on the microvilli. Other cells in the gastrointestinal tract that secrete hormones have a similar morphology. Gastrin is typical of a number of polypeptide hormones in that it shows both macroheterogeneity and microheteroge-neity. Macroheterogeneity refers to the occurrence in tissues and body fluids of peptide chains of various lengths; microhet-erogeneity refers to differences in molecular structure due to derivatization of single amino acid residues. Preprogastrin is processed into fragments of various sizes. Three main frag-ments contain 34, 17, and 14 amino acid residues. All have the same carboxyl terminal configuration (Table 26–6). These forms are also known as G 34, G 17, and G 14 gastrins, respec-tively. Another form is the carboxyl terminal tetrapeptide, and there is also a large form that is extended at the amino terminal and contains more than 45 amino acid residues. One form of derivatization is sulfation of the tyrosine that is the sixth amino acid residue from the carboxyl terminal. Approximately equal amounts of nonsulfated and sulfated forms are present in blood and tissues, and they are equally active. Another deriva-tization is amidation of the carboxyl terminal phenylalanine. What is the physiologic significance of this marked hetero-geneity? Some differences in activity exist between the various components, and the proportions of the components also dif-fer in the various tissues in which gastrin is found. This sug-gests that different forms are tailored for different actions. However, all that can be concluded at present is that G 17 is the principal form with respect to gastric acid secretion. The carboxyl terminal tetrapeptide has all the activities of gastrin but only 10% of the strength of G 17. G 14 and G 17 have half-lives of 2 to 3 min in the circula-tion, whereas G 34 has a half-life of 15 min. Gastrins are inac-tivated primarily in the kidney and small intestine. In large doses, gastrin has a variety of actions, but its princi-pal physiologic actions are stimulation of gastric acid and pep-sin secretion and stimulation of the growth of the mucosa of the stomach and small and large intestines (trophic action). Gastrin secretion is affected by the contents of the stomach, the rate of discharge of the vagus nerves, and bloodborne fac-tors (Table 26–7). Atropine does not inhibit the gastrin response to a test meal in humans, because the transmitter secreted by the postganglionic vagal fibers that innervate the G cells is gastrin-releasing polypeptide (GRP; see below) rather than acetylcholine. Gastrin secretion is also increased by the presence of the products of protein digestion in the stomach, particularly amino acids, which act directly on the G cells. Phenylalanine and tryptophan are particularly effective. Acid in the antrum inhibits gastrin secretion, partly by a direct action on G cells and partly by release of somatostatin, a relatively potent inhibitor of gastrin secretion. The effect of acid is the basis of a negative feedback loop regulating gastrin secretion. Increased secretion of the hormone increases acid secretion, but the acid then feeds back to inhibit further gas-trin secretion. In conditions such as pernicious anemia in which the acid-secreting cells of the stomach are damaged, gastrin secretion is chronically elevated. CHOLECYSTOKININ Cholecystokinin (CCK) is secreted by cells in the mucosa of the upper small intestine. It has a plethora of actions in the gastrointestinal system, but the most important appear to be the stimulation of pancreatic enzyme secretion, the contrac-tion of the gallbladder (the action for which it was named), and relaxation of the sphincter of Oddi, which allows both bile and pancreatic juice to flow into the intestinal lumen. Like gastrin, CCK shows both macroheterogeneity and microheterogeneity. Prepro-CCK is processed into many frag-ments. A large CCK contains 58 amino acid residues (CCK 58). In addition, there are CCK peptides that contain 39 amino acid residues (CCK 39) and 33 amino acid residues (CCK 33), several forms that contain 12 (CCK 12) or slightly more amino acid residues, and a form that contains 8 amino acid residues (CCK 8). All of these forms have the same 5 amino acids at the carboxyl terminal as gastrin (Table 26–6). 444 SECTION V Gastrointestinal Physiology TABLE 26–6. Structures of some of the hormonally active polypeptides secreted by cells in the human gastrointestinal tract.a Gastrin Family GIP Secretin Family Other Polypeptides CCK Gastrin 39 34 Glucagon Secretin VIP Motilin Substance P GRP Guanylin aHomologous amino acid residues are enclosed by the lines that generally cross from one polypeptide to another. Arrows indicate points of cleavage to form smaller variants. Tys, tyrosine sulfate. All gastrins occur in unsulfated (gastrin I) and sulfated (gastrin II) forms. Glicentin, an additional member of the secretin family, is a C-terminally extended relative of glucagon. Tyr Tyr His His His Phe Arg Val Pro n s A o r P o r P l a V r e S r e S r e S a l A e l I Gln Glu Gln Asp Asp Pro Lys Leu Thr Gln Gly Gly Gly Ala Ile Pro Pro Cys Ala Thr Thr Thr Val Phe Gln Ala Glu Arg (pyro)Glu Phe Phe Phe Phe Thr Gln Gly Ile Lys Leu Ile Thr Thr Thr Tyr Phe Gly Cys Ala Gly Ser Ser Ser Asp Gly Phe Gly Ala Pro Pro Asp Asp Glu Asn Glu Gly Thr Tyr Ser Gln Tyr Tyr Leu Tyr Leu Leu Val Ala Gly Gly Ser Ser Ser Thr Gln Met-NH2 Leu Ala Arg Pro Ile Lys Arg Arg Arg Thr Cys Met Pro Ala Tyr Leu Leu Met Lys Thr Ser His Met Leu Arg Arg Gln Met Gly Ile Leu Asp Asp Glu Lys Glu Tyr Cys Val Val Lys Ser o r P s y L n l G y l G Lys Ala Ile Arg Ala Met Glu Arg Asn Asp His Arg Arg Ala Arg Gly Leu Pro Gln Ala Leu Val Asn Asn s i H s y L s y L n l G n l G n l G r e S n l G p r T y l G s y L g r A p s A p s A s y L n s A Leu Lys Phe Phe Leu Tyr Gln Ala Asp Gln Va l a V u e L u e L l a V l Pro Gly Asn Gln Gln A y l G n s Ser Pro Trp T s i H r e S y l G p r u e L e l I u e L u e L u e L p r T s i H Arg Leu Leu Met Val-NH2 Le M u et-NH2 Ile Glu Ala Asn Asn-NH2 Ser Glu Glu Thr Asp Glu Lys Arg Glu Gly Asp Glu Lys Tys Ala Lys Met Tys Asn Gly Gly Asp Trp Trp Trp Met Met Lys Asp Asp His Phe-NH2 Phe-NH2 Asn Ile Thr Gln → → → → → → → CHAPTER 26 Overview of Gastrointestinal Function & Regulation 445 The carboxyl terminal tetrapeptide (CCK 4) also exists in tissues. The carboxyl terminal is amidated, and the tyrosine that is the seventh amino acid residue from the carboxyl terminal is sulfated. Unlike gastrin, the nonsulfated form of CCK has not been found in tissues. However, derivatization of other amino acid residues in CCK can occur. The half-life of circulating CCK is about 5 minutes, but little is known about its metabolism. In addition to its secretion by I cells in the upper intestine, CCK is found in nerves in the distal ileum and colon. It is also found in neurons in the brain, especially the cerebral cortex, and in nerves in many parts of the body (see Chapter 7). In the brain, it may be involved in the regulation of food intake, and it appears to be related to the production of anxiety and analgesia. The CCK secreted in the duodenum and jejunum is probably mostly CCK 8 and CCK 12, although CCK 58 is also present in the intestine and circulating blood in some species. The enteric and pancreatic nerves contain primarily CCK 4. CCK 58 and CCK 8 are found in the brain. In addition to its primary actions, CCK augments the action of secretin in producing secretion of an alkaline pan-creatic juice. It also inhibits gastric emptying, exerts a trophic effect on the pancreas, increases the synthesis of enterokinase, and may enhance the motility of the small intestine and colon. There is some evidence that, along with secretin, it augments the contraction of the pyloric sphincter, thus preventing the reflux of duodenal contents into the stomach. Gastrin and CCK stimulate glucagon secretion, and since the secretion of FIGURE 26–22 Sites of production of the five gastrointestinal hormones along the length of the gastrointestinal tract. The width of the bars reflects the relative abundance at each location. Fundus Antrum Duodenum Jejunum Ileum Colon Gastrin CCK Secretin GIP Motilin TABLE 26–7 Stimuli that affect gastrin secretion. Stimuli that increase gastrin secretion Luminal Peptides and amino acids Distention Neural Increased vagal discharge via GRP Bloodborne Calcium Epinephrine Stimuli that inhibit gastrin secretion Luminal Acid Somatostatin Bloodborne Secretin, GIP, VIP, glucagon, calcitonin 446 SECTION V Gastrointestinal Physiology both gastrointestinal hormones is increased by a protein meal, either or both may be the “gut factor” that stimulates glucagon secretion (see Chapter 21). Two CCK receptors have been identified. CCK-A receptors are primarily located in the periphery, whereas both CCK-A and CCK-B receptors are found in the brain. Both activate PLC, causing increased pro-duction of IP3 and DAG (see Chapter 2). The secretion of CCK is increased by contact of the intesti-nal mucosa with the products of digestion, particularly pep-tides and amino acids, and also by the presence in the duodenum of fatty acids containing more than 10 carbon atoms. There are also two protein releasing factors that acti-vate CCK secretion, known as CCK-releasing peptide and monitor peptide, which derive from the intestinal mucosa and pancreas, respectively. Because the bile and pancreatic juice that enter the duodenum in response to CCK further the digestion of protein and fat, and the products of this digestion stimulate further CCK secretion, a sort of positive feedback operates in the control of the secretion of this hormone. How-ever, the positive feedback is terminated when the products of digestion move on to the lower portions of the gastrointesti-nal tract, and also because CCK-releasing peptide and moni-tor peptide are degraded by proteolytic enzymes once these are no longer occupied in digesting dietary proteins. SECRETIN Secretin occupies a unique position in the history of physiolo-gy. In 1902, Bayliss and Starling first demonstrated that the ex-citatory effect of duodenal stimulation on pancreatic secretion was due to a bloodborne factor. Their research led to the iden-tification of the first hormone, secretin. They also suggested that many chemical agents might be secreted by cells in the body and pass in the circulation to affect organs some distance away. Starling introduced the term hormone to categorize such “chemical messengers.” Modern endocrinology is the proof of the correctness of this hypothesis. Secretin is secreted by S cells that are located deep in the glands of the mucosa of the upper portion of the small intes-tine. The structure of secretin (Table 26–6) is different from that of CCK and gastrin, but very similar to that of glucagon, GLI, VIP, and GIP. Only one form of secretin has been iso-lated, and the fragments of the molecule that have been tested to date are inactive. Its half-life is about 5 minutes, but little is known about its metabolism. Secretin increases the secretion of bicarbonate by the duct cells of the pancreas and biliary tract. It thus causes the secre-tion of a watery, alkaline pancreatic juice. Its action on pan-creatic duct cells is mediated via cAMP. It also augments the action of CCK in producing pancreatic secretion of digestive enzymes. It decreases gastric acid secretion and may cause contraction of the pyloric sphincter. The secretion of secretin is increased by the products of pro-tein digestion and by acid bathing the mucosa of the upper small intestine. The release of secretin by acid is another exam-ple of feedback control: Secretin causes alkaline pancreatic juice to flood into the duodenum, neutralizing the acid from the stomach and thus inhibiting further secretion of the hormone. GIP GIP contains 42 amino acid residues (Table 26–6) and is pro-duced by K cells in the mucosa of the duodenum and jejunum. Its secretion is stimulated by glucose and fat in the duodenum, and because in large doses it inhibits gastric secretion and mo-tility, it was named gastric inhibitory peptide. However, it now appears that it does not have significant gastric inhibiting ac-tivity when administered in smaller amounts comparable to those seen after a meal. In the meantime, it was found that GIP stimulates insulin secretion. Gastrin, CCK, secretin, and glu-cagon also have this effect, but GIP is the only one of these that stimulates insulin secretion when administered in doses that produce blood levels comparable to those produced by oral glucose. For this reason, it is often called glucose-dependent insulinotropic polypeptide. The glucagon derivative GLP-1 (7–36) (see Chapter 21) also stimulates insulin secretion and is said to be more potent in this regard than GIP. Therefore, it may also be a physiologic B cell-stimulating hormone of the gastrointestinal tract. The integrated action of gastrin, CCK, secretin, and GIP in facilitating digestion and utilization of absorbed nutrients is summarized in Figure 26–23. FIGURE 26–23 Integrated action of gastrointestinal hormones in regulating digestion and utilization of absorbed nutrients. The dashed arrows indicate inhibition. The exact identity of the hormonal factor or factors from the intestine that inhibit(s) gastric acid secretion and motility is unsettled, but it may be peptide YY. Food in stomach Gastrin secretion Increased motility Increased acid secretion Food and acid into duodenum CCK and secretin secretion GIP GLP-1 (7–26) secretion Insulin secretion Pancreatic and biliary secretion Intestinal digestion of food Peptide YY? CHAPTER 26 Overview of Gastrointestinal Function & Regulation 447 VIP VIP contains 28 amino acid residues (Table 26–6). It is found in nerves in the gastrointestinal tract and thus is not itself a hor-mone, despite its similarities to secretin. Prepro-VIP contains both VIP and a closely related polypeptide (PHM-27 in hu-mans, PHI-27 in other species). VIP is also found in blood, in which it has a half-life of about 2 minutes. In the intestine, it markedly stimulates intestinal secretion of electrolytes and hence of water. Its other actions include relaxation of intestinal smooth muscle, including sphincters; dilation of peripheral blood vessels; and inhibition of gastric acid secretion. It is also found in the brain and many autonomic nerves (see Chapter 7), where it often occurs in the same neurons as acetylcholine. It potentiates the action of acetylcholine in salivary glands. However, VIP and acetylcholine do not coexist in neurons that innervate other parts of the gastrointestinal tract. VIP-secret-ing tumors (VIPomas) have been described in patients with se-vere diarrhea. MOTILIN Motilin is a polypeptide containing 22 amino acid residues that is secreted by enterochromaffin cells and Mo cells in the stom-ach, small intestine, and colon. It acts on G protein-coupled re-ceptors on enteric neurons in the duodenum and colon and on injection produces contraction of smooth muscle in the stom-ach and intestines. Its circulating level increases at intervals of approximately 100 min in the interdigestive state, and it is a major regulator of the migrating motor complexes (MMCs) (Figure 26–24) that control gastrointestinal motility between meals. Conversely, when a meal is ingested, secretion of motil-in is suppressed until digestion and absorption are complete. The antibiotic erythromycin binds to motilin receptors, and derivatives of this compound may be of value in treating pa-tients in whom gastrointestinal motility is decreased. SOMATOSTATIN Somatostatin, the growth-hormone-inhibiting hormone origi-nally isolated from the hypothalamus, is secreted as a paracrine by D cells in the pancreatic islets (see Chapter 21) and by similar D cells in the gastrointestinal mucosa. It exists in tissues in two forms, somatostatin 14 and somatostatin 28, and both are se-creted. Somatostatin inhibits the secretion of gastrin, VIP, GIP, secretin, and motilin. Its secretion is stimulated by acid in the lumen, and it probably acts in a paracrine fashion to mediate the inhibition of gastrin secretion produced by acid. It also inhibits FIGURE 26–24 Migrating motor complexes (MMCs). Note that the complexes move down the gastrointestinal tract at a regular rate dur-ing fasting, that they are completely inhibited by a meal, and that they resume 90–120 minutes after the meal. (Reproduced with permission from Chang EB, Sitrin MD, Black DD: Gastrointestinal, Hepatobiliary, and Nutritional Physiology. Lippincott-Raven, 1996.) Phase I -Phase II -Phase III - Regular spike potentials and contractions No spike potentials, no contractions Irregular spike potentials and contractions Phases of MMC III II I MEAL Stomach Propagation rate (5 cm/m) Distal ileum ~90 min Resumption of MMCs 448 SECTION V Gastrointestinal Physiology pancreatic exocrine secretion; gastric acid secretion and motili-ty; gallbladder contraction; and the absorption of glucose, ami-no acids, and triglycerides. OTHER GASTROINTESTINAL PEPTIDES PEPTIDE YY The structure of peptide YY is discussed in Chapter 21. It also inhibits gastric acid secretion and motility and is a good can-didate to be the gastric inhibitory peptide (Figure 26–23). Its release from the jejunum is stimulated by fat. OTHERS Ghrelin is secreted primarily by the stomach and appears to play an important role in the central control of food intake. It also stimulates growth hormone secretion by acting directly on receptors in the pituitary (see Chapter 24). Substance P (Table 26–6) is found in endocrine and nerve cells in the gastrointestinal tract and may enter the circulation. It increases the motility of the small intestine. The neurotrans-mitter GRP contains 27 amino acid residues, and the 10 amino acid residues at its carboxyl terminal are almost identical to those of amphibian bombesin. It is present in the vagal nerve endings that terminate on G cells and is the neurotransmitter producing vagally mediated increases in gastrin secretion. Glu-cagon from the gastrointestinal tract may be responsible (at least in part) for the hyperglycemia seen after pancreatectomy. Guanylin is a gastrointestinal polypeptide that binds to gua-nylyl cyclase. It is made up of 15 amino acid residues (Table 26–6) and is secreted by cells of the intestinal mucosa. Stimula-tion of guanylyl cyclase increases the concentration of intracell-ular cyclic 3',5'-guanosine monophosphate (cGMP), and this in turn causes increased secretion of Cl– into the intestinal lumen. Guanylin appears to act predominantly in a paracrine fashion, and it is produced in cells from the pylorus to the rectum. In an interesting example of molecular mimicry, the heat-stable enterotoxin of certain diarrhea-producing strains of E. coli has a structure very similar to guanylin and activates guanylin recep-tors in the intestine. Guanylin receptors are also found in the kidneys, the liver, and the female reproductive tract, and guany-lin may act in an endocrine fashion to regulate fluid movement in these tissues as well, and particularly to integrate the actions of the intestine and kidneys. THE ENTERIC NERVOUS SYSTEM Two major networks of nerve fibers are intrinsic to the gas-trointestinal tract: the myenteric plexus (Auerbach’s plexus), between the outer longitudinal and middle circular muscle lay-ers, and the submucous plexus (Meissner’s plexus), between the middle circular layer and the mucosa (Figure 26–1). Collec-tively, these neurons constitute the enteric nervous system. The system contains about 100 million sensory neurons, inter-neurons, and motor neurons in humans—as many as are found in the whole spinal cord—and the system is probably best viewed as a displaced part of the central nervous system (CNS) that is concerned with the regulation of gastrointestinal func-tion. It is sometimes referred to as the “little brain” for this rea-son. It is connected to the CNS by parasympathetic and sympathetic fibers but can function autonomously without these connections (see below). The myenteric plexus innervates the longitudinal and circular smooth muscle layers and is con-cerned primarily with motor control, whereas the submucous plexus innervates the glandular epithelium, intestinal endocrine cells, and submucosal blood vessels and is primarily involved in the control of intestinal secretion. The neurotransmitters in the system include acetylcholine, the amines norepinephrine and serotonin, the amino acid γ-aminobutyrate (GABA), the purine adenosine triphosphate (ATP), the gases NO and CO, and many different peptides and polypeptides (Table 26–8). Some of these peptides also act in a paracrine fashion, and some enter the bloodstream, becoming hormones. Not surprisingly, most of them are also found in the brain. EXTRINSIC INNERVATION The intestine receives a dual extrinsic innervation from the au-tonomic nervous system, with parasympathetic cholinergic ac-tivity generally increasing the activity of intestinal smooth TABLE 26–8 Principal peptides found in the enteric nervous system. CGRP CCK Endothelin-2 Enkephalins Galanin GRP Neuropeptide Y Neurotensin Peptide YY PACAP Somatostatin Substance P TRH VIP CHAPTER 26 Overview of Gastrointestinal Function & Regulation 449 muscle and sympathetic noradrenergic activity generally de-creasing it while causing sphincters to contract. The pregangli-onic parasympathetic fibers consist of about 2000 vagal efferents and other efferents in the sacral nerves. They generally end on cholinergic nerve cells of the myenteric and submucous plexuses. The sympathetic fibers are postganglionic, but many of them end on postganglionic cholinergic neurons, where the norepinephrine they secrete inhibits acetylcholine secretion by activating α2 presynaptic receptors. Other sympathetic fibers appear to end directly on intestinal smooth muscle cells. The electrical properties of intestinal smooth muscle are discussed in Chapter 5. Still other fibers innervate blood vessels, where they produce vasoconstriction. It appears that the intestinal blood vessels have a dual innervation: They have an extrinsic noradrenergic innervation and an intrinsic innervation by fi-bers of the enteric nervous system. VIP and NO are among the mediators in the intrinsic innervation, which seems, among other things, to be responsible for the hyperemia that accompa-nies digestion of food. It is unsettled whether the blood vessels have an additional cholinergic innervation. GASTROINTESTINAL (SPLANCHNIC) CIRCULATION A final general point that should be made about the gas-trointestinal tract relates to its unusual circulatory features. The blood flow to the stomach, intestines, pancreas, and liver is arranged in a series of parallel circuits, with all the blood from the intestines and pancreas draining via the portal vein to the liver (Figure 26–25). The blood from the intestines, pancreas, and spleen drains via the hepatic portal vein to the liver and from the liver via the hepatic veins to the inferior vena cava. The viscera and the liver receive about 30% of the cardiac output via the celiac, superior mesenteric, and inferi-or mesenteric arteries. The liver receives about 1300 mL/min from the portal vein and 500 mL/min from the hepatic artery during fasting, and the portal supply increases still further af-ter meals. CHAPTER SUMMARY ■The gastrointestinal system evolved as a portal to permit con-trolled nutrient uptake in multicellular organisms. It is func-tionally continuous with the outside environment and is defended by a well-developed mucosal immune system. Never-theless, the gut usually lives in harmony with an extensive com-mensal microflora, particularly in the colon. ■Digestive secretions serve to chemically alter the components of meals (particularly macromolecules) such that their constitu-ents can be absorbed across the epithelium. Meal components are acted on sequentially by saliva, gastric juice, pancreatic juice, and bile, which contain enzymes, ions, water, and other special-ized components. ■The intestine and the organs that drain into it secrete about 8 L of fluid per day, which are added to water consumed in food and beverages. Most of this fluid is reabsorbed, leaving only approx-imately 200 mL to be lost to the stool. Fluid secretion and absorption are both dependent on the active epithelial transport of ions, nutrients, or both. ■Gastrointestinal functions are regulated in an integrated fashion by endocrine, paracrine, and neurocrine mechanisms. Hor-mones and paracrine factors are released from enteroendocrine cells in response to signals coincident with the intake of meals. ■The enteric nervous system conveys information from the cen-tral nervous system to the gastrointestinal tract, but also often can activate programmed responses of secretion and motility in an autonomous fashion. ■The intestine has an unusual circulation, in that the majority of its venous outflow does not return directly to the heart, but rath-er is directed initially to the liver via the portal vein. FIGURE 26–25 Schematic of the splanchnic circulation under fasting conditions. Note that even during fasting, the liver receives the majority of its blood supply via the portal vein. Vena cava Hepatic veins 1300 mL/min 500 mL/min 700 mL/min 700 mL/min Aorta Rest of body Inferior mesenteric artery Superior mesenteric artery Branches of the hepatic artery also supply the stomach, pancreas and small intestine Spleen Stomach Pancreas Colon Small intestine 400 mL/min Liver Celiac artery Portal vein Heart H e p a ti c a rt e r y 450 SECTION V Gastrointestinal Physiology MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Water is absorbed in the jejunum, ileum, and colon and excreted in the feces. Arrange these in order of the amount of water absorbed or excreted from greatest to smallest. A) colon, jejunum, ileum, feces B) feces, colon, ileum, jejunum C) jejunum, ileum, colon, feces D) colon, ileum, jejunum, feces E) feces, jejunum, ileum, colon 2. Drugs and toxins that increase the cAMP content of the intesti-nal mucosa cause diarrhea because they A) increase Na+–K+ cotransport in the small intestine. B) increase K+ secretion into the colon. C) inhibit K+ absorption in the crypts of Lieberkühn. D) increase Na+ absorption in the small intestine. E) increase Cl– secretion into the intestinal lumen. 3. A patient with a tumor secreting abnormal amounts of gastrin (gastrinoma) would be most likely to exhibit which of the following? A) decreased chief cell exocytosis B) duodenal ulceration C) increased gastric pH in the period between meals D) a reduced incidence of gastroesophageal reflux disease E) protein malabsorption 4. Which of the following has the highest pH? A) gastric juice B) hepatic bile C) pancreatic juice D) saliva E) secretions of the intestinal glands 5. Which of the following would not be produced by total pancreatectomy? A) vitamin E deficiency B) hyperglycemia C) metabolic acidosis D) weight gain E) decreased absorption of amino acids CHAPTER RESOURCES Baron TH, Morgan DE: Current concepts: Acute necrotizing pancreatitis. N Engl J Med 1999;340:1412. 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Lewis JH (editor): A Pharmacological Approach to Gastrointestinal Disorders. Williams & Wilkins, 1994. Meier PJ, Stieger B: Molecular mechanisms of bile formation. News Physiol Sci 2000;15:89. Montecucco C, Rappuoli R: Living dangerously: How Helicobacter pylori survives in the human stomach. Nat Rev Mol Cell Biol 2001;2:457. Nakazato M: Guanylin family: New intestinal peptides regulating electrolyte and water homeostasis. J Gastroenterol 2001;36:219. Rabon EC, Reuben MA: The mechanism and structure of the gastric H+, K+–ATPase. Annu Rev Physiol 1990;52:321. Sachs G, Zeng N, Prinz C: Pathophysiology of isolated gastric endocrine cells. Annu Rev Physiol 1997;59:234. Sellin JH: SCFAs: The enigma of weak electrolyte transport in the colon. News Physiol Sci 1999;14:58. Specian RD, Oliver MG: Functional biology of intestinal goblet cells. Am J Med 1991;260:C183. Topping DL, Clifton PM: Short-chain fatty acids and human colonic function: Select resistant starch and nonstarch polysaccharides. Physiol Rev 2001;81:1031. Trauner M, Meier PJ, Boyer JL: Molecular mechanisms of cholestasis. N Engl J Med 1998;339:1217. Walsh JH (editor): Gastrin. Raven Press, 1993. Williams JA, Blevins GT Jr: Cholecystokinin and regulation of pancreatic acinar cell function. Physiol Rev 1993;73:701. Wolfe MM, Lichtenstein DR, Singh G: Gastrointestinal toxicity of nonsteroidal anti-inflammatory drugs. N Engl J Med 1999;340:1888. Wright EM: The intestinal Na+/glucose cotransporter. Annu Rev Physiol 1993;55:575. Young JA, van Lennep EW: The Morphology of Salivary Glands. Academic Press, 1978. Zoetendal EG et al: Molecular ecological analysis of the gastrointestinal microbiota: A review. J Nutr 2004;134:465. 451 C H A P T E R 27 Digestion, Absorption, & Nutritional Principles O B J E C T I V E S After studying this chapter, you should be able to: ■Understand how nutrients are delivered to the body and the chemical processes needed to convert them to a form suitable for absorption. ■List the major dietary carbohydrates and define the luminal and brush border pro-cesses that produce absorbable monosaccharides as well as the transport mecha-nisms that provide for the uptake of these hydrophilic molecules. ■Understand the process of protein assimilation, and the ways in which it is compa-rable to, or converges from, that used for carbohydrates. ■Define the stepwise processes of lipid digestion and absorption, the role of bile acids in solubilizing the products of lipolysis, and the consequences of fat malabsorption. ■Identify the source and functions of short-chain fatty acids in the colon. ■Delineate the mechanisms of uptake for vitamins and minerals. ■Understand basic principles of energy metabolism and nutrition. INTRODUCTION The gastrointestinal system is the portal through which nutri-tive substances, vitamins, minerals, and fluids enter the body. Proteins, fats, and complex carbohydrates are broken down into absorbable units (digested), principally in the small intestine. The products of digestion and the vitamins, miner-als, and water cross the mucosa and enter the lymph or the blood (absorption). The digestive and absorptive processes are the subject of this chapter. Digestion of the major foodstuffs is an orderly process involving the action of a large number of digestive enzymes (Table 27–1). Enzymes from the salivary glands attack carbo-hydrates (and fats in some species); enzymes from the stom-ach attack proteins and fats; and enzymes from the exocrine portion of the pancreas attack carbohydrates, proteins, lipids, DNA, and RNA. Other enzymes that complete the digestive process are found in the luminal membranes and the cyto-plasm of the cells that line the small intestine. The action of the enzymes is aided by the hydrochloric acid secreted by the stomach and the bile secreted by the liver. Most substances pass from the intestinal lumen into the enterocytes and then out of the enterocytes to the interstitial fluid. The processes responsible for movement across the luminal cell membrane are often quite different from those responsible for movement across the basal and lateral cell membranes to the interstitial fluid. 452 SECTION V Gastrointestinal Physiology DIGESTION & ABSORPTION: CARBOHYDRATES DIGESTION The principal dietary carbohydrates are polysaccharides, dis-accharides, and monosaccharides. Starches (glucose poly-mers) and their derivatives are the only polysaccharides that are digested to any degree in the human gastrointestinal tract. Amylopectin, which constitutes 80–90% of dietary starch, is a branched molecule, whereas amylose is a straight chain with only 1:4α linkages (Figure 27-1). The disaccharides lactose (milk sugar) and sucrose (table sugar) are also ingested, along with the monosaccharides fructose and glucose. In the mouth, starch is attacked by salivary α-amylase. However, the optimal pH for this enzyme is 6.7, and its action is inhibited by the acidic gastric juice when food enters the stomach. In the small intestine, both the salivary and the pan-creatic α-amylase also act on the ingested polysaccharides. Both the salivary and the pancreatic α-amylases hydrolyze 1:4α linkages but spare 1:6α linkages and terminal 1:4α link-ages. Consequently, the end products of α-amylase digestion are oligosaccharides: the disaccharide maltose; the trisaccha-ride maltotriose; and α-limit dextrins, polymers of glucose containing an average of about eight glucose molecules with 1:6α linkages (Figure 27–1). The oligosaccharidases responsible for the further digestion of the starch derivatives are located in the brush border of small intestinal epithelial cells (Figure 27–1). Some of these enzymes have more than one substrate. Isomaltase is mainly responsible for hydrolysis of 1:6α linkages. Along with maltase and sucrase, it also breaks down maltotriose and maltose. Sucrase and isomaltase are initially synthesized as a single glycoprotein chain which is inserted into the brush border membrane. It is then hydrolyzed by pancreatic proteases into sucrase and iso-maltase subunits. Sucrase hydrolyzes sucrose into a molecule of glucose and a molecule of fructose. In addition, two disaccharidases are present in the brush border: lactase, which hydrolyzes lactose to glucose and galactose, and trehalase, which hydrolyzes treha-lose, a 1:1α-linked dimer of glucose, into two glucose molecules. Deficiency of one or more of the brush border oligosaccha-ridases may cause diarrhea, bloating, and flatulence after ingestion of sugar (Clinical Box 27–1). The diarrhea is due to the increased number of osmotically active oligosaccharide molecules that remain in the intestinal lumen, causing the volume of the intestinal contents to increase. In the colon, TABLE 27–1 Normal transport of substances by the intestine and location of maximum absorption or secretion.a Small Intestine Absorption of: Upperb Mid Lower Colon Sugars (glucose, galactose, etc) ++ +++ ++ 0 Amino acids ++ ++ ++ 0 Water-soluble and fat-soluble vitamins except vitamin B12 +++ ++ 0 0 Betaine, dimethylglycine, sarcosine + ++ ++ ? Antibodies in newborns + ++ +++ ? Pyrimidines (thymine and uracil) + + ? ? Long-chain fatty acid absorption and conversion to triglyceride +++ ++ + 0 Bile acids + + +++ Vitamin B12 0 + +++ 0 Na+ +++ ++ +++ +++ K+ + + + Sec Ca2+ +++ ++ + ? Fe2+ +++ + + ? Cl– +++ ++ + + SO4 2– ++ + 0 ? aAmount of absorption is graded + to +++. Sec, secreted when luminal K+ is low. bUpper small intestine refers primarily to jejunum, although the duodenum is similar in most cases studied (with the notable exception that the duodenum secretes HCO3 – and shows little net absorption or secretion of NaCl). CHAPTER 27 Digestion, Absorption, & Nutritional Principles 453 bacteria break down some of the oligosaccharides, further increasing the number of osmotically active particles. The bloating and flatulence are due to the production of gas (CO2 and H2) from disaccharide residues in the lower small intes-tine and colon. ABSORPTION Hexoses are rapidly absorbed across the wall of the small in-testine (Table 27–1). Essentially all the hexoses are removed before the remains of a meal reach the terminal part of the il-eum. The sugar molecules pass from the mucosal cells to the blood in the capillaries draining into the portal vein. The transport of most hexoses is dependent on Na+ in the intestinal lumen; a high concentration of Na+ on the mucosal surface of the cells facilitates and a low concentration inhibits sugar influx into the epithelial cells. This is because glucose and Na+ share the same cotransporter, or symport, the sodium-dependent glucose transporter (SGLT, Na+ glucose cotransporter) (Figure 27–2). The members of this family of transporters, SGLT 1 and SGLT 2, resemble the glucose trans-porters responsible for facilitated diffusion (see Chapter 21) in that they cross the cell membrane 12 times and have their –COOH and –NH2 terminals on the cytoplasmic side of the membrane. However, there is no homology to the glucose transporter (GLUT) series of transporters. SGLT-1 is respon-sible for uptake of dietary glucose from the gut. The related transporter, SGLT 2, is responsible for glucose transport out of the renal tubules (see Chapter 38). Because the intracellular Na+ concentration is low in intes-tinal cells as it is in other cells, Na+ moves into the cell along its concentration gradient. Glucose moves with the Na+ and is released in the cell (Figure 27–2). The Na+ is transported into the lateral intercellular spaces, and the glucose is transported by GLUT 2 into the interstitium and thence to the capillaries. Thus, glucose transport is an example of secondary active transport (see Chapter 2); the energy for glucose transport is provided indirectly, by the active transport of Na+ out of the cell. This maintains the concentration gradient across the luminal border of the cell, so that more Na+ and consequently more glucose enter. When the Na+/glucose cotransporter is congenitally defective, the resulting glucose/galactose mal-absorption causes severe diarrhea that is often fatal if glucose and galactose are not promptly removed from the diet. The use of glucose and its polymers to retain Na+ in diarrheal dis-ease was discussed in Chapter 26. SGLT-1 also transports galactose, but fructose utilizes a dif-ferent mechanism. Its absorption is independent of Na+ or the FIGURE 27–1 Left: Structure of amylose and amylopectin, which are polymers of glucose (indicated by circles). These molecules are partially digested by the enzyme amylase, yielding the products shown at the bottom of the figure. Right: Brush border hydrolases responsible for the se-quential digestion of the products of luminal starch digestion (1, linear oligomers; 2, alpha-limit dextrins). Amylase Glucose α1,4 bond α1,6 bond Amylose Amylopectin Maltose Maltotriose Glucose oligomers α-limit dextrin Maltose Maltotriose Glucoamylase Sucrase Isomaltase 1 α-limit dextrin Glucoamylase Isomaltase + + Glucoamylase Sucrase Isomaltase 2 454 SECTION V Gastrointestinal Physiology transport of glucose and galactose; it is transported instead by facilitated diffusion from the intestinal lumen into the entero-cytes by GLUT 5 and out of the enterocytes into the intersti-tium by GLUT 2. Some fructose is converted to glucose in the mucosal cells. Insulin has little effect on intestinal transport of sugars. In this respect, intestinal absorption resembles glucose reabsorp-tion in the proximal convoluted tubules of the kidneys (see Chapter 38); neither process requires phosphorylation, and both are essentially normal in diabetes but are depressed by the drug phlorizin. The maximal rate of glucose absorption from the intestine is about 120 g/h. PROTEINS & NUCLEIC ACIDS PROTEIN DIGESTION Protein digestion begins in the stomach, where pepsins cleave some of the peptide linkages. Like many of the other enzymes concerned with protein digestion, pepsins are secreted in the form of inactive precursors (proenzymes) and activated in the CLINICAL BOX 27–1 Lactose Intolerance In most mammals and in many races of humans, intestinal lactase activity is high at birth, then declines to low levels during childhood and adulthood. The low lactase levels are associated with intolerance to milk (lactose intolerance). Most Europeans and their American descendants retain sufficient intestinal lactase activity in adulthood; the inci-dence of lactase deficiency in northern and western Euro-peans is only about 15%. However, the incidence in blacks, American Indians, Asians, and Mediterranean populations is 70–100%. When such individuals ingest dairy products, they are unable to digest lactose sufficiently, and so symp-toms such as bloating, pain, gas, and diarrhea are produced by the unabsorbed osmoles that are subsequently digested by colonic bacteria. Milk intolerance can be ameliorated by administration of commercial lactase preparations, but this is expensive. Yogurt is better tolerated than milk in intoler-ant individuals because it contains its own bacterial lactase. FIGURE 27–2 Brush border digestion and assimilation of the disaccharides sucrose (panel 1) and lactose (panel 2). SGLT-1, sodium-glucose cotransporter-1. 1 Sucrase Isomaltase Na+ Sucrose Cytosol Glucose Fructose 2 Na+ Cytosol Glucose Galactose Lactose Lactase SGLT-1 SGLT-1 GLUT5 Brush border membrane CHAPTER 27 Digestion, Absorption, & Nutritional Principles 455 gastrointestinal tract. The pepsin precursors are called pepsi-nogens and are activated by gastric acid. Human gastric mu-cosa contains a number of related pepsinogens, which can be divided into two immunohistochemically distinct groups, pepsinogen I and pepsinogen II. Pepsinogen I is found only in acid-secreting regions, whereas pepsinogen II is also found in the pyloric region. Maximal acid secretion correlates with pepsinogen I levels. Pepsins hydrolyze the bonds between aromatic amino acids such as phenylalanine or tyrosine and a second amino acid, so the products of peptic digestion are polypeptides of very diverse sizes. Because pepsins have a pH optimum of 1.6 to 3.2, their action is terminated when the gastric contents are mixed with the alkaline pancreatic juice in the duodenum and jejunum. The pH of the intestinal contents in the duodenal bulb is 2.0 to 4.0, but in the rest of the duodenum it is about 6.5. In the small intestine, the polypeptides formed by digestion in the stomach are further digested by the powerful proteolytic enzymes of the pancreas and intestinal mucosa. Trypsin, the chymotrypsins, and elastase act at interior peptide bonds in the peptide molecules and are called endopeptidases. The forma-tion of the active endopeptidases from their inactive precursors occurs only when they have reached their site of action, secon-dary to the action of the brush border hydrolase, enterokinase (Figure 27–3). The powerful protein-splitting enzymes of the pancreatic juice are secreted as inactive proenzymes. Trypsino-gen is converted to the active enzyme trypsin by enterokinase when the pancreatic juice enters the duodenum. Enterokinase contains 41% polysaccharide, and this high polysaccharide content apparently prevents it from being digested itself before it can exert its effect. Trypsin converts chymotrypsinogens into chymotrypsins and other proenzymes into active enzymes (Figure 27–3). Trypsin can also activate trypsinogen; therefore, once some trypsin is formed, there is an auto-catalytic chain reaction. Enterokinase deficiency occurs as a congenital abnor-mality and leads to protein malnutrition. The carboxypeptidases of the pancreas are exopeptidases that hydrolyze the amino acids at the carboxyl ends of the polypeptides (Figure 27–4). Some free amino acids are liber-ated in the intestinal lumen, but others are liberated at the cell surface by the aminopeptidases, carboxypeptidases, endopep-tidases, and dipeptidases in the brush border of the mucosal cells. Some di- and tripeptides are actively transported into the intestinal cells and hydrolyzed by intracellular peptidases, with the amino acids entering the bloodstream. Thus, the final digestion to amino acids occurs in three locations: the intestinal lumen, the brush border, and the cytoplasm of the mucosal cells. ABSORPTION At least seven different transport systems transport amino ac-ids into enterocytes. Five of these require Na+ and cotransport amino acids and Na+ in a fashion similar to the cotransport of Na+ and glucose (Figure 27–3). Two of these five also require Cl–. In two systems, transport is independent of Na+. FIGURE 27–3 Mechanism to avoid activation of pancreatic proteases until they are in the duodenal lumen. Enterokinase Trypsinogen Trypsinogen Trypsin Trypsin Chymotrypsinogen Chymotrypsin Proelastase Elastase Procarboxypeptidase A Carboxy-peptidase A Carboxy-peptidase B Procarboxypeptidase B Epithelium Lumen Pancreatic juice 456 SECTION V Gastrointestinal Physiology The di- and tripeptides are transported into enterocytes by a system known as PepT1 (or peptide transporter 1) that requires H+ instead of Na+ (Figure 27–5). There is very little absorption of larger peptides. In the enterocytes, amino acids released from the peptides by intracellular hydrolysis plus the amino acids absorbed from the intestinal lumen and brush border are transported out of the enterocytes along their basolateral borders by at least five transport systems. From there, they enter the hepatic portal blood. Absorption of amino acids is rapid in the duodenum and jejunum but slow in the ileum. Approximately 50% of the digested protein comes from ingested food, 25% from proteins in digestive juices, and 25% from desquamated mucosal cells. Only 2–5% of the protein in the small intestine escapes diges-tion and absorption. Some of this is eventually digested by bac-terial action in the colon. Almost all of the protein in the stools is not of dietary origin but comes from bacteria and cellular debris. Evidence suggests that the peptidase activities of the brush border and the mucosal cell cytoplasm are increased by resection of part of the ileum and that they are independently altered in starvation. Thus, these enzymes appear to be subject to homeostatic regulation. In humans, a congenital defect in the mechanism that transports neutral amino acids in the intestine and renal tubules causes Hartnup disease. A congenital defect in the transport of basic amino acids causes cystinuria. How-ever, most patients do not experience nutritional deficiencies of these amino acids because peptide transport compensates. In infants, moderate amounts of undigested proteins are also absorbed. The protein antibodies in maternal colostrum are largely secretory immunoglobulins (IgAs), the production of which is increased in the breast in late pregnancy. They cross the mammary epithelium by transcytosis and enter the circulation of the infant from the intestine, providing passive immunity against infections. Absorption is by endocytosis and subsequent exocytosis. Protein absorption declines with age, but adults still absorb small quantities. Foreign proteins that enter the circulation pro-voke the formation of antibodies, and the antigen–antibody reaction occurring on subsequent entry of more of the same protein may cause allergic symptoms. Thus, absorption of pro-teins from the intestine may explain the occurrence of allergic symptoms after eating certain foods. The incidence of food allergy in children is said to be as high as 8%. Certain foods are more allergenic than others. Crustaceans, mollusks, and fish are common offenders, and allergic responses to legumes, cows’ milk, and egg white are also relatively frequent. Absorption of protein antigens, particularly bacterial and viral proteins, takes place in large microfold cells or M cells, specialized intestinal epithelial cells that overlie aggregates of lymphoid tissue (Peyer’s patches). These cells pass the antigens FIGURE 27–4 Luminal digestion of peptides by pancreatic endopeptidases and exopeptidases. Individual amino acids are shown as squares. Chymotrypsin Elastase Peptide with C-terminal neutral AA Carboxypeptidase A Short peptides free neutral and basic AA’s Carboxypeptidase B Peptide with C-terminal basic AA Trypsin Ser Arg Ser Arg Large peptides FIGURE 27–5 Disposition of short peptides in intestinal epithelial cells. Peptides are absorbed together with a proton sup-plied by an apical sodium/hydrogen exchanger (NHE) by the peptide transporter 1 (PepT1). Absorbed peptides are digested by cytosolic proteases, and any amino acids that are surplus to the needs of the ep-ithelial cell are transported into the bloodstream by a series of basolat-eral transport proteins. Na+ H+ H+ 3Na+ 2K+ Di-, tripeptides Cytosolic digestion Basolateral amino acid transporters NHE PEPT1 CHAPTER 27 Digestion, Absorption, & Nutritional Principles 457 to the lymphoid cells, and lymphocytes are activated. The activated lymphoblasts enter the circulation, but they later return to the intestinal mucosa and other epithelia, where they secrete IgA in response to subsequent exposures to the same antigen. This secretory immunity is an important defense mechanism (see Chapter 3). NUCLEIC ACIDS Nucleic acids are split into nucleotides in the intestine by the pancreatic nucleases, and the nucleotides are split into the nu-cleosides and phosphoric acid by enzymes that appear to be lo-cated on the luminal surfaces of the mucosal cells. The nucleosides are then split into their constituent sugars and pu-rine and pyrimidine bases. The bases are absorbed by active transport. LIPIDS FAT DIGESTION A lingual lipase is secreted by Ebner’s glands on the dorsal sur-face of the tongue in some species, and the stomach also se-cretes a lipase (Table 27–1). They are of little quantitative significance for lipid digestion other than in the setting of pan-creatic insufficiency, however. Most fat digestion therefore begins in the duodenum, pan-creatic lipase being one of the most important enzymes involved. This enzyme hydrolyzes the 1- and 3-bonds of the tri-glycerides (triacylglycerols) with relative ease but acts on the 2-bonds at a very low rate, so the principal products of its action are free fatty acids and 2-monoglycerides (2-monoacylglycer-ols). It acts on fats that have been emulsified (see below). Its activity is facilitated when an amphipathic helix that covers the active site like a lid is bent back. Colipase, a protein with a molecular weight of about 11,000, is also secreted in the pan-creatic juice, and when this molecule binds to the –COOH-terminal domain of the pancreatic lipase, opening of the lid is facilitated. Colipase is secreted in an inactive proform (Table 27–1) and is activated in the intestinal lumen by trypsin. Another pancreatic lipase that is activated by bile salts has been characterized. This 100,000-kDa cholesterol esterase represents about 4% of the total protein in pancreatic juice. In adults, pancreatic lipase is 10–60 times more active, but unlike pancreatic lipase, this bile salt-activated lipase catalyzes the hydrolysis of cholesterol esters, esters of fat-soluble vita-mins, and phospholipids, as well as triglycerides. A very simi-lar enzyme is found in human milk. Fats are relatively insoluble, which limits their ability to cross the unstirred layer and reach the surface of the mucosal cells. However, they are finely emulsified in the small intestine by the detergent action of bile salts, lecithin, and monoglycer-ides. When the concentration of bile salts in the intestine is high, as it is after contraction of the gallbladder, lipids and bile salts interact spontaneously to form micelles (Figure 26–16). These cylindrical aggregates, which are discussed in more detail in Chapter 29, take up lipids, and although their lipid concentration varies, they generally contain fatty acids, monoglycerides, and cholesterol in their hydrophobic centers. Micellar formation further solubilizes the lipids and provides a mechanism for their transport to the enterocytes. Thus, the micelles move down their concentration gradient through the unstirred layer to the brush border of the mucosal cells. The lipids diffuse out of the micelles, and a saturated aqueous solution of the lipids is maintained in contact with the brush border of the mucosal cells (Figure 26–16). STEATORRHEA Pancreatectomized animals and patients with diseases that de-stroy the exocrine portion of the pancreas have fatty, bulky, clay-colored stools (steatorrhea) because of the impaired di-gestion and absorption of fat. The steatorrhea is due mostly to lipase deficiency. However, acid inhibits the lipase, and the lack of alkaline secretion from the pancreas also contributes by lowering the pH of the intestine contents. In some cases, hy-persecretion of gastric acid can cause steatorrhea. Another cause of steatorrhea is defective reabsorption of bile salts in the distal ileum (see Chapter 29). FAT ABSORPTION Traditionally, lipids were thought to enter the enterocytes by passive diffusion, but some evidence now suggests that carri-ers are involved. Inside the cells, the lipids are rapidly esteri-fied, maintaining a favorable concentration gradient from the lumen into the cells (Figure 27–6). There are also carriers that export certain lipids back into the lumen, thereby limiting their oral availability. This is the case for plant sterols as well as cholesterol. The fate of the fatty acids in enterocytes depends on their size. Fatty acids containing less than 10 to 12 carbon atoms are water-soluble enough that they pass through the enterocyte unmodified and are actively transported into the portal blood. They circulate as free (unesterified) fatty acids. The fatty acids containing more than 10 to 12 carbon atoms are too insoluble for this. They are reesterified to triglycerides in the enterocytes. In addition, some of the absorbed cholesterol is esterified. The triglycerides and cholesterol esters are then coated with a layer of protein, cholesterol, and phospholipid to form chylomicrons. These leave the cell and enter the lym-phatics, because they are too large to pass through the junc-tions between capillary endothelial cells (Figure 27–6). In mucosal cells, most of the triglyceride is formed by the acylation of the absorbed 2-monoglycerides, primarily in the smooth endoplasmic reticulum. However, some of the tri-glyceride is formed from glycerophosphate, which in turn is a product of glucose catabolism. Glycerophosphate is also 458 SECTION V Gastrointestinal Physiology converted into glycerophospholipids that participate in chy-lomicron formation. The acylation of glycerophosphate and the formation of lipoproteins occur in the rough endoplasmic reticulum. Carbohydrate moieties are added to the proteins in the Golgi apparatus, and the finished chylomicrons are extruded by exocytosis from the basal or lateral aspects of the cell. Absorption of long-chain fatty acids is greatest in the upper parts of the small intestine, but appreciable amounts are also absorbed in the ileum. On a moderate fat intake, 95% or more of the ingested fat is absorbed. The processes involved in fat absorption are not fully mature at birth, and infants fail to absorb 10–15% of ingested fat. Thus, they are more susceptible to the ill effects of disease processes that reduce fat absorption. SHORT-CHAIN FATTY ACIDS IN THE COLON Increasing attention is being focused on short-chain fatty ac-ids (SCFAs) that are produced in the colon and absorbed from it. SCFAs are two- to five-carbon weak acids that have an av-erage normal concentration of about 80 mmol/L in the lumen. About 60% of this total is acetate, 25% propionate, and 15% butyrate. They are formed by the action of colonic bacteria on complex carbohydrates, resistant starches, and other compo-nents of the dietary fiber, that is, the material that escapes di-gestion in the upper gastrointestinal tract and enters the colon. Absorbed SCFAs are metabolized and make a significant contribution to the total caloric intake. In addition, they exert a trophic effect on the colonic epithelial cells, combat inflam-mation, and are absorbed in part by exchange for H+, helping to maintain acid–base equilibrium. SCFAs are absorbed by specific transporters present in colonic epithelial cells. SCFAs also promote the absorption of Na+, although the exact mech-anism for coupled Na+–SCFA absorption is unsettled. ABSORPTION OF VITAMINS & MINERALS VITAMINS Absorption of the fat-soluble vitamins A, D, E, and K is defi-cient if fat absorption is depressed because of lack of pancreat-ic enzymes or if bile is excluded from the intestine by obstruction of the bile duct. Most vitamins are absorbed in the upper small intestine, but vitamin B12 is absorbed in the ileum. This vitamin binds to intrinsic factor, a protein secreted by the stomach, and the complex is absorbed across the ileal mucosa (see Chapter 26). Vitamin B12 absorption and folate absorption are Na+-inde-pendent, but all seven of the remaining water-soluble vita-mins—thiamin, riboflavin, niacin, pyridoxine, pantothenate, biotin, and ascorbic acid—are absorbed by carriers that are Na+ cotransporters. CALCIUM A total of 30–80% of ingested calcium is absorbed. The absorp-tive process and its relation to 1,25-dihydroxycholecalciferol are discussed in Chapter 23. Through this vitamin D deriva-tive, Ca2+ absorption is adjusted to body needs; absorption is increased in the presence of Ca2+ deficiency and decreased in the presence of Ca2+ excess. Ca2+ absorption is also facilitated by protein. It is inhibited by phosphates and oxalates because these anions form insoluble salts with Ca2+ in the intestine. Magnesium absorption is also facilitated by protein. IRON In adults, the amount of iron lost from the body is relatively small. The losses are generally unregulated, and total body stores of iron are regulated by changes in the rate at which it is absorbed from the intestine. Men lose about 0.6 mg/d, largely in the stools. Women have a variable, larger loss averaging about twice this value because of the additional iron lost dur-ing menstruation. The average daily iron intake in the United States and Europe is about 20 mg, but the amount absorbed is equal only to the losses. Thus, the amount of iron absorbed is normally about 3–6% of the amount ingested. Various dietary factors affect the availability of iron for absorption; for exam-ple, the phytic acid found in cereals reacts with iron to form insoluble compounds in the intestine, as do phosphates and oxalates. Most of the iron in the diet is in the ferric (Fe3+) form, whereas it is the ferrous (Fe2+) form that is absorbed. Fe3+ reductase activity is associated with the iron transporter in the FIGURE 27–6 Intracellular handling of the products of lipid digestion. Absorbed fatty acids (FA) and monoglycerides (MG) are re-esterified to form triglyceride (TG) in the smooth endoplasmic reticu-lum. Apoproteins synthesized in the rough endoplasmic reticulum are coated around lipid cores, and the resulting chylomicrons are secreted from the basolateral pole of epithelial cells by exocytosis. Rough ER Golgi Smooth ER FA/MG Chylomicrons Synthesis of TG and phospholipids Synthesis of apolipoproteins Apolipoprotein glycosylation Exocytosis TG CHAPTER 27 Digestion, Absorption, & Nutritional Principles 459 brush borders of the enterocytes (Figure 27–7). Gastric secre-tions dissolve the iron and permit it to form soluble com-plexes with ascorbic acid and other substances that aid its reduction to the Fe2+ form. The importance of this function in humans is indicated by the fact that iron deficiency anemia is a troublesome and relatively frequent complication of par-tial gastrectomy. Almost all iron absorption occurs in the duodenum. Trans-port of Fe2+ into the enterocytes occurs via divalent metal transporter 1 (DMT1) (Figure 27–7). Some is stored in ferritin, and the remainder is transported out of the enterocytes by a basolateral transporter named ferroportin 1. A protein called hephaestin (Hp) is associated with ferroportin 1. It is not a transporter itself, but it facilitates basolateral transport. In the plasma, Fe2+ is converted to Fe3+ and bound to the iron trans-port protein transferrin. This protein has two iron-binding sites. Normally, transferrin is about 35% saturated with iron, and the normal plasma iron level is about 130 μg/dL (23 μmol/ L) in men and 110 μg/dL (19 μmol/L) in women. Heme (see Chapter 32) binds to an apical transport protein in enterocytes and is carried into the cytoplasm. In the cyto-plasm, HO2, a subtype of heme oxygenase, removes Fe2+ from the porphyrin and adds it to the intracellular Fe2+ pool. Seventy percent of the iron in the body is in hemoglobin, 3% in myoglobin, and the rest in ferritin, which is present not only in enterocytes, but also in many other cells. Apoferritin is a globular protein made up of 24 subunits. Ferritin is readily visible under the electron microscope and has been used as a tracer in studies of phagocytosis and related phenomena. Fer-ritin molecules in lysosomal membranes may aggregate in deposits that contain as much as 50% iron. These deposits are called hemosiderin. Intestinal absorption of iron is regulated by three factors: recent dietary intake of iron, the state of the iron stores in the body, and the state of erythropoiesis in the bone marrow. The normal operation of the factors that maintain iron balance is essential for health (Clinical Box 27–2). NUTRITIONAL PRINCIPLES & ENERGY METABOLISM The animal organism oxidizes carbohydrates, proteins, and fats, producing principally CO2, H2O, and the energy necessary for life processes (Clinical Box 27–3). CO2, H2O, and energy are also produced when food is burned outside the body. However, in the body, oxidation is not a one-step, semiexplosive reaction but a complex, slow, stepwise process called catabolism, which liberates energy in small, usable amounts. Energy can be stored in the body in the form of special energy-rich phosphate com-pounds and in the form of proteins, fats, and complex carbohy-drates synthesized from simpler molecules. Formation of these substances by processes that take up rather than liberate energy is called anabolism. This chapter consolidates consideration of endocrine function by providing a brief summary of the pro-duction and utilization of energy and the metabolism of carbo-hydrates, proteins, and fats. METABOLIC RATE The amount of energy liberated by the catabolism of food in the body is the same as the amount liberated when food is burned outside the body. The energy liberated by catabolic processes in the body is used for maintaining body functions, digesting and metabolizing food, thermoregulation, and physical activity. It appears as external work, heat, and energy storage: Energy output = External work + Energy storage + Heat FIGURE 27–7 Absorption of iron. Fe3+ is converted to Fe2+ by ferric reductase, and Fe2+ is transported into the enterocyte by the apical membrane iron transporter DMT1. Heme is transported into the enterocyte by a separate heme transporter (HT), and HO2 releases Fe2+ from the heme. Some of the intracellular Fe2+ is converted to Fe3+ and bound to ferritin. The rest binds to the basolateral Fe2+ transporter ferroportin (FP) and is transported to the interstitial fluid. The transport is aided by hephaestin (Hp). In plasma, Fe2+ is converted to Fe3+ and bound to the iron transport protein transferrin (TF). Enterocyte Intestinal lumen Brush border Blood Fe3+-ferritin Fe3+−TF Fe2+ Fe2+ Fe3+ Fe2+ Fe2+ Heme Heme Shed reductase HT DMT1 HO2 FP Hp Fe2+ Fe3+ 460 SECTION V Gastrointestinal Physiology The amount of energy liberated per unit of time is the met-abolic rate. Isotonic muscle contractions perform work at a peak efficiency approximating 50%: Essentially all of the energy of isometric contractions appears as heat, because little or no external work (force mul-tiplied by the distance that the force moves a mass) is done (see Chapter 5). Energy is stored by forming energy-rich compounds. The amount of energy storage varies, but in fast-ing individuals it is zero or negative. Therefore, in an adult individual who has not eaten recently and who is not moving (or growing, reproducing, or lactating), all of the energy out-put appears as heat. CALORIES The standard unit of heat energy is the calorie (cal), defined as the amount of heat energy necessary to raise the temperature of 1 g of water 1 degree, from 15 °C to 16 °C. This unit is also called the gram calorie, small calorie, or standard calorie. The unit commonly used in physiology and medicine is the Calo-rie (kilocalorie; kcal), which equals 1000 cal. CALORIMETRY The energy released by combustion of foodstuffs outside the body can be measured directly (direct calorimetry) by oxidiz-ing the compounds in an apparatus such as a bomb calorim-eter, a metal vessel surrounded by water inside an insulated container. The food is ignited by an electric spark. The change in the temperature of the water is a measure of the calories produced. Similar measurements of the energy released by combustion of compounds in living animals and humans are much more complex, but calorimeters have been constructed that can physically accommodate human beings. The heat CLINICAL BOX 27–2 Disorders of Iron Uptake Iron deficiency causes anemia. Conversely, iron overload causes hemosiderin to accumulate in the tissues, produc-ing hemosiderosis. Large amounts of hemosiderin can damage tissues, causing hemochromatosis. This syndrome is characterized by pigmentation of the skin, pancreatic damage with diabetes (“bronze diabetes"), cirrhosis of the liver, a high incidence of hepatic carcinoma, and gonadal atrophy. Hemochromatosis may be hereditary or acquired. The most common cause of the hereditary forms is a mu-tated HFE gene that is common in the Caucasian popula-tion. It is located on the short arm of chromosome 6 and is closely linked to the human leukocyte antigen-A (HLA-A) locus. It is still unknown precisely how mutations in HFE cause hemochromatosis, but individuals who are homoge-nous for HFE mutations absorb excess amounts of iron be-cause HFE normally inhibits expression of the duodenal transporters that participate in iron uptake. If the abnor-mality is diagnosed before excessive amounts of iron accu-mulate in the tissues, life expectancy can be prolonged by repeated withdrawal of blood. Acquired hemochromatosis occurs when the iron-regulating system is overwhelmed by excess iron loads due to chronic destruction of red blood cells, liver disease, or repeated transfusions in diseases such as intractable anemia. Efficiency Work done Total energy expended ------------------------------------------------------= CLINICAL BOX 27–3 Obesity Obesity is the most common and most expensive nutritional problem in the United States. A convenient and reliable indi-cator of body fat is the body mass index (BMI), which is body weight (in kilograms) divided by the square of height (in meters). Values above 25 are abnormal. Individuals with val-ues of 25–30 are overweight, and those with values > 30 are obese. In the United States, 55% of the population are over-weight and 22% are obese. The incidence of obesity is also in-creasing in other countries. Indeed, the Worldwatch Institute has estimated that although starvation continues to be a problem in many parts of the world, the number of over-weight people in the world is now as great as the number of underfed. Obesity is a problem because of its complications. It is associated with accelerated atherosclerosis and an in-creased incidence of gallbladder and other diseases. Its asso-ciation with type 2 diabetes is especially striking. As weight increases, insulin resistance increases and frank diabetes ap-pears. At least in some cases, glucose tolerance is restored when weight is lost. In addition, the mortality rates from many kinds of cancer are increased in obese individuals. The causes of the high incidence of obesity in the general popula-tion are probably multiple. Studies of twins raised apart show a definite genetic component. It has been pointed out that through much of human evolution, famines were common, and mechanisms that permitted increased energy storage as fat had survival value. Now, however, food is plentiful in many countries, and the ability to gain and retain fat has become a liability. As noted above, the fundamental cause of obesity is still an excess of energy intake in food over energy expendi-ture. If human volunteers are fed a fixed high-calorie diet, some gain weight more rapidly than others, but the slower weight gain is due to increased energy expenditure in the form of small, fidgety movements (nonexercise activity thermogenesis; NEAT). Body weight generally increases at a slow but steady rate throughout adult life. Decreased physi-cal activity is undoubtedly a factor in this increase, but de-creased sensitivity to leptin may also play a role. CHAPTER 27 Digestion, Absorption, & Nutritional Principles 461 produced by their bodies is measured by the change in tem-perature of the water in the walls of the calorimeter. The caloric values of the common foodstuffs, as measured in a bomb calorimeter, are found to be 4.1 kcal/g of carbohy-drate, 9.3 kcal/g of fat, and 5.3 kcal/g of protein. In the body, similar values are obtained for carbohydrate and fat, but the oxidation of protein is incomplete, the end products of pro-tein catabolism being urea and related nitrogenous com-pounds in addition to CO2 and H2O (see below). Therefore, the caloric value of protein in the body is only 4.1 kcal/g. INDIRECT CALORIMETRY Energy production can also be calculated by measuring the products of the energy-producing biologic oxidations; that is, CO2, H2O, and the end products of protein catabolism pro-duced, but this is difficult. However, O2 is not stored, and ex-cept when an O2 debt is being incurred, the amount of O2 consumption per unit of time is proportionate to the energy liberated by metabolism. Consequently, measurement of O2 consumption (indirect calorimetry) is used to determine the metabolic rate. RESPIRATORY QUOTIENT (RQ) The respiratory quotient (RQ) is the ratio in the steady state of the volume of CO2 produced to the volume of O2 consumed per unit of time. It should be distinguished from the respiratory ex-change ratio (R), which is the ratio of CO2 to O2 at any given time whether or not equilibrium has been reached. R is affected by factors other than metabolism. RQ and R can be calculated for reactions outside the body, for individual organs and tissues, and for the whole body. The RQ of carbohydrate is 1.00, and that of fat is about 0.70. This is because H and O are present in carbohydrate in the same proportions as in water, whereas in the various fats, extra O2 is necessary for the formation of H2O. Carbohydrate: C6H12O6 + 6O2 → 6CO2 + 6H2O (glucose) RQ = 6/6 = 1.00 Fat: 2C51H98O6 + 145O2 → 102CO2 + 98H2O (tripalmitin) RQ = 102/145 = 0.703 Determining the RQ of protein in the body is a complex process, but an average value of 0.82 has been calculated. The approximate amounts of carbohydrate, protein, and fat being oxidized in the body at any given time can be calculated from the RQ and the urinary nitrogen excretion. RQ and R for the whole body differ in various conditions. For example, during hyperventilation, R rises because CO2 is being blown off. Dur-ing strenuous exercise, R may reach 2.00 because CO2 is being blown off and lactic acid from anaerobic glycolysis is being converted to CO2 (see below). After exercise, R may fall for a while to 0.50 or less. In metabolic acidosis, R rises because res-piratory compensation for the acidosis causes the amount of CO2 expired to rise (see Chapter 39). In severe acidosis, R may be greater than 1.00. In metabolic alkalosis, R falls. The O2 consumption and CO2 production of an organ can be calculated at equilibrium by multiplying its blood flow per unit of time by the arteriovenous differences for O2 and CO2 across the organ, and the RQ can then be calculated. Data on the RQ of individual organs are of considerable interest in drawing inferences about the metabolic processes occurring in them. For example, the RQ of the brain is regularly 0.97– 0.99, indicating that its principal but not its only fuel is carbo-hydrate. During secretion of gastric juice, the stomach has a negative R because it takes up more CO2 from the arterial blood than it puts into the venous blood (see Chapter 26). MEASURING THE METABOLIC RATE In determining the metabolic rate, O2 consumption is usually measured with some form of oxygen-filled spirometer and a CO2-absorbing system. Such a device is illustrated in Figure 27–8. The spirometer bell is connected to a pen that writes on a rotating drum as the bell moves up and down. The slope of a line joining the ends of each of the spirometer excursions is propor-tional to the O2 consumption. The amount of O2 (in milliliters) consumed per unit of time is corrected to standard temperature and pressure (see Chapter 35) and then converted to energy pro-duction by multiplying by 4.82 kcal/L of O2 consumed. FACTORS AFFECTING THE METABOLIC RATE The metabolic rate is affected by many factors (Table 27–2). The most important is muscular exertion. O2 consumption is elevated not only during exertion but also for as long after-ward as is necessary to repay the O2 debt (see Chapter 5). Re-cently ingested foods also increase the metabolic rate because of their specific dynamic action (SDA). The SDA of a food is the obligatory energy expenditure that occurs during its as-similation into the body. It takes 30 kcal to assimilate the amount of protein sufficient to raise the metabolic rate 100 kcal; 6 kcal to assimilate a similar amount of carbohydrate; and 5 kcal to assimilate a similar amount of fat. The cause of the SDA, which may last up to 6 h, is uncertain. Another factor that stimulates metabolism is the environ-mental temperature. The curve relating the metabolic rate to the environmental temperature is U-shaped. When the envi-ronmental temperature is lower than body temperature, heat-producing mechanisms such as shivering are activated and the metabolic rate rises. When the temperature is high enough to raise the body temperature, metabolic processes generally accelerate, and the metabolic rate rises about 14% for each degree Celsius of elevation. 462 SECTION V Gastrointestinal Physiology The metabolic rate determined at rest in a room at a com-fortable temperature in the thermoneutral zone 12 to 14 h after the last meal is called the basal metabolic rate (BMR). This value falls about 10% during sleep and up to 40% during prolonged starvation. The rate during normal daytime activi-ties is, of course, higher than the BMR because of muscular activity and food intake. The maximum metabolic rate reached during exercise is often said to be 10 times the BMR, but trained athletes can increase their metabolic rate as much as 20-fold. The BMR of a man of average size is about 2000 kcal/d. Large animals have higher absolute BMRs, but the ratio of BMR to body weight in small animals is much greater. One variable that correlates well with the metabolic rate in differ-ent species is the body surface area. This would be expected, since heat exchange occurs at the body surface. The actual relation to body weight (W) would be BMR = 3.52W0.67 However, repeated measurements by numerous investiga-tors have come up with a higher exponent, averaging 0.75: BMR = 3.52W0.75 Thus, the slope of the line relating metabolic rate to body weight is steeper than it would be if the relation were due solely to body area (Figure 27–9). The cause of the greater slope has been much debated but remains unsettled. For clinical use, the BMR is usually expressed as a percent-age increase or decrease above or below a set of generally used standard normal values. Thus, a value of +65 means that the individual’s BMR is 65% above the standard for that age and sex. The decrease in metabolic rate is part of the explanation of why, when an individual is trying to lose weight, weight loss is initially rapid and then slows down. ENERGY BALANCE The first law of thermodynamics, the principle that states that energy is neither created nor destroyed when it is converted from one form to another, applies to living organisms as well as inanimate systems. One may therefore speak of an energy FIGURE 27–8 Diagram of a modified Benedict apparatus, a recording spirometer used for measuring human O2 consumption, and the record obtained with it. The slope of the line AB is proportionate to the O2 consumption. V: one-way check valve. A B Pulleys Oxygen bell Water seal Inhalation tube Exhalation tube Mouthpiece Pulley Rotating drum Volume Time v v Breathing chamber CO2 absorber TABLE 27–2 Factors affecting the metabolic rate. Muscular exertion during or just before measurement Recent ingestion of food High or low environmental temperature Height, weight, and surface area Sex Age Growth Reproduction Lactation Emotional state Body temperature Circulating levels of thyroid hormones Circulating epinephrine and norepinephrine levels CHAPTER 27 Digestion, Absorption, & Nutritional Principles 463 balance between caloric intake and energy output. If the calor-ic content of the food ingested is less than the energy output, that is, if the balance is negative, endogenous stores are uti-lized. Glycogen, body protein, and fat are catabolized, and the individual loses weight. If the caloric value of the food intake exceeds energy loss due to heat and work and the food is prop-erly digested and absorbed, that is, if the balance is positive, energy is stored, and the individual gains weight. To balance basal output so that the energy-consuming tasks essential for life can be performed, the average adult must take in about 2000 kcal/d. Caloric requirements above the basal level depend on the individual’s activity. The average sedentary stu-dent (or professor) needs another 500 kcal, whereas a lumber-jack needs up to 3000 additional kcal per day. NUTRITION The aim of the science of nutrition is the determination of the kinds and amounts of foods that promote health and well-be-ing. This includes not only the problems of undernutrition but those of overnutrition, taste, and availability (Clinical Box 27–4). However, certain substances are essential constituents of any human diet. Many of these compounds have been mentioned in previous sections of this chapter, and a brief summary of the essential and desirable dietary components is presented below. ESSENTIAL DIETARY COMPONENTS An optimal diet includes, in addition to sufficient water (see Chapter 38), adequate calories, protein, fat, minerals, and vitamins. CALORIC INTAKE & DISTRIBUTION As noted above, the caloric value of the dietary intake must be approximately equal to the energy expended if body weight is to be maintained. In addition to the 2000 kcal/d necessary to meet basal needs, 500 to 2500 kcal/d (or more) are required to meet the energy demands of daily activities. The distribution of the calories among carbohydrate, pro-tein, and fat is determined partly by physiologic factors and partly by taste and economic considerations. A daily protein FIGURE 27–9 Correlation between metabolic rate and body weight, plotted on logarithmic scales. The slope of the colored line is 0.75. The black line represents the way surface area increases with weight for geometrically similar shapes and has a slope of 0.67. (Modified from Kleiber M and reproduced with permission from McMahon TA: Size and shape in biology. Science 1973;179:1201. Copyright © 1973 by the American Association for the Advancement of Science.) 105 104 103 102 101 100 10−3 10−2 10−1 100 101 102 103 104 Body weight (kg) Heat production (kcal/d) Mouse Rat Guinea pig Macaque Cats Rabbits Sheep Steer Goat Chimpanzee Cow Elephant CLINICAL BOX 27–4 The Malabsorption Syndrome The digestive and absorptive functions of the small intestine are essential for life. However, the digestive and absorptive capacity of the intestine is larger than needed for normal function (the anatomic reserve). Removal of short segments of the jejunum or ileum generally does not cause severe symptoms, and compensatory hypertrophy and hyperplasia of the remaining mucosa occur. However, when more than 50% of the small intestine is resected or bypassed, the ab-sorption of nutrients and vitamins is so compromised that it is very difficult to prevent malnutrition and wasting (malab-sorption). Resection of the ileum also prevents the absorp-tion of bile acids, and this leads in turn to deficient fat ab-sorption. It also causes diarrhea because the unabsorbed bile salts enter the colon, where they activate chloride secretion (see Chapter 26). Other complications of intestinal resection or bypass include hypocalcemia, arthritis, hyperuricemia, and possibly fatty infiltration of the liver, followed by cirrho-sis. Various disease processes can also impair absorption without a loss of intestinal length. The pattern of deficiencies that results is sometimes called the malabsorption syn-drome. This pattern varies somewhat with the cause, but it can include deficient absorption of amino acids, with marked body wasting and, eventually, hypoproteinemia and edema. Carbohydrate and fat absorption are also depressed. Be-cause of the defective fat absorption, the fat-soluble vita-mins (vitamins A, D, E, and K) are not absorbed in adequate amounts. One of the most interesting conditions causing the malabsorption syndrome is the autoimmune disease celiac disease. This disease occurs in genetically predisposed indi-viduals who have the major histocompatibility complex (MHC) class II antigen HLA-DQ2 or DQ8 (see Chapter 3). In these individuals gluten and closely related proteins cause intestinal T cells to mount an inappropriate im-mune response that damages the intestinal epithelial cells and results in a loss of villi and a flattening of the mucosa. The proteins are found in wheat, rye, barley, and to a lesser extent in oats—but not in rice or corn. When grains contain-ing gluten are omitted from the diet, bowel function is gen-erally restored to normal. 464 SECTION V Gastrointestinal Physiology intake of 1 g/kg body weight to supply the eight nutritionally essential amino acids and other amino acids is desirable. The source of the protein is also important. Grade I proteins, the animal proteins of meat, fish, dairy products, and eggs, con-tain amino acids in approximately the proportions required for protein synthesis and other uses. Some of the plant pro-teins are also grade I, but most are grade II because they sup-ply different proportions of amino acid and some lack one or more of the essential amino acids. Protein needs can be met with a mixture of grade II proteins, but the intake must be large because of the amino acid wastage. Fat is the most compact form of food, since it supplies 9.3 kcal/g. However, often it is also the most expensive. Indeed, internationally there is a reasonably good positive correlation between fat intake and standard of living. In the past, Western diets have contained large amounts (100 g/d or more). The evi-dence indicating that a high unsaturated/saturated fat ratio in the diet is of value in the prevention of atherosclerosis and the current interest in preventing obesity may change this. In Cen-tral and South American Indian communities where corn (car-bohydrate) is the dietary staple, adults live without ill effects for years on a very low fat intake. Therefore, provided that the needs for essential fatty acids are met, a low-fat intake does not seem to be harmful, and a diet low in saturated fats is desirable. Carbohydrate is the cheapest source of calories and pro-vides 50% or more of the calories in most diets. In the average middle-class American diet, approximately 50% of the calo-ries come from carbohydrate, 15% from protein, and 35% from fat. When calculating dietary needs, it is usual to meet the protein requirement first and then split the remaining cal-ories between fat and carbohydrate, depending on taste, income, and other factors. For example, a 65-kg man who is moderately active needs about 2800 kcal/d. He should eat at least 65 g of protein daily, supplying 267 (65 × 4.1) kcal. Some of this should be grade I protein. A reasonable figure for fat intake is 50 to 60 g. The rest of the caloric requirement can be met by supplying carbohydrate. MINERAL REQUIREMENTS A number of minerals must be ingested daily for the mainte-nance of health. Besides those for which recommended daily di-etary allowances have been set, a variety of different trace elements should be included. Trace elements are defined as ele-ments found in tissues in minute amounts. Those believed to be essential for life, at least in experimental animals, are listed in Ta-ble 27–3. In humans, iron deficiency causes anemia. Cobalt is part of the vitamin B12 molecule, and vitamin B12 deficiency leads to megaloblastic anemia (see Chapter 32). Iodine deficien-cy causes thyroid disorders (see Chapter 20). Zinc deficiency causes skin ulcers, depressed immune responses, and hypogo-nadal dwarfism. Copper deficiency causes anemia and changes in ossification. Chromium deficiency causes insulin resistance. Fluorine deficiency increases the incidence of dental caries. Conversely, some minerals can be toxic when present in the body in excess. For example, severe iron overload causes hemo-chromatosis, copper excess causes brain damage (Wilson dis-ease), and aluminum poisoning in patients with renal failure who are receiving dialysis treatment causes a rapidly progressive dementia that resembles Alzheimer disease (see Chapter 19). Sodium and potassium are also essential minerals, but list-ing them is academic, because it is very difficult to prepare a sodium-free or potassium-free diet. A low-salt diet is well tol-erated for prolonged periods because of the compensatory mechanisms that conserve Na+. VITAMINS Vitamins were discovered when it was observed that diets adequate in calories, essential amino acids, fats, and miner-als failed to maintain health. The term vitamin has now come to refer to any organic dietary constituent necessary for life, health, and growth that does not function by supply-ing energy. Because there are minor differences in metabolism between mammalian species, some substances are vitamins in one spe-cies and not in another. The sources and functions of the major vitamins in humans are listed in Table 27–4. Most vitamins have important functions in intermediary metabolism or the special metabolism of the various organ systems. Those that are water-soluble (vitamin B complex, vitamin C) are easily absorbed, but the fat-soluble vitamins (vitamins A, D, E, and K) are poorly absorbed in the absence of bile or pancreatic lipase. Some dietary fat intake is necessary for their absorption, and in obstructive jaundice or disease of the exocrine pancreas, defi-ciencies of the fat-soluble vitamins can develop even if their intake is adequate. Vitamin A and vitamin D are bound to transfer proteins in the circulation. The α-tocopherol form of vitamin E is normally bound to chylomicrons. In the liver, it is transferred to very low density lipoprotein (VLDL) and distrib-uted to tissues by an α-tocopherol transfer protein. When this protein is abnormal due to mutation of its gene in humans, there is cellular deficiency of vitamin E and the development of a condition resembling Friedreich ataxia. Two Na+-dependent L-ascorbic acid transporters have recently been isolated. One is found in the kidneys, intestines, and liver, and the other in the brain and eyes. TABLE 27–3 Trace elements believed essential for life. Arsenic Manganese Chromium Molybdenum Cobalt Nickel Copper Selenium Fluorine Silicon Iodine Vanadium Iron Zinc CHAPTER 27 Digestion, Absorption, & Nutritional Principles 465 TABLE 27–4 Vitamins essential or probably essential to human nutrition.a Vitamin Action Deficiency Symptoms Sources Chemistry A (A1, A2) Constituents of visual pig-ments (see Chapter 12: Vi-sion); necessary for fetal development and for cell development throughout life Night blindness, dry skin Yellow vegeta-bles and fruit B complex Thiamin (vitamin B1) Cofactor in decarboxyla-tions Beriberi, neuritis Liver, unrefined cereal grains Riboflavin (vitamin B2) Constituent of flavopro-teins Glossitis, cheilosis Liver, milk Niacin Constituent of NAD+ and NADP+ Pellagra Yeast, lean meat, liver Pyridoxine (vitamin B6) Forms prosthetic group of certain decarboxylases and transaminases. Con-verted in body into pyri-doxal phosphate and pyridoxamine phosphate Convulsions, hyperirri-tability Yeast, wheat, corn, liver Pantothenic acid Constituent of CoA Dermatitis, enteritis, al-opecia, adrenal insuffi-ciency Eggs, liver, yeast Biotin Catalyzes CO2 “fixation” (in fatty acid synthesis, etc) Dermatitis, enteritis Egg yolk, liver, to-matoes Folates (folic acid) and relat-ed compounds Coenzymes for “1-car-bon” transfer; involved in methylating reactions Sprue, anemia. Neural tube defects in children born to folate-deficient women Leafy green vege-tables Cyanocobal-amin (vitamin B12) Coenzyme in amino acid metabolism. Stimulates erythropoiesis Pernicious anemia (see Chapter 26: Overview of Gastrointestinal Function & Regulation) Liver, meat, eggs, milk Complex of four substituted pyrrole rings around a cobalt atom (see Chapter 26: Overview of Gastro-intestinal Function & Regulation) C Maintains prosthetic metal ions in their re-duced form; scavenges free radicals Scurvy Citrus fruits, leafy green vegetables (continued) Vitamin A1 alcohol (retinol). C C H2 C C (CH CH3 CH C CH3 H2C H2C H3C CH3 CH)2 CH2OH +N S N N CH2 CH3 NH2 CH3 CH2CH2OH C C H H C C C C H3C H3C CH2(CHOH)3 CH2OH N N C C N C O N H C O COOH N Can be synthesized in body from tryptophan. CH2OH HO H3C N CH2OH HO C C H H CH3 CH3 CH2CH2COOH C C N H H OH O O C C N H H C N H H S CH H2C (CH2)4COOH Folic acid NH2 OH N N N N CH2 H C O CHNH COOH CH2 CH2 HOOC NH O CO C H OH C HO H CH2OH C C Ascorbic acid (synthesized in most mammals except guinea pigs and primates, including humans). OH 466 SECTION V Gastrointestinal Physiology The diseases caused by deficiency of each of the vitamins are listed in Table 27–4. It is worth remembering, however, particularly in view of the advertising campaigns for vitamin pills and supplements, that very large doses of the fat-soluble vitamins are definitely toxic. Hypervitaminosis A is charac-terized by anorexia, headache, hepatosplenomegaly, irritabil-ity, scaly dermatitis, patchy loss of hair, bone pain, and hyperostosis. Acute vitamin A intoxication was first described by Arctic explorers, who developed headache, diarrhea, and dizziness after eating polar bear liver. The liver of this animal is particularly rich in vitamin A. Hypervitaminosis D is asso-ciated with weight loss, calcification of many soft tissues, and eventual renal failure. Hypervitaminosis K is characterized by gastrointestinal disturbances and anemia. Large doses of water-soluble vitamins have been thought to be less likely to cause problems because they can be rapidly cleared from the body. However, it has been demonstrated that ingestion of megadoses of pyridoxine (vitamin B6) can produce peripheral neuropathy. CHAPTER SUMMARY ■A typical mixed meal consists of carbohydrates, proteins, and lipids (the latter largely in the form of triglycerides). Each must be digested to allow its uptake into the body. Specific transport-ers carry the products of digestion into the body. ■In the process of carbohydrate assimilation, the epithelium can only transport monomers, whereas for proteins, short peptides can be absorbed in addition to amino acids. ■The protein assimilation machinery, which rests heavily on the proteases in pancreatic juice, is arranged such that these en-zymes are not activated until they reach their substrates in the small intestinal lumen. This is accomplished by the restricted localization of an activating enzyme, enterokinase. ■Lipids face special challenges to assimilation given their hydro-phobicity. Bile acids solubilize the products of lipolysis in micelles and accelerate their ability to diffuse to the epithelial surface. The assimilation of triglycerides is enhanced by this mechanism, whereas that of cholesterol and fat-soluble vitamins absolutely requires it. ■The catabolism of nutrients provides energy to the body in a controlled fashion, via stepwise oxidations and other reactions. ■A balanced diet is important for health, and certain substances obtained from the diet are essential to life. The caloric value of dietary intake must be approximately equal to energy expendi-ture for homeostasis. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Maximum absorption of short-chain fatty acids produced by bacteria occurs in the A) stomach. B) duodenum. C) jejunum. D) ileum. E) colon. 2. Calcium absorption is increased by A) hypercalcemia. B) oxalates in the diet. C) iron overload. D) 1,25-dihydroxycholecalciferol. E) increased Na+ absorption. D group Increase intestinal ab-sorption of calcium and phosphate (see Chapter 21: Hormonal Control of Calcium & Phosphate Metabolism & the Physi-ology of Bone) Rickets Fish liver Family of sterols (see Chapter 21: Hormonal Con-trol of Calcium & Phosphate Metabolism & the Physiology of Bone) E group Antioxidants; cofactors in electron transport in cy-tochrome chain? Ataxia and other symp-toms and signs of spinocerebellar dys-function Milk, eggs, meat, leafy vegetables K group Catalyze γ carboxylation of glutamic acid residues on various proteins con-cerned with blood clot-ting Hemorrhagic phenom-ena Leafy green vege-tables aCholine is synthesized in the body in small amounts, but it has recently been added to the list of essential nutrients. TABLE 27–4 Vitamins essential or probably essential to human nutrition.a (Continued) Vitamin Action Deficiency Symptoms Sources Chemistry CH3 CH3 H2 H3C O α-Tocopherol; β- and γ-tocopherol also active. HO (CH2)3 (CH2)3 (CH2)3 CH3 CH3 CH CH CH CH3 CH3 CH3 H2 O O CH3 Vitamin K3; a large number of similar compounds have biological activity. CHAPTER 27 Digestion, Absorption, & Nutritional Principles 467 3. A decrease in which of the following would be expected in a child exhibiting a congenital absence of enterokinase? A) incidence of pancreatitis B) glucose absorption C) bile acid reabsorption D) gastric pH E) protein assimilation 4. In Hartnup disease (a defect in the transport of neutral amino acids), patients do not become deficient in these amino acids due to the activity of A) PepT1. B) brush border peptidases. C) Na+,K+ ATPase. D) cystic fibrosis transmembrane conductance regulator (CFTR). E) trypsin. 5. A newborn baby is brought to the pediatrician suffering from severe diarrhea that worsens with meals. The symptoms diminish when nutrients are delivered intravenously. The child most likely has a mutation in which of the following intestinal transporters? A) Na+,K+ ATPase B) NHE3 C) SGLT1 D) H+,K+ ATPase E) NKCC1 CHAPTER RESOURCES Andrews NC: Disorders of iron metabolism. N Engl J Med 1999;341:1986. Chong L, Marx J (editors): Lipids in the limelight. Science 2001;294:1861. Farrell RJ, Kelly CP: Celiac sprue. N Engl J Med 2002;346:180. Hofmann AF: Bile acids: The good, the bad, and the ugly. News Physiol Sci 1999;14:24. Levitt MD, Bond JH: Volume, composition and source of intestinal gas. Gastroenterology 1970;59:921. Mann NS, Mann SK: Enterokinase. Proc Soc Exp Biol Med 1994;206:114. Meier PJ, Stieger B: Molecular mechanisms of bile formation. News Physiol Sci 2000;15:89. Topping DL, Clifton PM: Short-chain fatty acids and human colonic function: Select resistant starch and nonstarch polysaccharides. Physiol Rev 2001;81:1031. Wright EM: The intestinal Na+/glucose cotransporter. Annu Rev Physiol 1993;55:575. This page intentionally left blank 469 C H A P T E R 28 Gastrointestinal Motility O B J E C T I V E S After studying this chapter, you should be able to: ■List the major forms of motility in the gastrointestinal tract and their roles in diges-tion and excretion. ■Distinguish between peristalsis and segmentation. ■Explain the electrical basis of gastrointestinal contractions and the role of basic electrical activity in governing motility patterns. ■Describe how gastrointestinal motility changes during fasting. ■Understand how food is swallowed and transferred to the stomach. ■Define the factors that govern gastric emptying and the abnormal response of vomiting. ■Define how the motility patterns of the colon subserve its function to desiccate and evacuate the stool. INTRODUCTION The digestive and absorptive functions of the gastrointestinal system outlined in the previous chapter depend on a variety of mechanisms that soften the food, propel it through the length of the gastrointestinal tract (Table 28–1), and mix it with hepatic bile stored in the gallbladder and digestive enzymes secreted by the salivary glands and pancreas. Some of these mechanisms depend on intrinsic properties of the intestinal smooth muscle. Others involve the operation of reflexes involving the neurons intrinsic to the gut, reflexes involving the central nervous system (CNS), paracrine effects of chemi-cal messengers, and gastrointestinal hormones. GENERAL PATTERNS OF MOTILITY PERISTALSIS Peristalsis is a reflex response that is initiated when the gut wall is stretched by the contents of the lumen, and it occurs in all parts of the gastrointestinal tract from the esophagus to the rec-tum. The stretch initiates a circular contraction behind the stimulus and an area of relaxation in front of it (Figure 28–1). The wave of contraction then moves in an oral-to-caudal direc-tion, propelling the contents of the lumen forward at rates that vary from 2 to 25 cm/s. Peristaltic activity can be increased or decreased by the autonomic input to the gut, but its occurrence is independent of the extrinsic innervation. Indeed, progres-sion of the contents is not blocked by removal and resuture of a segment of intestine in its original position and is blocked only if the segment is reversed before it is sewn back into place. Peristalsis is an excellent example of the integrated activity of the enteric nervous system. It appears that local stretch releases serotonin, which activates sensory neurons that activate the myenteric plexus. Cholinergic neurons passing in a retrograde direction in this plexus activate neurons that release substance P and acetylcholine, causing smooth muscle contraction. At the same time, cholinergic neurons passing in an anterograde 470 SECTION V Gastrointestinal Physiology direction activate neurons that secrete NO, vasoactive intesti-nal polypeptide (VIP), and adenosine triphosphate (ATP), producing the relaxation ahead of the stimulus. SEGMENTATION & MIXING When the meal is present, the enteric nervous system pro-motes a motility pattern that is related to peristalsis, but is de-signed to retard the movement of the intestinal contents along the length of the intestinal tract to provide time for digestion and absorption (Figure 28–1). This motility pattern is known as segmentation, and it provides for ample mixing of the intes-tinal contents (known as chyme) with the digestive juices. A segment of bowel contracts at both ends, and then a second contraction occurs in the center of the segment to force the chyme both backward and forward. Unlike peristalsis, there-fore, retrograde movement of the chyme occurs routinely in the setting of segmentation. This mixing pattern persists for as long as nutrients remain in the lumen to be absorbed. It pre-sumably reflects programmed activity of the bowel dictated by the enteric nervous system, and can occur independent of cen-tral input, although the latter can modulate it. BASIC ELECTRICAL ACTIVITY & REGULATION OF MOTILITY Except in the esophagus and the proximal portion of the stom-ach, the smooth muscle of the gastrointestinal tract has sponta-neous rhythmic fluctuations in membrane potential between about –65 and –45 mV. This basic electrical rhythm (BER) is initiated by the interstitial cells of Cajal, stellate mesenchymal pacemaker cells with smooth muscle-like features that send long multiply branched processes into the intestinal smooth muscle. In the stomach and the small intestine, these cells are located in the outer circular muscle layer near the myenteric plexus; in the colon, they are at the submucosal border of the circular muscle layer. In the stomach and small intestine, there is a descending gradient in pacemaker frequency, and as in the heart, the pace-maker with the highest frequency usually dominates. The BER itself rarely causes muscle contraction, but spike potentials superimposed on the most depolarizing portions of the BER waves do increase muscle tension (Figure 28–2). The depolarizing portion of each spike is due to Ca2+ influx, and the repolarizing portion is due to K+ efflux. Many polypep-tides and neurotransmitters affect the BER. For example, ace-tylcholine increases the number of spikes and the tension of the smooth muscle, whereas epinephrine decreases the num-ber of spikes and the tension. The rate of the BER is about 4/min in the stomach. It is about 12/min in the duodenum and falls to about 8/min in the distal ileum. In the colon, the BER rate rises from about 2/min at the cecum to about 6/min at the sigmoid. The function of the BER is to coordinate peristaltic and other motor activity; contractions occur only during the depolarizing part of the waves. After vagotomy or transection of the stomach wall, for example, peristalsis in the stomach becomes irregular and chaotic. MIGRATING MOTOR COMPLEX During fasting between periods of digestion, the pattern of elec-trical and motor activity in gastrointestinal smooth muscle be-comes modified so that cycles of motor activity migrate from the stomach to the distal ileum. Each cycle, or migrating motor complex (MMC), starts with a quiescent period (phase I), con-tinues with a period of irregular electrical and mechanical activ-ity (phase II), and ends with a burst of regular activity (phase III). The MMCs are initiated by motilin, migrate aborally at a rate of about 5 cm/min, and occur at intervals of approximately 90 min. Gastric secretion, bile flow, and pancreatic secretion in-crease during each MMC. They likely serve to clear the stomach and small intestine of luminal contents in preparation for the next meal. They are immediately stopped by ingestion of food TABLE 28–1 Mean lengths of various segments of the gastrointestinal tract as measured by intubation in living humans. Segment Length (cm) Pharynx, esophagus, and stomach 65 Duodenum 25 Jejunum and ileum 260 Colon 110 Data from Hirsch JE, Ahrens EH Jr, Blankenhorn DH: Measurement of human intesti-nal length in vivo and some causes of variation. Gastroenterology 1956;31:274. FIGURE 28–1 Patterns of gastrointestinal motility and propulsion. An isolated contraction moves contents orally and abo-rally. Segmentation mixes contents over a short stretch of intestine, as indicated by the time sequence from left to right. In the diagram on the left, the vertical arrows indicate the sites of subsequent contrac-tion. Peristalsis involves both contraction and relaxation, and moves contents aborally. Isolated contraction Segmentation Peristalsis Contraction Relaxation CHAPTER 28 Gastrointestinal Motility 471 (which suppresses motilin release via mechanisms that have not yet been elucidated), with a return to peristalsis and the other forms of BER and spike potentials. SEGMENT-SPECIFIC PATTERNS OF MOTILITY MOUTH & ESOPHAGUS In the mouth, food is mixed with saliva and propelled into the esophagus. Peristaltic waves in the esophagus move the food into the stomach. MASTICATION Chewing (mastication) breaks up large food particles and mixes the food with the secretions of the salivary glands. This wetting and homogenizing action aids swallowing and subse-quent digestion. Large food particles can be digested, but they cause strong and often painful contractions of the esophageal musculature. Particles that are small tend to disperse in the ab-sence of saliva and also make swallowing difficult because they do not form a bolus. The number of chews that is optimal de-pends on the food, but usually ranges from 20 to 25. Edentulous patients are generally restricted to a soft diet and have considerable difficulty eating dry food. SWALLOWING Swallowing (deglutition) is a reflex response that is triggered by afferent impulses in the trigeminal, glossopharyngeal, and vagus nerves (Figure 28–3). These impulses are integrated in the nucleus of the tractus solitarius and the nucleus ambiguus. The efferent fibers pass to the pharyngeal musculature and the tongue via the trigeminal, facial, and hypoglossal nerves. Swal-lowing is initiated by the voluntary action of collecting the oral contents on the tongue and propelling them backward into the pharynx. This starts a wave of involuntary contraction in the pharyngeal muscles that pushes the material into the esopha-gus. Inhibition of respiration and glottic closure are part of the reflex response. A peristaltic ring contraction of the esoph-ageal muscle forms behind the material, which is then swept down the esophagus at a speed of approximately 4 cm/s. When humans are in an upright position, liquids and semisolid foods generally fall by gravity to the lower esophagus ahead of the peristaltic wave. LOWER ESOPHAGEAL SPHINCTER Unlike the rest of the esophagus, the musculature of the gas-troesophageal junction (lower esophageal sphincter; LES) is tonically active but relaxes on swallowing. The tonic activity of the LES between meals prevents reflux of gastric contents into the esophagus. The LES is made up of three components (Fig-ure 28–4). The esophageal smooth muscle is more prominent FIGURE 28–2 Basic electrical rhythm (BER) of gastrointestinal smooth muscle. Top: Morphology, and relation to muscle contraction. Bottom: Stimulatory effect of acetylcholine and inhibitory effect of epinephrine. (Modified and reproduced with permission from Chang EB, Sitrin MD, Black DD: Gastrointestinal, Hepatobiliary, and Nutritional Physiology. Lippincott-Raven, 1996.) Electrical recording Mechanical recording (tension) Electrical recording Mechanical recording <15 <50 <50 mV <15 mV 1.5 g 1.5 g mV Spike potentials 10 s 10 s Epinephrine Acetylcholine BER 472 SECTION V Gastrointestinal Physiology at the junction with the stomach (intrinsic sphincter). Fibers of the crural portion of the diaphragm, a skeletal muscle, sur-round the esophagus at this point (extrinsic sphincter) and ex-ert a pinchcock-like action on the esophagus. In addition, the oblique or sling fibers of the stomach wall create a flap valve that helps close off the esophagogastric junction and prevent regurgitation when intragastric pressure rises. The tone of the LES is under neural control. Release of ace-tylcholine from vagal endings causes the intrinsic sphincter to contract, and release of NO and VIP from interneurons inner-vated by other vagal fibers causes it to relax. Contraction of the crural portion of the diaphragm, which is innervated by the phrenic nerves, is coordinated with respiration and con-tractions of chest and abdominal muscles. Thus, the intrinsic and extrinsic sphincters operate together to permit orderly flow of food into the stomach and to prevent reflux of gastric contents into the esophagus (Clinical Box 28–1). FIGURE 28–3 Movement of food through the pharynx and upper esophagus during swallowing. (a) The tongue pushes the food bo-lus to the back of the mouth. (b) The soft palate elevates to prevent food from entering the nasal passages. (c) The epiglottis covers the glottis to prevent food from entering the trachea and the upper esophageal sphincter relaxes. (d) Food descends into the esophagus. Soft palate Hard palate Tongue Glottis Trachea Esophagus (a) (b) (c) Epiglottis Food Upper esophageal sphincter (d) Pharynx FIGURE 28–4 Esophagogastric junction. Note that the lower esophageal sphincter (intrinsic sphincter) is supplemented by the crural por-tion of the diaphragm (extrinsic sphincter), and that the two are anchored to each other by the phrenoesophageal ligament. (Reproduced with permission, from Mittal RK, Balaban DH: The esophagogastric junction. N Engl J Med 1997;336:924. Copyright © 1997 by the Massachusetts Medical Society. All rights reserved.) Costal part Crural part Diaphragm Internal External Lower esophageal sphincter Phrenoesophageal ligament Sling fibers Stomach Squamocolumnar junction Longitudinal muscle Circular muscle CHAPTER 28 Gastrointestinal Motility 473 AEROPHAGIA & INTESTINAL GAS Some air is unavoidably swallowed in the process of eating and drinking (aerophagia). Some of the swallowed air is regurgi-tated (belching), and some of the gases it contains are ab-sorbed, but much of it passes on to the colon. Here, some of the oxygen is absorbed, and hydrogen, hydrogen sulfide, car-bon dioxide, and methane formed by the colonic bacteria from carbohydrates and other substances are added to it. It is then expelled as flatus. The smell is largely due to sulfides. The volume of gas normally found in the human gastrointestinal tract is about 200 mL, and the daily production is 500 to 1500 mL. In some individuals, gas in the intestines causes cramps, borborygmi (rumbling noises), and abdominal discomfort. STOMACH Food is stored in the stomach; mixed with acid, mucus, and pepsin; and released at a controlled, steady rate into the duodenum. GASTRIC MOTILITY & EMPTYING When food enters the stomach, the fundus and upper portion of the body relax and accommodate the food with little if any increase in pressure (receptive relaxation). Peristalsis then begins in the lower portion of the body, mixing and grinding the food and permitting small, semiliquid portions of it to pass through the pylorus and enter the duodenum. Receptive relaxation is vagally mediated and triggered by movement of the pharynx and esophagus. Peristaltic waves controlled by the gastric BER begin soon thereafter and sweep toward the pylorus. The contraction of the distal stomach caused by each wave is sometimes called antral systole and can last up to 10 s. Waves occur three to four times per minute. In the regulation of gastric emptying, the antrum, pylorus, and upper duodenum apparently function as a unit. Contrac-tion of the antrum is followed by sequential contraction of the pyloric region and the duodenum. In the antrum, partial con-traction ahead of the advancing gastric contents prevents solid masses from entering the duodenum, and they are mixed and crushed instead. The more liquid gastric contents are squirted a bit at a time into the small intestine. Normally, regurgitation from the duodenum does not occur, because the contraction of the pyloric segment tends to persist slightly longer than that of the duodenum. The prevention of regurgi-tation may also be due to the stimulating action of cholecysto-kinin (CCK) and secretin on the pyloric sphincter. REGULATION OF GASTRIC MOTILITY & EMPTYING The rate at which the stomach empties into the duodenum de-pends on the type of food ingested. Food rich in carbohydrate leaves the stomach in a few hours. Protein-rich food leaves more slowly, and emptying is slowest after a meal containing fat (Figure 28–5). The rate of emptying also depends on the os-motic pressure of the material entering the duodenum. Hy-perosmolality of the duodenal contents is sensed by “duodenal osmoreceptors” that initiate a decrease in gastric emptying which is probably neural in origin. Fats, carbohydrates, and acid in the duodenum inhibit gas-tric acid and pepsin secretion and gastric motility via neural and hormonal mechanisms. The hormone involved is proba-bly peptide YY. CCK has also been implicated as an inhibitor of gastric emptying (Clinical Box 28–2). VOMITING Vomiting is an example of central regulation of gut motility functions. Vomiting starts with salivation and the sensation of nausea. Reverse peristalsis empties material from the upper part of the small intestine into the stomach. The glottis closes, preventing aspiration of vomitus into the trachea. The breath is held in mid inspiration. The muscles of the abdominal wall contract, and because the chest is held in a fixed position, the CLINICAL BOX 28–1 Motor Disorders of the Esophagus Achalasia (literally, failure to relax) is a condition in which food accumulates in the esophagus and the organ be-comes massively dilated. It is due to increased resting LES tone and incomplete relaxation on swallowing. The myen-teric plexus of the esophagus is deficient at the LES in this condition and the release of NO and VIP is defective. It can be treated by pneumatic dilation of the sphincter or inci-sion of the esophageal muscle (myotomy). Inhibition of acetylcholine release by injection of botulinum toxin into the LES is also effective and produces relief that lasts for several months. The opposite condition is LES incompe-tence, which permits reflux of acid gastric contents into the esophagus (gastroesophageal reflux disease). This com-mon condition causes heartburn and esophagitis and can lead to ulceration and stricture of the esophagus due to scarring. In severe cases, the intrinsic sphincter, the extrin-sic sphincter, and sometimes both are weak, but less severe cases are caused by intermittent periods of poorly under-stood decreases in the neural drive to both sphincters. The condition can be treated by inhibition of acid secretion with H2 receptor blockers or omeprazole (see Chapter 26). Surgical treatment in which a portion of the fundus of the stomach is wrapped around the lower esophagus so that the LES is inside a short tunnel of stomach (fundoplica-tion) can also be tried, although in many patients who un-dergo this procedure the symptoms eventually return. 474 SECTION V Gastrointestinal Physiology contraction increases intra-abdominal pressure. The lower esophageal sphincter and the esophagus relax, and the gastric contents are ejected. The “vomiting center” in the reticular formation of the medulla (Figure 28–6) consists of various scattered groups of neurons in this region that control the dif-ferent components of the vomiting act. Irritation of the mucosa of the upper gastrointestinal tract is one trigger for vomiting. Impulses are relayed from the mucosa to the medulla over visceral afferent pathways in the sympathetic nerves and vagi. Other causes of vomiting can FIGURE 28–5 Effect of protein and fat on the rate of emptying of the human stomach. Subjects were fed 300-mL liquid meals. (Reproduced with permission from Brooks FP: Integrative lecture. Response of the GI tract to a meal. Undergraduate Teaching Project. American Gastroenterological Association, 1974.) Mean ± S.E. 300 200 100 0 20 60 100 Protein Protein + lipid Standard meal (inert pectin) Time after feeding (min) (300-mL liquid meals) Volume of test meal emptied (mL) CLINICAL BOX 28–2 Consequences of Gastric Bypass Surgery Patients who are morbidly obese often undergo a surgical procedure in which the stomach is stapled so that most of it is bypassed, and thus the reservoir function of the stomach is lost. As a result, such patients must eat frequent small meals. If larger meals are taken, because of rapid absorption of glu-cose from the intestine and the resultant hyperglycemia and abrupt rise in insulin secretion, gastrectomized patients sometimes develop hypoglycemic symptoms about 2 h after meals. Weakness, dizziness, and sweating after meals, due in part to hypoglycemia, are part of the picture of the “dump-ing syndrome,” a distressing syndrome that develops in pa-tients in whom portions of the stomach have been removed or the jejunum has been anastomosed to the stomach. An-other cause of the symptoms is rapid entry of hypertonic meals into the intestine; this provokes the movement of so much water into the gut that significant hypovolemia and hypotension are produced. FIGURE 28–6 Neural pathways leading to the initiation of vomiting in response to various stimuli. Pain Sights Anticipation Motion Vertigo Drugs eg, opiates, chemotherapy Hormones eg, pregnancy Ipecac Cytotoxic drugs Irritants Vagus nerve Higher centers Gastric mucosa Nucleus tractus solitarius Pharyngeal stimulation Glossopharyngeal nerve Brain stem vomiting center Area postrema chemoreceptor trigger zone Cerebellum Labyrinth Programmed vomiting response CHAPTER 28 Gastrointestinal Motility 475 arise centrally. For example, afferents from the vestibular nuclei mediate the nausea and vomiting of motion sickness. Other afferents presumably reach the vomiting control areas from the diencephalon and limbic system, because emetic responses to emotionally charged stimuli also occur. Thus, we speak of “nauseating smells” and “sickening sights.” Chemoreceptor cells in the medulla can also initiate vomit-ing when they are stimulated by certain circulating chemical agents. The chemoreceptor trigger zone in which these cells are located (Figure 28–6) is in the area postrema, a V-shaped band of tissue on the lateral walls of the fourth ventricle near the obex. This structure is one of the circumventricular organs (see Chapter34) and is not protected by the blood– brain barrier. Lesions of the area postrema have little effect on the vomiting response to gastrointestinal irritation or motion sickness, but abolish the vomiting that follows injection of apomorphine and a number of other emetic drugs. Such lesions also decrease vomiting in uremia and radiation sick-ness, both of which may be associated with endogenous pro-duction of circulating emetic substances. Serotonin (5-HT) released from enterochromaffin cells in the small intestine appears to initiate impulses via 5-HT3 receptors that trigger vomiting. In addition, there are dopa-mine D2 receptors and 5-HT3 receptors in the area postrema and adjacent nucleus of the solitary tract. 5-HT3 antagonists such as ondansetron and D2 antagonists such as chlorproma-zine and haloperidol are effective antiemetic agents. Cortico-steroids, cannabinoids, and benzodiazepines, alone or in combination with 5-HT3 and D2 antagonists, are also useful in treatment of the vomiting produced by chemotherapy. The mechanisms of action of corticosteroids and cannabinoids are unknown, whereas the benzodiazepines probably reduce the anxiety associated with chemotherapy. SMALL INTESTINE In the small intestine, the intestinal contents are mixed with the secretions of the mucosal cells and with pancreatic juice and bile. INTESTINAL MOTILITY The MMCs that pass along the intestine at regular intervals in the fasting state and their replacement by peristaltic and other contractions controlled by the BER are described above. In the small intestine, there are an average of 12 BER cycles/min in the proximal jejunum, declining to 8/min in the distal ileum. There are three types of smooth muscle contractions: peristal-tic waves, segmentation contractions, and tonic contractions. Peristalsis is described above. It propels the intestinal con-tents (chyme) toward the large intestines. Segmentation con-tractions (Figure 28–1), also described above, move the chyme to and fro and increase its exposure to the mucosal sur-face. These contractions are initiated by focal increases in Ca2+ influx with waves of increased Ca2+ concentration spreading from each focus. Tonic contractions are relatively prolonged contractions that in effect isolate one segment of the intestine from another. Note that these last two types of contractions slow transit in the small intestine to the point that the transit time is actually longer in the fed than in the fasted state. This permits longer contact of the chyme with the enterocytes and fosters absorption (Clinical Box 28–3). COLON The colon serves as a reservoir for the residues of meals that can-not be digested or absorbed (Figure 28–7). Motility in this seg-ment is likewise slowed to allow the colon to absorb water, Na+, and other minerals. By removal of about 90% of the fluid, it con-verts the 1000 to 2000 mL of isotonic chyme that enters it each day from the ileum to about 200 to 250 mL of semisolid feces. MOTILITY OF THE COLON The ileum is linked to the colon by a structure known as the ileocecal valve, which restricts reflux of colonic contents, and particularly the large numbers of commensal bacteria, into the relatively sterile ileum. The portion of the ileum containing the ileocecal valve projects slightly into the cecum, so that in-creases in colonic pressure squeeze it shut, whereas increases in ileal pressure open it. It is normally closed. Each time a peri-staltic wave reaches it, it opens briefly, permitting some of the ileal chyme to squirt into the cecum. When food leaves the stomach, the cecum relaxes and the passage of chyme through the ileocecal valve increases (gastroileal reflex). This is pre-sumably a vagal reflex. CLINICAL BOX 28–3 Ileus When the intestines are traumatized, there is a direct inhi-bition of smooth muscle, which causes a decrease in intesti-nal motility. It is due in part to activation of opioid receptors and is relieved by opioid-blocking drugs. When the perito-neum is irritated, reflex inhibition occurs due to increased discharge of noradrenergic fibers in the splanchnic nerves. Both types of inhibition operate to cause paralytic (ady-namic) ileus after abdominal operations. Because of the diffuse decrease in peristaltic activity in the small intestine, its contents are not propelled into the colon, and it be-comes irregularly distended by pockets of gas and fluid. Intestinal peristalsis returns in 6 to 8 h, followed by gastric peristalsis, but colonic activity takes 2 to 3 d to return. Ady-namic ileus can be relieved by passing a tube through the nose down to the small intestine and aspirating the fluid and gas for a few days until peristalsis returns. 476 SECTION V Gastrointestinal Physiology The movements of the colon include segmentation contrac-tions and peristaltic waves like those occurring in the small intestine. Segmentation contractions mix the contents of the colon and, by exposing more of the contents to the mucosa, facilitate absorption. Peristaltic waves propel the contents toward the rectum, although weak antiperistalsis is some-times seen. A third type of contraction that occurs only in the colon is the mass action contraction, in which there is simul-taneous contraction of the smooth muscle over large conflu-ent areas. These contractions move material from one portion of the colon to another (Clinical Box 28–4). They also move material into the rectum, and rectal distention initiates the defecation reflex (see below). The movements of the colon are coordinated by the BER of the colon. The frequency of this wave, unlike the wave in the small intestine, increases along the colon, from about 2/min at the ileocecal valve to 6/min at the sigmoid. TRANSIT TIME IN THE SMALL INTESTINE & COLON The first part of a test meal reaches the cecum in about 4 h, and all the undigested portions have entered the colon in 8 or 9 h. On average, the first remnants of the meal traverse the first third of the colon in 6 h, the second third in 9 h, and reach the terminal part of the colon (the sigmoid colon) in 12 h. From the sigmoid colon to the anus, transport is much slower (Clin-ical Box 28–5). When small colored beads are fed with a meal, an average of 70% of them are recovered in the stool in 72 h, but total recovery requires more than a week. Transit time, pressure fluctuations, and changes in pH in the gastrointesti-nal tract can be observed by monitoring the progress of a small pill that contains sensors and a miniature radio transmitter. DEFECATION Distention of the rectum with feces initiates reflex contrac-tions of its musculature and the desire to defecate. In humans, the sympathetic nerve supply to the internal (involuntary) anal sphincter is excitatory, whereas the parasympathetic sup-ply is inhibitory. This sphincter relaxes when the rectum is distended. The nerve supply to the external anal sphincter, a skeletal muscle, comes from the pudendal nerve. The sphinc-ter is maintained in a state of tonic contraction, and moderate distention of the rectum increases the force of its contraction (Figure 28–8). The urge to defecate first occurs when rectal pressure increases to about 18 mm Hg. When this pressure reaches 55 mm Hg, the external as well as the internal sphinc-ter relaxes and there is reflex expulsion of the contents of the rectum. This is why reflex evacuation of the rectum can occur even in the setting of spinal injury. Before the pressure that relaxes the external anal sphincter is reached, voluntary defecation can be initiated by straining. Normally, the angle between the anus and the rectum is approximately 90 degrees (Figure 28–9), and this plus con-traction of the puborectalis muscle inhibit defecation. With straining, the abdominal muscles contract, the pelvic floor is lowered 1 to 3 cm, and the puborectalis muscle relaxes. The anorectal angle is reduced to 15 degrees or less. This is com-bined with relaxation of the external anal sphincter and FIGURE 28–7 The human colon. Ascending colon Descending colon Ileum Tenia coli Haustra Sigmoid colon Internal External anal sphincter Rectum Appendix Cecum Hepatic flexure Transverse colon Splenic flexure CLINICAL BOX 28–4 Hirschsprung Disease Some children present with a genetically determined con-dition of abnormal colonic motility known as Hirschsprung disease or aganglionic megacolon, which is characterized by abdominal distention, anorexia, and lassitude. The dis-ease is typically diagnosed in infancy, and affects as many as 1 in 5000 live births. It is due to a congenital absence of the ganglion cells in both the myenteric and submucous plexuses of a segment of the distal colon, as a result of fail-ure of the normal cranial-to-caudal migration of neural crest cells during development. The action of endothelins on the endothelin B receptor (see Chapter 7) are necessary for normal migration of certain neural crest cells, and knockout mice lacking endothelin B receptors developed megacolon. In addition, one cause of congenital agangli-onic megacolon in humans appears to be a mutation in the endothelin B receptor gene. The absence of peristalsis in patients with this disorder causes feces to pass the agangli-onic region with difficulty, and children with the disease may defecate as infrequently as once every 3 wk. The symptoms can be relieved completely if the aganglionic portion of the colon is resected and the portion of the colon above it anastomosed to the rectum. CHAPTER 28 Gastrointestinal Motility 477 defecation occurs. Defecation is therefore a spinal reflex that can be voluntarily inhibited by keeping the external sphincter contracted or facilitated by relaxing the sphincter and con-tracting the abdominal muscles. Distention of the stomach by food initiates contractions of the rectum and, frequently, a desire to defecate. The response is called the gastrocolic reflex, and may be amplified by an action of gastrin on the colon. Because of the CLINICAL BOX 28–5 Constipation Constipation refers to a pathological decrease in bowel movements. It was previously considered to reflect changes in motility, but the recent success of a drug designed to en-hance chloride secretion for the treatment of chronic consti-pation suggests alterations in the balance between secretion and absorption in the colon could also contribute to symp-tom generation. Patients with persistent constipation, and particularly those with a recent change in bowel habits, should be examined carefully to rule out underlying organic disease. However, many normal humans defecate only once every 2–3 d, even though others defecate once a day and some as often as three times a day. Furthermore, the only symptoms caused by constipation are slight anorexia and mild abdominal discomfort and distention. These symptoms are not due to absorption of “toxic substances,” because they are promptly relieved by evacuating the rectum and can be reproduced by distending the rectum with inert ma-terial. In western societies, the amount of misinformation and undue apprehension about constipation probably ex-ceeds that about any other health topic. Symptoms other than those described above that are attributed by the lay public to constipation are due to anxiety or other causes. FIGURE 28–8 Responses to distention of the rectum by pressures less than 55 mm Hg. Distention produces passive tension due to stretching of the wall of the rectum, and additional active ten-sion when the smooth muscle in the wall contracts. (Reproduced with permission from Davenport HW: A Digest of Digestion, 2nd ed. Year Book, 1978.) Pressures Distention of rectum Active Passive Some More Still more Rectum Internal anal sphincter External anal sphincter FIGURE 28–9 Sagittal view of the anorectal area at rest (above) and during straining (below). Note the reduction of the anorectal an-gle and lowering of the pelvic floor during straining. (Modified and reproduced with permission from Lembo A, Camilleri, M: Chronic constipation. N Engl J Med 2003;349:1360.) A AT REST Pubis Puborectalis External anal sphincter Internal anal sphincter Anorectal angle Coccyx B DURING STRAINING Anorectal angle Descent of the pelvic floor 478 SECTION V Gastrointestinal Physiology response, defecation after meals is the rule in children. In adults, habit and cultural factors play a large role in determin-ing when defecation occurs. CHAPTER SUMMARY ■The regulatory factors that govern gastrointestinal secretion also regulate its motility to soften the food, mix it with secre-tions, and propel it along the length of the tract. ■Two major patterns of motility are peristalsis and segmentation, which serve to propel or retard/mix the luminal contents, re-spectively. Peristalsis involves coordinated contractions and relaxations above and below the food bolus. ■The membrane potential of the majority of gastrointestinal smooth muscle undergoes rhythmic fluctuations that sweep along the length of the gut. The rhythm varies in different gut segments and is established by pacemaker cells known as inter-stitial cells of Cajal. This basic electrical rhythm provides for sites of muscle contraction when stimuli superimpose spike potentials on the depolarizing portion of the BER waves. ■In the period between meals, the intestine is relatively quiescent, but every 90 min or so it is swept through by a large peristaltic wave triggered by the hormone motilin. This migrating motor complex presumably serves a “housekeeping” function. ■Swallowing is triggered centrally and is coordinated with a peri-staltic wave along the length of the esophagus that drives the food bolus to the stomach, even against gravity. Relaxation of the lower esophageal sphincter is timed to just precede the arriv-al of the bolus, thereby limiting reflux of the gastric contents. Nevertheless, gastroesophageal reflux disease is one of the most common gastrointestinal complaints ■The stomach accommodates the meal by a process of receptive relaxation. This permits an increase in volume without a signif-icant increase in pressure. The stomach then serves to mix the meal and to control its delivery to downstream segments. ■Luminal contents move slowly through the colon, which en-hances water recovery. Distension of the rectum causes reflex contraction of the internal anal sphincter and the desire to def-ecate. After toilet training, defecation can be delayed till a con-venient time via voluntary contraction of the external anal sphincter. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. In infants, defecation often follows a meal. The cause of colonic contractions in this situation is A) histamine. B) increased circulating levels of CCK. C) the gastrocolic reflex. D) increased circulating levels of somatostatin. E) the enterogastric reflex. 2. The symptoms of the dumping syndrome (discomfort after meals in patients with intestinal short circuits such as anastomo-sis of the jejunum to the stomach) are caused in part by A) increased blood pressure. B) increased secretion of glucagon. C) increased secretion of CCK. D) hypoglycemia. E) hyperglycemia. 3. Gastric pressures seldom rise above the levels that breach the lower esophageal sphincter, even when the stomach is filled with a meal, due to which of the following processes? A) peristalsis B) gastroileal reflex C) segmentation D) stimulation of the vomiting center E) receptive relaxation 4. The migrating motor complex is triggered by which of the following? A) motilin B) NO C) CCK D) somatostatin E) secretin 5. A patient with achalasia would be expected to exhibit a decrease in which of the following? A) esophageal peristalsis B) expression of neuronal NO synthase at the esophageal/ gastric junction C) acetylcholine receptors D) substance P release E) contraction of the crural diaphragm CHAPTER RESOURCES Barrett KE: Gastrointestinal Physiology. McGraw-Hill, 2006. Cohen S, Parkman HP: Heartburn—A serious symptom. N Engl J Med 1999;340:878. Itoh Z: Motilin and clinical application. Peptides 1997;18:593. Lembo A, Camilleri M: Chronic constipation. N Engl J Med 2003;349:1360. Levitt MD, Bond JH: Volume, composition and source of intestinal gas. Gastroenterology 1970;59:921. Mayer EA, Sun XP, Willenbucher RF: Contraction coupling in colonic smooth muscle. Annu Rev Physiol 1992;54:395. Mittal RK, Balaban DH: The esophagogastric junction. N Engl J Med 1997;336:924. Sanders KM, Warm SM: Nitric oxide as a mediator of noncholinergic neurotransmission. Am J Physiol 1992;262:G379. 479 C H A P T E R 29 Transport & Metabolic Functions of the Liver O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the major functions of the liver with respect to metabolism, detoxifica-tion, and excretion of hydrophobic substances. ■Understand the functional anatomy of the liver and the relative arrangements of hepatocytes, cholangiocytes, endothelial cells, and Kupffer cells. ■Define the characteristics of the hepatic circulation and its role in subserving the liver’s functions. ■Identify the plasma proteins that are synthesized by the liver. ■Describe the formation of bile, its constituents, and its role in the excretion of cholesterol and bilirubin. ■Outline the mechanisms by which the liver contributes to whole body ammonia homeostasis and the consequences of the failure of these mechanisms, particular-ly for brain function. ■Identify the mechanisms that permit normal functioning of the gallbladder and the basis of gallstone disease. INTRODUCTION The liver is the largest gland in the body. It is essential for life because it conducts a vast array of biochemical and metabolic functions, including ridding the body of substances that would otherwise be injurious if allowed to accumulate, and excreting drug metabolites. It is also the first port of call for most nutrients absorbed across the gut wall, supplies most of the plasma proteins, and synthesizes the bile that optimizes the absorption of fats as well as serving as an excretory fluid. The liver and associated biliary system have therefore evolved an array of structural and physiologic features that underpin this broad range of critical functions. THE LIVER FUNCTIONAL ANATOMY An important function of the liver is to serve as a filter be-tween the blood coming from the gastrointestinal tract and the blood in the rest of the body. Blood from the intestines and other viscera reach the liver via the portal vein. This blood per-colates in sinusoids between plates of hepatic cells and eventu-ally drains into the hepatic veins, which enter the inferior vena cava. During its passage through the hepatic plates, it is exten-sively modified chemically. Bile is formed on the other side at each plate. The bile passes to the intestine via the hepatic duct (Figure 29–1). In each hepatic lobule, the plates of hepatic cells are usually only one cell thick. Large gaps occur between the endothelial cells, and plasma is in intimate contact with the cells (Figure 29–2). Hepatic artery blood also enters the sinusoids. The central veins coalesce to form the hepatic veins, which drain into the inferior vena cava. The average transit time for blood 480 SECTION V Gastrointestinal Physiology across the liver lobule from the portal venule to the central hepatic vein is about 8.4 s. Additional details of the features of the hepatic micro- and macrocirculation, which are critical to organ function, are provided below. Numerous macrophages (Kupffer cells) are anchored to the endothelium of the sinu-soids and project into the lumen. The functions of these phagocytic cells are discussed in Chapter 3. Each liver cell is also apposed to several bile canaliculi (Fig-ure 29–2). The canaliculi drain into intralobular bile ducts, and these coalesce via interlobular bile ducts to form the right and left hepatic ducts. These ducts join outside the liver to form the common hepatic duct. The cystic duct drains the gallbladder. The hepatic duct unites with the cystic duct to form the common bile duct (Figure 29–1). The common bile duct enters the duodenum at the duodenal papilla. Its orifice is surrounded by the sphincter of Oddi, and it usually unites with the main pancreatic duct just before entering the duodenum. The sphincter is usually closed, but when the gastric contents enter the duodenum, cholecystokinin (CCK) is released and the gastrointestinal hormone relaxes the sphincter and makes the gallbladder contract. The walls of the extrahepatic biliary ducts and the gallblad-der contain fibrous tissue and smooth muscle. They are lined by a layer of columnar cells with scattered mucous glands. In the gallbladder, the surface is extensively folded; this increases its surface area and gives the interior of the gallbladder a honey-combed appearance. The cystic duct is also folded to form the so-called spiral valves. This arrangement is believed to increase the turbulence of bile as it flows out of the gallbladder, thereby reducing the risk that it will precipitate and form gallstones. HEPATIC CIRCULATION Large gaps occur between endothelial cells in the walls of he-patic sinusoids, and the sinusoids are highly permeable. The way the intrahepatic branches of the hepatic artery and portal vein converge on the sinusoids and drain into the central lob-ular veins of the liver is shown in Figure 29–1. The functional unit of the liver is the acinus. Each acinus is at the end of a vas-cular stalk containing terminal branches of portal veins, he-patic arteries, and bile ducts. Blood flows from the center of FIGURE 29–1 Top: Organization of the liver. CV, central vein. PS, portal space containing branches of bile duct (green), portal vein (blue), and hepatic artery (red). Bottom: Arrangement of plates of liv-er cells, sinusoids, and bile ducts in a liver lobule, showing centripetal flow of blood in sinusoids to central vein and centrifugal flow of bile in bile canaliculi to bile ducts. (Reproduced with permission from Fawcett DW: Bloom and Fawcett, A Textbook of Histology, 11th ed. Saunders, 1986.) Central vein Sinusoids Bile canaliculi Bile duct Branch of the hepatic artery Branch of the portal vein CV CV CV PS PS PS PS PS PS PS PS PS PS Liver lobule FIGURE 29–2 Hepatocyte. Note the relation of the cell to bile canaliculi and sinusoids. Note also the wide openings between the en-dothelial cells next to the hepatocyte. (Reproduced with permission from Fawcett DW: Bloom and Fawcett, A Textbook of Histology, 11th ed. Saunders, 1986.) Kupffer cell Hepatic sinusoid Space of Disse Lipoprotein Agranular reticulum Bile canaliculus Golgi complex Lysosome Granular reticulum Lysosomes Microbody Mitochondrion Golgi complex CHAPTER 29 Transport & Metabolic Functions of the Liver 481 this functional unit to the terminal branches of the hepatic veins at the periphery (Figure 29–3). This is why the central portion of the acinus, sometimes called zone 1, is well oxygen-ated, the intermediate zone (zone 2) is moderately well oxy-genated, and the peripheral zone (zone 3) is least well oxygenated and most susceptible to anoxic injury. The hepatic veins drain into the inferior vena cava. The acini have been lik-ened to grapes or berries, each on a vascular stem. The human liver contains about 100,000 acini. Portal venous pressure is normally about 10 mm Hg in humans, and hepatic venous pressure is approximately 5 mm Hg. The mean pressure in the hepatic artery branches that converge on the sinusoids is about 90 mm Hg, but the pres-sure in the sinusoids is lower than the portal venous pressure, so a marked pressure drop occurs along the hepatic arterioles. This pressure drop is adjusted so that there is an inverse rela-tionship between hepatic arterial and portal venous blood flow. This inverse relationship may be maintained in part by the rate at which adenosine is removed from the region around the arterioles. According to this hypothesis, adenosine is produced by metabolism at a constant rate. When portal flow is reduced, it is washed away more slowly, and the local accumulation of adenosine dilates the terminal arterioles. In the period between meals, moreover, many of the sinusoids are collapsed. Following a meal, on the other hand, when por-tal flow to the liver from the intestine increases considerably, these “reserve” sinusoids are recruited. This arrangement means that portal pressures do not increase linearly with por-tal flow until all sinusoids have been recruited. This may be important to prevent fluid loss from the highly permeable liver under normal conditions. Indeed, if hepatic pressures are increased in disease states (such as the hardening of the liver that is seen in cirrhosis), many liters of fluid can accumulate in the peritoneal cavity as ascites. The intrahepatic portal vein radicles have smooth muscle in their walls that is innervated by noradrenergic vasoconstric-tor nerve fibers reaching the liver via the third to eleventh thoracic ventral roots and the splanchnic nerves. The vaso-constrictor innervation of the hepatic artery comes from the hepatic sympathetic plexus. No known vasodilator fibers reach the liver. When systemic venous pressure rises, the por-tal vein radicles are dilated passively and the amount of blood in the liver increases. In congestive heart failure, this hepatic venous congestion may be extreme. Conversely, when diffuse noradrenergic discharge occurs in response to a drop in sys-temic blood pressure, the intrahepatic portal radicles con-strict, portal pressure rises, and blood flow through the liver is brisk, bypassing most of the organ. Most of the blood in the liver enters the systemic circulation. Constriction of the hepatic arterioles diverts blood from the liver, and constric-tion of the mesenteric arterioles reduces portal inflow. In severe shock, hepatic blood flow may be reduced to such a degree that patchy necrosis of the liver takes place. FUNCTIONS OF THE LIVER The liver has many complex functions that are summarized in Table 29–1. Several will be touched upon briefly here. METABOLISM & DETOXIFICATION It is beyond the scope of this volume to touch upon all of the metabolic functions of the liver. Instead, we will describe here those aspects most closely aligned to gastrointestinal physiol-ogy. First, the liver plays key roles in carbohydrate metabo-lism, including glycogen storage, conversion of galactose and fructose to glucose, and gluconeogenesis, as well as many of the reactions covered in Chapter 1. The substrates for these re-actions derive from the products of carbohydrate digestion and absorption that are transported from the intestine to the liver in the portal blood. The liver also plays a major role in maintaining the stability of blood glucose levels in the post-prandial period, removing excess glucose from the blood and returning it as needed—the so-called glucose buffer function of the liver. In liver failure, hypoglycemia is commonly seen. Similarly, the liver contributes to fat metabolism. It supports a high rate of fatty acid oxidation for energy supply to the liver itself and other organs. Amino acids and two carbon frag-ments derived from carbohydrates are also converted in the liver to fats for storage. The liver also synthesizes most of the lipoproteins required by the body and preserves cholesterol homeostasis by synthesizing this molecule and also converting excess cholesterol to bile acids. The liver also detoxifies the blood of substances originating from the gut or elsewhere in the body (Clinical Box 29–1). Part of this function is physical in nature—bacteria and other FIGURE 29–3 Concept of the acinus as the functional unit of the liver. In each acinus, blood in the portal venule and hepatic arteri-ole enters the center of the acinus and flows outward to the hepatic venule. (Reproduced with permission from Lautt WW, Greenway CV: Conceptual review of the hepatic vascular bed. Hepatology 1987;7:952. Copyright © 1987 by The American Association for the Study of Liver Diseases.) Terminal hepatic arteriole Terminal hepatic venule Terminal portal venule Terminal bile duct Terminal hepatic venule 482 SECTION V Gastrointestinal Physiology particulates are trapped in and broken down by the strategically-located Kupffer cells. The remaining reactions are biochemical, and mediated in their first stages by the large number of cyto-chrome P450 enzymes expressed in hepatocytes. These convert xenobiotics and other toxins to inactive, less lipophilic metabo-lites. Detoxification reactions are divided into phase I (oxidation, hydroxylation, and other reactions mediated by cytochrome P450s) and phase II (esterification). Ultimately, metabolites are secreted into the bile for elimination via the gastrointestinal tract. In this regard, in addition to disposing of drugs, the liver is responsible for metabolism of essentially all steroid hormones. Liver disease can therefore result in the apparent overactivity of the relevant hormone systems. SYNTHESIS OF PLASMA PROTEINS The principal proteins synthesized by the liver are listed in Ta-ble 29–1. Albumin is quantitatively the most significant, and accounts for the majority of plasma oncotic pressure. Many of the products are acute-phase proteins, proteins synthesized and secreted into the plasma on exposure to stressful stimuli (see Chapter 3). Others are proteins that transport steroids and other hormones in the plasma, and still others are clotting factors. Following blood loss, the liver replaces the plasma proteins in days to weeks. The only major class of plasma pro-teins not synthesized by the liver are the immunoglobulins. BILE Bile is made up of the bile acids, bile pigments, and other sub-stances dissolved in an alkaline electrolyte solution that re-sembles pancreatic juice (Table 29–2). About 500 mL is secreted per day. Some of the components of the bile are reab-sorbed in the intestine and then excreted again by the liver (enterohepatic circulation). In addition to its role in diges-tion and absorption of fats (Chapter 27), bile (and subsequent-ly the feces) is the major excretory route for lipid-soluble waste products. TABLE 29–1 Principal functions of the liver. Formation and secretion of bile Nutrient and vitamin metabolism Glucose and other sugars Amino acids Lipids Fatty acids Cholesterol Lipoproteins Fat-soluble vitamins Water-soluble vitamins Inactivation of various substances Toxins Steroids Other hormones Synthesis of plasma proteins Acute-phase proteins Albumin Clotting factors Steroid-binding and other hormone-binding proteins Immunity Kupffer cells CLINICAL BOX 29–1 Hepatic Encephalopathy The clinical importance of hepatic ammonia metabolism is seen in liver failure, when increased levels of circulating ammonia cause the condition of hepatic encephalopathy. Initially, patients may seem merely confused, but if un-treated, the condition can progress to coma and irrevers-ible changes in cognition. The disease results not only from the loss of functional hepatocytes, but also shunting of por-tal blood around the hardened liver, meaning that less am-monia is removed from the blood by the remaining hepatic mass. Additional substances that are normally detoxified by the liver likely also contribute to the mental status changes. The condition can be minimized by reducing the load of ammonia coming to the liver from the colon (eg, by feeding the nonabsorbable carbohydrate, lactulose, which is converted into short-chain fatty acids in the colonic lumen and thereby traps luminal ammonia in its ionized form). However, in severe disease, the only truly effective treatment is to perform a liver transport, although the pau-city of available organs means that there is great interest in artificial liver assist devices that could clean the blood. TABLE 29–2 Comparison of human hepatic duct bile and gallbladder bile. Hepatic Duct Bile Gallbladder Bile Percentage of solids 2–4 10–12 Bile acids (mmol/L) 10–20 50–200 pH 7.8–8.6 7.0–7.4 CHAPTER 29 Transport & Metabolic Functions of the Liver 483 The glucuronides of the bile pigments, bilirubin and biliverdin, are responsible for the golden yellow color of bile. The formation of these breakdown products of hemoglobin is discussed in detail in Chapter 32, and their excretion is dis-cussed in the following text. BILIRUBIN METABOLISM & EXCRETION Most of the bilirubin in the body is formed in the tissues by the break down of hemoglobin (see Chapter 32 and Figure 29–4). The bilirubin is bound to albumin in the circulation. Some of it is tightly bound, but most of it can dissociate in the liver, and free bilirubin enters liver cells via a member of the organic anion transporting polypeptide (OATP) family, and then becomes bound to cytoplasmic proteins (Figure 29–5). It is next conju-gated to glucuronic acid in a reaction catalyzed by the enzyme glucuronyl transferase (UDP-glucuronosyltransferase). This enzyme is located primarily in the smooth endoplasmic reticu-lum. Each bilirubin molecule reacts with two uridine diphos-phoglucuronic acid (UDPG) molecules to form bilirubin diglucuronide. This glucuronide, which is more water-soluble than the free bilirubin, is then transported against a concentra-tion gradient most likely by an active transporter known as mul-tidrug resistance protein-2 (MRP-2) into the bile canaliculi. A small amount of the bilirubin glucuronide escapes into the blood, where it is bound less tightly to albumin than is free bili-rubin, and is excreted in the urine. Thus, the total plasma biliru-bin normally includes free bilirubin plus a small amount of conjugated bilirubin. Most of the bilirubin glucuronide passes via the bile ducts to the intestine. The intestinal mucosa is relatively impermeable to conju-gated bilirubin but is permeable to unconjugated bilirubin and to urobilinogens, a series of colorless derivatives of biliru-bin formed by the action of bacteria in the intestine. Conse-quently, some of the bile pigments and urobilinogens are reabsorbed in the portal circulation. Some of the reabsorbed substances are again excreted by the liver (enterohepatic cir-culation), but small amounts of urobilinogens enter the gen-eral circulation and are excreted in the urine. JAUNDICE When free or conjugated bilirubin accumulates in the blood, the skin, scleras, and mucous membranes turn yellow. This yellow-ness is known as jaundice (icterus) and is usually detectable when the total plasma bilirubin is greater than 2 mg/dL (34 μmol/L). Hyperbilirubinemia may be due to (1) excess production of bili-rubin (hemolytic anemia, etc; see Chapter 32), (2) decreased up-take of bilirubin into hepatic cells, (3) disturbed intracellular protein binding or conjugation, (4) disturbed secretion of con-jugated bilirubin into the bile canaliculi, or (5) intrahepatic or FIGURE 29–4 Conversion of heme to bilirubin is a two-step reaction catalyzed by heme oxygenase and biliverdin reductase. M, methyl; P, propionate; V, vinyl. N H N H N N H M V M P P M M V O O N H N H N H N H M V M P P M M V O O H H N N N N COO− COO− H3C CH3 CH3 H2C CH3 CH2 Fe Heme Heme oxygenase NADPH + O2 CO + Fe3+ + NADP+ Biliverdin reductase NADPH NADP+ Biliverdin Bilirubin FIGURE 29–5 Handling of bilirubin by hepatocytes. Albumin (Alb)-bound bilirubin (B) enters the space of Disse adjacent to the baso-lateral membrane of hepatocytes, and bilirubin is selectively transport-ed into the hepatocyte. Here, it is conjugated with glucuronic acid (G). The conjugates are secreted into bile via the multidrug resistance pro-tein 2 (MRP-2). Some unconjugated and conjugated bilirubin also reflux-es into the plasma. OATP, organic anion transporting polypeptide. UDP-G UDP B BG BG2 UDP glucuronyl transferase OATP Alb + B Alb B Reflux to plasma Hepatocyte Canaliculus Space of Disse BG BG2 MRP2 484 SECTION V Gastrointestinal Physiology extrahepatic bile duct obstruction. When it is due to one of the first three processes, the free bilirubin rises. When it is due to disturbed secretion of conjugated bilirubin or bile duct obstruc-tion, bilirubin glucuronide regurgitates into the blood, and it is predominantly the conjugated bilirubin in the plasma that is elevated. OTHER SUBSTANCES CONJUGATED BY GLUCURONYL TRANSFERASE The glucuronyl transferase system in the smooth endoplasmic reticulum catalyzes the formation of the glucuronides of a variety of substances in addition to bilirubin. As discussed above, the list includes steroids (see Chapter 22) and various drugs. These other compounds can compete with bilirubin for the enzyme system when they are present in appreciable amounts. In addition, sev-eral barbiturates, antihistamines, anticonvulsants, and other compounds cause marked proliferation of the smooth endoplas-mic reticulum in the hepatic cells, with a concurrent increase in hepatic glucuronyl transferase activity. Phenobarbital has been used successfully for the treatment of a congenital disease in which there is a relative deficiency of glucuronyl transferase (type 2 UDP-glucuronosyltransferase deficiency). OTHER SUBSTANCES EXCRETED IN THE BILE Cholesterol and alkaline phosphatase are excreted in the bile. In patients with jaundice due to intra- or extrahepatic obstruc-tion of the bile duct, the blood levels of these two substances usually rise. A much smaller rise is generally seen when the jaundice is due to nonobstructive hepatocellular disease. Adrenocortical and other steroid hormones and a number of drugs are excreted in the bile and subsequently reabsorbed (enterohepatic circulation). AMMONIA METABOLISM & EXCRETION The liver is critical for ammonia handling in the body. Ammo-nia levels must be carefully controlled because it is toxic to the central nervous system (CNS), and freely permeable across the blood–brain barrier. The liver is the only organ in which the complete urea cycle (also known as the Krebs-Henseleit cycle) is expressed (Figure 29–6). This converts circulating ammonia to urea, which can then be excreted in the urine (Figure 29–7). Ammonia in the circulation comes primarily from the colon and kidneys with lesser amounts deriving from the breakdown of red blood cells and from metabolism in the muscles. As it passes through the liver, the vast majority of FIGURE 29–6 The urea cycle, which converts ammonia to urea, takes place in the mitochondria and cytosol of hepatocytes. To circulation ATP HCO3 − NH4 + H2N C O O P O− O− O ADP Carbamoyl phosphate Mitochondrion H2N C O NH CH COO− (CH2)3 NH3 + Citrulline COO− NH2 + NH3 + OOC CH2 CH NH C NH (CH2)3 CH COO− Aspartate AMP 2 3 4 Arginine succinate H2N C NH (CH2)3 CH COO− NH2 + NH3 + Arginine H3N (CH2)2 CH COO− NH3 + + NH3 H2N C O NH2 Fumarate H2O Arginase 1 Carbamoyl synthetase 3 Arginine succinate lyase Hepatocyte Net reaction 2NH3 + CO2 = Urea + H2O 2 Arginosuccinate synthetase Cytosol Urea cycle Urea Ornithine 1 4 1 P CHAPTER 29 Transport & Metabolic Functions of the Liver 485 ammonia in the circulation is cleared into the hepatocytes. There, it is converted in the mitochondria to carbamoyl phos-phate, which in turn reacts with ornithine to generate citrul-line. A series of subsequent cytoplasmic reactions eventually produce arginine, and this can be dehydrated to urea and ornithine. The latter returns to the mitochondria to begin another cycle, and urea, as a small molecule, diffuses readily back out into the sinusoidal blood. It is then filtered in the kidneys and lost from the body in the urine. THE BILIARY SYSTEM BILE FORMATION Bile contains substances that are actively secreted into it across the canalicular membrane, such as bile acids, phosphatidyl-choline, conjugated bilirubin, cholesterol, and xenobiotics. Each of these enters the bile by means of a specific canalicular transporter. It is the active secretion of bile acids, however, that is believed to be the primary driving force for the initial formation of canalicular bile. Because they are osmotically ac-tive, the canalicular bile is transiently hypertonic. However, the tight junctions that join adjacent hepatocytes are relatively permeable and thus a number of additional substances pas-sively enter the bile from the plasma by diffusion. These sub-stances include water, glucose, calcium, glutathione, amino acids, and urea. Phosphatidylcholine that enters the bile forms mixed micelles with the bile acids and cholesterol. The ratio of bile acids:phosphatidylcholine:cholesterol in canalicular bile is approximately 10:3:1. Deviations from this ratio may cause cholesterol to precipitate, leading to one type of gallstones (Figure 29–8). The bile is then transferred to progressively larger bile ductules and ducts, where it undergoes modification of its composition. The bile ductules are lined by cholangiocytes, specialized columnar epithelial cells. Their tight junctions are less permeable than those of the hepatocytes, although they remain freely permeable to water and thus bile remains iso-tonic. The ductules scavenge plasma constituents, such as glu-cose and amino acids, and return them to the circulation by active transport. Glutathione is also hydrolyzed to its constit-uent amino acids by an enzyme, gamma glutamyltranspepti-dase (GGT), expressed on the apical membrane of the cholangiocytes. Removal of glucose and amino acids is likely important to prevent bacterial overgrowth of the bile, particu-larly during gallbladder storage (see below). The ductules also secrete bicarbonate in response to secretin in the postprandial period, as well as IgA and mucus for protection. FUNCTIONS OF THE GALLBLADDER In normal individuals, bile flows into the gallbladder when the sphincter of Oddi is closed (ie, the period in between meals). In the gallbladder, the bile is concentrated by absorption of water. The degree of this concentration is shown by the in-crease in the concentration of solids (Table 29–2); liver bile is 97% water, whereas the average water content of gallbladder bile is 89%. However, because the bile acids are a micellar FIGURE 29–7 Whole body ammonia homeostasis in health. The majority of ammonia produced by the body is excreted by the kidneys in the form of urea. P o rt a l ci r c u l a ti o n S y s t e m i c c ir c u la ti o n Urea NH3 NH4 + H+ Proteins + Amino acids Urinary excretion as urea Fecal excretion as ammonium ion Urea NH3 25% 75% 15% 85% Systemic circulation FIGURE 29–8 Cholesterol solubility in bile as a function of the proportions of lecithin, bile salts, and cholesterol. In bile that has a composition described by any point below line ABC (eg, point P), cholesterol is solely in micellar solution; points above line ABC de-scribe bile in which there are cholesterol crystals as well. (Reproduced with permission from Small DM: Gallstones. N Engl J Med 1968;279:588.) Percent lecithin Percent cholesterol 100 80 0 20 40 60 80 100 0 20 40 60 80 100 0 20 40 60 Micellar liquid Two or more phases (cholesterol crystals and micellar liquid) B A C Percent bile salt P 486 SECTION V Gastrointestinal Physiology solution, the micelles simply become larger, and since osmo-larity is a colligative property, bile remains isotonic. However, bile becomes slightly acidic as sodium ions are exchanged for protons (although the overall concentration of sodium ions rises with a concomitant loss of chloride and bicarbonate as the bile is concentrated). When the bile duct and cystic duct are clamped, the intra-biliary pressure rises to about 320 mm of bile in 30 min, and bile secretion stops. However, when the bile duct is clamped and the cystic duct is left open, water is reabsorbed in the gall-bladder, and the intrabiliary pressure rises only to about 100 mm of bile in several hours. REGULATION OF BILIARY SECRETION When food enters the mouth, the resistance of the sphincter of Oddi decreases under both neural and hormonal influences (Figure 29–9). Fatty acids and amino acids in the duodenum release CCK, which causes gallbladder contraction. The production of bile is increased by stimulation of the vagus nerves and by the hormone secretin, which increases the water and HCO3 – content of bile. Substances that increase the secretion of bile are known as choleretics. Bile acids them-selves are among the most important physiologic choleretics. EFFECTS OF CHOLECYSTECTOMY The periodic discharge of bile from the gallbladder aids diges-tion but is not essential for it. Cholecystectomized patients maintain good health and nutrition with a constant slow dis-charge of bile into the duodenum, although eventually the bile duct becomes somewhat dilated, and more bile tends to enter the duodenum after meals than at other times. Cholecystecto-mized patients can even tolerate fried foods, although they gen-erally must avoid foods that are particularly high in fat content. VISUALIZING THE GALLBLADDER Exploration of the right upper quadrant with an ultrasonic beam (ultrasonography) and computed tomography (CT) have become the most widely used methods for visualizing the gallbladder and detecting gallstones. A third method of diag-nosing gallbladder disease is nuclear cholescintigraphy. When administered intravenously, technetium-99m-labeled derivatives of iminodiacetic acid are excreted in the bile and provide excellent gamma camera images of the gallbladder and bile ducts. The response of the gallbladder to CCK can then be observed following intravenous administration of the hormone. The biliary tree can also be visualized by injecting contrast media from an endoscope channel maneuvered into the sphincter of Oddi, in a procedure known as endoscopic retrograde cholangiopancreatography (ERCP). It is even pos-sible to insert small instruments with which to remove gall-stone fragments that may be obstructing the flow of bile, the flow of pancreatic juice, or both (Clinical Box 29–2). FIGURE 29–9 Neurohumoral control of gallbladder contraction and biliary secretion. Dorsal vagal complex Vagal efferents ACh CCK Vagal afferent ACh and CCK cause smooth muscle contraction CCK CCK Sphincter of Oddi Gall-bladder V i a b l o o d s tr e a m Duodenum Nutrients NO VIP CLINICAL BOX 29–2 Gallstones Cholelithiasis, that is, the presence of gallstones, is a com-mon condition. Its incidence increases with age, so that in the United States, for example, 20% of the women and 5% of the men between the ages of 50 and 65 have gallstones. The stones are of two types: calcium bilirubinate stones and cholesterol stones. In the United States and Europe, 85% of the stones are cholesterol stones. Three factors ap-pear to be involved in the formation of cholesterol stones. One is bile stasis; stones form in the bile that is seques-trated in the gallbladder rather than the bile that is flowing in the bile ducts. A second is supersaturation of the bile with cholesterol. Cholesterol is very insoluble in bile, and it is maintained in solution in micelles only at certain concen-trations of bile salts and lecithin. At concentrations above line ABC in Figure 29–8, the bile is supersaturated and con-tains small crystals of cholesterol in addition to micelles. However, many normal individuals who do not develop gallstones also have supersaturated bile. The third factor is a mix of nucleation factors that favors formation of stones from the supersaturated bile. Outside the body, bile from patients with cholelithiasis forms stones in 2 to 3 d, whereas it takes more than 2 wk for stones to form in bile from normal individuals. The exact nature of the nucleation factors is unsettled, although glycoproteins in gallbladder mucus have been implicated. In addition, it is unsettled whether stones form as a result of excess production of components that favor nucleation or decreased production of antinucleation components that prevent stones from forming in normal individuals. CHAPTER 29 Transport & Metabolic Functions of the Liver 487 CHAPTER SUMMARY ■The liver conducts a huge number of metabolic reactions and serves to detoxify and dispose of many exogenous substances, as well as metabolites endogenous to the body that would be harm-ful if allowed to accumulate. ■The structure of the liver is such that it can filter large volumes of blood and remove even hydrophobic substances that are protein-bound. This function is provided for by a fenestrated endothelium. The liver also receives essentially all venous blood from the intestine prior to its delivery to the remainder of the body. ■The liver serves to buffer blood glucose, synthesize the majority of plasma proteins, contribute to lipid metabolism, and preserve cholesterol homeostasis. ■Bilirubin is an end product of heme metabolism that is glucu-ronidated by the hepatocyte to permit its excretion in bile. Bili-rubin and its metabolites impart color to the bile and stools. ■The liver removes ammonia from the blood and converts it to urea for excretion by the kidneys. An accumulation of ammonia as well as other toxins causes hepatic encephalopathy in the setting of liver failure. ■Bile contains substances actively secreted across the canalicular membrane by hepatocytes, and notably bile acids, phosphatidyl-choline, and cholesterol. The composition of bile is modified as it passes through the bile ducts and is stored in the gallbladder. Gallbladder contraction is regulated to coordinate bile availabil-ity with the timing of meals. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Removal of the entire colon would be expected to cause A) death. B) megaloblastic anemia. C) severe malnutrition. D) a decrease in the blood level of ammonia in patients with cirrhosis of the liver. E) decreased urinary urobilinogen. 2. After complete hepatectomy, a rise would be expected in the blood level of A) glucose. B) fibrinogen. C) 25-hydroxycholecalciferol. D) conjugated bilirubin. E) estrogens. 3. Which of the following cell types protects against sepsis secon-dary to translocation of intestinal bacteria? A) hepatic stellate cell B) cholangiocyte C) Kupffer cell D) hepatocyte E) gallbladder epithelial cell 4. P450s (CYPs) are found in many parts of the body. In which of the following do they not play an important role? A) bile acid formation B) carcinogenesis C) steroid hormone formation D) detoxification of drugs E) glycogen synthesis CHAPTER RESOURCES Ankoma-Sey V: Hepatic regeneration—Revising the myth of Prometheus. News Physiol Sci 1999;14:149. Arias JM, et al (editors): The Liver: Biology and Pathology, 3rd ed. Raven Press, 1994. Chong L, Marx J (editors): Lipids in the limelight. Science 2001;294:1861. Hofmann AF: Bile acids: The good, the bad, and the ugly. News Physiol Sci 1999;14:24. Lee WM: Drug-induced hepatoxicity. N Engl J Med 2003;349:474. Meier PJ, Stieger B: Molecular mechanisms of bile formation. News Physiol Sci 2000;15:89. Michalopoulos GK, DeFrances MC: Liver regeneration. Science 1997;276:60. Trauner M, Meier PJ, Boyer JL: Molecular mechanisms of cholestasis. N Engl J Med 1998;339:1217. This page intentionally left blank 489 C H A P T E R SECTION VI CARDIOVASCULAR PHYSIOLOGY 30 Origin of the Heartbeat & the Electrical Activity of the Heart O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the structure and function of the conduction system of the heart and compare the action potentials in each part. ■Describe the way the electrocardiogram (ECG) is recorded, the waves of the ECG, and the relationship of the ECG to the electrical axis of the heart. ■Name the common cardiac arrhythmias and describe the processes that produce them. ■List the principal early and late ECG manifestations of myocardial infarction and ex-plain the early changes in terms of the underlying ionic events that produce them. ■Describe the ECG changes and the changes in cardiac function produced by alter-ations in the ionic composition of the body fluids. INTRODUCTION The parts of the heart normally beat in orderly sequence: Con-traction of the atria (atrial systole) is followed by contraction of the ventricles (ventricular systole), and during diastole all four chambers are relaxed. The heartbeat originates in a specialized cardiac conduction system and spreads via this system to all parts of the myocardium. The structures that make up the con-duction system (Figure 30–1) are the sinoatrial node (SA node), the internodal atrial pathways, the atrioventricular node (AV node), the bundle of His and its branches, and the Purkinje system. The various parts of the conduction system and, under abnormal conditions, parts of the myocardium, are capable of spontaneous discharge. However, the SA node nor-mally discharges most rapidly, with depolarization spreading from it to the other regions before they discharge spontane-ously. The SA node is therefore the normal cardiac pacemaker, with its rate of discharge determining the rate at which the heart beats. Impulses generated in the SA node pass through the atrial pathways to the AV node, through this node to the bundle of His, and through the branches of the bundle of His via the Purkinje system to the ventricular muscle. 490 SECTION VI Cardiovascular Physiology ORIGIN & SPREAD OF CARDIAC EXCITATION ANATOMIC CONSIDERATIONS In the human heart, the SA node is located at the junction of the superior vena cava with the right atrium. The AV node is located in the right posterior portion of the interatrial septum (Figure 30–1). There are three bundles of atrial fibers that con-tain Purkinje-type fibers and connect the SA node to the AV node: the anterior internodal tract of Bachman, the middle in-ternodal tract of Wenckebach, and the posterior internodal tract of Thorel. Conduction also occurs through atrial myo-cytes, but it is more rapid in these bundles. The AV node is continuous with the bundle of His, which gives off a left bun-dle branch at the top of the interventricular septum and con-tinues as the right bundle branch. The left bundle branch divides into an anterior fascicle and a posterior fascicle. The branches and fascicles run subendocardially down either side of the septum and come into contact with the Purkinje system, whose fibers spread to all parts of the ventricular myocardium. The histology of cardiac muscle is described in Chapter 5. The conduction system is composed, for the most part, of modified cardiac muscle that has fewer striations and indis-tinct boundaries. The SA node and, to a lesser extent, the AV node also contain small round cells with few organelles, which are connected by gap junctions. These are probably the actual pacemaker cells, and therefore they are called P cells. The atrial muscle fibers are separated from those of the ven-tricles by a fibrous tissue ring, and normally the only conduct-ing tissue between the atria and ventricles is the bundle of His. The SA node develops from structures on the right side of the embryo and the AV node from structures on the left. This is why in the adult the right vagus is distributed mainly to the SA node and the left vagus mainly to the AV node. Similarly, the sympathetic innervation on the right side is distributed primarily to the SA node and the sympathetic innervation on the left side primarily to the AV node. On each side, most sym-pathetic fibers come from the stellate ganglion. Noradrenergic fibers are epicardial, whereas the vagal fibers are endocardial. However, connections exist for reciprocal inhibitory effects of the sympathetic and parasympathetic innervation of the heart on each other. Thus, acetylcholine acts presynaptically to reduce norepinephrine release from the sympathetic nerves, and conversely, neuropeptide Y released from noradrenergic endings may inhibit the release of acetylcholine. PROPERTIES OF CARDIAC MUSCLE The electrical responses of cardiac muscle and nodal tissue and the ionic fluxes that underlie them are discussed in detail in Chapter 5 and are briefly reviewed here for comparison with the pacemaker cells below. Myocardial fibers have a resting membrane potential of approximately –90 mV (Figure 30–2A). The individual fibers are separated by membranes, but depo-larization spreads radially through them as if they were a syn-cytium because of the presence of gap junctions. The transmembrane action potential of single cardiac muscle cells FIGURE 30–1 Conducting system of the heart. Left: Anatomical depiction of the human heart with additional focus on areas of the con-duction system. Right: Typical transmembrane action potentials for the SA and AV nodes, other parts of the conduction system, and the atrial and ventricular muscles are shown along with the correlation to the extracellularly recorded electrical activity, that is, the electrocardiogram (ECG). The action potentials and ECG are plotted on the same time axis but with different zero points on the vertical scale. LAF, left anterior fascicle. 0.2 0.4 0.6 Q R S P ECG T U Action potential Time (s) Aorta Superior vena cava Sinoatrial node Internodal pathways Atrioventricular node Bundle of His Right bundle branch Purkinje system Left posterior fascicle Ventricular muscle Purkinje fibers Bundle branches Common bundle AV node Atrial muscle SA node LAF CHAPTER 30 Origin of the Heartbeat & the Electrical Activity of the Heart 491 is characterized by rapid depolarization (phase 0), an initial rapid repolarization (phase 1), a plateau (phase 2), and a slow repolarization process (phase 3) that allows return to the rest-ing membrane potential (phase 4). The initial depolarization is due to Na+ influx through rapidly opening Na+ channels (the Na+ current, INa). The inactivation of Na+ channels contrib-utes to the rapid repolarization phase. Ca2+ influx through more slowly opening Ca2+ channels (the Ca2+ current, ICa) produces the plateau phase, and repolarization is due to net K+ efflux through multiple types of K+ channels. Recorded extra-cellularly, the summed electrical activity of all the cardiac mus-cle fibers is the electrocardiogram (ECG). The timing of the discharge of the individual units relative to the ECG is shown in Figure 30–1. PACEMAKER POTENTIALS Rhythmically discharging cells have a membrane potential that, after each impulse, declines to the firing level. Thus, this prepotential or pacemaker potential (Figure 30–2B) triggers the next impulse. At the peak of each impulse, IK begins and brings about repolarization. IK then declines, and a channel that can pass both Na+ and K+ is activated. Because this chan-nel is activated following hyperpolarization, it is referred to as an “h” channel; however, because of its unusual (funny) acti-vation this has also been dubbed an “f” channel. As Ih increas-es, the membrane begins to depolarize, forming the first part of the prepotential. Ca2+ channels then open. These are of two types in the heart, the T (for transient) channels and the L (for long-lasting) channels. The calcium current (ICa) due to opening of T channels completes the prepotential, and ICa due to opening of L channels produces the impulse. Other ion channels are also involved, and there is evidence that local Ca2+ release from the sarcoplasmic reticulation (Ca2+ sparks) occurs during the prepotential. The action potentials in the SA and AV nodes are largely due to Ca2+, with no contribution by Na+ influx. Consequently, there is no sharp, rapid depolarizing spike before the plateau, as there is in other parts of the conduction system and the atrial and ventricular fibers. In addition, prepotentials are nor-mally prominent only in the SA and AV nodes. However, “latent pacemakers” are present in other portions of the con-duction system that can take over when the SA and AV nodes are depressed or conduction from them is blocked. Atrial and ventricular muscle fibers do not have prepotentials, and they discharge spontaneously only when injured or abnormal. When the cholinergic vagal fibers to nodal tissue are stimu-lated, the membrane becomes hyperpolarized and the slope of the prepotentials is decreased (Figure 30–3) because the ace-tylcholine released at the nerve endings increases the K+ con-ductance of nodal tissue. This action is mediated by M2 muscarinic receptors, which, via the βγ subunit of a G protein, open a special set of K+ channels. The resulting IKAch slows the depolarizing effect of Ih. In addition, activation of the M2 FIGURE 30–2 Comparison of action potentials in ventricular muscle and diagram of the membrane potential of pacemaker tissue. A) Phases of action potential in ventricular myocyte (0–4, see text for details) are superimposed with principal changes in current that contribute to changes in membrane potential. B) The principal current responsible for each part of the potential of pacemaker tissue is shown under or beside the component. L, long-lasting; T, transient. Other ion channels contribute to the electrical response. Note that the resting membrane potential of pacemaker tissue is somewhat lower than that of atrial and ventricular muscle. 0 +20 –90 0 0 A B –40 MV MV –60 ↑INA ↓IK ↓INA ↓ICa ↑ICA↑IK ↑IK ↑ICaL ↑ICaT ↑Ih ↓IK 4 3 2 1 FIGURE 30–3 Effect of sympathetic (noradrenergic) and vagal (cholinergic) and sympathetic (noradrenergic) stimulation on the membrane potential of the SA node. Note the reduced slope of the prepotential after vagal stimulation and the increased spontaneous discharge after sympathetic stimulation. Sympathetic stimulation Vagal stimulation 0 60 mV 0 60 mV 492 SECTION VI Cardiovascular Physiology receptors decreases cyclic adenosine 3',5'-monophosphate (cAMP) in the cells, and this slows the opening of the Ca2+ channels. The result is a decrease in firing rate. Strong vagal stimulation may abolish spontaneous discharge for some time. Conversely, stimulation of the sympathetic cardiac nerves speeds the depolarizating effect of Ih, and the rate of sponta-neous discharge increases (Figure 30–3). Norepinephrine secreted by the sympathetic endings binds to β1 receptors, and the resulting increase in intracellular cAMP facilitates the opening of L channels, increasing ICa and the rapidity of the depolarization phase of the impulse. The rate of discharge of the SA node and other nodal tissue is influenced by temperature and by drugs. The discharge fre-quency is increased when the temperature rises, and this may contribute to the tachycardia associated with fever. Digitalis depresses nodal tissue and exerts an effect like that of vagal stimulation, particularly on the AV node. SPREAD OF CARDIAC EXCITATION Depolarization initiated in the SA node spreads radially through the atria, then converges on the AV node. Atrial de-polarization is complete in about 0.1 s. Because conduction in the AV node is slow (Table 30–1), a delay of about 0.1 s (AV nodal delay) occurs before excitation spreads to the ventricles. It is interesting to note here that when there is a lack of contri-bution of INa in the depolarization (phase 0) of the action po-tential, a marked loss of conduction is observed. This delay is shortened by stimulation of the sympathetic nerves to the heart and lengthened by stimulation of the vagi. From the top of the septum, the wave of depolarization spreads in the rap-idly conducting Purkinje fibers to all parts of the ventricles in 0.08–0.1 s. In humans, depolarization of the ventricular mus-cle starts at the left side of the interventricular septum and moves first to the right across the mid portion of the septum. The wave of depolarization then spreads down the septum to the apex of the heart. It returns along the ventricular walls to the AV groove, proceeding from the endocardial to the epicar-dial surface (Figure 30–4). The last parts of the heart to be de-polarized are the posterobasal portion of the left ventricle, the pulmonary conus, and the uppermost portion of the septum. THE ELECTROCARDIOGRAM Because the body fluids are good conductors (ie, because the body is a volume conductor), fluctuations in potential that represent the algebraic sum of the action potentials of myocar-dial fibers can be recorded extracellularly. The record of these potential fluctuations during the cardiac cycle is the electro-cardiogram (ECG). The ECG may be recorded by using an active or exploring electrode connected to an indifferent electrode at zero poten-tial (unipolar recording) or by using two active electrodes (bipolar recording). In a volume conductor, the sum of the potentials at the points of an equilateral triangle with a current source in the center is zero at all times. A triangle with the heart at its center (Einthoven’s triangle) can be approximated by placing electrodes on both arms and on the left leg. These are the three standard limb leads used in electrocardiography. If these electrodes are connected to a common terminal, an indifferent electrode that stays near zero potential is obtained. Depolarization moving toward an active electrode in a volume conductor produces a positive deflection, whereas depolariza-tion moving in the opposite direction produces a negative deflection. The names of the various waves and segments of the ECG in humans are shown in Figure 30–5. By convention, an upward deflection is written when the active electrode becomes posi-tive relative to the indifferent electrode, and a downward deflection is written when the active electrode becomes nega-tive. The P wave is produced by atrial depolarization, the QRS complex by ventricular depolarization, and the T wave by ven-tricular repolarization. The U wave is an inconstant finding, believed to be due to slow repolarization of the papillary mus-cles. The intervals between the various waves of the ECG and the events in the heart that occur during these intervals are shown in Table 30–2. BIPOLAR LEADS Bipolar leads were used before unipolar leads were developed. The standard limb leads—leads I, II, and III—each record the differences in potential between two limbs. Because current flows only in the body fluids, the records obtained are those that would be obtained if the electrodes were at the points of attachment of the limbs, no matter where on the limbs the electrodes are placed. In lead I, the electrodes are connected so that an upward deflection is inscribed when the left arm be-comes positive relative to the right (left arm positive). In lead II, the electrodes are on the right arm and left leg, with the leg positive; and in lead III, the electrodes are on the left arm and left leg, with the leg positive. UNIPOLAR (V) LEADS An additional nine unipolar leads, that is, leads that record the potential difference between an exploring electrode and an TABLE 30–1 Conduction speeds in cardiac tissue. Tissue Conduction Rate (m/s) SA node 0.05 Atrial pathways 1 AV node 0.05 Bundle of His 1 Purkinje system 4 Ventricular muscle 1 CHAPTER 30 Origin of the Heartbeat & the Electrical Activity of the Heart 493 FIGURE 30–4 Normal spread of electrical activity in the heart. A) Conducting system of the heart. B) Sequence of cardiac excitation. Top: Anatomical position of electrical activity. Bottom: corresponding electrocardiogram. The yellow color denotes areas that are depolarized. (Reproduced with permission from Goldman MJ: Principles of Clinical Electrocardiography, 12th ed. Originally published by Appleton & Lange. Copyright © 1986 by McGraw-Hill.) Electrocardiogram Time Time Time Time Time Superior vena cava Sinoatrial node Right bundle branch Purkinje fibers Right ventricle Right atrium Left ventricle Interventricular septum Left atrium Inferior vena cava Atrioventricular node Bundle of His Left bundle branch Atrial excitation Begins SA node AV node Atrial relaxation Complete Begins Complete Ventricular excitation Ventricular relaxation A B 494 SECTION VI Cardiovascular Physiology indifferent electrode, are commonly used in clinical electro-cardiography. There are six unipolar chest leads (precordial leads) designated V1–V6 (Figure 30–6) and three unipolar limb leads: VR (right arm), VL (left arm), and VF (left foot). Augmented limb leads, designated by the letter a (aVR, aVL, aVF), are generally used. The augmented limb leads are re-cordings between one limb and the other two limbs. This in-creases the size of the potentials by 50% without any change in configuration from the nonaugmented record. Unipolar leads can also be placed at the tips of catheters and inserted into the esophagus or heart. NORMAL ECG The ECG of a normal individual is shown in Figure 30–7. The sequence in which the parts of the heart are depolarized (Figure 30–4) and the position of the heart relative to the electrodes are the important considerations in interpreting the configurations of the waves in each lead. The atria are located posteriorly in the chest. The ventricles form the base and anterior surface of the heart, and the right ventricle is anterolateral to the left. Thus, aVR “looks at” the cavities of the ventricles. Atrial depolariza-tion, ventricular depolarization, and ventricular repolarization move away from the exploring electrode, and the P wave, QRS complex, and T wave are therefore all negative (downward) de-flections; aVL and aVF look at the ventricles, and the deflections are therefore predominantly positive or biphasic. There is no Q wave in V1 and V2, and the initial portion of the QRS complex is a small upward deflection because ventricular depolarization first moves across the midportion of the septum from left to right toward the exploring electrode. The wave of excitation then moves down the septum and into the left ventricle away from the electrode, producing a large S wave. Finally, it moves back along the ventricular wall toward the electrode, producing the return to the isoelectric line. Conversely, in the left ventric-ular leads (V4–V6) there may be an initial small Q wave (left to right septal depolarization), and there is a large R wave (septal and left ventricular depolarization) followed in V4 and V5 by a moderate S wave (late depolarization of the ventricular walls moving back toward the AV junction). There is considerable variation in the position of the nor-mal heart, and the position affects the configuration of the electrocardiographic complexes in the various leads. BIPOLAR LIMB LEADS & THE CARDIAC VECTOR Because the standard limb leads are records of the potential dif-ferences between two points, the deflection in each lead at any instant indicates the magnitude and direction in the axis of the lead of the electromotive force generated in the heart (cardiac FIGURE 30–5 Waves of the ECG. TABLE 30–2 ECG intervals. Normal Durations Intervals Average Range Events in the Heart during Interval PR intervala 0.18b 0.12–0.20 Atrial depolarization and conduction through AV node QRS duration 0.08 to 0.10 Ventricular depolarization and atrial repolarization QT interval 0.40 to 0.43 Ventricular depolariza-tion plus ventricular re-polarization ST interval (QT minus QRS) 0.32 . . . Ventricular repolariza-tion (during T wave) aMeasured from the beginning of the P wave to the beginning of the QRS complex. bShortens as heart rate increases from average of 0.18 s at a rate of 70 beats/min to 0.14 s at a rate of 130 beats/min. 1.0 0.5 −0.5 0 0 0.2 0.4 0.6 P R S Q T U PR segment PR interval QT interval ST segment QRS duration mV Isoelectric line Time (s) FIGURE 30–6 Unipolar electrocardiographic leads. aVR aVF aVL V1 V2 V3 V5 V6 V4 CHAPTER 30 Origin of the Heartbeat & the Electrical Activity of the Heart 495 vector or axis). The vector at any given moment in the two di-mensions of the frontal plane can be calculated from any two standard limb leads (Figure 30–8) if it is assumed that the three electrode locations form the points of an equilateral triangle (Einthoven’s triangle) and that the heart lies in the center of the triangle. These assumptions are not completely warranted, but calculated vectors are useful approximations. An approximate mean QRS vector (“electrical axis of the heart”) is often plotted by using the average QRS deflection in each lead, as shown in Figure 30–8. This is a mean vector as opposed to an instanta-neous vector, and the average QRS deflections should be mea-sured by integrating the QRS complexes. However, they can be approximated by measuring the net differences between the positive and negative peaks of the QRS. The normal direction of the mean QRS vector is generally said to be –30 to +110 de-grees on the coordinate system shown in Figure 30–8. Left or right axis deviation is said to be present if the calculated axis falls to the left of –30 degrees or to the right of +110 degrees, re-spectively. Right axis deviation suggests right ventricular hy-pertrophy, and left axis deviation may be due to left ventricular hypertrophy, but there are better and more reliable electrocar-diographic criteria for ventricular hypertrophy. VECTORCARDIOGRAPHY If the tops of the arrows representing all the instantaneous car-diac vectors in the frontal plane during the cardiac cycle are connected, from first to last, the line connecting them forms a FIGURE 30–7 Normal ECG. (Reproduced with permission from Goldman MJ: Principles of Clinical Electrocardiography, 12th ed. Originally published by Appleton & Lange. Copyright © 1986 by McGraw-Hill.) I aVR aVL II III aVF V1 V2 V3 V5 V6 V1 V2 V3 V4 V5 V6 V4 496 SECTION VI Cardiovascular Physiology series of three loops: one for the P wave, one for the QRS com-plex, and one for the T wave. This can be done electronically and the loops, called vectorcardiograms, are projected on the face of an oscilloscope. His BUNDLE ELECTROGRAM In patients with heart block, the electrical events in the AV node, bundle of His, and Purkinje system are frequently stud-ied with a catheter containing an electrode at its tip that is passed through a vein to the right side of the heart and manip-ulated into a position close to the tricuspid valve. Three or more standard electrocardiographic leads are recorded simul-taneously. The record of the electrical activity obtained with the catheter (Figure 30–9) is the His bundle electrogram (HBE). It normally shows an A deflection when the AV node is activated, an H spike during transmission through the His bundle, and a V deflection during ventricular depolarization. With the HBE and the standard electrocardiographic leads, it is possible to accurately time three intervals: (1) the PA inter-val, the time from the first appearance of atrial depolarization to the A wave in the HBE, which represents conduction time from the SA node to the AV node; (2) the AH interval, from the A wave to the start of the H spike, which represents the AV nodal conduction time; and (3) the HV interval, the time from the start of the H spike to the start of the QRS deflection in the ECG, which represents conduction in the bundle of His and the bundle branches. The approximate normal values for these intervals in adults are PA, 27 ms; AH, 92 ms; and HV, 43 ms. These values illustrate the relative slowness of conduction in the AV node (Table 30–1). MONITORING The ECG is often recorded continuously in hospital coronary care units, with alarms arranged to sound at the onset of life-threatening arrhythmias. Using a small portable tape recorder (Holter monitor), it is also possible to record the ECG in am-bulatory individuals as they go about their normal activities. The recording is later played back at high speed and analyzed. Long-term continuous records can be obtained. Recordings obtained with monitors have proved valuable in the diagnosis of arrhythmias and in planning the treatment of patients re-covering from myocardial infarctions. FIGURE 30–8 Cardiac vector. Left: Einthoven’s triangle. Perpendiculars dropped from the midpoints of the sides of the equilateral triangle intersect at the center of electrical activity. RA, right arm; LA, left arm; LL, left leg. Center: Calculation of mean QRS vector. In each lead, distances equal to the height of the R wave minus the height of the largest negative deflection in the QRS complex are measured off from the midpoint of the side of the triangle representing that lead. An arrow drawn from the center of electrical activity to the point of intersection of perpendiculars extended from the distances measured off on the sides represents the magnitude and direction of the mean QRS vector. Right: Reference axes for determining the direction of the vector. II III I 15 10 5 0 −5 +5 −0 +5mm +11 −1 +10mm +16 −1 +15mm mm − + + + − − −120∞ +120∞ 180∞ 0∞ +60∞ −60∞ − + − + + − Lead I Lead III Lead II RA LA LL FIGURE 30–9 Normal His bundle electrogram (HBE) with simultaneously recorded ECG. A H V ECG His bundle electrogram CHAPTER 30 Origin of the Heartbeat & the Electrical Activity of the Heart 497 CLINICAL APPLICATIONS: CARDIAC ARRHYTHMIAS NORMAL CARDIAC RATE In the normal human heart, each beat originates in the SA node (normal sinus rhythm, NSR). The heart beats about 70 times a minute at rest. The rate is slowed (bradycardia) during sleep and accelerated (tachycardia) by emotion, exercise, fever, and many other stimuli. In healthy young individuals breathing at a normal rate, the heart rate varies with the phases of respiration: It accelerates during inspiration and decelerates during expira-tion, especially if the depth of breathing is increased. This sinus arrhythmia (Figure 30–10) is a normal phenomenon and is due primarily to fluctuations in parasympathetic output to the heart. During inspiration, impulses in the vagi from the stretch receptors in the lungs inhibit the cardio-inhibitory area in the medulla oblongata. The tonic vagal discharge that keeps the heart rate slow decreases, and the heart rate rises. Disease processes affecting the sinus node lead to marked bradycardia accompanied by dizziness and syncope (sick sinus syndrome). ABNORMAL PACEMAKERS The AV node and other portions of the conduction system can, in abnormal situations, become the cardiac pacemaker. In addi-tion, diseased atrial and ventricular muscle fibers can have their membrane potentials reduced and discharge repetitively. As noted above, the discharge rate of the SA node is more rapid than that of the other parts of the conduction system, and this is why the SA node normally controls the heart rate. When conduction from the atria to the ventricles is completely inter-rupted, complete (third-degree) heart block results, and the ventricles beat at a low rate (idioventricular rhythm) indepen-dently of the atria (Figure 30–11). The block may be due to dis-ease in the AV node (AV nodal block) or in the conducting system below the node (infranodal block). In patients with AV nodal block, the remaining nodal tissue becomes the pace-maker and the rate of the idioventricular rhythm is approxi-mately 45 beats/min. In patients with infranodal block due to disease in the bundle of His, the ventricular pacemaker is located more peripherally in the conduction system and the ventricular rate is lower; it averages 35 beats/min, but in indi-vidual cases it can be as low as 15 beats/min. In such individu-als, there may also be periods of asystole lasting a minute or more. The resultant cerebral ischemia causes dizziness and fainting (Stokes–Adams syndrome). Causes of third-degree heart block include septal myocardial infarction and damage to the bundle of His during surgical correction of congenital inter-ventricular septal defects. When conduction between the atria and ventricles is slowed but not completely interrupted, incomplete heart block is present. In the form called first-degree heart block, all the atrial impulses reach the ventricles but the PR interval is abnor-mally long. In the form called second-degree heart block, not all atrial impulses are conducted to the ventricles. For example, a ventricular beat may follow every second or every third atrial beat (2:1 block, 3:1 block, etc). In another form of incomplete heart block, there are repeated sequences of beats in which the PR interval lengthens progressively until a ventricular beat is dropped (Wenckebach phenomenon). The PR interval of the cardiac cycle that follows each dropped beat is usually normal or only slightly prolonged (Figure 30–11). Sometimes one branch of the bundle of His is interrupted, causing right or left bundle branch block. In bundle branch block, excitation passes normally down the bundle on the intact side and then sweeps back through the muscle to acti-vate the ventricle on the blocked side. The ventricular rate is therefore normal, but the QRS complexes are prolonged and deformed (Figure 30–11). Block can also occur in the anterior or posterior fascicle of the left bundle branch, producing the condition called hemiblock or fascicular block. Left anterior hemiblock produces abnormal left axis deviation in the ECG, whereas left posterior hemiblock produces abnormal right axis deviation. It is not uncommon to find combinations of fascicu-lar and branch blocks (bifascicular or trifascicular block). The His bundle electrogram permits detailed analysis of the site of block when there is a defect in the conduction system. IMPLANTED PACEMAKERS When there is marked bradycardia in patients with sick sinus syndrome or third-degree heart block, an electronic pacemak-er is frequently implanted. These devices, which have become sophisticated and reliable, are useful in patients with sinus node dysfunction, AV block, and bifascicular or trifascicular block. They are useful also in patients with severe neurogenic FIGURE 30–10 Sinus arrhythmia in a young man and an old man. Each subject breathed five times per minute. With each inspira-tion the RR interval (the interval between R waves) declined, indicating an increase in heart rate. Note the marked reduction in the magnitude of the arrhythmia in the older man. These records were obtained after β-adrenergic blockade, but would have been generally similar in its absence. (Reproduced with permission from Pfeifer MA et al: Differential changes of autonomic nervous system function with age in man. Am J Med 1983;75:249.) Time (sec) Heart rate (BPM) RR interval (msec) 60 45 30 15 75 60 50 40 900 1100 1300 1500 22-year-old normal male 79-year-old normal male 498 SECTION VI Cardiovascular Physiology syncope in whom carotid sinus stimulation produces pauses of more than 3 s between heartbeats. ECTOPIC FOCI OF EXCITATION Normally, myocardial cells do not discharge spontaneously, and the possibility of spontaneous discharge of the His bundle and Purkinje system is low because the normal pacemaker dis-charge of the SA node is more rapid than their rate of sponta-neous discharge. However, in abnormal conditions, the His– Purkinje fibers or the myocardial fibers may discharge spon-taneously. In these conditions, increased automaticity of the heart is said to be present. If an irritable ectopic focus dis-charges once, the result is a beat that occurs before the expect-ed next normal beat and transiently interrupts the cardiac rhythm (atrial, nodal, or ventricular extrasystole or prema-ture beat). If the focus discharges repetitively at a rate higher than that of the SA node, it produces rapid, regular tachycar-dia (atrial, ventricular, or nodal paroxysmal tachycardia or atrial flutter). REENTRY A more common cause of paroxysmal arrhythmias is a defect in conduction that permits a wave of excitation to propagate con-tinuously within a closed circuit (circus movement). For exam-ple, if a transient block is present on one side of a portion of the conducting system, the impulse can go down the other side. If the block then wears off, the impulse may conduct in a retro-grade direction in the previously blocked side back to the origin and then descend again, establishing a circus movement. An ex-ample of this in a ring of tissue is shown in Figure 30–12. If the reentry is in the AV node, the reentrant activity depolarizes the atrium, and the resulting atrial beat is called an echo beat. In ad-dition, the reentrant activity in the node propagates back down to the ventricle, producing paroxysmal nodal tachycardia. Cir-cus movements can also become established in the atrial or ven-tricular muscle fibers. In individuals with an abnormal extra bundle of conducting tissue connecting the atria to the ventri-cles (bundle of Kent), the circus activity can pass in one direc-tion through the AV node and in the other direction through the bundle, thus involving both the atria and the ventricles. FIGURE 30–11 Heart block. PR = 0.16 s Normal complex PR = 0.38 s First-degree heart block Second-degree heart block (2:1 heart block) Complete heart block. Atrial rate, 107; ventricular rate, 43 Second-degree heart block (Wenckebach phenomenon) aVF aVF aVF Two V leads in left bundle branch block V6 V5 FIGURE 30–12 Depolarization of a ring of cardiac tissue. Normally, the impulse spreads in both directions in the ring (left) and the tissue immediately behind each branch of the impulse is refracto-ry. When a transient block occurs on one side (center), the impulse on the other side goes around the ring, and if the transient block has now worn off (right), the impulse passes this area and continues to circle indefinitely (circus movement). CHAPTER 30 Origin of the Heartbeat & the Electrical Activity of the Heart 499 ATRIAL ARRHYTHMIAS Excitation spreading from an independently discharging fo-cus in the atria stimulates the AV node prematurely and is conducted to the ventricles. The P waves of atrial extrasystoles are abnormal, but the QRST configurations are usually nor-mal (Figure 30–13). The excitation may depolarize the SA node, which must repolarize and then depolarize to the firing level before it can initiate the next normal beat. Consequently, a pause occurs between the extrasystole and the next normal beat that is usually equal in length to the interval between the normal beats preceding the extrasystole, and the rhythm is “reset” (see below). Atrial tachycardia occurs when an atrial focus discharges regularly or there is reentrant activity producing atrial rates up to 220/min. Sometimes, especially in digitalized patients, some degree of atrioventricular block is associated with the tachycardia (paroxysmal atrial tachycardia with block). In atrial flutter, the atrial rate is 200 to 350/min (Figure 30–13). In the most common form of this arrhythmia, there is large counterclockwise circus movement in the right atrium. This produces a characteristic sawtooth pattern of flutter waves due to atrial contractions. It is almost always associated with 2:1 or greater AV block, because in adults the AV node cannot con-duct more than about 230 impulses per minute. In atrial fibrillation, the atria beat very rapidly (300–500/ min) in a completely irregular and disorganized fashion. Because the AV node discharges at irregular intervals, the ven-tricles beat at a completely irregular rate, usually 80 to 160/min (Figure 30–13). The condition can be paroxysmal or chronic, and in some cases there appears to be a genetic predisposition. The cause of atrial fibrillation is still a matter of debate, but in most cases it appears to be due to multiple concurrently circu-lating reentrant excitation waves in both atria. However, some cases of paroxysmal atrial fibrillation seem to be produced by discharge of one or more ectopic foci. Many of these foci appear to be located in the pulmonary veins as much as 4 cm from the heart. Atrial muscle fibers extend along the pulmo-nary veins and are the origin of these discharges. CONSEQUENCES OF ATRIAL ARRHYTHMIAS Occasional atrial extrasystoles occur from time to time in most normal humans and have no pathologic significance. In parox-ysmal atrial tachycardia and flutter, the ventricular rate may be so high that diastole is too short for adequate filling of the ven-tricles with blood between contractions. Consequently, cardiac output is reduced and symptoms of heart failure appear. Heart failure may also complicate atrial fibrillation when the ventric-ular rate is high. Acetylcholine liberated at vagal endings de-presses conduction in the atrial musculature and AV node. This is why stimulating reflex vagal discharge by pressing on the eyeball (oculocardiac reflex) or massaging the carotid si-nus often converts tachycardia and sometimes converts atrial flutter to normal sinus rhythm. Alternatively, vagal stimulation increases the degree of AV block, abruptly lowering the ven-tricular rate. Digitalis also depresses AV conduction and is used to lower a rapid ventricular rate in atrial fibrillation. VENTRICULAR ARRHYTHMIAS Premature beats that originate in an ectopic ventricular focus usually have bizarrely shaped prolonged QRS complexes (Fig-ure 30–14) because of the slow spread of the impulse from the focus through the ventricular muscle to the rest of the ventri-cle. They are usually incapable of exciting the bundle of His, and retrograde conduction to the atria therefore does not oc-cur. In the meantime, the next succeeding normal SA nodal impulse depolarizes the atria. The P wave is usually buried in the QRS of the extrasystole. If the normal impulse reaches the ventricles, they are still in the refractory period following de-polarization from the ectopic focus. FIGURE 30–13 Atrial arrhythmias. The illustration shows an atrial premature beat with its P wave superimposed on the T wave of the preceding beat (arrow); atrial tachycardia; atrial flutter with 4:1 AV block; and atrial fibrillation with a totally irregular ventricular rate. (Tracings reproduced with permission from Goldschlager N, Goldman MJ: Principles of Clinical Electrocardiography, 13th ed. Originally published by Appleton & Lange. Copyright © 1989 by McGraw-Hill.) Atrial fibrillation Atrial flutter Atrial tachycardia Atrial extrasystole V1 V1 II II 500 SECTION VI Cardiovascular Physiology However, the second succeeding impulse from the SA node produces a normal beat. Thus, ventricular premature beats are followed by a compensatory pause that is often longer than the pause after an atrial extrasystole. Furthermore, ventricular premature beats do not interrupt the regular discharge of the SA node, whereas atrial premature beats often interrupt and “reset” the normal rhythm. Atrial and ventricular premature beats are not strong enough to produce a pulse at the wrist if they occur early in diastole, when the ventricles have not had time to fill with blood and the ventricular musculature is still in its relatively refractory period. They may not even open the aortic and pul-monary valves, in which case there is, in addition, no second heart sound. Paroxysmal ventricular tachycardia (Figure 30–14) is in effect a series of rapid, regular ventricular depolarizations usually due to a circus movement involving the ventricles. Torsade de pointes is a form of ventricular tachycardia in which the QRS morphology varies (Figure 30–15). Tachycar-dias originating above the ventricles (supraventricular tachy-cardias such as paroxysmal nodal tachycardia) can be distinguished from paroxysmal ventricular tachycardia by use of the HBE; in supraventricular tachycardias, a His bundle H deflection is present, whereas in ventricular tachycardias, there is none. Ventricular premature beats are common and, in the absence of ischemic heart disease, usually benign. Ven-tricular tachycardia is more serious because cardiac output is decreased, and ventricular fibrillation is an occasional com-plication of ventricular tachycardia. In ventricular fibrillation (Figure 30–15), the ventricular muscle fibers contract in a totally irregular and ineffective way because of the very rapid discharge of multiple ventricular FIGURE 30–14 Top: Ventricular premature beats (VPB). The lines under the tracing illustrate the compensatory pause and show that the duration of the premature beat plus the preceding normal beat is equal to the duration of two normal beats. Bottom: Ventricular tachycardia. N N N P N Comp. pause VPB FIGURE 30–15 Record obtained from an implanted cardioverter–defibrillator in a 12-year-old boy with congenital long QT syndrome who collapsed while answering a question in school. Top: Normal sinus rhythm with long QT interval. Middle: Torsade de pointes. Bottom: Ventricular fibrillation with discharge of defibrillator, as programmed 7.5 s after the start of ventricular tachycardia, converting the heart to normal sinus rhythm. The boy recovered consciousness in 2 min and had no neurologic sequelae. (Reproduced with permission from Moss AJ, Daubert JP: Images in clinical medicine. Internal ventricular fibrillation. N Engl J Med 2000;342:398.) Sinus rhythm Torsade de pointes Ventricular fibrillation and sinus rhythm Discharge CHAPTER 30 Origin of the Heartbeat & the Electrical Activity of the Heart 501 ectopic foci or a circus movement. The fibrillating ventricles, like the fibrillating atria, look like a quivering “bag of worms.” Ventricular fibrillation can be produced by an electric shock or an extrasystole during a critical interval, the vulnerable period. The vulnerable period coincides in time with the midportion of the T wave; that is, it occurs at a time when some of the ventricular myocardium is depolarized, some is incompletely repolarized, and some is completely repolarized. These are excellent conditions in which to establish reentry and a circus movement. The fibrillating ventricles cannot pump blood effectively, and circulation of the blood stops. Therefore, in the absence of emergency treatment, ventricular fibrillation that lasts more than a few minutes is fatal. The most frequent cause of sudden death in patients with myocar-dial infarcts is ventricular fibrillation. LONG QT SYNDROME An indication of vulnerability of the heart during repolariza-tion is the fact that in patients in whom the QT interval is pro-longed, cardiac repolarization is irregular and the incidence of ventricular arrhythmias and sudden death increases. The syn-drome can be caused by a number of different drugs, by elec-trolyte abnormalities, and by myocardial ischemia. It can also be congenital. Mutations of eight different genes have been re-ported to cause the syndrome. Six cause reduced function of various K+ channels by alterations in their structure; one in-hibits a K+ channel by reducing the amount of the ankyrin iso-form that links it to the cytoskeleton; and one increases the function of the cardiac Na+ channel. ACCELERATED AV CONDUCTION An interesting condition seen in some otherwise normal indi-viduals who are prone to attacks of paroxysmal atrial arrhyth-mias is accelerated AV conduction (Wolff–Parkinson–White syndrome). Normally, the only conducting pathway between the atria and the ventricles is the AV node. Individuals with Wolff–Parkinson–White syndrome have an additional aber-rant muscular or nodal tissue connection (bundle of Kent) be-tween the atria and ventricles. This conducts more rapidly than the slowly conducting AV node, and one ventricle is excited early. The manifestations of its activation merge with the nor-mal QRS pattern, producing a short PR interval and a pro-longed QRS deflection slurred on the upstroke (Figure 30–16), with a normal interval between the start of the P wave and the end of the QRS complex (“PJ interval”). The paroxysmal atrial tachycardias seen in this syndrome often follow an atrial prema-ture beat. This beat conducts normally down the AV node but spreads to the ventricular end of the aberrant bundle, and the impulse is transmitted retrograde to the atrium. A circus move-ment is thus established. Less commonly, an atrial premature beat finds the AV node refractory but reaches the ventricles via the bundle of Kent, setting up a circus movement in which the impulse passes from the ventricles to the atria via the AV node. In some instances, the Wolff–Parkinson–White syndrome is familial. In two such families, there is a mutation in a gene that codes for an AMP-activated protein kinase. Presumably, this kinase is normally involved in suppressing abnormal atri-oventricular pathways during fetal development. Attacks of paroxysmal supraventricular tachycardia, usually nodal tachycardia, are seen in individuals with short PR inter-vals and normal QRS complexes (Lown–Ganong–Levine syndrome). In this condition, depolarization presumably passes from the atria to the ventricles via an aberrant bundle that bypasses the AV node but enters the intraventricular con-ducting system distal to the node. ANTIARRHYTHMIC DRUGS Many different drugs have been developed that are used in the treatment of arrhythmias because they slow conduction in the conduction system and the myocardium. This depresses ec-topic activity and reduces the discrepancy between normal and reentrant paths so that reentry does not occur. However, it has now become clear that in some patients any of these drugs can be proarrhythmic rather than antiarrhythmic— that is, they can also cause various arrhythmias. Therefore, they are increasingly being replaced by radiofrequency cathe-ter ablation for the treatment of arrhythmias. RADIOFREQUENCY CATHETER ABLATION OF REENTRANT PATHWAYS Catheters with electrodes at the tip can now be inserted into the chambers of the heart and its environs and used to map the ex-act location of an ectopic focus or accessory bundle that is re-sponsible for the production of reentry and supraventricular tachycardia. The pathway can then be ablated by passing FIGURE 30–16 Accelerated AV conduction. Top: Normal si-nus beat. Middle: Short PR interval; wide, slurred QRS complex; normal PJ interval (Wolff–Parkinson–White syndrome). Bottom: Short PR interval, normal QRS complex (Lown–Ganong–Levine syndrome). (Reproduced with permission from Goldschlager N, Goldman MJ: Principles of Clinical Electrocardiography, 13th ed. Originally published by Appleton & Lange. Copyright © 1989 by McGraw-Hill.) P J 502 SECTION VI Cardiovascular Physiology radiofrequency current with the catheter tip placed close to the bundle or focus. In skilled hands, this form of treatment can be very effective and is associated with few complications. It is par-ticularly useful in conditions that cause supraventricular tachy-cardias, including Wolff–Parkinson–White syndrome and atrial flutter. It has also been used with success to ablate foci in the pulmonary veins causing paroxysmal atrial fibrillation. ELECTROCARDIOGRAPHIC FINDINGS IN OTHER CARDIAC & SYSTEMIC DISEASES MYOCARDIAL INFARCTION When the blood supply to part of the myocardium is inter-rupted, profound changes take place in the myocardium that lead to irreversible changes and death of muscle cells. The ECG is very useful for diagnosing ischemia and locating areas of infarction. The underlying electrical events and the result-ing electrocardiographic changes are complex, and only a brief review can be presented here. The three major abnormalities that cause electrocardio-graphic changes in acute myocardial infarction are summa-rized in Table 30–3. The first change—abnormally rapid repolarization after discharge of the infarcted muscle fibers as a result of accelerated opening of K+ channels—develops sec-onds after occlusion of a coronary artery in experimental ani-mals. It lasts only a few minutes, but before it is over the resting membrane potential of the infarcted fibers declines because of the loss of intracellular K+. Starting about 30 min later, the infarcted fibers also begin to depolarize more slowly than the surrounding normal fibers. All three of these changes cause current flow that pro-duces elevation of the ST segment in electrocardiographic leads recorded with electrodes over the infarcted area (Fig-ure 30–17). Because of the rapid repolarization in the infarct, the membrane potential of the area is greater than it is in the normal area during the latter part of repolarization, making the normal region negative relative to the infarct. Extracellu-larly, current therefore flows out of the infarct into the normal area (since, by convention, current flow is from positive to neg-ative). This current flows toward electrodes over the injured area, causing increased positivity between the S and T waves of the ECG. Similarly, the delayed depolarization of the infarcted cells causes the infarcted area to be positive relative to the healthy tissue (Table 30–3) during the early part of repolariza-tion, and the result is also ST segment elevation. The remain-ing change—the decline in resting membrane potential during diastole—causes a current flow into the infarct during ventric-ular diastole. The result of this current flow is a depression of the TQ segment of the ECG. However, the electronic arrange-ment in electrocardiographic recorders is such that a TQ seg-ment depression is recorded as an ST segment elevation. Thus, the hallmark of acute myocardial infarction is elevation of the ST segments in the leads overlying the area of infarction (Figure 30–17). Leads on the opposite side of the heart show ST segment depression. After some days or weeks, the ST segment abnormalities subside. The dead muscle and scar tissue become electrically silent. The infarcted area is therefore negative relative to the normal myocardium during systole, and it fails to contribute its share of positivity to the electrocardiographic complexes. The manifestations of this negativity are multiple and subtle. Common changes include the appearance of a Q wave in some of the leads in which it was not previously present and an increase in the size of the normal Q wave in some of the other leads, although so-called non-Q-wave infarcts are also seen. These infarcts tend to be less severe, but there is a high incidence of subsequent reinfarction. Another finding in infarction of the anterior left ventricle is “failure of progres-sion of the R wave”; that is, the R wave fails to become succes-sively larger in the precordial leads as the electrode is moved from right to left over the left ventricle. If the septum is infarcted, the conduction system may be damaged, causing bundle branch block or other forms of heart block. Myocardial infarctions are often complicated by serious ventricular arrhythmias, with the threat of ventricular fibrilla-tion and death. In experimental animals, and presumably in humans, ventricular arrhythmias occur during three periods. During the first 30 min of an infarction, arrhythmias due to reentry are common. There follows a period relatively free from arrhythmias, but, starting 12 h after infarction, arrhyth-mias occur as a result of increased automaticity. Arrhythmias occurring 3 d to several weeks after infarction are once again usually due to reentry. It is worth noting in this regard that infarcts that damage the epicardial portions of the myocar-dium interrupt sympathetic nerve fibers, producing denerva-tion super-sensitivity to catecholamines in the area beyond the infarct. Alternatively, endocardial lesions can selectively interrupt vagal fibers, leaving the actions of sympathetic fibers unopposed. TABLE 30–3 Summary of the three major abnormalities of membrane polarization associated with acute myocardial infarction. Defect in Infarcted Cells Current Flow Resultant ECG Change in Leads Over Infarct Rapid repolarization Out of infarct ST segment elevation Decreased resting membrane potential Into infarct TQ segment depression (mani-fested as ST segment elevation) Delayed depolarization Out of infarct ST segment elevation CHAPTER 30 Origin of the Heartbeat & the Electrical Activity of the Heart 503 EFFECTS OF CHANGES IN THE IONIC COMPOSITION OF THE BLOOD Changes in ECF Na+ and K+ concentration would be expected to affect the potentials of the myocardial fibers, because the electrical activity of the heart depends upon the distribution of these ions across the muscle cell membranes. Clinically, a fall in the plasma level of Na+ may be associated with low-voltage electrocardio-graphic complexes, but changes in the plasma K+ level produce severe cardiac abnormalities. Hyperkalemia is a very dangerous and potentially lethal condition because of its effects on the heart. As the plasma K+ level rises, the first change in the ECG is the FIGURE 30–17 Diagrammatic illustration of serial electrocardiographic patterns in anterior infarction. A) Normal tracing. B) Very early pattern (hours after infarction): ST segment elevation in I, aVL, and V3–6; reciprocal ST depression in II, III, and aVF. C) Later pattern (many hours to a few days): Q waves have appeared in I, aVL, and V5–6. QS complexes are present in V3–4. This indicates that the major transmural infarction is underlying the area recorded by V3–4; ST segment changes persist but are of lesser degree, and the T waves are beginning to invert in the leads in which the ST segments are elevated. D) Late established pattern (many days to weeks): The Q waves and QS complexes persist, the ST segments are isoelectric, and the T waves are symmetric and deeply inverted in leads that had ST elevation and tall in leads that had ST depression. This pat-tern may persist for the remainder of the patient’s life. E) Very late pattern: This may occur many months to years after the infarction. The abnormal Q waves and QS complexes persist. The T waves have gradually returned to normal. (Reproduced with permission from Goldschlager N, Goldman MJ: Principles of Clinical Electrocardiography, 13th ed. Originally published by Appleton & Lange. Copyright © 1989 by McGraw-Hill.) I II III aVR A B C D E aVL aVF V1–2 V3–4 V5–6 504 SECTION VI Cardiovascular Physiology appearance of tall peaked T waves, a manifestation of altered re-polarization (Figure 30–18). At higher K+ levels, paralysis of the atria and prolongation of the QRS complexes occur. Ventricular arrhythmias may develop. The resting membrane potential of the muscle fibers decreases as the extracellular K+ concentration in-creases. The fibers eventually become unexcitable, and the heart stops in diastole. Conversely, a decrease in the plasma K+ level causes prolongation of the PR interval, prominent U waves, and, occasionally, late T wave inversion in the precardial leads. If the T and U waves merge, the apparent QT interval is often prolonged; if the T and U waves are separated, the true QT interval is seen to be of normal duration. Hypokalemia is a serious condition, but it is not as rapidly fatal as hyperkalemia. Increases in extracellular Ca2+ concentration enhance myo-cardial contractility. When large amounts of Ca2+ are infused into experimental animals, the heart relaxes less during diastole and eventually stops in systole (calcium rigor). However, in clinical conditions associated with hypercalcemia, the plasma calcium level is rarely if ever high enough to affect the heart. Hypocalcemia causes prolongation of the ST segment and con-sequently of the QT interval, a change that is also produced by phenothiazines and tricyclic antidepressant drugs and by vari-ous diseases of the central nervous system. CHAPTER SUMMARY ■Contractions in the heart are controlled via a well-regulated electrical signaling cascade that originates in pacemaker cells in the sinoatrial (SA) node and is passed via internodal atrial path-ways to the atrioventrical (AV) node, the bundle of His, the Purkinje system, and to all parts of the ventricle. ■Most cardiac cells have an action potential that includes a rapid depolarization, an initial rapid repolarization, a plateau, and a slow repolarization process to return to resting potential. These changes are defined by sequential activation and inactivation of Na+, Ca2+, and K+ channels. ■Pacemaker cells have a slightly different sequence of events. After repolarization to the resting potential, there is a slow de-polarization that occurs due to a channel that can pass both Na+ and K+. As this “funny” current continues to depolarize the cell, Ca2+ channels are activated to rapidly depolarize the cell. The hyperpolarization phase is again dominated by K+ current. ■Spread of the electrical signal from cell to cell is via gap junc-tions. The rate of spread is dependent on anatomical features, but also can be altered (to a certain extent) via neural input. ■The electrocardiogram (ECG) is an algebraic sum of the electri-cal activity in the heart. The normal ECG includes well-defined waves and segments, including the P wave (atrial depolarization), FIGURE 30–18 Correlation of plasma K+ level and the ECG, assuming that the plasma Ca2+ level is normal. The diagrammed com-plexes are left ventricular epicardial leads. (Reproduced with permission from Goldman MJ: Principles of Clinical Electrocardiography, 12th ed. Originally published by Appleton & Lange. Copyright © 1986 by McGraw-Hill.) R T U T U P Normal tracing (plasma K+ 4–5.5 meq/L). PR interval = 0.16 s; QRS interval = 0.06 s; QT interval = 0.4 s (normal for an assumed heart rate of 60). Hyperkalemia (plasma K+ ±7.0 meq/L). The PR and QRS intervals are within normal limits. Very tall, slender peaked T waves are now present. Hyperkalemia (plasma K+ ±8.5 meq/L). There is no evidence of atrial activity; the QRS complex is broad and slurred and the QRS interval has widened to 0.2 s. The T waves remain tall and slender. Further elevation of the plasma K+ level may result in ventricular tachycardia and ventricular fibrillation. Hypokalemia (plasma K+ ±3.5 meq/L). PR interval = 0.2 s; QRS interval = 0.06 s; ST segment depression. A prominent U wave is now present immediately following the T. The actual QT interval remains 0.4 s. If the U wave is erroneously considered a part of the T, a falsely prolonged QT interval of 0.6 s will be measured. Hypokalemia (plasma K+ ±2.5 meq/L). The PR interval is lengthened to 0.32 s; the ST segment is depressed; the T wave is inverted; a prominent U wave is seen. The true QT interval remains normal. CHAPTER 30 Origin of the Heartbeat & the Electrical Activity of the Heart 505 the QRS complex (ventricular depolarization), and the T wave (ventricular hyperpolarization). Various arrhythmias can be de-tected in irregular ECG recordings. ■Because of the contribution of ionic movement to cardiac mus-cle contraction, heart tissue is sensitive to ionic composition of the blood. Most serious are increases in [K+] that can produce severe cardiac abnormalities, including paralysis of the atria and ventricular arrhythmias. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Which part of the ECG (eg, Figure 30–5) corresponds to ventric-ular repolarization? A) the P wave B) the QRS duration C) the T wave D) the U wave E) the PR interval 2. Which of the following normally has a slowly depolarizing “prepotential”? A) sinoatrial node B) atrial muscle cells C) bundle of His D) Purkinje fibers E) ventricular muscle cells 3. In second-degree heart block A) the ventricular rate is lower than the atrial rate. B) the ventricular ECG complexes are distorted. C) there is a high incidence of ventricular tachycardia. D) stroke volume is decreased. E) cardiac output is increased. 4. Currents caused by opening of which of the following channels contribute to the repolarization phase of the action potential of ventricular muscle fibers? A) Na+ channels B) Cl– channels C) Ca2+ channels D) K+ channels E) HCO3 – channels 5. In complete heart block A) fainting may occur because the atria are unable to pump blood into the ventricles. B) ventricular fibrillation is common. C) the atrial rate is lower than the ventricular rate. D) fainting may occur because of prolonged periods during which the ventricles fail to contract. CHAPTER RESOURCES Hile B: Ionic Channels of Excitable Membranes, 3rd ed. Sinauer Associates, Inc., 2001. Jackson WF: Ion channels and vascular tone. Hypertension 2000;35:173. Jessup M, Brozena S: Heart failure. N Engl J Med 2003;348:2007. Morady F: Radiofrequency ablation as treatment for cardiac arrhythmias. N Engl J Med 1999;340:534. Nabel EG: Genomic medicine: cardiovascular disease. N Engl J Med 2003;349:60. Roder DM: Drug-induced prolongation of the Q-T interval. N Engl J Med 2004;350:1013. Rowell LB: Human Cardiovascular Control. Oxford University Press, 1993. Wagner GS: Marriott’s Practical Electrocardiography, 10th ed. Lippincott Williams and Wilkins, 2000. This page intentionally left blank 507 C H A P T E R 31 The Heart as a Pump O B J E C T I V E S After studying this chapter, you should be able to: ■Describe how the sequential pattern of contraction and relaxation in the heart results in a normal pattern of blood flow. ■Understand the pressure, volume, and flow changes that occur during the cardiac cycle. ■Explain the basis of the arterial pulse, heart sounds, and murmurs. ■Delineate the ways by which cardiac output can be up-regulated in the setting of specific physiologic demands for increased oxygen supply to the tissues, such as exercise. ■Describe how the pumping action of the heart can be compromised in the setting of specific disease states. INTRODUCTION Of course, the electrical activity of the heart discussed in the previous chapter is designed to subserve the heart’s primary physiological role—to pump blood through the lungs, where gas exchange can occur, and thence to the remainder of the body (Clinical Box 31–1). This is accomplished when the orderly depolarization process described in the previous chapter triggers a wave of contraction that spreads through the myocardium. In single muscle fibers, contraction starts just after depolarization and lasts until about 50 ms after repolarization is completed (see Figure 5–15). Atrial systole starts after the P wave of the electrocardiogram (ECG); ven-tricular systole starts near the end of the R wave and ends just after the T wave. In this chapter, we will consider how these changes in contraction produce sequential changes in pres-sures and flows in the heart chambers and blood vessels, and thereby propel blood appropriately as needed by whole body demands for oxygen and nutrients. As an aside, it should be noted that the term systolic pressure in the vascular system refers to the peak pressure reached during systole, not the mean pressure; similarly, the diastolic pressure refers to the lowest pressure during diastole. MECHANICAL EVENTS OF THE CARDIAC CYCLE EVENTS IN LATE DIASTOLE Late in diastole, the mitral (bicuspid) and tricuspid valves be-tween the atria and ventricles (atrioventricular [AV] valves) are open and the aortic and pulmonary valves are closed. Blood flows into the heart throughout diastole, filling the atria and ventricles. The rate of filling declines as the ventricles be-come distended, and, especially when the heart rate is low, the cusps of the AV valves drift toward the closed position (Figure 31–1). The pressure in the ventricles remains low. About 70% of the ventricular filling occurs passively during diastole. 508 SECTION VI Cardiovascular Physiology ATRIAL SYSTOLE Contraction of the atria propels some additional blood into the ventricles. Contraction of the atrial muscle narrows the or-ifices of the superior and inferior vena cava and pulmonary veins, and the inertia of the blood moving toward the heart tends to keep blood in it. However, despite these inhibitory in-fluences, there is some regurgitation of blood into the veins. VENTRICULAR SYSTOLE At the start of ventricular systole, the AV valves close. Ventric-ular muscle initially shortens relatively little, but intraventric-ular pressure rises sharply as the myocardium presses on the blood in the ventricle (Figure 31–2). This period of isovolu-metric (isovolumic, isometric) ventricular contraction lasts about 0.05 s, until the pressures in the left and right ventricles exceed the pressures in the aorta (80 mm Hg; 10.6 kPa) and pulmonary artery (10 mm Hg) and the aortic and pulmonary valves open. During isovolumetric contraction, the AV valves bulge into the atria, causing a small but sharp rise in atrial pressure (Figure 31–3). When the aortic and pulmonary valves open, the phase of ventricular ejection begins. Ejection is rapid at first, slowing down as systole progresses. The intraventricular pressure rises to a maximum and then declines somewhat before ventricular systole ends. Peak pressures in the left and right ventricles are about 120 and 25 mm Hg, respectively. Late in systole, pres-sure in the aorta actually exceeds that in the left ventricle, but for a short period momentum keeps the blood moving for-ward. The AV valves are pulled down by the contractions of the ventricular muscle, and atrial pressure drops. The amount of blood ejected by each ventricle per stroke at rest is 70 to 90 mL. The end-diastolic ventricular volume is about 130 mL. Thus, about 50 mL of blood remains in each ventricle at the end of systole (end-systolic ventricular volume), and the ejection fraction, the percent of the end-diastolic ventricular volume that is ejected with each stroke, is about 65%. The ejection fraction is a valuable index of ventricular function. It can be measured by injecting radionuclide-labeled red blood cells and imaging the cardiac blood pool at the end of diastole and the end of systole (equilibrium radionuclide angiocardio-graphy), or by computed tomography. CLINICAL BOX 31–1 Heart Failure Heart failure occurs when the heart is unable to put out an amount of blood that is adequate for the needs of the tissues. It can be acute and associated with sudden death, or chronic. The failure may involve primarily the right ventricle (cor pulmo-nale), but much more commonly it involves the larger, thicker left ventricle or both ventricles. Heart failure may also be sys-tolic or diastolic. In systolic failure, stroke volume is reduced because ventricular contraction is weak. This causes an increase in the end-systolic ventricular volume, so that the ejection fraction falls from 65% to as low as 20%. The initial response to failure is activation of the genes that cause cardiac myocytes to hypertrophy, and thickening of the ventricular wall (cardiac re-modeling). The incomplete filling of the arterial system leads to increased discharge of the sympathetic nervous system and increased secretion of renin and aldosterone, so Na+ and water are retained. These responses are initially compensatory, but eventually the failure worsens and the ventricles dilate. output may be elevated in absolute terms but still be inade-quate to meet the needs of the tissues (high-output failure). Treatment of congestive heart failure is aimed at improving cardiac contractility, treating the symptoms, and decreasing the load on the heart. Currently, the most effective treatment in general use is inhibition of the production of angiotensin II with angiotensin-converting enzyme (ACE) inhibitors. Block-ade of the effects of angiotensin II on AT1 receptors with non-peptide antagonists is also of value. Blocking the production of angiotensin II or its effects also reduces the circulating al-dosterone level and decreases blood pressure, reducing the afterload against which the heart pumps. The effects of aldos-terone can be further reduced by administering aldosterone receptor blockers. Reducing venous tone with nitrates or hy-dralazine increases venous capacity so that the amount of blood returned to the heart is reduced, lowering the preload. Diuretics reduce the fluid overload. Drugs that block β-adre-nergic receptors have been shown to decrease mortality and morbidity. Digitalis derivatives such as digoxin have classi-cally been used to treat congestive heart failure because of their ability to increase intracellular Ca2+ and hence exert a positive inotropic effect, but they are now used in a secon-dary role to treat systolic dysfunction and slow the ventricular rate in patients with atrial fibrillation. In diastolic failure, the ejection fraction is initially main-tained, but the elasticity of the myocardium is reduced so fill-ing during diastole is reduced. This leads to inadequate stroke volume and the same cardiac remodeling and Na+ and water retention that occur in systolic failure. It should be noted that the inadequate cardiac output in failure may be relative rather than absolute. When a large arterior venous fistula is present, in thyrotoxicosis and in thiamine deficiency, cardiac CHAPTER 31 The Heart as a Pump 509 EARLY DIASTOLE Once the ventricular muscle is fully contracted, the already falling ventricular pressures drop more rapidly. This is the pe-riod of protodiastole, which lasts about 0.04 s. It ends when the momentum of the ejected blood is overcome and the aortic and pulmonary valves close, setting up transient vibrations in the blood and blood vessel walls. After the valves are closed, pressure continues to drop rapidly during the period of iso-volumetric ventricular relaxation. Isovolumetric relaxation ends when the ventricular pressure falls below the atrial pres-sure and the AV valves open, permitting the ventricles to fill. Filling is rapid at first, then slows as the next cardiac contrac-tion approaches. Atrial pressure continues to rise after the end of ventricular systole until the AV valves open, then drops and slowly rises again until the next atrial systole. FIGURE 31–1 Divisions of the cardiac cycle: A) systole and B) diastole. The phases of the cycle are identical in both halves of the heart. The direction in which the pressure difference favors flow is denoted by an arrow; note, however, that flow will not actually occur if valve prevents it. Systole A B Blood flows out of ventricle Ventricles contract Atria relaxed Ventricles relaxed Atria relaxed Ventricles relaxed Atria relaxed Ventricles relaxed Atria contract Ventricles contract Atria relaxed Closed Closed Closed Open Diastole Blood flows into ventricles Atrial contraction AV valves: Aortic and pulmonary valves: Closed Closed Open Closed Open Closed AV valves: Aortic and pulmonary valves: Isovolumetric ventricular contraction Isovolumetric ventricular relaxation Ventricular ejection Ventricular filling 510 SECTION VI Cardiovascular Physiology PERICARDIUM The myocardium is covered by a fibrous layer known as the epi-cardium. This, in turn, is surrounded by the pericardium, which separates the heart from the rest of the thoracic viscera. The space between the epicardium and pericardium (the pericardial sac) normally contains 5 to 30 mL of clear fluid, which lubricates the heart and permits it to contract with minimal friction. TIMING Although events on the two sides of the heart are similar, they are somewhat asynchronous. Right atrial systole precedes left atrial systole, and contraction of the right ventricle starts after that of the left (see Chapter 30). However, since pulmonary ar-terial pressure is lower than aortic pressure, right ventricular ejection begins before that of the left. During expiration, the pulmonary and aortic valves close at the same time; but during inspiration, the aortic valve closes slightly before the pulmo-nary. The slower closure of the pulmonary valve is due to low-er impedance of the pulmonary vascular tree. When measured over a period of minutes, the outputs of the two ventricles are, of course, equal, but transient differences in output during the respiratory cycle occur in normal individuals. LENGTH OF SYSTOLE & DIASTOLE Cardiac muscle has the unique property of contracting and re-polarizing faster when the heart rate is high (see Chapter 5), and the duration of systole decreases from 0.27 s at a heart rate of 65 to 0.16 s at a rate of 200 beats/min (Table 31–1). The shortening is due mainly to a decrease in the duration of sys-tolic ejection. However, the duration of systole is much more fixed than that of diastole, and when the heart rate is in-creased, diastole is shortened to a much greater degree. For ex-ample, at a heart rate of 65, the duration of diastole is 0.62 s, whereas at a heart rate of 200, it is only 0.14 s. This fact has im-portant physiologic and clinical implications. It is during dias-tole that the heart muscle rests, and coronary blood flow to the subendocardial portions of the left ventricle occurs only dur-ing diastole (see Chapter 34). Furthermore, most of the ven-tricular filling occurs in diastole. At heart rates up to about 180, filling is adequate as long as there is ample venous return, and cardiac output per minute is increased by an increase in rate. However, at very high heart rates, filling may be compro-mised to such a degree that cardiac output per minute falls. Because it has a prolonged action potential, cardiac muscle cannot contract in response to a second stimulus until near the end of the initial contraction (see Figure 5–15). Therefore, cardiac muscle cannot be tetanized like skeletal muscle. The highest rate at which the ventricles can contract is theoreti-cally about 400/min, but in adults the AV node will not con-duct more than about 230 impulses/min because of its long refractory period. A ventricular rate of more than 230 is seen only in paroxysmal ventricular tachycardia (see Chapter 30). Exact measurement of the duration of isovolumetric ven-tricular contraction is difficult in clinical situations, but it is relatively easy to measure the duration of total electrome-chanical systole (QS2), the preejection period (PEP), and the left ventricular ejection time (LVET) by recording the ECG, phonocardiogram, and carotid pulse simultaneously. QS2 is the period from the onset of the QRS complex to the closure of the aortic valves, as determined by the onset of the second heart sound. LVET is the period from the beginning of the carotid pressure rise to the dicrotic notch (see below). PEP is the difference between QS2 and LVET and represents the time for the electrical as well as the mechanical events that precede systolic ejection. The ratio PEP/LVET is normally about 0.35, and it increases without a change in QS2 when left ventricular performance is compromised in a variety of cardiac diseases. ARTERIAL PULSE The blood forced into the aorta during systole not only moves the blood in the vessels forward but also sets up a pressure wave that travels along the arteries. The pressure wave expands the arterial walls as it travels, and the expansion is palpable as the pulse. The rate at which the wave travels, which is independent of and much higher than the velocity of blood flow, is about 4 m/s in the aorta, 8 m/s in the large arteries, and 16 m/s in the small arteries of young adults. Consequently, the pulse is felt in the radial artery at the wrist about 0.1 s after the peak of systolic ejection into the aorta (Figure 31–3). With advancing age, the arteries become more rigid, and the pulse wave moves faster. FIGURE 31–2 Pressure–volume loop of the left ventricle. During diastole, the ventricle fills and pressure increases from d to a. Pressure then rises sharply from a to b during isovolumetric con-traction and from b to c during ventricular ejection. At c, the aortic valves close and pressure falls during isovolumetric relaxation from c back to d. (Reproduced with permission from McPhee SJ, Lingappa VR, Ganong WF [editors]: Pathophysiology of Disease, 4th ed. McGraw-Hill, 2003.) c d a b Diastolic pressure-volume relationship Isovolumic pressure-volume curve Pressure (mm Hg) 200 100 50 130 Volume (mL) 0 CHAPTER 31 The Heart as a Pump 511 FIGURE 31–3 Events of the cardiac cycle at a heart rate of 75 beats/min. The phases of the cardiac cycle identified by the numbers at the bottom are as follows: 1, atrial systole; 2, isovolumetric ventricular contraction; 3, ventricular ejection; 4, isovolumetric ventricular relaxation; 5, ventricular filling. Note that late in systole, aortic pressure actually exceeds left ventricular pressure. However, the momentum of the blood keeps it flowing out of the ventricle for a short period. The pressure relationships in the right ventricle and pulmonary artery are similar. Atr. syst., atrial systole; Ventric. syst., ventricular systole. Time (s) Electrocardiogram Heart sounds (phonocardiogram) Aortic pressure (at o, the aortic valve opens; at c, it closes) Left ventricular pressure ( ) Left atrial pressure ( ) (right is similar) Left ventricular volume (at c’, the mitral valve closes; at o’, it opens) Jugular venous pressure, showing a, c, and v waves Carotid pressure (n = dicrotic notch) Radial pressure Plumonary arterial pressure Right ventricular pressure 0 Atr. syst. Ventric. syst. Diastole 0.2 P Q 4 1 o c' o' a 1 2 3 4 5 c v n c 2 3 S R T U 0.4 0.6 0.8 120 Pressure (mm Hg) Ventricular volume (mL) Aortic blood flow (L/min) Pressure (mm Hg) 130 65 25 15 0 30 15 0 0 80 40 0 Phases of cardiac cycle 512 SECTION VI Cardiovascular Physiology The strength of the pulse is determined by the pulse pressure and bears little relation to the mean pressure. The pulse is weak (“thready”) in shock. It is strong when stroke volume is large; for example, during exercise or after the administration of his-tamine. When the pulse pressure is high, the pulse waves may be large enough to be felt or even heard by the individual (pal-pitation, “pounding heart”). When the aortic valve is incompe-tent (aortic insufficiency), the pulse is particularly strong, and the force of systolic ejection may be sufficient to make the head nod with each heartbeat. The pulse in aortic insufficiency is called a collapsing, Corrigan, or water-hammer pulse. The dicrotic notch, a small oscillation on the falling phase of the pulse wave caused by vibrations set up when the aortic valve snaps shut (Figure 31–3), is visible if the pressure wave is recorded but is not palpable at the wrist. The pulmonary artery pressure curve also has a dicrotic notch produced by the closure of the pulmonary valves. ATRIAL PRESSURE CHANGES & THE JUGULAR PULSE Atrial pressure rises during atrial systole and continues to rise during isovolumetric ventricular contraction when the AV valves bulge into the atria. When the AV valves are pulled down by the contracting ventricular muscle, pressure falls rapidly and then rises as blood flows into the atria until the AV valves open early in diastole. The return of the AV valves to their relaxed po-sition also contributes to this pressure rise by reducing atrial ca-pacity. The atrial pressure changes are transmitted to the great veins, producing three characteristic waves in the record of jug-ular pressure (Figure 31–3). The a wave is due to atrial systole. As noted above, some blood regurgitates into the great veins when the atria contract. In addition, venous inflow stops, and the resultant rise in venous pressure contributes to the a wave. The c wave is the transmitted manifestation of the rise in atrial pres-sure produced by the bulging of the tricuspid valve into the atria during isovolumetric ventricular contraction. The v wave mir-rors the rise in atrial pressure before the tricuspid valve opens during diastole. The jugular pulse waves are superimposed on the respiratory fluctuations in venous pressure. Venous pressure falls during inspiration as a result of the increased negative intra-thoracic pressure and rises again during expiration. HEART SOUNDS Two sounds are normally heard through a stethoscope during each cardiac cycle. The first is a low, slightly prolonged “lub” (first sound), caused by vibrations set up by the sudden clo-sure of the AV valves at the start of ventricular systole (Figure 31–3). The second is a shorter, high-pitched “dup” (second sound), caused by vibrations associated with closure of the aortic and pulmonary valves just after the end of ventricular systole. A soft, low-pitched third sound is heard about one third of the way through diastole in many normal young indi-viduals. It coincides with the period of rapid ventricular filling and is probably due to vibrations set up by the inrush of blood. A fourth sound can sometimes be heard immediately before the first sound when atrial pressure is high or the ventricle is stiff in conditions such as ventricular hypertrophy. It is due to ventricular filling and is rarely heard in normal adults. The first sound has a duration of about 0.15 s and a fre-quency of 25 to 45 Hz. It is soft when the heart rate is low, because the ventricles are well filled with blood and the leaflets of the AV valves float together before systole. The second sound lasts about 0.12 s, with a frequency of 50 Hz. It is loud and sharp when the diastolic pressure in the aorta or pulmonary artery is elevated, causing the respective valves to shut briskly at the end of systole. The interval between aortic and pulmonary valve closure during inspiration is frequently long enough for the second sound to be reduplicated (physiologic splitting of the second sound). Splitting also occurs in various diseases. The third sound, when present, has a duration of 0.1 s. MURMURS Murmurs, or bruits, are abnormal sounds heard in various parts of the vascular system. The two terms are used inter-changeably, though “murmur” is more commonly used to de-note noise heard over the heart than over blood vessels. As discussed in detail in Chapter 32, blood flow is laminar, non-turbulent, and silent up to a critical velocity; above this veloc-ity and beyond an obstruction, blood flow is turbulent and creates sounds. Blood flow speeds up when an artery or a heart valve is narrowed. Examples of vascular sounds outside the heart are the bruit heard over a large, highly vascular goiter, the bruit heard over a carotid artery when its lumen is narrowed and distorted by atherosclerosis, and the murmurs heard over an aneurysmal dilation of one of the large arteries, an arteriovenous (A-V) fistula, or a patent ductus arteriosus. TABLE 31–1 Variation in length of action potential and associated phenomena with cardiac rate.a Heart Rate 75/min Heart Rate 200/min Skeletal Muscle Duration, each cardiac cycle 0.80 0.30 … Duration of systole 0.27 0.16 … Duration of action potential 0.25 0.15 0.007 Duration of absolute refractory period 0.20 0.13 0.004 Duration of relative refractory period 0.05 0.02 0.003 Duration of diastole 0.53 0.14 … aAll values are in seconds. Courtesy of AC Barger and GS Richardson. CHAPTER 31 The Heart as a Pump 513 The major—but certainly not the only—cause of cardiac mur-murs is disease of the heart valves. When the orifice of a valve is narrowed (stenosis), blood flow through it is accelerated and turbulent. When a valve is incompetent, blood flows through it backward (regurgitation or insufficiency), again through a narrow orifice that accelerates flow. The timing (systolic or dia-stolic) of a murmur due to any particular valve (Table 31–2) can be predicted from a knowledge of the mechanical events of the cardiac cycle. Murmurs due to disease of a particular valve can generally be heard best when the stethoscope is directly over the valve. There are also other aspects of the duration, character, accentuation, and transmission of the sound that help to locate its origin in one valve or another. One of the loudest murmurs is that produced when blood flows backward in diastole through a hole in a cusp of the aortic valve. Most murmurs can be heard only with the aid of the stethoscope, but this high-pitched musi-cal diastolic murmur is sometimes audible to the unaided ear several feet from the patient. In patients with congenital interventricular septal defects, flow from the left to the right ventricle causes a systolic mur-mur. Soft murmurs may also be heard in patients with intera-trial septal defects, although they are not a constant finding. Soft systolic murmurs are also common in individuals, espe-cially children, who have no cardiac disease. Systolic murmurs are also heard in anemic patients as a result of the low viscosity of the blood and associated rapid flow (see Chapter 32). ECHOCARDIOGRAPHY Wall movement and other aspects of cardiac function can be evaluated by the noninvasive technique of echocardiography. Pulses of ultrasonic waves are emitted from a transducer that also functions as a receiver to detect waves reflected back from various parts of the heart. Reflections occur wherever acoustic impedance changes, and a recording of the echoes displayed against time on an oscilloscope provides a record of the movements of the ven-tricular wall, septum, and valves during the cardiac cycle. When combined with Doppler techniques, echocardiography can be used to measure velocity and volume of flow through valves. It has considerable clinical usefulness, particularly in evaluating and planning therapy in patients with valvular lesions. CARDIAC OUTPUT METHODS OF MEASUREMENT In experimental animals, cardiac output can be measured with an electromagnetic flow meter placed on the ascending aorta. Two methods of measuring output that are applicable to hu-mans, in addition to Doppler combined with echocardiography, are the direct Fick method and the indicator dilution method. The Fick principle states that the amount of a substance taken up by an organ (or by the whole body) per unit of time is equal to the arterial level of the substance minus the venous level (A-V difference) times the blood flow. This principle can be applied, of course, only in situations in which the arte-rial blood is the sole source of the substance taken up. The principle can be used to determine cardiac output by measur-ing the amount of O2 consumed by the body in a given period and dividing this value by the A-V difference across the lungs. Because systemic arterial blood has the same O2 content in all parts of the body, the arterial O2 content can be measured in a sample obtained from any convenient artery. A sample of venous blood in the pulmonary artery is obtained by means of a cardiac catheter. It has now become commonplace to insert a long catheter through a forearm vein and to guide its tip into the heart with the aid of a fluoroscope. The procedure is gen-erally benign. Catheters can be inserted through the right atrium and ventricle into the small branches of the pulmonary artery. An example of the calculation of cardiac output using a typical set of values is as follows: Output of left ventricle = O2 consumption (mL/min) [A02] – [V02] = 250 mL/min 190 mL/L arterial blood – 140 mL/L venous blood in pulmonary artery = 250 mL/min 50 mL/L = 5 L/min In the indicator dilution technique, a known amount of a substance such as a dye or, more commonly, a radioactive iso-tope is injected into an arm vein and the concentration of the indicator in serial samples of arterial blood is determined. The output of the heart is equal to the amount of indicator injected divided by its average concentration in arterial blood after a single circulation through the heart (Figure 31–4). The indicator must, of course, be a substance that stays in the bloodstream during the test and has no harmful or hemody-namic effects. In practice, the log of the indicator concentra-tion in the serial arterial samples is plotted against time as the concentration rises, falls, and then rises again as the indicator recirculates. The initial decline in concentration, linear on a semilog plot, is extrapolated to the abscissa, giving the time for first passage of the indicator through the circulation. The cardiac output for that period is calculated (Figure 31–4) and then converted to output per minute. TABLE 31–2 Heart murmurs. Valve Abnormality Timing of Murmur Aortic or pulmonary Stenosis Systolic Insufficiency Diastolic Mitral or tricuspid Stenosis Diastolic Insufficiency Systolic 514 SECTION VI Cardiovascular Physiology A popular indicator dilution technique is thermodilution, in which the indicator used is cold saline. The saline is injected into the right atrium through one channel of a dou-ble-lumen catheter, and the temperature change in the blood is recorded in the pulmonary artery, using a thermistor in the other, longer side of the catheter. The temperature change is inversely proportionate to the amount of blood flowing through the pulmonary artery; that is, to the extent that the cold saline is diluted by blood. This technique has two impor-tant advantages: (1) the saline is completely innocuous; and (2) the cold is dissipated in the tissues so recirculation is not a problem, and it is easy to make repeated determinations. CARDIAC OUTPUT IN VARIOUS CONDITIONS The amount of blood pumped out of the heart per beat, the stroke volume, is about 70 mL from each ventricle in a resting man of average size in the supine position. The output of the heart per unit of time is the cardiac output. In a resting, supine man, it averages about 5.0 L/min (70 mL × 72 beats/min). There is a correlation between resting cardiac output and body surface area. The output per minute per square meter of body surface (the cardiac index) averages 3.2 L. The effects of various condi-tions on cardiac output are summarized in Table 31–3. FACTORS CONTROLLING CARDIAC OUTPUT Predictably, changes in cardiac output that are called for by physiologic conditions can be produced by changes in cardiac rate or stroke volume or both (Figure 31–5). The cardiac rate is controlled primarily by the autonomic nerves, with sympathetic stimulation increasing the rate and parasympathetic stimula-tion decreasing it (see Chapter 30). Stroke volume is also deter-mined in part by neural input, with sympathetic stimuli making the myocardial muscle fibers contract with greater strength at any given length and parasympathetic stimuli having the oppo-site effect. When the strength of contraction increases without an increase in fiber length, more of the blood that normally re-mains in the ventricles is expelled; that is, the ejection fraction increases. The cardiac accelerator action of the catecholamines liberated by sympathetic stimulation is referred to as their chro-notropic action, whereas their effect on the strength of cardiac contraction is called their inotropic action. FIGURE 31–4 Determination of cardiac output by indicator (dye) dilution. 5.0 4.0 3.0 2.0 1.0 0.8 0.6 0.4 0.3 0.2 0.1 0 4 8 12 16 20 Time (s) 24 28 32 36 Rest Exercise mg/L F = F = flow E = amount of indicator injected C = instantaneous concentration of indicator in arterial blood In the rest example above, Cdt α ο ∫ E Flow in 39 s (time of first passage) = 5 mg injection 1.6 mg/L (avg concentration) Flow = 3.1 L in 39 s Flow (cardiac output)/min = 3.1 × = 4.7 L 60 39 For the exercise example, Flow in 9 s = = 3.3 L 5 mg 1.51 mg/L Flow/min = 3.3 × = 22.0 L 60 9 TABLE 31–3 Effect of various conditions on cardiac output. Condition or Factora No change Sleep Moderate changes in environmental temperature Increase Anxiety and excitement (50–100%) Eating (30%) Exercise (up to 700%) High environmental temperature Pregnancy Epinephrine Decrease Sitting or standing from lying position (20–30%) Rapid arrhythmias Heart disease aApproximate percent changes are shown in parentheses. CHAPTER 31 The Heart as a Pump 515 The force of contraction of cardiac muscle depends on its preloading and its afterloading. These factors are illustrated in Figure 31–6, in which a muscle strip is stretched by a load (the preload) that rests on a platform. The initial phase of the con-traction is isometric; the elastic component in series with the contractile element is stretched, and tension increases until it is sufficient to lift the load. The tension at which the load is lifted is the afterload. The muscle then contracts isotonically without developing further tension. In vivo, the preload is the degree to which the myocardium is stretched before it contracts and the afterload is the resistance against which blood is expelled. RELATION OF TENSION TO LENGTH IN CARDIAC MUSCLE The length–tension relationship in cardiac muscle (see Figure 5–17) is similar to that in skeletal muscle (see Figure 5–11); as the muscle is stretched, the developed tension increases to a maximum and then declines as stretch becomes more extreme. Starling pointed this out when he stated that the “energy of contraction is proportional to the initial length of the cardiac muscle fiber” (Starling’s law of the heart or the Frank–Star-ling law). For the heart, the length of the muscle fibers (ie, the extent of the preload) is proportional to the end-diastolic vol-ume. The relation between ventricular stroke volume and end-diastolic volume is called the Frank–Starling curve. Regulation of cardiac output as a result of changes in cardiac muscle fiber length is sometimes called heterometric regula-tion, whereas regulation due to changes in contractility inde-pendent of length is sometimes called homometric regulation. FACTORS AFFECTING END-DIASTOLIC VOLUME Alterations in systolic and diastolic function have different ef-fects on the heart. When systolic contractions are reduced, there is a primary reduction in stroke volume. Diastolic func-tion also affects stroke volume, but in a different way. An increase in intrapericardial pressure limits the extent to which the ventricle can fill (eg, as a result of infection or pres-sure from a tumor), as does a decrease in ventricular compli-ance; that is, an increase in ventricular stiffness produced by myocardial infarction, infiltrative disease, and other abnormali-ties. Atrial contractions aid ventricular filling. Factors affecting the amount of blood returning to the heart likewise influence the degree of cardiac filling during diastole. An increase in total blood volume increases venous return (Clinical Box 31–2). Constriction of the veins reduces the size of the venous reser-voirs, decreasing venous pooling and thus increasing venous return. An increase in the normal negative intrathoracic pres-sure increases the pressure gradient along which blood flows to the heart, whereas a decrease impedes venous return. Standing decreases venous return, and muscular activity increases it as a result of the pumping action of skeletal muscle. The effects of systolic and diastolic dysfunction on the pressure–volume loop of the left ventricle are summarized in Figure 31–7. MYOCARDIAL CONTRACTILITY The contractility of the myocardium exerts a major influence on stroke volume. When the sympathetic nerves to the heart are stimulated, the whole length–tension curve shifts upward and to the left (Figure 31–8). The positive inotropic effect of norepi-nephrine liberated at the nerve endings is augmented by circu-lating norepinephrine, and epinephrine has a similar effect. Conversely, there is a negative inotropic effect of vagal stimula-tion on both atrial and (to a lesser extent) ventricular muscle. Changes in cardiac rate and rhythm also affect myocardial contractility (known as the force–frequency relation, Figure 31–8). Ventricular extrasystoles condition the myocardium in such a way that the next succeeding contraction is stronger FIGURE 31–5 Interactions between the components that regulate cardiac output and arterial pressure. Solid arrows indicate increases, and the dashed arrow indicates a decrease. FIGURE 31–6 Model for contraction of afterloaded muscles. A: Rest. B: Partial contraction of the contractile element of the muscle (CE), with stretching of the series elastic element (SE) but no shorten-ing. C: Complete contraction, with shortening. (Reproduced with permission from Sonnenblick EH in: The Myocardial Cell: Structure, Function and Modification. Briller SA, Conn HL [editors]. University Pennsylvania Press, 1966.) Arterial pressure Peripheral resistance Cardiac output Stroke volume Heart rate Myocardial fiber shortening Left ventricular size Afterload Preload Contractility Shortening Tension Load Time Stimulation CE L L = Load SE A CE L SE CE L SE B C 516 SECTION VI Cardiovascular Physiology than the preceding normal contraction. This postextrasystolic potentiation is independent of ventricular filling, since it occurs in isolated cardiac muscle and is due to increased availability of intracellular Ca2+. A sustained increment in contractility can be produced therapeutically by delivering paired electrical stimuli to the heart in such a way that the second stimulus is delivered shortly after the refractory period of the first. It has also been shown that myocardial contractility increases as the heart rate increases, although this effect is relatively small. Catecholamines exert their inotropic effect via an action on cardiac β1-adrenergic receptors and Gs, with resultant activation of adenylyl cyclase and increased intracellular cyclic adenosine CLINICAL BOX 31–2 Shock Circulatory shock comprises a collection of different entities that share certain common features; however, the feature that is common to all the entities is inadequate tissue perfusion with a relatively or absolutely inadequate cardiac output. The cardiac output may be inadequate because the amount of fluid in the vascular system is inadequate to fill it (hypovolemic shock). Alternatively, it may be inadequate in the relative sense because the size of the vascular system is increased by vasodi-lation even though the blood volume is normal (distributive, vasogenic, or low-resistance shock). Shock may also be caused by inadequate pumping action of the heart as a result of myocardial abnormalities (cardiogenic shock), and by inad-equate cardiac output as a result of obstruction of blood flow in the lungs or heart (obstructive shock). Hypovolemic shock is also called “cold shock.” It is character-ized by hypotension; a rapid, thready pulse; cold, pale, clammy skin; intense thirst; rapid respiration; and restlessness or, alterna-tively, torpor. None of these findings, however, are invariably present. Hypovolemic shock is commonly subdivided into cate-gories on the basis of cause. Of these, it is useful to consider the effects of hemorrhage in some detail because of the multiple compensatory reactions that come into play to defend extracel-lular fluid (ECF) volume. Thus, the decline in blood volume pro-duced by bleeding decreases venous return, and cardiac output falls. The heart rate is increased, and with severe hemorrhage, a fall in blood pressure always occurs. With moderate hemor-rhage (5–15 mL/kg body weight), pulse pressure is reduced but mean arterial pressure may be normal. The blood pressure changes vary from individual to individual, even when exactly the same amount of blood is lost. The skin is cool and pale and may have a grayish tinge because of stasis in the capillaries and a small amount of cyanosis. Inadequate perfusion of the tissues leads to increased anaerobic glycolysis, with the production of large amounts of lactic acid. In severe cases, the blood lactate level rises from the normal value of about 1 mmol/L to 9 mmol/ L or more. The resulting lactic acidosis depresses the myocar-dium, decreases peripheral vascular responsiveness to catechol-amines, and may be severe enough to cause coma. When blood volume is reduced and venous return is decreased, moreover, stimulation of arterial baroreceptors is reduced, increasing sympathetic output. Even if there is no drop in mean arterial pressure, the decrease in pulse pressure decreases the rate of discharge in the arterial baroreceptors, and reflex tachycardia and vasoconstriction result. With more severe blood loss, tachycardia is replaced by bradycardia; this occurs while shock is still reversible. With even greater hemorrhage, the heart rate rises again. The bradycardia is presumably due to unmasking a vagally medi-ated depressor reflex, and the response may have evolved as a mechanism for stopping further blood loss. Vasoconstriction is generalized, sparing only the vessels of the brain and heart. A widespread reflex venoconstriction also helps maintain the fill-ing pressure of the heart. In the kidneys, both afferent and ef-ferent arterioles are constricted, but the efferent vessels are constricted to a greater degree. The glomerular filtration rate is depressed, but renal plasma flow is decreased to a greater ex-tent, so that the filtration fraction increases. Na+ retention is marked, and the nitrogenous products of metabolism are re-tained in the blood (azotemia or uremia). If the hypotension is prolonged, renal tubular damage may be severe (acute renal failure). After a moderate hemorrhage, the circulating plasma volume is restored in 12 to 72 h. Preformed albumin also enters rapidly from extravascular stores, but most of the tissue fluids that are mobilized are protein-free. After the initial influx of preformed albumin, the rest of the plasma protein losses are replaced, presumably by hepatic synthesis, over a period of 3 to 4 d. Erythropoietin appears in the circulation, and the reticu-locyte count increases, reaching a peak in 10 d. The red cell mass is restored to normal in 4 to 8 wk. The treatment of shock is aimed at correcting the cause and helping the physiologic compensatory mechanisms to restore an adequate level of tissue perfusion. If the primary cause of the shock is blood loss, the treatment should include early and rapid transfusion of adequate amounts of compatible whole blood. In shock due to burns and other conditions in which there is hemoconcentration, plasma is the treatment of choice to re-store the fundamental defect, the loss of plasma. Concentrated human serum albumin and other hypertonic solutions expand the blood volume by drawing fluid out of the interstitial spaces. They are valuable in emergency treatment but have the disad-vantage of further dehydrating the tissues of an already dehy-drated patient. CHAPTER 31 The Heart as a Pump 517 3',5'-monophosphate (cAMP). Xanthines such as caffeine and theophylline that inhibit the breakdown of cAMP are predictably positively inotropic. The positively inotropic effect of digitalis and related drugs (Figure 31–8), on the other hand, is due to their inhibitory effect on the Na+–K+ ATPase in the myocardium (see Chapter 5). Hypercapnia, hypoxia, acidosis, and drugs such as quinidine, procainamide, and barbiturates depress myocardial contractility. The contractility of the myocardium is also reduced in heart failure (intrinsic depression). The causes of this depres-sion are not fully understood, but may reflect down-regulation of β-adrenergic receptors and associated signaling pathways and impaired calcium liberation from the sarcoplasmic reticulum. In acute heart failure, such as that associated with sepsis, this response could be considered an appropriate adaptation to a situ-ation where energy supply to the heart is limited, thereby reduc-ing energy expenditure and avoiding cell death. FIGURE 31–7 Effect of systolic and diastolic dysfunction on the pressure–volume loop of the left ventricle. Left: Systolic dysfunction shifts the isovolumic pressure–volume curve (see Figure 31–2) to the right, decreasing the stroke volume from b–c to b'–c'. Right: Diastolic dys-function increases end-diastolic volume and shifts the diastolic pressure–volume relationship upward and to the left. This reduces the stroke vol-ume from b–c to b'–c'. (Reproduced with permission from McPhee SJ, Lingappa VR, Ganong WF [editors]: Pathophysiology of Disease, 4th ed. McGraw-Hill, 2003.) 200 100 50 130 c b' d' d a a' c' b 200 100 c c' b' a' a d' d b 50 130 Pressure (mm Hg) Volume (mL) Pressure (mm Hg) Volume (mL) 0 0 FIGURE 31–8 Effect of changes in myocardial contractility on the Frank–Starling curve. The curve shifts downward and to the right as contractility is decreased. The major factors influencing contractility are summarized on the right. The dashed lines indicate portions of the ven-tricular function curves where maximum contractility has been exceeded; that is, they identify points on the “descending limb” of the Frank– Starling curve. EDV, end-diastolic volume. (Reproduced with permission from Braunwald E, Ross J, Sonnenblick EH: Mechanisms of contraction of the normal and failing heart. N Engl J Med 1967;277:794. Courtesy of Little, Brown.) Ventricular EDV Force-frequency relation Contractile state of myocardium Loss of myocardium Digitalis, other inotropic agents Circulating catecholamines Pharmacologic depressants Intrinsic depression Hypoxia Hypercapnia Acidosis Sympathetic and parasympathetic nerve impulses Stroke volume 518 SECTION VI Cardiovascular Physiology INTEGRATED CONTROL OF CARDIAC OUTPUT The mechanisms listed above operate in an integrated way to maintain cardiac output. For example, during muscular exer-cise, there is increased sympathetic discharge, so that myocar-dial contractility is increased and the heart rate rises. The increase in heart rate is particularly prominent in normal in-dividuals, and there is only a modest increase in stroke volume (see Table 31–4 and Clinical Box 31–3). However, patients with transplanted hearts are able to increase their cardiac out-put during exercise in the absence of cardiac innervation through the operation of the Frank–Starling mechanism (Figure 31–9). Circulating catecholamines also contribute. If venous return increases and there is no change in sympathetic tone, venous pressure rises, diastolic inflow is greater, ventricular end-diastolic pressure increases, and the heart muscle con-tracts more forcefully. During muscular exercise, venous re-turn is increased by the pumping action of the muscles and the increase in respiration (see Chapter 33). In addition, because of vasodilation in the contracting muscles, peripheral resis-tance and, consequently, afterload are decreased. The end CLINICAL BOX 31–3 Circulatory Changes during Exercise The blood flow of resting skeletal muscle is low (2–4 mL/100 g/ min). When a muscle contracts, it compresses the vessels in it if it develops more than 10% of its maximal tension; when it de-velops more than 70% of its maximal tension, blood flow is completely stopped. Between contractions, however, flow is so greatly increased that blood flow per unit of time in a rhythmi-cally contracting muscle is increased as much as 30-fold. Local mechanisms maintaining a high blood flow in exercising mus-cle include a fall in tissue PO2, a rise in tissue PCO2, and accumu-lation of K+ and other vasodilator metabolites. The tempera-ture rises in active muscle, and this further dilates the vessels. Dilation of the arterioles and precapillary sphincters causes a 10- to 100-fold increase in the number of open capillaries. The average distance between the blood and the active cells—and the distance O2 and metabolic products must diffuse—is thus greatly decreased. The dilation increases the cross-sectional area of the vascular bed, and the velocity of flow therefore decreases. The systemic cardiovascular response to exercise that pro-vides for the additional blood flow to contracting muscle de-pends on whether the muscle contractions are primarily iso-metric or primarily isotonic with the performance of external work. With the start of an isometric muscle contraction, the heart rate rises, probably as a result of psychic stimuli acting on the medulla oblongata. The increase is largely due to de-creased vagal tone, although increased discharge of the car-diac sympathetic nerves plays some role. Within a few seconds of the onset of an isometric muscle contraction, systolic and diastolic blood pressures rise sharply. Stroke volume changes relatively little, and blood flow to the steadily contracting mus-cles is reduced as a result of compression of their blood vessels. The response to exercise involving isotonic muscle contraction is similar in that there is a prompt increase in heart rate, but dif-ferent in that a marked increase in stroke volume occurs. In ad-dition, there is a net fall in total peripheral resistance due to vasodilation in exercising muscles. Consequently, systolic blood pressure rises only moderately, whereas diastolic pres-sure usually remains unchanged or falls. The difference in response to isometric and isotonic exer-cise is explained in part by the fact that the active muscles are tonically contracted during isometric exercise and conse-quently contribute to increased total peripheral resistance. Cardiac output is increased during isotonic exercise to values that may exceed 35 L/min, the amount being proportionate to the increase in O2 consumption. The maximal heart rate achieved during exercise decreases with age. In children, it rises to 200 or more beats/min; in adults it rarely exceeds 195 beats/min, and in elderly individuals the rise is even smaller. Both at rest and at any given level of exercise, trained ath-letes have a larger stroke volume and lower heart rate than untrained individuals and they tend to have larger hearts. Training increases the maximal oxygen consumption (VO2max) that can be produced by exercise in an individual. VO2max averages about 38 mL/kg/min in active healthy men and about 29 mL/kg/min in active healthy women. It is lower in sedentary individuals. VO2max is the product of maximal cardiac output and maximal O2 extraction by the tissues, and both increase with training. A great increase in venous return also takes place with ex-ercise, although the increase in venous return is not the pri-mary cause of the increase in cardiac output. Venous return is increased by the activity of the muscle and thoracic pumps; by mobilization of blood from the viscera; by increased pres-sure transmitted through the dilated arterioles to the veins; and by noradrenergically mediated venoconstriction, which decreases the volume of blood in the veins. Blood mobilized from the splanchnic area and other reservoirs may increase the amount of blood in the arterial portion of the circulation by as much as 30% during strenuous exercise. After exercise, the blood pressure may transiently drop to subnormal levels, presumably because accumulated metabolites keep the mus-cle vessels dilated for a short period. However, the blood pressure soon returns to the pre-exercise level. The heart rate returns to normal more slowly. CHAPTER 31 The Heart as a Pump 519 result in both normal and transplanted hearts is thus a prompt and marked increase in cardiac output. One of the differences between untrained individuals and trained athletes is that the athletes have lower heart rates, greater end-systolic ventricular volumes, and greater stroke volumes at rest. Therefore, they can potentially achieve a given increase in cardiac output by further increases in stroke volume without increasing their heart rate to as great a degree as an untrained individual. OXYGEN CONSUMPTION BY THE HEART The basal O2 consumption by the myocardium is about 2 mL/ 100 g/min. This value is considerably higher than that of rest-ing skeletal muscle. O2 consumption by the beating heart is about 9 mL/100 g/min at rest. Increases occur during exercise and in a number of different states. Cardiac venous O2 tension is low, and little additional O2 can be extracted from the blood in the coronaries, so increases in O2 consumption require in-creases in coronary blood flow. The regulation of coronary flow is discussed in Chapter 34. O2 consumption by the heart is determined primarily by the intramyocardial tension, the contractile state of the myo-cardium, and the heart rate. Ventricular work per beat corre-lates with O2 consumption. The work is the product of stroke volume and mean arterial pressure in the pulmonary artery or the aorta (for the right and left ventricle, respectively). Because aortic pressure is 7 times greater than pulmonary artery pressure, the stroke work of the left ventricle is approx-imately 7 times the stroke work of the right. In theory, a 25% increase in stroke volume without a change in arterial pres-sure should produce the same increase in O2 consumption as a 25% increase in arterial pressure without a change in stroke volume. However, for reasons that are incompletely under-stood, pressure work produces a greater increase in O2 con-sumption than volume work. In other words, an increase in afterload causes a greater increase in cardiac O2 consumption than does an increase in preload. This is why angina pectoris due to deficient delivery of O2 to the myocardium is more common in aortic stenosis than in aortic insufficiency. In aor-tic stenosis, intraventricular pressure must be increased to force blood through the stenotic valve, whereas in aortic insufficiency, regurgitation of blood produces an increase in stroke volume with little change in aortic impedance. It is worth noting that the increase in O2 consumption pro-duced by increased stroke volume when the myocardial fibers are stretched is an example of the operation of the law of Laplace. This law, which is discussed in detail in Chapter 32, states that the tension developed in the wall of a hollow viscus is proportionate to the radius of the viscus, and the radius of a dilated heart is increased. O2 consumption per unit time increases when the heart rate is increased by sympathetic stimulation because of the increased number of beats and the increased velocity and strength of each contraction. However, this is somewhat offset by the decrease in end-systolic volume and hence in the radius of the heart. TABLE 31–4 Changes in cardiac function with exercise. Note that stroke volume levels off, then falls somewhat (as a result of the shortening of diastole) when the heart rate rises to high values. Work (kg-m/min) O2 Usage (mL/min) Pulse Rate (per min) Cardiac Output (L/min) Stroke Volume (mL) A-V O2 Difference (mL/dL) Rest 267 64 6.4 100 4.3 288 910 104 13.1 126 7.0 540 1430 122 15.2 125 9.4 900 2143 161 17.8 110 12.3 1260 3007 173 20.9 120 14.5 Reproduced with permission from Asmussen E, Nielsen M: The cardiac output in rest and work determined by the acetylene and the dye injection methods. Acta Physiol Scand 1952;27:217. FIGURE 31–9 Cardiac responses to moderate supine exercise in normal humans and patients with transplanted and hence denervated hearts. (Reproduced with permission from Kent KM, Cooper T: The denervated heart. N Engl J Med 1974;291:1017.) Normal Transplant O2 consumption Cardiac output Heart rate Stroke volume Exercise Time Exercise 520 SECTION VI Cardiovascular Physiology CHAPTER SUMMARY ■Blood flows into the atria and then the ventricles of the heart during diastole and atrial systole, and is ejected during systole when the ventricles contract and pressure exceeds the pressures in the pulmonary artery and aorta. ■Careful timing of the opening and closing of the atrioventricular (AV), pulmonary, and aortic valves allows blood to move in an ap-propriate direction through the heart with minimal regurgitation. ■The proportion of blood leaving the ventricles in each cardiac cycle is called the ejection fraction and is a sensitive indicator of cardiac health. ■The arterial pulse represents a pressure wave set up when blood is forced into the aorta; it travels much faster than the blood itself. ■Heart sounds reflect the normal vibrations set up by abrupt valve closures; heart murmurs can arise from abnormal flow often (although not exclusively) caused by diseased valves. ■Changes in cardiac output reflect variations in heart rate, stroke volume, or both; these are controlled, in turn, by neural and hor-monal input to cardiac myocytes. ■Cardiac output is strikingly increased during exercise. ■In heart failure, the ejection fraction of the heart is reduced due to impaired contractility in systole or reduced filling during diastole; this results in inadequate blood supplies to meet the body’s needs. Initially, this is manifested only during exercise, but eventually the heart will not be able to supply sufficient blood flow even at rest. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. The second heart sound is caused by A) closure of the aortic and pulmonary valves. B) vibrations in the ventricular wall during systole. C) ventricular filling. D) closure of the mitral and tricuspid valves. E) retrograde flow in the vena cava. 2. The fourth heart sound is caused by A) closure of the aortic and pulmonary valves. B) vibrations in the ventricular wall during systole. C) ventricular filling. D) closure of the mitral and tricuspid valves. E) retrograde flow in the vena cava. 3. The dicrotic notch on the aortic pressure curve is caused by A) closure of the mitral valve. B) closure of the tricuspid valve. C) closure of the aortic valve. D) closure of the pulmonary valve. E) rapid filling of the left ventricle. 4. During exercise, a man consumes 1.8 L of oxygen per minute. His arterial O2 content is 190 mL/L, and the O2 content of his mixed venous blood is 134 mL/L. His cardiac output is approxi-mately A) 3.2 L/min. B) 16 L/min. C) 32 L/min. D) 54 L/min. E) 160 mL/min. 5. The work performed by the left ventricle is substantially greater than that performed by the right ventricle, because in the left ventricle A) the contraction is slower. B) the wall is thicker. C) the stroke volume is greater. D) the preload is greater. E) the afterload is greater. 6. Starling’s law of the heart A) does not operate in the failing heart. B) does not operate during exercise. C) explains the increase in heart rate produced by exercise. D) explains the increase in cardiac output that occurs when venous return is increased. E) explains the increase in cardiac output when the sympa-thetic nerves supplying the heart are stimulated. CHAPTER RESOURCES Leach JK, Priola DV, Grimes LA, Skipper BJ: Shortening deactivation of cardiac muscle: Physiological mechanisms and clinical implications. J Investig Med 1999;47:369. Overgaard CB, Dzavik V: Inotropes and vasopressors: Review of physiology and clinical use in cardiovascular disease. Circulation 2008;118:1047. Rudiger A., Singer M: Mechanisms of sepsis-induced cardiac dysfunction. Crit Care Med 2007;35:1599. 521 C H A P T E R 6 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the components of blood and lymph, their origins, and the role of hemo-globin in transporting oxygen in red blood cells. ■Understand the molecular basis of blood groups and the reasons for transfusion reactions. ■Delineate the process of hemostasis that restricts blood loss when vessels are damaged, and the adverse consequences of intravascular thrombosis. ■Identify the types of blood and lymphatic vessels that make up the circulatory system and the regulation and function of their primary constituent cell types. ■Describe how physical principles dictate the flow of blood and lymph around the body. ■Understand the basis of methods used to measure blood flow and blood pressure in various vascular segments. ■Understand the basis of disease states where components of the blood and vascu-lature are abnormal, dysregulated, or both. INTRODUCTION The circulatory system supplies O2 and substances absorbed from the gastrointestinal tract to the tissues, returns CO2 to the lungs and other products of metabolism to the kidneys, func-tions in the regulation of body temperature, and distributes hormones and other agents that regulate cell function. The blood, the carrier of these substances, is pumped through a closed system of blood vessels by the heart. From the left ventri-cle, blood is pumped through the arteries and arterioles to the capillaries, where it equilibrates with the interstitial fluid. The capillaries drain through venules into the veins and back to the right atrium. Some tissue fluids enter another system of closed vessels, the lymphatics, which drain lymph via the thoracic duct and the right lymphatic duct into the venous system. The circu-lation is controlled by multiple regulatory systems that function in general to maintain adequate capillary blood flow when pos-sible in all organs, but particularly in the heart and brain. Blood flows through the circulation primarily because of the forward motion imparted to it by the pumping of the heart, although in the case of the systemic circulation, dia-stolic recoil of the walls of the arteries, compression of the veins by skeletal muscles during exercise, and the negative pressure in the thorax during inspiration also move the blood forward. The resistance to flow depends to a minor degree on the viscosity of the blood but mostly on the diam-eter of the vessels, principally the arterioles. The blood flow to each tissue is regulated by local chemical and general neu-ral and humoral mechanisms that dilate or constrict the ves-sels of the tissue. All the blood flows through the lungs, but the systemic circulation is made up of numerous different circuits in parallel (Figure 32–1). The arrangement permits wide variations in regional blood flow without changing total systemic flow. 522 SECTION VI Cardiovascular Physiology This chapter is concerned with blood and lymph and with the multiple functions of the cells they contain. It will also address general principles that apply to all parts of the circula-tion and with pressure and flow in the systemic circulation. The homeostatic mechanisms operating to adjust flow are the subject of Chapter 33. The special characteristics of pulmo-nary and renal circulation are discussed in Chapters 35 and 38. Likewise, the role of blood as the carrier of many immune effector cells will not be discussed here, but rather will be cov-ered in Chapter 33. BLOOD AS A CIRCULATORY FLUID Blood consists of a protein-rich fluid known as plasma, in which are suspended cellular elements: white blood cells, red blood cells, and platelets. The normal total circulating blood volume is about 8% of the body weight (5600 mL in a 70-kg man). About 55% of this volume is plasma. BONE MARROW In the adult, red blood cells, many white blood cells, and plate-lets are formed in the bone marrow. In the fetus, blood cells are also formed in the liver and spleen, and in adults such ex-tramedullary hematopoiesis may occur in diseases in which the bone marrow becomes destroyed or fibrosed. In children, blood cells are actively produced in the marrow cavities of all the bones. By age 20, the marrow in the cavities of the long bones, except for the upper humerus and femur, has become inactive (Figure 32–2). Active cellular marrow is called red marrow; inactive marrow that is infiltrated with fat is called yellow marrow. The bone marrow is actually one of the largest organs in the body, approaching the size and weight of the liver. It is also one of the most active. Normally, 75% of the cells in the mar-row belong to the white blood cell-producing myeloid series and only 25% are maturing red cells, even though there are over 500 times as many red cells in the circulation as there are white cells. This difference in the marrow reflects the fact that the average life span of white cells is short, whereas that of red cells is long. Hematopoietic stem cells (HSCs) are bone marrow cells that are capable of producing all types of blood cells. They dif-ferentiate into one or another type of committed stem cells (progenitor cells). These in turn form the various differenti-ated types of blood cells. There are separate pools of progenitor cells for megakaryocytes, lymphocytes, erythrocytes, eosino-phils, and basophils; neutrophils and monocytes arise from a common precursor. The bone marrow stem cells are also the source of osteoclasts (see Chapter 23), Kupffer cells (see Chap-ter 29), mast cells, dendritic cells, and Langerhans cells. The HSCs are few in number but are capable of completely replac-ing the bone marrow when injected into a host whose own bone marrow has been completely destroyed. The HSCs are derived from uncommitted, totipotent stem cells that can be stimulated to form any cell in the body. Adults have a few of these, but they are more readily obtained from the blastocysts of embryos. There is not surprisingly immense interest in stem cell research due to its potential to regenerate diseased tissues, but ethical issues are involved, and debate on these issues will undoubtedly continue. WHITE BLOOD CELLS Normally, human blood contains 4000 to 11,000 white blood cells per microliter (Table 32–1). Of these, the granulocytes (polymorphonuclear leukocytes, PMNs) are the most nu-merous. Young granulocytes have horseshoe-shaped nuclei that become multilobed as the cells grow older (Figure 32–3). FIGURE 32–1 Diagram of the circulation in the adult. HEAD, ARMS BRAIN LUNGS RIGHT HEART LEFT HEART LIVER KIDNEYS TRUNK, LEGS SPLEEN, GI TRACT CORONARY VESSELS HEPATIC ARTERY PORTAL VEIN FIGURE 32–2 Changes in red bone marrow cellularity with age. 100% equals the degree of cellularity at birth. (Reproduced with permission from Whitby LEH, Britton CJC: Disorders of the Blood, 10th ed. Churchill Livingstone, 1969.) 100 50 0 10 30 50 70 Vertebra Sternum Rib Femur Tibia Years Cellularity (%) CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 523 Most of them contain neutrophilic granules (neutrophils), but a few contain granules that stain with acidic dyes (eosino-phils), and some have basophilic granules (basophils). The other two cell types found normally in peripheral blood are lymphocytes, which have large round nuclei and scanty cyto-plasm, and monocytes, which have abundant agranular cyto-plasm and kidney-shaped nuclei (Figure 32–3). Acting together, these cells provide the body with powerful defenses against tumors and viral, bacterial, and parasitic infections that was discussed in Chapter 3. PLATELETS Platelets are small, granulated bodies that aggregate at sites of vascular injury. They lack nuclei and are 2–4 μm in diameter (Figure 32–3). There are about 300,000/μL of circulating blood, and they normally have a half-life of about 4 d. The megakaryo-cytes, giant cells in the bone marrow, form platelets by pinching off bits of cytoplasm and extruding them into the circulation. Between 60% and 75% of the platelets that have been extruded from the bone marrow are in the circulating blood, and the re-mainder are mostly in the spleen. Splenectomy causes an in-crease in the platelet count (thrombocytosis). RED BLOOD CELLS The red blood cells (erythrocytes) carry hemoglobin in the circulation. They are biconcave disks (Figure 32–4) that are manufactured in the bone marrow. In mammals, they lose their nuclei before entering the circulation. In humans, they survive in the circulation for an average of 120 d. The average normal red blood cell count is 5.4 million/μL in men and 4.8 million/μL in women. Each human red blood cell is about 7.5 μm in diameter and 2 μm thick, and each contains ap-proximately 29 pg of hemoglobin (Table 32–2). There are thus about 3 × 1013 red blood cells and about 900 g of hemoglobin in the circulating blood of an adult man (Figure 32–5). The feedback control of erythropoiesis by erythropoietin is discussed in Chapter 39, and the role of IL-1, IL-3, IL-6 (inter-leukin), and GM-CSF (granulocyte-macrophage colony-stim-ulating factor) in development of the relevant erythroid stem cells is shown in Figure 32–3. ROLE OF THE SPLEEN The spleen is an important blood filter that removes aged or ab-normal red cells. It also contains many platelets and plays a sig-nificant role in the immune system. Abnormal red cells are removed if they are not as flexible as normal red cells and conse-quently are unable to squeeze through the slits between the endo-thelial cells that line the splenic sinuses (see Clinical Box 32–1). HEMOGLOBIN The red, oxygen-carrying pigment in the red blood cells of vertebrates is hemoglobin, a protein with a molecular weight of 64,450. Hemoglobin is a globular molecule made up of four subunits (Figure 32–6). Each subunit contains a heme moiety conjugated to a polypeptide. Heme is an iron-containing por-phyrin derivative (Figure 32–7). The polypeptides are referred to collectively as the globin portion of the hemoglobin mole-cule. There are two pairs of polypeptides in each hemoglobin molecule. In normal adult human hemoglobin (hemoglobin A), the two polypeptides are called α chains, each of which contains 141 amino acid residues, and β chains, each of which contains 146 amino acid residues. Thus, hemoglobin A is des-ignated α2β2. Not all the hemoglobin in the blood of normal adults is hemoglobin A. About 2.5% of the hemoglobin is he-moglobin A2, in which β chains are replaced by δ chains (α2δ2). The δ chains also contain 146 amino acid residues, but 10 individual residues differ from those in the β chains. There are small amounts of hemoglobin A derivatives closely associated with hemoglobin A that represent glycated hemoglobins. One of these, hemoglobin A1c (HbA1c), has a glucose attached to the terminal valine in each β chain and is of special interest because it increases in the blood of patients with poorly controlled diabetes mellitus (see Chapter 21). REACTIONS OF HEMOGLOBIN Hemoglobin binds O2 to form oxyhemoglobin, O2 attaching to the Fe2+ in the heme. The affinity of hemoglobin for O2 is affected by pH, temperature, and the concentration in the red cells of 2,3-bisphosphoglycerate (2,3-BPG). 2,3-BPG and H+ TABLE 32–1 Normal values for the cellular elements in human blood. Cell Cells/μL (average) Approximate Normal Range Percentage of Total White Cells Total white blood cells 9000 4000–11,000 ... Granulocytes Neutrophils 5400 3000–6000 50–70 Eosinophils 275 150–300 1–4 Basophils 35 0–100 0.4 Lymphocytes 2750 1500–4000 20–40 Monocytes 540 300–600 2–8 Erythrocytes Females 4.8 × 106 . . . . . . Males 5.4 × 106 . . . . . . Platelets 300,000 200,000– 500,000 . . . 524 SECTION VI Cardiovascular Physiology compete with O2 for binding to deoxygenated hemoglobin, decreasing the affinity of hemoglobin for O2 by shifting the positions of the four peptide chains (quaternary structure). The details of the oxygenation and deoxygenation of hemo-globin and the physiologic role of these reactions in O2 trans-port are discussed in Chapter 36. When blood is exposed to various drugs and other oxidizing agents in vitro or in vivo, the ferrous iron (Fe2+) that is nor-mally in the molecule is converted to ferric iron (Fe3+), form-ing methemoglobin. Methemoglobin is dark-colored, and when it is present in large quantities in the circulation, it causes a dusky discoloration of the skin resembling cyanosis (see Chapter 36). Some oxidation of hemoglobin to methemo-globin occurs normally, but an enzyme system in the red cells, the dihydronicotinamide adenine dinucleotide (NADH)-methemoglobin reductase system, converts methemoglobin FIGURE 32–3 Development of various formed elements of the blood from bone marrow cells. Cells below the horizontal line are found in normal peripheral blood. The principal sites of action of erythropoietin (erythro) and the various colony-stimulating factors (CSF) that stimulate the differentiation of the components are indicated. G, granulocyte; M, macrophage; IL, interleukin; thrombo, thrombopoietin; SCF, stem cell factor. IL-1 IL-6 IL-3 Hemopoietic stem cell Bone marrow lymphocyte precursor GM-CSF GM-CSF IL-5 GM-CSF thrombo GM-CSF erythro Megakaryocyte M-CSF G-CSF Neutrophil Eosinophil Basophil Polymorphonuclear cells Tissue macrophage B T Lymphocytes Platelets Red blood cell Juvenile Monocyte Segmented Monocyte Reticulocyte Late normoblast IL-4 IL-3 C o m m i t t e d s t e m c e l l s (progenitor cell) Thymus Bursal equiv. GM-CSF G-CSF SCF CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 525 back to hemoglobin. Congenital absence of this system is one cause of hereditary methemoglobinemia. Carbon monoxide reacts with hemoglobin to form car-bon monoxyhemoglobin (carboxyhemoglobin). The affin-ity of hemoglobin for O2 is much lower than its affinity for carbon monoxide, which consequently displaces O2 on hemoglobin, reducing the oxygen-carrying capacity of blood (see Chapter 36). HEMOGLOBIN IN THE FETUS The blood of the human fetus normally contains fetal hemo-globin (hemoglobin F). Its structure is similar to that of he-moglobin A except that the β chains are replaced by γ chains; that is, hemoglobin F is α2γ2. The γ chains also contain 146 amino acid residues but have 37 that differ from those in the β chain. Fetal hemoglobin is normally replaced by adult hemo-globin soon after birth (Figure 32–8). In certain individuals, it fails to disappear and persists throughout life. In the body, its O2 content at a given PO2 is greater than that of adult hemo-globin because it binds 2,3-BPG less avidly. Hemoglobin F is critical to facilitate movement of O2 from the maternal to the fetal circulation, particularly at later stages of gestation where oxygen demand increases (see Chapter 34). In young embryos there are, in addition, ζ and ε chains, forming Gower 1 FIGURE 32–4 Human red blood cells and fibrin fibrils. Blood was placed on a polyvinyl chloride surface, fixed, and photographed with a scanning electron microscope. Reduced from ×2590. (Courtesy of NF Rodman.) TABLE 32–2 Characteristics of human red cells.a Male Female Hematocrit (Hct) (%) 47 42 Red blood cells (RBC) (106/μL) 5.4 4.8 Hemoglobin (Hb) (g/dL) 16 14 Mean corpuscular volume (MCV) (fL) = Hct × 10 RBC (106/μL) 87 87 Mean corpuscular hemoglobin (MCH) (pg) = Hb × 10 RBC (106/μL) 29 29 Mean corpuscular hemoglobin concentration (MCHC) (g/dL) = Hb × 100 Hct 34 34 Mean cell diameter (MCD) (μm) = Mean diameter of 500 cells in smear 7.5 7.5 aCells with MCVs > 95 fL are called macrocytes; cells with MCVs < 80 fL are called microcytes; cells with MCHs < 25 g/dL are called hypochromic. CLINICAL BOX 32–1 Red Cell Fragility Red blood cells, like other cells, shrink in solutions with an osmotic pressure greater than that of normal plasma. In so-lutions with a lower osmotic pressure they swell, become spherical rather than disk-shaped, and eventually lose their hemoglobin (hemolysis). The hemoglobin of hemolyzed red cells dissolves in the plasma, coloring it red. A 0.9% so-dium chloride solution is isotonic with plasma. When os-motic fragility is normal, red cells begin to hemolyze when suspended in 0.5% saline; 50% lysis occurs in 0.40–0.42% saline, and lysis is complete in 0.35% saline. In hereditary spherocytosis (congenital hemolytic icterus), the cells are spherocytic in normal plasma and hemolyze more readily than normal cells in hypotonic sodium chloride solutions. Abnormal spherocytes are also trapped and destroyed in the spleen, meaning that hereditary spherocytosis is one of the most common causes of hereditary hemolytic ane-mia. The spherocytosis is caused by mutations in proteins that make up the membrane skeleton of the erythrocyte, which normally maintain the shape and flexibility of the red cell membrane, including spectrin, the transmembrane protein band 3, and the linker protein, ankyrin. The condi-tion can be cured by splenectomy, but this is not without other risks. Red cells can also be lysed by drugs and infec-tions. The susceptibility of red cells to hemolysis by these agents is increased by deficiency of the enzyme glucose 6-phosphate dehydrogenase (G6PD), which catalyzes the ini-tial step in the oxidation of glucose via the hexose mono-phosphate pathway (see Chapter 1). This pathway gener-ates dihydronicotinamide adenine dinucleotide phosphate (NADPH), which is needed for the maintenance of normal red cell fragility. Severe G6PD deficiency also inhibits the killing of bacteria by granulocytes and predisposes to se-vere infections. 526 SECTION VI Cardiovascular Physiology hemoglobin (ζ2ε2) and Gower 2 hemoglobin (α2ε2). There are two copies of the α globin gene on human chromosome 16. In addition, there are five globin genes in tandem on chromo-some 11 that encode β, γ, and δ globin chains and the two chains normally found only during fetal life. Switching from one form of hemoglobin to another during development seems to be regulated largely by oxygen availability, with rela-tive hypoxia favoring the production of hemoglobin F both via direct effects on globin gene expression, as well as up-regulat-ed production of erythropoietin. SYNTHESIS OF HEMOGLOBIN The average normal hemoglobin content of blood is 16 g/dL in men and 14 g/dL in women, all of it in red cells. In the body of a 70-kg man, there are about 900 g of hemoglobin, and 0.3 g of hemoglobin is destroyed and 0.3 g synthesized every hour (Figure 32–5). The heme portion of the hemoglobin molecule is synthesized from glycine and succinyl-CoA (see Clinical Box 32–2). CATABOLISM OF HEMOGLOBIN When old red blood cells are destroyed by tissue macrophages, the globin portion of the hemoglobin molecule is split off, and the heme is converted to biliverdin. The enzyme involved is a subtype of heme oxygenase (see Figure 29–4), and CO is formed in the process. CO may be an intercellular messenger, like NO (see Chapters 2 and 3). FIGURE 32–5 Red cell formation and destruction. RBC, red blood cells. Circulation 3 × 1013 red blood cells 900 g hemoglobin 1 × 1010 RBC 0.3 g hemoglobin per hour 1 × 1010 RBC 0.3 g hemoglobin per hour Tissue macrophage system Bone marrow Iron Diet Amino acids Bile pigments in stool, urine Small amount of iron FIGURE 32–6 Diagrammatic representation of a molecule of hemoglobin A, showing the four subunits. There are two α and two β polypeptide chains, each containing a heme moiety. These moieties are represented by the disks. (Reproduced with permission from Harper HA et al: Physiologische Chemie. Springer, 1975.) 1 nm COO + − COO − NH3 + NH3 + NH3 α β α β FIGURE 32–7 Reaction of heme with O2. The abbreviations M, V, and P stand for the groups shown on the molecule on the left. N CH HC N N N N HC H2C CH N CH3 CH3 CH3 CH2 CH2 COOH CH3 CH=CH2 CH=CH2 CH2 COOH Fe Polypeptide chain (imidazole) Heme (imidazole) Deoxygenated hemoglobin N CH HC N N N O2 + O2 HC CH N M V P M M M V P Fe Polypeptide chain Oxyhemoglobin CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 527 In humans, most of the biliverdin is converted to bilirubin (Figure 32–9) and excreted in the bile (see Chapter 29). The iron from the heme is reused for hemoglobin synthesis. Exposure of the skin to white light converts bilirubin to lumirubin, which has a shorter half-life than bilirubin. Photo-therapy (exposure to light) is of value in treating infants with jaundice due to hemolysis. Iron is essential for hemoglobin synthesis; if blood is lost from the body and the iron defi-ciency is not corrected, iron deficiency anemia results. The metabolism of iron is discussed in Chapter 27. BLOOD TYPES The membranes of human red cells contain a variety of blood group antigens, which are also called agglutinogens. The most important and best known of these are the A and B anti-gens, but there are many more. THE ABO SYSTEM The A and B antigens are inherited as mendelian dominants, and individuals are divided into four major blood types on this basis. Type A individuals have the A antigen, type B have the B, type AB have both, and type O have neither. The A and B antigens are complex oligosaccharides that differ in their terminal sugar. An H gene codes for a fucose transferase that adds a terminal fucose, forming the H antigen that is usually present in individuals of all blood types (Figure 32–10). Indi-viduals who are type A also express a second transferase that catalyzes placement of a terminal N-acetylgalactosamine on the H antigen, whereas individuals who are type B express a transferase that places a terminal galactose. Individuals who are type AB have both transferases. Individuals who are type O have neither, so the H antigen persists. Antibodies against red cell agglutinogens are called aggluti-nins. Antigens very similar to A and B are common in intesti-nal bacteria and possibly in foods to which newborn individuals are exposed. Therefore, infants rapidly develop antibodies against the antigens not present in their own cells. Thus, type A individuals develop anti-B antibodies, type B individuals develop anti-A antibodies, type O individuals develop both, and type AB individuals develop neither (Table 32–4). When the plasma of a type A individual is mixed with type B red cells, the anti-B antibodies cause the type B red CLINICAL BOX 32–2 Abnormalities of Hemoglobin Production There are two major types of inherited disorders of hemoglo-bin in humans: the hemoglobinopathies, in which abnor-mal globin polypeptide chains are produced, and the thalas-semias and related disorders, in which the chains are normal in structure but produced in decreased amounts or absent because of defects in the regulatory portion of the globin genes. Mutant genes that cause the production of abnormal hemoglobins are widespread, and over 1000 abnormal he-moglobins have been described in humans. In one of the most common examples, hemoglobin S, the α chains are normal but the β chains have a single substitution of a valine residue for one glutamic acid, leading to sickle cell anemia (Table 32–3). When an abnormal gene inherited from one parent dictates formation of an abnormal hemoglobin (ie, when the individual is heterozygous), half the circulating he-moglobin is abnormal and half is normal. When identical ab-normal genes are inherited from both parents, the individual is homozygous and all the hemoglobin is abnormal. It is the-oretically possible to inherit two different abnormal hemo-globins, one from the father and one from the mother. Stud-ies of the inheritance and geographic distribution of abnormal hemoglobins have made it possible in some cases to decide where the mutant gene originated and approxi-mately how long ago the mutation occurred. In general, harmful mutations tend to die out, but mutant genes that confer traits with survival value persist and spread in the population. Many of the abnormal hemoglobins are harm-less; however, some have abnormal O2 equilibriums, while others cause anemia. For example, hemoglobin S polymer-izes at low O2 tensions, and this causes the red cells to be-come sickle-shaped, hemolyze, and form aggregates that block blood vessels. The sickle cell gene is an example of a gene that has persisted and spread in the population due to its beneficial effect when present in heterozygous form. It originated in Africa, and confers resistance to one type of malaria. In some parts of Africa, 40% of the population is het-erozygous for hemoglobin S. There is a corresponding preva-lence of 10% among African Americans in the United States. Hemoglobin F decreases the polymerization of deoxygen-ated hemoglobin S, and hydroxyurea stimulates production of hemoglobin F in children and adults. It has proved to be a very valuable agent for the treatment of sickle cell disease. In patients with severe sickle cell disease, bone marrow trans-plantation has also been shown to have some benefit. FIGURE 32–8 Development of human hemoglobin chains. 50 40 30 20 10 0 3 6 3 6 Birth α chain δ chain β chain (adult) and ζ chains (embryonic) γ chain (fetal) Gestation (months) Age (months) ∋ Globin chain synthesis (% of total) 528 SECTION VI Cardiovascular Physiology cells to clump (agglutinate), as shown in Figure 32–11. The other agglutination reactions produced by mismatched plasma and red cells are summarized in Table 32–4. Blood typing is performed by mixing an individual’s red blood cells with antisera containing the various agglutinins on a slide and seeing whether agglutination occurs. TRANSFUSION REACTIONS Dangerous hemolytic transfusion reactions occur when blood is transfused into an individual with an incompatible blood type; that is, an individual who has agglutinins against the red cells in the transfusion. The plasma in the transfusion is usually so diluted in the recipient that it rarely causes agglu-tination even when the titer of agglutinins against the recipi-ent’s cells is high. However, when the recipient’s plasma has agglutinins against the donor’s red cells, the cells agglutinate and hemolyze. Free hemoglobin is liberated into the plasma. The severity of the resulting transfusion reaction may vary from an asymptomatic minor rise in the plasma bilirubin level to severe jaundice and renal tubular damage leading to anuria and death. Incompatibilities in the ABO blood group system are sum-marized in Table 32–4. Persons with type AB blood are “uni-versal recipients” because they have no circulating agglutinins and can be given blood of any type without developing a trans-fusion reaction due to ABO incompatibility. Type O individu-als are “universal donors” because they lack A and B antigens, and type O blood can be given to anyone without producing a transfusion reaction due to ABO incompatibility. This does not mean, however, that blood should ever be transfused with-out being cross-matched except in the most extreme emergen-cies, since the possibility of reactions or sensitization due to incompatibilities in systems other than ABO systems always TABLE 32–3 Partial amino acid composition of normal human β chain, and some hemoglobins with abnormal β chains.a Positions on Polypeptide Chain of Hemoglobin Hemoglobin 1 2 3 6 7 26 63 67 121 146 A (normal) Val-His-Leu Glu-Glu Glu His Val Glu His S (sickle cell) Val C Lys GSan Jose Gly E Lys MSaskatoon Tyr MMilwaukee Glu OArabia Lys aOther hemoglobins have abnormal α chains. Abnormal hemoglobins that are very similar electrophoretically but differ slightly in composition are indicated by the same let-ter and a subscript indicating the geographic location where they were first discovered; hence, MSaskatoon and MMilwaukee. FIGURE 32–9 Bilirubin. The abbreviations M, V, and P stand for the groups shown on the molecule on the left in Figure 32–7. FIGURE 32–10 Antigens of the ABO system on the surface of red blood cells. N OH OH HC NH H N H2C CH N M V M V P M M P F G G G C = fucose = glucose = N-acetylgalactosamine = galactose = lipid bilayer = ceramide F G G G G C H antigen F G G G G C G A antigen F G G G G G C B antigen CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 529 exists. In cross-matching, donor red cells are mixed with recip-ient plasma on a slide and checked for agglutination. It is advisable to check the action of the donor’s plasma on the recipient cells in addition, even though, as noted above, this is rarely a source of trouble. A procedure that has recently become popular is to with-draw the patient’s own blood in advance of elective surgery and then infuse this blood back (autologous transfusion) if a transfusion is needed during the surgery. With iron treatment, 1000 to 1500 mL can be withdrawn over a 3-wk period. The popularity of banking one’s own blood is due primarily to fear of transmission of infectious diseases by heterologous transfu-sions, but of course another advantage is elimination of the risk of transfusion reactions. INHERITANCE OF A & B ANTIGENS The A and B antigens are inherited as mendelian allelomor-phs, A and B being dominants. For example, an individual with type B blood may have inherited a B antigen from each parent or a B antigen from one parent and an O from the oth-er; thus, an individual whose phenotype is B may have the genotype BB (homozygous) or BO (heterozygous). When the blood types of the parents are known, the possi-ble genotypes of their children can be stated. When both par-ents are type B, they could have children with genotype BB (B antigen from both parents), BO (B antigen from one parent, O from the other heterozygous parent), or OO (O antigen from both parents, both being heterozygous). When the blood types of a mother and her child are known, typing can prove that a man cannot be the father, although it cannot prove that he is the father. The predictive value is increased if the blood typing of the parties concerned includes identifica-tion of antigens other than the ABO agglutinogens. With the use of DNA fingerprinting (see Chapter 1), the exclusion rate for paternity rises to close to 100%. FIGURE 32–11 Red cell agglutination in incompatible plasma. Anti-B Anti-A Type A Type B Type AB TABLE 32–4 Summary of ABO system. Blood Type Agglutinins in Plasma Frequency in United States (%) Plasma Agglutinates Red Cells of Type: O Anti-A, anti-B 45 A, B, AB A Anti-B 41 B, AB B Anti-A 10 A, AB AB None 4 None 530 SECTION VI Cardiovascular Physiology OTHER AGGLUTINOGENS In addition to the ABO system of antigens in human red cells, there are systems such as the Rh, MNSs, Lutheran, Kell, Kidd, and many others. There are over 500 billion possible known blood group phenotypes, and because undiscovered antigens undoubtedly exist, it has been calculated that the number of phenotypes is actually in the trillions. The number of blood groups in animals is as large as it is in humans. An interesting question is why this degree of poly-morphism developed and has persisted through evolution. Certain diseases are more common in individuals with one blood type or another, but the differences are not great. One, the Duffy antigen, is a chemokine receptor. Many of the oth-ers seem to be cell recognition molecules, but the significance of a recognition code of this complexity is unknown. THE RH GROUP Aside from the antigens of the ABO system, those of the Rh system are of the greatest clinical importance. The Rh factor, named for the rhesus monkey because it was first studied us-ing the blood of this animal, is a system composed primarily of the C, D, and E antigens, although it actually contains many more. Unlike the ABO antigens, the system has not been de-tected in tissues other than red cells. D is by far the most anti-genic component, and the term Rh-positive as it is generally used means that the individual has agglutinogen D. The D protein is not glycosylated, and its function is unknown. The Rh-negative individual has no D antigen and forms the anti-D agglutinin when injected with D-positive cells. The Rh typ-ing serum used in routine blood typing is anti-D serum. Eighty-five percent of Caucasians are D-positive and 15% are D-negative; over 99% of Asians are D-positive. Unlike the an-tibodies of the ABO system, anti-D antibodies do not develop without exposure of a D-negative individual to D-positive red cells by transfusion or entrance of fetal blood into the mater-nal circulation. However, D-negative individuals who have received a transfusion of D-positive blood (even years previ-ously) can have appreciable anti-D titers and thus may devel-op transfusion reactions when transfused again with D-positive blood. HEMOLYTIC DISEASE OF THE NEWBORN Another complication due to Rh incompatibility arises when an Rh-negative mother carries an Rh-positive fetus. Small amounts of fetal blood leak into the maternal circulation at the time of delivery, and some mothers develop significant titers of anti-Rh agglutinins during the postpartum period. During the next pregnancy, the mother’s agglutinins cross the placen-ta to the fetus. In addition, there are some cases of fetal–ma-ternal hemorrhage during pregnancy, and sensitization can occur during pregnancy. In any case, when anti-Rh aggluti-nins cross the placenta to an Rh-positive fetus, they can cause hemolysis and various forms of hemolytic disease of the new-born (erythroblastosis fetalis). If hemolysis in the fetus is se-vere, the infant may die in utero or may develop anemia, severe jaundice, and edema (hydrops fetalis). Kernicterus, a neurologic syndrome in which unconjugated bilirubin is de-posited in the basal ganglia, may also develop, especially if birth is complicated by a period of hypoxia. Bilirubin rarely penetrates the brain in adults, but it does in infants with eryth-roblastosis, possibly in part because the blood–brain barrier is more permeable in infancy. However, the main reasons that the concentration of unconjugated bilirubin is very high in this condition are that production is increased and the biliru-bin-conjugating system is not yet mature. About 50% of Rh-negative individuals are sensitized (develop an anti-Rh titer) by transfusion of Rh-positive blood. Because sensitization of Rh-negative mothers by carrying an Rh-positive fetus generally occurs at birth, the first child is usu-ally normal. However, hemolytic disease occurs in about 17% of the Rh-positive fetuses born to Rh-negative mothers who have previously been pregnant one or more times with Rh-positive fetuses. Fortunately, it is usually possible to prevent sensitiza-tion from occurring the first time by administering a single dose of anti-Rh antibodies in the form of Rh immune globulin during the postpartum period. Such passive immunization does not harm the mother and has been demonstrated to pre-vent active antibody formation by the mother. In obstetric clin-ics, the institution of such treatment on a routine basis to unsensitized Rh-negative women who have delivered an Rh-positive baby has reduced the overall incidence of hemolytic disease by more than 90%. In addition, fetal Rh typing with material obtained by amniocentesis or chorionic villus sam-pling is now possible, and treatment with a small dose of Rh immune serum will prevent sensitization during pregnancy. PLASMA The fluid portion of the blood, the plasma, is a remarkable so-lution containing an immense number of ions, inorganic mol-ecules, and organic molecules that are in transit to various parts of the body or aid in the transport of other substances. Normal plasma volume is about 5% of body weight, or roughly 3500 mL in a 70-kg man. Plasma clots on standing, remaining fluid only if an anticoagulant is added. If whole blood is al-lowed to clot and the clot is removed, the remaining fluid is called serum. Serum has essentially the same composition as plasma, except that its fibrinogen and clotting factors II, V, and VIII (Table 32–5) have been removed and it has a higher serotonin content because of the breakdown of platelets dur-ing clotting. PLASMA PROTEINS The plasma proteins consist of albumin, globulin, and fibrin-ogen fractions. Most capillary walls are relatively imperme-able to the proteins in plasma, and the proteins therefore exert CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 531 an osmotic force of about 25 mm Hg across the capillary wall (oncotic pressure; see Chapter 1) that pulls water into the blood. The plasma proteins are also responsible for 15% of the buffering capacity of the blood (see Chapter 39) because of the weak ionization of their substituent COOH and NH2 groups. At the normal plasma pH of 7.40, the proteins are mostly in the anionic form (see Chapter 1). Plasma proteins may have specific functions (eg, antibodies and the proteins concerned with blood clotting), whereas others function as carriers for various hormones, other solutes, and drugs. ORIGIN OF PLASMA PROTEINS Circulating antibodies are manufactured by lymphocytes. Most of the other plasma proteins are synthesized in the liver. These proteins and their principal functions are listed in Table 32–6. Data on the turnover of albumin show that its synthesis plays an important role in the maintenance of normal levels. In normal adult humans, the plasma albumin level is 3.5 to 5.0 g/dL, and the total exchangeable albumin pool is 4.0 to 5.0 g/ kg body weight; 38–45% of this albumin is intravascular, and much of the rest of it is in the skin. Between 6% and 10% of the exchangeable pool is degraded per day, and the degraded albumin is replaced by hepatic synthesis of 200 to 400 mg/kg/ d. The albumin is probably transported to the extravascular areas by vesicular transport across the walls of the capillaries (see Chapter 2). Albumin synthesis is carefully regulated. It is decreased during fasting and increased in conditions such as nephrosis in which there is excessive albumin loss. HYPOPROTEINEMIA Plasma protein levels are maintained during starvation until body protein stores are markedly depleted. However, in pro-longed starvation and in malabsorption syndromes due to intes-tinal diseases, plasma protein levels are low (hypoproteinemia). They are also low in liver disease, because hepatic protein syn-thesis is depressed, and in nephrosis, because large amounts of albumin are lost in the urine. Because of the decrease in the plas-ma oncotic pressure, edema tends to develop. Rarely, there is congenital absence of one or another plasma protein. An exam-ple of congenital protein deficiency is the congenital form of afibrinogenemia, characterized by defective blood clotting. HEMOSTASIS Hemostasis is the process of forming clots in the walls of dam-aged blood vessels and preventing blood loss while maintaining blood in a fluid state within the vascular system. A collection of complex interrelated systemic mechanisms operates to main-tain a balance between coagulation and anticoagulation. RESPONSE TO INJURY When a small blood vessel is transected or damaged, the injury initiates a series of events (Figure 32–12) that lead to the for-mation of a clot. This seals off the damaged region and pre-vents further blood loss. The initial event is constriction of the vessel and formation of a temporary hemostatic plug of plate-lets that is triggered when platelets bind to collagen and aggre-gate. This is followed by conversion of the plug into the definitive clot. The constriction of an injured arteriole or small artery may be so marked that its lumen is obliterated, at least temporarily. The vasoconstriction is due to serotonin and oth-er vasoconstrictors liberated from platelets that adhere to the walls of the damaged vessels. THE CLOTTING MECHANISM The loose aggregation of platelets in the temporary plug is bound together and converted into the definitive clot by fibrin. Fibrin formation involves a cascade of enzymatic reactions and a series of numbered clotting factors (Table 32–5). The funda-mental reaction is conversion of the soluble plasma protein fibrinogen to insoluble fibrin (Figure 32–13). The process TABLE 32–5 System for naming blood-clotting factors. Factora Names I Fibrinogen II Prothrombin III Thromboplastin IV Calcium V Proaccelerin, labile factor, accelerator globulin VII Proconvertin, SPCA, stable factor VIII Antihemophilic factor (AHF), antihemophilic factor A, antihemophilic globulin (AHG) IX Plasma thromboplastic component (PTC), Christmas factor, antihemophilic factor B X Stuart–Prower factor XI Plasma thromboplastin antecedent (PTA), antihemo-philic factor C XII Hageman factor, glass factor XIII Fibrin-stabilizing factor, Laki–Lorand factor HMW-K High-molecular-weight kininogen, Fitzgerald factor Pre-Ka Prekallikrein, Fletcher factor Ka Kallikrein PL Platelet phospholipid aFactor VI is not a separate entity and has been dropped. 532 SECTION VI Cardiovascular Physiology involves the release of two pairs of polypeptides from each fi-brinogen molecule. The remaining portion, fibrin monomer, then polymerizes with other monomer molecules to form fi-brin. The fibrin is initially a loose mesh of interlacing strands. It is converted by the formation of covalent cross-linkages to a dense, tight aggregate (stabilization). This latter reaction is cat-alyzed by activated factor XIII and requires Ca2+. The conversion of fibrinogen to fibrin is catalyzed by thrombin. Thrombin is a serine protease that is formed from its circulating precursor, prothrombin, by the action of acti-vated factor X. It has additional actions, including activation of platelets, endothelial cells, and leukocytes via so-called pro-teinase activated receptors, which are G protein-coupled. Factor X can be activated by either of two systems, known as intrinsic and extrinsic (Figure 32–13). The initial reaction in the intrinsic system is conversion of inactive factor XII to active factor XII (XIIa). This activation, which is catalyzed by high-molecular-weight kininogen and kallikrein (see Chapter 33), can be brought about in vitro by exposing the blood to glass, or in vivo by collagen fibers underlying the endothelium. Active factor XII then activates factor XI, and active factor XI activates factor IX. Activated factor IX forms a complex with TABLE 32–6 Some of the proteins synthesized by the liver: Physiologic functions and properties. Name Principal Function Binding Characteristics Serum or Plasma Concentration Albumin Binding and carrier protein; osmotic regulator Hormones, amino acids, steroids, vita-mins, fatty acids 4500–5000 mg/dL Orosomucoid Uncertain; may have a role in inflammation Trace; rises in inflammation α1-Antiprotease Trypsin and general protease inhibitor Proteases in serum and tissue secretions 1.3–1.4 mg/dL α-Fetoprotein Osmotic regulation; binding and carrier proteina Hormones, amino acids Found normally in fetal blood α2-Macroglobulin Inhibitor of serum endoproteases Proteases 150–420 mg/dL Antithrombin-III Protease inhibitor of intrinsic coagulation system 1:1 binding to proteases 17–30 mg/dL Ceruloplasmin Transport of copper Six atoms copper/mol 15–60 mg/dL C-reactive protein Uncertain; has role in tissue inflammation Complement C1q < 1 mg/dL; rises in inflamma-tion Fibrinogen Precursor to fibrin in hemostasis 200–450 mg/dL Haptoglobin Binding, transport of cell-free hemoglobin Hemoglobin 1:1 binding 40–180 mg/dL Hemopexin Binds to porphyrins, particularly heme for heme recycling 1:1 with heme 50–100 mg/dL Transferrin Transport of iron Two atoms iron/mol 3.0–6.5 mg/dL Apolipoprotein B Assembly of lipoprotein particles Lipid carrier Angiotensinogen Precursor to pressor peptide angiotensin II Proteins, coagulation factors II, VII, IX, X Blood clotting 20 mg/dL Antithrombin C, protein C Inhibition of blood clotting Insulinlike growth factor I Mediator of anabolic effects of growth hor-mone IGF-I receptor Steroid hormone-binding globulin Carrier protein for steroids in bloodstream Steroid hormones 3.3 mg/dL Thyroxine-binding globulin Carrier protein for thyroid hormone in bloodstream Thyroid hormones 1.5 mg/dL Transthyretin (thyroid-binding prealbumin) Carrier protein for thyroid hormone in bloodstream Thyroid hormones 25 mg/dL aThe function of alpha-fetoprotein is uncertain, but because of its structural homology to albumin it is often assigned these functions. CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 533 active factor VIII, which is activated when it is separated from von Willebrand factor. The complex of IXa and VIIIa activate factor X. Phospholipids from aggregated platelets (PL) and Ca2+ are necessary for full activation of factor X. The extrinsic system is triggered by the release of tissue thromboplastin, a protein–phospholipid mixture that activates factor VII. Tissue thromboplastin and factor VII activate factors IX and X. In the presence of PL, Ca2+, and factor V, activated factor X catalyzes the conversion of prothrombin to thrombin. The extrinsic pathway is inhibited by a tissue factor pathway inhibitor that forms a quaternary structure with tissue thromboplastin (TPL), factor VIIa, and factor Xa. ANTICLOTTING MECHANISMS The tendency of blood to clot is balanced in vivo by reactions that prevent clotting inside the blood vessels, break down any clots that do form, or both. These reactions include the interac-tion between the platelet-aggregating effect of thromboxane A2 and the antiaggregating effect of prostacyclin, which causes clots to form at the site when a blood vessel is injured but keeps the vessel lumen free of clot (see Chapter 33 and Clinical Box 32–3). Antithrombin III is a circulating protease inhibitor that binds to serine proteases in the coagulation system, blocking their activity as clotting factors. This binding is facilitated by heparin, a naturally occurring anticoagulant that is a mixture of sulfated polysaccharides with molecular weights averaging 15,000–18,000. The clotting factors that are inhibited are the active forms of factors IX, X, XI, and XII. The endothelium of the blood vessels also plays an active role in preventing the extension of clots. All endothelial cells except those in the cerebral microcirculation produce throm-bomodulin, a thrombin-binding protein, on their surfaces. In circulating blood, thrombin is a procoagulant that activates factors V and VIII, but when it binds to thrombomodulin, it becomes an anticoagulant in that the thrombomodulin– thrombin complex activates protein C (Figure 32–14). Acti-vated protein C (APC), along with its cofactor protein S, inac-tivates factors V and VIII and inactivates an inhibitor of tissue plasminogen activator, increasing the formation of plasmin. Plasmin (fibrinolysin) is the active component of the plas-minogen (fibrinolytic) system (Figure 32–14). This enzyme lyses fibrin and fibrinogen, with the production of fibrinogen degradation products (FDP) that inhibit thrombin. Plasmin is formed from its inactive precursor, plasminogen, by the action of thrombin and tissue-type plasminogen activator (t-PA). It is also activated by urokinase-type plasminogen activator (u-PA). If the t-PA gene or the u-PA gene is knocked out in mice, some fibrin deposition occurs and clot lysis is slowed. However, when both are knocked out, spontaneous fibrin deposition is extensive. Human plasminogen consists of a 560-amino-acid heavy chain and a 241-amino-acid light chain. The heavy chain, with glutamate at its amino terminal, is folded into five loop structures, each held together by three disulfide bonds (Fig-ure 32–15). These loops are called kringles because of their FIGURE 32–12 Summary of reactions involved in hemostasis. The dashed arrow indicates inhibition. (Modified from Deykin D: Thrombogenesis, N Engl J Med 1967;267:622.) Temporary hemostatic plug Definitive hemostatic plug Injury to wall of blood vessel Contraction Collagen Tissue thromboplastin Platelet reactions Loose platelet aggregation Activation of coagulation Thrombin Limiting reactions FIGURE 32–13 The clotting mechanism. a, active form of clot-ting factor. TPL, tissue thromboplastin; TFI, tissue factor pathway in-hibitor. For other abbreviations, see Table 32–5. INTRINSIC SYSTEM HMW kininogen Kallikrein XII XIIa HMW kininogen XI XIa EXTRINSIC SYSTEM TFI VII VIIA IX IXa PL Ca2+ VIIIa Ca2+ PL TPL VIII X Xa PL Ca2+ Va V Prothrombin Thrombin Fibrinogen Fibrin Stabilization XIIIa XIII TPL 534 SECTION VI Cardiovascular Physiology resemblance to a Danish pastry of the same name. The kringles are lysine-binding sites by which the molecule attaches to fibrin and other clot proteins, and they are also found in prothrom-bin. Plasminogen is converted to active plasmin when t-PA hydrolyzes the bond between Arg 560 and Val 561. Plasminogen receptors are located on the surfaces of many different types of cells and are plentiful on endothelial cells. When plasminogen binds to its receptors, it becomes acti-vated, so intact blood vessel walls are provided with a mecha-nism that discourages clot formation. Human t-PA is now produced by recombinant DNA tech-niques for clinical use in myocardial infarction and stroke. Streptokinase, a bacterial enzyme, is also fibrinolytic and is also used in the treatment of early myocardial infarction (see Chapter 34). CLINICAL BOX 32–3 Abnormalities of Hemostasis In addition to clotting abnormalities due to platelet disorders, hemorrhagic diseases can be produced by selective deficien-cies of most of the clotting factors (Table 32–7). Hemophilia A, which is caused by factor VIII deficiency, is relatively com-mon. The disease has been treated with factor VIII-rich prepa-rations made from plasma, or, more recently, factor VIII pro-duced by recombinant DNA techniques. von Willebrand factor deficiency likewise causes a bleeding disorder (von Willebrand disease) by reducing platelet adhesion and by lowering plasma factor VIII. The condition can be congenital or acquired. The large von Willebrand molecule is subject to cleavage and resulting inactivation by the plasma metallo-protease ADAM 13 in vascular areas where fluid shear stress is elevated. Finally, when absorption of vitamin K is depressed along with absorption of other fat-soluble vitamins (see Chapter 27), the resulting clotting factor deficiencies may cause the development of a significant bleeding tendency. Formation of clots inside blood vessels is called thrombo-sis to distinguish it from the normal extravascular clotting of blood. Thromboses are a major medical problem. They are particularly prone to occur where blood flow is sluggish be-cause the slow flow permits activated clotting factors to accu-mulate instead of being washed away. They also occur in ves-sels when the intima is damaged by atherosclerotic plaques, and over areas of damage to the endocardium. They fre-quently occlude the arterial supply to the organs in which they form, and bits of thrombus (emboli) sometimes break off and travel in the bloodstream to distant sites, damaging other organs. An example is obstruction of the pulmonary ar-tery or its branches by thrombi from the leg veins (pulmo-nary embolism). Congenital absence of protein C leads to uncontrolled intravascular coagulation and, in general, death in infancy. If this condition is diagnosed and treatment is insti-tuted, the coagulation defect disappears. Resistance to acti-vated protein C is another cause of thrombosis, and this con-dition is common. It is due to a point mutation in the gene for factor V, which prevents activated protein C from inactivating the factor. Mutations in protein S and antithrombin III may less commonly increase the incidence of thrombosis. Disseminated intravascular coagulation is another seri-ous complication of septicemia, extensive tissue injury, and other diseases in which fibrin is deposited in the vascular sys-tem and many small- and medium-sized vessels are throm-bosed. The increased consumption of platelets and coagula-tion factors causes bleeding to occur at the same time. The cause of the condition appears to be increased generation of thrombin due to increased TPL activity without adequate tis-sue factor inhibitory pathway activity. FIGURE 32–14 The fibrinolytic system and its regulation by protein C. FIGURE 32–15 Structure of human plasminogen. Note the Glu at the amino terminal, the Asn at the carboxyl terminal, and five uniquely shaped loop structures (kringles). Hydrolysis by t-PA at the ar-row separates the carboxyl terminal light chain from the amino terminal heavy chain but leaves the disulfide bonds intact. This activates the mol-ecule. (Modified and reproduced with permission from Bachman F, in: Thrombosis and Hemostasis. Verstraete M et al [editors]. Leuven University Press, 1987.) Endothelial cell Thrombomodulin Thrombin Protein C Activated protein C (APC) VIlIa Inactive VIIIa Va Inactive Va Inactivates inhibitor of tissue plasminogen activator (t-PA) Plasminogen Plasmin Lyses fibrin + Protein S Thrombin t-PA, u-PA t-PA Asn Glu Ser CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 535 ANTICOAGULANTS As noted above, heparin is a naturally occurring anticoagulant that facilitates the action of antithrombin III. Low-molecular-weight fragments with an average molecular weight of 5000 have been produced from unfractionated heparin, and these low-molecular-weight heparins are seeing increased clinical use because they have a longer half-life and produce a more predict-able anticoagulant response than unfractionated heparin. The highly basic protein protamine forms an irreversible complex with heparin and is used clinically to neutralize heparin. In vivo, a plasma Ca2+ level low enough to interfere with blood clotting is incompatible with life, but clotting can be pre-vented in vitro if Ca2+ is removed from the blood by the addi-tion of substances such as oxalates, which form insoluble salts with Ca2+, or chelating agents, which bind Ca2+. Coumarin derivatives such as dicumarol and warfarin are also effective anticoagulants. They inhibit the action of vitamin K, which is a necessary cofactor for the enzyme that catalyzes the conversion of glutamic acid residues to γ-carboxyglutamic acid residues. Six of the proteins involved in clotting require conversion of a num-ber of glutamic acid residues to γ-carboxyglutamic acid residues before being released into the circulation, and hence all six are vitamin K-dependent. These proteins are factors II (prothrom-bin), VII, IX, and X, protein C, and protein S (see above). LYMPH Lymph is tissue fluid that enters the lymphatic vessels. It drains into the venous blood via the thoracic and right lym-phatic ducts. It contains clotting factors and clots on standing in vitro. In most locations, it also contains proteins that traverse capillary walls and return to the blood via the lymph. Its protein content is generally lower than that of plasma, which contains about 7 g/dL, but lymph protein content varies with the region from which the lymph drains (Table 32–8). Water-insoluble fats are absorbed from the intestine into the lymphatics, and the lymph in the thoracic duct after a meal is milky because of its high fat content (see Chapter 27). Lym-phocytes enter the circulation principally through the lym-phatics, and there are appreciable numbers of lymphocytes in thoracic duct lymph. STRUCTURAL FEATURES OF THE CIRCULATION Here, we will first describe the two major cell types that make up the blood vessels and then how they are arranged into the various vessel types that subserve the needs of the circulation. ENDOTHELIUM Located between the circulating blood and the media and ad-ventitia of the blood vessels, the endothelial cells constitute a large and important organ. They respond to flow changes, stretch, a variety of circulating substances, and inflammatory mediators. They secrete growth regulators and vasoactive sub-stances (see below and Chapter 33). TABLE 32–7 Examples of diseases due to deficiency of clotting factors. Deficiency of Factor: Clinical Syndrome Cause I Afibrinogenemia Depletion during pregnancy with premature separation of placenta; also congenital (rare) II Hypoprothrom-binemia (hemor-rhagic tendency in liver disease) Decreased hepatic synthesis, usually secondary to vitamin K deficiency V Parahemophilia Congenital VII Hypoconvertinemia Congenital VIII Hemophilia A (classic hemophilia) Congenital defect due to vari-ous abnormalities of the gene on X chromosome that codes for factor VIII; disease is there-fore inherited as sex-linked characteristic IX Hemophilia B (Christmas disease) Congenital X Stuart–Prower factor deficiency Congenital XI PTA deficiency Congenital XII Hageman trait Congenital TABLE 32–8 Probable approximate protein content of lymph in humans. Source of Lymph Protein Content (g/dL) Choroid plexus 0 Ciliary body 0 Skeletal muscle 2 Skin 2 Lung 4 Gastrointestinal tract 4.1 Heart 4.4 Liver 6.2 Data largely from JN Diana. 536 SECTION VI Cardiovascular Physiology VASCULAR SMOOTH MUSCLE The smooth muscle in blood vessel walls has been one of the most-studied forms of visceral smooth muscle because of its importance in the regulation of blood pressure and hyperten-sion. The membranes of the muscle cells contain various types of K+, Ca2+, and Cl– channels. Contraction is produced pri-marily by the myosin light chain mechanism described in Chapter 5. However, vascular smooth muscle also undergoes prolonged contractions that determine vascular tone. These may be due in part to the latch-bridge mechanism (see Chap-ter 5), but other factors also play a role. Some of the molecular mechanisms that appear to be involved in contraction and re-laxation are shown in Figure 32–16. Vascular smooth muscle cells provide an interesting exam-ple of the way high and low cytosolic Ca2+ can have different and even opposite effects (see Chapter 2). In these cells, influx of Ca2+ via voltage-gated Ca2+ channels produces a dif-fuse increase in cytosolic Ca2+ that initiates contraction. However, the Ca2+ influx also initiates Ca2+ release from the sarcoplasmic reticulum via ryanodine receptors (see Chapter 5), and the high local Ca2+ concentration produced by these Ca2+ sparks increases the activity of Ca2+-activated K+ chan-nels in the cell membrane. These are also known as big K or BK channels because K+ flows through them at a high rate. The increased K+ efflux increases the membrane potential, shutting off voltage-gated Ca2+ channels and producing relaxation. The site of action of the Ca2+ sparks is the β1-sub-unit of the BK channel, and mice in which this subunit is knocked out develop increased vascular tone and blood pres-sure. Obviously, therefore, the sensitivity of the β1 subunit to Ca2+ sparks plays an important role in the control of vascular tone. ARTERIES & ARTERIOLES The characteristics of the various types of blood vessels are listed in Table 32–9. The walls of all arteries are made up of an outer layer of connective tissue, the adventitia; a middle layer of smooth muscle, the media; and an inner layer, the intima, made up of the endothelium and underlying connective tissue (Figure 32–17). The walls of the aorta and other arteries of large diameter contain a relatively large amount of elastic tis-sue, primarily located in the inner and external elastic laminas. They are stretched during systole and recoil on the blood dur-ing diastole. The walls of the arterioles contain less elastic tis-sue but much more smooth muscle. The muscle is innervated by noradrenergic nerve fibers, which function as constrictors, and in some instances by cholinergic fibers, which dilate the FIGURE 32–16 Some of the established and postulated mechanisms involved in the contraction and relaxation of vascular smooth muscle. A, agonist; AA, arachidonic acid; BK, Ca+-activated K+ channel; G, heterotrimeric G protein; MLC, myosin light chain; MLCK, myosin light chain kinase; PLD, phospholipase D; R, receptor; SF, sarcoplasmic reticulum; VGCC, voltage-gated Ca2+ channel; RR, ryanodine receptors. For other abbrevia-tions, see Chapter 2 and Appendix. (Modified from Khahl R: Mechanisms of vascular smooth muscle contraction. Council for High Blood Pressure Newsletter, Spring 2001.) VGCC BK AA CaM Ca2+ Ca2+ sparks MLCK active MLCK inactive ADP P -MLC MLC MLC phosphatase ATP PI Contraction Actin CaP- P CaD- P Actin Actin-CaP Actin-CaD MAPK Myosin CPI-17- P CPI-17 Rho-kinase SR A K+ R PC G GTP GDP PLD DAG PS Choline PKC RR PKC Raf MEK Cytoplasm Interstitial fluid CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 537 vessels. The arterioles are the major site of the resistance to blood flow, and small changes in their caliber cause large changes in the total peripheral resistance. CAPILLARIES The arterioles divide into smaller muscle-walled vessels, sometimes called metarterioles, and these in turn feed into capillaries (Figure 32–18). The openings of the capillaries are surrounded on the upstream side by minute smooth muscle precapillary sphincters. It is unsettled whether the metarteri-oles are innervated, and it appears that the precapillary sphincters are not. However, they can of course respond to lo-cal or circulating vasoconstrictor substances. The capillaries are about 5 μm in diameter at the arterial end and 9 μm in di-ameter at the venous end. When the sphincters are dilated, the diameter of the capillaries is just sufficient to permit red blood cells to squeeze through in “single file.” As they pass through the capillaries, the red cells become thimble- or parachute-shaped, with the flow pushing the center ahead of the edges. This configuration appears to be due simply to the pressure in the center of the vessel whether or not the edges of the red blood cell are in contact with the capillary walls. The total area of all the capillary walls in the body exceeds 6300 m2 in the adult. The walls, which are about 1 μm thick, are made up of a single layer of endothelial cells. The structure of the walls varies from organ to organ. In many beds, including those in skeletal, cardiac, and smooth muscle, the junctions between the endothelial cells (Figure 32–19) permit the pas-sage of molecules up to 10 nm in diameter. It also appears that plasma and its dissolved proteins are taken up by endocytosis, transported across the endothelial cells, and discharged by exo-cytosis (vesicular transport; see Chapter 2). However, this pro-cess can account for only a small portion of the transport across the endothelium. In the brain, the capillaries resemble the cap-illaries in muscle, but the junctions between endothelial cells are tighter, and transport across them is largely limited to small molecules. In most endocrine glands, the intestinal villi, and parts of the kidneys, the cytoplasm of the endothelial cells is attenuated to form gaps called fenestrations. These fenestra-tions are 20 to 100 nm in diameter and may permit the passage of larger molecules, although they appear to be closed by a thin membrane. An exception to this, however, is found in the liver, where the sinusoidal capillaries are extremely porous, the endo-thelium is discontinuous, and gaps occur between endothelial cells that are not closed by membranes (see Figure 29–2). Some of the gaps are 600 nm in diameter, and others may be as large TABLE 32–9 Characteristics of various types of blood vessels in humans. All Vessels of Each Type Vessel Lumen Diameter Wall Thickness Approximate Total Cross-Sectional Area (cm2) Percentage of Blood Volume Containeda Aorta 2.5 cm 2 mm 4.5 2 Artery 0.4 cm 1 mm 20 8 Arteriole 30 μm 20 μm 400 1 Capillary 5 μm 1 μm 4500 5 Venule 20 μm 2 μm 4000 Vein 0.5 cm 0.5 mm 40 54 Vena cava 3 cm 1.5 mm 18 aIn systemic vessels; there is an additional 12% in the heart and 18% in the pulmo-nary circulation. FIGURE 32–17 Structure of normal muscle artery. (Reproduced with permission from Ross R, Glomset JA: The pathogenesis of atherosclerosis. N Engl J Med 1976;295:369.) Endothelium Internal elastic lamina External elastic lamina Intima Media Adventitia FIGURE 32–18 The microcirculation. Arterioles give rise to metarterioles, which give rise to capillaries. The capillaries drain via short collecting venules to the venules. The walls of the arteries, arte-rioles, and small venules contain relatively large amounts of smooth muscle. There are scattered smooth muscle cells in the walls of the metarterioles, and the openings of the capillaries are guarded by mus-cular precapillary sphincters. The diameters of the various vessels are also shown. (Courtesy of JN Diana.) Artery > 50 μm Precapillary sphincter Arterial end of capillary 5 μm Metarteriole 10–15 μm True capillary Arteriole 20–50 μm Venous end of capillary 9 μm Small venule 20 μm Collecting venule 538 SECTION VI Cardiovascular Physiology as 3000 nm. They therefore permit the passage of large mole-cules, including plasma proteins, which is important for hepatic function (see Chapter 29). The permeabilities of capillaries in various parts of the body, expressed in terms of their hydraulic conductivity, are summarized in Table 32–10. Capillaries and postcapillary venules have pericytes around their endothelial cells (Figure 32–19). These cells have long processes that wrap around the vessels. They are contractile and release a wide variety of vasoactive agents. They also syn-thesize and release constituents of the basement membrane and extracellular matrix. One of their physiologic functions appears to be regulation of flow through the junctions between endothelial cells, particularly in the presence of inflammation. They are closely related to the mesangial cells in the renal glomeruli (see Chapter 38). LYMPHATICS The lymphatics serve to collect plasma and its constituents that have exuded from the capillaries into the interstitial space. They drain from the body tissues via a system of vessels that coalesce and eventually enter the right and left subclavian veins at their junctions with the respective internal jugular veins. The lymph vessels contain valves and regularly traverse lymph nodes along their course. The ultrastructure of the small lymph vessels differs from that of the capillaries in sev-eral details: No fenestrations are visible in the lymphatic endo-thelium; very little if any basal lamina is present under the endothelium; and the junctions between endothelial cells are open, with no tight intercellular connections. ARTERIOVENOUS ANASTOMOSES In the fingers, palms, and ear lobes, short channels connect ar-terioles to venules, bypassing the capillaries. These arteriove-nous (A-V) anastomoses, or shunts, have thick, muscular walls and are abundantly innervated, presumably by vasocon-strictor nerve fibers. VENULES & VEINS The walls of the venules are only slightly thicker than those of the capillaries. The walls of the veins are also thin and easily distended. They contain relatively little smooth muscle, but considerable venoconstriction is produced by activity in the noradrenergic nerves to the veins and by circulating vasocon-strictors such as endothelins. Variations in venous tone are important in circulatory adjustments. The intima of the limb veins is folded at intervals to form venous valves that prevent retrograde flow. The way these valves function was first demonstrated by William Harvey in the 17th century. No valves are present in the very small veins, the great veins, or the veins from the brain and viscera. FIGURE 32–19 Cross-sections of capillaries. Left: Type of capillary found in muscle. Right: Fenestrated type of capillary. (Reproduced with permission from Fawcett DW: Bloom and Fawcett, Textbook of Histology, 11th ed. Saunders, 1986.) Vesicles Basal lamina Pericyte Pericyte Fenestrations or pores Interdigitated junction TABLE 32–10 Hydraulic conductivity of capillaries in various parts of the body. Organ Conductivitya Type of Endothelium Brain (excluding circumventricular organs) 3 Skin 100 Continuous Skeletal muscle 250 Lung 340 Heart 860 Gastrointestinal tract (intestinal mucosa) 13,000 Fenestrated Glomerulus in kidney 15,000 aUnits of conductivity are 10–13 cm3 s–1 dyne–1. Data courtesy of JN Diana. CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 539 ANGIOGENESIS When tissues grow, blood vessels must proliferate if the tissue is to maintain a normal blood supply. Therefore, angiogenesis, the formation of new blood vessels, is important during fetal life and growth to adulthood. It is also important in adulthood for processes such as wound healing, formation of the corpus lu-teum after ovulation, and formation of new endometrium after menstruation. Abnormally, it is important in tumor growth; if tumors do not develop a blood supply, they do not grow. During embryonic development, a network of leaky capil-laries is formed in tissues from angioblasts: this process is sometimes called vasculogenesis. Vessels then branch off from nearby vessels, hook up with the capillaries, and provide them with smooth muscle, which brings about their matura-tion. Angiogenesis in adults is presumably similar, but con-sists of new vessel formation by branching from pre-existing vessels rather than from angioblasts. Many factors are involved in angiogenesis. A key compound is the protein growth factor vascular endothelial growth fac-tor (VEGF). This factor exists in multiple isoforms, and there are three VEGF receptors that are tyrosine kinases, which also cooperate with nonkinase co-receptors known as neuropilins in some cell types. VEGF appears to be primarily responsible for vasculogenesis, whereas the budding of vessels that connect to the immature capillary network is regulated by other as yet uni-dentified factors. Some of the VEGF isoforms and receptors may play a more prominent role in the formation of lymphatic vessels (lymphangiogenesis) than that of blood vessels. The actions of VEGF and related factors have received con-siderable attention in recent years because of the requirement for angiogenesis in the development of tumors. VEGF antago-nists and other angiogenesis inhibitors have now entered clin-ical practice as adjunctive therapies for many malignancies and are being tested as first line therapies as well. BIOPHYSICAL CONSIDERATIONS FOR CIRCULATORY PHYSIOLOGY FLOW, PRESSURE, & RESISTANCE Blood always flows, of course, from areas of high pressure to areas of low pressure, except in certain situations when mo-mentum transiently sustains flow (see Figure 31–3). The rela-tionship between mean flow, mean pressure, and resistance in the blood vessels is analogous in a general way to the relation-ship between the current, electromotive force, and resistance in an electrical circuit expressed in Ohm’s law: Flow in any portion of the vascular system is equal to the effective perfusion pressure in that portion divided by the resistance. The effective perfusion pressure is the mean intralu-minal pressure at the arterial end minus the mean pressure at the venous end. The units of resistance (pressure divided by flow) are dyne·s/cm5. To avoid dealing with such complex units, resistance in the cardiovascular system is sometimes expressed in R units, which are obtained by dividing pressure in mm Hg by flow in mL/s (see also Table 34–1). Thus, for example, when the mean aortic pressure is 90 mm Hg and the left ventricular output is 90 mL/s, the total peripheral resistance is METHODS FOR MEASURING BLOOD FLOW Blood flow can be measured by cannulating a blood vessel, but this has obvious limitations. Various noninvasive devices have therefore been developed to measure flow. Most commonly, blood velocity can be measured with Doppler flow meters. Ultrasonic waves are sent into a vessel diagonally, and the waves reflected from the red and white blood cells are picked up by a downstream sensor. The frequency of the reflected waves is higher by an amount that is proportionate to the rate of flow toward the sensor because of the Doppler effect. Indirect methods for measuring the blood flow of various organs in humans include adaptations of the Fick and indicator dilution techniques described in Chapter 31. One example is the use of the Kety N2O method for measuring cerebral blood flow (see Chapter 34). Another is determination of the renal blood flow by measuring the clearance of para-aminohippuric acid (see Chapter 38). A considerable amount of data on blood flow in the extremities has been obtained by plethysmography (Figure 32–20). The forearm, for example, is sealed in a water-tight chamber (plethysmograph). Changes in the volume of the forearm, reflecting changes in the amount of blood and interstitial fluid it contains, displace the water, and this dis-placement is measured with a volume recorder. When the venous drainage of the forearm is occluded, the rate of increase in the volume of the forearm is a function of the arterial blood flow (venous occlusion plethysmography). APPLICABILITY OF PHYSICAL PRINCIPLES TO FLOW IN BLOOD VESSELS Physical principles and equations that describe the behavior of perfect fluids in rigid tubes have often been used indiscrimi-nately to explain the behavior of blood in blood vessels. Blood vessels are not rigid tubes, and the blood is not a perfect fluid but a two-phase system of liquid and cells. Therefore, the be-havior of the circulation deviates, sometimes markedly, from that predicted by these principles. However, the physical Current (I) Electromotive force (E) Resistance (R) --------------------------------------------------------= Flow (F) Pressure (P) Resistance (R) ----------------------------------= 90 mm Hg 90 mL/s -------------------------1 R unit = 540 SECTION VI Cardiovascular Physiology principles are of value when used as an aid to understanding what goes on in the body. LAMINAR FLOW The flow of blood in straight blood vessels, like the flow of liq-uids in narrow rigid tubes, is normally laminar. Within the blood vessels, an infinitely thin layer of blood in contact with the wall of the vessel does not move. The next layer within the vessel has a low velocity, the next a higher velocity, and so forth, velocity being greatest in the center of the stream (Figure 32–21). Laminar flow occurs at velocities up to a cer-tain critical velocity. At or above this velocity, flow is turbu-lent. Laminar flow is silent, but turbulent flow creates sounds. The probability of turbulence is also related to the diameter of the vessel and the viscosity of the blood. This probability can be expressed by the ratio of inertial to viscous forces as follows: where Re is the Reynolds number, named for the man who de-scribed the relationship; ρ is the density of the fluid; D is the diameter of the tube under consideration; V is the velocity of the flow; and η is the viscosity of the fluid. The higher the val-ue of Re, the greater the probability of turbulence. When D is in cm, V is in cm/s–1, and η is in poise; flow is usually not tur-bulent if Re is less than 2000. When Re is more than 3000, tur-bulence is almost always present. Laminar flow can be disturbed at the branching points of arteries, and the resulting turbulence may increase the likelihood that atherosclerotic plaques will be deposited. Constriction of an artery likewise increases the velocity of blood flow through the constriction, producing turbulence and sound beyond the constriction (Figure 32–22). Examples are bruits heard over arteries con-stricted by atherosclerotic plaques and the sounds of Korot-koff heard when measuring blood pressure (see below). In humans, the critical velocity is sometimes exceeded in the ascending aorta at the peak of systolic ejection, but it is usually exceeded only when an artery is constricted. Turbu-lence occurs more frequently in anemia because the viscosity of the blood is lower. This may be the explanation of the sys-tolic murmurs that are common in anemia. SHEAR STRESS & GENE ACTIVATION Flowing blood creates a force on the endothelium that is par-allel to the long axis of the vessel. This shear stress (γ) is pro-portionate to viscosity (η) times the shear rate (dy/dr), which FIGURE 32–20 Plethysmography. FIGURE 32–21 Diagram of the velocities of concentric laminas of a viscous fluid flowing in a tube, illustrating the parabolic distribution of velocities. Rubber sleeve Water Volume recorder Vessel wall Flow FIGURE 32–22 Top: Effect of constriction (C) on the profile of velocities in a blood vessel. The arrows indicate direction of velocity components, and their length is proportionate to their magnitude. Bottom: Range of velocities at each point along the vessel. In the area of turbulence, there are many different anterograde (A) and some ret-rograde (R) velocities. (Modified and reproduced with permission from Richards KE: Doppler echocardiography in diagnosis and quantification of vascular disease. Mod Concepts Cardiovasc Dis 1987;56:43. By permission of the American Heart Association.) Re ρDV η -------------= • Laminar Laminar Turbulent High velocity Upstream A 0 R Velocity C CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 541 is the rate at which the axial velocity increases from the vessel wall toward the lumen. γ = η (dy/dr) Change in shear stress and other physical variables, such as cyclic strain and stretch, produce marked changes in the expression of genes by endothelial cells. The genes that are activated include those that produce growth factors, integrins, and related molecules (Table 32–11). AVERAGE VELOCITY When considering flow in a system of tubes, it is important to distinguish between velocity, which is displacement per unit time (eg, cm/s), and flow, which is volume per unit time (eg, cm3/s). Velocity (V) is proportional to flow (Q) divided by the area of the conduit (A): Therefore, Q = A × V, and if flow stays constant, velocity increases in direct proportion to any decrease in A (Figure 32–22). The average velocity of fluid movement at any point in a system of tubes in parallel is inversely proportional to the total cross-sectional area at that point. Therefore, the average velocity of the blood is high in the aorta, declines steadily in the smaller vessels, and is lowest in the capillaries, which have 1000 times the total cross-sectional area of the aorta (Table 32–9). The average velocity of blood flow increases again as the blood enters the veins and is relatively high in the vena cava, although not so high as in the aorta. Clinically, the velocity of the circulation can be measured by injecting a bile salt preparation into an arm vein and timing the first appear-ance of the bitter taste it produces (Figure 32–23). The aver-age normal arm-to-tongue circulation time is 15 s. POISEUILLE–HAGEN FORMULA The relationship between the flow in a long narrow tube, the viscosity of the fluid, and the radius of the tube is expressed mathematically in the Poiseuille–Hagen formula: where F = flow PA – PB = pressure difference between two ends of the tube η = viscosity r = radius of tube L = length of tube Because flow is equal to pressure difference divided by resistance (R), Because flow varies directly and resistance inversely with the fourth power of the radius, blood flow and resistance in TABLE 32–11 Genes in human, bovine, and rabbit endothelial cells that are affected by shear stress, and transcription factors involved.a Gene Transcription Factors Endothelin-1 AP-1 VCAM-1 AP-1, NF-κB ACE SSRE, AP-1, Egr-1 Tissue factor SP1, Egr-1 TM AP-1 PDGF-α SSRE, Egr-1 PDGF-β SSRE ICAM-1 SSRE, AP-1, NF-κB TGF-β SSRE, AP-1, NF-κB Egr-1 SREs c-fos SSRE c-jun SSRE, AP-1 NOS 3 SSRE, AP-1, NF-κB MCP-1 SSRE, AP-1, NF-κB aAcronyms are expanded in the Appendix. Modified from Braddock M et al: Fluid shear stress modulation of gene expression in endothelial cells. News Physiol Sci 1998;13:241. V Q A ----= • FIGURE 32–23 Pathway traversed by the injected material when the arm-to-tongue circulation time is measured. Site of injection (antecubital vein) Site of end point (tongue) F PA ( PB) π 8 ---⎝⎠ ⎛⎞ 1 η ---⎝⎠ ⎛⎞ r4 L ----⎝ ⎠ ⎛ ⎞ × × × – = R 8ηL πr4 ----------= 542 SECTION VI Cardiovascular Physiology vivo are markedly affected by small changes in the caliber of the vessels. Thus, for example, flow through a vessel is dou-bled by an increase of only 19% in its radius; and when the radius is doubled, resistance is reduced to 6% of its previous value. This is why organ blood flow is so effectively regulated by small changes in the caliber of the arterioles and why varia-tions in arteriolar diameter have such a pronounced effect on systemic arterial pressure. VISCOSITY & RESISTANCE The resistance to blood flow is determined not only by the ra-dius of the blood vessels (vascular hindrance) but also by the viscosity of the blood. Plasma is about 1.8 times as viscous as water, whereas whole blood is 3 to 4 times as viscous as water. Thus, viscosity depends for the most part on the hematocrit, that is, the percentage of the volume of blood occupied by red blood cells. The effect of viscosity in vivo deviates from that predicted by the Poiseuille–Hagen formula. In large vessels, increases in hematocrit cause appreciable increases in viscosi-ty. However, in vessels smaller than 100 μm in diameter—that is, in arterioles, capillaries, and venules—the viscosity change per unit change in hematocrit is much less than it is in large-bore vessels. This is due to a difference in the nature of flow through the small vessels. Therefore, the net change in viscos-ity per unit change in hematocrit is considerably smaller in the body than it is in vitro (Figure 32–24). This is why hematocrit changes have relatively little effect on the peripheral resistance except when the changes are large. In severe polycythemia, the increase in resistance does increase the work of the heart. Con-versely, in marked anemia, peripheral resistance is decreased, in part because of the decline in viscosity. Of course, the de-crease in hemoglobin decreases the O2-carrying ability of the blood, but the improved blood flow due to the decrease in vis-cosity partially compensates for this. Viscosity is also affected by the composition of the plasma and the resistance of the cells to deformation. Clinically sig-nificant increases in viscosity are seen in diseases in which plasma proteins such as the immunoglobulins are markedly elevated as well as when red blood cells are abnormally rigid (hereditary spherocytosis). CRITICAL CLOSING PRESSURE In rigid tubes, the relationship between pressure and flow of ho-mogeneous fluids is linear, but in thin-walled blood vessels in vivo it is not. When the pressure in a small blood vessel is re-duced, a point is reached at which no blood flows, even though the pressure is not zero (Figure 32–25). This is because the ves-sels are surrounded by tissues that exert a small but definite pressure on them, and when the intraluminal pressure falls be-low the tissue pressure, they collapse. In inactive tissues, for ex-ample, the pressure in many capillaries is low because the precapillary sphincters and metarterioles are constricted, and many of these capillaries are collapsed. The pressure at which flow ceases is called the critical closing pressure. LAW OF LAPLACE The relationship between distending pressure and tension is shown diagrammatically in Figure 32–26. It is perhaps surpris-ing that structures as thin-walled and delicate as the capillaries are not more prone to rupture. The principal reason for their rel-ative invulnerability is their small diameter. The protective effect of small size in this case is an example of the operation of the law of Laplace, an important physical principle with several other applications in physiology. This law states that tension in the wall of a cylinder (T) is equal to the product of the transmural pressure (P) and the radius (r) divided by the wall thickness (w): T = Pr/w FIGURE 32–24 Effect of changes in hematocrit on the relative viscosity of blood measured in a glass viscometer and in the hind leg of a dog. In each case, the middle line represents the mean and the upper and lower lines the standard deviation. (Reproduced with permission from Whittaker SRF, Winton FR: The apparent viscosity of blood flowing in the isolated hind limb of the dog, and its variation with corpuscular concentration. J Physiol [Lond] 1933;78:338.) 14 12 10 8 6 4 2 20 40 60 80% Hematocrit Glass viscometer Hind limb Relative viscosity FIGURE 32–25 Relation of pressure to flow in thin-walled blood vessels. Pressure Critical closing pressure Blood vessels Flow CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 543 The transmural pressure is the pressure inside the cylinder minus the pressure outside the cylinder, but because tissue pressure in the body is low, it can generally be ignored and P equated to the pressure inside the viscus. In a thin-walled vis-cus, w is very small and it too can be ignored, but it becomes a significant factor in vessels such as arteries. Therefore, in a thin-walled viscus, P = T divided by the two principal radii of curvature of the viscus: In a sphere, r1 = r2, so In a cylinder such as a blood vessel, one radius is infinite, so Consequently, the smaller the radius of a blood vessel, the lower the tension in the wall necessary to balance the distend-ing pressure. In the human aorta, for example, the tension at normal pressures is about 170,000 dynes/cm, and in the vena cava it is about 21,000 dynes/cm; but in the capillaries, it is approximately 16 dynes/cm. The law of Laplace also makes clear a disadvantage faced by dilated hearts. When the radius of a cardiac chamber is increased, a greater tension must be developed in the myocar-dium to produce any given pressure; consequently, a dilated heart must do more work than a nondilated heart. In the lungs, the radii of curvature of the alveoli become smaller during expiration, and these structures would tend to collapse because of the pull of surface tension if the tension were not reduced by the surface-tension-lowering agent, surfactant (see Chapter 35). Another example of the operation of this law is seen in the urinary bladder (see Chapter 38). RESISTANCE & CAPACITANCE VESSELS In vivo, the veins are an important blood reservoir. Normally, they are partially collapsed and oval in cross-section. A large amount of blood can be added to the venous system before the veins become distended to the point where further increments in volume produce a large rise in venous pressure. The veins are therefore called capacitance vessels. The small arteries and arterioles are referred to as resistance vessels because they are the principal site of the peripheral resistance (see below). At rest, at least 50% of the circulating blood volume is in the systemic veins, 12% is in the heart cavities, and 18% is in the low-pressure pulmonary circulation. Only 2% is in the aorta, 8% in the arteries, 1% in the arterioles, and 5% in the capillar-ies (Table 32–9). When extra blood is administered by trans-fusion, less than 1% of it is distributed in the arterial system (the “high-pressure system”), and all the rest is found in the systemic veins, pulmonary circulation, and heart chambers other than the left ventricle (the “low-pressure system”). ARTERIAL & ARTERIOLAR CIRCULATION The pressure and velocities of the blood in the various parts of the systemic circulation are summarized in Figure 32–27. The general relationships in the pulmonary circulation are similar, but the pressure in the pulmonary artery is 25/10 mm Hg or less. FIGURE 32–26 Relationship between distending pressure (P) and wall tension (T) in a hollow viscus. P T T P T 1 r1 ----1 r2 ----+ ⎝ ⎠ ⎛ ⎞ = P 2T r ------= P T r ----= FIGURE 32–27 Diagram of the changes in pressure and velocity as blood flows through the systemic circulation. TA, total cross-sectional area of the vessels, which increases from 4.5 cm2 in the aorta to 4500 cm2 in the capillaries (Table 32–9). RR, relative resistance, which is highest in the arterioles. 120 80 40 0 0 Systolic Diastolic Velocity TA RR Vena cava Veins Venules Capillaries Arterioles Arteries Aorta Mean velocity (cm/s) Pressure (mm Hg) 544 SECTION VI Cardiovascular Physiology VELOCITY & FLOW OF BLOOD Although the mean velocity of the blood in the proximal por-tion of the aorta is 40 cm/s, the flow is phasic, and velocity ranges from 120 cm/s during systole to a negative value at the time of the transient backflow before the aortic valve closes in diastole. In the distal portions of the aorta and in the large ar-teries, velocity is also much greater in systole than it is in dias-tole. However, the vessels are elastic, and forward flow is continuous because of the recoil during diastole of the vessel walls that have been stretched during systole (Figure 32–28). Pulsatile flow appears to maintain optimal function of the tis-sues, apparently via distinct effects on gene transcription. If an organ is perfused with a pump that delivers a nonpulsatile flow, inflammatory markers are produced, there is a gradual rise in vascular resistance, and ultimately tissue perfusion fails. ARTERIAL PRESSURE The pressure in the aorta and in the brachial and other large ar-teries in a young adult human rises to a peak value (systolic pressure) of about 120 mm Hg during each heart cycle and falls to a minimum (diastolic pressure) of about 70 mm Hg. The ar-terial pressure is conventionally written as systolic pressure over diastolic pressure, for example, 120/70 mm Hg. One millimeter of mercury equals 0.133 kPa, so in SI units (see Appendix) this value is 16.0/9.3 kPa. The pulse pressure, the difference be-tween the systolic and diastolic pressures, is normally about 50 mm Hg. The mean pressure is the average pressure throughout the cardiac cycle. Because systole is shorter than diastole, the mean pressure is slightly less than the value halfway between sys-tolic and diastolic pressure. It can actually be determined only by integrating the area of the pressure curve (Figure 32–29); how-ever, as an approximation, mean pressure equals the diastolic pressure plus one-third of the pulse pressure. The pressure falls very slightly in the large- and medium-sized arteries because their resistance to flow is small, but it falls rapidly in the small arteries and arterioles, which are the main sites of the peripheral resistance against which the heart pumps. The mean pressure at the end of the arterioles is 30 to 38 mm Hg. Pulse pressure also declines rapidly to about 5 mm Hg at the ends of the arterioles (Figure 32–26). The magni-tude of the pressure drop along the arterioles varies consider-ably depending on whether they are constricted or dilated. EFFECT OF GRAVITY The pressures in Figure 32–28 are those in blood vessels at heart level. The pressure in any vessel below heart level is in-creased and that in any vessel above heart level is decreased by the effect of gravity. The magnitude of the gravitational effect is 0.77 mm Hg/cm of vertical distance above or below the heart at the density of normal blood. Thus, in an adult human in the upright position, when the mean arterial pressure at heart level is 100 mm Hg, the mean pressure in a large artery in the head (50 cm above the heart) is 62 mm Hg (100 – [0.77 × 50]) and the pressure in a large artery in the foot (105 cm be-low the heart) is 180 mm Hg (100 + [0.77 × 105]). The effect of gravity on venous pressure is similar (Figure 32–30). METHODS OF MEASURING BLOOD PRESSURE If a cannula is inserted into an artery, the arterial pressure can be measured directly with a mercury manometer or a suitably calibrated strain gauge. When an artery is tied off beyond the point at which the cannula is inserted, an end pressure is re-corded, flow in the artery is interrupted, and all the kinetic en-ergy of flow is converted into pressure energy. If, alternatively, FIGURE 32–28 Changes in blood flow during the cardiac cycle in the dog. Diastole is followed by systole starting at 0.1 and again at 0.5 s. Flow patterns in humans are similar. Ao, aorta; PA, pul-monary artery; PV, pulmonary vein; IVC, inferior vena cava; RA, renal ar-tery. (Reproduced with permission from Milnor WR: Pulsatile blood flow. N Engl J Med 1972;287:27.) 100 0 0 10 0 50 0 5 0.2 0.4 0.6 Time (s) RA IVC PV PA Ao Flow (mL/s) FIGURE 32–29 Brachial artery pressure curve of a normal young human, showing the relation of systolic and diastolic pressure to mean pressure. The shaded area above the mean pres-sure line is equal to the shaded area below it. 0 1 2 3 4 120 70 Systolic pressure Mean pressure Diastolic pressure Time (s) Pressure (mm Hg) CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 545 a T tube is inserted into a vessel and the pressure is measured in the side arm of the tube, the recorded side pressure, under conditions where pressure drop due to resistance is negligible, is lower than the end pressure by the kinetic energy of flow. This is because in a tube or a blood vessel the total energy—the sum of the kinetic energy of flow and the potential energy—is constant (Bernoulli’s principle). It is worth noting that the pressure drop in any segment of the arterial system is due both to resistance and to conversion of potential into kinetic energy. The pressure drop due to energy lost in overcoming resistance is irreversible, since the energy is dissipated as heat; but the pressure drop due to con-version of potential to kinetic energy as a vessel narrows is reversed when the vessel widens out again (Figure 32–31). Bernoulli’s principle also has a significant application in pathophysiology. According to the principle, the greater the velocity of flow in a vessel, the lower the lateral pressure dis-tending its walls. When a vessel is narrowed, the velocity of flow in the narrowed portion increases and the distending pressure decreases. Therefore, when a vessel is narrowed by a pathologic process such as an atherosclerotic plaque, the lat-eral pressure at the constriction is decreased and the narrow-ing tends to maintain itself. AUSCULTATORY METHOD The arterial blood pressure in humans is routinely measured by the auscultatory method. An inflatable cuff (Riva–Rocci cuff) attached to a mercury manometer (sphygmomanome-ter) is wrapped around the arm and a stethoscope is placed over the brachial artery at the elbow. The cuff is rapidly inflat-ed until the pressure is well above the expected systolic pres-sure in the brachial artery. The artery is occluded by the cuff, and no sound is heard with the stethoscope. The pressure in the cuff is then lowered slowly. At the point at which systolic pressure in the artery just exceeds the cuff pressure, a spurt of blood passes through with each heartbeat and, synchronously with each beat, a tapping sound is heard below the cuff. The cuff pressure at which the sounds are first heard is the systolic pressure. As the cuff pressure is lowered further, the sounds become louder, then dull and muffled. These are the sounds of Korotkoff. Finally, in most individuals, they disappear. When direct and indirect blood pressure measurements are made si-multaneously, the diastolic pressure in resting adults corre-lates best with the pressure at which the sound disappears. However, in adults after exercise and in children, the diastolic pressure correlates best with the pressure at which the sounds become muffled. This is also true in diseases such as hyperthy-roidism and aortic insufficiency. The sounds of Korotkoff are produced by turbulent flow in the brachial artery. When the artery is narrowed by the cuff, the velocity of flow through the constriction exceeds the criti-cal velocity and turbulent flow results (Figure 32–22). At cuff pressures just below the systolic pressure, flow through the artery occurs only at the peak of systole, and the intermittent turbulence produces a tapping sound. As long as the pressure in the cuff is above the diastolic pressure in the artery, flow is FIGURE 32–30 Effects of gravity on arterial and venous pressure. The scale on the right indicates the increment (or decre-ment) in mean pressure in a large artery at each level. The mean pressure in all large arteries is approximately 100 mm Hg when they are at the level of the left ventricle. The scale on the left indicates the increment in venous pressure at each level due to gravity. The manometers on the left of the figure indicate the height to which a column of blood in a tube would rise if connected to an ankle vein (A), the femoral vein (B), or the right atrium (C), with the subject in the standing position. The approximate pressures in these loca-tions in the recumbent position; that is, when the ankle, thigh, and right atrium are at the same level, are A, 10 mm Hg; B, 7.5 mm Hg; and C, 4.6 mm Hg. 0 20 40 60 80 A B C −80 −60 −40 −20 20 40 60 80 0 Increment in venous pressure due to gravity (mm Hg) Increment or decrement in mean arterial pressure (mm Hg) FIGURE 32–31 Bernoulli’s principle. When fluid flows through the narrow portion of the tube, the kinetic energy of flow is in-creased as the velocity increases, and the potential energy is reduced. Consequently, the measured pressure (P) is lower than it would have been at that point if the tube had not been narrowed. The dashed line indicates what the pressure drop due to frictional forces would have been if the tube had been of uniform diameter. P Flow 546 SECTION VI Cardiovascular Physiology interrupted at least during part of diastole, and the intermit-tent sounds have a staccato quality. When the cuff pressure is near the arterial diastolic pressure, the vessel is still con-stricted, but the turbulent flow is continuous. Continuous sounds have a muffled rather than a staccato quality. NORMAL ARTERIAL BLOOD PRESSURE The blood pressure in the brachial artery in young adults in the sitting position at rest is approximately 120/70 mm Hg. Because the arterial pressure is the product of the cardiac out-put and the peripheral resistance, it is affected by conditions that affect either or both of these factors. Emotion increases the cardiac output and peripheral resistance, and about 20% of hypertensive patients have blood pressures that are higher in the doctor’s office than at home, going about their regular dai-ly activities (“white coat hypertension”). Blood pressure nor-mally falls up to 20 mm Hg during sleep. This fall is reduced or absent in hypertension. There is general agreement that blood pressure rises with advancing age, but the magnitude of this rise is uncertain because hypertension is a common disease and its incidence increases with advancing age (see Clinical Box 32–4). Individ-uals who have systolic blood pressures < 120 mm Hg at age 50 to 60 and never develop clinical hypertension still have sys-tolic pressures that rise throughout life (Figure 32–32). This rise may be the closest approximation to the rise in normal individuals. Individuals with mild hypertension that is untreated show a significantly more rapid rise in systolic pres-sure. In both groups, diastolic pressure also rises, but then starts to fall in middle age as the stiffness of arteries increases. Consequently, pulse pressure rises with advancing age. It is interesting that systolic and diastolic blood pressures are lower in young women than in young men until age 55 to 65, after which they become comparable. Because there is a positive correlation between blood pressure and the incidence of heart attacks and strokes (see below), the lower blood pres-sure before menopause in women may be one reason that, on average, they live longer than men. CAPILLARY CIRCULATION At any one time, only 5% of the circulating blood is in the cap-illaries, but this 5% is in a sense the most important part of the blood volume because it is the only pool from which O2 and nu-trients can enter the interstitial fluid and into which CO2 and waste products can enter the bloodstream. Exchange across the capillary walls is essential to the survival of the tissues. METHODS OF STUDY It is difficult to obtain accurate measurements of capillary pressures and flows. Capillary pressure has been estimated by determining the amount of external pressure necessary to FIGURE 32–32 Effects of age and sex on arterial pressure components in humans. Data are from a large group of individuals who were studied every 2 y throughout their adult lives. Group 1: Indi-viduals who had systolic blood pressures < 120 mm Hg at age 50 to 60. Group 4: Individuals who had systolic blood pressure ≥ 160 mm Hg at age 50 to 60, that is, individuals with mild, untreated hypertension. The red line shows the values for women, and the blue line shows the val-ues for men. (Modified and reproduced with permission from Franklin SS et al: Hemodynamic patterns of age-related changes in blood pressure: The Framingham Heart Study. Circulation 1997;96:308.) 4 1 4 1 Systolic Diastolic Pulse Systolic pressure (mm Hg) 175 165 155 145 135 125 115 105 Diastolic pressure (mm Hg) 65 70 75 80 85 90 95 Pulse pressure (mm Hg) 35 45 55 65 75 85 95 Age (y) 4 1 4 1 4 1 CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 547 occlude the capillaries or the amount of pressure necessary to make saline start to flow through a micropipette inserted so that its tip faces the arteriolar end of the capillary. CAPILLARY PRESSURE & FLOW Capillary pressures vary considerably, but typical values in hu-man nail bed capillaries are 32 mm Hg at the arteriolar end and 15 mm Hg at the venous end. The pulse pressure is ap-proximately 5 mm Hg at the arteriolar end and zero at the venous end. The capillaries are short, but blood moves slowly (about 0.07 cm/s) because the total cross-sectional area of the capillary bed is large. Transit time from the arteriolar to the venular end of an average-sized capillary is 1 to 2 s. CLINICAL BOX 32–4 Hypertension Hypertension is a sustained elevation of the systemic arterial pressure. It is most commonly due to increased peripheral re-sistance and is a very common abnormality in humans. It can be produced by many diseases (Table 32–12) and causes a number of serious disorders. When the resistance against which the left ventricle must pump (afterload) is elevated for a long period, the cardiac muscle hypertrophies. The initial response is activation of immediate-early genes in the ven-tricular muscle, followed by activation of a series of genes in-volved in growth during fetal life. Left ventricular hypertro-phy is associated with a poor prognosis. The total O2 consumption of the heart, already increased by the work of expelling blood against a raised pressure (see Chapter 31), is increased further because there is more muscle. Therefore, any decrease in coronary blood flow has more serious conse-quences in hypertensive patients than it does in normal indi-viduals, and degrees of coronary vessel narrowing that do not produce symptoms when the size of the heart is normal may produce myocardial infarction when the heart is enlarged. In other, less common forms of hypertension, the cause is known. A review of these is helpful because it emphasizes ways disordered physiology can lead to disease. Pathology that compromises the renal blood supply leads to renal hy-pertension, as does narrowing (coarctation) of the thoracic aorta, which both increases renin secretion and increases pe-ripheral resistance. Pheochromocytomas, adrenal medullary tumors that secrete norepinephrine and epinephrine, can cause sporadic or sustained hypertension (see Chapter 22). Estrogens increase angiotensinogen secretion, and contra-ceptive pills containing large amounts of estrogen cause hy-pertension (pill hypertension) on this basis (see Chapter 25). Increased secretion of aldosterone or other mineralocorti-coids causes renal Na+ retention, which leads to hyperten-sion. A primary increase in plasma mineralocorticoids inhibits renin secretion. For unknown reasons, plasma renin is also low in 10–15% of patients with essential hypertension and normal circulating mineralocortical levels (low renin hyper-tension). Mutations in a number of single genes are also known to cause hypertension. These cases of monogenic hy-pertension are rare but informative. One of these is glucocor-ticoid-remediable aldosteronism (GRA), in which a hybrid gene encodes an adrenocorticotropic hormone (ACTH)-sensi-tive aldosterone synthase, with resulting hyperaldosteronism (see Chapter 22). 11-β hydroxylase deficiency also causes hy-pertension by increasing the secretion of deoxycorticoster-one (see Chapter 22). Normal blood pressure is restored when ACTH secretion is inhibited by administering a glucocorticoid. Mutations that decrease 11-β hydroxysteroid dehydrogenase cause loss of specificity of the mineralocorticoid receptors (see Chapter 22) with stimulation of them by cortisol and, in pregnancy, by the elevated circulating levels of progesterone. Finally, mutations of the genes for ENaCs that reduce degra-dation of the β or γ subunits increase ENaC activity and lead to excess renal Na+ retention and hypertension (Liddle syn-drome; see Chapter 38). The incidence of atherosclerosis increases in hypertension, and myocardial infarcts are common even when the heart is not enlarged. Eventually, the ability to compensate for the high peripheral resistance is exceeded, and the heart fails. Hyperten-sive individuals are also predisposed to thromboses of cerebral vessels and cerebral hemorrhage. An additional complication is renal failure. However, the incidence of heart failure, strokes, and renal failure can be markedly reduced by active treatment of hypertension, even when the hypertension is relatively mild. In about 88% of patients with elevated blood pressure, the cause of the hypertension is unknown, and they are said to have essential hypertension. At present, essential hyperten-sion is treatable but not curable. Effective lowering of the blood pressure can be produced by drugs that block α-adrenergic receptors, either in the periphery or in the central nervous sys-tem; drugs that block β-adrenergic receptors; drugs that inhibit the activity of angiotensin-converting enzyme; and calcium channel blockers that relax vascular smooth muscle. Essential hypertension is probably polygenic in origin, and environmen-tal factors are also involved. 548 SECTION VI Cardiovascular Physiology EQUILIBRATION WITH INTERSTITIAL FLUID As noted above, the capillary wall is a thin membrane made up of endothelial cells. Substances pass through the junctions be-tween endothelial cells and through fenestrations when they are present. Some also pass through the cells by vesicular transport. The factors other than vesicular transport that are respon-sible for transport across the capillary wall are diffusion and filtration (see Chapter 1). Diffusion is quantitatively much more important. O2 and glucose are in higher concentration in the bloodstream than in the interstitial fluid and diffuse into the interstitial fluid, whereas CO2 diffuses in the oppo-site direction. The rate of filtration at any point along a capillary depends on a balance of forces sometimes called the Starling forces, after the physiologist who first described their operation in detail. One of these forces is the hydrostatic pressure gradi-ent (the hydrostatic pressure in the capillary minus the hydro-static pressure of the interstitial fluid) at that point. The interstitial fluid pressure varies from one organ to another, and there is considerable evidence that it is subatmospheric (about –2 mm Hg) in subcutaneous tissue. It is, however, pos-itive in the liver and kidneys and as high as 6 mm Hg in the brain. The other force is the osmotic pressure gradient across the capillary wall (colloid osmotic pressure of plasma minus colloid osmotic pressure of interstitial fluid). This component is directed inward. Thus: Fluid movement = k[(Pc – Pi) – (πc – πi)] where k = capillary filtration coefficient Pc = capillary hydrostatic pressure Pi = interstitial hydrostatic pressure πc = capillary colloid osmotic pressure πi = interstitial colloid osmotic pressure πi is usually negligible, so the osmotic pressure gradient (πc – πi) usually equals the oncotic pressure. The capillary filtration coefficient takes into account, and is proportional to, the permeability of the capillary wall and the area available for filtration. The magnitude of the Starling forces along a typi-cal muscle capillary is shown in Figure 32–33. Fluid moves into the interstitial space at the arteriolar end of the capillary and into the capillary at the venular end. In other capillaries, the balance of Starling forces may be different. For example, fluid moves out of almost the entire length of the capillaries in the renal glomeruli. On the other hand, fluid moves into the capillaries through almost their entire length in the intes-tines. About 24 L of fluid is filtered through the capillaries per day. This is about 0.3% of the cardiac output. About 85% of the filtered fluid is reabsorbed into the capillaries, and the remainder returns to the circulation via the lymphatics. It is worth noting that small molecules often equilibrate with the tissues near the arteriolar end of each capillary. In this sit-uation, total diffusion can be increased by increasing blood flow; that is, exchange is flow-limited (Figure 32–34). Con-versely, transfer of substances that do not reach equilibrium with the tissues during their passage through the capillaries is said to be diffusion-limited. TABLE 32–12 Estimated frequency of various forms of hypertension in the general hypertensive population. Percentage of Population Essential hypertension 88 Renal hypertension Renovascular 2 Parenchymal 3 Endocrine hypertension Primary aldosteronism 5 Cushing syndrome 0.1 Pheochromocytoma 0.1 Other adrenal forms 0.2 Estrogen treatment (“pill hypertension”) 1 Miscellaneous (Liddle syndrome, coarctation of the aorta, etc) 0.6 Reproduced with permission from McPhee SJ, Lingappa V, Ganong WF: Pathophysi-ology of Disease, 4th ed. McGraw-Hill, 2003. FIGURE 32–33 Schematic representation of pressure gradients across the wall of a muscle capillary. The numbers at the arteriolar and venular ends of the capillary are the hydrostatic pressures in mm Hg at these locations. The arrows indicate the approximate mag-nitude and direction of fluid movement. In this example, the pressure differential at the arteriolar end of the capillary is 11 mm Hg ([37 – 1] – 25) outward; at the opposite end, it is 9 mm Hg (25 – [17 – 1]) inward. Interstitial space Arteriole Venule Oncotic P = 25 Interstitial P = 1 37 17 CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 549 ACTIVE & INACTIVE CAPILLARIES In resting tissues, most of the capillaries are collapsed. In ac-tive tissues, the metarterioles and the precapillary sphinc-ters dilate. The intracapillary pressure rises, overcoming the critical closing pressure of the vessels, and blood flows through all of the capillaries. Relaxation of the smooth mus-cle of the metarterioles and precapillary sphincters is due to the action of vasodilator metabolites formed in active tissue (see Chapter 33). After noxious stimulation, substance P released by the axon reflex (see Chapter 34) increases capillary permeability. Bradykinin and histamine also increase capillary permeabil-ity. When capillaries are stimulated mechanically, they empty (white reaction; see Chapter 34), probably due to contraction of the precapillary sphincters. VENOUS CIRCULATION Blood flows through the blood vessels, including the veins, pri-marily because of the pumping action of the heart. However, venous flow is aided by the heartbeat, the increase in the negative intrathoracic pressure during each inspiration, and contractions of skeletal muscles that compress the veins (muscle pump). VENOUS PRESSURE & FLOW The pressure in the venules is 12 to 18 mm Hg. It falls steadily in the larger veins to about 5.5 mm Hg in the great veins out-side the thorax. The pressure in the great veins at their en-trance into the right atrium (central venous pressure) averages 4.6 mm Hg, but fluctuates with respiration and heart action. Peripheral venous pressure, like arterial pressure, is affected by gravity. It is increased by 0.77 mm Hg for each centimeter below the right atrium and decreased by a like amount for each centimeter above the right atrium the pressure is mea-sured (Figure 32–30). Thus, on a proportional basis, gravity has a greater effect on venous than on arterial pressures. When blood flows from the venules to the large veins, its average velocity increases as the total cross-sectional area of the vessels decreases. In the great veins, the velocity of blood is about one fourth that in the aorta, averaging about 10 cm/s. THORACIC PUMP During inspiration, the intrapleural pressure falls from –2.5 to –6 mm Hg. This negative pressure is transmitted to the great veins and, to a lesser extent, the atria, so that central venous pressure fluctuates from about 6 mm Hg during expiration to approximately 2 mm Hg during quiet inspiration. The drop in venous pressure during inspiration aids venous return. When the diaphragm descends during inspiration, intra-abdominal pressure rises, and this also squeezes blood toward the heart because backflow into the leg veins is prevented by the venous valves. EFFECTS OF HEARTBEAT The variations in atrial pressure are transmitted to the great veins, producing the a, c, and v waves of the venous pressure-pulse curve (see Chapter 31). Atrial pressure drops sharply during the ejection phase of ventricular systole because the atrioventricular valves are pulled downward, increasing the capacity of the atria. This action sucks blood into the atria from the great veins. The sucking of the blood into the atria during systole contributes appreciably to the venous return, especially at rapid heart rates. Close to the heart, venous flow becomes pulsatile. When the heart rate is slow, two periods of peak flow are detectable, one during ventricular systole, due to pulling down of the atrioventricular valves, and one in early diastole, during the period of rapid ventricular filling (Figure 32–28). MUSCLE PUMP In the limbs, the veins are surrounded by skeletal muscles, and contraction of these muscles during activity compresses the veins. Pulsations of nearby arteries may also compress veins. Because the venous valves prevent reverse flow, the blood moves toward the heart. During quiet standing, when the full effect of gravity is manifest, venous pressure at the ankle is 85– 90 mm Hg (Figure 32–30). Pooling of blood in the leg veins re-duces venous return, with the result that cardiac output is re-duced, sometimes to the point where fainting occurs. Rhythmic contractions of the leg muscles while the person is standing FIGURE 32–34 Flow-limited and diffusion-limited exchange across capillary walls. A and V indicate the arteriolar and venular ends of the capillary. Substance X equilibrates with the tissues (move-ment into the tissues equals movement out) well before the blood leaves the capillary, whereas substance Y does not equilibrate. If other factors stay constant, the amount of X entering the tissues can be in-creased only by increasing blood flow; that is, it is flow-limited. The movement of Y is diffusion-limited. Y X A V Distance along capillary Concentration in capillary blood 550 SECTION VI Cardiovascular Physiology serve to lower the venous pressure in the legs to less than 30 mm Hg by propelling blood toward the heart. This heartward move-ment of the blood is decreased in patients with varicose veins because their valves are incompetent. These patients may de-velop stasis and ankle edema. However, even when the valves are incompetent, muscle contractions continue to produce a basic heartward movement of the blood because the resistance of the larger veins in the direction of the heart is less than the resistance of the small vessels away from the heart. VENOUS PRESSURE IN THE HEAD In the upright position, the venous pressure in the parts of the body above the heart is decreased by the force of gravity. The neck veins collapse above the point where the venous pressure is close to zero. However, the dural sinuses have rigid walls and cannot collapse. The pressure in them in the standing or sitting position is therefore subatmospheric. The magnitude of the negative pressure is proportional to the vertical distance above the top of the collapsed neck veins, and in the superior sagittal sinus may be as much as –10 mm Hg. This fact must be kept in mind by neurosurgeons. Neurosurgical procedures are sometimes performed with the patient seated. If one of the sinuses is opened during such a procedure it sucks air, causing air embolism. AIR EMBOLISM Because air, unlike fluid, is compressible, its presence in the cir-culation has serious consequences. The forward movement of the blood depends on the fact that blood is incompressible. Large amounts of air fill the heart and effectively stop the cir-culation, causing sudden death because most of the air is com-pressed by the contracting ventricles rather than propelled into the arteries. Small amounts of air are swept through the heart with the blood, but the bubbles lodge in the small blood vessels. The surface capillarity of the bubbles markedly increases the resistance to blood flow, and flow is reduced or abolished. Blockage of small vessels in the brain leads to serious and even fatal neurologic abnormalities. Treatment with hyperbaric ox-ygen (see Chapter 37) is of value because the pressure reduces the size of the gas emboli. In experimental animals, the amount of air that produces fatal air embolism varies considerably, de-pending in part on the rate at which it enters the veins. Some-times as much as 100 mL can be injected without ill effects, whereas at other times as little as 5 mL is lethal. MEASURING VENOUS PRESSURE Central venous pressure can be measured directly by insert-ing a catheter into the thoracic great veins. Peripheral venous pressure correlates well with central venous pressure in most conditions. To measure peripheral venous pressure, a needle attached to a manometer containing sterile saline is inserted into an arm vein. The peripheral vein should be at the level of the right atrium (a point half the chest diameter from the back in the supine position). The values obtained in millimeters of saline can be converted into millimeters of mercury (mm Hg) by dividing by 13.6 (the density of mercury). The amount by which peripheral venous pressure exceeds central venous pressure increases with the distance from the heart along the veins. The mean pressure in the antecubital vein is normally 7.1 mm Hg, compared with a mean pressure of 4.6 mm Hg in the central veins. A fairly accurate estimate of central venous pressure can be made without any equipment by simply noting the height to which the external jugular veins are distended when the sub-ject lies with the head slightly above the heart. The vertical distance between the right atrium and the place the vein col-lapses (the place where the pressure in it is zero) is the venous pressure in mm of blood. Central venous pressure is decreased during negative pres-sure breathing and shock. It is increased by positive pressure breathing, straining, expansion of the blood volume, and heart failure. In advanced congestive heart failure or obstruc-tion of the superior vena cava, the pressure in the antecubital vein may reach values of 20 mm Hg or more. LYMPHATIC CIRCULATION & INTERSTITIAL FLUID VOLUME LYMPHATIC CIRCULATION Fluid efflux normally exceeds influx across the capillary walls, but the extra fluid enters the lymphatics and drains through them back into the blood. This keeps the interstitial fluid pres-sure from rising and promotes the turnover of tissue fluid. The normal 24-h lymph flow is 2 to 4 L. Lymphatic vessels can be divided into two types: initial lymphatics and collecting lymphatics (Figure 32–35). The former lack valves and smooth muscle in their walls, and they are found in regions such as the intestine or skeletal muscle. Tissue fluid appears to enter them through loose junctions between the endothelial cells that form their walls. The fluid in them apparently is massaged by muscle contractions of the organs and contraction of arterioles and venules, with which they are often associated. They drain into the collecting lym-phatics, which have valves and smooth muscle in their walls and contract in a peristaltic fashion, propelling the lymph along the vessels. Flow in the collecting lymphatics is further aided by movements of skeletal muscle, the negative intrathoracic pressure during inspiration, and the suction effect of high-velocity flow of blood in the veins in which the lymphatics terminate. However, the contractions are the principal factor propelling the lymph. CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 551 OTHER FUNCTIONS OF THE LYMPHATIC SYSTEM Appreciable quantities of protein enter the interstitial fluid in the liver and intestine, and smaller quantities enter from the blood in other tissues. The macromolecules enter the lym-phatics, presumably at the junctions between the endothelial cells, and the proteins are returned to the bloodstream via the lymphatics. The amount of protein returned in this fashion in 1 d is equal to 25–50% of the total circulating plasma protein. The transport of absorbed long-chain fatty acids and choles-terol from the intestine via the lymphatics has been discussed in Chapter 27. INTERSTITIAL FLUID VOLUME The amount of fluid in the interstitial spaces depends on the capillary pressure, the interstitial fluid pressure, the oncotic pressure, the capillary filtration coefficient, the number of active capillaries, the lymph flow, and the total extracellular fluid (ECF) volume. The ratio of precapillary to postcapillary venular resistance is also important. Precapillary constric-tion lowers filtration pressure, whereas postcapillary con-striction raises it. Changes in any of these variables lead to changes in the volume of interstitial fluid. Factors promoting an increase in this volume are summarized in Table 32–13. Edema is the accumulation of interstitial fluid in abnormally large amounts. In active tissues, capillary pressure rises, often to the point where it exceeds the oncotic pressure throughout the length of the capillary. In addition, osmotically active metabolites may temporarily accumulate in the interstitial fluid because they cannot be washed away as rapidly as they are formed. To the extent that they accumulate, they exert an osmotic effect that decreases the magnitude of the osmotic gradient due to the oncotic pressure. The amount of fluid leaving the capil-laries is therefore markedly increased and the amount enter-ing them reduced. Lymph flow is increased, decreasing the degree to which the fluid would otherwise accumulate, but exercising muscle, for example, still increases in volume by as much as 25%. Interstitial fluid tends to accumulate in dependent parts because of the effect of gravity. In the upright position, the capillaries in the legs are protected from the high arterial pressure by the arterioles, but the high venous pressure is transmitted to them through the venules. Skeletal muscle con-tractions keep the venous pressure low by pumping blood toward the heart (see above) when the individual moves about; however, if one stands still for long periods, fluid accu-mulates and edema eventually develops. The ankles also swell during long trips when travelers sit for prolonged periods with their feet in a dependent position. Venous obstruction may contribute to the edema in these situations. Whenever there is abnormal retention of salt in the body, water is also retained. The salt and water are distributed throughout the ECF, and since the interstitial fluid volume is therefore increased, there is a predisposition to edema. Salt and water retention is a factor in the edema seen in heart fail-ure, nephrosis, and cirrhosis, but there are also variations in the mechanisms that govern fluid movement across the capillary walls in these diseases. In congestive heart failure, for exam-ple, venous pressure is usually elevated, with a consequent ele-vation in capillary pressure. In cirrhosis of the liver, oncotic FIGURE 32–35 Initial lymphatics draining into collecting lymphatics in the mesentery. Note the close association with arcad-ing arterioles, indicated by the single red lines. (Reproduced with permission from Schmid Schönbein GW, Zeifach BW: Fluid pump mechanisms in initial lymphatics. News Physiol Sci 1994;9:67.) Collecting lymphatic Valve Arcading arteriole Initial lymphatics TABLE 32–13 Causes of increased interstitial fluid volume and edema. Increased filtration pressure Venular constriction Increased venous pressure (heart failure, incompetent valves, venous obstruction, increased total ECF volume, effect of gravity, etc) Decreased osmotic pressure gradient across capillary Decreased plasma protein level Accumulation of osmotically active substances in interstitial space Increased capillary permeability Substance P Histamine and related substances Kinins, etc Inadequate lymph flow 552 SECTION VI Cardiovascular Physiology pressure is low because hepatic synthesis of plasma proteins is depressed; and in nephrosis, oncotic pressure is low because large amounts of protein are lost in the urine. Another cause of edema is inadequate lymphatic drainage. Edema caused by lymphatic obstruction is called lymphedema, and the edema fluid has a high protein content. If it persists, it causes a chronic inflammatory condition that leads to fibrosis of the interstitial tissue. One cause of lymphedema is radical mastectomy, during which removal of the axillary lymph nodes leads to reduced lymph drainage. In filariasis, parasitic worms migrate into the lymphatics and obstruct them. Fluid accumu-lation plus tissue reaction lead in time to massive swelling, usu-ally of the legs or scrotum (elephantiasis). CHAPTER SUMMARY ■Blood consists of a suspension of red blood cells (erythrocytes), white blood cells, and platelets in a protein-rich fluid known as plasma. ■Blood cells arise in the bone marrow and are subject to regular renewal; the majority of plasma proteins are synthesized by the liver. ■Hemoglobin, stored in red blood cells, transports oxygen to peripheral tissues. Fetal hemoglobin is specialized to facilitate diffusion of oxygen from mother to fetus during development. Mutated forms of hemoglobin lead to red cell abnormalities and anemia. ■Complex oligosaccharide structures, specific to groups of indi-viduals, form the basis of the ABO blood group system. AB blood group oligosaccharides, as well as other blood group mol-ecules, can trigger the production of antibodies in naïve individ-uals following inappropriate transfusions, with potentially serious consequences due to erythrocyte agglutination. ■Blood flows from the heart to arteries and arterioles, thence to capillaries, and eventually to venules and veins and back to the heart. Each segment of the vasculature has specific contractile properties and regulatory mechanisms that subserve physiolog-ic function. Physical principles of pressure, wall tension, and vessel caliber govern the flow of blood through each segment of the circulation. ■Transfer of oxygen and nutrients from the blood to tissues, as well as collection of metabolic wastes, occurs exclusively in the capillary beds. ■Fluid also leaves the circulation across the walls of capillaries. Some is reabsorbed; the remainder enters the lymphatic system, which eventually drains into the subclavian veins to return fluid to the bloodstream. ■Hypertension is an increase in mean blood pressure that is usu-ally chronic and is common in humans. Hypertension can result in serious health consequences if left untreated. The majority of hypertension is of unknown cause, but several gene mutations underlie rare forms of the disease and are informative about mechanisms that control the dynamics of the circulatory system and its integration with other organs. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Which of the following has the highest total cross-sectional area in the body? A) arteries B) arterioles C) capillaries D) venules E) veins 2. Lymph flow from the foot is A) increased when an individual rises from the supine to the standing position. B) increased by massaging the foot. C) increased when capillary permeability is decreased. D) decreased when the valves of the leg veins are incompetent. E) decreased by exercise. 3. The pressure in a capillary in skeletal muscle is 35 mm Hg at the arteriolar end and 14 mm Hg at the venular end. The interstitial pressure is 0 mm Hg. The colloid osmotic pressure is 25 mm Hg in the capillary and 1 mm Hg in the interstitium. The net force producing fluid movement across the capillary wall at its arteri-olar end is A) 3 mm Hg out of the capillary. B) 3 mm Hg into the capillary. C) 10 mm Hg out of the capillary. D) 11 mm Hg out of the capillary. E) 11 mm Hg into the capillary. 4. The velocity of blood flow A) is higher in the capillaries than the arterioles. B) is higher in the veins than in the venules. C) is higher in the veins than the arteries. D) falls to zero in the descending aorta during diastole. E) is reduced in a constricted area of a blood vessel. 5. When the radius of the resistance vessels is increased, which of the following is increased? A) systolic blood pressure B) diastolic blood pressure C) viscosity of the blood D) hematocrit E) capillary blood flow 6. When the viscosity of the blood is increased, which of the fol-lowing is increased? A) mean blood pressure B) radius of the resistance vessels C) radius of the capacitance vessels D) central venous pressure E) capillary blood flow 7. A pharmacologist discovers a drug that stimulates the produc-tion of VEGF receptors. He is excited because his drug might be of value in the treatment of A) coronary artery disease. B) cancer. C) emphysema. D) diabetes insipidus. E) dysmenorrhea. CHAPTER 32 Blood as a Circulatory Fluid & the Dynamics of Blood & Lymph Flow 553 CHAPTER RESOURCES de Montalembert M: Management of sickle cell disease. Brit Med J 2008;337:626. Miller JL: Signaled expression of fetal hemoglobin during development. Transfusion 2005;45:1229. Perrotta S, Gallagher PG, Mohandas N: Hereditary spherocytosis. Lancet 2008;372:1411. Semenza GL: Vasculogenesis, angiogenesis, and arteriogenesis: Mechanisms of blood vessel formation and remodeling. J Cell Biochem 2007;102:840. This page intentionally left blank 555 C H A P T E R 6 33 Cardiovascular Regulatory Mechanisms O B J E C T I V E S After studying this chapter, you should be able to: ■Outline the neural mechanisms that control arterial blood pressure and heart rate, including the receptors, afferent and efferent pathways, central integrating path-ways, and effector mechanisms involved. ■Describe the direct effects of CO2 and hypoxia on the vasomotor areas in the medulla oblongata. ■Describe how the process of autoregulation contributes to control of vascular caliber. ■Identify the paracrine factors and hormones that regulate vascular tone, their sources, and their mechanisms of action. INTRODUCTION In humans and other mammals, multiple cardiovascular regu-latory mechanisms have evolved. These mechanisms increase the blood supply to active tissues and increase or decrease heat loss from the body by redistributing the blood. In the face of challenges such as hemorrhage, they maintain the blood flow to the heart and brain. When the challenge faced is severe, flow to these vital organs is maintained at the expense of the circulation to the rest of the body. Circulatory adjustments are effected by altering the output of the pump (the heart), changing the diameter of the resistance vessels (primarily the arterioles), or altering the amount of blood pooled in the capacitance vessels (the veins). Regulation of cardiac output is discussed in Chapter 31. The caliber of the arterioles is adjusted in part by autoregulation (Table 33–1). It is also increased in active tissues by locally produced vasodila-tor metabolites, is affected by substances secreted by the endo-thelium, and is regulated systemically by circulating vasoactive substances and the nerves that innervate the arterioles. The cal-iber of the capacitance vessels is also affected by circulating vasoactive substances and by vasomotor nerves. The systemic regulatory mechanisms synergize with the local mechanisms and adjust vascular responses throughout the body. The terms vasoconstriction and vasodilation are gener-ally used to refer to constriction and dilation of the resistance vessels. Changes in the caliber of the veins are referred to spe-cifically as venoconstriction or venodilation. NEURAL CONTROL OF THE CARDIOVASCULAR SYSTEM NEURAL REGULATORY MECHANISMS Although the arterioles and the other resistance vessels are most densely innervated, all blood vessels except capillaries and venules contain smooth muscle and receive motor nerve fibers from the sympathetic division of the autonomic nervous system. The fibers to the resistance vessels regulate tissue blood flow and arterial pressure. The fibers to the venous capacitance vessels vary the volume of blood “stored” in the veins. The in-nervation of most veins is sparse, but the splanchnic veins are well innervated. Venoconstriction is produced by stimuli that also activate the vasoconstrictor nerves to the arterioles. The 556 SECTION VI Cardiovascular Physiology resultant decrease in venous capacity increases venous return, shifting blood to the arterial side of the circulation. INNERVATION OF THE BLOOD VESSELS Sympathetic noradrenergic fibers end on blood vessels in all parts of the body to mediate vasoconstriction. In addition to their vasoconstrictor innervation, resistance vessels in skeletal muscles are innervated by vasodilator fibers, which, although they travel with the sympathetic nerves, are cholinergic (sym-pathetic cholinergic vasodilator system). There is no tonic ac-tivity in the vasodilator fibers, but the vasoconstrictor fibers to most vascular beds have some tonic activity. When the sympa-thetic nerves are cut (sympathectomy), the blood vessels di-late. In most tissues, vasodilation is produced by decreasing the rate of tonic discharge in the vasoconstrictor nerves, although in skeletal muscles it can also be produced by activating the sympathetic cholinergic vasodilator system (Table 33–1). CARDIAC INNERVATION Impulses in the sympathetic nerves to the heart increase the cardiac rate (chronotropic effect), rate of transmission in the cardiac conductive tissue (dromotropic effect), and the force of contraction (inotropic effect). They also inhibit the effects of vagal parasympathetic stimulation, probably by release of neuropeptide Y, which is a cotransmitter in the sympathetic endings. Impulses in vagal fibers decrease heart rate. A mod-erate amount of tonic discharge takes place in the cardiac sym-pathetic nerves at rest, but there is a good deal of tonic vagal discharge (vagal tone) in humans and other large animals. Af-ter the administration of parasympatholytic drugs such as at-ropine, the heart rate in humans increases from 70, its normal resting value, to 150 to 180 beats/min because the sympathetic tone is unopposed. In humans in whom both noradrenergic and cholinergic systems are blocked, the heart rate is approxi-mately 100 beats/min. CARDIOVASCULAR CONTROL The cardiovascular system is under neural influences coming from several parts of the brain (see Figure 17–6), which in turn receive feedback from sensory receptors in the vasculature (eg, baroreceptors). A simplified model of the feedback control circuit is shown in Figure 33–1. An increase in neural output from the brain stem to sympathetic nerves leads to a decrease in blood vessel diameter (arteriolar constriction) and increas-es in stroke volume and heart rate, which contribute to a rise in blood pressure. This in turn causes an increase in barore-ceptor activity, which signals the brain stem to reduce the neu-ral output to sympathetic nerves. Venoconstriction and a decrease in the stores of blood in the venous reservoirs usually accompany increases in arteri-olar constriction, although changes in the capacitance vessels do not always parallel changes in the resistance vessels. In the presence of an increase in sympathetic nerve activity to the heart and vasculature, there is usually an associated decrease in the activity of vagal fibers to the heart. Conversely, a decrease in sympathetic activity causes vasodilation, a fall in blood pressure, and an increase in the storage of blood in the venous reservoirs. There is usually a concomitant decrease in heart rate, but this is mostly due to stimulation of the vagal innervation of the heart. MEDULLARY CONTROL OF THE CARDIOVASCULAR SYSTEM One of the major sources of excitatory input to sympathetic nerves controlling the vasculature is neurons located near the pial surface of the medulla in the rostral ventrolateral medulla TABLE 33–1 Summary of factors affecting the caliber of the arterioles. Constriction Dilation Local factors Decreased local temperature Increased CO2 and decreased O2 Autoregulation Increased K+, adenosine, lactate, etc Decreased local pH Increased local temperature Endothelial products Endothelin-1 NO Locally released platelet serotonin Kinins Thromboxane A2 Prostacyclin Circulating hormones Epinephrine (except in skele-tal muscle and liver) Epinephrine in skeletal muscle and liver Norepinephrine CGRPα AVP Substance P Angiotensin II Histamine Circulating Na+-K+ ATPase inhibitor ANP Neuropeptide Y VIP Neural factors Increased discharge of sympathetic nerves Decreased discharge of sympathetic nerves Activation of sympathetic cholinergic vasodilator nerves to skeletal muscle CHAPTER 33 Cardiovascular Regulatory Mechanisms 557 (RVLM; Figure 33–2). This region is sometimes called a vasomo-tor area. The axons of RVLM neurons course dorsally and medi-ally and then descend in the lateral column of the spinal cord to the thoracolumbar intermediolateral gray column (IML). They contain phenylethanolamine-N-methyltransferase (PNMT; see Chapter 7), but it appears that the excitatory transmitter they se-crete is glutamate rather than epinephrine. Neurovascular com-pression of the RVLM has been linked to some cases of essential hypertension in humans (see Clinical Box 33–1). The activity of RVLM neurons is determined by many fac-tors (see Table 33–2). They include not only the very impor-tant fibers from arterial and venous baroreceptors, but also fibers from other parts of the nervous system and from the carotid and aortic chemoreceptors. In addition, some stimuli act directly on the vasomotor area. There are descending tracts to the vasomotor area from the cerebral cortex (particularly the limbic cortex) that relay in the hypothalamus. These fibers are responsible for the blood FIGURE 33–1 Feedback control of blood pressure. Brain stem excitatory input to sympathetic nerves to the heart and vascula-ture increases heart rate and stroke volume and reduces vessel diam-eter. Together these increase blood pressure, which activates the baroreceptor reflex to reduce the activity in the brain stem. Baroreceptors Heart rate Stroke volume Vessel diameter Blood pressure Brain stem FIGURE 33–2 Basic pathways involved in the medullary control of blood pressure. The vagal efferent pathways that slow the heart are not shown. The putative neurotransmitters in the pathways are indicated in parentheses. Glu, glutamate; GABA,γ-aminobutyric acid; Ach, acetyl-choline; NE, norepinephrine; IML, intermediolateral gray column; NTS, nucleus of the tractus solitarius; CVLM, IVLM, RVLM, caudal, intermediate, and rostral ventrolateral medulla; IX and X, glossopharyngeal and vagus nerves. Baroreceptor afferents (Glu) (GABA) Bulbospinal pathway (Glu) IX X CVLM IVLM RVLM NTS Medulla (Glu) IML Thoracic cord Arteriole or venule Adrenal medulla Postganglionic sympathetic neuron (NE) Heart Preganglionic sympathetic neuron (Ach) Aortic arch Carotid sinus 558 SECTION VI Cardiovascular Physiology pressure rise and tachycardia produced by emotions such as sexual excitement and anger. The connections between the hypothalamus and the vasomotor area are reciprocal, with afferents from the brain stem closing the loop. Inflation of the lungs causes vasodilation and a decrease in blood pressure. This response is mediated via vagal afferents from the lungs that inhibit vasomotor discharge. Pain usually causes a rise in blood pressure via afferent impulses in the reticular formation converging in the RVLM. However, pro-longed severe pain may cause vasodilation and fainting. The activity in afferents from exercising muscles probably exerts a similar pressor effect via pathway to the RVLM. The pressor response to stimulation of somatic afferent nerves is called the somatosympathetic reflex. Unlike the vasculature, the heart is controlled by both sym-pathetic and parasympathetic (vagal) nerves. The medulla is also a major site of origin of excitatory input to cardiac vagal motor neurons in the nucleus ambiguus (Figure 33–3). Table 33–3 is a summary of conditions that affect the heart rate. In general, stimuli that increase the heart rate also increase blood pressure, whereas those that decrease the heart rate lower blood pressure. However, there are exceptions, such as the production of hypotension and tachycardia by stimulation of atrial stretch receptors and the production of hypertension and bradycardia by increased intracranial pressure. BARORECEPTORS The baroreceptors are stretch receptors in the walls of the heart and blood vessels. The carotid sinus and aortic arch re-ceptors monitor the arterial circulation. Receptors are also lo-cated in the walls of the right and left atria at the entrance of the superior and inferior venae cavae and the pulmonary veins, as well as in the pulmonary circulation. These receptors in the low-pressure part of the circulation are referred to col-lectively as the cardiopulmonary receptors. The carotid sinus is a small dilation of the internal carotid artery just above the bifurcation of the common carotid into external and internal carotid branches (Figure 33–4). Barore-ceptors are located in this dilation. They are also found in the CLINICAL BOX 33–1 Essential Hypertension & Neurovascular Compression of the RVLM In about 88% of patients with elevated blood pressure, the cause of the hypertension is unknown, and they are said to have essential hypertension. There are data available to support the view that neurovascular compression of the RVLM is associated with essential hypertension in some subjects. In the 1970s, Dr. Peter Jannetta, a neurosurgeon in Pittsburgh, PA, developed a technique for “microvascu-lar decompression” of the medulla to treat trigeminal neu-ralgia and hemifacial spasm, which he attributed to pulsa-tile compression of the vertebral and posterior inferior cerebellar arteries impinging on the fifth and seventh cra-nial nerves. Moving the arteries away from the nerves led to reversal of the neurologic symptoms in many cases. Some of these patients were also hypertensive, and they showed reductions in blood pressure postoperatively. Later, a few human studies claimed that surgical decom-pression of the RVLM could sometimes relieve hyperten-sion. There are several reports of patients with a schwan-noma or meningioma lying close to the RVLM whose hypertension has been reversed by surgical decompres-sion. Magnetic resonance angiography (MRA) has been used to compare the incidence of neurovascular compres-sion in hypertensive and normotensive individuals and to correlate indices of sympathetic nerve activity with the presence or absence of compression. Some of these stud-ies showed a higher incidence of coexistence of neurovas-cular compression with essential hypertension than in other forms of hypertension or normotension, but others showed the presences of a compression in normotensive subjects. On the other hand, there was a strong positive re-lationship between the presence of neurovascular com-pression and increased sympathetic activity. TABLE 33–2 Factors affecting the activity of the RVLM. Direct stimulation CO2 Hypoxia Excitatory inputs Cortex via hypothalamus Mesencephalic periaqueductal gray Brain stem reticular formation Pain pathways Somatic afferents (somatosympathetic reflex) Carotid and aortic chemoreceptors Inhibitory inputs Cortex via hypothalamus Caudal ventrolateral medulla Caudal medullary raphé nuclei Lung inflation afferents Carotid, aortic, and cardiopulmonary baroreceptors CHAPTER 33 Cardiovascular Regulatory Mechanisms 559 wall of the arch of the aorta. The receptors are located in the adventitia of the vessels. The afferent nerve fibers from the carotid sinus form a distinct branch of the glossopharyngeal nerve, the carotid sinus nerve. The fibers from the aortic arch form a branch of the vagus nerve, the aortic depressor nerve. The baroreceptors are stimulated by distention of the struc-tures in which they are located, and so they discharge at an increased rate when the pressure in these structures rises. Their afferent fibers pass via the glossopharyngeal and vagus nerves to the medulla. Most of them end in the nucleus of the tractus solitarius (NTS), and the excitatory transmitter they secrete is glutamate (Figure 33–2). Excitatory (glutamate) projections extend from the NTS to the caudal ventrolateral medulla (CVLM), where they stimulate γ-aminobutyrate (GABA)-secreting inhibitory neurons that project to the RVLM. Excita-tory projections also extend from the NTS to the vagal motor neurons in the nucleus ambiguus and dorsal motor nucleus (Figure 33–3). Thus, increased baroreceptor discharge inhibits the tonic discharge of sympathetic nerves and excites the vagal innervation of the heart. These neural changes produce vasodi-lation, venodilation, a drop in blood pressure, bradycardia, and a decrease in cardiac output. FIGURE 33–3 Basic pathways involved in the medullary control of heart rate by the vagus nerves. NTS neurons (dashed lines) project to and inhibit cardiac preganglionic parasympathetic neurons primarily in the nucleus ambiguus. Some are also located in the dorsal motor nucleus of the vagus; however, this nucleus primarily contains vagal motor neurons that project to the gastrointestinal tract. AP, area postrema; Pyr, pyramid; XII, hypoglossal nucleus. Dorsal motor nucleus AP NTS Nucleus ambiguus Vagus nerve Heart Pyr XII TABLE 33–3 Factors affecting heart rate. Heart rate accelerated by: Decreased activity of arterial baroreceptors Increased activity of atrial stretch receptors Inspiration Excitement Anger Most painful stimuli Hypoxia Exercise Thyroid hormones Fever Heart rate slowed by: Increased activity of arterial baroreceptors Expiration Fear Grief Stimulation of pain fibers in trigeminal nerve Increased intracranial pressure FIGURE 33–4 Baroreceptor areas in the carotid sinus and aortic arch. X, sites where receptors are located. The carotid and aor-tic bodies, which contain chemoreceptors, are also shown. X X X X X X X X X X X X Carotid body External carotid artery Carotid sinus Aortic body Internal carotid artery Common carotid artery X X X X X X X X X X X X Left common carotid artery Left subclavian artery Aortic body Innominate artery Aortic arch (viewed from behind) 560 SECTION VI Cardiovascular Physiology BARORECEPTOR NERVE ACTIVITY Baroreceptors are more sensitive to pulsatile pressure than to constant pressure. A decline in pulse pressure without any change in mean pressure decreases the rate of baroreceptor dis-charge and provokes a rise in systemic blood pressure and tachycardia. At normal blood pressure levels (about 100 mm Hg mean pressure), a burst of action potentials appears in a single baroreceptor fiber during systole, but there are few action po-tentials in early diastole (Figure 33–5). At lower mean pres-sures, this phasic change in firing is even more dramatic with activity only occurring during systole. At these lower pres-sures, the overall firing rate is considerably reduced. The threshold for eliciting activity in the carotid sinus nerve is about 50 mm Hg; maximal activity occurs at about 200 mm Hg. When one carotid sinus is isolated and perfused and the other baroreceptors are denervated, there is no discharge in the afferent fibers from the perfused sinus and no drop in the animal’s arterial pressure or heart rate when the perfusion pressure is below 30 mm Hg (Figure 33–6). At carotid sinus perfusion pressures of 70–110 mm Hg, there is a near linear relationship between perfusion pressure and the fall in sys-temic blood pressure and heart rate. At perfusion pressures above 150 mm Hg there is no further increase in response, pre-sumably because the rate of baroreceptor discharge and the degree of inhibition of sympathetic nerve activity are maximal. From the foregoing discussion, it is apparent that the barore-ceptors on the arterial side of the circulation, their afferent connections to the medullary cardiovascular areas, and the efferent pathways from these areas constitute a reflex feedback mechanism that operates to stabilize blood pressure and heart rate. Any drop in systemic arterial pressure decreases the inhibitory discharge in the buffer nerves, and there is a com-pensatory rise in blood pressure and cardiac output. Any rise in pressure produces dilation of the arterioles and decreases cardiac output until the blood pressure returns to its previous normal level. BARORECEPTOR RESETTING In chronic hypertension, the baroreceptor reflex mechanism is “reset” to maintain an elevated rather than a normal blood pressure. In perfusion studies on hypertensive experimental animals, raising the pressure in the isolated carotid sinus low-ers the elevated systemic pressure, and decreasing the perfu-sion pressure raises the elevated pressure (Figure 33–6). Little is known about how and why this occurs, but resetting occurs rapidly in experimental animals. It is also rapidly reversible, both in experimental animals and in clinical situations. ROLE OF BARORECEPTORS IN SHORT-TERM CONTROL OF BLOOD PRESSURE The changes in pulse rate and blood pressure that occur in hu-mans on standing up or lying down are due for the most part to baroreceptor reflexes. The function of the receptors can be tested by monitoring changes in heart rate as a function of in-creasing arterial pressure during infusion of the α-adrenergic agonist phenylephrine. A normal response is shown in Figure 33–7; from a systolic pressure of about 120 to 150 mm Hg, there is a linear relation between pressure and lowering of the heart rate (greater RR interval). Baroreceptors are very impor-tant in short-term control of arterial pressure. Activation of the reflex allows for rapid adjustments in blood pressure in FIGURE 33–5 Discharges (vertical lines) in a single afferent nerve fiber from the carotid sinus at various levels of mean arterial pressures, plotted against changes in aortic pressure with time. Baroreceptors are very sensitive to changes in pulse pressure as shown by the record of phasic aortic pressure. (Reproduced with permission from Berne RM, Levy MN: Cardiovascular Physiology, 3rd ed. Mosby, 1977.) 2.0 1.5 1.0 Time (s) 0.5 0 200 125 100 75 50 Phasic aortic pressure Mean arterial pressures (mm Hg) FIGURE 33–6 Fall in systemic blood pressure produced by raising the pressure in the isolated carotid sinus to various values. Solid line: Response in a normal monkey. Dashed line: Re-sponse in a hypertensive monkey, demonstrating baroreceptor reset-ting (arrow). 80 70 60 50 40 30 20 10 0 50 100 150 200 Pressure in carotid sinus (mm Hg) % fall in systemic blood pressure CHAPTER 33 Cardiovascular Regulatory Mechanisms 561 response to abrupt changes in blood volume, cardiac output, or peripheral resistance during exercise. Blood pressure initially rises dramatically after bilateral sec-tion of baroreceptor nerves or bilateral lesions of the NTS. However, after a period of time, mean blood pressure returns to near control levels, but there are huge fluctuations in pres-sure during the course of a day. Removal of the baroreceptor reflex prevents an individual from responding to stimuli that cause abrupt changes in blood volume, cardiac output, or peripheral resistance, including exercise and postural changes. A long-term change in blood pressure resulting from loss of baroreceptor reflex control is called neurogenic hypertension. ATRIAL STRETCH RECEPTORS The stretch receptors in the atria are of two types: those that dis-charge primarily during atrial systole (type A), and those that discharge primarily late in diastole, at the time of peak atrial fill-ing (type B). The discharge of type B baroreceptors is increased when venous return is increased and decreased by positive-pressure breathing, indicating that these baroreceptors respond primarily to distention of the atrial walls. The reflex circulatory adjustments initiated by increased discharge from most if not all of these receptors include vasodilation and a fall in blood pres-sure. However, the heart rate is increased rather than decreased. CARDIOPULMONARY RECEPTORS Receptors in the endocardial surfaces of the ventricles are ac-tivated during ventricular distention. The response is a vagal bradycardia and hypotension, comparable to a baroreceptor reflex. Left ventricular stretch receptors may play a role in the maintenance of vagal tone that keeps the heart rate low at rest. Various chemicals are known to elicit reflexes due to activa-tion of cardiopulmonary chemoreceptors and may play a role in various cardiovascular disorders (see Clinical Box 33–2). VALSALVA MANEUVER The function of the receptors can also be tested by monitoring the changes in pulse and blood pressure that occur in response FIGURE 33–7 Baroreflex-mediated lowering of the heart rate during infusion of phenylephrine in a human subject. Note that the values for the RR interval of the electrocardiogram, which are plotted on the vertical axis, are inversely proportionate to the heart rate. (Reproduced with permission from Kotrly K et al: Effects of fentanyl-diazepam-nitrous oxide anaesthesia on arterial baroreflex control of heart rate in man. Br J Anaesth 1986;58:406.) 1800 1600 1400 1200 1000 800 600 80 100 120 140 160 Systolic pressure (mm Hg) Slope = 33.3 ms mm Hg−1 Threshold = 124 mm Hg RR interval (ms) CLINICAL BOX 33–2 Cardiopulmonary Chemosensitive Receptors For nearly 150 years, it has been known that activation of chemosensitive vagal C fibers in the cardiopulmonary re-gion (eg, juxtacapillary region of alveoli, ventricles, atria, great veins, and pulmonary artery) causes profound brady-cardia, hypotension, and a brief period of apnea followed by rapid shallow breathing. This response pattern is called the Bezold–Jarisch reflex and was named after the individuals who first reported these findings. This reflex can be elicited by a variety of substances including capsaicin, serotonin, phenylbiguanide, and veratridine in cats, rabbits, and ro-dents. Although originally viewed as a pharmacologic curi-osity, there is a growing body of evidence supporting the view that the Bezold–Jarisch reflex is activated during cer-tain pathophysiologic conditions. For example, this reflex may be activated during myocardial ischemia and reperfu-sion as a result of increased production of oxygen radicals and by agents used as radio-contrast for coronary angiogra-phy. This can contribute to the hypotension that is fre-quently a stubborn complication of this disease. Activation of cardiopulmonary chemosensitive receptors may also be part of a defense mechanism protecting individuals from toxic chemical hazards. Activation of cardiopulmonary re-flexes may help reduce the amount of inspired pollutants that get absorbed into the blood, protecting vital organs from potential toxicity of these pollutants, and facilitating the elimination of the pollutants. Finally, the syndrome of cardiac slowing with hypotension (vasovagal syncope) has also been attributed to activation of the Bezold–Jarisch re-flex. Vasovagal syncope can occur after prolonged upright posture that results in pooling of blood in the lower extrem-ities and diminished intracardiac blood volume (also called postural syncope). This phenomenon is exaggerated if combined with dehydration. The resultant arterial hypoten-sion is sensed in the carotid sinus baroreceptors, and affer-ent fibers from these receptors trigger autonomic signals that increase cardiac rate and contractility. However, pres-sure receptors in the wall of the left ventricle respond by sending signals that trigger paradoxical bradycardia and de-creased contractility, resulting in sudden marked hypoten-sion. The individual also feels lightheaded and may experi-ence a brief episode of loss of consciousness. 562 SECTION VI Cardiovascular Physiology to brief periods of straining (forced expiration against a closed glottis: the Valsalva maneuver). Valsalva maneuvers occur regularly during coughing, defecation, and heavy lifting. The blood pressure rises at the onset of straining (Figure 33–8) be-cause the increase in intrathoracic pressure is added to the pressure of the blood in the aorta. It then falls because the high intrathoracic pressure compresses the veins, decreasing venous return and cardiac output. The decreases in arterial pressure and pulse pressure inhibit the baroreceptors, causing tachycar-dia and a rise in peripheral resistance. When the glottis is opened and the intrathoracic pressure returns to normal, car-diac output is restored but the peripheral vessels are constrict-ed. The blood pressure therefore rises above normal, and this stimulates the baroreceptors, causing bradycardia and a drop in pressure to normal levels. In sympathectomized patients, heart rate changes still occur because the baroreceptors and the vagi are intact. How-ever, in patients with autonomic insufficiency, a syndrome in which autonomic function is widely disrupted, the heart rate changes are absent. For reasons that are still obscure, patients with primary hyperaldosteronism also fail to show the heart rate changes and the blood pressure rise when the intratho-racic pressure returns to normal. Their response to the Val-salva maneuver returns to normal after removal of the aldosterone-secreting tumor. PERIPHERAL CHEMORECEPTOR REFLEX Peripheral arterial chemoreceptors in the carotid and aortic bodies (Figure 33–2) have very high rates of blood flow. These receptors are primarily activated by a reduction in partial pres-sure of oxygen (PaO2), but they also respond to an increase in the partial pressure of carbon dioxide (PaCO2) and pH. Chemoreceptors exert their main effects on respiration; how-ever, their activation also leads to vasoconstriction. Heart rate changes are variable and depend on various factors, including changes in respiration. A direct effect of chemoreceptor activa-tion is to increase vagal nerve activity. However, hypoxia also produces hyperpnea and increased catecholamine secretion from the adrenal medulla, both of which produce tachycardia and an increase in cardiac output. Hemorrhage that produces hypotension leads to chemoreceptor stimulation due to de-creased blood flow to the chemoreceptors and consequent stagnant anoxia of these organs. Chemoreceptor discharge may also contribute to the production of Mayer waves. These should not be confused with Traube–Hering waves, which are fluctuations in blood pressure synchronized with respiration. The Mayer waves are slow, regular oscillations in arterial pres-sure that occur at the rate of about one per 20–40 s during hy-potension. Under these conditions, hypoxia stimulates the chemoreceptors. The stimulation raises the blood pressure, which improves the blood flow in the receptor organs and eliminates the stimulus to the chemoreceptors, so that the pres-sure falls and a new cycle is initiated. DIRECT EFFECTS ON THE RVLM When intracranial pressure is increased, the blood supply to RVLM neurons is compromised, and the local hypoxia and hypercapnia increase their discharge. The resultant rise in sys-temic arterial pressure (Cushing reflex) tends to restore the blood flow to the medulla and over a considerable range, the blood pressure rise is proportional to the increase in intracra-nial pressure. The rise in blood pressure causes a reflex de-crease in heart rate via the arterial baroreceptors. This is why bradycardia rather than tachycardia is characteristically seen in patients with increased intracranial pressure. A rise in arterial PCO2 stimulates the RVLM, but the direct peripheral effect of hypercapnia is vasodilation. Therefore, the peripheral and central actions tend to cancel each other out. Moderate hyperventilation, which significantly lowers the CO2 tension of the blood, causes cutaneous and cerebral FIGURE 33–8 Diagram of the response to straining (the Valsalva maneuver) in a normal man, recorded with a needle in the brachial artery. Blood pressure rises at the onset of straining because increased intrathoracic pressure is added to the pressure of the blood in the aorta. It then falls because the high intrathoracic pressure compresses veins, decreasing venous return and cardiac output. (Courtesy of M Mcllroy.) Esophageal pressure (cm H2O) Arterial pressure (mm Hg) +40 0 200 0 −40 Start Stop 10 s CHAPTER 33 Cardiovascular Regulatory Mechanisms 563 vasoconstriction in humans, but there is little change in blood pressure. Exposure to high concentrations of CO2 is associ-ated with marked cutaneous and cerebral vasodilation, but vasoconstriction occurs elsewhere and usually there is a slow rise in blood pressure. LOCAL REGULATION AUTOREGULATION The capacity of tissues to regulate their own blood flow is re-ferred to as autoregulation. Most vascular beds have an intrin-sic capacity to compensate for moderate changes in perfusion pressure by changes in vascular resistance, so that blood flow remains relatively constant. This capacity is well developed in the kidneys (see Chapter 38), but it has also been observed in the mesentery, skeletal muscle, brain, liver, and myocardium. It is probably due in part to the intrinsic contractile response of smooth muscle to stretch (myogenic theory of autoregula-tion). As the pressure rises, the blood vessels are distended and the vascular smooth muscle fibers that surround the vessels contract. If it is postulated that the muscle responds to the ten-sion in the vessel wall, this theory could explain the greater de-gree of contraction at higher pressures; the wall tension is proportional to the distending pressure times the radius of the vessel (law of Laplace; see Chapter 32), and the maintenance of a given wall tension as the pressure rises would require a de-crease in radius. Vasodilator substances tend to accumulate in active tissues, and these “metabolites” also contribute to auto-regulation (metabolic theory of autoregulation). When blood flow decreases, they accumulate and the vessels dilate; when blood flow increases, they tend to be washed away. VASODILATOR METABOLITES The metabolic changes that produce vasodilation include, in most tissues, decreases in O2 tension and pH. These changes cause relaxation of the arterioles and precapillary sphincters. A local fall in O2 tension, in particular, can initiate a program of vasodilatory gene expression secondary to production of hypoxia-inducible factor-1α (HIF-1α), a transcription factor with multiple targets. Increases in CO2 tension and osmolality also dilate the vessels. The direct dilator action of CO2 is most pronounced in the skin and brain. The neurally mediated vasoconstrictor effects of systemic as opposed to local hypoxia and hypercapnia have been discussed above. A rise in temper-ature exerts a direct vasodilator effect, and the temperature rise in active tissues (due to the heat of metabolism) may con-tribute to the vasodilation. K+ is another substance that accu-mulates locally, and has demonstrated dilator activity secondary to the hyperpolarization of vascular smooth muscle cells. Lactate may also contribute to the dilation. In injured tis-sues, histamine released from damaged cells increases capil-lary permeability. Thus, it is probably responsible for some of the swelling in areas of inflammation. Adenosine may play a vasodilator role in cardiac muscle but not in skeletal muscle. It also inhibits the release of norepinephrine. LOCALIZED VASOCONSTRICTION Injured arteries and arterioles constrict strongly. The constric-tion appears to be due in part to the local liberation of seroto-nin from platelets that stick to the vessel wall in the injured area. Injured veins also constrict. A drop in tissue temperature causes vasoconstriction, and this local response to cold plays a part in temperature regula-tion (see Chapter 18). SUBSTANCES SECRETED BY THE ENDOTHELIUM ENDOTHELIAL CELLS As noted in Chapter 32, the endothelial cells constitute a large and important tissue. They secrete many growth factors and va-soactive substances. The vasoactive substances include prosta-glandins and thromboxanes, nitric oxide, and endothelins. PROSTACYCLIN & THROMBOXANE A2 Prostacyclin is produced by endothelial cells and thrombox-ane A2 by platelets from their common precursor arachidonic acid via the cyclooxygenase pathway. Thromboxane A2 pro-motes platelet aggregation and vasoconstriction, whereas prostacyclin inhibits platelet aggregation and promotes vaso-dilation. The balance between platelet thromboxane A2 and prostacyclin fosters localized platelet aggregation and conse-quent clot formation (see Chapter 32) while preventing ex-cessive extension of the clot and maintaining blood flow around it. The thromboxane A2–prostacyclin balance can be shifted toward prostacyclin by administration of low doses of aspirin. Aspirin produces irreversible inhibition of cyclooxygenase by acetylating a serine residue in its active site. Obviously, this reduces production of both thromboxane A2 and prostacyclin. However, endothelial cells produce new cyclooxygenase in a matter of hours, whereas platelets cannot manufacture the enzyme, and the level rises only as new platelets enter the circu-lation. This is a slow process because platelets have a half-life of about 4 days. Therefore, administration of small amounts of aspirin for prolonged periods reduces clot formation and has been shown to be of value in preventing myocardial infarctions, unstable angina, transient ischemic attacks, and stroke. NITRIC OXIDE A chance observation two decades ago led to the discovery that the endothelium plays a key role in vasodilation. Many 564 SECTION VI Cardiovascular Physiology different stimuli act on the endothelial cells to produce endo-thelium-derived relaxing factor (EDRF), a substance that is now known to be nitric oxide (NO). NO is synthesized from arginine (Figure 33–9) in a reaction catalyzed by nitric oxide synthase (NO synthase, NOS). Three isoforms of NOS have been identified: NOS 1, found in the nervous system; NOS 2, found in macrophages and other immune cells; and NOS 3, found in endothelial cells. NOS 1 and NOS 3 are activated by agents that increase intracellular Ca2+ concentrations, includ-ing the vasodilators acetylcholine and bradykinin. The NOS in immune cells is not activated by Ca2+ but is induced by cyto-kines. The NO that is formed in the endothelium diffuses to smooth muscle cells, where it activates soluble guanylyl cy-clase, producing cyclic 3,5-guanosine monophosphate (cG-MP; see Figure 33–9), which in turn mediates the relaxation of vascular smooth muscle. NO is inactivated by hemoglobin. Adenosine, atrial natriuretic peptide (ANP), and histamine via H2 receptors produce relaxation of vascular smooth muscle that is independent of the endothelium. However, acetylcho-line, histamine via H1 receptors, bradykinin, vasoactive intesti-nal peptide (VIP), substance P, and some other polypeptides act via the endothelium, and various vasoconstrictors that act directly on vascular smooth muscle would produce much greater constriction if their effects were not limited by their ability simultaneously to cause release of NO. When flow to a tissue is suddenly increased by arteriolar dilation, the large arteries to the tissue also dilate. This flow-induced dilation is due to local release of NO. Products of platelet aggregation also cause release of NO, and the resulting vasodilation helps keep blood vessels with an intact endothelium patent. This is in contrast to injured blood vessels, where the endothelium is damaged at the site of injury and platelets therefore aggregate and produce vasoconstriction (see Chapter 32). Further evidence for a physiologic role of NO is the obser-vation that mice lacking NOS 3 are hypertensive. This sug-gests that tonic release of NO is necessary to maintain normal blood pressure. NO is also involved in vascular remodeling and angiogene-sis, and NO may be involved in the pathogenesis of athero-sclerosis. It is interesting in this regard that some patients with heart transplants develop an accelerated form of atherosclero-sis in the vessels of the transplant, and there is reason to believe that this is triggered by endothelial damage. Nitroglyc-erin and other nitrovasodilators that are of great value in the treatment of angina act by stimulating guanylyl cyclase in the same manner as NO. Penile erection is also produced by release of NO, with con-sequent vasodilation and engorgement of the corpora caver-nosa (see Chapter 25). This accounts for the efficacy of drugs such as Viagra, which slow the breakdown of cGMP. OTHER FUNCTIONS OF NO NO is present in the brain and, acting via cGMP, it is impor-tant in brain function (see Chapter 7). It is necessary for the antimicrobial and cytotoxic activity of various inflammatory cells, although the net effect of NO in inflammation and tissue injury depends on the amount and kinetics of release, which in turn may depend on the specific NOS isoform involved. In the gastrointestinal tract, it is important in the relaxation of smooth muscle. Other functions of NO are mentioned in oth-er parts of this book. CARBON MONOXIDE The production of carbon monoxide (CO) from heme is shown in Figure 29–4. HO2, the enzyme that catalyzes the re-action, is present in cardiovascular tissues, and there is grow-ing evidence that CO as well as NO produces local dilation in blood vessels. Interestingly, hydrogen sulfide is likewise emerging as a third gaseotransmitter that regulates vascular tone, although the relative roles of NO, CO, and H2S have yet to be established. ENDOTHELINS Endothelial cells also produce endothelin-1, one of the most potent vasoconstrictor agents yet isolated. Endothelin-1 (ET-1), endothelin-2 (ET-2), and endothelin-3 (ET-3) are the members of a family of three similar 21-amino-acid polypep-tides (Figure 33–10). Each is encoded by a different gene. The unique structure of the endothelins resembles that of the sa-rafotoxins, polypeptides found in the venom of a snake, the Israeli burrowing asp. FIGURE 33–9 Synthesis of NO from arginine in endothelial cells and its action via stimulation of soluble guanylyl cyclase and generation of cGMP to produce relaxation in vascular smooth muscle cells. The endothelial form of nitric oxide synthase (NOS) is ac-tivated by increased intracellular Ca2+ concentration, and an increase is produced by acetylcholine (Ach), bradykinin, or shear stress acting on the cell membrane. Thiol, tetrahydrobiopterin, FAD, and FMN are requisite cofactors. Ach Bradykinin Shear stress Ca2+ NOS L-Arginine + O2 + NADPH Citruline + NO + NADP Thiol Tetrahydro- biopterin FAD FMN GTP cGMP Smooth muscle relaxation Soluble guanylyl cyclase CHAPTER 33 Cardiovascular Regulatory Mechanisms 565 ENDOTHELIN-1 In endothelial cells, the product of the endothelin-1 gene is processed to a 39-amino-acid prohormone, big endothelin-1, which has about 1% of the activity of endothelin-1. The pro-hormone is cleaved at a tryptophan-valine (Trp-Val) bond to form endothelin-1 by endothelin-converting enzyme. Small amounts of big endothelin-1 and endothelin-1 are secreted into the blood, but for the most part, they are secreted locally and act in a paracrine fashion. Two different endothelin receptors have been cloned, both of which are coupled via G proteins to phospholipase C (see Chapter 2). The ETA receptor, which is specific for endothe-lin-1, is found in many tissues and mediates the vasoconstric-tion produced by endothelin-1. The ETB receptor responds to all three endothelins, and is coupled to Gi. It may mediate vasodilation, and it appears to mediate the developmental effects of the endothelins (see below). REGULATION OF SECRETION Endothelin-1 is not stored in secretory granules, and most reg-ulatory factors alter the transcription of its gene, with changes in secretion occurring promptly thereafter. Factors activating and inhibiting the gene are summarized in Table 33–4. CARDIOVASCULAR FUNCTIONS As noted above, endothelin-1 appears to be primarily a para-crine regulator of vascular tone. However, endothelin-1 is not increased in hypertension, and in mice in which one allele of the endothelin-1 gene is knocked out, blood pressure is actu-ally elevated rather than reduced. The concentration of circu-lating endothelin-1 is, however, elevated in congestive heart failure and after myocardial infarction, so it may play a role in the pathophysiology of these diseases. OTHER FUNCTIONS OF ENDOTHELINS Endothelin-1 is found in the brain and kidneys as well as the endothelial cells. Endothelin-2 is produced primarily in the kidneys and intestine. Endothelin-3 is present in the blood and is found in high concentrations in the brain. It is also found in the kidneys and gastrointestinal tract. In the brain, endothelins are abundant and, in early life, are produced by both astrocytes and neurons. They are found in the dorsal root ganglia, ventral horn cells, the cortex, the hypothalamus, and cerebellar Purkinje cells. They also play a role in regulating transport across the blood–brain barrier. There are endothelin receptors on mesangial cells (see Chapter 38), and the polypeptide participates in tubuloglomerular feedback. Mice that have both alleles of the endothelin-1 gene deleted have severe craniofacial abnormalities and die of respiratory failure at birth. They also have megacolon (Hirschsprung FIGURE 33–10 Structure of human endothelins and one of the snake venom sarafotoxins. The amino acid residues that differ from endothelin-1 are indicated in pink. L Y Y T F T K W H L I I C C S D C E K D M L S S C Y F V Endothelin-1 W H L I I C C D C E K D C Y V Endothelin-3 M F D K S T W H Q V I C C D C E K D C Y Sarafotoxin b W L W H L I I C C S D C E K D S S C Y F V Endothelin-2 TABLE 33–4 Regulation of endothelin-1 secretion via transcription of its gene. Stimulators Angiotensin II Catecholamines Growth factors Hypoxia Insulin Oxidized LDL HDL Shear stress Thrombin Inhibitors NO ANP PGE2 Prostacyclin 566 SECTION VI Cardiovascular Physiology disease), apparently because the cells that normally form the myenteric plexus fail to migrate to the distal colon. In addition, endothelins play a role in closing the ductus arteriosus at birth. SYSTEMIC REGULATION BY HORMONES Many circulating hormones affect the vascular system. The vasodilator hormones include kinins, VIP, and ANP. Circu-lating vasoconstrictor hormones include vasopressin, norepi-nephrine, epinephrine, and angiotensin II. KININS Two related vasodilator peptides called kinins are found in the body. One is the nonapeptide bradykinin, and the other is the decapeptide lysylbradykinin, also known as kallidin (Figure 33–11). Lysylbradykinin can be converted to bradykinin by aminopeptidase. Both peptides are metabolized to inactive fragments by kininase I, a carboxypeptidase that removes the carboxyl terminal arginine (Arg). In addition, the dipeptidyl-carboxypeptidase kininase II inactivates bradykinin and ly-sylbradykinin by removing phenylalanine-arginine (Phe-Arg) from the carboxyl terminal. Kininase II is the same enzyme as angiotensin-converting enzyme (see Chapter 39), which re-moves histidine-leucine (His-Leu) from the carboxyl terminal end of angiotensin I. Bradykinin and lysylbradykinin are formed from two pre-cursor proteins: high-molecular-weight kininogen and low-molecular-weight kininogen (Figure 33–12). They are formed by alternative splicing of a single gene located on chromosome 3. Proteases called kallikreins release the pep-tides from their precursors. They are produced in humans by a family of three genes located on chromosome 19. There are two types of kallikreins: plasma kallikrein, which circulates in an inactive form, and tissue kallikrein, which appears to be located primarily on the apical membranes of cells con-cerned with transcellular electrolyte transport. Tissue kal-likrein is found in many tissues, including sweat and salivary glands, the pancreas, the prostate, the intestine, and the kid-neys. Tissue kallikrein acts on high-molecular-weight kinino-gen to form bradykinin and low-molecular-weight kininogen to form lysylbradykinin. When activated, plasma kallikrein acts on high-molecular-weight kininogen to form bradykinin. Inactive plasma kallikrein (prekallikrein) is converted to the active form, kallikrein, by active factor XII, the factor that initiates the intrinsic blood clotting cascade. Kallikrein also activates factor XII in a positive feedback loop, and high-molecular-weight kininogen has a factor XII-activating action (see Figure 32–13). The actions of both kinins resemble those of histamine. They are primarily tissue hormones, although small amounts are also found in the circulating blood. They cause contrac-tion of visceral smooth muscle, but they relax vascular smooth muscle via NO, lowering blood pressure. They also increase capillary permeability, attract leukocytes, and cause pain upon injection under the skin. They are formed during active secretion in sweat glands, salivary glands, and the exo-crine portion of the pancreas, and they are probably responsi-ble for the increase in blood flow when these tissues are actively secreting their products. Two bradykinin receptors, B1 and B2, have been identified. Their amino acid residues are 36% identical, and both are coupled to G proteins. The B1 receptor may mediate the pain-producing effects of the kinins, but little is known about its distribution and function. The B2 receptor has strong homol-ogy to the H2 receptor and is found in many different tissues. NATRIURETIC HORMONES There is a family of natriuretic peptides involved in vascular regulation, including atrial natriuretic peptide (ANP) secreted by the heart, brain natriuretic peptide (BNP), and C-type natriuretic peptide (CNP). They are released in response to hypervolemia. ANP and BNP circulate, whereas CNP acts pre-dominantly in a paracrine fashion. In general, these peptides antagonize the action of various vasoconstrictor agents and lower blood pressure. ANP and BNP also serve to coordinate the control of vascular tone with fluid and electrolyte homeo-stasis via actions on the kidney. FIGURE 33–11 Kinins. Lysylbradykinin (top) can be converted to bradykinin (bottom) by aminopeptidase. The peptides are inacti-vated by kininase I (KI) or kininase II (KII) at the sites indicated by the short arrows. Lys Aminopeptidase KII KI KII KI Arg Phe Pro Ser Phe Gly Pro Phe Pro Ser Phe Gly Pro Pro Pro Arg Arg Arg Arg FIGURE 33–12 Formation of kinins from high-molecular-weight (HMW) and low-molecular-weight (LMW) kininogens. XII XIIa Clotting Plasma kallikrein Prekallikrein Bradykinin HMW kininogen LMW kininogen Tissue kallikrein Lysylbradykinin CHAPTER 33 Cardiovascular Regulatory Mechanisms 567 CIRCULATING VASOCONSTRICTORS Vasopressin is a potent vasoconstrictor, but when it is injected in normal individuals, there is a compensating decrease in car-diac output, so that there is little change in blood pressure. Its role in blood pressure regulation is discussed in Chapter 18. Norepinephrine has a generalized vasoconstrictor action, whereas epinephrine dilates the vessels in skeletal muscle and the liver. The relative unimportance of circulating norepineph-rine, as opposed to norepinephrine released from vasomotor nerves, is pointed out in Chapter 22, where the cardiovascular actions of catecholamines are discussed in detail. Angiotensin II has a generalized vasoconstrictor action. It is formed by the action of angiotensin converting enzyme (ACE) on angiotensin I, which itself is liberated by the action of renin from the kidney on circulating angiotensinogen (see Chapter 39). Renin secretion, in turn, is increased when the blood pres-sure falls or extracellular fluid (ECF) volume is reduced, and angiotensin II therefore helps to maintain blood pressure. Angiotensin II also increases water intake and stimulates aldosterone secretion, and increased formation of angiotensin II is part of a homeostatic mechanism that operates to main-tain ECF volume (see Chapter 22). In addition, there are ren-nin–angiotensin systems in many different organs, and there may be one in the walls of blood vessels. Angiotensin II pro-duced in blood vessel walls could be important in some forms of clinical hypertension. The role of angiotensin II in cardio-vascular regulation is also amply demonstrated in the wide-spread use of so-called ACE inhibitors as antihypertensive medications. Urotensin-II, a polypeptide first isolated from the spinal cord of fish, is present in human cardiac and vascular tissue. It is one of the most potent mammalian vasoconstrictors known, but its pathophysiogic and physiologic roles are cur-rently the subject of intense interest. CHAPTER SUMMARY ■RVLM neurons project to the thoracolumbar IML and release glutamate on preganglionic sympathetic neurons that innervate the heart and vasculature. ■The NTS is the major excitatory input to cardiac vagal motor neurons in the nucleus ambiguus. ■Carotid sinus and aortic depressor baroreceptors are innervated by branches of the 9th and 10th cranial nerves, respectively (glossopharyngeal and aortic depressor nerves). These receptors are most sensitive to changes in pulse pressure but also respond to changes in mean arterial pressure. ■Baroreceptor nerves terminate in the NTS and release glutamate. NTS neurons project to the CVLM and nucleus ambiguus and release glutamate. CVLM neurons project to RVLM and release GABA. This leads to a reduction in sympathetic activity and an increase in vagal activity (ie, the baroreceptor reflex). ■Activation of peripheral chemoreceptors in the carotid and aor-tic bodies by a reduction in PaO2 or an increase in PaCO2 leads to an increase in vasoconstriction. Heart rate changes are vari-able and depend on a number of factors including changes in respiration. ■In addition to various neural inputs, RVLM neurons are directly activated by hypoxia and hypercapnia. ■Most vascular beds have an intrinsic capacity to respond to changes in blood pressure within a certain range by altering vascular resistance to maintain stable blood flow. This property is known as autoregulation. ■Local factors such as oxygen tension, pH, temperature, and metabolic products contribute to vascular regulation; many produce vasodilation to restore blood flow. ■The endothelium is an important source of vasoactive media-tors that act to either contract or relax vascular smooth muscle. ■Three gaseous mediators—NO, CO, and H2S—are important regulators of vasodilation. ■Endothelins and angiotensin II induce vasoconstriction and may be involved in the pathogenesis of some forms of hypertension. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. When a pheochromocytoma (tumor of the adrenal medulla) suddenly discharges a large amount of epinephrine into the cir-culation, the patient’s heart rate would be expected to A) increase, because the increase in blood pressure stimulates the carotid and aortic baroreceptors. B) increase, because epinephrine has a direct chronotropic effect on the heart. C) increase, because of increased tonic parasympathetic discharge to the heart. D) decrease, because the increase in blood pressure stimulates the carotid and aortic chemoreceptors. E) decrease, because of increased tonic parasympathetic discharge to the heart. 2. Activation of the baroreceptor reflex A) is primarily involved in short-term regulation of systemic blood pressure. B) leads to an increase in heart rate because of inhibition of the vagal cardiac motor neurons. C) inhibits neurons in the CVLM. D) excites neurons in the RVLM. E) all of the above 3. Sympathetic nerve activity would be expected to increase A) if glutamate receptors were blocked in the NTS. B) if GABA receptors were blocked in the RVLM. C) if there was a compression of the RVLM. D) during hypoxia. E) for all of the above. 4. Why is the dilator response to injected acetylcholine changed to a constrictor response when the endothelium is damaged? A) More Na+ is generated. B) More bradykinin is generated. C) The damage lowers the pH of the remaining layers of the artery. D) The damage augments the production of endothelin by the endothelium. E) The damage interferes with the production of NO by the endothelium. 568 SECTION VI Cardiovascular Physiology CHAPTER RESOURCES Ahluwalia A, MacAllister RJ, Hobbs AJ: Vascular actions of natriuretic peptides. Cyclic GMP-dependent and -independent mechanisms. Basic Res Cardiol 2004;99:83. Benarroch EE: Central Autonomic Network. Functional Organization and Clinical Correlations. Futura Publishing, 1997. Chapleau MW, Abboud F (editors): Neuro-cardiovascular regulation: From molecules to man. Ann NY Acad Sci 2001;940. de Burgh Daly M: Peripheral Arterial Chemoreceptors and Respiratory-Cardiovascular Integration. Clarendon Press, 1997. Haddy FJ, Vanhouttee PM, Feletou M: Role of potassium in regulating blood flow and blood pressure. Am J Physiol Regul Integr Comp Physiol 2006;290:R546. Loewy AD, Spyer KM (editors): Central Regulation of Autonomic Function. Oxford University Press, 1990. Marshall JM: Peripheral chemoreceptors and cardiovascular regulation. Physiol Rev 1994;74:543. Paffett ML, Walker BR: Vascular adaptations to hypoxia: Molecular and cellular mechanisms regulating vascular tone. Essays Biochem 2007;43:105. Squire LR, Bloom FE, Spitzer NC, du Lac S, Ghosh A, Berg D (editors): Fundamental Neuroscience, 3rd ed. Academic Press, 2008. Trouth CO, Millis RM, Kiwull-Schöne HF, Schläfke ME: Ventral Brainstem Mechanisms and Control of Respiration and Blood Pressure. Marcel Dekker, 1995. 569 C H A P T E R 34 Circulation Through Special Regions O B J E C T I V E S After studying this chapter, you should be able to: ■Define the special features of the circulation in the brain, coronary vessels, skin, and fetus, and how these are regulated. ■Describe how cerebrospinal fluid (CSF) is formed and reabsorbed, and its role in protecting the brain from injury. ■Understand how the blood–brain barrier impedes the entry of specific substances into the brain. ■Delineate how the oxygen needs of the contracting myocardium are met by the coronary arteries and the consequences of their occlusion. ■List the vascular reactions of the skin and the reflexes that mediate them. ■Understand how the fetus is supplied with oxygen and nutrients in utero, and the circulatory events required for a transition to independent life after birth. INTRODUCTION The distribution of the cardiac output to various parts of the body at rest in a normal man is shown in Table 34–1. The general principles described in preceding chapters apply to the circulation of all these regions, but the vascular supplies of many organs have additional special features that are impor-tant to their physiology. The portal circulation of the anterior pituitary is discussed in Chapter 24, the pulmonary circula-tion in Chapter 35, the renal circulation in Chapter 38, and the circulation of the splanchnic area, particularly the intes-tines and liver, in Chapters 26 and 29. This chapter is con-cerned with the special circulations of the brain, the heart, and the skin, as well as the placenta and fetus. CEREBRAL CIRCULATION: ANATOMIC CONSIDERATIONS VESSELS The principal arterial inflow to the brain in humans is via four arteries: two internal carotids and two vertebrals. In humans, the carotid arteries are quantitatively the most significant. The vertebral arteries unite to form the basilar artery, and the basi-lar artery and the carotids form the circle of Willis below the hypothalamus. The circle of Willis is the origin of the six large vessels supplying the cerebral cortex. Substances injected into one carotid artery are distributed almost exclusively to the ce-rebral hemisphere on that side. Normally no crossing over oc-curs, probably because the pressure is equal on both sides. Even when it is not, the anastomotic channels in the circle do not permit a very large flow. Occlusion of one carotid artery, particularly in older patients, often causes serious symptoms of cerebral ischemia. There are precapillary anastomoses be-tween the cerebral vessels, but flow through these channels is generally insufficient to maintain the circulation and prevent infarction when a cerebral artery is occluded. Venous drainage from the brain by way of the deep veins and dural sinuses empties principally into the internal jugular veins in humans, although a small amount of venous blood 570 SECTION VI Cardiovascular Physiology drains through the ophthalmic and pterygoid venous plex-uses, through emissary veins to the scalp, and down the sys-tem of paravertebral veins in the spinal canal. The cerebral vessels have a number of unique anatomic fea-tures. In the choroid plexuses, there are gaps between the endothelial cells of the capillary wall, but the choroid epithe-lial cells that separate them from the cerebrospinal fluid (CSF) are connected to one another by tight junctions. The capillar-ies in the brain substance resemble nonfenestrated capillaries in muscle (see Chapter 32), but there are tight junctions between the endothelial cells that limit the passage of sub-stances through the junctions. In addition, there are relatively few vesicles in the endothelial cytoplasm, and presumably lit-tle vesicular transport. However, multiple transport systems are present in the capillary cells. The brain capillaries are sur-rounded by the endfeet of astrocytes (Figure 34–1). These endfeet are closely applied to the basal lamina of the capillar-ies, but they do not cover the entire capillary wall, and gaps of about 20 nm occur between endfeet (Figure 34–2). However, the endfeet induce the tight junctions in the capillaries (see Chapter 32). The protoplasm of astrocytes is also found around synapses, where it appears to isolate the synapses in the brain from one another. INNERVATION Three systems of nerves innervate the cerebral blood vessels. Postganglionic sympathetic neurons have their cell bodies in the superior cervical ganglia, and their endings contain nor-epinephrine. Many also contain neuropeptide Y. Cholinergic neurons that probably originate in the sphenopalatine ganglia also innervate the cerebral vessels, and the postganglionic cho-linergic neurons on the blood vessels contain acetylcholine. Many also contain vasoactive intestinal peptide (VIP) and peptide histidyl methionine (PHM-27) (see Chapter 7). These nerves end primarily on large arteries. Sensory nerves are found on more distal arteries. They have their cell bodies in the trigeminal ganglia and contain substance P, neurokinin A, and calcitonin gene-related peptide (CGRP). Substance P, CGRP, VIP, and PHM-27 cause vasodilation, whereas TABLE 34–1 Resting blood flow and O2 consumption of various organs in a 63-kg adult man with a mean arterial blood pressure of 90 mm Hg and an O2 consumption of 250 mL/min. Blood Flow Arteriovenous Oxygen Difference (mL/L) Oxygen Consumption Resistance (R units)a Percentage of Total Region Mass (kg) mL/min mL/100 g/min mL/min mL/100 g/min Absolute per kg Cardiac Output Oxygen Consumption Liver 2.6 1500 57.7 34 51 2.0 3.6 9.4 27.8 20.4 Kidneys 0.3 1260 420.0 14 18 6.0 4.3 1.3 23.3 7.2 Brain 1.4 750 54.0 62 46 3.3 7.2 10.1 13.9 18.4 Skin 3.6 462 12.8 25 12 0.3 11.7 42.1 8.6 4.8 Skeletal muscle 31.0 840 2.7 60 50 0.2 6.4 198.4 15.6 20.0 Heart muscle 0.3 250 84.0 114 29 9.7 21.4 6.4 4.7 11.6 Rest of body 23.8 336 1.4 129 44 0.2 16.1 383.2 6.2 17.6 Whole body 63.0 5400 8.6 46 250 0.4 1.0 63.0 100.0 100.0 aR units are pressure (mm Hg) divided by blood flow (mL/s). Reproduced with permission from Bard P (editor): Medical Physiology, 11th ed. Mosby, 1961. FIGURE 34–1 Relation of fibrous astrocyte (3) to a capillary (2) and neuron (4) in the brain. The endfeet of the astrocyte process-es form a discontinuous membrane around the capillary (1). Astrocyte processes also envelop the neuron. (Adapted from Krstic RV: Die Gewebe des Menschen und der Säugetiere. Springer, 1978.) 2 3 1 4 CHAPTER 34 Circulation Through Special Regions 571 neuropeptide Y is a vasoconstrictor. Touching or pulling on the cerebral vessels causes pain. CEREBROSPINAL FLUID FORMATION & ABSORPTION CSF fills the ventricles and subarachnoid space. In humans, the volume of CSF is about 150 mL and the rate of CSF production is about 550 mL/d. Thus the CSF turns over about 3.7 times a day. In experiments on animals, it has been estimated that 50– 70% of the CSF is formed in the choroid plexuses and the re-mainder is formed around blood vessels and along ventricular walls. Presumably, the situation in humans is similar. The CSF in the ventricles flows through the foramens of Magendie and Luschka to the subarachnoid space and is absorbed through the arachnoid villi into veins, primarily the cerebral venous si-nuses. The villi consist of projections of the fused arachnoid membrane and endothelium of the sinuses into the venous si-nuses. Similar, smaller villi project into veins around spinal nerve routes. These projections may contribute to the outflow of CSF into venous blood by a process known as bulk flow, which is unidirectional. However, recent studies suggest that, at least in animals, a more important route for CSF reabsorp-tion into the bloodstream in health is via the cribriform plate above the nose and thence into the cervical lymphatics. How-ever, reabsorption via one-way valves (of uncertain structural basis) in the arachnoid villi may assume a greater role if CSF pressure is elevated. Likewise, when CSF builds up abnormally, aquaporin water channels may be expressed in the choroid plexus and brain microvessels as a compensatory adaptation. CSF is formed continuously by the choroid plexus in two stages. First, plasma is passively filtered across the choroidal capillary endothelium. Next, secretion of water and ions across the choroidal epithelium provides for active control of CSF composition and quantity. Bicarbonate, chloride, and potassium ions enter the CSF via channels in the epithelial cell apical membranes. Aquaporins provide for water movement to balance osmotic gradients. The composition of CSF (Table 34–2) is essentially the same as that of brain extracellular fluid (ECF), which in living humans makes up 15% of the brain volume. In adults, free communication appears to take place between the brain interstitial fluid and CSF, although the dif-fusion distances from some parts of the brain to the CSF are appreciable. Consequently, equilibration may take some time to occur, and local areas of the brain may have extracellular microenvironments that are transiently different from CSF. Lumbar CSF pressure is normally 70 to 180 mm H2O. Up to pressures well above this range, the rate of CSF formation is independent of intraventricular pressure. However, absorp-tion is proportional to the pressure (Figure 34–3). At a pres-sure of 112 mm H2O, which is the average normal CSF pressure, filtration and absorption are equal. Below a pressure of approximately 68 mm H2O, absorption stops. Large amounts of fluid accumulate when the capacity for CSF reab-sorption is decreased (external hydrocephalus, communi-cating hydrocephalus). Fluid also accumulates proximal to the block and distends the ventricles when the foramens of Luschka and Magendie are blocked or there is obstruction within the ventricular system (internal hydrocephalus, non-communicating hydrocephalus). FIGURE 34–2 Transport across cerebral capillaries. Glial endfoot Glucose, etc Tight junction Mitochondrion Nucleus Lipid-soluble diffusion, carrier-mediated transport TABLE 34–2 Concentration of various substances in human CSF and plasma. Substance CSF Plasma Ratio CSF/Plasma Na+ (meq/kg H2O) 147.0 150.0 0.98 K+ (meq/kg H2O) 2.9 4.6 0.62 Mg2+ (meq/kg H2O) 2.2 1.6 1.39 Ca2+ (meq/kg H2O) 2.3 4.7 0.49 Cl– (meq/kg H2O) 113.0 99.0 1.14 HCO3 – (meq/L) 25.1 24.8 1.01 PCO2 (mm Hg) 50.2 39.5 1.28 pH 7.33 7.40 . . . Osmolality (mosm/kg H2O) 289.0 289.0 1.00 Protein (mg/dL) 20.0 6000.0 0.003 Glucose (mg/dL) 64.0 100.0 0.64 Inorganic P (mg/dL) 3.4 4.7 0.73 Urea (mg/dL) 12.0 15.0 0.80 Creatinine (mg/dL) 1.5 1.2 1.25 Uric acid (mg/dL) 1.5 5.0 0.30 Cholesterol (mg/dL) 0.2 175.0 0.001 572 SECTION VI Cardiovascular Physiology PROTECTIVE FUNCTION The most critical role for CSF (and the meninges) is to protect the brain. The dura is attached firmly to bone. Normally, there is no “subdural space,” with the arachnoid being held to the dura by the surface tension of the thin layer of fluid between the two membranes. As shown in Figure 34–4, the brain itself is supported within the arachnoid by the blood vessels and nerve roots and by the multiple fine fibrous arachnoid trabeculae. The brain weighs about 1400 g in air, but in its “water bath” of CSF it has a net weight of only 50 g. The buoyancy of the brain in the CSF permits its relatively flimsy attachments to suspend it very effectively. When the head receives a blow, the arachnoid slides on the dura and the brain moves, but its motion is gently checked by the CSF cushion and by the arachnoid trabeculae. The pain produced by spinal fluid deficiency illustrates the importance of CSF in supporting the brain. Removal of CSF dur-ing lumbar puncture can cause a severe headache after the fluid is removed, because the brain hangs on the vessels and nerve roots, and traction on them stimulates pain fibers. The pain can be relieved by intrathecal injection of sterile isotonic saline. HEAD INJURIES Without the protection of the spinal fluid and the meninges, the brain would probably be unable to withstand even the mi-nor traumas of everyday living; but with the protection afford-ed, it takes a fairly severe blow to produce cerebral damage. The brain is damaged most commonly when the skull is frac-tured and bone is driven into neural tissue (depressed skull fracture), when the brain moves far enough to tear the delicate bridging veins from the cortex to the bone, or when the brain is accelerated by a blow on the head and is driven against the skull or the tentorium at a point opposite where the blow was struck (contrecoup injury). THE BLOOD–BRAIN BARRIER The tight junctions between capillary endothelial cells in the brain and between the epithelial cells in the choroid plexus ef-fectively prevent proteins from entering the brain in adults and slow the penetration of some smaller molecules as well. An example is the slow penetration of urea (Figure 34–5). This uniquely limited exchange of substances into the brain is re-ferred to as the blood–brain barrier, a term most commonly used to encompass this barrier overall and more specifically the barrier in the choroid epithelium between blood and CSF. Passive diffusion across the tight cerebral capillaries is very limited, and little vesicular transport takes place. However, there are numerous carrier-mediated and active transport sys-tems in the cerebral capillaries. These move substances out of as well as into the brain, though movement out of the brain is generally more free than movement into it. PENETRATION OF SUBSTANCES INTO THE BRAIN Water, CO2, and O2 penetrate the brain with ease, as do the lipid-soluble free forms of steroid hormones, whereas their FIGURE 34–3 CSF formation and absorption in humans at various CSF pressures. Note that at 112 mm CSF, formation and ab-sorption are equal, and at 68 mm CSF, absorption is zero. (Modified and reproduced with permission from Cutler RWP, et al: Formation and absorption of cerebrospinal fluid in man. Brain 1968;91:707.) 1.6 1.2 0.8 0.4 0 0 68 100 112 200 Outflow pressure (mm CSF) Formation Flow (mL/min) Absorption FIGURE 34–4 Investing membranes of the brain, showing their relation to the skull and to brain tissue. (Reproduced with permission from Wheater PR et al: Functional Histology. Churchill Livingstone, 1979.) Subarachnoid space Arachnoid trabeculae Arachnoid Artery Pia mater Brain Perivascular spaces Dura mater Inner table of skull Subdural (potential) space Trabecular bone Outer table of skull CHAPTER 34 Circulation Through Special Regions 573 protein-bound forms and, in general, all proteins and polypep-tides do not. The rapid passive penetration of CO2 contrasts with the regulated transcellular penetration of H+ and HCO3 – and has physiologic significance in the regulation of respira-tion (see Chapter 37). Glucose is the major ultimate source of energy for nerve cells. Its diffusion across the blood–brain barrier would be very slow, but the rate of transport into the CSF is markedly enhanced by the presence of specific transporters, including the glucose transporter 1 (GLUT 1). The brain contains two forms of GLUT 1: GLUT 1 55K and GLUT 1 45K. Both are encoded by the same gene, but they differ in the extent to which they are glycosylated. GLUT 1 55K is present in high concentration in brain capillaries (Figure 34–6). Infants with congenital GLUT 1 deficiency develop low CSF glucose concentrations in the pres-ence of normal plasma glucose, and they have seizures and delayed development. In addition, transporters for thyroid hor-mones; several organic acids; choline; nucleic acid precursors; and neutral, basic, and acidic amino acids are present at the blood–brain barrier. A variety of drugs and peptides actually cross the cerebral capillaries but are promptly transported back into the blood by a multidrug nonspecific transporter in the apical mem-branes of the endothelial cells. This P-glycoprotein is a mem-ber of the family of adenosine triphosphate (ATP) binding cassettes that transport various proteins and lipids across cell membranes (see Chapter 2). In the absence of this transporter in mice, much larger proportions of systemically adminis-tered doses of various chemotherapeutic drugs, analgesics, and opioid peptides are found in the brain than in controls. If pharmacologic agents that inhibit this transporter can be developed, they could be of value in the treatment of brain tumors and other central nervous system (CNS) diseases in which it is difficult to introduce adequate amounts of thera-peutic agents into the brain. CIRCUMVENTRICULAR ORGANS When dyes that bind to proteins in the plasma are injected, they stain many tissues but spare most of the brain. However, four small areas in or near the brain stem do take up the stain. These areas are (1) the posterior pituitary (neurohypophysis) and the adjacent ventral part of the median eminence of the hypothalamus, (2) the area postrema, (3) the organum vas-culosum of the lamina terminalis (OVLT, supraoptic crest), and (4) the subfornical organ (SFO). These areas are referred to collectively as the circumventric-ular organs (Figure 34–7). All have fenestrated capillaries, and because of their permeability they are said to be “outside the blood–brain barrier.” Some of them function as neurohemal organs; that is, areas in which polypeptides secreted by neu-rons enter the circulation. Others contain receptors for many different peptides and other substances, and function as chemoreceptor zones in which substances in the circulating blood can act to trigger changes in brain function without penetrating the blood–brain barrier. For example, the area postrema is a chemoreceptor trigger zone that initiates vomit-ing in response to chemical changes in the plasma (see Chap-ter 28). It is also concerned with cardiovascular control, and in many species circulating angiotensin II acts on the area pos-trema to produce a neurally mediated increase in blood pres-sure. Angiotensin II also acts on the SFO and possibly on the OVLT to increase water intake. In addition, it appears that the OVLT is the site of the osmoreceptor controlling vasopressin secretion (see Chapter 39), and evidence suggests that circulat-ing interleukin-1 (IL-1) produces fever by acting here too. The subcommissural organ (Figure 34–7) is closely associ-ated with the pineal gland and histologically resembles the cir-cumventricular organs. However, it does not have fenestrated capillaries, is not highly permeable, and has no established FIGURE 34–5 Penetration of urea into muscle, brain, spinal cord, and CSF. Urea was administered by constant infusion. 1.0 0.8 0.6 0.4 0.2 0 30 60 90 120 150 180 Min after start of infusion Tissue Plasma concentration Muscle Brain CSF FIGURE 34–6 Localization of the various GLUT transporters in the brain. (Adapted from Maher F, Vannucci SJ, Simpson IA: Glucose transporter proteins in brain. FASEB J 1994;8:1003.) GLUT 3 GLUT 3 GLUT 1 55K GLUT 1 55K GLUT 1 45K GLUT 1 45K GLUT 5 GLUT 5 Endothelial cell Astroglia Neuron Oligodendroglia Microglia Microvessel - Lumen -574 SECTION VI Cardiovascular Physiology function. Conversely, the pineal and the anterior pituitary do have fenestrated capillaries and are outside the blood–brain barrier, but both are endocrine glands and are not part of the brain. FUNCTION OF THE BLOOD–BRAIN BARRIER The blood–brain barrier strives to maintain the constancy of the environment of the neurons in the central nervous system (see Clinical Box 34–1). Even minor variations in the concen-trations of K+, Ca2+, Mg2+, H+, and other ions can have far-reaching consequences. The constancy of the composition of the ECF in all parts of the body is maintained by multiple ho-meostatic mechanisms (see Chapters 1 and 39), but because of the sensitivity of the cortical neurons to ionic change, it is not surprising that an additional defense has evolved to protect them. Other functions of the blood–brain barrier include pro-tection of the brain from endogenous and exogenous toxins in the blood and prevention of the escape of neurotransmitters into the general circulation. DEVELOPMENT OF THE BLOOD–BRAIN BARRIER In experimental animals, many small molecules penetrate the brain more readily during the fetal and neonatal period than they do in the adult. On this basis, it is often stated that the blood–brain barrier is immature at birth. Humans are more mature at birth than rats and various other experimental ani-mals, and detailed data on passive permeability of the human blood–brain barrier are not available. However, in severely jaundiced infants with high plasma levels of free bilirubin and an immature hepatic bilirubin-conjugating system, free biliru-bin enters the brain and, in the presence of asphyxia, damages the basal ganglia (kernicterus). The counterpart of this situa-tion in later life is the Crigler–Najjar syndrome in which there is a congenital deficiency of glucuronyl transferase. These indi-viduals can have very high free bilirubin levels in the blood and develop encephalopathy. In other conditions, free bilirubin le-vels are generally not high enough to produce brain damage. CEREBRAL BLOOD FLOW & ITS REGULATION KETY METHOD According to the Fick principle (see Chapter 31), the blood flow of any organ can be measured by determining the amount of a given substance (Qx) removed from the blood-stream by the organ per unit of time and dividing that value by the difference between the concentration of the substance in arterial blood and the concentration in the venous blood from the organ ([Ax] – [Vx]). Thus: FIGURE 34–7 Circumventricular organs. The neurohypophys-is (NH), organum vasculosum of the lamina terminalis (OVLT, organum vasculosum of the lamina terminalis), subfornical organ (SFO), and area postrema (AP) are shown projected on a sagittal section of the hu-man brain. SCO, subcommissural organ; PI, pineal. AP NH PI OVLT SFO SCO CLINICAL BOX 34–1 Clinical Implications of the Blood–Brain Barrier Physicians must know the degree to which drugs penetrate the brain in order to treat diseases of the nervous system intelligently. For example, it is clinically relevant that the amines dopamine and serotonin penetrate brain tissue to a very limited degree but their corresponding acid precur-sors, L-dopa and 5-hydroxytryptophan, respectively, enter with relative ease (see Chapters 7 and 16). Another impor-tant clinical consideration is the fact that the blood–brain barrier tends to break down in areas of infection or injury. Tumors develop new blood vessels, and the capillaries that are formed lack contact with normal astrocytes. Therefore, there are no tight junctions, and the vessels may even be fenestrated. The lack of a barrier helps in identifying the lo-cation of tumors; substances such as radioactive iodine-la-beled albumin penetrate normal brain tissue very slowly, but they enter tumor tissue, making the tumor stand out as an island of radioactivity in the surrounding normal brain. The blood–brain barrier can also be temporarily disrupted by sudden marked increases in blood pressure or by intra-venous injection of hypertonic fluids. Cerebral blood flow (CBF) Qx Ax [ ] Vx [ ] – ----------------------------= CHAPTER 34 Circulation Through Special Regions 575 This can be applied clinically using inhaled nitrous oxide (N2O) (Kety method). The average cerebral blood flow in young adults is 54 mL/100 g/min. The average adult brain weighs about 1400 g, so the flow for the whole brain is about 756 mL/min. Note that the Kety method provides an average value for perfused areas of brain because it gives no informa-tion about regional differences in blood flow. It also can only measure flow to perfused parts of the brain. If the blood flow to a portion of the brain is occluded, the measured flow does not change because the nonperfused area does not take up any N2O. In spite of the marked local fluctuations in brain blood flow with neural activity, the cerebral circulation is regulated in such a way that total blood flow remains relatively constant. The factors involved in regulating the flow are summarized in Figure 34–8. ROLE OF INTRACRANIAL PRESSURE In adults, the brain, spinal cord, and spinal fluid are encased, along with the cerebral vessels, in a rigid bony enclosure. The cranial cavity normally contains a brain weighing approximate-ly 1400 g, 75 mL of blood, and 75 mL of spinal fluid. Because brain tissue and spinal fluid are essentially incompressible, the volume of blood, spinal fluid, and brain in the cranium at any time must be relatively constant (Monro–Kellie doctrine). More importantly, the cerebral vessels are compressed whenev-er the intracranial pressure rises. Any change in venous pres-sure promptly causes a similar change in intracranial pressure. Thus, a rise in venous pressure decreases cerebral blood flow both by decreasing the effective perfusion pressure and by com-pressing the cerebral vessels. This relationship helps to compen-sate for changes in arterial blood pressure at the level of the head. For example, if the body is accelerated upward (positive g), blood moves toward the feet and arterial pressure at the level of the head decreases. However, venous pressure also falls and intracranial pressure falls, so that the pressure on the vessels de-creases and blood flow is much less severely compromised than it would otherwise be. Conversely, during acceleration down-ward, force acting toward the head (negative g) increases arterial pressure at head level, but intracranial pressure also rises, so that the vessels are supported and do not rupture. The cerebral ves-sels are protected during the straining associated with defeca-tion or delivery in the same way. AUTOREGULATION As seen in other vascular beds, autoregulation is prominent in the brain (Figure 34–9). This process, by which the flow to many tissues is maintained at relatively constant levels despite variations in perfusion pressure, is discussed in Chapter 32. In the brain, autoregulation maintains a normal cerebral blood flow at arterial pressures of 65 to 140 mm Hg. ROLE OF VASOMOTOR & SENSORY NERVES The innervation of large cerebral blood vessels by postgangli-onic sympathetic and parasympathetic nerves and the addi-tional distal innervation by sensory nerves have been described above. The nerves may also modulate tone indirectly, via the release of paracine substances from astrocytes. The precise role of these nerves, however, remains a matter of debate. It has been argued that noradrenergic discharge occurs when the blood pressure is markedly elevated. This reduces the resultant passive increase in blood flow and helps protect the blood– brain barrier from the disruption that could otherwise occur (see above). Thus, vasomotor discharges affect autoregulation. With sympathetic stimulation, the constant-flow, or plateau, part of the pressure-flow curve is extended to the right (Figure 34–9); that is, greater increases in pressure can occur without an increase in flow. On the other hand, the vasodilator hydral-azine and the angiotensin-converting enzyme (ACE) inhibitor captopril reduce the length of the plateau. Finally, neurovascu-lar coupling may adjust local perfusion in response to changes in brain activity (see below). FIGURE 34–8 Diagrammatic summary of the factors affecting overall cerebral blood flow. Brain, spinal cord, and spinal fluid Vertebral column Cranium Intracranial pressure Local con-striction and dilation of cerebral arterioles Mean arterial pressure at brain level Viscosity of blood Mean venous pressure at brain level FIGURE 34–9 Autoregulation of cerebral blood flow (CBF) during steady-state conditions. The blue line shows the alteration produced by sympathetic stimulation during autoregulation. 100 50 70 140 Arterial pressure (mm Hg) CBF 576 SECTION VI Cardiovascular Physiology BLOOD FLOW IN VARIOUS PARTS OF THE BRAIN A major advance in recent decades has been the development of techniques for monitoring regional blood flow in living, conscious humans. Among the most valuable methods are positron emission tomography (PET) and related techniques in which a short-lived radioisotope is used to label a com-pound and the compound is injected. The arrival and clear-ance of the tracer are monitored by scintillation detectors placed over the head. Because blood flow is tightly coupled to brain metabolism, local uptake of 2-deoxyglucose is also a good index of blood flow (see below and Chapter 1). If the 2-deoxyglucose is labeled with a short-half-life positron emitter such as 18F, 11O, or 15O, its concentration in any part of the brain can be monitored. Another valuable technique involves magnetic resonance imaging (MRI). MRI is based on detecting resonant signals from different tissues in a magnetic field. Functional mag-netic resonance imaging (fMRI) measures the amount of blood in a tissue area. When neurons become active, their increased discharge alters the local field potential. A still unsettled mechanism triggers an increase in local blood flow and oxygen. The increase in oxygenated blood is detected by fMRI. PET scanning can be used to measure not only blood flow but the concentration of molecules, such as dopamine, in various regions of the living brain. On the other hand, fMRI does not involve the use of radioactivity. Consequently, it can be used at frequent intervals to measure changes in regional blood flow in a single individual. In resting humans, the average blood flow in gray matter is 69 mL/100 g/min compared with 28 mL/100 g/min in white matter. A striking feature of cerebral function is the marked variation in local blood flow with changes in brain activity. An example is shown in Figure 34–10. In subjects who are awake but at rest, blood flow is greatest in the premotor and frontal regions. This is the part of the brain that is believed to be con-cerned with decoding and analyzing afferent input and with intellectual activity. During voluntary clenching of the right hand, flow is increased in the hand area of the left motor cortex and the corresponding sensory areas in the postcentral gyrus. Especially when the movements being performed are sequen-tial, the flow is also increased in the supplementary motor area. When subjects talk, there is a bilateral increase in blood flow in the face, tongue, and mouth-sensory and motor areas and the upper premotor cortex in the categorical (usually the left) hemisphere. When the speech is stereotyped, Broca’s and Wer-nicke’s areas do not show increased flow, but when the speech is creative—that is, when it involves ideas—flow increases in both these areas. Reading produces widespread increases in blood flow. Problem solving, reasoning, and motor ideation without movement produce increases in selected areas of the premotor and frontal cortex. In anticipation of a cognitive task, many of the brain areas that will be activated during the task are acti-vated beforehand, as if the brain produces an internal model of the expected task. In right-handed individuals, blood flow to the left hemisphere is greater when a verbal task is being per-formed and blood flow to the right hemisphere is greater when a spatial task is being performed (see Clinical Box 34–2). BRAIN METABOLISM & OXYGEN REQUIREMENTS UPTAKE & RELEASE OF SUBSTANCES BY THE BRAIN If the cerebral blood flow is known, it is possible to calculate the consumption or production by the brain of O2, CO2, glu-cose, or any other substance present in the bloodstream by multiplying the cerebral blood flow by the difference between the concentration of the substance in arterial blood and its concentration in cerebral venous blood (Table 34–3). When calculated in this fashion, a negative value indicates that the brain is producing the substance. OXYGEN CONSUMPTION O2 consumption by the human brain (cerebral metabolic rate for O2, CMRO2) averages approximately 20% of the total body resting O2 consumption (Table 34–1). The brain is FIGURE 34–10 Activity in the human brain at five different horizontal levels while a subject generates a verb that is appropriate for each noun presented by an examiner. This mental task activates the frontal cortex (slices 1–4), anterior cingulate gyrus (slice 1), and poster-ior temporal lobe (slice 3) on the left side and the cerebellum (slices 4 and 5) on the right side. Light purple, moderate activation; dark purple, marked activation. (Based on PET scans in Posner MI, Raichle ME: Images of Mind. Scientific American Library, 1994.) CHAPTER 34 Circulation Through Special Regions 577 extremely sensitive to hypoxia, and occlusion of its blood sup-ply produces unconsciousness in a period as short as 10 s. The vegetative structures in the brain stem are more resistant to hypoxia than the cerebral cortex, and patients may recover from accidents such as cardiac arrest and other conditions causing fairly prolonged hypoxia with normal vegetative func-tions but severe, permanent intellectual deficiencies. The basal ganglia use O2 at a very high rate, and symptoms of Parkinson disease as well as intellectual deficits can be produced by chronic hypoxia. The thalamus and the inferior colliculus are also very susceptible to hypoxic damage (see Clinical Box 34–3). ENERGY SOURCES Glucose is the major ultimate source of energy for the brain; under normal conditions, 90% of the energy needed to main-tain ion gradients across cell membranes and transmit electri-cal impulses comes from this source. Glucose enters the brain via GLUT 1 in cerebral capillaries (see above). Other trans-porters then distribute it to neurons and glial cells. Glucose is taken up from the blood in large amounts, and the RQ (respiratory quotient; see Chapter 21) of cerebral tis-sue is 0.95–0.99 in normal individuals. Importantly, insulin is not required for most cerebral cells to utilize glucose. In gen-eral, glucose utilization at rest parallels blood flow and O2 consumption. This does not mean that the total source of energy is always glucose. During prolonged starvation, appre-ciable utilization of other substances occurs. Indeed, evidence indicates that as much as 30% of the glucose taken up under normal conditions is converted to amino acids, lipids, and proteins, and that substances other than glucose are metabo-lized for energy during convulsions. Some utilization of amino acids from the circulation may also take place even though the amino acid arteriovenous difference across the brain is normally minute. The consequences of hypoglycemia in terms of neural func-tion are discussed in Chapter 21. GLUTAMATE & AMMONIA REMOVAL The brain’s uptake of glutamate is approximately balanced by its output of glutamine. Glutamate entering the brain takes up ammonia and leaves as glutamine. The glutamate– glutamine conversion in the brain—the opposite of the reac-tion in the kidney that produces some of the ammonia enter-ing the tubules—serves as a detoxifying mechanism to keep the brain free of ammonia. Ammonia is very toxic to nerve cells, and ammonia intoxication is believed to be a major cause of the bizarre neurologic symptoms in hepatic coma (see Chapter 29). CORONARY CIRCULATION ANATOMIC CONSIDERATIONS The two coronary arteries that supply the myocardium arise from the sinuses behind two of the cusps of the aortic valve at the root of the aorta (Figure 34–11). Eddy currents keep the valves away from the orifices of the arteries, and they are patent throughout the cardiac cycle. Most of the venous blood returns to the heart through the coronary sinus and anterior CLINICAL BOX 34–2 Changes in Cerebral Blood Flow in Disease Several disease states are now known to be associated with localized or general changes in cerebral blood flow, as re-vealed by PET scanning and fMRI techniques. For example, epileptic foci are hyperemic during seizures, whereas flow is reduced in other parts of the brain. Between seizures, flow is sometimes reduced in the foci that generate the sei-zures. Parietooccipital flow is decreased in patients with symptoms of agnosia (see Chapter 14). In Alzheimer dis-ease, the earliest change is decreased metabolism and blood flow in the superior parietal cortex, with later spread to the temporal and finally the frontal cortex. The pre- and postcentral gyri, basal ganglia, thalamus, brain stem, and cerebellum are relatively spared. In Huntington disease, blood flow is reduced bilaterally in the caudate nucleus, and this alteration in flow occurs early in the disease. In manic depressives (but interestingly, not in patients with unipolar depression), there is a general decrease in cortical blood flow when the patients are depressed. In schizophre-nia, some evidence suggests decreased blood flow in the frontal lobes, temporal lobes, and basal ganglia. Finally, during the aura in patients with migraine, a bilateral de-crease in blood flow starts in the occipital cortex and spreads anteriorly to the temporal and parietal lobes. TABLE 34–3 Utilization and production of substances by the adult human brain in vivo. Substance Uptake (+) or Output (–) per 100 g of Brain/min Total/min Substances utilized Oxygen +3.5 mL +49 mL Glucose +5.5 mg +77 mg Glutamate +0.4 mg +5.6 mg Substances produced Carbon dioxide –3.5 mL –49 mL Glutamine –0.6 mL –8.4 mg Substances not used or produced in the fed state: lactate, pyruvate, total ketones, and α-ketoglutarate. 578 SECTION VI Cardiovascular Physiology cardiac veins (Figure 34–12), which drain into the right atri-um. In addition, there are other vessels that empty directly into the heart chambers. These include arteriosinusoidal ves-sels, sinusoidal capillary-like vessels that connect arterioles to the chambers; thebesian veins that connect capillaries to the chambers; and a few arterioluminal vessels that are small ar-teries draining directly into the chambers. A few anastomoses occur between the coronary arterioles and extracardiac arteri-oles, especially around the mouths of the great veins. Anasto-moses between coronary arterioles in humans only pass particles less than 40 μm in diameter, but evidence indicates that these channels enlarge and increase in number in patients with coronary artery disease. PRESSURE GRADIENTS & FLOW IN THE CORONARY VESSELS The heart is a muscle that, like skeletal muscle, compresses its blood vessels when it contracts. The pressure inside the left ventricle is slightly higher than in the aorta during systole (Table 34–4). Consequently, flow occurs in the arteries sup-plying the subendocardial portion of the left ventricle only during diastole, although the force is sufficiently dissipated in the more superficial portions of the left ventricular myocardi-um to permit some flow in this region throughout the cardiac cycle. Because diastole is shorter when the heart rate is high, left ventricular coronary flow is reduced during tachycardia. On the other hand, the pressure differential between the aorta and the right ventricle, and the differential between the aorta and the atria, are somewhat greater during systole than during diastole. Consequently, coronary flow in those parts of the heart is not appreciably reduced during systole. Flow in the right and left coronary arteries is shown in Figure 34–13. Be-cause no blood flow occurs during systole in the subendocar-dial portion of the left ventricle, this region is prone to ischemic damage and is the most common site of myocardial infarction. Blood flow to the left ventricle is decreased in CLINICAL BOX 34–3 Stroke When the blood supply to a part of the brain is interrupted, ischemia damages or kills the cells in the area, producing the signs and symptoms of a stroke. There are two general types of strokes: hemorrhagic and ischemic. Hemorrhagic stroke occurs when a cerebral artery or arteriole ruptures, sometimes but not always at the site of a small aneurysm. Ischemic stroke occurs when flow in a vessel is compro-mised by atherosclerotic plaques on which thrombi form. Thrombi may also be produced elsewhere (eg, in the atria in patients with atrial fibrillation) and pass to the brain as emboli where they then lodge and interrupt flow. In the past, little could be done to modify the course of a stroke and its consequences. However, it has now become clear that in the penumbra, the area surrounding the most se-vere brain damage, ischemia reduces glutamate uptake by astrocytes, and the increase in local glutamate causes exci-totoxic damage and death to neurons (see Chapter 7). In experimental animals, and perhaps in humans, drugs that prevent this excitotoxic damage significantly reduce the ef-fects of strokes. In addition, clot-lysing drugs such as tissue-type plasminogen activator (t-PA) (see Chapter 32) are of benefit in ischemic strokes. Both antiexcitotoxic treatment and t-PA must be given early in the course of a stroke to be of maximum benefit, and this is why stroke has become a condition in which rapid diagnosis and treatment have be-come important. In addition, of course, it is important to determine if a stroke is thrombotic or hemorrhagic, since clot lysis is contraindicated in the latter. FIGURE 34–11 Coronary arteries and their principal branches in humans. (Reproduced with permission from Ross G: The cardiovascular system. In: Essentials of Human Physiology. Ross G [editor]. Copyright © 1978 by Year Book Medical Publishers.) FIGURE 34–12 Diagram of the coronary circulation. Marginal branch Right coronary artery Left coronary artery Circumflex branch Anterior descending branch Septal branches Marginal branch Posterior descending branch Extracoronary arteries Coronary arteries Arterioles Capillaries Arterioles Arteriosinusoidal vessels Heart chambers Thebesian veins Coronary sinus or anterior cardiac veins Arterioluminal vessels Veins CHAPTER 34 Circulation Through Special Regions 579 patients with stenotic aortic valves because the pressure in the left ventricle must be much higher than that in the aorta to eject the blood. Consequently, the coronary vessels are severely com-pressed during systole. Patients with this disease are particularly prone to develop symptoms of myocardial ischemia, in part be-cause of this compression and in part because the myocardium requires more O2 to expel blood through the stenotic aortic valve. Coronary flow is also decreased when the aortic diastolic pres-sure is low. The rise in venous pressure in conditions such as con-gestive heart failure reduces coronary flow because it decreases effective coronary perfusion pressure (see Clinical Box 34–4). Coronary blood flow has been measured by inserting a cath-eter into the coronary sinus and applying the Kety method to the heart on the assumption that the N2O content of coronary venous blood is typical of the entire myocardial effluent. Coro-nary flow at rest in humans is about 250 mL/min (5% of the cardiac output). A number of techniques utilizing radionu-clides, radioactive tracers that can be detected with radiation detectors over the chest, have been used to study regional blood flow in the heart and to detect areas of ischemia and infarct as well as to evaluate ventricular function. Radionuclides such as thallium-201 (201T1) are pumped into cardiac muscle cells by Na, K ATPase and equilibrate with the intracellular K+ pool. For the first 10–15 min after intravenous injection, 201T1 distri-bution is directly proportional to myocardial blood flow, and areas of ischemia can be detected by their low uptake. The uptake of this isotope is often determined soon after exercise and again several hours later to bring out areas in which exertion leads to compromised flow. Conversely, radiopharmaceuticals such as technetium-99m stannous pyrophosphate (99mTc-PYP) are selectively taken up by infarcted tissue by an incompletely understood mechanism and make infarcts stand out as “hot spots” on scintigrams of the chest. Coronary angiography can be combined with measurement of 133Xe washout (see above) to provide detailed analysis of coronary blood flow. Radiopaque contrast medium is first injected into the coronary arteries, and x-rays are used to outline their distribution. The angiographic camera is then replaced with a scintillation camera, and 133Xe washout is measured. VARIATIONS IN CORONARY FLOW At rest, the heart extracts 70–80% of the O2 from each unit of blood delivered to it (Table 34–1). O2 consumption can be in-creased significantly only by increasing blood flow. Therefore, it is not surprising that blood flow increases when the metab-olism of the myocardium is increased. The caliber of the cor-onary vessels, and consequently the rate of coronary blood flow, is influenced not only by pressure changes in the aorta but also by chemical and neural factors. The coronary circula-tion also shows considerable autoregulation. CHEMICAL FACTORS The close relationship between coronary blood flow and myo-cardial O2 consumption indicates that one or more of the products of metabolism cause coronary vasodilation. Factors suspected of playing this role include O2 lack and increased lo-cal concentrations of CO2, H+, K+, lactate, prostaglandins, ad-enine nucleotides, and adenosine. Likely several or all of these vasodilator metabolites act in an integrated fashion, redun-dant fashion, or both. Asphyxia, hypoxia, and intracoronary injections of cyanide all increase coronary blood flow 200– 300% in denervated as well as intact hearts, and the feature common to these three stimuli is hypoxia of the myocardial fi-bers. A similar increase in flow is produced in the area sup-plied by a coronary artery if the artery is occluded and then released. This reactive hyperemia is similar to that seen in the TABLE 34–4 Pressure in aorta and left and right ventricles (vent) in systole and diastole. Pressure (mm Hg) in Pressure Differential (mm Hg) between Aorta and Aorta Left Vent Right Vent Left Vent Right Vent Systole 120 121 25 –1 95 Diastole 80 0 0 80 80 FIGURE 34–13 Blood flow in the left and right coronary arteries during various phases of the cardiac cycle. Systole occurs between the two vertical dashed lines. (Reproduced with permission from Berne RM, Levy MN: Physiology, 2nd ed. Mosby, 1988.) 120 100 80 100 60 40 20 0 15 10 5 0 80 0.2 0.4 0.6 0.8 1.0 Time (s) Left coronary Right coronary Phasic coronary blood flow (mL/min) Aortic pressure (mm Hg) 580 SECTION VI Cardiovascular Physiology skin (see below). Evidence suggests that in the heart it is due to release of adenosine. NEURAL FACTORS The coronary arterioles contain α-adrenergic receptors, which mediate vasoconstriction, and β-adrenergic receptors, which mediate vasodilation. Activity in the noradrenergic nerves to the heart and injections of norepinephrine cause coronary vasodilation. However, norepinephrine increases the heart rate and the force of cardiac contraction, and the vasodilation is due to production of vasodilator metabolites in the myocardium secondary to the increase in its activity. When the inotropic and chronotropic effects of noradrenergic discharge are blocked by a β-adrenergic blocking drug, stimulation of the noradrenergic nerves or injection of norepinephrine in unanesthetized ani-mals elicits coronary vasoconstriction. Thus, the direct effect of noradrenergic stimulation is constriction rather than dilation of the coronary vessels. On the other hand, stimulation of vagal fibers to the heart dilates the coronaries. When the systemic blood pressure falls, the overall effect of the reflex increase in noradrenergic discharge is increased coronary blood flow secondary to the metabolic changes in the myocardium at a time when the cutaneous, renal, and splanchnic vessels are constricted. In this way the circulation of the heart, like that of the brain, is preserved when flow to other organs is compromised. CUTANEOUS CIRCULATION The amount of heat lost from the body is regulated to a large extent by varying the amount of blood flowing through the skin. The fingers, toes, palms, and earlobes contain well-innervated anastomotic connections between arterioles and venules (arteriovenous anastomoses; see Chapter 32). Blood flow in response to thermoregulatory stimuli can vary from 1 to as much as 150 mL/100 g of skin/min, and it has been pos-tulated that these variations are possible because blood can be shunted through the anastomoses. The subdermal capillary and venous plexus is a blood reservoir of some importance, and the skin is one of the few places where the reactions of blood vessels can be observed visually. WHITE REACTION When a pointed object is drawn lightly over the skin, the stroke lines become pale (white reaction). The mechanical stimulus apparently initiates contraction of the precapillary sphincters, and blood drains out of the capillaries and small veins. The response appears in about 15 s. TRIPLE RESPONSE When the skin is stroked more firmly with a pointed instru-ment, instead of the white reaction there is reddening at the site that appears in about 10 s (red reaction). This is followed in a few minutes by local swelling and diffuse, mottled reddening CLINICAL BOX 34–4 Coronary Artery Disease When flow through a coronary artery is reduced to the point that the myocardium it supplies becomes hypoxic, angina pectoris develops (see Chapter 31). If the myocardial is-chemia is severe and prolonged, irreversible changes occur in the muscle, and the result is myocardial infarction. Many individuals have angina only on exertion, and blood flow is normal at rest. Others have more severe restriction of blood flow and have anginal pain at rest as well. Partially occluded coronary arteries can be constricted further by vasospasm, producing myocardial infarction. However, it is now clear that the most common cause of myocardial infarction is rup-ture of an atherosclerotic plaque, or hemorrhage into it, which triggers the formation of a coronary-occluding clot at the site of the plaque. The electrocardiographic changes in myocardial infarction are discussed in Chapter 30. When myocardial cells actually die, they leak enzymes into the cir-culation, and measuring the rises in serum enzymes and isoenzymes produced by infarcted myocardial cells also plays an important role in the diagnosis of myocardial infarc-tion. The enzymes most commonly measured today are the MB isomer of creatine kinase (CK-MB), troponin T, and tropo-nin I. Myocardial infarction is a very common cause of death in developed countries because of the widespread occur-rence of atherosclerosis. In addition, there is a relation be-tween atherosclerosis and circulating levels of lipopro-tein(a) (Lp[a]). Lp(a) has an outer coat count of apo(a). It interferes with fibrinolysis by down-regulating plasmin gen-eration (see Chapter 32). There is also a strong positive corre-lation between atherosclerosis and circulating levels of ho-mocysteine. This substance damages endothelial cells. It is converted to nontoxic methionine in the presence of folate and vitamin B12, and clinical trials are under way to deter-mine whether supplements of folate and B12 lower the inci-dence of coronary disease. It now appears that atherosclero-sis has an important inflammatory component as well. The lesions of the disease contain inflammatory cells, and there is a positive correlation between increased levels of C-reactive protein and other inflammatory markers in the circulation and subsequent myocardial infarction. Treatment of myocar-dial infarction aims to restore flow to the affected area as rapidly as possible while minimizing reperfusion injury. Needless to say, it should be started as promptly as possible to avoid irreversible changes in heart function. CHAPTER 34 Circulation Through Special Regions 581 around the injury. The initial redness is due to capillary dilation, a direct response of the capillaries to pressure. The swelling (wheal) is local edema due to increased permeability of the cap-illaries and postcapillary venules, with consequent extravasa-tion of fluid. The redness spreading out from the injury (flare) is due to arteriolar dilation. This three-part response—the red reaction, wheal, and flare—is called the triple response and is part of the normal reaction to injury (see Chapter 3). It persists after total sympathectomy. On the other hand, the flare is absent in locally anesthetized skin and in denervated skin after the sen-sory nerves have degenerated, but it is present immediately after nerve block or section above the site of the injury. This, plus other evidence, indicates that it is due to an axon reflex, a re-sponse in which impulses initiated in sensory nerves by the in-jury are relayed antidromically down other branches of the sensory nerve fibers (Figure 34–14). This is the one situation in the body in which there is substantial evidence for a physiologic effect due to antidromic conduction. The transmitter released at the central termination of the sensory C fiber neurons is sub-stance P (see Chapter 7), and substance P and CGRP are present in all parts of the neurons. Both dilate arterioles and, in addi-tion, substance P causes extravasation of fluid. Effective non-peptide antagonists to substance P have now been developed, and they reduce the extravasation. Thus, it appears that these peptides produce the wheal. REACTIVE HYPEREMIA A response of the blood vessels that occurs in many organs but is visible in the skin is reactive hyperemia, an increase in the amount of blood in a region when its circulation is reestab-lished after a period of occlusion. When the blood supply to a limb is occluded, the cutaneous arterioles below the occlusion dilate. When the circulation is reestablished, blood flowing into the dilated vessels makes the skin become fiery red. O2 in the atmosphere can diffuse a short distance through the skin, and reactive hyperemia is prevented if the circulation of the limb is occluded in an atmosphere of 100% O2. Therefore, the arteriolar dilation is apparently due to a local effect of hypoxia. GENERALIZED RESPONSES Noradrenergic nerve stimulation and circulating epinephrine and norepinephrine constrict cutaneous blood vessels. No known vasodilator nerve fibers extend to the cutaneous ves-sels, and thus vasodilation is brought about by a decrease in constrictor tone as well as the local production of vasodilator metabolites. Skin color and temperature also depend on the state of the capillaries and venules. A cold blue or gray skin is one in which the arterioles are constricted and the capillaries dilated; a warm red skin is one in which both are dilated. Because painful stimuli cause diffuse noradrenergic dis-charge, a painful injury causes generalized cutaneous vaso-constriction in addition to the local triple response. When the body temperature rises during exercise, the cutaneous blood vessels dilate in spite of continuing noradrenergic discharge in other parts of the body. Dilation of cutaneous vessels in response to a rise in hypothalamic temperature overcomes other reflex activity. Cold causes cutaneous vasoconstriction; however, with severe cold, superficial vasodilation may super-vene. This vasodilation is the cause of the ruddy complexion seen on a cold day. Shock is more profound in patients with elevated tempera-tures because of cutaneous vasodilation, and patients in shock should not be warmed to the point that their body tempera-ture rises. This is sometimes a problem because well-meaning laymen have read in first-aid books that “injured patients should be kept warm,” and they pile blankets on accident vic-tims who are in shock. PLACENTAL & FETAL CIRCULATION UTERINE CIRCULATION The blood flow of the uterus parallels the metabolic activity of the myometrium and endometrium and undergoes cyclic fluctuations that correlate with the menstrual cycle in non-pregnant women. The function of the spiral and basilar arter-ies of the endometrium in menstruation is discussed in Chapter 25. During pregnancy, blood flow increases rapidly as the uterus increases in size (Figure 34–15). Vasodilator metab-olites are undoubtedly produced in the uterus, as they are in other active tissues. In early pregnancy, the arteriovenous O2 difference across the uterus is small, and it has been suggested that estrogens act on the blood vessels to increase uterine blood flow in excess of tissue O2 needs. However, even though uterine blood flow increases 20-fold during pregnancy, the FIGURE 34–14 Axon reflex. Orthodromic conduction Antidromic conduction Direction taken by impulses Sensory neuron Spinal cord Endings in skin Endings near arteriole 582 SECTION VI Cardiovascular Physiology size of the conceptus increases much more, changing from a single cell to a fetus plus a placenta that weighs 4 to 5 kg at term in humans. Consequently, more O2 is extracted from the uterine blood during the latter part of pregnancy, and the O2 saturation of uterine blood falls. Corticotrophin-releasing hormone appears to play an important role in up-regulating uterine blood flow, as well as in the eventual timing of birth. PLACENTA The placenta is the “fetal lung” (Figures 34–16 and 34–17). Its maternal portion is in effect a large blood sinus. Into this “lake” project the villi of the fetal portion containing the small branch-es of the fetal umbilical arteries and vein (Figure 34–16). O2 is taken up by the fetal blood and CO2 is discharged into the ma-ternal circulation across the walls of the villi in a fashion analo-gous to O2 and CO2 exchange in the lungs (see Chapter 36). However, the cellular layers covering the villi are thicker and less permeable than the alveolar membranes in the lungs, and exchange is much less efficient. The placenta is also the route by which all nutritive materials enter the fetus and by which fetal wastes are discharged to the maternal blood. FETAL CIRCULATION The arrangement of the circulation in the fetus is shown dia-grammatically in Figure 34–17. Fifty-five percent of the fetal cardiac output goes through the placenta. The blood in the um-bilical vein in humans is believed to be about 80% saturated with O2, compared with 98% saturation in the arterial circula-tion of the adult. The ductus venosus (Figure 34–18) diverts some of this blood directly to the inferior vena cava, and the re-mainder mixes with the portal blood of the fetus. The portal and systemic venous blood of the fetus is only 26% saturated, and the saturation of the mixed blood in the inferior vena cava is approximately 67%. Most of the blood entering the heart through the inferior vena cava is diverted directly to the left atrium via the patent foramen ovale. Most of the blood from the superior vena cava enters the right ventricle and is expelled into the pulmonary artery. The resistance of the collapsed lungs is high, and the pressure in the pulmonary artery is sev-eral mm Hg higher than it is in the aorta, so that most of the blood in the pulmonary artery passes through the ductus arte-riosus to the aorta. In this fashion, the relatively unsaturated blood from the right ventricle is diverted to the trunk and lower body of the fetus, while the head of the fetus receives the better-oxygenated blood from the left ventricle. From the aorta, some of the blood is pumped into the umbilical arteries and back to the placenta. The O2 saturation of the blood in the lower aorta and umbilical arteries of the fetus is approximately 60%. FETAL RESPIRATION The tissues of fetal and newborn mammals have a remarkable but poorly understood resistance to hypoxia. However, the O2 saturation of the maternal blood in the placenta is so low that the fetus might suffer hypoxic damage if fetal red cells did not FIGURE 34–15 Changes in uterine blood flow and the amount of O2 in uterine venous blood during pregnancy. (After Barcroft H. Modified and redrawn with permission from Keele CA, Neil E: Samson Wright’s Applied Physiology, 12th ed. Oxford University Press, 1971.) Parturition Parturition Uterine blood flow Fetal weight Systemic venous blood Uterine venous blood Time after conception O2 saturation Relative units FIGURE 34–16 Diagram of a section through the human placenta, showing the way the fetal villi project into the maternal sinuses. (Reproduced with permission from Benson RC: Handbook of Obstetrics and Gynecology, 8th ed. Originally published by Appleton & Lange. Copyright © 1983 McGraw-Hill.) Amnion Septum Umbilical arteries Umbilical vein Umbilical cord Chorion Villus Intervillous space Spiral arteriole Basal plate Chorionic plate Endometrium Myometrium CHAPTER 34 Circulation Through Special Regions 583 have a greater O2 affinity than adult red cells (Figure 34–19). The fetal red cells contain fetal hemoglobin (hemoglobin F), whereas the adult cells contain adult hemoglobin (hemoglobin A). The cause of the difference in O2 affinity between the two is that hemoglobin F binds 2, 3-DPG less effectively than he-moglobin A does. The decrease in O2 affinity due to the bind-ing of 2, 3-DPG is discussed in Chapter 32). Some hemoglobin A is present in blood during fetal life (see Chapter 32). After birth, production of hemoglobin F nor-mally ceases, and by the age of 4 mo 90% of the circulating hemoglobin is hemoglobin A. FIGURE 34–17 Diagram of the circulation in the fetus, the newborn infant, and the adult. DA, ductus arteriosus; FO, foramen ovale. (Redrawn and reproduced with permission from Born GVR et al: Changes in the heart and lungs at birth. Cold Spring Harbor Symp Quant Biol 1954;19:102.) Placenta Body L heart Lungs R heart Body L heart Lungs R heart Body L heart Lungs R heart DA DA FO FETUS NEWBORN ADULT FIGURE 34–18 Circulation in the fetus. Most of the oxygenat-ed blood reaching the heart via the umbilical vein and inferior vena cava is diverted through the foramen ovale and pumped out the aorta to the head, while the deoxygenated blood returned via the superior vena cava is mostly pumped through the pulmonary artery and ductus arteriosus to the feet and the umbilical arteries. Pulmonary artery Left atrium Left ventricle Aorta Umbilical arteries Portal vein Umbilical vein Superior vena cava Foramen ovale Right atrium Right ventricle Ductus arteriosus Ductus venosus From placenta Inferior vena cava To placenta FIGURE 34–19 Dissociation curves of hemoglobin in human maternal and fetal blood. 22 20 18 16 14 12 10 8 6 4 2 0 10 20 30 40 50 60 70 80 90 100 PO2 (mm Hg) O2 content (mL/dL) Fetus Mother 584 SECTION VI Cardiovascular Physiology CHANGES IN FETAL CIRCULATION & RESPIRATION AT BIRTH Because of the patent ductus arteriosus and foramen ovale (Figure 34–18), the left heart and right heart pump in parallel in the fetus rather than in series as they do in the adult. At birth, the placental circulation is cut off and the peripheral re-sistance suddenly rises. The pressure in the aorta rises until it exceeds that in the pulmonary artery. Meanwhile, because the placental circulation has been cut off, the infant becomes in-creasingly asphyxial. Finally, the infant gasps several times, and the lungs expand. The markedly negative intrapleural pressure (–30 to –50 mm Hg) during the gasps contributes to the expansion of the lungs, but other factors are likely also in-volved. The sucking action of the first breath plus constriction of the umbilical veins squeezes as much as 100 mL of blood from the placenta (the “placental transfusion”). Once the lungs are expanded, the pulmonary vascular resis-tance falls to less than 20% of the value in utero, and pulmonary blood flow increases markedly. Blood returning from the lungs raises the pressure in the left atrium, closing the foramen ovale by pushing the valve that guards it against the interatrial sep-tum. The ductus arteriosus constricts within a few hours after birth, producing functional closure, and permanent anatomic closure follows in the next 24–48 h due to extensive intimal thickening. The mechanism producing the initial constriction is not completely understood, but the increase in arterial O2 tension plays an important role. Relatively high concentrations of vasodilators are present in the ductus in utero—especially prostaglandin F2a—and synthesis of these prostaglandins is blocked by inhibition of cyclooxygenase at birth. In many pre-mature infants the ductus fails to close spontaneously, but closure can be produced by infusion of drugs that inhibit cyclooxygenase. NO may also be involved in maintaining ductal patency in this setting. CHAPTER SUMMARY ■Cerebrospinal fluid is produced predominantly in the choroid plexus of the brain, in part via active transport mechanisms in the choroid epithelial cells. Fluid is reabsorbed into the blood-stream to maintain appropriate pressure in the setting of contin-uous production. ■The permeation of circulating substances into the brain is tight-ly controlled. Water, CO2, and O2 permeate freely. Other sub-stances (such as glucose) require specific transport mechanisms, whereas entry of macromolecules is negligible. The effectiveness of the blood–brain barrier in preventing entry of xenobiotics is bolstered by active efflux mediated by P-glycoprotein. ■The coronary circulation supplies oxygen to the contracting myocardium. Metabolic products and neural input induce vaso-dilation as needed for oxygen demand. Blockage of coronary arteries may lead to irreversible injury to heart tissue. ■Control of cutaneous blood flow is a key facet of temperature regulation, and is underpinned by varying levels of shunting through arteriovenous anastomoses. Hypoxia, axon reflexes, and sympathetic input are all important determinants of flow through the cutaneous vasculature. ■The fetal circulation cooperates with that of the placenta and uterus to deliver oxygen and nutrients to the growing fetus, as well as carrying away waste products. Unique anatomic features of the fetal circulation as well as biochemical properties of fetal hemoglobin serve to ensure adequate O2 supply, particularly to the head. At birth, the foramen ovale and the ductus arteriosus close such that the neonatal lungs now serve as the site for oxy-gen exchange. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Blood in which of the following vessels normally has the lowest PO2? A) maternal artery B) maternal uterine vein C) maternal femoral vein D) umbilical artery E) umbilical vein 2. The pressure differential between the heart and the aorta is least in the A) left ventricle during systole. B) left ventricle during diastole. C) right ventricle during systole. D) right ventricle during diastole. E) left atrium during systole. 3. Injection of tissue plasminogen activator (t-PA) would probably be most beneficial A) after at least 1 y of uncomplicated recovery following occlu-sion of a coronary artery. B) after at least 2 mo of rest and recuperation following occlu-sion of a coronary artery. C) during the second week after occlusion of a coronary artery. D) during the second day after occlusion of a coronary artery. E) during the second hour after occlusion of a coronary artery. 4. Which of the following organs has the greatest blood flow per 100 g of tissue? A) brain B) heart muscle C) skin D) liver E) kidneys 5. Which of the following does not dilate arterioles in the skin? A) increased body temperature B) epinephrine C) bradykinin D) substance P E) vasopressin CHAPTER 34 Circulation Through Special Regions 585 6. A baby boy is brought to the hospital because of convulsions. In the course of a workup, his body temperature and plasma glu-cose are found to be normal, but his cerebrospinal fluid glucose is 12 mg/dL (normal, 65 mg/dL). A possible explanation of his condition is A) constitutive activation of GLUT 3 in neurons. B) SGLT 1 deficiency in astrocytes. C) GLUT 5 deficiency in cerebral capillaries. D) GLUT 1 55K deficiency in cerebral capillaries. E) GLUT 1 45K deficiency in microglia. CHAPTER RESOURCES Begley DJ, Bradbury MW, Kreater J (editors): The Blood–Brain Barrier and Drug Delivery to the CNS. Marcel Dekker, 2000. Birmingham K (editor): The heart. Nature 2002;415:197. Duncker DJ, Bache RJ: Regulation of coronary blood flow during exercise. Physiol Rev 2008;88:1009. Hamel E: Perivascular nerves and the regulation of cerebrovascular tone. J Appl Physiol 2006;100:1059. Johanson CE, et al: Multiplicity of cerebrospinal fluid functions: New challenges in health and disease. Cerebrospinal Fluid Res 2008;5:10. Ward JPT: Oxygen sensing in context. Biochim Biophys Acta 2008;1777:1. This page intentionally left blank 587 C H A P T E R SECTION VII RESPIRATORY PHYSIOLOGY 35 Pulmonary Function O B J E C T I V E S After studying this chapter, you should be able to: ■Define partial pressure and calculate the partial pressure of each of the important gases in the atmosphere at sea level. ■List the passages through which air passes from the exterior to the alveoli, and describe the cells that line each of them. ■List the major muscles involved in respiration, and state the role of each. ■Define the basic measures of lung volume and give approximate values for each in a normal adult. ■Define compliance, and give examples of diseases in which it is abnormal. ■Describe the chemical composition and function of surfactant. ■List the factors that determine alveolar ventilation. ■Define diffusion capacity, and compare the diffusion of O2 with that of CO2 in the lungs. ■Compare the pulmonary and systemic circulations, listing the main differences between them. ■Describe basic lung defense and metabolic functions. INTRODUCTION Respiration, as the term is generally used, includes two pro-cesses: external respiration, the absorption of O2 and removal of CO2 from the body as a whole; and internal respi-ration, the utilization of O2 and production of CO2 by cells and the gaseous exchanges between the cells and their fluid medium. Aspects of external respiratory physiology are pre-sented throughout this section. In this chapter, the processes responsible for the uptake of O2 and excretion of CO2 in the lungs are explored. The next chapter is concerned with the transport of O2 and CO2 to and from the tissues. The final chapter in this section examines some key factors that regu-late respiration. Throughout each chapter, clinical implica-tions of specific physiology will be presented. 588 SECTION VII Respiratory Physiology PROPERTIES OF GASES The pressure of a gas is proportional to its temperature and the number of moles per volume: where P = Pressure n = Number of moles R = Gas constant T = Absolute temperature V = Volume PARTIAL PRESSURES Unlike liquids, gases expand to fill the volume available to them, and the volume occupied by a given number of gas mol-ecules at a given temperature and pressure is (ideally) the same regardless of the composition of the gas. Therefore, the pres-sure exerted by any one gas in a mixture of gases (its partial pressure) is equal to the total pressure times the fraction of the total amount of gas it represents. The composition of dry air is 20.98% O2, 0.04% CO2, 78.06% N2, and 0.92% other inert constituents such as argon and helium. The barometric pressure (PB) at sea level is 760 mm Hg (1 atmosphere). The partial pressure (indicated by the symbol P) of O2 in dry air is therefore 0.21 × 760, or 160 mm Hg at sea level. The PN2 and the other inert gases is 0.79 × 760, or 600 mm Hg; and the PCO2 is 0.0004 × 760, or 0.3 mm Hg. The water vapor in the air in most climates reduces these per-centages, and therefore the partial pressures, to a slight degree. Air equilibrated with water is saturated with water vapor, and inspired air is saturated by the time it reaches the lungs. The PH2O at body temperature (37 °C) is 47 mm Hg. Therefore, the partial pressures at sea level of the other gases in the air reach-ing the lungs are PO2, 149 mm Hg; PCO2, 0.3 mm Hg; and PN2 (including the other inert gases), 564 mm Hg. Gas diffuses from areas of high pressure to areas of low pressure, with the rate of diffusion depending on the concen-tration gradient and the nature of the barrier between the two areas. When a mixture of gases is in contact with and permit-ted to equilibrate with a liquid, each gas in the mixture dis-solves in the liquid to an extent determined by its partial pressure and its solubility in the fluid. The partial pressure of a gas in a liquid is the pressure that, in the gaseous phase in equilibrium with the liquid, would produce the concentration of gas molecules found in the liquid. METHODS OF QUANTITATING RESPIRATORY PHENOMENA Modern spirometers permit direct measurement of gas intake and output. Since gas volumes vary with temperature and pressure and since the amount of water vapor in them varies, these devices have the ability to correct respiratory measure-ments involving volume to a stated set of standard conditions. The four most commonly used standards and their abbrevia-tions are shown in Table 35–1. It should be noted that correct measurements are highly dependent on the ability for the practitioner to properly encourage the patient to fully utilize the device. Modern techniques for gas analysis make possible rapid, reliable measurements of the composition of gas mix-tures and the gas content of body fluids. For example, O2 and CO2 electrodes, small probes sensitive to O2 or CO2, can be in-serted into the airway or into blood vessels or tissues and the PO2 and PCO2 recorded continuously. Chronic assessment of oxygenation is carried out noninvasively with a pulse oxime-ter, which is usually attached to the ear. ANATOMY OF THE LUNGS THE RESPIRATORY SYSTEM The respiratory system is made up of a gas-exchanging organ (the lungs) and a “pump” that ventilates the lungs. The pump consists of the chest wall; the respiratory muscles, which in-crease and decrease the size of the thoracic cavity; the areas in the brain that control the muscles; and the tracts and nerves that connect the brain to the muscles. At rest, a normal human breathes 12 to 15 times a minute. About 500 mL of air per breath, or 6 to 8 L/min, is inspired and expired. This air mixes with the gas in the alveoli, and, by simple diffusion, O2 enters the blood in the pulmonary capillaries while CO2 enters the al-veoli. In this manner, 250 mL of O2 enters the body per minute and 200 mL of CO2 is excreted. Traces of other gases, such as methane from the intestines, are also found in expired air. Alcohol and acetone are expired when present in appreciable quantities in the body. Indeed, over 250 different volatile substances have been identified in human breath. AIR PASSAGES After passing through the nasal passages and pharynx, where it is warmed and takes up water vapor, the inspired air passes P nRT V ---------- (from equation of state of ideal gas) = TABLE 35–1 Standard conditions to which measurements involving gas volumes are corrected. STPD 0 °C, 760 mm Hg, dry (standard temperature and pressure, dry) BTPS Body temperature and pressure, saturated with water vapor ATPD Ambient temperature and pressure, dry ATPS Ambient temperature and pressure, saturated with water vapor CHAPTER 35 Pulmonary Function 589 down the trachea and through the bronchioles, respiratory bronchioles, and alveolar ducts to the alveoli, where gas ex-change occurs (Figure 35–1). Between the trachea and the alve-olar sacs, the airways divide 23 times. The first 16 generations of passages form the conducting zone of the airways that trans-ports gas from and to the exterior. They are made up of bron-chi, bronchioles, and terminal bronchioles. The remaining seven generations form the transitional and respiratory zones where gas exchange occurs; they are made up of respiratory bronchioles, alveolar ducts, and alveoli. These multiple divi-sions greatly increase the total cross-sectional area of the air-ways, from 2.5 cm2 in the trachea to 11,800 cm2 in the alveoli (Figure 35–2). Consequently, the velocity of air flow in the small airways declines to very low values. The alveoli are surrounded by pulmonary capillaries (Figure 35–1). In most areas, air and blood are separated only by the alveolar epithelium and the capillary endothelium, so they are about 0.5 μm apart (Figure 35–3). Humans have 300 million alveoli, and the total area of the alveolar walls in contact with capillaries in both lungs is about 70 m2. The alveoli are lined by two types of epithelial cells. Type I cells are flat cells with large cytoplasmic extensions and are FIGURE 35–1 Structure of the respiratory system. A) The respiratory system is diagrammed with a transparent lung to emphasize the flow of air into and out of the system. B) Enlargement of boxed area from (A) shows transition from conducting airway to the respiratory airway, with em-phasis on the anatomy of the alveoli. Red and blue represent oxygenated and deoxygenated blood, respectively. (Continued) Terminal bronchiole A B Branch of pulmonary vein Branch of pulmonary artery Smooth muscle Respiratory bronchiole Alveoli Capillary Trachea Left pulmonary artery Left main bronchus Bronchiole Heart Pulmonary veins 590 SECTION VII Respiratory Physiology the primary lining cells of the alveoli, covering approximately 95% of the alveolar epithelial surface area. Type II cells (gran-ular pneumocytes) are thicker and contain numerous lamel-lar inclusion bodies. A primary function of these cells is to secrete surfactant; however, they are also important in alveo-lar repair as well as other cellular physiology. Although these cells make up approximately 5% of the surface area, they rep-resent approximately 60% of the epithelial cells in the alveoli. The alveoli also contain other specialized cells, including pul-monary alveolar macrophages (PAMs, or AMs), lymphocytes, plasma cells, neuroendocrine cells, and mast cells. The mast cells contain heparin, various lipids, histamine, and various proteases that participate in allergic reactions (see Chapter 3). THE BRONCHI & THEIR INNERVATION The trachea and bronchi have cartilage in their walls but rela-tively little smooth muscle. They are lined by a ciliated epithe-lium that contains mucous and serous glands. Cilia are present as far as the respiratory bronchioles, but glands are absent from the epithelium of the bronchioles and terminal bronchi-oles, and their walls do not contain cartilage. However, their walls contain more smooth muscle, of which the largest amount relative to the thickness of the wall is present in the terminal bronchioles. The walls of the bronchi and bronchioles are innervated by the autonomic nervous system. Muscarinic receptors are abundant, and cholinergic discharge causes bronchoconstric-tion. The bronchial epithelium and smooth muscle contain β2-adrenergic receptors. Many of these are not innervated. Some may be located on cholinergic endings, where they inhibit acetylcholine release. The β2 receptors mediate broncho-dilation. They increase bronchial secretion, while α1 adrener-gic receptors inhibit secretion. There is, in addition, a noncholinergic, nonadrenergic innervation of the bronchi-oles that produces bronchodilation, and evidence suggests that vasoactive intestinal polypeptide (VIP) is the mediator responsible for the dilation. ANATOMY OF BLOOD FLOW IN THE LUNG Both the pulmonary circulation and the bronchial circula-tion contribute to blood flow in the lung. In the pulmonary circulation, almost all the blood in the body passes via the pul-monary artery to the pulmonary capillary bed, where it is oxygenated and returned to the left atrium via the pulmonary FIGURE 35–1 (Continued) C) The branching patterns of the air-way during the transition form conducting to respiratory airway are drawn (not all divisions are drawn, and drawings are not to scale). Name of branches Number of tubes in branch 1 4 2 8 16 32 6 x 104 5 x 105 8 x 106 Trachea Bronchi Bronchioles Terminal bronchioles Respiratory bronchioles Alveolar ducts Alveolar sacs Respiratory zone Conducting zone C FIGURE 35–2 Total airway cross-sectional area as a function of airway generation. Note the extremely rapid increase in total cross-sectional area in the respiratory zone. As a result, forward veloc-ity of gas during inspiration falls to a very low level in this zone. (Reproduced with permission from West JB: Respiratory Physiology: The Essentials, 4th ed. Williams & Wilkins, 1991.) 500 400 300 200 100 Total cross section area (cm2) 23 5 10 15 20 0 Airway generation Terminal bronchioles Conducting zone Respiratory zone CHAPTER 35 Pulmonary Function 591 veins (Figure 35–4). The separate and much smaller bronchial circulation includes the bronchial arteries that come from sys-temic arteries. They form capillaries, which drain into bron-chial veins or anastomose with pulmonary capillaries or veins (Figure 35–5). The bronchial veins drain into the azygos vein. The bronchial circulation nourishes the trachea down to the terminal bronchioles and also supplies the pleura and hilar lymph nodes. It should be noted that lymphatic channels are more abundant in the lungs than in any other organ. MECHANICS OF RESPIRATION INSPIRATION & EXPIRATION The lungs and the chest wall are elastic structures. Normally, no more than a thin layer of fluid is present between the lungs and the chest wall (intrapleural space). The lungs slide easily on the chest wall, but resist being pulled away from it in the same way that two moist pieces of glass slide on each other but resist separation. The pressure in the “space” between the lungs and chest wall (intrapleural pressure) is subatmospheric (Figure 35–6). The lungs are stretched when they expand at birth, and at the end of quiet expiration their tendency to re-coil from the chest wall is just balanced by the tendency of the FIGURE 35–3 Portion of an interalveolar septum in the adult human lung. A) A cross-section of the respiratory zone shows the relation-ship between capillaries and the airway epithelium. Only 4 of the 18 alveoli are labeled. B) Enlargement of the boxed area from (A) displaying in-timate relationship between capillaries, the interstitium, and the alveolar epithelium. C) Electron micrograph displaying area depicted in (B). The pulmonary capillary (cap) in the septum contains plasma with red blood cells apposed to the thin epithelial cells that line the alveoli. Note the closely apposed endothelial wall and pulmonary epithelium, separated at places by connective tissue fibers (cf); en, nucleus of endothelial cell; epl, nucleus of type I alveolar epithelial cell; a, alveolar space; ma, alveolar macrophage. (Reproduced with permission from (A, B) Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology: The Mechanisms of Body Function, 11th ed. McGraw-Hill, 2008; and (C) Burri PA: Development and growth of the human lung. In: Handbook of Physiology, Section 3, The Respiratory System. Fishman AP, Fisher AB [editors]. American Physiological Society, 1985.) Respiratory bronchiole Alveolar duct Alveolus pore Alveolus Alveolus Alveolus Capillaries Capillary endothelium Alveolar air Type II cell Type I cell Alveolar air Interstitium Plasma in capillary Basement membrane B A Erythrocyte Erythrocyte a ma a cf cf cap epI en C 592 SECTION VII Respiratory Physiology chest wall to recoil in the opposite direction. If the chest wall is opened, the lungs collapse; and if the lungs lose their elastic-ity, the chest expands and becomes barrel-shaped. Inspiration is an active process. The contraction of the inspiratory muscles increases intrathoracic volume. The intrapleural pressure at the base of the lungs, which is nor-mally about –2.5 mm Hg (relative to atmospheric) at the start of inspiration, decreases to about –6 mm Hg. The lungs are pulled into a more expanded position. The pressure in the airway becomes slightly negative, and air flows into the lungs. At the end of inspiration, the lung recoil begins to pull the chest back to the expiratory position, where the recoil pressures of the lungs and chest wall balance. The pressure in the airway becomes slightly positive, and air flows out of the lungs. Expiration during quiet breathing is passive in the sense that no muscles that decrease intratho-racic volume contract. However, some contraction of the inspiratory muscles occurs in the early part of expiration. This contraction exerts a braking action on the recoil forces and slows expiration. Strong inspiratory efforts reduce intrapleural pressure to values as low as –30 mm Hg, producing correspondingly greater degrees of lung inflation. When ventilation is increased, the extent of lung deflation is also increased by active contraction of expiratory muscles that decrease intrathoracic volume. FIGURE 35–4 Pulmonary and systemic circulations. Repre-sentative areas of blood flow are labeled with corresponding blood pressure (mm Hg). (Modified from Comroe JH Jr.: Physiology of Respiration, 2nd ed. Year Book, 1974.) FIGURE 35–5 Relationship between the bronchial and pulmonary circulations. The pulmonary artery supplies pulmonary capillary network A. The bronchial artery supplies capillary networks B, C, and D. Blue-colored areas represent blood of low O2 content. (Reproduced with permission from Murray JF: The Normal Lung. Saunders, 1986.) 120/80 24 9 25 0 120 0 8 2 14 12 12 14 120 80 12 12 10 20 30 Pulmonary artery A B C D Bronchial artery Bronchial vein Azygos vein Broncho-pulmonary vein Pulmonary vein Bronchopulmonary arterial anastomosis FIGURE 35–6 Pressure in the alveoli and the plural space relative to atmospheric pressure during inspiration and expiration. The dashed line indicates what the intrapleural pressure would be in the absence of airway and tissue resistance; the actual curve (solid line) is skewed to the left by the resistance. Volume of breath during inspiration/expiration is graphed for comparison. Pressure in alveoli Inspi-ration Expi-ration +2 +1 0 0.6 0.4 0.2 0 0 1 2 3 4 −1 −2 Intrapleural pressure −3 −4 −5 −6 Volume of breath Volume (L) Pressure (mm Hg) Time (s) CHAPTER 35 Pulmonary Function 593 LUNG VOLUMES The amount of air that moves into the lungs with each inspira-tion (or the amount that moves out with each expiration) is called the tidal volume. The air inspired with a maximal inspira-tory effort in excess of the tidal volume is the inspiratory reserve volume. The volume expelled by an active expiratory effort after passive expiration is the expiratory reserve volume, and the air left in the lungs after a maximal expiratory effort is the residual volume. Normal values for these lung volumes, and names ap-plied to combinations of them, are shown in Figure 35–7. The space in the conducting zone of the airways occupied by gas that does not exchange with blood in the pulmonary vessels is the respiratory dead space. The forced vital capacity (FVC), the largest amount of air that can be expired after a maximal inspira-tory effort, is frequently measured clinically as an index of pul-monary function. It gives useful information about the strength of the respiratory muscles and other aspects of pulmonary func-tion. The fraction of the vital capacity expired during the first second of a forced expiration is referred to as FEV1 (formerly the timed vital capacity) (Figure 35–8). The FEV1 to FVC ratio (FEV1/FVC) is a useful tool in the diagnosis of airway disease (Clinical Box 35–1). The amount of air inspired per minute (pulmonary ventilation, respiratory minute volume) is nor-mally about 6 L (500 mL/ breath × 12 breaths/min). The maxi-mal voluntary ventilation (MVV) is the largest volume of gas that can be moved into and out of the lungs in 1 min by volun-tary effort. The normal MVV is 125 to 170 L/min. FIGURE 35–7 Lung volumes and capacity measurements. Top left: A cartoon figure representing lung space divided into lung volumes. Dead space refers to areas where gas exchange does not occur; all other spaces are defined in the accompanying table. Top right: Spirometer recordings are shown with marked lung volumes and capacities. Table at bottom defines individual measurements and values from the top graphs. Note that residual volume, total lung capacity, and function residual capacity cannot be measure with a spirometer.(Right figure reproduced with permission from Widmaier EP, Raff H, Strang KT: Vander’s Human Physiology: The Mechanisms of Body Function, 11th ed. McGraw-Hill, 2008.) Inspiratory reserve volume Maximum possible inspiration Tidal volume Maximum voluntary expiration Expiratory reserve volume Residual volume Functional residual capacity Total lung capacity Inspiratory capacity Vital capacity Expiration Lung volume (ml) Inspiration 1000 2000 3000 4000 5000 6000 0 1 3 4 8 2 5 7 6 Tidal volume (TV) Inspiratory reserve volume (IRV) Expiratory reserve volume (ERV) Residual volume (RV) Respiratory Volumes and Capacities for an Average Young Adult Male Respiratory Volumes Respiratory Capacities Measurement 500 ml 3000 ml 1200 ml 1200 ml Typical Value Amount of air inhaled or exhaled in one breath during relaxed, quiet breathing Amount of air in excess of tidal inspiration that can be inhaled with maximum effort Amount of air in excess of tidal expiration that can be exhaled with maximum effort Amount of air remaining in the lungs after maximum expiration; keeps alveoli inflated between breaths and mixes with fresh air on next inspiration Vital capacity (VC) 4700 ml Amount of air that can be exhaled with maximum effort after maximum inspiration (ERV + TV + IRV); used to assess strength of thoracic muscles as well as pulmonary function Maximum amount of air that can be inhaled after a normal tidal expiration (TV + IRV) Amount of air remaining in the lungs after a normal tidal expiration (RV + ERV) Maximum amount of air the lungs can contain (RV + VC) Definition 3500 ml 2400 ml 5900 ml Inspiratory capacity (IC) Functional residual capacity (FRC) Total lung capacity (TLC) 8 3 2 4 5 7 6 1 IRV TV ERV RV Dead space 594 SECTION VII Respiratory Physiology RESPIRATORY MUSCLES Movement of the diaphragm accounts for 75% of the change in intrathoracic volume during quiet inspiration. Attached around the bottom of the thoracic cage, this muscle arches over the liver and moves downward like a piston when it con-tracts. The distance it moves ranges from 1.5 cm to as much as 7 cm with deep inspiration (Figure 35–9). The diaphragm has three parts: the costal portion, made up of muscle fibers that are attached to the ribs around the bot-tom of the thoracic cage; the crural portion, made up of fibers that are attached to the ligaments along the vertebrae; and the central tendon, into which the costal and the crural fibers insert. The central tendon is also the inferior part of the peri-cardium. The crural fibers pass on either side of the esopha-gus and can compress it when they contract. The costal and crural portions are innervated by different parts of the phrenic nerve and can contract separately. For example, dur-ing vomiting and eructation, intra-abdominal pressure is increased by contraction of the costal fibers but the crural fibers remain relaxed, allowing material to pass from the stomach into the esophagus. The other important inspiratory muscles are the external intercostal muscles, which run obliquely downward and for-ward from rib to rib. The ribs pivot as if hinged at the back, so that when the external intercostals contract they elevate the lower ribs. This pushes the sternum outward and increases the anteroposterior diameter of the chest. The transverse diameter also increases, but to a lesser degree. Either the dia-phragm or the external intercostal muscles alone can main-tain adequate ventilation at rest. Transection of the spinal cord above the third cervical segment is fatal without artificial respiration, but transection below the fifth cervical segment is not, because it leaves the phrenic nerves that innervate the diaphragm intact; the phrenic nerves arise from cervical seg-ments 3–5. Conversely, in patients with bilateral phrenic nerve palsy but intact innervation of their intercostal muscles, respiration is somewhat labored but adequate to maintain life. The scalene and sternocleidomastoid muscles in the neck are accessory inspiratory muscles that help to elevate the thoracic cage during deep labored respiration. A decrease in intrathoracic volume and forced expiration result when the expiratory muscles contract. The internal intercostals have this action because they pass obliquely downward and posteriorly from rib to rib and therefore pull the rib cage downward when they contract. Contractions of the muscles of the anterior abdominal wall also aid expiration by pulling the rib cage downward and inward and by increas-ing the intra-abdominal pressure, which pushes the dia-phragm upward. GLOTTIS The abductor muscles in the larynx contract early in inspira-tion, pulling the vocal cords apart and opening the glottis. During swallowing or gagging, a reflex contraction of the ad-ductor muscles closes the glottis and prevents aspiration of food, fluid, or vomitus into the lungs. In unconscious or anes-thetized patients, glottic closure may be incomplete and vom-itus may enter the trachea, causing an inflammatory reaction in the lung (aspiration pneumonia). The laryngeal muscles are supplied by the vagus nerves. When the abductors are paralyzed, there is inspiratory stridor. When the adductors are paralyzed, food and fluid enter the trachea, causing aspiration pneumonia and edema. Bilateral cervical vagotomy in animals causes the slow development of fatal pulmonary congestion and edema. The edema is due at least in part to aspiration, although some edema develops even if a tracheostomy is performed before the vagotomy. BRONCHIAL TONE In general, the smooth muscle in the bronchial walls aids res-piration. The bronchi dilate during inspiration and constrict during expiration. Dilation is produced by sympathetic dis-charge and constriction by parasympathetic discharge. Stimu-lation of sensory receptors in the airways by irritants and chemicals such as sulfur dioxide produces reflex bronchocon-striction that is mediated via cholinergic pathways. Cool air also causes bronchoconstriction, and so does exercise, possi-bly because the increased respiration associated with it cools the airways. In addition, the bronchial muscles protect the bronchi during coughing. There is a circadian rhythm in bronchial tone, with maximal constriction at about 6:00 AM and maximal dilation at about 6:00 PM. Many chemical sub-stances including VIP, substance P, adenosine, and many cy-tokines and inflammatory modulators can affect bronchial tone, although their full roles in the physiologic regulation of bronchial tone is still unsettled. FIGURE 35–8 Volume of gas expired by a normal adult man during a forced expiration, demonstrating the FEV1 and the total vital capacity (VC). (Reproduced, with permission, from Crapo RO: Pulmonary-function testing. N Engl J Med 1994;331:25. Copyright © 1994, Massachusetts Medical Society.) 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 0 VC FEV1 Volume (L) Time (s) CHAPTER 35 Pulmonary Function 595 COMPLIANCE OF THE LUNGS & CHEST WALL The interaction between the recoil of the lungs and recoil of the chest can be demonstrated in living subjects through a spirometer that has a valve just beyond the mouthpiece. The mouthpiece contains a pressure-measuring device. After the subject inhales a given amount, the valve is shut, closing off the airway. The respiratory muscles are then relaxed while the pressure in the airway is recorded. The procedure is repeated after inhaling or actively exhaling various volumes. The curve of airway pressure obtained in this way, plotted against vol-ume, is the relaxation pressure curve of the total respiratory system (Figure 35–10). The pressure is zero at a lung volume that corresponds to the volume of gas in the lungs at the end of quiet expiration (functional residual capacity, or FRC; also known as relaxation volume). It is positive at greater vol-umes and negative at smaller volumes. The change in lung vol-ume per unit change in airway pressure (ΔV/ΔP) is the compliance (stretchability) of the lungs and chest wall. It is normally measured in the pressure range where the relaxation pressure curve is steepest, and the normal value is approxi-mately 0.2 L/cm H2O. However, compliance depends on lung volume; an individual with only one lung has approximately half the ΔV for a given ΔP. Compliance is also slightly greater when measured during deflation than when measured during inflation. Consequently, it is more informative to examine the whole pressure–volume curve. The curve is shifted downward and to the right (compliance is decreased) by pulmonary con-gestion and interstitial pulmonary fibrosis (Figure 35–11). Pulmonary fibrosis is a progressive restrictive airway disease of unknown cause in which there is stiffening and scarring of CLINICAL BOX 35–1 Airway Diseases That Alter Airflow Obstructive Disease: Asthma in the number of pulmonary alveolar macrophages, and these macrophages release a chemical substance that attracts leu-kocytes to the lungs. The leukocytes in turn release proteases including elastase, which attacks the elastic tissue in the lungs. At the same time, α1-antitrypsin, a plasma protein that normally inactivates elastase and other proteases, is itself in-hibited. The α1-antitrypsin is inactivated by oxygen radicals, and these are released by the leukocytes. The final result is a protease–antiprotease imbalance with increased destruction of lung tissue. Similar protease–antiprotease imbalance can occur through congenital deficiency α1-antitrypsin. Airflow Measurements of Obstructive & Restrictive Disease In a healthy normal adult male, FVC is approximately 5.0 L, FEV1 is approximately 4.0 L, and thus, the calculated FEV1/FVC is 80%. As would be expected, patients with obstructive or re-strictive diseases display reduced FVC, on the order of 3.0 L, and this measurement alone does not differentiate between the two. However, measurement of FEV1 can significantly vary be-tween the two diseases. In obstructive disorders, patients tend to show a slow, steady slope to the FVC, resulting in a small FEV1, on the order of 1.3 L. However, in the restrictive disorder patients, air flow tends to be fast at first, and then due to the loss of elasticity, quickly levels out to approach FVC. The resul-tant FEV1 is much greater, on the order of 2.8 L, even though FVC is equivalent. A quick calculation of FEV1/FVC for obstruc-tive (42%) versus restrictive (90%) patients defines the hallmark measurements in evaluating these two diseases. Obstructive disorders result in a marked decrease in both FVC and FEV1/ FVC, whereas restrictive disorders result in a loss of FVC without loss in FEV1/FVC. Asthma is characterized by episodic or chronic wheezing, cough, and a feeling of tightness in the chest as a result of bronchoconstriction. Although the disease is not fully under-stood, three airway abnormalities are present: airway ob-struction that is at least partially reversible, airway inflamma-tion, and airway hyperresponsiveness to a variety of stimuli. A link to allergy has long been recognized, and plasma IgE le-vels are often elevated. Proteins released from eosinophils in the inflammatory reaction may damage the airway epithe-lium and contribute to the hyperresponsiveness. Leuko-trienes are released from eosinophils and mast cells, and can enhance bronchoconstriction. Numerous other amines, neu-ropeptides, chemokines, and interleukins have effects on bronchial smooth muscle or produce inflammation, and they may be involved in asthma. Because β2-adrenergic receptors mediate bronchodilation, β2-adrenergic agonists have long been the mainstay of treat-ment for mild to moderate asthma attacks. Inhaled and sys-temic steroids are used even in mild to moderate cases to re-duce inflammation; they are very effective, but their side effects can be a problem. Agents that block synthesis of leu-kotrienes or their CysLT1 receptor have also proved useful in some cases. Restrictive Disease: Emphysema Emphysema is a degenerative and potentially fatal pulmo-nary disease that is characterized by a loss of lung elasticity and replacement of alveoli with large air sacs. This loss of elas-ticity prevents full expansion of the lung, or airway restric-tion, during breathing. The most common cause of emphy-sema is heavy cigarette smoking. The smoke causes an increase 596 SECTION VII Respiratory Physiology the lung. The curve is shifted upward and to the left (compli-ance is increased) in emphysema. It should be noted that com-pliance is a static measure of lung and chest recoil. The resistance of the lung and chest is the pressure difference re-quired for a unit of air flow; this measurement, which is dy-namic rather than static, also takes into account the resistance to air flow in the airways. ALVEOLAR SURFACE TENSION An important factor affecting the compliance of the lungs is the surface tension of the film of fluid that lines the alveoli. The mag-nitude of this component at various lung volumes can be mea-sured by removing the lungs from the body of an experimental FIGURE 35–9 X-ray of chest in full expiration (left) and full inspiration (right). The dashed white line on the right is an outline of the lungs in full expiration. Note the difference in intrathoracic volume. (Reproduced with permission from Comroe JH Jr.: Physiology of Respiration, 2nd ed., Year Book, 1974.) FIGURE 35–10 Intrapulmonary pressure and volume relationship, the relaxation pressure curve. The middle curve is the relaxation pressure curve of the total respiratory system; that is, the static pressure curve of values obtained when the lungs are inflated or deflated by various amounts and the intrapulmonary pressure (elastic recoil pressure) is measured with the airway closed. The relaxation volume is the point where the recoil of the chest and the recoil of the lungs balance. The slope of the curve is the compliance of the lungs and chest wall. The maximal in-spiratory and expiratory curves are the airway pressures that can be developed during maximal inspiratory and expiratory efforts. Total lung capacity Functional residual capacity Residual volume Liters Liters 6 5 4 3 2 1 Maximal inspiratory curve Maximal expiratory curve Relaxation pressure curve Relaxation volume Vital capacity +3 +2 +1 0 −1 −120 −80 −40 0 40 80 120 160 200 −2 Intrapulmonary pressure (mm Hg) Change from resting volume CHAPTER 35 Pulmonary Function 597 animal and distending them alternately with saline and with air while measuring the intrapulmonary pressure. Because saline reduces the surface tension to nearly zero, the pressure–volume curve obtained with saline measures only the tissue elasticity (Figure 35–12), whereas the curve obtained with air measures both tissue elasticity and surface tension. The difference be-tween the two curves, the elasticity due to surface tension, is much smaller at small than at large lung volumes. The surface tension is also much lower than the expected surface tension at a water–air interface of the same dimensions. SURFACTANT The low surface tension when the alveoli are small is due to the presence in the fluid lining the alveoli of surfactant, a lipid surface-tension-lowering agent. Surfactant is a mixture of di-palmitoylphosphatidylcholine (DPPC), other lipids, and pro-teins (Table 35–2). If the surface tension is not kept low when the alveoli become smaller during expiration, they collapse in accordance with the law of Laplace. In spherical structures like the alveoli, the distending pressure equals two times the ten-sion divided by the radius (P = 2T/r); if T is not reduced as r is reduced, the tension overcomes the distending pressure. Sur-factant also helps to prevent pulmonary edema. It has been calculated that if it were not present, the unopposed surface tension in the alveoli would produce a 20 mm Hg force favor-ing transudation of fluid from the blood into the alveoli. Surfactant is produced by type II alveolar epithelial cells (Figure 35–13). Typical lamellar bodies, membrane-bound organelles containing whorls of phospholipid, are formed in these cells and secreted into the alveolar lumen by exocytosis. Tubes of lipid called tubular myelin form from the extruded bodies, and the tubular myelin in turn forms the phospholipid film. Following secretion, the phospholipids of surfactant line up in the alveoli with their hydrophobic fatty acid tails facing the alveolar lumen. Surface tension is inversely proportional to their concentration per unit area. The surfactant molecules move further apart as the alveoli enlarge during inspiration, and surface tension increases, whereas it decreases when they move closer together during expiration. Some of the protein– lipid complexes in surfactant are taken up by endocytosis in type II alveolar cells and recycled. Formation of the phospholipid film is greatly facilitated by the proteins in surfactant. This material contains four unique proteins: surfactant protein (SP)-A, SP-B, SP-C, and SP-D. SP-A is a large glycoprotein and has a collagen-like domain within its structure. It has multiple functions, including regulation of the feedback uptake of surfactant by the type II alveolar FIGURE 35–11 Static expiratory pressure–volume curves of lungs in normal subjects and subjects with severe emphysema and pulmonary fibrosis. (Modified and reproduced with permission from Pride NB, Macklem PT: Lung mechanics in disease. In: Handbook of Physiology. Section 3, The Respiratory System. Vol III, part 2. Fishman AP [editor]. American Physiological Society, 1986.) FIGURE 35–12 Pressure–volume relations in the lungs of a cat after removal from the body. Saline: lungs inflated and deflated with saline to reduce surface tension, resulting in a measurement of tissue elasticity. Air: lungs inflated (Inf) and deflated (Def) with air re-sults in a measure of both tissue elasticity and surface tension. (Reproduced with permission from Morgan TE: Pulmonary surfactant. N Engl J Med 1971;284:1185.) 10 20 30 40 0 1 2 3 4 5 6 7 8 Transpulmonary pressure (cm H2O) Lung volume (L) Fibrosis Normal Emphysema Saline Air 100 50 0 10 20 30 40 Pressure (cm H2O) Volume (% maximum inflation) Def Inf TABLE 35–2 Approximate composition of surfactant. Component Percentage Composition Dipalmitoylphosphatidylcholine 62 Phosphatidylglycerol 5 Other phospholipids 10 Neutral lipids 13 Proteins 8 Carbohydrate 2 598 SECTION VII Respiratory Physiology epithelial cells that secrete it. SP-B and SP-C are smaller pro-teins, which facilitate formation of the monomolecular film of phospholipid. A mutation of the gene for SP-C has been reported to be associated with familial interstitial lung dis-ease. Like SP-A, SP-D is a glycoprotein. Its full function is uncertain. However, SP-A and SP-D are members of the col-lectin family of proteins that are involved in innate immunity in the conducting airway as well as in the alveoli. For other roles of surfactant, see Clinical Box 35–2. WORK OF BREATHING Work is performed by the respiratory muscles in stretching the elastic tissues of the chest wall and lungs (elastic work; approxi-mately 65% of the total work), moving inelastic tissues (viscous resistance; 7% of total), and moving air through the respiratory passages (airway resistance; 28% of total). Because pressure times volume (g/cm2 × cm3 = g × cm) has the same dimensions as work (force × distance), the work of breathing can be calcu-lated from the relaxation pressure curve (Figures 35–10 and 35–14). The total elastic work required for inspiration is repre-sented by the area ABCA in Figure 35–14. Note that the relax-ation pressure curve of the total respiratory system differs from that of the lungs alone. The actual elastic work required to in-crease the volume of the lungs alone is area ABDEA. The amount of elastic work required to inflate the whole respiratory system is less than the amount required to inflate the lungs alone because part of the work comes from elastic energy stored in the thorax. The elastic energy lost from the thorax (area AFGBA) is equal to that gained by the lungs (area AEDCA). FIGURE 35–13 Formation and metabolism of surfactant. Lamellar bodies (LB) are formed in type II alveolar epithelial cells and secreted by exocytosis into the fluid lining the alveoli. The released lamellar body material is converted to tubular myelin (TM), and the TM is the source of the phospholipid surface film (SF). Surfactant is taken up by endocytosis into alveolar macrophages and type II epithelial cells. N, nucleus; RER, rough endoplasmic reticulum; CB, composite body. (Reproduced with permission from Wright JR: Metabolism and turnover of lung surfactant. Am Rev Respir Dis 1987;136:426.) RER LB CB N N N TM Golgi Type II cell Type I cell Alveolar macrophage Air space SF Fatty acids Choline Glycerol Amino acids Etc CLINICAL BOX 35–2 Surfactant Surfactant is important at birth. The fetus makes respiratory movements in utero, but the lungs remain collapsed until birth. After birth, the infant makes several strong inspira-tory movements and the lungs expand. Surfactant keeps them from collapsing again. Surfactant deficiency is an im-portant cause of infant respiratory distress syndrome (IRDS, also known as hyaline membrane disease), the seri-ous pulmonary disease that develops in infants born before their surfactant system is functional. Surface tension in the lungs of these infants is high, and the alveoli are collapsed in many areas (atelectasis). An additional factor in IRDS is retention of fluid in the lungs. During fetal life, Cl– is se-creted with fluid by the pulmonary epithelial cells. At birth, there is a shift to Na+ absorption by these cells via the epi-thelial Na+ channels (ENaCs), and fluid is absorbed with the Na+. Prolonged immaturity of the ENaCs contributes to the pulmonary abnormalities in IRDS. Patchy atelectasis is also associated with surfactant defi-ciency in patients who have undergone cardiac surgery in-volving use of a pump oxygenator and interruption of the pulmonary circulation. In addition, surfactant deficiency may play a role in some of the abnormalities that develop following occlusion of a main bronchus, occlusion of one pulmonary artery, or long-term inhalation of 100% O2. Cig-arette smoking also decreases lung surfactant. FIGURE 35–14 Relaxation pressure curves in the lung. The relaxation pressure curves of the total respiratory system (PTR), the lungs (PL), and the chest (PW) are plotted together with standard vol-umes for functional residual capacity and tidal volume. The transmural pressure is intrapulmonary pressure minus intrapleural pressure in the case of the lungs, intrapleural pressure minus outside (barometric) pressure in the case of the chest wall, and intrapulmonary pressure mi-nus barometric pressure in the case of the total respiratory system. From these curves, the total and actual elastic work associated with breathing can be derived (see text). (Modified from Mines AH: Respiratory Physiology, 3rd ed. Raven Press, 1993.) 6 4 2 0 0 −20 +20 B C H G F A E D Transmural pressure (cm H2O) PW PL PTR Lung volume (L) CHAPTER 35 Pulmonary Function 599 The frictional resistance to air movement is relatively small during quiet breathing, but it does cause the intrapleural pres-sure changes to lead the lung volume changes during inspira-tion and expiration (Figure 35–6), producing a hysteresis loop rather than a straight line when pressure is plotted against vol-ume (Figure 35–15). In this diagram, area AXBYA represents the work done to overcome airway resistance and lung viscos-ity. If the air flow becomes turbulent during rapid respiration, the energy required to move the air is greater than when the flow is laminar. Estimates of the total work of quiet breathing range from 0.3 up to 0.8 kg-m/min. The value rises markedly during exer-cise, but the energy cost of breathing in normal individuals represents less than 3% of the total energy expenditure during exercise. The work of breathing is greatly increased in diseases such as emphysema, asthma, and congestive heart failure with dyspnea and orthopnea. The respiratory muscles have length– tension relations like those of other skeletal and cardiac mus-cles, and when they are severely stretched, they contract with less strength. They can also become fatigued and fail (pump failure), leading to inadequate ventilation. DIFFERENCES IN VENTILATION & BLOOD FLOW IN DIFFERENT PARTS OF THE LUNG In the upright position, ventilation per unit lung volume is greater at the base of the lung than at the apex. The reason for this is that at the start of inspiration, intrapleural pressure is less negative at the base than at the apex (Figure 35–16), and since the intrapulmonary intrapleural pressure difference is less than at the apex, the lung is less expanded. Conversely, at the apex, the lung is more expanded; that is, the percentage of maximum lung volume is greater. Because of the stiffness of the lung, the increase in lung volume per unit increase in pres-sure is smaller when the lung is initially more expanded, and ventilation is consequently greater at the base. Blood flow is also greater at the base than the apex. The relative change in blood flow from the apex to the base is greater than the relative change in ventilation, so the ventilation/perfusion ratio is low at the base and high at the apex. The ventilation and perfusion differences from the apex to the base of the lung have usually been attributed to gravity; they tend to disappear in the supine position, and the weight of the lung would be expected to make the intrapleural pres-sure lower at the base in the upright position. However, the inequalities of ventilation and blood flow in humans were found to persist to a remarkable degree in the weightlessness of space. Therefore, other factors also play a role in producing the inequalities. DEAD SPACE & UNEVEN VENTILATION Because gaseous exchange in the respiratory system occurs only in the terminal portions of the airways, the gas that occu-pies the rest of the respiratory system is not available for gas exchange with pulmonary capillary blood. Normally, the vol-ume (in mL) of this anatomic dead space is approximately equal to the body weight in pounds. As an example, in a man who weighs 150 lb (68 kg), only the first 350 mL of the 500 mL inspired with each breath at rest mixes with the air in the alve-oli. Conversely, with each expiration, the first 150 mL expired is gas that occupied the dead space, and only the last 350 mL is gas from the alveoli. Consequently, the alveolar ventilation; FIGURE 35–15 Pressure volume relationships in breathing. Diagrammatic representation of pressure and volume changes during quiet inspiration (line AXB) and expiration (line BZA). Line AYB is the compliance line. −2 0 −4 −6 500 Z Y X A Intrapleural pressure (mm Hg) Tidal volume (mL) C B FIGURE 35–16 Intrapleural pressures in the upright position and their effect on ventilation. Note that because intrapul-monary pressure is atmospheric, the more negative intrapleural pres-sure at the apex holds the lung in a more expanded position at the start of inspiration. Further increases in volume per unit increase in in-trapleural pressure are smaller than at the base because the expanded lung is stiffer. (Reproduced with permission from West JB: Ventilation/Blood Flow and Gas Exchange, 3rd ed. Blackwell, 1977.) +10 0 0 50% 100% –10 –20 –30 Lung volume Intrapleural pressure (cm H2O) –2.5 cm H2O –10 cm H2O Intrapleural pressure 600 SECTION VII Respiratory Physiology that is, the amount of air reaching the alveoli per minute, is less than the respiratory minute volume. Note in addition that because of the dead space, rapid shallow breathing produces much less alveolar ventilation than slow deep breathing at the same respiratory minute volume (Table 35–3). It is important to distinguish between the anatomic dead space (respiratory system volume exclusive of alveoli) and the total (physiologic) dead space (volume of gas not equilibrat-ing with blood; ie, wasted ventilation). In healthy individuals, the two dead spaces are identical and can be estimated by body weight. However, in disease states, no exchange may take place between the gas in some of the alveoli and the blood, and some of the alveoli may be overventilated. The volume of gas in non-perfused alveoli and any volume of air in the alveoli in excess of that necessary to arterialize the blood in the alveolar capil-laries is part of the dead space (nonequilibrating) gas volume. The anatomic dead space can be measured by analysis of the single-breath N2 curves (Figure 35–17). From mid-inspira-tion, the subject takes as deep a breath as possible of pure O2, then exhales steadily while the N2 content of the expired gas is continuously measured. The initial gas exhaled (phase I) is the gas that filled the dead space and that consequently contains no N2. This is followed by a mixture of dead space and alveolar gas (phase II) and then by alveolar gas (phase III). The volume of the dead space is the volume of the gas expired from peak inspiration to the midportion of phase II. Phase III of the single-breath N2 curve terminates at the closing volume (CV) and is followed by phase IV, during which the N2 content of the expired gas is increased. The CV is the lung volume above residual volume at which airways in the lower, dependent parts of the lungs begin to close off because of the lesser transmural pressure in these areas. The gas in the upper portions of the lungs is richer in N2 than the gas in the lower, dependent portions because the alveoli in the upper portions are more distended at the start of the inspira-tion of O2 and, consequently, the N2 in them is less diluted with O2. It is also worth noting that in most normal individu-als, phase III has a slight positive slope even before phase IV is reached. This indicates that even during phase III there is a gradual increase in the proportion of the expired gas coming from the relatively N2-rich upper portions of the lungs. The total dead space can be calculated from the PCO2 of expired air, the PCO2 of arterial blood, and the tidal volume. The tidal volume (VT) times the PCO2 of the expired gas (PECO2) equals the arterial PCO2 (PaCO2) times the difference between the tidal volume and the dead space (VD) plus the PCO2 of inspired air (PICO2) times VD (Bohr’s equation): PECO2 × VT = PaCO2 × (VT – VD) + PICO2 × VD The term PICO2 × VD is so small that it can be ignored and the equation solved for VD. If, for example, PECO2 = 28 mm Hg PaCO2 = 40 mm Hg VT = 500 mL then, Vd = 150 mL The equation can also be used to measure the anatomic dead space if one replaces PaCO2 with alveolar PCO2 (PACO2), which is the PCO2 of the last 10 mL of expired gas. PCO2 is an average of gas from different alveoli in proportion to their ventilation regardless of whether they are perfused. This is in contrast to PaCO2, which is gas equilibrated only with per-fused alveoli, and consequently, in individuals with unper-fused alveoli, is greater than PCO2. GAS EXCHANGE IN THE LUNGS SAMPLING ALVEOLAR AIR Theoretically, all but the first 150 mL expired from a healthy 150-lb man (ie, the dead space) with each expiration is the gas that was in the alveoli (alveolar air), but some mixing always occurs at the interface between the dead-space gas and the al-veolar air (Figure 35–17). A later portion of expired air is therefore the portion taken for analysis. Using modern appa-ratus with a suitable automatic valve, it is possible to collect the last 10 mL expired during quiet breathing. The composi-tion of alveolar gas is compared with that of inspired and ex-pired air in Figure 35–18. TABLE 35–3 Effect of variations in respiratory rate and depth on alveolar ventilation. Respiratory rate 30/min 10/min Tidal volume 200 mL 600 mL Minute volume 6 L 6 L Alveolar ventilation (200 – 150) × 30 = 1500 mL (600 – 150) × 10 = 4500 mL FIGURE 35–17 Single-breath N2 curve. From mid-inspiration, the subject takes a deep breath of pure O2 then exhales steadily. The changes in the N2 concentration of expired gas during expiration are shown, with the various phases of the curve indicated by roman nu-merals. Notably, region I is representative of the dead space (DS); from I–II is a mixture of DS and alveolar gas; the transition form III–IV is the closing volume (CV), and the end of IV is the residual volume (RV). 6 0 30 0 Lung volume (L) DS CV RV IV III II I N2 concentration (%) CHAPTER 35 Pulmonary Function 601 PAO2 can also be calculated from the alveolar gas equation: PAO2 = PIO2 – PACO2 FIO2 + 1– FIO2 R where FIO2 is the fraction of O2 molecules in the dry gas, PIO2 is the inspired PO2, and R is the respiratory exchange ratio; that is, the flow of CO2 molecules across the alveolar mem-brane per minute divided by the flow of O2 molecules across the membrane per minute. COMPOSITION OF ALVEOLAR AIR Oxygen continuously diffuses out of the gas in the alveoli into the bloodstream, and CO2 continuously diffuses into the alve-oli from the blood. In the steady state, inspired air mixes with the alveolar gas, replacing the O2 that has entered the blood and diluting the CO2 that has entered the alveoli. Part of this mixture is expired. The O2 content of the alveolar gas then falls and its CO2 content rises until the next inspiration. Be-cause the volume of gas in the alveoli is about 2 L at the end of expiration (functional residual capacity), each 350 mL incre-ment of inspired and expired air has relatively little effect on PO2 and PCO2. Indeed, the composition of alveolar gas re-mains remarkably constant, not only at rest but also under a variety of other conditions. DIFFUSION ACROSS THE ALVEOLOCAPILLARY MEMBRANE Gases diffuse from the alveoli to the blood in the pulmonary capillaries or vice versa across the thin alveolocapillary membrane made up of the pulmonary epithelium, the capil-lary endothelium, and their fused basement membranes (Figure 35–3). Whether or not substances passing from the al-veoli to the capillary blood reach equilibrium in the 0.75 s that blood takes to traverse the pulmonary capillaries at rest de-pends on their reaction with substances in the blood. Thus, for example, the anesthetic gas nitrous oxide (N2O) does not react and reaches equilibrium in about 0.1 s (Figure 35–19). In this situation, the amount of N2O taken up is not limited by diffu-sion but by the amount of blood flowing through the pulmo-nary capillaries; that is, it is flow-limited. On the other hand, carbon monoxide (CO) is taken up by hemoglobin in the red blood cells at such a high rate that the partial pressure of CO in the capillaries stays very low and equilibrium is not reached in the 0.75 s the blood is in the pulmonary capillaries. There-fore, the transfer of CO is not limited by perfusion at rest and instead is diffusion-limited. O2 is intermediate between N2O and CO; it is taken up by hemoglobin, but much less avidly than CO, and it reaches equilibrium with capillary blood in about 0.3 s. Thus, its uptake is perfusion-limited. The diffusing capacity of the lung for a given gas is directly proportionate to the surface area of the alveolocapillary mem-brane and inversely proportionate to its thickness. The diffus-ing capacity for CO (DLCO) is measured as an index of diffusing capacity because its uptake is diffusion-limited. DLCO is proportionate to the amount of CO entering the blood (VCO) divided by the partial pressure of CO in the alve-oli minus the partial pressure of CO in the blood entering the pulmonary capillaries. Except in habitual cigarette smokers, FIGURE 35–18 Partial pressures of gases (mm Hg) in various parts of the respiratory system and in the circulatory system. O2 CO2 H2O N2 158.0 0.3 5.7 596.0 O2 CO2 H2O N2 116.0 32.0 47.0 565.0 O2 CO2 H2O N2 100.0 40.0 47.0 573.0 Inspired air Expired gas O2 CO2 H2O N2 95.0 40.0 47.0 573.0 O2 CO2 H2O N2 40.0− 46.0+ 47.0 573.0 O2 CO2 H2O N2 40.0 46.0 47.0 573.0 Tissues Capillaries Dead space Physiologic shunt Left heart Veins Arteries Right heart Alveoli FIGURE 35–19 Uptake of various substances during the 0.75 s they are in transit through a pulmonary capillary. N2O is not bound in blood, so its partial pressure in blood rises rapidly to its partial pressure in the alveoli. Conversely, CO is avidly taken up by red blood cells, so its partial pressure reaches only a fraction of its partial pressure in the alveoli. O2 is intermediate between the two. 0.25 0 0.50 0.75 CO Alveolar level Time in capillary (s) O2 N2O Partial pressure ( ( 602 SECTION VII Respiratory Physiology this latter term is close to zero, so it can be ignored and the equation becomes: DLCO = V • CO PACO The normal value of DLCO at rest is about 25 mL/min/mm Hg. It increases up to threefold during exercise because of capillary dilation and an increase in the number of active cap-illaries. The PO2 of alveolar air is normally 100 mm Hg (Figure 35–18), and the PO2 of the blood entering the pulmonary capillaries is 40 mm Hg. The diffusing capacity for O2, like that for CO at rest, is about 25 mL/min/mm Hg, and the PO2 of blood is raised to 97 mm Hg, a value just under the alveolar PO2. This falls to 95 mm Hg in the aorta because of the physiologic shunt. DLO2 increases to 65 mL/min/mm Hg or more during exercise and is reduced in diseases such as sarcoidosis and beryllium poison-ing (berylliosis) that cause fibrosis of the alveolar walls. The PCO2 of venous blood is 46 mm Hg, whereas that of alveolar air is 40 mm Hg, and CO2 diffuses from the blood into the alveoli along this gradient. The PCO2 of blood leaving the lungs is 40 mm Hg. CO2 passes through all biological membranes with ease, and the diffusing capacity of the lung for CO2 is much greater than the capacity for O2. It is for this reason that CO2 retention is rarely a problem in patients with alveolar fibrosis even when the reduction in diffusing capacity for O2 is severe. PULMONARY CIRCULATION PULMONARY BLOOD VESSELS The pulmonary vascular bed resembles the systemic one, ex-cept that the walls of the pulmonary artery and its large branch-es are about 30% as thick as the wall of the aorta, and the small arterial vessels, unlike the systemic arterioles, are endothelial tubes with relatively little muscle in their walls. The walls of the postcapillary vessels also contain some smooth muscle. The pulmonary capillaries are large, and there are multiple anasto-moses, so that each alveolus sits in a capillary basket. PRESSURE, VOLUME, & FLOW With two quantitatively minor exceptions, the blood put out by the left ventricle returns to the right atrium and is ejected by the right ventricle, making the pulmonary vasculature unique in that it accommodates a blood flow that is almost equal to that of all the other organs in the body. One of the ex-ceptions is part of the bronchial blood flow. As shown in Fig-ure 35–5, there are anastomoses between the bronchial capillaries and the pulmonary capillaries and veins, and al-though some of the bronchial blood enters the bronchial veins, some enters the pulmonary capillaries and veins, by-passing the right ventricle. The other exception is blood that flows from the coronary arteries into the chambers of the left side of the heart. Because of the small physiologic shunt cre-ated by those two exceptions, the blood in systemic arteries has a PO2 about 2 mm Hg lower than that of blood that has equilibrated with alveolar air, and the saturation of hemoglo-bin is 0.5% less. The pressure in the various parts of the pulmonary portion of the pulmonary circulation is shown in Figure 35–4. The pressure gradient in the pulmonary system is about 7 mm Hg, compared with a gradient of about 90 mm Hg in the systemic circulation. Pulmonary capillary pressure is about 10 mm Hg, whereas the oncotic pressure is 25 mm Hg, so that an inward-directed pressure gradient of about 15 mm Hg keeps the alve-oli free of all but a thin film of fluid. When the pulmonary capillary pressure is more than 25 mm Hg—as it may be, for example, in “backward failure” of the left ventricle—pulmo-nary congestion and edema result. The volume of blood in the pulmonary vessels at any one time is about 1 L, of which less than 100 mL is in the capillar-ies. The mean velocity of the blood in the root of the pulmo-nary artery is the same as that in the aorta (about 40 cm/s). It falls off rapidly, then rises slightly again in the larger pulmo-nary veins. It takes a red cell about 0.75 s to traverse the pul-monary capillaries at rest and 0.3 s or less during exercise. EFFECT OF GRAVITY Gravity has a relatively marked effect on the pulmonary circu-lation. In the upright position, the upper portions of the lungs are well above the level of the heart, and the bases are at or be-low it. Consequently, in the upper part of the lungs, the blood flow is less, the alveoli are larger, and ventilation is less than at the base (Figure 35–20). The pressure in the capillaries at the top of the lungs is close to the atmospheric pressure in the al-veoli. Pulmonary arterial pressure is normally just sufficient to maintain perfusion, but if it is reduced or if alveolar pressure is increased, some of the capillaries collapse. Under these cir-cumstances, no gas exchange takes place in the affected alveoli and they become part of the physiologic dead space. In the middle portions of the lungs, the pulmonary arterial and capillary pressure exceeds alveolar pressure, but the pres-sure in the pulmonary venules may be lower than alveolar pressure during normal expiration, so they are collapsed. Under these circumstances, blood flow is determined by the pulmonary artery–alveolar pressure difference rather than the pulmonary artery–pulmonary vein difference. Beyond the constriction, blood “falls” into the pulmonary veins, which are compliant and take whatever amount of blood the constriction lets flow into them. This has been called the waterfall effect. Obviously, the compression of vessels produced by alveolar pressure decreases and pulmonary blood flow increases as the arterial pressure increases toward the base of the lung. In the lower portions of the lungs, alveolar pressure is lower than the pressure in all parts of the pulmonary circulation and blood flow is determined by the arterial–venous pressure CHAPTER 35 Pulmonary Function 603 difference. Examples of diseases affecting pulmonary circula-tion are given in Clinical Box 35–3. VENTILATION/PERFUSION RATIOS The ratio of pulmonary ventilation to pulmonary blood flow for the whole lung at rest is about 0.8 (4.2 L/min ventilation di-vided by 5.5 L/min blood flow). However, relatively marked differences occur in this ventilation/perfusion ratio in vari-ous parts of the normal lung as a result of the effect of gravity, and local changes in the ventilation/perfusion ratio are com-mon in disease. If the ventilation to an alveolus is reduced rel-ative to its perfusion, the PO2 in the alveolus falls because less O2 is delivered to it and the PCO2 rises because less CO2 is ex-pired. Conversely, if perfusion is reduced relative to ventila-tion, the PCO2 falls because less CO2 is delivered and the PO2 rises because less O2 enters the blood. These effects are sum-marized in Figure 35–21. As noted above, ventilation, as well as perfusion in the upright position, declines in a linear fashion from the bases to the apices of the lungs. However, the ventilation/perfusion ratios are high in the upper portions of the lungs. When wide-spread, nonuniformity of ventilation and perfusion in the lungs can cause CO2 retention and declines in systemic arte-rial PO2. PULMONARY RESERVOIR Because of their distensibility, the pulmonary veins are an im-portant blood reservoir. When a normal individual lies down, the pulmonary blood volume increases by up to 400 mL, and when the person stands up this blood is discharged into the general circulation. This shift is the cause of the decrease in vi-tal capacity in the supine position and is responsible for the occurrence of orthopnea in heart failure. REGULATION OF PULMONARY BLOOD FLOW It is unsettled whether pulmonary veins and pulmonary arter-ies are regulated separately, although constriction of the veins increases pulmonary capillary pressure and constriction of pulmonary arteries increases the load on the right side of the heart. Pulmonary blood flow is affected by both active and passive factors. There is an extensive autonomic innervation of the pulmonary vessels, and stimulation of the cervical sympathetic FIGURE 35–20 Diagram of normal differences in ventilation and perfusion of the lung in the upright position. Outlined areas are representative of changes in alveolar size (not actual size). Note the gradual change in alveolar size from top (apex) to bottom. Character-istic differences of alveoli at the apex of the lung are stated. (Modified from Levitsky, MG: Pulmonary Physiology, 6th ed. McGraw-Hill, 2003). At apex Intrapleural pressure more negative Greater transmural pressure Large alveoli Lower intravascular pressure Less blood flow So less ventilation and perfusion CLINICAL BOX 35–3 Diseases Affecting the Pulmonary Circulation Pulmonary Hypertension Sustained primary pulmonary hypertension can occur at any age. Like systemic arterial hypertension, it is a syndrome with multiple causes. However, the causes are different from those causing systemic hypertension. They include hypoxia, inhalation of cocaine, treatment with dexfenfluramine and related appetite-suppressing drugs that increase extracellu-lar serotonin, and systemic lupus erythematosus. Some cases are familial and appear to be related to mutations that in-crease the sensitivity of pulmonary vessels to growth factors or cause deformations in the pulmonary vascular system. All these conditions lead to increased pulmonary vascu-lar resistance. If appropriate therapy is not initiated, the in-creased right ventricular afterload can lead eventually to right heart failure and death. Treatment with vasodilators such as prostacyclin and prostacyclin analogs is effective. Until recently, these had to be administered by continuous intravenous infusion, but aerosolized preparations that ap-pear to be effective are now available. Pulmonary Embolization One of the normal functions of the lungs is to filter out small blood clots, and this occurs without any symptoms. When emboli block larger branches of the pulmonary ar-tery, they provoke a rise in pulmonary arterial pressure and rapid, shallow respiration (tachypnea). The rise in pulmo-nary arterial pressure may be due to reflex vasoconstriction via the sympathetic nerve fibers, but reflex vasoconstric-tion appears to be absent when large branches of the pul-monary artery are blocked. The tachypnea is a reflex re-sponse to activation of vagally innervated pulmonary receptors close to the vessel walls. These appear to be acti-vated at the site of the embolization. 604 SECTION VII Respiratory Physiology ganglia reduces pulmonary blood flow by as much as 30%. The vessels also respond to circulating humoral agents. Sev-eral of the receptors involved and their effect on pulmonary smooth muscle are summarized in Table 35–4. Many of the dilator responses are endothelium-dependent and presum-ably operate via release of nitric oxide (NO). Passive factors such as cardiac output and gravitational forces also have significant effects on pulmonary blood flow. Local adjustments of perfusion to ventilation are determined by local effects of O2 (or the lack of O2). With exercise, cardiac output increases and pulmonary arterial pressure rises pro-portionately with little or no vasodilation. More red cells move through the lungs without any reduction in the O2 satu-ration of the hemoglobin in them, and consequently, the total amount of O2 delivered to the systemic circulation is increased. Capillaries dilate, and previously underperfused capillaries are “recruited” to carry blood. The net effect is a marked increase in pulmonary blood flow with few, if any, alterations in autonomic outflow to the pulmonary vessels. When a bronchus or a bronchiole is obstructed, hypoxia develops in the underventilated alveoli beyond the obstruc-tion. The O2 deficiency apparently acts directly on vascular smooth muscle in the area to produce constriction, shunting blood away from the hypoxic area. Accumulation of CO2 leads to a drop in pH in the area, and a decline in pH also pro-duces vasoconstriction in the lungs, as opposed to the vasodi-lation it produces in other tissues. Conversely, reduction of the blood flow to a portion of the lung lowers the alveolar PCO2 in that area, and this leads to constriction of the bronchi supplying it, shifting ventilation away from the poorly per-fused area. Systemic hypoxia also causes the pulmonary arte-rioles to constrict, with a resultant increase in pulmonary arterial pressure. FIGURE 35–21 Effects of decreasing or increasing the ventilation/perfusion ratio (V • A/Q • ) on the PCO2 and PO2 in an alveolus. The drawings above the curve represent an alveolus and a pulmonary capillary, and the dark red areas indicate sites of blockage. With complete obstruction of the airway to the alveolus, PCO2 and PO2 approximate the values in mixed venous blood (V –). With complete block of perfusion, PCO2 and PO2 approximate the values in inspired air. (Reproduced with permission from West JB: Ventilation/Blood Flow and Gas Exchange, 3rd ed. Blackwell, 1977.) • • 50 0 50 100 150 O A _ V Decreasing VA/Q Increasing VA/Q Normal PO2 (mm Hg) PCO2 (mm Hg) • • TABLE 35–4 Receptors affecting smooth muscle in pulmonary arteries and veins. Receptor Subtype Response Endothelium Dependency Autonomic Adrenergic α1 Contraction No α2 Relaxation Yes β2 Relaxation Yes Muscarinic M3 Relaxation Yes Purinergic P2x Contraction No P2y Relaxation Yes Tachykinin NK1 Relaxation Yes NK2 Contraction No VIP ? Relaxation ? CGRP ? Relaxation No Humoral Adenosine A1 Contraction No A2 Relaxation No Angiotensin II AT1 Contraction No ANP ANPA Relaxation No ANPB Relaxation No Bradykinin B1? Relaxation Yes B2 Relaxation Yes Endothelin ETA Contraction No ETB Relaxation Yes Histamine H1 Relaxation Yes H2 Relaxation No 5-HT 5-HT1 Contraction No 5-HT1C Relaxation Yes Thromboxane TP Contraction No Vasopressin V1 Relaxation Yes Modified and reproduced with permission from Barnes PJ, Lin SF: Regulation of pul-monary vascular tone. Pharmacol Rev 1995;47:88. CHAPTER 35 Pulmonary Function 605 OTHER FUNCTIONS OF THE RESPIRATORY SYSTEM LUNG DEFENSE MECHANISMS The respiratory passages that lead from the exterior to the alveoli do more than serve as gas conduits. They humidify and cool or warm the inspired air so that even very hot or very cold air is at or near body temperature by the time it reaches the alveoli. Airway epithelial cells can secrete a variety of mole-cules that aid in lung defense. Secretory immunoglobulins (IgA), collectins (including Surfactant A and D), defensins and other peptides and proteases, reactive oxygen species, and reactive nitrogen species are all generated by airway epithelial cells. These secretions can act directly as antimicrobials to help keep the airway free of infection. Airway epithelial cells also secrete a variety of chemokines and cytokines that recruit the traditional immune cells and others to site of infections. Various mechanisms operate to prevent foreign matter from reaching the alveoli. The hairs in the nostrils strain out many particles larger than 10 μm in diameter. Most of the remaining particles of this size settle on mucous membranes in the nose and pharynx; because of their momentum, they do not follow the airstream as it curves downward into the lungs, and they impact on or near the tonsils and adenoids, large collections of immunologically active lymphoid tissue in the back of the pharynx. Particles 2 to 10 μm in diameter gener-ally fall on the walls of the bronchi as the air flow slows in the smaller passages. There they can initiate reflex bronchial con-striction and coughing. Alternatively, they can be moved away from the lungs by the “mucociliary escalator.” The epithelium of the respiratory passages from the anterior third of the nose to the beginning of the respiratory bronchioles is ciliated. The cilia are bathed in a periciliary fluid where they typically beat at rates of 10–15 Hz. On top of the periciliary layer and the beating cilia rests a mucus layer, a complex mixture of pro-teins and polysaccharides secreted from specialized cells, glands, or both in the conducting airway. This combination allows for the trapping of foreign particles (in the mucus) and their transport out of the airway (powered by ciliary beat). The ciliary mechanism is capable of moving particles away from the lungs at a rate of at least 16 mm/min. When ciliary motility is defective, as can occur from smoking, other envi-ronmental conditions, or genetic deficiency, mucus transport is virtually absent. This can lead to chronic sinusitis, recurrent lung infections, and bronchiectasis. Some of these symptoms are evident in cystic fibrosis (Clinical Box 35–4). The pulmonary alveolar macrophages (PAMs) are another important component of the pulmonary defense system. Like other macrophages , these cells come originally from the bone marrow. Particles less than 2 μm in diameter can evade the mucociliary escalator and reach the alveoli. PAMs are actively phagocytic and ingest these small particles. They also help process inhaled antigens for immunologic attack, and they secrete substances that attract granulocytes to the lungs as well as substances that stimulate granulocyte and monocyte formation in the bone marrow. When the PAMs ingest large amounts of the substances in cigarette smoke or other irri-tants, they may also release lysosomal products into the extra-cellular space to cause inflammation. METABOLIC & ENDOCRINE FUNCTIONS OF THE LUNGS In addition to their functions in gas exchange, the lungs have a number of metabolic functions. They manufacture surfac-tant for local use, as noted above. They also contain a fibrino-lytic system that lyses clots in the pulmonary vessels. They release a variety of substances that enter the systemic arterial blood (Table 35–5), and they remove other substances from the systemic venous blood that reach them via the pulmonary artery. Prostaglandins are removed from the circulation, but they are also synthesized in the lungs and released into the blood when lung tissue is stretched. CLINICAL BOX 35–4 Cystic Fibrosis Among Caucasians, cystic fibrosis is one of the most com-mon genetic disorders: 5% of the population carry a defec-tive gene, and the disease occurs in 1 of every 2000 births. The gene that is abnormal in cystic fibrosis is located on the long arm of chromosome 7 and encodes the cystic fi-brosis transmembrane conductance regulator (CFTR), a regulated Cl– channel located on the apical membrane of various secretary and reabsorptive epithelia. The number of reported mutations in the CFTR gene that cause cystic fibro-sis is large, and the severity of the defect varies with the mu-tation; however, this is not surprising in a gene encoding such a complex protein. The most common mutation caus-ing cystic fibrosis is loss of the phenylalanine residue at po-sition 508 of the protein (ΔF508). This hinders proper fold-ing of the molecule, leading to reduced membrane levels. One outcome of cystic fibrosis is repeated pulmonary in-fections, particularly with Pseudomonas aeruginosa, and progressive, eventually fatal destruction of the lungs. In this congenital recessive condition, the function of a Cl– chan-nel, the CFTR channel, is depressed by loss-of-function mu-tations in the gene that encodes it. One would expect Na+ reabsorption to be depressed as well, and indeed in sweat glands it is. However, in the lungs, it is enhanced, so that the Na+ and water move out of airways, leaving their other secretions inspissated and sticky. This results in a reduced periciliary layer that inhibits function of the mucociliary es-calator, and alters the local environment to reduce the ef-fectiveness of antimicrobial secretions. 606 SECTION VII Respiratory Physiology The lungs also activate one hormone; the physiologically inactive decapeptide angiotensin I is converted to the pressor, aldosterone-stimulating octapeptide angiotensin II in the pul-monary circulation. The reaction occurs in other tissues as well, but it is particularly prominent in the lungs. Large amounts of the angiotensin-converting enzyme responsible for this activation are located on the surface of the endothelial cells of the pulmonary capillaries. The converting enzyme also inactivates bradykinin. Circulation time through the pul-monary capillaries is less than 1 s, yet 70% of the angiotensin I reaching the lungs is converted to angiotensin II in a single trip through the capillaries. Four other peptidases have been identified on the surface of the pulmonary endothelial cells, but their full physiologic role is unsettled. Removal of serotonin and norepinephrine reduces the amounts of these vasoactive substances reaching the systemic circulation. However, many other vasoactive hormones pass through the lungs without being metabolized. These include epinephrine, dopamine, oxytocin, vasopressin, and angioten-sin II. In addition, various amines and polypeptides are secreted by neuroendocrine cells in the lungs. CHAPTER SUMMARY ■The pressure exerted by any one gas in a mixture of gases is de-fined as its partial pressure. Partial pressures (P) of gases in air at sea level are as follows: Po2 = 149 mm Hg; Pco2 = 0.3 mm Hg; PN2 (including other gases) = 564 mm Hg. ■Air enters the respiratory system in the upper airway, then pro-ceeds to the conducting airway and on to the respiratory airway that ends in the alveoli. In the upper airway, air is humidified and warmed. The cross sectional area of the airway gradually increases through the conducting zone, then rapidly increases during the transition from conducting to respiratory zones ■The epithelium that line the conducting airway include ciliated cells that keep particulates from reaching the respiratory zone. The epithelium that lines the alveoli consist of two cell types: alveolar type I cells and alveolar type II cells. Type I cells are flattened epithelial cells that provide approximately 95% of the alveolar surface area and are the site of gas exchange. Type II cells are cuboidal epithelial cells that secrete surfactants that line the alveolar surface. ■There are several important measures of lung volume, includ-ing: tidal volume; inspiratory volume; expiratory reserve vol-ume; forced vital capacity (FVC); the forced expiratory volume in one second (FEV1); respiratory minute volume and maximal voluntary ventilation. ■Lung compliance refers to the ability of lungs to stretch. How-ever, many normal factors affect lung compliance and it is best represented by a whole pressure-volume curve. ■Surfactant is a lipid-protein mixture that is in the fluid lining the alveolar epithelium. A primary function of surfactant is to in-crease surface tension in the alveoli to keep them from deflating. ■Both ventilation and perfusion are greater at the base of the lung and lower at the apex of the lung. The ventilation/perfusion ratio is lower at the base compared to the apex of the lung. ■Not all air that enters the airway is available for gas exchange. The regions where gas is not exchanged in the airway are termed “dead space.” The conducting airway represents anatomical dead space. Increased dead space can occur in response to disease that affects air exchange in the respiratory zone. ■The pressure gradient in the pulmonary circulation system is much less than that in the systemic circulation. Because pulmo-nary capillary pressure is much lower than oncotic pressure in the plasma, fluid remains in the plasma as it traverses the lung. ■The mucociliary escalator in the conducting airway helps to keep particulates out of the respiratory zone. ■There are a variety of biologically activated substances that are metabolized in the lung. These include substances that are made and function in the lung (eg, surfactant), substances that are released or removed from the blood (eg, prostaglandins), and substances that are activated as they pass through the lung (eg, angiotensin II). MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. On the summit of Mt. Everest, where the barometric pressure is about 250 mm Hg, the partial pressure of O2 is about A) 0.1 mm Hg. B) 0.5 mm Hg. C) 5 mm Hg. D) 50 mm Hg. E) 100 mm Hg. TABLE 35–5 Biologically active substances metabolized by the lungs. Synthesized and used in the lungs Surfactant Synthesized or stored and released into the blood Prostaglandins Histamine Kallikrein Partially removed from the blood Prostaglandins Bradykinin Adenine nucleotides Serotonin Norepinephrine Acetylcholine Activated in the lungs Angiotensin I → angiotensin II CHAPTER 35 Pulmonary Function 607 2. The forced vital capacity is A) the amount of air that normally moves into (or out of) the lung with each respiration. B) the amount of air that enters the lung but does not partici-pate in gas exchange. C) the largest amount of air expired after maximal expiratory effort. D) the largest amount of gas that can be moved into and out of the lungs in 1 min. 3. The tidal volume is A) the amount of air that normally moves into (or out of) the lung with each respiration. B) the amount of air that enters the lung but does not partici-pate in gas exchange. C) the largest amount of air expired after maximal expiratory effort. D) the largest amount of gas that can be moved into and out of the lungs in 1 min. 4. Which of the following is responsible for the movement of O2 from the alveoli into the blood in the pulmonary capillaries? A) active transport B) filtration C) secondary active transport D) facilitated diffusion E) passive diffusion 5. Which of the following causes relaxation of bronchial smooth muscle? A) leukotrienes B) vasoactive intestinal polypeptide C) acetylcholine D) cool air E) sulfur dioxide 6. Airway resistance A) is increased if the lungs are removed and inflated with saline. B) does not affect the work of breathing. C) is increased in paraplegic patients. D) is increased in asthma. E) makes up 80% of the work of breathing. 7. Surfactant lining the alveoli A) helps prevent alveolar collapse. B) is produced in alveolar type I cells and secreted into the alveolus. C) is increased in the lungs of heavy smokers. D) is a glycolipid complex. CHAPTER RESOURCES Barnes PJ: Chronic obstructive pulmonary disease. N Engl J Med 2000;343:269. Budhiraja R, Tudor RM, Hassoun PM: Endothelial dysfunction in pulmonary hypertension. Circulation 2004;88:159. Crystal RG, West JB (editors): The Lung: Scientific Foundations, 2nd ed. Raven Press, 1997. Fishman AP, et al (editors): Fishman’s Pulmonary Diseases and Disorders, 4th ed. McGraw-Hill, 2008. Levitzky MG: Pulmonary Physiology, 7th ed. McGraw-Hill, 2007. Prisk GK, Paiva M, West JB (editors): Gravity and the Lung: Lessons from Micrography. Marcel Dekker, 2001. West JB: Pulmonary Pathophysiology, 5th ed. McGraw-Hill, 1995. Wright JR: Immunoregulatory functions of surfactant proteins. Nat Rev Immunol 2005;5:58. This page intentionally left blank 609 C H A P T E R 36 Gas Transport & pH in the Lung O B J E C T I V E S After studying this chapter, you should be able to: ■Describe the manner in which O2 flows “downhill” from the lungs to the tissues and CO2 flows “downhill” from the tissues to the lungs. ■Describe the reactions of O2 with hemoglobin and the oxygen–hemoglobin disso-ciation curve. ■List the important factors affecting the affinity of hemoglobin for O2 and the phys-iologic significance of each. ■List the reactions that increase the amount of CO2 in the blood, and draw the CO2 dissociation curve for arterial and venous blood. ■List the principal buffers in blood and, using the Henderson–Hasselbalch equation, describe what is unique about the bicarbonate buffer system. ■Define alkalosis and acidosis and outline respiratory and renal compensatory mechanisms in response to alkalosis and acidosis. ■Define hypoxia and describe its four principal forms. ■List and explain the effects of carbon monoxide on the body. ■Describe the effects of hypercapnia and hypocapnia, and give examples of condi-tions that can cause them. INTRODUCTION The partial pressure gradients for O2 and CO2, plotted in graphic form in Figure 36–1, emphasize that they are the key to gas movement and that O2 “flows downhill” from the air through the alveoli and blood into the tissues, whereas CO2 “flows downhill” from the tissues to the alveoli. However, the amount of both of these gases transported to and from the tis-sues would be grossly inadequate if it were not that about 99% of the O2 that dissolves in the blood combines with the O2-carrying protein hemoglobin and that about 94.5% of the CO2 that dissolves enters into a series of reversible chemical reac-tions that convert it into other compounds. Thus, the pres-ence of hemoglobin increases the O2-carrying capacity of the blood 70-fold, and the reactions of CO2 increase the blood CO2 content 17-fold. In this chapter, physiologic details that underlie O2 and CO2 movement under various conditions are discussed. OXYGEN TRANSPORT OXYGEN DELIVERY TO THE TISSUES The O2 delivery system in the body consists of the lungs and the cardiovascular system. O2 delivery to a particular tissue depends on the amount of O2 entering the lungs, the adequa-cy of pulmonary gas exchange, the blood flow to the tissue, and the capacity of the blood to carry O2. The blood flow de-pends on the degree of constriction of the vascular bed in the tissue and the cardiac output. The amount of O2 in the blood is determined by the amount of dissolved O2, the amount of 610 SECTION VII Respiratory Physiology hemoglobin in the blood, and the affinity of the hemoglobin for O2. REACTION OF HEMOGLOBIN & OXYGEN The dynamics of the reaction of hemoglobin with O2 make it a particularly suitable O2 carrier. Hemoglobin is a protein made up of four subunits, each of which contains a heme moi-ety attached to a polypeptide chain. In normal adults, most of the hemoglobin molecules contain two α and two β chains. Heme (see Figure 32–7) is a porphyrin ring complex that in-cludes one atom of ferrous iron. Each of the four iron atoms in hemoglobin can reversibly bind one O2 molecule. The iron stays in the ferrous state, so that the reaction is oxygenation, not oxidation. It has been customary to write the reaction of hemoglobin with O2 as Hb + O2 ← → HbO2. Because it contains four deoxyhemoglobin (Hb) units, the hemoglobin molecule can also be represented as Hb4, and it actually reacts with four molecules of O2 to form Hb4O8. Hb4 + O2 ← → Hb4O2 Hb4O2 + O2 ← → Hb4O4 Hb4O4 + O2 ← → Hb4O6 Hb4O6 + O2 ← → Hb4O8 The reaction is rapid, requiring less than 0.01 s. The deoxy-genation (reduction) of Hb4O8 is also very rapid. The quaternary structure of hemoglobin determines its affinity for O2. In deoxyhemoglobin, the globin units are tightly bound in a tense (T) configuration, which reduces the affinity of the molecule for O2. When O2 is first bound, the bonds holding the globin units are released, producing a relaxed (R) configuration, which exposes more O2 binding sites. The net result is a 500-fold increase in O2 affinity. In tis-sue, these reactions are reversed, releasing O2. The transition from one state to another has been calculated to occur about 108 times in the life of a red blood cell. The oxygen–hemoglobin dissociation curve relates per-centage saturation of the O2 carrying power of hemoglobin to the PO2 (Figure 36–2). This curve has a characteristic sigmoid shape due to the T–R interconversion. Combination of the first heme in the Hb molecule with O2 increases the affinity of the second heme for O2, and oxygenation of the second increases the affinity of the third, and so on, so that the affinity of Hb for the fourth O2 molecule is many times that for the first. When blood is equilibrated with 100% O2 (PO2 = 760 mm Hg), the normal hemoglobin becomes 100% saturated. When fully saturated, each gram of normal hemoglobin contains 1.39 mL of O2. However, blood normally contains small amounts of inactive hemoglobin derivatives, and the mea-sured value in vivo is lower. The traditional figure is 1.34 mL of O2. The hemoglobin concentration in normal blood is about 15 g/dL (14 g/dL in women and 16 g/dL in men). Therefore, 1 dL of blood contains 20.1 mL (1.34 mL × 15) of O2 bound to hemoglobin when the hemoglobin is 100% satu-rated. The amount of dissolved O2 is a linear function of the PO2 (0.003 mL/dL blood/mm Hg PO2). In vivo, the hemoglobin in the blood at the ends of the pul-monary capillaries is about 97.5% saturated with O2 (PO2 = 97 mm Hg). Because of a slight admixture with venous blood that bypasses the pulmonary capillaries (physiologic shunt), the hemoglobin in systemic arterial blood is only 97% satu-rated. The arterial blood therefore contains a total of about 19.8 mL of O2 per dL: 0.29 mL in solution and 19.5 mL bound to hemoglobin. In venous blood at rest, the hemoglobin is 75% saturated and the total O2 content is about 15.2 mL/dL: 0.12 mL in solution and 15.1 mL bound to hemoglobin. Thus, at rest the tissues remove about 4.6 mL of O2 from each decil-iter of blood passing through them (Table 36–1); 0.17 mL of this total represents O2 that was in solution in the blood, and the remainder represents O2 that was liberated from hemo-globin. In this way, 250 mL of O2 per minute is transported from the blood to the tissues at rest. FIGURE 36–1 PO2 and PCO2 values in air, lungs, blood, and tissues. Note that both O2 and CO2 diffuse “downhill” along gradients of decreasing partial pressure. (Redrawn and reproduced with permission from Kinney JM: Transport of carbon dioxide in blood. Anesthesiology 1960;21:615.) 150 120 90 60 30 0 Air Lungs Blood Tissues (Arterial) Partial pressure (mm Hg) (Est) (Est) (Venous) PCO2 PO2 FIGURE 36–2 Oxygen–hemoglobin dissociation curve. pH 7.40, temperature 38 °C. Inset table notes the percentage of saturated hemoglobin to PO2 and dissolved O2. (Redrawn and reproduced with permission from Comroe JH Jr., et al: The Lung: Clinical Physiology and Pulmonary Function Tests, 2nd ed. Year Book, 1962.) 100 Percentage O2 saturation of hemoglobin 90 80 70 60 50 40 30 20 10 0 10 20 30 40 50 60 70 80 90 100 110 PO2 (mm Hg) PO2 (mm Hg) % Sat of Hb Dissolved O2 (mL/dL) 10 20 30 40 50 60 70 80 90 100 13.5 35 57 75 83.5 89 92.7 94.5 96.5 97.5 0.03 0.06 0.09 0.12 0.15 0.18 0.21 0.24 0.27 0.30 CHAPTER 36 Gas Transport & pH in the Lung 611 FACTORS AFFECTING THE AFFINITY OF HEMOGLOBIN FOR OXYGEN Three important conditions affect the oxygen–hemoglobin dis-sociation curve: the pH, the temperature, and the concentra-tion of 2,3-biphosphoglycerate (BPG; 2,3-BPG). A rise in temperature or a fall in pH shifts the curve to the right (Figure 36–3). When the curve is shifted in this direction, a higher PO2 is required for hemoglobin to bind a given amount of O2. Con-versely, a fall in temperature or a rise in pH shifts the curve to the left, and a lower PO2 is required to bind a given amount of O2. A convenient index for comparison of such shifts is the P50, the PO2 at which hemoglobin is half saturated with O2. The higher the P50, the lower the affinity of hemoglobin for O2. The decrease in O2 affinity of hemoglobin when the pH of blood falls is called the Bohr effect and is closely related to the fact that deoxygenated hemoglobin (deoxyhemoglobin) binds H+ more actively than does oxygenated hemoglobin (oxy-hemoglobin). The pH of blood falls as its CO2 content increases, so that when the PCO2 rises, the curve shifts to the right and the P50 rises. Most of the unsaturation of hemoglobin that occurs in the tissues is secondary to the decline in the PO2, but an extra 1–2% unsaturation is due to the rise in PCO2 and consequent shift of the dissociation curve to the right. 2,3-BPG is very plentiful in red cells. It is formed from 3-phosphoglyceraldehyde, which is a product of glycolysis via the Embden–Meyerhof pathway (Figure 36–4). It is a highly charged anion that binds to the β chains of deoxyhemoglobin. One mole of deoxyhemoglobin binds 1 mol of 2,3-BPG. In effect, HbO2 + 2,3-BPG ← → Hb – 2,3-BPG + O2 In this equilibrium, an increase in the concentration of 2,3-BPG shifts the reaction to the right, causing more O2 to be liberated. Because acidosis inhibits red cell glycolysis, the 2,3-BPG concentration falls when the pH is low. Conversely, thyroid hormones, growth hormones, and androgens can all increase the concentration of 2,3-BPG and the P50. Exercise has been reported to produce an increase in 2,3-BPG within 60 min, although the rise may not occur in trained athletes. The P50 is also increased during exercise, because the temperature rises in active tissues and CO2 and metabolites accumulate, lowering the pH. In addition, much more O2 is removed from each unit of blood flowing through active tissues because the tissues’ PO2 declines. Finally, at low PO2 values, the oxygen–hemoglobin dissociation curve is steep, and large amounts of O2 are liberated per unit drop in PO2. Some clinical features of hemoglobin are discussed in Clinical Box 36–1. MYOGLOBIN Myoglobin is an iron-containing pigment found in skeletal mus-cle. It resembles hemoglobin but binds 1 rather than 4 mol of O2 per mole. Its dissociation curve is a rectangular hyperbola rather than a sigmoid curve. Because its curve is to the left of the hemo-globin curve (Figure 36–5), it takes up O2 from hemoglobin in the blood. It releases O2 only at low PO2 values, but the PO2 in ex-ercising muscle is close to zero. The myoglobin content is great-est in muscles specialized for sustained contraction. The muscle blood supply is compressed during such contractions, and myo-globin may provide O2 when blood flow is cut off. TABLE 36–1 Gas content of blood. mL/dL of Blood Containing 15 g of Hemoglobin Arterial Blood (PO2 95 mm Hg; PCO2 40 mm Hg; Hb 97% Saturated) Venous Blood (PO2 40 mm Hg; PCO2 46 mm Hg; Hb 75% Saturated) Gas Dissolved Combined Dissolved Combined O2 0.29 19.5 0.12 15.1 CO2 2.62 46.4 2.98 49.7 N2 0.98 0 0.98 0 FIGURE 36–3 Effects of temperature and pH on the oxygen–hemoglobin dissociation curve. Both changes in temperature (left) and pH (right) can alter the affinity of hemoglobin for O2. Plasma pH can be estimated using the modified Henderson–Hasselbalch equation, as shown. (Redrawn and reproduced with permission from Comroe JH Jr., et al: The Lung: Clinical Physiology and Pulmonary Function Tests, 2nd ed. Year Book, 1962.) 100 80 60 40 20 0 20 40 60 80 100 80 60 40 20 0 20 40 60 80 10° 20° 38° 43° Effect of temperature 7.6 7.2 7.4 Effect of pH pH arterial blood ≅ 7.40 pH venous blood ≅ 7.36 pH = 6.10 + log [HCO3−] 0.0301 PCO2 612 SECTION VII Respiratory Physiology CARBON DIOXIDE TRANSPORT FATE OF CARBON DIOXIDE IN BLOOD The solubility of CO2 in blood is about 20 times that of O2; therefore, considerably more CO2 than O2 is present in simple solution at equal partial pressures. The CO2 that diffuses into red blood cells is rapidly hydrated to H2CO3 because of the presence of carbonic anhydrase. The H2CO3 dissociates to H+ and HCO3 –, and the H+ is buffered, primarily by hemoglobin, while the HCO3 – enters the plasma. Some of the CO2 in the red cells reacts with the amino groups of hemoglobin and oth-er proteins (R), forming carbamino compounds: Because deoxyhemoglobin binds more H+ than oxyhemo-globin does and forms carbamino compounds more readily, binding of O2 to hemoglobin reduces its affinity for CO2 (Haldane effect). Consequently, venous blood carries more CO2 than arterial blood, CO2 uptake is facilitated in the tis-sues, and CO2 release is facilitated in the lungs. About 11% of the CO2 added to the blood in the systemic capillaries is car-ried to the lungs as carbamino-CO2. CHLORIDE SHIFT Because the rise in the HCO3 – content of red cells is much greater than that in plasma as the blood passes through the capillaries, about 70% of the HCO3 – formed in the red cells en-ters the plasma. The excess HCO3 – leaves the red cells in ex-change for Cl– (Figure 36–6). This process is mediated by anion exchanger 1 (AE1; formerly called Band 3), a major membrane protein in the red blood cell. Because of this chlo-ride shift, the Cl– content of the red cells in venous blood is significantly greater than that in arterial blood. The chloride shift occurs rapidly and is essentially complete within 1 s. Note that for each CO2 molecule added to a red cell, there is an increase of one osmotically active particle in the cell— either an HCO3 – or a Cl– in the red cell (Figure 36–6). Conse-quently, the red cells take up water and increase in size. For FIGURE 36–4 Formation and catabolism of 2,3-BPG. Note that 2,3 BPG can be associated with the Embden–Meyerhoff pathway (see Chapter 1). H2C O — — P O OH OH H+ + HC COO− O — — P O OH OH 3-Phosphoglyceraldehyde Glucose 6-PO4 1,3-Biphosphoglycerate 2,3-Biphosphoglycerate (2,3-BPG) 3-Phosphoglycerate Pyruvate 2,3-BPG mutase 2,3-BPG phosphatase Phospho-glycerate kinase CO2 + R—N ← →R—N COOH H H H CLINICAL BOX 36–1 Hemoglobin & O2 Binding In Vivo Cyanosis Reduced hemoglobin has a dark color, and a dusky bluish discoloration of the tissues, called cyanosis, appears when the reduced hemoglobin concentration of the blood in the capillaries is more than 5 g/dL. Its occurrence depends on the total amount of hemoglobin in the blood, the degree of hemoglobin unsaturation, and the state of the capillary cir-culation. Cyanosis is most easily seen in the nail beds and mucous membranes and in the earlobes, lips, and fingers, where the skin is thin. Effects of 2,3-BPG on Fetal & Stored Blood The affinity of fetal hemoglobin (hemoglobin F) for O2, which is greater than that for adult hemoglobin (hemoglo-bin A), facilitates the movement of O2 from the mother to the fetus. The cause of this greater affinity is the poor bind-ing of 2,3-BPG by the γ polypeptide chains that replace β chains in fetal hemoglobin. Some abnormal hemoglobins in adults have low P50 values, and the resulting high O2 af-finity of the hemoglobin causes enough tissue hypoxia to stimulate increased red cell formation, with resulting poly-cythemia. It is interesting to speculate that these hemoglo-bins may not bind 2,3-BPG. Red cell 2,3-BPG concentration is increased in anemia and in a variety of diseases in which there is chronic hy-poxia. This facilitates the delivery of O2 to the tissues by raising the PO2 at which O2 is released in peripheral capil-laries. In banked blood that is stored, the 2,3-BPG level falls and the ability of this blood to release O2 to the tissues is reduced. This decrease, which obviously limits the benefit of the blood if it is transfused into a hypoxic patient, is less if the blood is stored in citrate–phosphate–dextrose solu-tion rather than the usual acid–citrate–dextrose solution. CHAPTER 36 Gas Transport & pH in the Lung 613 this reason, plus the fact that a small amount of fluid in the arterial blood returns via the lymphatics rather than the veins, the hematocrit of venous blood is normally 3% greater than that of the arterial blood. In the lungs, the Cl– moves out of the cells and they shrink. SUMMARY OF CARBON DIOXIDE TRANSPORT For convenience, the various fates of CO2 in the plasma and red cells are summarized in Table 36–2. The extent to which they increase the capacity of the blood to carry CO2 is indicat-ed by the difference between the lines indicating the dissolved CO2 and the total CO2 in the dissociation curves for CO2 shown in Figure 36–7. Of the approximately 49 mL of CO2 in each deciliter of arterial blood (Table 36–1), 2.6 mL is dissolved, 2.6 mL is in carbamino compounds, and 43.8 mL is in HCO3 –. In the tis-sues, 3.7 mL of CO2 per deciliter of blood is added; 0.4 mL stays in solution, 0.8 mL forms carbamino compounds, and 2.5 mL forms HCO3 –. The pH of the blood drops from 7.40 to 7.36. In the lungs, the processes are reversed, and the 3.7 mL of CO2 is discharged into the alveoli. In this fashion, 200 mL of CO2 per minute at rest and much larger amounts during exercise are transported from the tissues to the lungs and excreted. It is worth noting that this amount of CO2 is equiva-lent in 24 hours to over 12,500 mEq of H+. ACID–BASE BALANCE & GAS TRANSPORT The major source of acids in the blood under normal condi-tions is through cellular metabolism. The CO2 formed by me-tabolism in the tissues is in large part hydrated to H2CO3, and the total H+ load from this source is over 12,500 mEq/d. How-ever, most of the CO2 is excreted in the lungs, and the small quantities of the remaining H+ are excreted by the kidneys. FIGURE 36–5 Dissociation curve of hemoglobin and myoglobin. The myoglobin binding curve (B) lacks the sigmoidal shape of the hemoglobin binding curve (A) because of the single O2 binding site in each molecule. Myoglobin also has greater affinity for O2 than he-moglobin (curve shifted left) and thus can store O2 in muscle. FIGURE 36–6 Fate of CO2 in the red blood cell. Upon entering the red blood cell, CO2 is rapidly hydrated to H2CO3 by carbonic anhy-drase. H2CO3 is in equilibrium with H+ and its conjugate base, HCO3 –. H+ can interact with deoxyhemoglobin, whereas HCO3 – can be trans-ported outside of the cell via AE1 (Band 3). In effect, for each CO2 molecule that enters the red cell, there is an additional HCO3 – or Cl– in the cell. 100 80 60 40 20 0 40 80 120 B A PO2 (mm Hg) O2 saturation (%) A = Hemoglobin B = Myoglobin CO2 CI− CO2 + H2O H2CO3 Carbonic anhydrase HHb H+ + Hb− H+ + HCO3 − TABLE 36–2 Fate of CO2 in blood. In plasma 1. Dissolved 2. Formation of carbamino compounds with plasma protein 3. Hydration, H+ buffered, HCO3 – in plasma In red blood cells 1. Dissolved 2. Formation of carbamino-Hb 3. Hydration, H+ buffered, 70% of HCO3 – enters the plasma 4. Cl– shifts into cells; mOsm in cells increases FIGURE 36–7 CO2 dissociation curves. The arterial point (a) and the venous point (v) indicate the total CO2 content found in arterial blood and venous blood of normal resting humans. Note the low amount of CO2 that is dissolved (orange trace) compared to that which can be carried by other means (Table 36–2). (Modified and reproduced with permission from Schmidt RF, Thews G [editors]: Human Physiology. Springer, 1983.) 70 60 50 40 30 20 10 0 10 20 30 40 50 60 70 5 10 15 20 25 30 v Deoxygenated blood Oxygenated blood Dissolved CO2 CO2 concentration (mL/dL) CO2 concentration (mmol/L) a PCO2 (mm Hg) 614 SECTION VII Respiratory Physiology Fruits are the main dietary source of alkali. They contain Na+ and K+ salts of weak organic acids, and the anions of these salts are metabolized to CO2, leaving NaHCO3 and KHCO3 in the body. Such ingestion contributes little to changes in pH and a more common cause of alkalosis is loss of acid from the body as a result of vomiting of gastric juice rich in HCl. This is, of course, equivalent to adding alkali to the body. BUFFERING IN THE BLOOD Acid and base shifts in the blood are largely controlled by three main buffers in blood: (1) proteins, (2) hemoglobin, and (3) the carbonic acid–bicarbonate system. Plasma proteins are effective buffers because both their free carboxyl and their free amino groups dissociate: The second buffer system is provided by the dissociation of the imidazole groups of the histidine residues in hemoglobin: In the pH 7.0–7.7 range, the free carboxyl and amino groups of hemoglobin contribute relatively little to its buffer-ing capacity. However, the hemoglobin molecule contains 38 histidine residues, and on this basis—plus the fact that hemo-globin is present in large amounts—the hemoglobin in blood has six times the buffering capacity of the plasma proteins. In addition, the action of hemoglobin is unique because the imi-dazole groups of deoxyhemoglobin (Hb) dissociate less than those of oxyhemoglobin (HbO2), making Hb a weaker acid and therefore a better buffer than HbO2. Titration curves for Hb and HbO2 are shown in Figure 36–8. The third and major buffer system in blood is the carbonic acid–bicarbonate system: The Henderson–Hasselbalch equation for this system is The pK for this system in an ideal solution is low (about 3), and the amount of H2CO3 is small and hard to measure accu-rately. However, in the body, H2CO3 is in equilibrium with CO2: If the pK is changed to pK' (apparent ionization constant; distinguished from the true pK due to less than ideal condi-tions for the solution) and [CO2] is substituted for [H2CO3], the pK' is 6.1: The clinically relevant form of this equation is: since the amount of dissolved CO2 is proportional to the par-tial pressure of CO2 and the solubility coefficient of CO2 in mmol/L/mm Hg is 0.0301. [HCO3 –] cannot be measured directly, but pH and PCO2 can be measured with suitable accuracy with pH and PCO2 glass electrodes, and [HCO3 –] can then be calculated. The pK' of this system is still low relative to the pH of the blood, but the system is one of the most effective buffer systems in the body because the amount of dissolved CO2 is controlled by respiration. Additional control of the plasma concentration of HCO3 – is provided by the kidneys. When H+ is added to the blood, HCO3 – declines as more H2CO3 is formed. If the extra H2CO3 were not converted to CO2 and H2O and the CO2 excreted in the lungs, the H2CO3 concentration would rise. When enough H+ has been added to halve the plasma HCO3 –, the pH would have dropped from 7.4 to 6.0. However, not only is all the extra H2CO3 that is formed removed, but also the H+ rise stimulates respiration and therefore produces a drop in RCOOH → ←RCOO−+ H+ [RCOO−] pH = pK´RCOOH + log [RCOOH] RNH3+ → ←RNH2 + H+ [RNH2] pH = pK´RNH3 + log [RNH3+] H H C C NH+ N H N H N + H+ ← → R R C C C H HC H2CO3 → ←H+ + HCO3− pH = pK + log [HCO3−] [H2CO3] FIGURE 36–8 Titration curves for hemoglobin. Individual titration curves for deoxygenated hemoglobin (Hb) and oxygenated hemoglobin (HbO2) are shown. The arrow from a to c indicates the number of millimoles of H that can be added without pH shift. The arrow from a to b indicates the pH shift on deoxygenation. +1.0 +0.5 −0.5 0 7.30 7.40 7.50 7.60 7.70 pH c a b Hb mmol HbO2 mmol of H+ added to 1 mmol of HbO2 or Hb mmol of H+ removed from 1 mmol of HbO2 or Hb H2CO3 → ←CO2 + H2O pH = 6.10 + log [HCO3−] [CO2] [HCO3−] pH = 6.10 + log 0.0301 PCO2 CHAPTER 36 Gas Transport & pH in the Lung 615 PCO2, so that some additional H2CO3 is removed. The pH thus falls only to 7.2 or 7.3 (Figure 36–9). There are two additional factors that make the carbonic-acid-bicarbonate system such a good biological buffer. First, the reaction CO2 + H2O ← → H2CO3 proceeds slowly in either direction unless the enzyme carbonic anhydrase is present. There is no carbonic anhydrase in plasma, but there is an abundant supply in red blood cells. Second, the presence of hemoglobin in the blood increases the buffering of the system by binding free H+ produced by the hydration of CO2 and allowing for movement of the HCO3 – into the plasma. ACIDOSIS & ALKALOSIS The pH of the arterial plasma is normally 7.40 and that of venous plasma slightly lower. A decrease in pH below the norm (acidosis) is technically present whenever the arterial pH is below 7.40 and an increase in pH (alkalosis) is techni-cally present whenever pH is above 7.40. In practice, varia-tions of up to 0.05 pH unit occur without untoward effects. Acid–base disorders are split into four categories: respira-tory acidosis, respiratory alkalosis, metabolic acidosis, and metabolic alkalosis. In addition, these disorders can occur in combination. Some examples of acid–base disturbances are shown in Table 36–3. RESPIRATORY ACIDOSIS Any short-term rise in arterial PCO2 (ie, above 40 mm Hg) due to decreased ventilation results in respiratory acidosis. The CO2 that is retained is in equilibrium with H2CO3, which in turn is in equilibrium with HCO3 –, so that the plasma HCO3 – rises and a new equilibrium is reached at a lower pH. This can be indicated graphically on a plot of plasma HCO3 – concen-tration versus pH (Figure 36–10). The pH change observed at any increase in PCO2 during respiratory acidosis is dependent on the buffering capacity of the blood. The initial changes shown in Figure 36–10 are those that occur independently of any compensatory mechanism; that is, they are those of un-compensated respiratory acidosis. RESPIRATORY ALKALOSIS Any short-term decrease in ventilation that lowers PCO2 be-low what is needed for proper CO2 exchange (ie, below 35 mm Hg) results in respiratory alkalosis. The decreased CO2 shifts the equilibrium of the carbonic acid–bicarbonate system to ef-fectively lower the [H+] and increase the pH. As in respiratory acidosis, initial pH changes corresponding to respiratory alka-losis (Figure 36–10) are those that occur independently of any compensatory mechanism and are thus uncompensated res-piratory alkalosis. METABOLIC ACIDOSIS & ALKALOSIS Blood pH changes can also arise by nonrespiratory mechanism. Metabolic acidosis (or nonrespiratory acidosis) occurs when strong acids are added to blood. If, for example, a large amount of acid is ingested (eg, aspirin overdose), acids in the blood are FIGURE 36–9 Buffering by the H2CO3–HCO3 – system in blood. The bars are drawn as if buffering occurred in separate steps over time (left to right) in order to show the effect of the initial reaction, the reduction of H2CO3 to its previous value, and its further reduction by the increase in ventilation. In this case, [H2CO3] is actually the concentra-tion of dissolved CO2, so that the mEq/L values for it are arbitrary. 25 20 15 10 5 0 5 10 15 meq/L [HCO3 −] meq/L [H2CO3] [HCO3 −] [H2CO3] ratio 20 0.9 10 16 pH 7.4 6.0 7.1 7.3 Acid added TABLE 36–3 Plasma pH, HCO3 –, and PCO2 values in various typical disturbances of acid–base balance.a Arterial Plasma Condition pH HCO3 – (mEq/L) PCO2 (mm Hg) Cause Normal 7.40 24.1 40 Metabolic acidosis 7.28 18.1 40 NH4 Cl ingestion 6.96 5.0 23 Diabetic acidosis Metabolic alkalosis 7.50 30.1 40 NaHCO3 – ingestion 7.56 49.8 58 Prolonged vomiting Respiratory acidosis 7.34 25.0 48 Breathing 7% CO2 7.34 33.5 64 Emphysema Respiratory alkalosis 7.53 22.0 27 Voluntary hyper-ventilation 7.48 18.7 26 Three-week resi-dence at 4000-m altitude aIn the diabetic acidosis and prolonged vomiting examples, respiratory compensation for primary metabolic acidosis and alkalosis has occurred, and the Pco2 has shifted from 40 mm Hg. In the emphysema and high-altitude examples, renal compensation for primary respiratory acidosis and alkalosis has occurred and has made the devia-tions from normal of the plasma HCO3 – larger than they would otherwise be. 616 SECTION VII Respiratory Physiology quickly increased, lowering the available Hb–, Prot–, and HCO3 – buffers. The H2CO3 that is formed is converted to H2O and CO2, and the CO2 is rapidly excreted via the lungs. This is the situation in uncompensated metabolic acidosis (Figure 36–10). Note that in contrast to respiratory acidosis, PCO2 is unchanged and the shift toward metabolic acidosis occurs along the isobar line (Figure 36–11). When the free [H+] level falls as a result of addition of alkali, or more commonly, the re-moval of large amounts of acid (eg, following vomiting), meta-bolic alkalosis results. In uncompensated metabolic alkalosis the pH rises along the isobar line (Figures 36–10 and 36–11). RESPIRATORY & RENAL COMPENSATION Uncompensated acidosis and alkalosis as described above are seldom seen because of compensation systems. The two main compensatory systems are respiratory compensation and renal compensation. The respiratory system compensates for metabolic acidosis or alkalosis by altering ventilation, and consequently, the PCO2, which can directly change blood pH. Respiratory mechanisms tend to be fast. In response to metabolic acidosis, ventilation is increased, resulting in a decrease of PCO2 (eg, from 40 mm Hg to 20 mm Hg) and a subsequent increase in pH toward normal (Figure 36–11). In response to metabolic alkalosis, ventilation is decreased, PCO2 is increased, and a subsequent decrease in pH occurs. Because respiratory compensation is a quick response, the graphical representation in Figure 36–11 over-states the two-step adjustment in blood pH. In actuality, as soon as metabolic acidosis begins, respiratory compensation is invoked and pH is kept from the large shifts depicted. For complete compensation from respiratory or metabolic acidosis/alkalosis, renal compensatory mechanisms are invoked. The kidney responds to acidosis by actively secreting fixed acids while retaining filtered HCO3 –. In contrast, the kidney responds to alkalosis by decreasing H+ secretion and by decreasing the retention of filtered HCO3 –. Renal tubule cells in the kidney have active carbonic anhy-drase and thus can produce H+ and HCO3 – from CO2. In response to acidosis, these cells secrete H+ into the tubular fluid in exchange for Na+ while the HCO3 – is actively reabsorbed into the peritubular capillary; for each H+ secreted, one Na+ and one HCO3 – are added to the blood. The result of this renal compen-sation for respiratory acidosis is shown graphically in the shift from acute to chronic respiratory acidosis in Figure 36–10. Conversely, in response to alkalosis, the kidney decreases H+ secretion and depresses HCO3 – reabsorption. The kidney tends to reabsorb HCO3 – until the level in plasma exceeds 26–28 mEq/L (normal is 24 mEq/L). Above this threshold, HCO3 – appears in the urine. The result of this renal compensation for respiratory alkalosis is shown graphically in the shift from acute to chronic respiratory alkalosis in Figure 36–10. Clinical evalu-ations of acid–base status are discussed in Clinical Box 36–2. HYPOXIA Hypoxia is O2 deficiency at the tissue level. It is a more cor-rect term than anoxia, with there rarely being no O2 at all left in the tissues. FIGURE 36–10 Acid–base nomogram. Changes in the PCO2 (curved lines), plasma HCO3 –, and pH (or [H+]) of arterial blood in res-piratory and metabolic acidosis are shown. Note the shifts in HCO3 – and pH as acute respiratory acidosis and alkalosis are compensated, producing their chronic counterparts. (Reproduced with permission from Cogan MG, Rector FC Jr.: Acid–base disorders. In: The Kidney, 4th ed. Brenner BM, Rector FC Jr. [editors]. Saunders, 1991.) 60 56 52 48 44 40 36 32 Arterial plasma [HCO3−] (meq/L) 28 24 20 16 12 8 4 0 7.00 7.10 7.20 7.30 7.40 7.50 7.60 7.70 7.80 Arterial blood pH 100 90 80 70 60 50 40 35 30 25 20 Arterial blood [H+] (nmol/L) 120 100 90 80 70 60 50 40 20 15 10 35 30 25 Acute respiratory alkalosis Acute respiratory acidosis Normal Chronic respiratory alkalosis PCO2 (mm Hg) Metabolic acidosis Chronic respiratory acidosis Meta-bolic alkalosis FIGURE 36–11 Acid–base paths during metabolic acidosis. Changes in true plasma pH, HCO3 –, and PCO2 at rest, during metabolic acidosis and alkalosis, and following respiratory compensation are plotted. Metabolic acidosis or alkalosis causes changes in pH along the PCO2 isobar line. Respiratory compensation moves pH towards normal by altering PCO2. (This is called a Davenport diagram and is based on Davenport HW: The ABC of Acid–Base Chemistry, 6th ed. University of Chicago Press, 1974.) 34 32 30 28 26 24 22 20 18 16 14 12 10 7.2 7.3 7.4 7.5 7.6 pH Compensated metabolic alkalosis, PCO2 48 mm Hg Compensated metabolic acidosis, PCO2 21 mm Hg PCO2 40 mm Hg Uncompensated metabolic alkalosis, PCO2 40 mm Hg Uncompensated metabolic acidosis, PCO2 40 mm Hg Normal Plasma HCO3−(meq/L) CHAPTER 36 Gas Transport & pH in the Lung 617 Traditionally, hypoxia has been divided into four types. Numerous other classifications have been used, but the four-type system still has considerable utility if the definitions of the terms are kept clearly in mind. The four categories are (1) hypoxic hypoxia, in which the PO2 of the arterial blood is reduced; (2) anemic hypoxia, in which the arterial PO2 is nor-mal but the amount of hemoglobin available to carry O2 is reduced; (3) stagnant or ischemic hypoxia, in which the blood flow to a tissue is so low that adequate O2 is not deliv-ered to it despite a normal PO2 and hemoglobin concentra-tion; and (4) histotoxic hypoxia, in which the amount of O2 delivered to a tissue is adequate but, because of the action of a toxic agent, the tissue cells cannot make use of the O2 sup-plied to them. Some specific effects of hypoxia on cells and tissues are discussed in Clinical Box 36–3. HYPOXIC HYPOXIA By definition, hypoxic hypoxia is a condition of reduced arte-rial PO2. Hypoxic hypoxia is a problem in normal individuals at high altitudes and is a complication of pneumonia and a va-riety of other diseases of the respiratory system. EFFECTS OF DECREASED BAROMETRIC PRESSURE The composition of air stays the same, but the total baromet-ric pressure falls with increasing altitude (Figure 36–12). Therefore, the PO2 also falls. At 3000 m (approximately 10,000 ft) above sea level, the alveolar PO2 is about 60 mm Hg and there is enough hypoxic stimulation of the chemoreceptors CLINICAL BOX 36–2 Clinical Evaluation of Acid–Base Status In evaluating disturbances of acid–base balance, it is im-portant to know the pH and HCO3 – content of arterial plasma. Reliable pH determinations can be made with a pH meter and a glass pH electrode. Using pH and a direct mea-surement of the PCO2 with a CO2 electrode, HCO3 – concen-tration can be calculated. The PCO2 is 7 to 8 mm Hg higher and the pH 0.03 to 0.04 unit lower in venous than arterial plasma because venous blood contains the CO2 being car-ried from the tissues to the lungs. Therefore, the calculated HCO3 – concentration is about 2 mmol/L higher. However, if this is kept in mind, free-flowing venous blood can be sub-stituted for arterial blood in most clinical situations. A measurement that is of some value in the differential diagnosis of metabolic acidosis is the anion gap. This gap, which is something of a misnomer, refers to the difference between the concentration of cations other than Na+ and the concentration of anions other than Cl– and HCO3 – in the plasma. It consists for the most part of proteins in the anionic form, HPO4 2–, SO4 2–, and organic acids, and a nor-mal value is about 12 mEq/L. It is increased when the plasma concentration of K+, Ca2+, or Mg+ is decreased; when the concentration of or the charge on plasma pro-teins is increased; or when organic anions such as lactate or foreign anions accumulate in blood. It is decreased when cations are increased or when plasma albumin is de-creased. The anion gap is increased in metabolic acidosis due to ketoacidosis, lactic acidosis, and other forms of aci-dosis in which organic anions are increased. CLINICAL BOX 36–3 Effects of Hypoxia on Cells and Selected Tissues Effects on Cells Hypoxia causes the production of transcription factors (hypoxia-inducible factors; HIFs). These are made up of α and β subunits. In normally oxygenated tissues, the α sub-units are rapidly ubiquitinated and destroyed. However, in hypoxic cells, the α subunits dimerize with β subunits, and the dimers activate genes that produce angiogenic factors and erythropoietin. Effects on the Brain In hypoxic hypoxia and the other generalized forms of hy-poxia, the brain is affected first. A sudden drop in the in-spired PO2 to less than 20 mm Hg, which occurs, for exam-ple, when cabin pressure is suddenly lost in a plane flying above 16,000 m, causes loss of consciousness in 10 to 20 s and death in 4 to 5 min. Less severe hypoxia causes a variety of mental aberrations not unlike those produced by alcohol: impaired judgment, drowsiness, dulled pain sensibility, ex-citement, disorientation, loss of time sense, and headache. Other symptoms include anorexia, nausea, vomiting, tachy-cardia, and, when the hypoxia is severe, hypertension. The rate of ventilation is increased in proportion to the severity of the hypoxia of the carotid chemoreceptor cells. Respiratory Stimulation Dyspnea is by definition difficult or labored breathing in which the subject is conscious of shortness of breath; hyperpnea is the general term for an increase in the rate or depth of breathing regardless of the patient’s subjective sensations. Tachypnea is rapid, shallow breathing. In gen-eral, a normal individual is not conscious of respiration until ventilation is doubled, and breathing is not uncomfortable until ventilation is tripled or quadrupled. Whether or not a given level of ventilation is uncomfortable also appears to depend on a variety of other factors. Hypercapnia and, to a lesser extent, hypoxia cause dyspnea. An additional factor is the effort involved in moving the air in and out of the lungs (the work of breathing). 618 SECTION VII Respiratory Physiology to definitely increase ventilation. As one ascends higher, the alveolar PO2 falls less rapidly and the alveolar PCO2 declines somewhat because of the hyperventilation. The resulting fall in arterial PCO2 produces respiratory alkalosis. HYPOXIC SYMPTOMS BREATHING AIR A number of compensatory mechanisms operate over a period of time to increase altitude tolerance (acclimatization), but in unacclimatized subjects, mental symptoms such as irritability appear at about 3700 m. At 5500 m, the hypoxic symptoms are severe; and at altitudes above 6100 m (20,000 ft), conscious-ness is usually lost. HYPOXIC SYMPTOMS BREATHING OXYGEN The total atmospheric pressure becomes the limiting factor in altitude tolerance when breathing 100% O2. The partial pressure of water vapor in the alveolar air is constant at 47 mm Hg, and that of CO2 is normally 40 mm Hg, so that the lowest barometric pressure at which a normal alveolar PO2 of 100 mm Hg is possible is 187 mm Hg, the pressure at about 10,400 m (34,000 ft). At greater altitudes, the increased ventilation due to the decline in alveolar PO2 lowers the alveolar PCO2 somewhat, but the maximum alveolar PO2 that can be attained when breathing 100% O2 at the ambient barometric pressure of 100 mm Hg at 13,700 m is about 40 mm Hg. At about 14,000 m, consciousness is lost in spite of the administration of 100% O2. At 19,200 m, the barometric pressure is 47 mm Hg, and at or below this pressure the body fluids boil at body temperature. The point is largely academic, however, because any individual exposed to such a low pres-sure would be dead of hypoxia before the bubbles of steam could cause death. Of course, an artificial atmosphere can be created around an individual; in a pressurized suit or cabin supplied with O2 and a system to remove CO2, it is possible to ascend to any altitude and to live in the vacuum of interplanetary space. Some delayed effects of high altitude are discussed in Clinical Box 36–4. ACCLIMATIZATION Acclimatization to altitude is due to the operation of a variety of compensatory mechanisms. The respiratory alkalosis pro-duced by the hyperventilation shifts the oxygen–hemoglobin dissociation curve to the left, but a concomitant increase in red blood cell 2,3-BPG tends to decrease the O2 affinity of hemo-globin. The net effect is a small increase in P50. The decrease in O2 affinity makes more O2 available to the tissues. Howev-er, the value of the increase in P50 is limited because when the FIGURE 36–12 Composition of alveolar air in individuals breathing air (0–6100 m) and 100% O2 (6100–13,700 m). The minimal alveolar PO2 that an unacclimatized subject can tolerate without loss of consciousness is about 35–40 mm Hg. Note that with increasing altitude, the alveolar PCO2 drops because of the hyperventilation due to hypoxic stimulation of the carotid and aortic chemoreceptors. The fall in barometric pressure with increasing altitude is not linear, because air is compressible. 760 720 680 640 600 320 280 240 200 160 120 80 40 0 3000 6000 9000 12,000 15,000 18,000 21,000 0 Altitude (m) N2 O2 CO2 H2O Breathing air Breathing 100% O2 Life impossible without pressurization Highest permanent human habitations (5500 m) Loss of consciousness if unacclimatized breathing air Loss of consciousness breathing 100% O2 (19,200 m) Top of Mt. Everest (8854 m) Alveolar PO2 100 mm Hg (10,400 m) Alveolar PO2 40 mm Hg (13,700 m) Body fluids boil at 37° C Pressure (mm Hg) CHAPTER 36 Gas Transport & pH in the Lung 619 arterial PO2 is markedly reduced, the decreased O2 affinity also interferes with O2 uptake by hemoglobin in the lungs. The initial ventilatory response to increased altitude is rela-tively small, because the alkalosis tends to counteract the stimulating effect of hypoxia. However, ventilation steadily increases over the next 4 d (Figure 36–13) because the active transport of H+ into cerebrospinal fluid (CSF), or possibly a developing lactic acidosis in the brain, causes a fall in CSF pH that increases the response to hypoxia. After 4 d, the ventila-tory response begins to decline slowly, but it takes years of res-idence at higher altitudes for it to decline to the initial level. Associated with this decline is a gradual desensitization to the stimulatory effects of hypoxia. Erythropoietin secretion increases promptly on ascent to high altitude and then falls somewhat over the following 4 d as the ventilatory response increases and the arterial PO2 rises. The increase in circulating red blood cells triggered by the erythropoietin begins in 2 to 3 d and is sustained as long as the individual remains at high altitude. Compensatory changes also occur in the tissues. The mito-chondria, which are the site of oxidative reactions, increase in number, and myoglobin increases, which facilitates the move-ment of O2 into the tissues. The tissue content of cytochrome oxidase also increases. The effectiveness of the acclimatization process is indicated by the fact that permanent human habitations exist in the Andes and Himalayas at elevations above 5500 m (18,000 ft). The natives who live in these villages are barrel-chested and markedly polycythemic. They have low alveolar PO2 values, but in most other ways they are remarkably normal. DISEASES CAUSING HYPOXIC HYPOXIA Hypoxic hypoxia is the most common form of hypoxia seen clin-ically. The diseases that cause it can be roughly divided into those in which the gas exchange apparatus fails, those such as congen-ital heart disease in which large amounts of blood are shunted from the venous to the arterial side of the circulation, and those in which the respiratory pump fails. Lung failure occurs when conditions such as pulmonary fibrosis produce alveolar– capillary block, or there is ventilation–perfusion imbalance. Pump failure can be due to fatigue of the respiratory muscles in conditions in which the work of breathing is increased or to a CLINICAL BOX 36–4 Delayed Effects of High Altitude When they first arrive at a high altitude, many individuals de-velop transient “mountain sickness.” This syndrome devel-ops 8 to 24 h after arrival at altitude and lasts 4 to 8 d. It is characterized by headache, irritability, insomnia, breathless-ness, and nausea and vomiting. Its cause is unsettled, but it appears to be associated with cerebral edema. The low PO2 at high altitude causes arteriolar dilation, and if cerebral au-toregulation does not compensate, there is an increase in capillary pressure that favors increased transudation of fluid into brain tissue. Individuals who do not develop mountain sickness have a diuresis at high altitude, and urine volume is decreased in individuals who develop the condition. High-altitude illness includes not only mountain sickness but also two more serious syndromes that complicate it: high-altitude cerebral edema and high-altitude pulmo-nary edema. In high-altitude cerebral edema, the capillary leakage in mountain sickness progresses to frank brain swell-ing, with ataxia, disorientation, and in some cases coma and death due to herniation of the brain through the tentorium. High-altitude pulmonary edema is a patchy edema of the lungs that is related to the marked pulmonary hypertension that develops at high altitude. It has been argued that it oc-curs because not all pulmonary arteries have enough smooth muscle to constrict in response to hypoxia, and in the capillaries supplied by those arteries, the general rise in pulmonary arterial pressure causes a capillary pressure in-crease that disrupts their walls (stress failure). All forms of high-altitude illness are benefited by descent to lower altitude and by treatment with the diuretic aceta-zolamide. This drug inhibits carbonic anhydrase, producing increased HCO3 – excretion in the urine, stimulating respira-tion, increasing PaCO2, and reducing the formation of CSF. When cerebral edema is marked, large doses of glucocorti-coids are often administered as well. Their mechanism of action is unsettled. In high-altitude pulmonary edema, prompt treatment with O2 is essential—and, if available, use of a hyperbaric chamber. Portable hyperbaric cham-bers are now available in a number of mountain areas. Ni-fedipine, a Ca2+ channel blocker that lowers pulmonary ar-tery pressure, is also useful. FIGURE 36–13 Effect of acclimatization on the ventilatory response at various altitudes. V • E V • O2 is the ventilatory equivalent, the ratio of expired minute volume (V • E) to the O2 consumption (V • O2). (Reproduced with permission from Lenfant C, Sullivan K: Adaptation to high altitude. N Engl J Med 1971;284:1298.) 50 40 30 20 0 1000 2000 3000 Altitude (m) 4000 5000 6000 4 days’ acclimatization Acute exposure • • VE/VO2, mL min−1/mL min−1 620 SECTION VII Respiratory Physiology variety of mechanical defects such as pneumothorax or bronchi-al obstruction that limit ventilation. It can also be caused by ab-normalities of the neural mechanisms that control ventilation, such as depression of the respiratory neurons in the medulla by morphine and other drugs. Some specific causes of hypoxic hy-poxia are discussed in the following text. VENTILATION–PERFUSION IMBALANCE Patchy ventilation–perfusion imbalance is by far the most com-mon cause of hypoxic hypoxia in clinical situations. In disease processes that prevent ventilation of some of the alveoli, the ventilation–blood flow ratios in different parts of the lung de-termine the extent to which systemic arterial PO2 declines. If nonventilated alveoli are perfused, the nonventilated but per-fused portion of the lung is in effect a right-to-left shunt, dump-ing unoxygenated blood into the left side of the heart. Lesser degrees of ventilation–perfusion imbalance are more common. In the example illustrated in Figure 36–14, the underventilated alveoli (B) have a low alveolar PO2, whereas the overventilated alveoli (A) have a high alveolar PO2. However, the unsaturation of the hemoglobin of the blood coming from B is not completely compensated by the greater saturation of the blood coming from A, because hemoglobin is normally nearly saturated in the lungs and the higher alveolar PO2 adds only a little more O2 to the hemoglobin than it normally carries. Consequently, the ar-terial blood is unsaturated. On the other hand, the CO2 content of the arterial blood is generally normal in such situations, since extra loss of CO2 in overventilated regions can balance dimin-ished loss in underventilated areas. VENOUS-TO-ARTERIAL SHUNTS When a cardiovascular abnormality such as an interatrial septal defect permits large amounts of unoxygenated venous blood to bypass the pulmonary capillaries and dilute the oxygenated blood in the systemic arteries (“right-to-left shunt”), chronic hypoxic hypoxia and cyanosis (cyanotic congenital heart dis-ease) result. Administration of 100% O2 raises the O2 content of alveolar air and improves the hypoxia due to hypoventilation, impaired diffusion, or ventilation–perfusion imbalance (short of perfusion of totally unventilated segments) by increasing the amount of O2 in the blood leaving the lungs. However, in pa-tients with venous-to-arterial shunts and normal lungs, any beneficial effect of 100% O2 is slight and is due solely to an in-crease in the amount of dissolved O2 in the blood. FIGURE 36–14 Comparison of ventilation/blood flow relationships in health and disease. Left: “Ideal” ventilation/blood flow rela-tionship. Right: Nonuniform ventilation and uniform blood flow, uncompensated. V • A, alveolar ventilation; MV, respiratory minute volume. (Reproduced with permission from Comroe JH Jr., et al: The Lung: Clinical Physiology and Pulmonary Function Tests, 2nd ed. Year Book, 1962.) VA = 4.0 L . IDEAL MV = 6.0 L Uniform ventilation Uniform blood flow Mixed venous blood (A + B) Arterial blood (A + B) A B VA = 4.0 L . UNCOMPENSATED MV = 6.0 L Nonuniform ventilation Uniform blood flow Mixed venous blood (A + B) Arterial blood (A + B) A B Alveolar ventilation (L/min) Pulmonary blood flow (L/min) Ventilation/blood flow ratio Mixed venous O2 saturation (%) Arterial O2 saturation (%) Mixed venous O2 tension (mm Hg) Alveolar O2 tension (mm Hg) Arterial O2 tension (mm Hg) 2.0 2.5 0.8 75.0 97.4 40.0 104.0 104.0 2.0 2.5 0.8 75.0 97.4 40.0 104.0 104.0 4.0 5.0 0.8 75.0 97.4 40.0 104.0 104.0 A B A + B Alveolar ventilation (L/min) Pulmonary blood flow (L/min) Ventilation/blood flow ratio Mixed venous O2 saturation (%) Arterial O2 saturation (%) Mixed venous O2 tension (mm Hg) Alveolar O2 tension (mm Hg) Arterial O2 tension (mm Hg) 3.2 2.5 1.3 75.0 98.2 40.0 116.0 116.0 0.8 2.5 0.3 75.0 91.7 40.0 66.0 66.0 4.0 5.0 0.8 75.0 95.0 40.0 106.0 84.0 A B A + B CHAPTER 36 Gas Transport & pH in the Lung 621 OTHER FORMS OF HYPOXIA ANEMIC HYPOXIA Hypoxia due to anemia is not severe at rest unless the hemo-globin deficiency is marked, because red blood cell 2,3-BPG increases. However, anemic patients may have considerable difficulty during exercise because of limited ability to increase O2 delivery to the active tissues (Figure 36–15). CARBON MONOXIDE POISONING Small amounts of carbon monoxide (CO) are formed in the body, and this gas may function as a chemical messenger in the brain and elsewhere. In larger amounts, it is poisonous. Outside the body, it is formed by incomplete combustion of carbon. It was used by the Greeks and Romans to execute criminals, and today it causes more deaths than any other gas. CO poisoning has become less common in the United States, since natural gas, which does not contain CO, replaced artificial gases such as coal gas, which contains large amounts. However, the exhaust of gasoline engines is 6% or more CO. CO is toxic because it reacts with hemoglobin to form car-bon monoxyhemoglobin (carboxyhemoglobin, COHb), and COHb cannot take up O2 (Figure 36–15). Carbon monoxide poisoning is often listed as a form of anemic hypoxia because the amount of hemoglobin that can carry O2 is reduced, but the total hemoglobin content of the blood is unaffected by CO. The affin-ity of hemoglobin for CO is 210 times its affinity for O2, and COHb liberates CO very slowly. An additional difficulty is that when COHb is present the dissociation curve of the remaining HbO2 shifts to the left, decreasing the amount of O2 released. This is why an anemic individual who has 50% of the normal amount of HbO2 may be able to perform moderate work, whereas an individual whose HbO2 is reduced to the same level because of the formation of COHb is seriously incapacitated. Because of the affinity of CO for hemoglobin, progressive COHb formation occurs when the alveolar PCO is greater than 0.4 mm Hg. However, the amount of COHb formed depends on the duration of exposure to CO as well as the concentration of CO in the inspired air and the alveolar ventilation. CO is also toxic to the cytochromes in the tissues, but the amount of CO required to poison the cytochromes is 1000 times the lethal dose; tissue toxicity thus plays no role in clini-cal CO poisoning. The symptoms of CO poisoning are those of any type of hypoxia, especially headache and nausea, but there is little stim-ulation of respiration, since in the arterial blood, PO2 remains normal and the carotid and aortic chemoreceptors are not stim-ulated. The cherry-red color of COHb is visible in the skin, nail beds, and mucous membranes. Death results when about 70– 80% of the circulating hemoglobin is converted to COHb. The symptoms produced by chronic exposure to sublethal concen-trations of CO are those of progressive brain damage, including mental changes and, sometimes, a parkinsonism-like state. Treatment of CO poisoning consists of immediate termination of the exposure and adequate ventilation, by artificial respiration if necessary. Ventilation with O2 is preferable to ventilation with fresh air, since O2 hastens the dissociation of COHb. Hyperbaric oxygenation (see below) is useful in this condition. HYPOPERFUSION HYPOXIA Hypoperfusion hypoxia, or stagnant hypoxia, is due to slow circulation and is a problem in organs such as the kidneys and heart during shock. The liver and possibly the brain are dam-aged by hypoperfusion hypoxia in congestive heart failure. The blood flow to the lung is normally very large, and it takes prolonged hypotension to produce significant damage. How-ever, acute respiratory distress syndrome (ARDS) can develop when there is prolonged circulatory collapse. HISTOTOXIC HYPOXIA Hypoxia due to inhibition of tissue oxidative processes is most commonly the result of cyanide poisoning. Cyanide inhibits cytochrome oxidase and possibly other enzymes. Methylene blue or nitrites are used to treat cyanide poisoning. They act by forming methemoglobin, which then reacts with cyanide to form cyanmethemoglobin, a nontoxic compound. The extent of treatment with these compounds is, of course, limited by the amount of methemoglobin that can be safely formed. Hy-perbaric oxygenation may also be useful. OXYGEN TREATMENT OF HYPOXIA Administration of oxygen-rich gas mixtures is of very limited value in hypoperfusion, anemic, and histotoxic hypoxia FIGURE 36–15 Effects of anemia and CO on hemoglobin binding of O2. Normal oxyhemoglobin (14g/dL hemoglobin) disso-ciation curve compared with anemia (7 g/dL hemoglobin) and with oxyhemoglobin dissociation curves in CO poisoning (50% carboxy-hemoglobin). Note that the CO-poisoning curve is shifted to the left of the anemia curve. (Reproduced with permission from Leff AR, Schumacker PT: Respiratory Physiology: Basics and Applications. Saunders, 1993.) 160 140 120 100 80 60 40 20 0 0 5 10 15 20 Oxygen + hemoglobin (14 g/dL) Oxygen + hemoglobin (14 g/dL) with 50% carboxyhemoglobin Oxygen + hemoglobin (7 g/dL) Oxygen partial pressure (mm Hg) Blood oxygen content (mL/dL) 622 SECTION VII Respiratory Physiology because all that can be accomplished in this way is an increase in the amount of dissolved O2 in the arterial blood. This is also true in hypoxic hypoxia when it is due to shunting of unoxy-genated venous blood past the lungs. In other forms of hy-poxic hypoxia, O2 is of great benefit. Treatment regimens that deliver less than 100% O2 are of value both acutely and chron-ically, and administration of O2 24 h/d for 2 y in this fashion has been shown to significantly decrease the mortality of chronic obstructive pulmonary disease. O2 toxicity and thera-py are discussed in Clinical Box 36–5. HYPERCAPNIA & HYPOCAPNIA HYPERCAPNIA Retention of CO2 in the body (hypercapnia) initially stimu-lates respiration. Retention of larger amounts produces symp-toms due to depression of the central nervous system: confusion, diminished sensory acuity, and, eventually, coma with respiratory depression and death. In patients with these symptoms, the PCO2 is markedly elevated, severe respiratory acidosis is present, and the plasma HCO3 – may exceed 40 mEq/L. Large amounts of HCO3 – are excreted, but more HCO3 – is reabsorbed, raising the plasma HCO3 – and partially compensating for the acidosis. CO2 is so much more soluble than O2 that hypercapnia is rarely a problem in patients with pulmonary fibrosis. How-ever, it does occur in ventilation–perfusion inequality and when for any reason alveolar ventilation is inadequate in the various forms of pump failure. It is exacerbated when CO2 production is increased. For example, in febrile patients there is a 13% increase in CO2 production for each 1°C rise in temperature, and a high carbohydrate intake increases CO2 production because of the increase in the respiratory quotient. Normally, alveolar ventilation increases and the extra CO2 is expired, but it accumulates when ventilation is compromised. CLINICAL BOX 36–5 Administration of Oxygen & Its Potential Toxicity It is interesting that while O2 is necessary for life in aerobic or-ganisms, it is also toxic. Indeed, 100% O2 has been demon-strated to exert toxic effects not only in animals but also in bac-teria, fungi, cultured animal cells, and plants. The toxicity seems to be due to the production of reactive oxygen species includ-ing superoxide anion (O2 –) and H2O2. When 80–100% O2 is ad-ministered to humans for periods of 8 h or more, the respiratory passages become irritated, causing substernal distress, nasal congestion, sore throat, and coughing. with which these symptoms develop is proportional to the pressure at which the O2 is administered; for example, at 4 atmospheres, symptoms develop in half the subjects in 30 min, whereas at 6 atmospheres, convulsions develop in a few minutes. On the other hand, exposure to 100% O2 at 2 to 3 atmos-pheres can increase dissolved O2 in arterial blood to the point that arterial O2 tension is greater than 2000 mm Hg and tis-sue O2 tension is 400 mm Hg. If exposure is limited to 5 h or less at these pressures, O2 toxicity is not a problem. There-fore, hyperbaric O2 therapy in closed tanks is used to treat diseases in which improved oxygenation of tissues cannot be achieved in other ways. It is of demonstrated value in carbon monoxide poisoning, radiation-induced tissue injury, gas gangrene, very severe blood loss anemia, diabetic leg ulcers and other wounds that are slow to heal, and rescue of skin flaps and grafts in which the circulation is marginal. It is also the primary treatment for decompression sickness and air embolism. In hypercapnic patients in severe pulmonary failure, the CO2 level may be so high that it depresses rather than stimulates respiration. Some of these patients keep breathing only be-cause the carotid and aortic chemoreceptors drive the respira-tory center. If the hypoxic drive is withdrawn by administering O2, breathing may stop. During the resultant apnea, the arterial PO2 drops but breathing may not start again, as PCO2 further depresses the respiratory center. Therefore, O2 therapy in this situation must be started with care. Some infants treated with O2 for respiratory distress syn-drome develop a chronic condition characterized by lung cysts and densities (bronchopulmonary dysplasia). This syndrome may be a manifestation of O2 toxicity. Another complication in these infants is retinopathy of prematurity (retrolental fi-broplasia), the formation of opaque vascular tissue in the eyes, which can lead to serious visual defects. The retinal receptors mature from the center to the periphery of the retina, and they use considerable O2. This causes the retina to become vascular-ized in an orderly fashion. Oxygen treatment before maturation is complete provides the needed O2 to the photoreceptors, and consequently the normal vascular pattern fails to develop. Evidence indicates that this condition can be prevented or ameliorated by treatment with vitamin E, which exerts an anti-oxidant effect, and, in animals, by growth hormone inhibitors. Administration of 100% O2 at increased pressure acceler-ates the onset of O2 toxicity, with the production not only of tracheobronchial irritation but also of muscle twitching, ring-ing in the ears, dizziness, convulsions, and coma. The speed CHAPTER 36 Gas Transport & pH in the Lung 623 HYPOCAPNIA Hypocapnia is the result of hyperventilation. During volun-tary hyperventilation, the arterial PCO2 falls from 40 to as low as 15 mm Hg while the alveolar PO2 rises to 120 to 140 mm Hg. The more chronic effects of hypocapnia are seen in neurotic patients who chronically hyperventilate. Cerebral blood flow may be reduced 30% or more because of the direct constrictor effect of hypocapnia on the cerebral vessels. The cerebral ischemia causes light-headedness, dizziness, and paresthesias. Hypocapnia also increases cardiac output. It has a direct con-strictor effect on many peripheral vessels, but it depresses the vasomotor center, so that the blood pressure is usually unchanged or only slightly elevated. Other consequences of hypocapnia are due to the associ-ated respiratory alkalosis, the blood pH being increased to 7.5 or 7.6. The plasma HCO3 – level is low, but HCO3 – reabsorp-tion is decreased because of the inhibition of renal acid secre-tion by the low PCO2. The plasma total calcium level does not change, but the plasma Ca2+ level falls and hypocapnic indi-viduals develop carpopedal spasm, a positive Chvostek sign, and other signs of tetany. CHAPTER SUMMARY ■Partial pressure differences between air and blood for O2 and CO2 dictate a net flow of O2 into the blood and CO2 out of the blood in the pulmonary system. However, this flow is greatly en-hanced by the ability for hemoglobin to bind O2 and chemical re-actions that increase CO2 in the blood (eg, carbonic anhydrase). ■The amount of O2 in the blood is determined by the amount dis-solved (minor) and the amount bound (major) to hemoglobin. Each hemoglobin molecule contains four subunits that each can bind O2. Binding of the first O2 to hemoglobin increases the affin-ity for the second O2, and this pattern is continued until four O2 are bound. Hemoglobin O2 binding is also affected by pH, tem-perature, and the concentration of 2,3-biphospholycerate (2,3-BPG). ■CO2 in blood is rapidly converted into H2CO3 due to the activity of carbonic anhydrase. CO2 also readily forms carbamino com-pounds with blood proteins (including hemoglobin). The rapid net loss of CO2 allows more CO2 to dissolve in blood. ■The pH of plasma is 7.4. A decrease in plasma pH is termed acidosis and an increase of plasma pH is termed alkalosis. Acid and base shifts in the blood are controlled by proteins, including hemoglobin, and principally by the carbonic acid-bicarbonate buffering system. The carbonic acid-bicarbonate buffering system is effective because dissolved CO2 can be controlled by respiration. ■A short-term change in arterial PCO2 due to decreased ventila-tion results in respiratory acidosis. A short-term change in arte-rial PCO2 due to increased ventilation results in respiratory alkalosis. Metabolic acidosis occurs when strong acids are added to the blood, and metabolic alkalosis occurs when strong bases are added to (or strong acids are removed from) the blood. ■Respiratory compensation to acidosis or alkalosis involves quick changes in ventilation. Such changes effectively change the PCO2 in the blood plasma. Renal compensation mechanisms are much slower and involve H+ secretion or HCO3 – reabsorption. ■Hypoxia is a deficiency of O2 at the tissue level. Hypoxia has powerful consequences at the cellular, tissue, and organ level: It can alter cellular transcription factors and thus protein expres-sion; it can quickly alter brain function and produce symptoms similar to alcohol (eg, dizziness, impaired mental function, drowsiness, headache); and it can affect ventilation. Long-term hypoxia results in cell and tissue death. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Most of the CO2 transported in the blood is A) dissolved in plasma. B) in carbamino compounds formed from plasma proteins. C) in carbamino compounds formed from hemoglobin. D) bound to Cl–. E) in HCO3 –. 2. Which of the following has the greatest effect on the ability of blood to transport oxygen? A) capacity of the blood to dissolve oxygen B) amount of hemoglobin in the blood C) pH of plasma D) CO2 content of red blood cells E) temperature of the blood 3. Which of the following is not true of the system? CO2 + H2O ← →1 H2CO3 ← →2 H+ + HCO3 – A) Reaction 1 is catalyzed by carbonic anhydrase. B) Because of reaction 2, the pH of blood declines during breath holding. C) Reaction 1 occurs in the kidneys. D) Reaction 1 occurs primarily in plasma. E) The reactions move to the left when there is excess H+ in the tissues. 4. Uncompensated respiratory acidosis differs from uncompen-sated metabolic acidosis in that A) plasma pH change is always greater in uncompensated res-piratory acidosis compared to uncompensated metabolic acidosis. B) there are no compensation mechanisms for respiratory acidosis, whereas there is respiratory compensation for metabolic acidosis. C) uncompensated respiratory acidosis involves changes in plasma [HCO3 –], whereas plasma [HCO3 –] is unchanged in uncompensated metabolic acidosis. D) uncompensated respiratory acidosis is associated with a change in PCO2, whereas uncompensated metabolic acidosis occurs along the isobar line for PCO2. 5. O2 delivery to the tissues would be reduced to the greatest extent in A) a normal subject breathing 100% O2 on top of Mt. Everest. B) a normal subject running a marathon at sea level. C) a patient with carbon monoxide poisoning. D) a patient who has ingested cyanide. E) a patient with moderately severe metabolic acidosis. 624 SECTION VII Respiratory Physiology CHAPTER RESOURCES Crystal RG, West JB (editors): The Lung: Scientific Foundations, 2nd ed. Raven Press, 1997. Fishman AP, et al (editors): Fishman’s Pulmonary Diseases and Disorders, 4th ed. McGraw-Hill, 2008. Hackett PH, Roach RC: High-altitude illness. N Engl J Med 2001;345:107. Laffey JG, Kavanagh BP: Hypocapnia. N Engl J Med 2002;347:43. Levitzky, MG: Pulmonary Physiology, 7th ed. McGraw-Hill, 2007. Prisk GK, Paiva M, West JB (editors): Gravity and the Lung: Lessons from Micrography. Marcel Dekker, 2001. Voelkel NF: High-altitude pulmonary edema. N Engl J Med 2002;346:1607. West JB: Pulmonary Pathophysiology, 5th ed. McGraw-Hill, 1995. 625 C H A P T E R 37 Regulation of Respiration O B J E C T I V E S After studying this chapter, you should be able to: ■Locate the pre-Bötzinger complex and describe its role in producing spontaneous respiration. ■Identify the location and probable functions of the dorsal and ventral groups of respi-ratory neurons, the pneumotaxic center, and the apneustic center in the brain stem. ■List the specific respiratory functions of the vagus nerves and the respiratory receptors in the carotid body, the aortic body, and the ventral surface of the medulla oblongata. ■Describe and explain the ventilatory responses to increased CO2 concentrations in the inspired air. ■Describe and explain the ventilatory responses to decreased O2 concentrations in the inspired air. ■Describe the effects of each of the main non-chemical factors that influence respiration. ■Describe the effects of exercise on ventilation and O2 exchange in the tissues. ■Define periodic breathing and explain its occurrence in various disease states. INTRODUCTION Spontaneous respiration is produced by rhythmic discharge of motor neurons that innervate the respiratory muscles. This discharge is totally dependent on nerve impulses from the brain; breathing stops if the spinal cord is transected above the origin of the phrenic nerves. The rhythmic discharges from the brain that produce spontaneous respiration are reg-ulated by alterations in arterial PO2, PCO2, and H+ concentra-tion, and this chemical control of breathing is supplemented by a number of non-chemical influences. The physiological bases for these phenomena are discussed in this chapter. NEURAL CONTROL OF BREATHING CONTROL SYSTEMS Two separate neural mechanisms regulate respiration. One is responsible for voluntary control and the other for automatic control. The voluntary system is located in the cerebral cortex and sends impulses to the respiratory motor neurons via the corticospinal tracts. The automatic system is driven by a group of pacemaker cells in the medulla. Impulses from these cells activate motor neurons in the cervical and thoracic spinal cord that innervate inspiratory muscles. Those in the cervical cord activate the diaphragm via the phrenic nerves, and those in the thoracic spinal cord activate the external intercostal muscles. However, the impulses also reach the innervation of the inter-nal intercostal muscles and other expiratory muscles. The motor neurons to the expiratory muscles are inhib-ited when those supplying the inspiratory muscles are active, and vice versa. Although spinal reflexes contribute to this reciprocal innervation, it is due primarily to activity in descending pathways. Impulses in these descending pathways excite agonists and inhibit antagonists. The one exception to 626 SECTION VII Respiratory Physiology the reciprocal inhibition is a small amount of activity in phrenic axons for a short period after inspiration. The func-tion of this post-inspiratory output appears to be to brake the lung’s elastic recoil and make respiration smooth. MEDULLARY SYSTEMS The main components of the respiratory control pattern gen-erator responsible for automatic respiration are located in the medulla. Rhythmic respiration is initiated by a small group of synaptically coupled pacemaker cells in the pre-Bötzinger com-plex (pre-BÖTC) on either side of the medulla between the nu-cleus ambiguus and the lateral reticular nucleus (Figure 37–1). These neurons discharge rhythmically, and they produce rhyth-mic discharges in phrenic motor neurons that are abolished by sections between the pre-Bötzinger complex and these motor neurons. They also contact the hypoglossal nuclei, and the tongue is involved in the regulation of airway resistance. Neurons in the pre-Bötzinger complex discharge rhythmi-cally in brain slice preparations in vitro, and if the slices become hypoxic, discharge changes to one associated with gasping. Addition of cadmium to the slices causes occa-sional sigh-like discharge patterns. There are NK1 receptors and μ-opioid receptors on these neurons, and, in vivo, sub-stance P stimulates and opioids inhibit respiration. Depres-sion of respiration is a side effect that limits the use of opioids in the treatment of pain. However, it is now known that 5HT4 receptors are present in the pre-Bötzinger complex and treat-ment with 5HT4 agonists blocks the inhibitory effect of opiates on respiration in experimental animals, without inhibiting their analgesic effect. In addition, dorsal and ventral groups of respiratory neu-rons are present in the medulla (Figure 37–2). However, lesions of these neurons do not abolish respiratory activity, and they apparently project to the pre-Bötzinger pacemaker neurons. FIGURE 37–1 Pacemaker cells in the pre-Bötzinger complex (pre-BÖTC). Top: Anatomical diagram of the pre-BÖTC from a neona-tal rat. Bottom: Sample rhythmic discharge tracing of neurons in the pre-BÖTC complex from a brain slice of a neonatal rat. IO, inferior olive; LRN, lateral reticular nucleus; NA, nucleus ambiguus; XII, nucleus of 12th cranial nerve; 5SP, spinal nucleus of trigeminal nerve. (Modified from Feldman JC, Gray PA: Sighs and gasps in a dish. Nat Neurosci 2000;3:531.) −60 mV 20 mV 5 s 5SP XII NA IO LRN Pre-BOTC •• FIGURE 37–2 Respiratory neurons in the brain stem. Dorsal view of brain stem; cerebellum removed. The effects of various lesions and brain stem transections are shown; the spirometer tracings at the right indicate the depth and rate of breathing. If a lesion is introduced at D, breathing ceases. The effects of higher transections, with and without vagus nerves transection, are shown (see text for details). DRG, dorsal group of respiratory neurons; VRG, ventral group of respiratory neurons; NPBL, nucleus parabrachialis (pneumotaxic center); 4th vent, fourth ventricle; IC, inferior colliculus; CP, middle cerebellar peduncle. The roman numerals identify cranial nerves. (Modified from Mitchell RA, Berger A: State of the art: Review of neural regulation of respiration. Am Rev Respir Dis 1975;111:206.) 4th vent NPBL CP IC VRG DRG A B C D IX X XI XII Vagi intact Vagi cut CHAPTER 37 Regulation of Respiration 627 PONTINE & VAGAL INFLUENCES Although the rhythmic discharge of medullary neurons con-cerned with respiration is spontaneous, it is modified by neu-rons in the pons and afferents in the vagus from receptors in the airways and lungs. An area known as the pneumotaxic center in the medial parabrachial and Kölliker–Fuse nuclei of the dorsolateral pons contains neurons active during inspira-tion and neurons active during expiration. When this area is damaged, respiration becomes slower and tidal volume great-er, and when the vagi are also cut in anesthetized animals, there are prolonged inspiratory spasms that resemble breath holding (apneusis; section B in Figure 37–2). The normal function of the pneumotaxic center is unknown, but it may play a role in switching between inspiration and expiration. Stretching of the lungs during inspiration initiates impulses in afferent pulmonary vagal fibers. These impulses inhibit inspiratory discharge. This is why the depth of inspiration is increased after vagotomy (Figure 37–2) and apneusis develops if the vagi are cut after damage to the pneumotaxic center. Vagal feedback activity does not alter the rate of rise of the neural activity in respiratory motor neurons (Figure 37–3). When the activity of the inspiratory neurons is increased in intact animals, the rate and the depth of breathing are increased. The depth of respiration is increased because the lungs are stretched to a greater degree before the amount of vagal and pneumotaxic center inhibitory activity is sufficient to overcome the more intense inspiratory neuron discharge. The respiratory rate is increased because the after-discharge in the vagal and possibly the pneumotaxic afferents to the medulla is rapidly overcome. REGULATION OF RESPIRATORY ACTIVITY A rise in the PCO2 or H+ concentration of arterial blood or a drop in its PO2 increases the level of respiratory neuron ac-tivity in the medulla, and changes in the opposite direction have a slight inhibitory effect. The effects of variations in blood chemistry on ventilation are mediated via respiratory chemoreceptors—the carotid and aortic bodies and collec-tions of cells in the medulla and elsewhere that are sensitive to changes in the chemistry of the blood. They initiate impulses that stimulate the respiratory center. Superimposed on this basic chemical control of respiration, other afferents provide non-chemical controls that affect breathing in particular situ-ations (Table 37–1). CHEMICAL CONTROL OF BREATHING The chemical regulatory mechanisms adjust ventilation in such a way that the alveolar PCO2 is normally held constant, the effects of excess H+ in the blood are combated, and the PO2 is raised when it falls to a potentially dangerous level. The res-piratory minute volume is proportional to the metabolic rate, but the link between metabolism and ventilation is CO2, not O2. The receptors in the carotid and aortic bodies are stimu-lated by a rise in the PCO2 or H+ concentration of arterial blood or a decline in its PO2. After denervation of the carotid chemoreceptors, the response to a drop in PO2 is abolished; the predominant effect of hypoxia after denervation of the ca-rotid bodies is a direct depression of the respiratory center. The response to changes in arterial blood H+ concentration in the pH 7.3–7.5 range is also abolished, although larger changes exert some effect. The response to changes in arterial PCO2, on the other hand, is affected only slightly; it is reduced no more than 30–35%. FIGURE 37–3 Afferent vagal fibers inhibit inspiratory discharge. Superimposed records of two breaths: (A) with and (B) without feedback vagal afferent activity from stretch receptors in the lungs. Note that the rate of rise in phrenic nerve activity to the dia-phragm is unaffected but the discharge is prolonged in the absence of vagal input. A A B B 0 1 2 Time (s) Summed phrenic efferent activity Summed vagal afferent activity TABLE 37–1 Stimuli affecting the respiratory center. Chemical control CO2 (via CSF and brain interstitial fluid H+ concentration) O2 } (via carotid and aortic bodies) H+ Non-chemical control Vagal afferents from receptors in the airways and lungs Afferents from the pons, hypothalamus, and limbic system Afferents from proprioceptors Afferents from baroreceptors: arterial, atrial, ventricular, pulmonary 628 SECTION VII Respiratory Physiology CAROTID & AORTIC BODIES There is a carotid body near the carotid bifurcation on each side, and there are usually two or more aortic bodies near the arch of the aorta (Figure 37–4). Each carotid and aortic body (glomus) contains islands of two types of cells, type I and type II cells, surrounded by fenestrated sinusoidal capillaries. The type I or glomus cells are closely associated with cuplike end-ings of the afferent nerves (Figure 37–5). The glomus cells re-semble adrenal chromaffin cells and have dense-core granules containing catecholamines that are released upon exposure to hypoxia and cyanide. The cells are excited by hypoxia, and the principal transmitter appears to be dopamine, which excites the nerve endings by way of D2 receptors. The type II cells are glia-like, and each surrounds four to six type I cells. Their function is probably sustentacular. Outside the capsule of each body, the nerve fibers acquire a myelin sheath; however, they are only 2 to 5 μm in diameter and conduct at the relatively low rate of 7 to 12 m/s. Afferents from the carotid bodies ascend to the medulla via the carotid sinus and glossopharyngeal nerves, and fibers from the aortic bodies ascend in the vagi. Studies in which one carotid body has been isolated and perfused while recordings are being taken from its afferent nerve fibers show that there is a graded increase in impulse traffic in these afferent fibers as the PO2 of the perfusing blood is lowered (Figure 37–6) or the PCO2 is raised. Type I glomus cells have O2-sensitive K+ channels, whose conductance is reduced in proportion to the degree of hypoxia to which they are exposed. This reduces the K+ efflux, depolarizing the cell and causing Ca2+ influx, primar-ily via L-type Ca2+ channels. The Ca2+ influx triggers action potentials and transmitter release, with consequent excitation FIGURE 37–4 Location of carotid and aortic bodies. Carotid bodies are positioned near a major arterial baroreceptor, the carotid si-nus. Two aortic bodies are shown near the aortic arch. Carotid body Carotid sinus Common carotid arteries Aortic bodies Aortic arch Heart FIGURE 37–5 Organization of the carotid body. Type I (glo-mus) cells contain catecholamines. When exposed to hypoxia, they re-lease their catecholamines, which stimulate the cuplike endings of the carotid sinus nerve fibers in the glossopharyngeal nerve. The glia-like type II cells surround the type I cells and probably have a sustentacular function. FIGURE 37–6 Effect of PCO2 on afferent nerve firing. The rate of discharge of a single afferent fiber from the carotid body is plotted at several PO2 (circles) and fitted to a line. A sharp increase in firing rate is observed as PO2 falls below normal resting levels (ie, near 100 mm Hg). (Courtesy of S Sampson.) Type II cell Type I (glomus) cell Glossopharyngeal afferent axons 8 6 4 2 0 100 200 400 600 Arterial PO2 (mm Hg) Impulses/s CHAPTER 37 Regulation of Respiration 629 of the afferent nerve endings. The smooth muscle of pulmo-nary arteries contains similar O2-sensitive K+ channels, which mediate the vasoconstriction caused by hypoxia. This is in contrast to systemic arteries, which contain adenosine tri-phosphate (ATP) dependent K+ channels that permit more K+ efflux with hypoxia and consequently cause vasodilation instead of vasoconstriction. The blood flow in each 2-mg carotid body is about 0.04 mL/min, or 2000 mL/100 g of tissue/min compared with a blood flow 54 mL or 420 mL per 100 g/min in the brain and kidneys, respectively. Because the blood flow per unit of tissue is so enormous, the O2 needs of the cells can be met largely by dissolved O2 alone. Therefore, the receptors are not stimu-lated in conditions such as anemia or carbon monoxide poi-soning, in which the amount of dissolved O2 in the blood reaching the receptors is generally normal, even though the combined O2 in the blood is markedly decreased. The recep-tors are stimulated when the arterial PO2 is low or when, because of vascular stasis, the amount of O2 delivered to the receptors per unit time is decreased. Powerful stimulation is also produced by cyanide, which prevents O2 utilization at the tissue level. In sufficient doses, nicotine and lobeline activate the chemoreceptors. It has also been reported that infusion of K+ increases the discharge rate in chemoreceptor afferents, and because the plasma K+ level is increased during exercise, the increase may contribute to exercise-induced hyperpnea. Because of their anatomic location, the aortic bodies have not been studied in as great detail as the carotid bodies. Their responses are probably similar but of lesser magnitude. In humans in whom both carotid bodies have been removed but the aortic bodies left intact, the responses are essentially the same as those following denervation of both carotid and aor-tic bodies in animals: little change in ventilation at rest, but the ventilatory response to hypoxia is lost and the ventilatory response to CO2 is reduced by 30%. Neuroepithelial bodies composed of innervated clusters of amine-containing cells are found in the airways. These cells have an outward K+ current that is reduced by hypoxia, and this would be expected to produce depolarization. However, the function of these hypoxia-sensitive cells is uncertain because, as noted above, removal of the carotid bodies alone abolishes the respiratory response to hypoxia. CHEMORECEPTORS IN THE BRAIN STEM The chemoreceptors that mediate the hyperventilation pro-duced by increases in arterial PCO2 after the carotid and aortic bodies are denervated are located in the medulla oblongata and consequently are called medullary chemoreceptors. They are separate from the dorsal and ventral respiratory neurons and are located on the ventral surface of the medulla (Figure 37–7). Recent evidence indicates that additional chemoreceptors are located in the vicinity of the solitary tract nuclei, the locus cer-uleus, and the hypothalamus. The chemoreceptors monitor the H+ concentration of cere-brospinal fluid (CSF), including the brain interstitial fluid. CO2 readily penetrates membranes, including the blood– brain barrier, whereas H+ and HCO3 – penetrate slowly. The CO2 that enters the brain and CSF is promptly hydrated. The H2CO3 dissociates, so that the local H+ concentration rises. The H+ concentration in brain interstitial fluid parallels the arterial PCO2. Experimentally produced changes in the PCO2 of CSF have minor, variable effects on respiration as long as the H+ concentration is held constant, but any increase in spi-nal fluid H+ concentration stimulates respiration. The magni-tude of the stimulation is proportional to the rise in H+ concentration. Thus, the effects of CO2 on respiration are mainly due to its movement into the CSF and brain interstitial fluid, where it increases the H+ concentration and stimulates receptors sensitive to H+. VENTILATORY RESPONSES TO CHANGES IN ACID–BASE BALANCE In metabolic acidosis due, for example, to the accumulation of acid ketone bodies in the circulation in diabetes mellitus, there is pronounced respiratory stimulation (Kussmaul breathing). The hyperventilation decreases alveolar PCO2 (“blows off CO2”) and thus produces a compensatory fall in blood H+ concentration. Conversely, in metabolic alkalosis due, for ex-ample, to protracted vomiting with loss of HCl from the body, ventilation is depressed and the arterial PCO2 rises, raising the H+ concentration toward normal. If there is an increase in ventilation that is not secondary to a rise in arterial H+ con-centration, the drop in PCO2 lowers the H+ concentration be-low normal (respiratory alkalosis); conversely, hypoventilation that is not secondary to a fall in plasma H+ concentration causes respiratory acidosis. FIGURE 37–7 Rostral (R) and caudal (C) chemosensitive areas on the ventral surface of the medulla. Pons C C R R VII V VIII IX X XI XII VI Pyramid 630 SECTION VII Respiratory Physiology VENTILATORY RESPONSES TO CO2 The arterial PCO2 is normally maintained at 40 mm Hg. When arterial PCO2 rises as a result of increased tissue metabolism, ventilation is stimulated and the rate of pulmonary excretion of CO2 increases until the arterial PCO2 falls to normal, shut-ting off the stimulus. The operation of this feedback mecha-nism keeps CO2 excretion and production in balance. When a gas mixture containing CO2 is inhaled, the alveolar PCO2 rises, elevating the arterial PCO2 and stimulating ventila-tion as soon as the blood that contains more CO2 reaches the medulla. CO2 elimination is increased, and the alveolar PCO2 drops toward normal. This is why relatively large increments in the PCO2 of inspired air (eg, 15 mm Hg) produce relatively slight increments in alveolar PCO2 (eg, 3 mm Hg). However, the PCO2 does not drop to normal, and a new equilibrium is reached at which the alveolar PCO2 is slightly elevated and the hyperventilation persists as long as CO2 is inhaled. The essen-tially linear relationship between respiratory minute volume and the alveolar PCO2 is shown in Figure 37–8. Of course, this linearity has an upper limit. When the PCO2 of the inspired gas is close to the alveolar PCO2, elimination of CO2 becomes difficult. When the CO2 content of the inspired gas is more than 7%, the alveolar and arterial PCO2 begin to rise abruptly in spite of hyperventilation. The resultant accumula-tion of CO2 in the body (hypercapnia) depresses the central nervous system, including the respiratory center, and produces headache, confusion, and eventually coma (CO2 narcosis). VENTILATORY RESPONSE TO OXYGEN LACK When the O2 content of the inspired air is decreased, respira-tory minute volume is increased. The stimulation is slight when the PO2 of the inspired air is more than 60 mm Hg, and marked stimulation of respiration occurs only at lower PO2 val-ues (Figure 37–9). However, any decline in arterial PO2 below 100 mm Hg produces increased discharge in the nerves from the carotid and aortic chemoreceptors. There are two reasons why this increase in impulse traffic does not increase ventila-tion to any extent in normal individuals until the PO2 is less than 60 mm Hg. Because Hb is a weaker acid than HbO2, there is a slight decrease in the H+ concentration of arterial blood when the arterial PO2 falls and hemoglobin becomes less satu-rated with O2. The fall in H+ concentration tends to inhibit res-piration. In addition, any increase in ventilation that does FIGURE 37–8 Responses of normal subjects to inhaling O2 and approximately 2, 4, and 6% CO2. The relatively linear increase in respiratory minute volume in response to increased CO2 is due to an increase in both the depth and rate of respiration. (Reproduced with permission from Lambertsen CJ in: Medical Physiology, 13th ed. Mountcastle VB [editor]. Mosby, 1974.) 32 28 24 20 16 12 8 4 38 40 42 44 46 48 50 ± 1 SE Alveolar PCO2 (mm Hg) Respiratory minute volume (L /min) FIGURE 37–9 Top: Average respiratory minute volume during the first half hour of exposure to gases containing various amounts of O2. Marked changes in ventilation occur at PO2 values lower than 60 mm Hg. The horizontal line in each case indicates the mean; the verti-cal bar indicates one standard deviation. Bottom: Alveolar PO2 and PCO2 values when breathing air at various barometric pressures. The two graphs are aligned so that the PO2 of the inspired gas mixtures in the upper graph correspond to the PO2 at the various barometric pres-sures in the lower graph. (Courtesy of RH Kellogg.) 40 30 20 10 0 120 100 80 60 40 20 0 21 160 20 152 %O2 in insp gas PO2 in insp gas 15 114 10 76 5 38 Ventilation (L /min) Pressure (mm Hg) 760 700 600 500 400 300 200 Alveolar PCO2 Alveolar PO2 Barometric pressure (mm Hg) CHAPTER 37 Regulation of Respiration 631 occur lowers the alveolar PCO2, and this also tends to inhibit respiration. Therefore, the stimulatory effects of hypoxia on ventilation are not clearly manifest until they become strong enough to override the counterbalancing inhibitory effects of a decline in arterial H+ concentration and PCO2. The effects on ventilation of decreasing the alveolar PO2 while holding the alveolar PCO2 constant are shown in Figure 37–10. When the alveolar PCO2 is stabilized at a level 2 to 3 mm Hg above normal, there is an inverse relationship between ventilation and the alveolar PO2 even in the 90 to 110 mm Hg range; but when the alveolar PCO2 is fixed at lower than normal values, there is no stimulation of ventilation by hypoxia until the alveolar PO2 falls below 60 mm Hg. EFFECTS OF HYPOXIA ON THE CO2 RESPONSE CURVE When the converse experiment is performed—that is, when the alveolar PO2 is held constant while the response to varying amounts of inspired CO2 is tested—a linear response is ob-tained (Figure 37–11). When the CO2 response is tested at dif-ferent fixed PO2 values, the slope of the response curve changes, with the slope increased when alveolar PO2 is decreased. In oth-er words, hypoxia makes the individual more sensitive to in-creases in arterial PCO2. However, the alveolar PCO2 level at which the curves in Figure 37–11 intersect is unaffected. In the normal individual, this threshold value is just below the normal alveolar PCO2, indicating that normally there is a very slight but definite “CO2 drive” of the respiratory area. EFFECT OF H+ ON THE CO2 RESPONSE The stimulatory effects of H+ and CO2 on respiration appear to be additive and not, like those of CO2 and O2, complexly in-terrelated. In metabolic acidosis, the CO2 response curves are similar to those in Figure 37–11, except that they are shifted to the left. In other words, the same amount of respiratory stim-ulation is produced by lower arterial PCO2 levels. It has been calculated that the CO2 response curve shifts 0.8 mm Hg to the left for each nanomole rise in arterial H+. About 40% of the ventilatory response to CO2 is removed if the increase in arte-rial H+ produced by CO2 is prevented. As noted above, the re-maining 60% is probably due to the effect of CO2 on spinal fluid or brain interstitial fluid H+ concentration. BREATH HOLDING Respiration can be voluntarily inhibited for some time, but eventually the voluntary control is overridden. The point at which breathing can no longer be voluntarily inhibited is called the breaking point. Breaking is due to the rise in arterial PCO2 and the fall in PO2. Individuals can hold their breath longer after removal of the carotid bodies. Breathing 100% oxygen before breath holding raises alveolar PO2 initially, so that the breaking point is delayed. The same is true of hyper-ventilating room air, because CO2 is blown off and arterial PCO2 is lower at the start. Reflex or mechanical factors appear to influence the breaking point, since subjects who hold their breath as long as possible and then breathe a gas mixture low in O2 and high in CO2 can hold their breath for an additional 20 s or more. Psychological factors also play a role, and sub-jects can hold their breath longer when they are told their per-formance is very good than when they are not. FIGURE 37–10 Ventilation at various alveolar PO2 values when PCO2 is held constant at 49, 44, or 37 mm Hg. Note the dramatic effect on the ventilatory response to PO2 when PCO is increased above normal. (Data from Loeschke HH and Gertz KH.) 0 10 20 30 40 50 60 20 40 60 80 100 120 140 Ventilation (L/min, BTPS) PAO2(mm Hg) PACO249 PACO244 PACO237 FIGURE 37–11 Fan of lines showing CO2 response curves at various fixed values of alveolar PO2. Decreased PAO2 results in a more sensitive response to PACO2. Ventilation (L/min BTPS) 50 40 25 0 50 75 100 PAO2100 PAO255 PAO240 PACO2 (mm Hg) 632 SECTION VII Respiratory Physiology NON-CHEMICAL INFLUENCES ON RESPIRATION RESPONSES MEDIATED BY RECEPTORS IN THE AIRWAYS & LUNGS Receptors in the airways and lungs are innervated by myeli-nated and unmyelinated vagal fibers. The unmyelinated fibers are C fibers. The receptors innervated by myelinated fibers are commonly divided into slowly adapting receptors and rapid-ly adapting receptors on the basis of whether sustained stim-ulation leads to prolonged or transient discharge in their afferent nerve fibers (Table 37–2). The other group of recep-tors presumably consists of the endings of C fibers, and they are divided into pulmonary and bronchial subgroups on the basis of their location. The shortening of inspiration produced by vagal afferent activity (Figure 37–3) is mediated by slowly adapting recep-tors, as are the Hering–Breuer reflexes. The Hering–Breuer inflation reflex is an increase in the duration of expiration pro-duced by steady lung inflation, and the Hering–Breuer defla-tion reflex is a decrease in the duration of expiration produced by marked deflation of the lung. Because the rapidly adapting receptors are stimulated by chemicals such as histamine, they have been called irritant receptors. Activation of rapidly adapting receptors in the trachea causes coughing, broncho-constriction, and mucus secretion, and activation of rapidly adapting receptors in the lung may produce hyperpnea. Because the C fiber endings are close to pulmonary vessels, they have been called J (juxtacapillary) receptors. They are stimulated by hyperinflation of the lung, but they respond as well to intravenous or intracardiac administration of chemi-cals such as capsaicin. The reflex response that is produced is apnea followed by rapid breathing, bradycardia, and hypoten-sion (pulmonary chemoreflex). A similar response is pro-duced by receptors in the heart (Bezold–Jarisch reflex or the coronary chemoreflex). The physiologic role of this reflex is uncertain, but it probably occurs in pathologic states such as pulmonary congestion or embolization, in which it is pro-duced by endogenously released substances. COUGHING & SNEEZING Coughing begins with a deep inspiration followed by forced expiration against a closed glottis. This increases the intra-pleural pressure to 100 mm Hg or more. The glottis is then suddenly opened, producing an explosive outflow of air at ve-locities up to 965 km (600 mi) per hour. Sneezing is a similar expiratory effort with a continuously open glottis. These re-flexes help expel irritants and keep airways clear. Other as-pects of innervation are considered in a special case (Clinical Box 37–1). AFFERENTS FROM PROPRIOCEPTORS Carefully controlled experiments have shown that active and passive movements of joints stimulate respiration, presumably TABLE 37–2 Airway and lung receptors. Vagal Innervation Type Location in Interstitium Stimulus Response Myelinated Slowly adapting Among airway smooth muscle cells (?) Lung inflation Inspiratory time shortening Hering–Breuer inflation and deflation reflexes Bronchodilation Tachycardia Hyperpnea Rapidly adapting Among airway epithelial cells Lung hyperinflation Cough Exogenous and endogenous substances (eg, histamine, prostaglandins) Bronchoconstriction Mucus secretion Unmyelinated C fibers Pulmonary C fibers Bronchial C fibers Close to blood vessels Lung hyperinflation Apnea followed by rapid breathing Exogenous and endogenous substanc-es (eg, capsaicin, bradykinin, serotonin) Bronchoconstriction Bradycardia Hypotension Mucus secretion Modified and reproduced with permission from Berger AJ, Hornbein TF: Control of respiration. In: Textbook of Physiology, 21st ed. Vol. 2. Patton HD, et al (editors). Saunders, 1989. CHAPTER 37 Regulation of Respiration 633 because impulses in afferent pathways from proprioceptors in muscles, tendons, and joints stimulate the inspiratory neurons. This effect probably helps increase ventilation during exercise. Other afferents are considered in Clinical Box 37–2. RESPIRATORY COMPONENTS OF VISCERAL REFLEXES Inhibition of respiration and closure of the glottis during vomiting, swallowing, and sneezing not only prevent the aspi-ration of food or vomitus into the trachea but, in the case of vomiting, fix the chest so that contraction of the abdominal muscles increases the intra-abdominal pressure. Similar glot-tic closure and inhibition of respiration occur during volun-tary and involuntary straining. Hiccup is a spasmodic contraction of the diaphragm and other inspiratory muscles that produces an inspiration during which the glottis suddenly closes. The glottic closure is responsible for the characteristic sensation and sound. Hic-cups occur in the fetus in utero as well as throughout extrau-terine life. Their function is unknown. Most attacks of hiccups are usually of short duration, and they often respond to breath holding or other measures that increase arterial PCO2. Intractable hiccups, which can be debilitating, some-times respond to dopamine antagonists and perhaps to some centrally acting analgesic compounds. Yawning is a peculiar “infectious” respiratory act whose physiologic basis and significance are uncertain. Like hiccup-ing, it occurs in utero, and it occurs in fish and tortoises as well as mammals. The view that it is needed to increase O2 intake has been discredited. Underventilated alveoli have a tendency to collapse, and it has been suggested that the deep inspiration and stretching them open prevents the develop-ment of atelectasis. However, in actual experiments, no atelectasis-preventing effect of yawning could be demon-strated. Yawning increases venous return to the heart, which may benefit the circulation. It has been suggested that yawn-ing is a nonverbal signal used for communication between monkeys in a group, and one could argue that on a different level, the same thing is true in humans. RESPIRATORY EFFECTS OF BARORECEPTOR STIMULATION Afferent fibers from the baroreceptors in the carotid sinuses, aortic arch, atria, and ventricles relay to the respiratory neu-rons, as well as the vasomotor and cardioinhibitory neurons in the medulla. Impulses in them inhibit respiration, but the in-hibitory effect is slight and of little physiologic importance. The hyperventilation in shock is due to chemoreceptor stimu-lation caused by acidosis and hypoxia secondary to local stag-nation of blood flow, and is not baroreceptor-mediated. The activity of inspiratory neurons affects blood pressure and heart rate, and activity in the vasomotor and cardiac areas in the medulla may have minor effects on respiration. CLINICAL BOX 37–1 Lung Innervation & Patients with Heart–Lung Transplants Transplantation of the heart and lungs is now an established treatment for severe pulmonary disease and other condi-tions. In individuals with transplants, the recipient’s right atrium is sutured to the donor heart, and the donor heart does not reinnervate, so the resting heart rate is elevated. The donor trachea is sutured to the recipient’s just above the carina, and afferent fibers from the lungs do not regrow. Con-sequently, healthy patients with heart–lung transplants pro-vide an opportunity to evaluate the role of lung innervation in normal physiology. Their cough responses to stimulation of the trachea are normal because the trachea remains inner-vated, but their cough responses to stimulation of the smaller airways are absent. Their bronchi tend to be dilated to a greater degree than normal. In addition, they have the nor-mal number of yawns and sighs, indicating that these do not depend on innervation of the lungs. Finally, they lack Hering– Breuer reflexes, but their pattern of breathing at rest is nor-mal, indicating that these reflexes do not play an important role in the regulation of resting respiration in humans. CLINICAL BOX 37–2 Afferents from “Higher Centers” Pain and emotional stimuli affect respiration, so there must also be afferents from the limbic system and hypothalamus to the respiratory neurons in the brain stem. In addition, even though breathing is not usually a conscious event, both inspiration and expiration are under voluntary con-trol. The pathways for voluntary control pass from the neo-cortex to the motor neurons innervating the respiratory muscles, bypassing the medullary neurons. Because voluntary and automatic control of respiration are separate, automatic control is sometimes disrupted without loss of voluntary control. The clinical condition that results has been called Ondine’s curse. In German legend, Ondine was a water nymph who had an unfaithful mortal lover. The king of the water nymphs punished the lover by casting a curse on him that took away all his auto-matic functions. In this state, he could stay alive only by staying awake and remembering to breathe. He eventually fell asleep from sheer exhaustion, and his respiration stopped. Patients with this intriguing condition generally have bulbar poliomyelitis or disease processes that com-press the medulla. 634 SECTION VII Respiratory Physiology EFFECTS OF SLEEP Respiration is less rigorously controlled during sleep than in the waking state, and brief periods of apnea occur in normal sleeping adults. Changes in the ventilatory response to hy-poxia vary. If the PCO2 falls during the waking state, various stimuli from proprioceptors and the environment maintain respiration, but during sleep, these stimuli are decreased and a decrease in PCO2 can cause apnea. During rapid eye move-ment (REM) sleep, breathing is irregular and the CO2 re-sponse is highly variable. RESPIRATORY ABNORMALITIES ASPHYXIA In asphyxia produced by occlusion of the airway, acute hyper-capnia and hypoxia develop together. Stimulation of respira-tion is pronounced, with violent respiratory efforts. Blood pressure and heart rate rise sharply, catecholamine secretion is increased, and blood pH drops. Eventually the respiratory ef-forts cease, the blood pressure falls, and the heart slows. As-phyxiated animals can still be revived at this point by artificial respiration, although they are prone to ventricular fibrillation, probably because of the combination of hypoxic myocardial damage and high circulating catecholamine levels. If artificial respiration is not started, cardiac arrest occurs in 4 to 5 min. DROWNING Drowning is asphyxia caused by immersion, usually in water. In about 10% of drownings, the first gasp of water after the los-ing struggle not to breathe triggers laryngospasm, and death results from asphyxia without any water in the lungs. In the re-maining cases, the glottic muscles eventually relax and fluid enters the lungs. Fresh water is rapidly absorbed, diluting the plasma and causing intravascular hemolysis. Ocean water is markedly hypertonic and draws fluid from the vascular system into the lungs, decreasing plasma volume. The immediate goal in the treatment of drowning is, of course, resuscitation, but long-term treatment must also take into account the circula-tory effects of the water in the lungs. PERIODIC BREATHING The acute effects of voluntary hyperventilation demonstrate the interaction of the chemical mechanisms regulating respi-ration. When a normal individual hyperventilates for 2 to 3 min, then stops and permits respiration to continue without exerting any voluntary control over it, a period of apnea oc-curs. This is followed by a few shallow breaths and then by an-other period of apnea, followed again by a few breaths (periodic breathing). The cycles may last for some time be-fore normal breathing is resumed (Figure 37–12). The apnea apparently is due to a lack of CO2 because it does not occur following hyperventilation with gas mixtures containing 5% CO2. During the apnea, the alveolar PO2 falls and the PCO2 ris-es. Breathing resumes because of hypoxic stimulation of the carotid and aortic chemoreceptors before the CO2 level has re-turned to normal. A few breaths eliminate the hypoxic stimu-lus, and breathing stops until the alveolar PO2 falls again. Gradually, however, the PCO2 returns to normal, and normal breathing resumes. Changes in breathing patterns can be symptomatic of disease (Clinical Box 37–3). EFFECTS OF EXERCISE Exercise provides a physiological example to explore many of the control systems discussed above. Of course, many cardio-vascular and respiratory mechanisms must operate in an inte-grated fashion if the O2 needs of the active tissue are to be met and the extra CO2 and heat removed from the body during ex-ercise. Circulatory changes increase muscle blood flow while maintaining adequate circulation in the rest of the body. In addition, there is an increase in the extraction of O2 from the blood in exercising muscles and an increase in ventilation. This provides extra O2, eliminates some of the heat, and ex-cretes extra CO2. A focus on regulation of ventilation and tis-sue O2 is presented below, as many other aspects of regulation have been presented in previous chapters. CHANGES IN VENTILATION During exercise, the amount of O2 entering the blood in the lungs is increased because the amount of O2 added to each FIGURE 37–12 Changes in breathing and composition of alveolar air after forced hyperventilation for 2 min. Bars in bottom indicate breathing, whereas blank spaces are indicative of apnea. 0 40 80 120 160 Partial pressure (mm Hg) 1 0 2 3 4 5 6 Breathing pattern Alveolar PO2 Alveolar PCO2 Time after stopping hyperventilation (min) CHAPTER 37 Regulation of Respiration 635 unit of blood and the pulmonary blood flow per minute are in-creased. The PO2 of blood flowing into the pulmonary capil-laries falls from 40 to 25 mm Hg or less, so that the alveolar– capillary PO2 gradient is increased and more O2 enters the blood. Blood flow per minute is increased from 5.5 L/min to as much as 20 to 35 L/min. The total amount of O2 entering the blood therefore increases from 250 mL/min at rest to values as high as 4000 mL/min. The amount of CO2 removed from each unit of blood is increased, and CO2 excretion increases from 200 mL/min to as much as 8000 mL/min. The increase in O2 uptake is proportional to work load, up to a maximum. Above this maximum, O2 consumption levels off and the blood lac-tate level continues to rise (Figure 37–13). The lactate comes from muscles in which aerobic resynthesis of energy stores cannot keep pace with their utilization, and an oxygen debt is being incurred. Ventilation increases abruptly with the onset of exercise, which is followed after a brief pause by a further, more gradual increase (Figure 37–14). With moderate exercise, the increase is due mostly to an increase in the depth of respiration; this is accompanied by an increase in the respiratory rate when the exercise is more strenuous. Ventilation abruptly decreases when exercise ceases, which is followed after a brief pause by a CLINICAL BOX 37–3 Periodic Breathing in Disease Cheyne–Stokes Respiration Periodic breathing occurs in various disease states and is often called Cheyne–Stokes respiration. It is seen most commonly in patients with congestive heart failure and uremia, but it occurs also in patients with brain disease and during sleep in some normal individuals. Some of the pa-tients with Cheyne–Stokes respiration have increased sen-sitivity to CO2. The increased response is apparently due to disruption of neural pathways that normally inhibit respira-tion. In these individuals, CO2 causes relative hyperventila-tion, lowering the arterial PCO2. During the resultant apnea, the arterial PCO2 again rises to normal, but the respiratory mechanism again overresponds to CO2. Breathing ceases, and the cycle repeats. Another cause of periodic breathing in patients with car-diac disease is prolongation of the lung-to-brain circulation time, so that it takes longer for changes in arterial gas ten-sions to affect the respiratory area in the medulla. When in-dividuals with a slower circulation hyperventilate, they lower the PCO2 of the blood in their lungs, but it takes longer than normal for the blood with a low PCO2 to reach the brain. During this time, the PCO2 in the pulmonary capil-lary blood continues to be lowered, and when this blood reaches the brain, the low PCO2 inhibits the respiratory area, producing apnea. In other words, the respiratory control system oscillates because the negative feedback loop from lungs to brain is abnormally long. Sleep Apnea Episodes of apnea during sleep can be central in origin; that is, due to failure of discharge in the nerves producing respiration, or they can be due to airway obstruction (ob-structive sleep apnea). This can occur at any age and is produced when the pharyngeal muscles relax during sleep. In some cases, failure of the genioglossus muscles to con-tract during inspiration contributes to the blockage; these muscles pull the tongue forward, and when they do not contract the tongue falls back and obstructs the airway. After several increasingly strong respiratory efforts, the pa-tient wakes up, takes a few normal breaths, and falls back to sleep. Not surprisingly, the apneic episodes are most common during REM sleep, when the muscles are most hy-potonic. The symptoms are loud snoring, morning head-aches, fatigue, and daytime sleepiness. When severe and prolonged, the condition apparently causes hypertension and its complications. In addition, the incidence of motor vehicle accidents in sleep apnea patients is 7 times greater than it is in the general driving population. FIGURE 37–13 Relation between work load, blood lactate level, and O2 uptake. I–VI, increasing work loads produced by in-creasing the speed and grade of a treadmill on which the subjects worked. (Reproduced with permission from Mitchell JH, Blomqvist G: Maximal oxygen uptake. N Engl J Med 1971;284:1018.) FIGURE 37–14 Diagrammatic representation of changes in ventilation during exercise. See text for details. 4 3 2 1 0 Rest I II III IV V VI 0 3 6 9 12 Maximal work loads Work load Submaximal work loads Blood lactate Blood lactate (meq/L) O2 uptake O2 uptake (L /min) Exercise Rest Recovery Time Ventilation (L/min) 636 SECTION VII Respiratory Physiology more gradual decline to pre-exercise values. The abrupt increase at the start of exercise is presumably due to psychic stimuli and afferent impulses from proprioceptors in muscles, tendons, and joints. The more gradual increase is presumably humoral, even though arterial pH, PCO2, and PO2 remain con-stant during moderate exercise. The increase in ventilation is proportional to the increase in O2 consumption, but the mech-anisms responsible for the stimulation of respiration are still the subject of much debate. The increase in body temperature may play a role. Exercise increases the plasma K+ level, and this increase may stimulate the peripheral chemoreceptors. In addition, it may be that the sensitivity of the neurons control-ling the response to CO2 is increased or that the respiratory fluctuations in arterial PCO2 increase so that, even though the mean arterial PCO2 does not rise, it is CO2 that is responsible for the increase in ventilation. O2 also seems to play some role, despite the lack of a decrease in arterial PO2, since during the performance of a given amount of work, the increase in venti-lation while breathing 100% O2 is 10–20% less than the increase while breathing air. Thus, it currently appears that a number of different factors combine to produce the increase in ventilation seen during moderate exercise. When exercise becomes more vigorous, buffering of the increased amounts of lactic acid that are produced liberates more CO2, and this further increases ventilation. The response to graded exercise is shown in Figure 37–15. With increased production of acid, the increases in ventilation and CO2 production remain proportional, so alveolar and arterial CO2 change relatively little (isocapnic buffering). Because of the hyperventilation, alveolar PO2 increases. With further accumulation of lactic acid, the increase in ventilation out-strips CO2 production and alveolar PCO2 falls, as does arterial PCO2. The decline in arterial PCO2 provides respiratory com-pensation for the metabolic acidosis produced by the addi-tional lactic acid. The additional increase in ventilation produced by the acidosis is dependent on the carotid bodies and does not occur if they are removed. The respiratory rate after exercise does not reach basal levels until the O2 debt is repaid. This may take as long as 90 min. The stimulus to ventilation after exercise is not the arte-rial PCO2, which is normal or low, or the arterial PO2, which is normal or high, but the elevated arterial H+ concentration due to the lactic acidemia. The magnitude of the O2 debt is the amount by which O2 consumption exceeds basal con-sumption from the end of exertion until the O2 consumption has returned to pre-exercise basal levels. During repayment of the O2 debt, the O2 concentration in muscle myoglobin rises slightly. ATP and phosphorylcreatine are resynthesized, and lactic acid is removed. Eighty percent of the lactic acid is con-verted to glycogen and 20% is metabolized to CO2 and H2O. Because of the extra CO2 produced by the buffering of lac-tic acid during strenuous exercise, the ratio of CO2 to O2 (res-piratory exchange ratio; R) rises, reaching 1.5 to 2.0. After exertion, while the O2 debt is being repaid, the R falls to 0.5 or less. CHANGES IN THE TISSUES Maximum O2 uptake during exercise is limited by the maxi-mum rate at which O2 is transported to the mitochondria in the exercising muscle. However, this limitation is not normally due to deficient O2 uptake in the lungs, and hemoglobin in arterial blood is saturated even during the most severe exercise. During exercise, the contracting muscles use more O2, and the tissue PO2 and the PO2 in venous blood from exercising muscle fall nearly to zero. More O2 diffuses from the blood, the blood PO2 of the blood in the muscles drops, and more O2 is removed from hemoglobin. Because the capillary bed of contracting muscle is dilated and many previously closed cap-illaries are open, the mean distance from the blood to the tis-sue cells is greatly decreased; this facilitates the movement of O2 from blood to cells. The oxygen–hemoglobin dissociation curve is steep in the PO2 range below 60 mm Hg, and a rela-tively large amount of O2 is supplied for each drop of 1 mm Hg in PO2 (see Figure 36–2). Additional O2 is supplied because, as a result of the accumulation of CO2 and the rise in temperature in active tissues—and perhaps because of a rise in red blood cell 2,3-biphosphoglycerate (2,3-BPG)—the dis-sociation curve shifts to the right. The net effect is a threefold FIGURE 37–15 Physiologic responses to work rate during exercise. Changes in alveolar PCO2, alveolar PO, ventilation (V ˙E), CO2 production (V ˙CO2), O2 consumption (V ˙O2), arterial HCO3 –, and arterial pH with graded increases in work by an adult male on a bicycle ergo-meter. Resp comp, respiratory compensation. See text for details. (Reproduced with permission from Wasserman K, Whipp BJ, Casaburi R: Respiratory control during exercise. In: Handbook of Physiology. Section 3, The Respiratory System.Vol II, part 2. Fishman AP [editor]. American Physiological Society, 1986.] VCO2 • VCO2 (L/min-STPD) • VO2 (L/min-STPD) • VO2 • VE • 50 120 0 25 15 7.40 7.30 3 0 0 P ACO2 (mm Hg) P AO2 (mm Hg) HCO3− HCO3− VE (L/min-BTPS) • pH pH 150 70 2 2 min 0 Isocapnic buffering Resp comp 15 30 45 60 75 90 105 120 135 150 165 180 0 Work rate (watts) CHAPTER 37 Regulation of Respiration 637 increase in O2 extraction from each unit of blood (see Figure 36–3). Because this increase is accompanied by a 30-fold or greater increase in blood flow, it permits the metabolic rate of muscle to rise as much as 100-fold during exercise. EXERCISE TOLERANCE & FATIGUE What determines the maximum amount of exercise that can be performed by an individual? Obviously, exercise tolerance has a time as well as an intensity dimension. For example, a fit young man can produce a power output on a bicycle of about 700 watts for 1 min, 300 watts for 5 min, and 200 watts for 40 min. It used to be argued that the limiting factors in exercise performance were the rate at which O2 could be delivered to the tissues or the rate at which O2 could enter the body in the lungs. These factors play a role, but it is clear that other factors also contribute and that exercise stops when the sensation of fatigue progresses to the sensation of exhaustion. Fatigue is produced in part by bombardment of the brain by neural im-pulses from muscles, and the decline in blood pH produced by lactic acidosis also makes one feel tired, as do the rise in body temperature, dyspnea, and, perhaps, the uncomfortable sensa-tions produced by activation of the J receptors in the lungs. CHAPTER SUMMARY ■Breathing is under both voluntary control (located in the cere-bral cortex) and automatic control (driven by pacemaker cells in the medulla). There is a reciprocal innervation to expiratory and inspiratory muscles in that motor neurons supplying expiratory muscles are inactive when motor neurons supplying inspiratory muscles are active, and vice versa. ■The pre-Bötzinger complex on either side of the medulla con-tains synaptically coupled pacemaker cells that allow for rhyth-mic generation of breathing. The spontaneous activity of these neurons can be altered by neurons in the pneumotaxic center, although the full regulatory function of these neurons on nor-mal breathing is not understood. ■Breathing patterns are sensitive to chemicals in the blood through activation of respiratory chemoreceptors. There are chemoreceptors in the carotid and aortic bodies and in collec-tions of cells in the medulla. These chemoreceptors respond to changes in PO2 and PCO2 as well as H+ to regulate breathing. ■Receptors in the airway are additionally innervated by slowly adapting and rapidly adapting myelinated vagal fibers. Slowly adapting receptors can be activated by lung inflation. Rapidly adapting receptors, or irritant receptors, can be activated by chem-icals such as histamine and result in cough or even hyperpnea. ■Receptors in the airway are also innervated by unmyelinated va-gal fibers (C fibers) that are typically found next to pulmonary vessels. They are stimulated by hyperinflation (or exogenous substances including capsaicin) and lead to the pulmonary chemoreflex. The physiologic role for this response is not fully understood. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. The main respiratory control neurons A) send out regular bursts of impulses to expiratory muscles during quiet respiration. B) are unaffected by stimulation of pain receptors. C) are located in the pons. D) send out regular bursts of impulses to inspiratory muscles during quiet respiration. E) are unaffected by impulses from the cerebral cortex. 2. Intravenous lactic acid increases ventilation. The receptors responsible for this effect are located in the A) medulla oblongata. B) carotid bodies. C) lung parenchyma. D) aortic baroreceptors. E) trachea and large bronchi. 3. Spontaneous respiration ceases after A) transection of the brain stem above the pons. B) transection of the brain stem at the caudal end of the medulla. C) bilateral vagotomy. D) bilateral vagotomy combined with transection of the brain stem at the superior border of the pons. E) transection of the spinal cord at the level of the first thoracic segment. 4. The following physiologic events that occur in vivo are listed in random order: (1) decreased CSF pH; (2) increased arterial PCO2; (3) increased CSF PCO2; (4) stimulation of medullary chemoreceptors; (5) increased alveolar PCO2. What is the usual sequence in which they occur when they affect respiration? A) 1, 2, 3, 4, 5 B) 4, 1, 3, 2, 5 C) 3, 4, 5, 1, 2 D) 5, 2, 3, 1, 4 E) 5, 3, 2, 4, 1 5. The following events that occur in the carotid bodies when they are exposed to hypoxia are listed in random order: (1) depolar-ization of type I glomus cells; (2) excitation of afferent nerve endings; (3) reduced conductance of hypoxia-sensitive K+ chan-nels in type I glomus cells; (4) Ca2+ entry into type I glomus cells; (5) decreased K+ efflux. What is the usual sequence in which they occur on exposure to hypoxia? A) 1, 3, 4, 5, 2 B) 1, 4, 2, 5, 3 C) 3, 4, 5, 1, 2 D) 3, 1, 4, 5, 2 E) 3, 5, 1, 4, 2 6. Stimulation of the central (proximal) end of a cut vagus nerve would be expected to A) increase heart rate. B) stimulate inspiration. C) inhibit coughing. D) raise blood pressure. E) cause apnea. 638 SECTION VII Respiratory Physiology 7. Injection of a drug that stimulates the carotid bodies would be expected to cause A) a decrease in the pH of arterial blood. B) a decrease in the PCO2 of arterial blood. C) an increase in the HCO3 – concentration of arterial blood. D) an increase in urinary Na+ excretion. E) an increase in plasma Cl–. 8. Variations in which of the following components of blood or CSF do not affect respiration? A) arterial HCO3 – concentration B) arterial H+ concentration C) arterial Na+ concentration D) CSF CO2 concentration E) CSF H+ concentration CHAPTER RESOURCES Barnes PJ: Chronic obstructive pulmonary disease. N Engl J Med 2000;343:269. Crystal RG, West JB (editors): The Lung: Scientific Foundations, 2nd ed. Lippincott-Raven, 1997. Fishman AP, et al (editors): Fishman’s Pulmonary Diseases and Disorders, 4th ed. McGraw-Hill, 2008. Hackett PH, Roach RC: High-altitude illness. N Engl J Med 2001;345:107. Jones NL, Killian KJ: Exercise limitation in health and disease. N Engl J Med 2000;343:632. Laffey JG, Kavanagh BP: Hypocapnia. N Engl J Med 2002;347:43. Levitzky, MG: Pulmonary Physiology, 7th ed. McGraw Hill, 2007. Prisk GK, Paiva M, West JB (editors): Gravity and the Lung: Lessons from Micrography. Marcel Dekker, 2001. Putnam RW, Dean JB, Ballantyne D (editors): Central chemosensitivity. Respir Physiol 2001;129:1. Rekling JC, Feldman JL: Pre-Bötzinger complex and pacemaker neurons: hypothesized site and kernel for respiratory rhythm generation. Annu Rev Physiol 1998;60:385. Tobin MJ: Advances in mechanical ventilation. N Engl J Med 2001;344:1986. Voelkel NF: High-altitude pulmonary edema. N Engl J Med 2002;346:1607. Ware LB, Matthay MA: The acute respiratory distress syndrome. N Engl J Med 2000;342:1334. West JB: Pulmonary Pathophysiology, 5th ed. McGraw-Hill, 1995. 639 C H A P T E R SECTION VIII RENAL PHYSIOLOGY 38 Renal Function & Micturition O B J E C T I V E S After reading this chapter, you should be able to: ■Describe the morphology of a typical nephron and its blood supply. ■Define autoregulation and list the major theories advanced to explain autoregula-tion in the kidneys. ■Define glomerular filtration rate, describe how it can be measured, and list the major factors affecting it. ■Outline tubular handling of Na+ and water. ■Discuss tubular reabsorption and secretion of glucose and K+. ■Describe how the countercurrent mechanism in the kidney operates to produce hypertonic or hypotonic urine. ■List the major classes of diuretics and how each operates to increase urine flow. ■Describe the voiding reflex and draw a cystometrogram. INTRODUCTION In the kidneys, a fluid that resembles plasma is filtered through the glomerular capillaries into the renal tubules (glo-merular filtration). As this glomerular filtrate passes down the tubules, its volume is reduced and its composition altered by the processes of tubular reabsorption (removal of water and solutes from the tubular fluid) and tubular secretion (secretion of solutes into the tubular fluid) to form the urine that enters the renal pelvis. A comparison of the composition of the plasma and an average urine specimen illustrates the magnitude of some of these changes (Table 38–1). It empha-sizes the manner by which water and important electrolytes and metabolites are conserved while wastes are eliminated in the urine. Furthermore, the composition of the urine can be varied to maintain whole body fluid homeostasis (extracellu-lar fluid [ECF]). This is achieved via many homeostatic regu-latory mechanisms that function to change the amount of water and solutes in the urine. From the renal pelvis, the urine passes to the bladder and is expelled to the exterior by the process of urination, or micturition. The kidneys are also endocrine organs, making kinins (see Chapter 33) and 1, 25-dihydroxycholecalciferol (see Chapter 23), and making and secreting renin (see Chapter 39). 640 SECTION VIII Renal Physiology FUNCTIONAL ANATOMY THE NEPHRON Each individual renal tubule and its glomerulus is a unit (nephron). The size of the kidneys between species varies, as does the number of nephrons they contain. Each human kid-ney has approximately 1.3 million nephrons. The specific structures of the nephron are shown in diagrammatic fashion in Figure 38–1. The glomerulus, which is about 200 μm in diameter, is formed by the invagination of a tuft of capillaries into the dilated, blind end of the nephron (Bowman’s capsule). The capillaries are supplied by an afferent arteriole and drained by a slightly smaller efferent arteriole (Figure 38–2), and it is from the glomerulus that the filtrate is formed. Two cellular layers separate the blood from the glomerular filtrate in Bow-man’s capsule: the capillary endothelium and the specialized epithelium of the capsule. The endothelium of the glomerular capillaries is fenestrated, with pores that are 70 to 90 nm in diameter. The endothelium of the glomerular capillaries is completely surrounded by the glomerular basement mem-brane along with specialized cells called podocytes. Podocytes have numerous pseudopodia that interdigitate (Figure 38–2) to form filtration slits along the capillary wall. The slits are approximately 25 nm wide, and each is closed by a thin mem-brane. The glomerular basement membrane, the basal lamina, does not contain visible gaps or pores. Stellate cells called mes-angial cells are located between the basal lamina and the endothelium. They are similar to cells called pericytes, which are found in the walls of capillaries elsewhere in the body. Mes-angial cells are especially common between two neighboring capillaries, and in these locations the basal membrane forms a sheath shared by both capillaries (Figure 38–2). The mesangial cells are contractile and play a role in the regulation of glomer-ular filtration. Mesangial cells secrete the extracellular matrix, take up immune complexes, and are involved in the progres-sion of glomerular disease. Functionally, the glomerular membrane permits the free passage of neutral substances up to 4 nm in diameter and almost totally excludes those with diameters greater than 8 nm. However, the charges on molecules as well as their diam-eters affect their passage into Bowman’s capsule. The total area of glomerular capillary endothelium across which filtra-tion occurs in humans is about 0.8 m2. The general features of the cells that make up the walls of the tubules are shown in Figure 38–1; however, there are cell sub-types in all segments, and the anatomic differences between them correlate with differences in function. The human proximal convoluted tubule is about 15 mm long and 55 μm in diameter. Its wall is made up of a single layer of cells that interdigitate with one another and are united by apical tight junctions. Between the bases of the cells are exten-sions of the extracellular space called the lateral intercellular spaces. The luminal edges of the cells have a striate brush bor-der due to the presence of many microvilli. The convoluted proximal tubule straightens and the next portion of each nephron is the loop of Henle. The descending portion of the loop and the proximal portion of the ascending limb are made up of thin, permeable cells. On the other hand, the thick portion of the ascending limb (Figure 38–1) is made up of thick cells containing many mitochondria. The neph-rons with glomeruli in the outer portions of the renal cortex have short loops of Henle (cortical nephrons), whereas those with glomeruli in the juxtamedullary region of the cortex (juxtamedullary nephrons) have long loops extending down TABLE 38–1 Typical urinary and plasma concentrations of some physiologically important substances. Concentration in Substance Urine (U) Plasma (P) U/P Ratio Glucose (mg/dL) 0 100 0 Na+ (mEq/L) 90 140 0.6 Urea (mg/dL) 900 15 60 Creatinine (mg/dL) 150 1 150 FIGURE 38–1 Diagram of a juxtamedullary nephron. The main histologic features of the cells that make up each portion of the tubule are also shown. Proximal convoluted tubule Distal convoluted tubule Collecting duct Cortex Outer medulla Inner medulla Loop of Henle, thin descending limb Loop of Henle, thick ascending limb Glomerulus CHAPTER 38 Renal Function & Micturition 641 into the medullary pyramids. In humans, only 15% of the nephrons have long loops. The thick end of the ascending limb of the loop of Henle reaches the glomerulus of the nephron from which the tubule arose and nestles between its afferent and efferent arterioles. Specialized cells at the end form the macula densa, which is close to the efferent and particularly the afferent arteriole (Figure 38–2). The macula, the neighboring lacis cells, and the renin-secreting juxtaglomerular cells in the afferent arte-riole form the juxtaglomerular apparatus (see Figure 39–9). The distal convoluted tubule, which starts at the macula densa, is about 5 mm long. Its epithelium is lower than that of the proximal tubule, and although a few microvilli are present, there is no distinct brush border. The distal tubules coalesce to form collecting ducts that are about 20 mm long and pass through the renal cortex and medulla to empty into the pelvis of the kidney at the apexes of the medullary pyra-mids. The epithelium of the collecting ducts is made up of principal cells (P cells) and intercalated cells (I cells). The P cells, which predominate, are relatively tall and have few organelles. They are involved in Na+ reabsorption and vaso-pressin-stimulated water reabsorption. The I cells, which are present in smaller numbers and are also found in the distal tubules, have more microvilli, cytoplasmic vesicles, and FIGURE 38–2 Structural details of glomerulus. A) Section through vascular pole, showing capillary loops. B) Relation of mesangial cells and podocytes to glomerular capillaries. C) Detail of the way podocytes form filtration slits on the basal lamina, and the relation of the lamina to the capillary endothelium. D) Enlargement of the rectangle in C to show the podocyte processes. The fuzzy material on their surfaces is glomerular polyanion. A Proximal tubule Capsule Red blood cells Glomerular basal lamina Bowman's space Juxtaglomerular cells Nerve fibers Smooth muscle Macula densa Distal tubule Mesangial cell Podocyte processes Afferent arteriole Efferent arteriole B Capillary Capillary Capillary Capillary Podocyte Podocyte process Basal lamina Cytoplasm of endothelial cell Mesangial cell C Podocyte Endothelium Endothelium Basal lamina Basal lamina D Foot processes of podocytes Filtration slit Bowman's space Basal lamina Fenestrations Capillary lumen 642 SECTION VIII Renal Physiology mitochondria. They are concerned with acid secretion and HCO3 – transport. The total length of the nephrons, including the collecting ducts, ranges from 45 to 65 mm. Cells in the kidneys that appear to have a secretory function include not only the juxtaglomerular cells but also some of the cells in the interstitial tissue of the medulla. These cells are called type I medullary interstitial cells. They contain lipid droplets and probably secrete prostaglandins, predominantly PGE2. PGE2 is also secreted by the cells in the collecting ducts; prostacyclin (PGI2) and other prostaglandins are secreted by the arterioles and glomeruli. BLOOD VESSELS The renal circulation is diagrammed in Figure 38–3. The af-ferent arterioles are short, straight branches of the interlobu-lar arteries. Each divides into multiple capillary branches to form the tuft of vessels in the glomerulus. The capillaries coa-lesce to form the efferent arteriole, which in turn breaks up into capillaries that supply the tubules (peritubular capillar-ies) before draining into the interlobular veins. The arterial segments between glomeruli and tubules are thus technically a portal system, and the glomerular capillaries are the only cap-illaries in the body that drain into arterioles. However, there is relatively little smooth muscle in the efferent arterioles. The capillaries draining the tubules of the cortical nephrons form a peritubular network, whereas the efferent arterioles from the juxtamedullary glomeruli drain not only into a peri-tubular network, but also into vessels that form hairpin loops (the vasa recta). These loops dip into the medullary pyramids alongside the loops of Henle (Figure 38–3). The descending vasa recta have a nonfenestrated endothelium that contains a FIGURE 38–3 Renal circulation. Interlobar arteries divide into arcuate arteries, which give off interlobular arteries in the cortex. The inter-lobular arteries provide an afferent arteriole to each glomerulus. The efferent arteriole from each glomerulus breaks up into capillaries that supply blood to the renal tubules. Venous blood enters interlobular veins, which in turn flow via arcuate veins to the interlobar veins. (Modified from Boron WF, Boulpaep EL: Medical Physiology. Saunders, 2003.) Renal cortex Superficial glomeruli Interlobular vein Peritubular capillary bed Arcuate vein Arcuate artery Ascending vasa recta Descending vasa recta Loop of Henle Interlobar vein Interlobar artery Renal medulla (pyramid) Efferent arteriole Afferent arteriole Interlobular artery Juxtamedullary glomerulus CHAPTER 38 Renal Function & Micturition 643 facilitated transporter for urea, and the ascending vasa recta have a fenestrated endothelium, consistent with their function in conserving solutes. The efferent arteriole from each glomerulus breaks up into capillaries that supply a number of different nephrons. Thus, the tubule of each nephron does not necessarily receive blood solely from the efferent arteriole of the same nephron. In humans, the total surface of the renal capillaries is approxi-mately equal to the total surface area of the tubules, both being about 12 m2. The volume of blood in the renal capillar-ies at any given time is 30 to 40 mL. LYMPHATICS The kidneys have an abundant lymphatic supply that drains via the thoracic duct into the venous circulation in the thorax. CAPSULE The renal capsule is thin but tough. If the kidney becomes edematous, the capsule limits the swelling, and the tissue pres-sure (renal interstitial pressure) rises. This decreases the glo-merular filtration rate and is claimed to enhance and prolong anuria in acute renal failure. INNERVATION OF THE RENAL VESSELS The renal nerves travel along the renal blood vessels as they en-ter the kidney. They contain many postganglionic sympathetic efferent fibers and a few afferent fibers. There also appears to be a cholinergic innervation via the vagus nerve, but its function is uncertain. The sympathetic preganglionic innervation comes primarily from the lower thoracic and upper lumbar segments of the spinal cord, and the cell bodies of the postganglionic neu-rons are in the sympathetic ganglion chain, in the superior mes-enteric ganglion, and along the renal artery. The sympathetic fibers are distributed primarily to the afferent and efferent arte-rioles, the proximal and distal tubules, and the juxtaglomerular cells (see Chapter 39). In addition, there is a dense noradrener-gic innervation of the thick ascending limb of the loop of Henle. Nociceptive afferents that mediate pain in kidney disease parallel the sympathetic efferents and enter the spinal cord in the thoracic and upper lumbar dorsal roots. Other renal affer-ents presumably mediate a renorenal reflex by which an increase in ureteral pressure in one kidney leads to a decrease in efferent nerve activity to the contralateral kidney, and this decrease permits an increase in its excretion of Na+ and water. RENAL CIRCULATION BLOOD FLOW In a resting adult, the kidneys receive 1.2 to 1.3 L of blood per minute, or just under 25% of the cardiac output. Renal blood flow can be measured with electromagnetic or other types of flow meters, or it can be determined by applying the Fick principle (see Chapter 33) to the kidney; that is, by measur-ing the amount of a given substance taken up per unit of time and dividing this value by the arteriovenous difference for the substance across the kidney. Because the kidney filters plasma, the renal plasma flow equals the amount of a sub-stance excreted per unit of time divided by the renal arterio-venous difference as long as the amount in the red cells is unaltered during passage through the kidney. Any excreted substance can be used if its concentration in arterial and re-nal venous plasma can be measured and if it is not metabo-lized, stored, or produced by the kidney and does not itself affect blood flow. Renal plasma flow can be measured by infusing p-amino-hippuric acid (PAH) and determining its urine and plasma concentrations. PAH is filtered by the glomeruli and secreted by the tubular cells, so that its extraction ratio (arterial con-centration minus renal venous concentration divided by arte-rial concentration) is high. For example, when PAH is infused at low doses, 90% of the PAH in arterial blood is removed in a single circulation through the kidney. It has therefore become commonplace to calculate the “renal plasma flow” by dividing the amount of PAH in the urine by the plasma PAH level, ignoring the level in renal venous blood. Peripheral venous plasma can be used because its PAH concentration is essen-tially identical to that in the arterial plasma reaching the kid-ney. The value obtained should be called the effective renal plasma flow (ERPF) to indicate that the level in renal venous plasma was not measured. In humans, ERPF averages about 625 mL/min. Example: Concentration of PAH in urine (UPAH): 14 mg/mL Urine flow (V • ): 0.9 mL/min Concentration of PAH in plasma (PPAH): 0.02 mg/mL = 630 mL/min It should be noted that the ERPF determined in this way is the clearance of PAH. The concept of clearance is discussed in detail below. ERPF can be converted to actual renal plasma flow (RPF): Average PAH extraction ratio: 0.9 ERPF UPAHV ˙ PPAH -------------------Clearance of PAH CPAH ( ) = = ERPF 14 0.9 × 0.02 -------------------= ERP Extraction ration -----------------------------------------630 0.9 --------Actual RPF 700 mL/min = = = 644 SECTION VIII Renal Physiology From the renal plasma flow, the renal blood flow can be cal-culated by dividing by 1 minus the hematocrit: Hematocrit (Hct): 45% Renal blood flow = RPF × 1 1–Hct = 700 × 1 0.55 = 1273 mL/min PRESSURE IN RENAL VESSELS The pressure in the glomerular capillaries has been measured directly in rats and has been found to be considerably lower than predicted on the basis of indirect measurements. When the mean systemic arterial pressure is 100 mm Hg, the glomerular capillary pressure is about 45 mm Hg. The pressure drop across the glomerulus is only 1 to 3 mm Hg, but a further drop occurs in the efferent arteriole so that the pressure in the peritubular capillaries is about 8 mm Hg. The pressure in the renal vein is about 4 mm Hg. Pressure gradients are similar in squirrel mon-keys and presumably in humans, with a glomerular capillary pressure that is about 40% of systemic arterial pressure. REGULATION OF THE RENAL BLOOD FLOW Norepinephrine (noradrenaline) constricts the renal vessels, with the greatest effect of injected norepinephrine being exert-ed on the interlobular arteries and the afferent arterioles. Do-pamine is made in the kidney and causes renal vasodilation and natriuresis. Angiotensin II exerts a constrictor effect on both the afferent and efferent arterioles. Prostaglandins in-crease blood flow in the renal cortex and decrease blood flow in the renal medulla. Acetylcholine also produces renal vaso-dilation. A high-protein diet raises glomerular capillary pres-sure and increases renal blood flow. FUNCTIONS OF THE RENAL NERVES Stimulation of the renal nerves increases renin secretion by a direct action of released norepinephrine on β1-adrenergic re-ceptors on the juxtaglomerular cells (see Chapter 39) and it in-creases Na+ reabsorption, probably by a direct action of norepinephrine on renal tubular cells. The proximal and distal tubules and the thick ascending limb of the loop of Henle are richly innervated. When the renal nerves are stimulated to in-creasing extents in experimental animals, the first response is an increase in the sensitivity of the juxtaglomerular cells (Table 38–2), followed by increased renin secretion, then in-creased Na+ reabsorption, and finally, at the highest threshold, renal vasoconstriction with decreased glomerular filtration and renal blood flow. It is still unsettled whether the effect on Na+ reabsorption is mediated via α- or β-adrenergic recep-tors, and it may be mediated by both. The physiologic role of the renal nerves in Na+ metabolism is also unsettled, in part because most renal functions appear to be normal in patients with transplanted kidneys, and it takes some time for trans-planted kidneys to acquire a functional innervation. Strong stimulation of the sympathetic noradrenergic nerves to the kidneys causes a marked decrease in renal blood flow. This effect is mediated by α1-adrenergic receptors and to a lesser extent by postsynaptic α2-adrenergic receptors. Some tonic dis-charge takes place in the renal nerves at rest in animals and humans. When systemic blood pressure falls, the vasoconstric-tor response produced by decreased discharge in the barorecep-tor nerves includes renal vasoconstriction. Renal blood flow is decreased during exercise and, to a lesser extent, on rising from the supine position. AUTOREGULATION OF RENAL BLOOD FLOW When the kidney is perfused at moderate pressures (90–220 mm Hg in the dog), the renal vascular resistance varies with the pressure so that renal blood flow is relatively constant (Figure 38–4). Autoregulation of this type occurs in other or-gans, and several factors contribute to it (see Chapter 33). Re-nal autoregulation is present in denervated and in isolated, perfused kidneys, but is prevented by the administration of drugs that paralyze vascular smooth muscle. It is probably produced in part by a direct contractile response to stretch of the smooth muscle of the afferent arteriole. NO may also be involved. At low perfusion pressures, angiotensin II also ap-pears to play a role by constricting the efferent arterioles, thus TABLE 38–2 Renal responses to graded renal nerve stimulation. Renal Nerve Stimulation Frequency (Hz) RSRa UNAV GFR RBFa 0.25 No effect on basal values; augments RSR mediated by nonneural stimuli. 0 0 0 0.50 Increased without chang-ing UNAV, GFR, or RBF. 0 0 0 1.0 Increased with decreased without changing GFR or RBF. ↓ 0 0 2.50 Increased with decreased UNAV, GFR, and RBF. ↓ ↓ ↓ aRSR, renin secretion rate; , urinary sodium excretion; RBF, renal blood flow. Reproduced from DiBona GF: Neural control of renal function: Cardiovascular impli-cations. Hypertension 1989;13:539. By permission of the American Heart Association. CHAPTER 38 Renal Function & Micturition 645 maintaining the glomerular filtration rate. This is believed to be the explanation of the renal failure that sometimes develops in patients with poor renal perfusion who are treated with drugs that inhibit angiotensin-converting enzyme. REGIONAL BLOOD FLOW & OXYGEN CONSUMPTION The main function of the renal cortex is filtration of large vol-umes of blood through the glomeruli, so it is not surprising that the renal cortical blood flow is relatively great and little oxygen is extracted from the blood. Cortical blood flow is about 5 mL/g of kidney tissue/min (compared with 0.5 mL/g/min in the brain), and the arteriovenous oxygen difference for the whole kidney is only 14 mL/L of blood, compared with 62 mL/L for the brain and 114 mL/L for the heart (see Table 34–1). The PO2 of the cortex is about 50 mm Hg. On the other hand, maintenance of the osmotic gradient in the medulla requires a relatively low blood flow. It is not surprising, therefore, that the blood flow is about 2.5 mL/g/min in the outer medulla and 0.6 mL/g/min in the inner medulla. However, metabolic work is being done, par-ticularly to reabsorb Na+ in the thick ascending limb of Henle, so relatively large amounts of O2 are extracted from the blood in the medulla. The PO2 of the medulla is about 15 mm Hg. This makes the medulla vulnerable to hypoxia if flow is reduced fur-ther. NO, prostaglandins, and many cardiovascular peptides in this region function in a paracrine fashion to maintain the bal-ance between low blood flow and metabolic needs. GLOMERULAR FILTRATION MEASURING GFR The glomerular filtration rate (GFR) can be measured in in-tact experimental animals and humans by measuring the ex-cretion and plasma level of a substance that is freely filtered through the glomeruli and neither secreted nor reabsorbed by the tubules. The amount of such a substance in the urine per unit of time must have been provided by filtering exactly the number of milliliters of plasma that contained this amount. Therefore, if the substance is designated by the letter X, the GFR is equal to the concentration of X in urine (UX) times the urine flow per unit of time (V • ) divided by the arterial plasma level of X (PX), or UXV • /PX. This value is called the clearance of X (CX). PX is, of course, the same in all parts of the arterial circulation, and if X is not metabolized to any extent in the tis-sues, the level of X in peripheral venous plasma can be substi-tuted for the arterial plasma level. SUBSTANCES USED TO MEASURE GFR In addition to the requirement that it be freely filtered and nei-ther reabsorbed nor secreted in the tubules, a substance suit-able for measuring the GFR should be nontoxic and not metabolized by the body. Inulin, a polymer of fructose with a molecular weight of 5200 that is found in Jerusalem artichokes (Helianthus tuberosus), meets these criteria in humans and most animals and is extensively used to measure GFR. In prac-tice, a loading dose of inulin is administered intravenously, followed by a sustaining infusion to keep the arterial plasma level constant. After the inulin has equilibrated with body flu-ids, an accurately timed urine specimen is collected and a plas-ma sample obtained halfway through the collection. Plasma and urinary inulin concentrations are determined and the clearance calculated: UIN = 35 mg/mL V • = 0.9 mL/min PIN = 0.25 mg/mL CIN = UIN V • = 35 × 0.9 PIN 0.25 CIN = 126 mL/min In dogs, cats, rabbits, and a number of other mammalian spe-cies, clearance of creatinine (CCr) can also be used to determine the precise GFR, but in primates, including humans, some crea-tinine is secreted by the tubules and some may be reabsorbed. In addition, plasma creatinine determinations are inaccurate at low creatinine levels because the method for determining creatinine measures small amounts of other plasma constituents. In spite of this, the clearance of endogenous creatinine is frequently mea-sured in patients. The values agree quite well with the GFR val-ues measured with inulin because, although the value for UCrV • is high as a result of tubular secretion, the value for PCr is also high as a result of nonspecific chromogens, and the errors thus tend to cancel. Endogenous creatinine clearance is easy to meas-ure and is a worthwhile index of renal function, but when pre-cise measurements of GFR are needed it seems unwise to rely on a method that owes what accuracy it has to compensating errors. FIGURE 38–4 Autoregulation in the kidneys. 800 600 400 200 0 70 140 210 Arterial pressure (mm Hg) Glomerular filtration Renal blood flow mL/min 646 SECTION VIII Renal Physiology NORMAL GFR The GFR in a healthy person of average size is approximately 125 mL/min. Its magnitude correlates fairly well with surface area, but values in women are 10% lower than those in men even after correction for surface area. A rate of 125 mL/min is 7.5 L/h, or 180 L/d, whereas the normal urine volume is about 1 L/d. Thus, 99% or more of the filtrate is normally reab-sorbed. At the rate of 125 mL/min, in 1 day the kidneys filter an amount of fluid equal to 4 times the total body water, 15 times the ECF volume, and 60 times the plasma volume. CONTROL OF GFR The factors governing filtration across the glomerular capillaries are the same as those governing filtration across all other capillar-ies (see Chapter 32), that is, the size of the capillary bed, the per-meability of the capillaries, and the hydrostatic and osmotic pressure gradients across the capillary wall. For each nephron: GFR = Kf [(PGC – PT) – (πGC – πT)] Kf, the glomerular ultrafiltration coefficient, is the product of the glomerular capillary wall hydraulic conductivity (ie, its permeability) and the effective filtration surface area. PGC is the mean hydrostatic pressure in the glomerular capillaries, PT the mean hydrostatic pressure in the tubule (Bowman’s space), πGC the oncotic pressure of the plasma in the glomer-ular capillaries, and πT the oncotic pressure of the filtrate in the tubule (Bowman’s space). PERMEABILITY The permeability of the glomerular capillaries is about 50 times that of the capillaries in skeletal muscle. Neutral substances with effective molecular diameters of less than 4 nm are freely fil-tered, and the filtration of neutral substances with diameters of more than 8 nm approaches zero (Figure 38–5). Between these values, filtration is inversely proportionate to diameter. Howev-er, sialoproteins in the glomerular capillary wall are negatively charged, and studies with anionically charged and cationically charged dextrans indicate that the negative charges repel nega-tively charged substances in blood, with the result that filtration of anionic substances 4 nm in diameter is less than half that of neutral substances of the same size. This probably explains why albumin, with an effective molecular diameter of approximately 7 nm, normally has a glomerular concentration only 0.2% of its plasma concentration rather than the higher concentration that would be expected on the basis of diameter alone; circulating al-bumin is negatively charged. Filtration of cationic substances is greater than that of neutral substances. The amount of protein in the urine is normally less than 100 mg/d, and most of this is not filtered but comes from shed tubular cells. The presence of significant amounts of albumin in the urine is called albuminuria. In nephritis, the negative charges in the glomerular wall are dissipated, and albumin-uria can occur for this reason without an increase in the size of the “pores” in the membrane. SIZE OF THE CAPILLARY BED Kf can be altered by the mesangial cells, with contraction of these cells producing a decrease in Kf that is largely due to a reduction in the area available for filtration. Contraction of points where the capillary loops bifurcate probably shifts flow away from some of the loops, and elsewhere, contracted mesangial cells dis-tort and encroach on the capillary lumen. Agents that have been shown to affect the mesangial cells are listed in Table 38–3. An-giotensin II is an important regulator of mesangial contraction, and there are angiotensin II receptors in the glomeruli. In addi-tion, some evidence suggests that mesangial cells make renin. FIGURE 38–5 Effect of electric charge on the fractional clearance of dextran molecules of various sizes in rats. The nega-tive charges in the glomerular membrane retard the passage of negative-ly charged molecules (anionic dextran) and facilitate the passage of positively charged molecules (cationic dextran). (Reproduced with permission from Brenner BM, Beeuwkes R: The renal circulations. Hosp Pract [July] 1978;13:35.) TABLE 38–3 Agents causing contraction or relaxation of mesangial cells. Contraction Relaxation Endothelins ANP Angiotensin II Dopamine Vasopressin PGE2 Norepinephrine cAMP Platelet-activating factor Platelet-derived growth factor Thromboxane A2 PGF2 Leukotrienes C4 and D4 Histamine 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 4.0 6.0 8.0 Anionic Neutral Cationic Effective molecular diameter (nm) Fractional clearance CHAPTER 38 Renal Function & Micturition 647 HYDROSTATIC & OSMOTIC PRESSURE The pressure in the glomerular capillaries is higher than that in other capillary beds because the afferent arterioles are short, straight branches of the interlobular arteries. Furthermore, the vessels “downstream” from the glomeruli, the efferent arteri-oles, have a relatively high resistance. The capillary hydrostatic pressure is opposed by the hydrostatic pressure in Bowman’s capsule. It is also opposed by the oncotic pressure gradient across the glomerular capillaries (πGC – πT). πT is normally negligible, and the gradient is essentially equal to the oncotic pressure of the plasma proteins. The actual pressures in one strain of rats are shown in Figure 38–6. The net filtration pressure (PUF) is 15 mm Hg at the afferent end of the glomerular capillaries, but it falls to zero— that is, filtration equilibrium is reached—proximal to the effer-ent end of the glomerular capillaries. This is because fluid leaves the plasma and the oncotic pressure rises as blood passes through the glomerular capillaries. The calculated change in Δπ along an idealized glomerular capillary is also shown in Figure 38–6. It is apparent that portions of the glomerular cap-illaries do not normally contribute to the formation of the glo-merular ultrafiltrate; that is, exchange across the glomerular capillaries is flow-limited rather than diffusion-limited. It is also apparent that a decrease in the rate of rise of the Δ curve produced by an increase in renal plasma flow would increase filtration because it would increase the distance along the capil-lary in which filtration was taking place. There is considerable species variation in whether filtration equilibrium is reached, and some uncertainties are inherent in the measurement of Kf. It is uncertain whether filtration equi-librium is reached in humans. CHANGES IN GFR Variations in the factors discussed in the preceding para-graphs and listed in Table 38–4 have predictable effects on the GFR. Changes in renal vascular resistance as a result of auto-regulation tend to stabilize filtration pressure, but when the mean systemic arterial pressure drops below the autoregulato-ry range (Figure 38–4), GFR drops sharply. The GFR tends to be maintained when efferent arteriolar constriction is greater than afferent constriction, but either type of constriction de-creases blood flow to the tubules. FILTRATION FRACTION The ratio of the GFR to the RPF, the filtration fraction, is nor-mally 0.16 to 0.20. The GFR varies less than the RPF. When there is a fall in systemic blood pressure, the GFR falls less than the RPF because of efferent arteriolar constriction, and conse-quently the filtration fraction rises. TUBULAR FUNCTION GENERAL CONSIDERATIONS The amount of any substance (X) that is filtered is the product of the GFR and the plasma level of the substance (ClnPX). The tubular cells may add more of the substance to the filtrate (tu-bular secretion), may remove some or all of the substance from the filtrate (tubular reabsorption), or may do both. The amount of the substance excreted per unit of time (UXV • ) FIGURE 38–6 Hydrostatic pressure (PGC) and osmotic pressure (πGC) in a glomerular capillary in the rat. PT, pressure in Bowman’s capsule; PUF, net filtration pressure. πT is normally negligi-ble, so Δπ = πGC. ΔP = PGC – PT. (Reproduced with permission from Mercer PF, Maddox DA, Brenner BM: Current concepts of sodium chloride and water transport by the mammalian nephron. West J Med 1974;120:33.) (mm Hg) Afferent end Efferent end 45 10 20 15 45 10 35 0 PUF = PGC − PT − πGC PGC PT πGC PUF Dimensionless distance along idealized glomerular capillary Pressure (mm Hg) 60 40 20 0 0 1 ΔP Δπ TABLE 38–4 Factors affecting the GFR. Changes in renal blood flow Changes in glomerular capillary hydrostatic pressure Changes in systemic blood pressure Afferent or efferent arteriolar constriction Changes in hydrostatic pressure in Bowman’s capsule Ureteral obstruction Edema of kidney inside tight renal capsule Changes in concentration of plasma proteins: dehydration, hypoproteinemia, etc (minor factors) Changes in Kf Changes in glomerular capillary permeability Changes in effective filtration surface area 648 SECTION VIII Renal Physiology equals the amount filtered plus the net amount transferred by the tubules. This latter quantity is conveniently indicated by the symbol TX (Figure 38–7). The clearance of the substance equals the GFR if there is no net tubular secretion or reabsorp-tion, exceeds the GFR if there is net tubular secretion, and is less than the GFR if there is net tubular reabsorption. Much of our knowledge about glomerular filtration and tubular function has been obtained by using micropuncture techniques. Micropipettes can be inserted into the tubules of the living kidney and the composition of aspirated tubular fluid determined by the use of microchemical techniques. In addition, two pipettes can be inserted in a tubule and the tubule perfused in vivo. Alternatively, isolated perfused seg-ments of tubules can be studied in vitro, and tubular cells can be grown and studied in culture. MECHANISMS OF TUBULAR REABSORPTION & SECRETION Small proteins and some peptide hormones are reabsorbed in the proximal tubules by endocytosis. Other substances are se-creted or reabsorbed in the tubules by passive diffusion between cells and through cells by facilitated diffusion down chemical or electrical gradients or active transport against such gradients. Movement is by way of ion channels, exchangers, cotransport-ers, and pumps. Many of these have now been cloned, and their regulation is being studied. It is important to note that the pumps and other units in the luminal membrane are different from those in the basolateral membrane. It is this different distribution that makes possible net movement of solutes across the epithelia. Like transport systems elsewhere, renal active transport sys-tems have a maximal rate, or transport maximum (Tm), at which they can transport a particular solute. Thus, the amount of a particular solute transported is proportional to the amount present up to the Tm for the solute, but at higher concentrations, the transport mechanism is saturated and there is no apprecia-ble increment in the amount transported. However, the Tms for some systems are high, and it is difficult to saturate them. It should also be noted that the tubular epithelium, like that of the small intestine, is a leaky epithelium in that the tight junctions between cells permit the passage of some water and electrolytes. The degree to which leakage by this paracellular pathway contributes to the net flux of fluid and solute into and out of the tubules is controversial since it is difficult to measure, but current evidence seems to suggest that it is a significant fac-tor in the proximal tubule. One indication of this is that para-cellin-1, a protein localized to tight junctions, is related to Mg2+ reabsorption, and a loss-of-function mutation of its gene causes severe Mg2+ and Ca2+ loss in the urine. The effects of tubular reabsorption and secretion on sub-stances of major physiologic interest are summarized in Table 38–5. Na+ REABSORPTION The reabsorption of Na+ and Cl– plays a major role in body electrolyte and water homeostasis. In addition, Na+ transport is coupled to the movement of H+, glucose, amino acids, or-ganic acids, phosphate, and other electrolytes and substances across the tubule walls. The principal cotransporters and ex-changers in the various parts of the nephron are listed in Table 38–6. In the proximal tubules, the thick portion of the ascend-ing limb of the loop of Henle, the distal tubules, and the col-lecting ducts, Na+ moves by cotransport or exchange from the tubular lumen into the tubular epithelial cells down its con-centration and electrical gradients, and is then actively pumped from these cells into the interstitial space. Na+ is pumped into the interstitium by Na, K ATPase in the basolat-eral membrane. Thus, Na+ is actively transported out of all parts of the renal tubule except the thin portions of the loop of Henle. The operation of the ubiquitous Na+ pump is consid-ered in detail in Chapter 2. It extrudes three Na+ in exchange for two K+ that are pumped into the cell. The tubular cells along the nephron are connected by tight junctions at their luminal edges, but there is space between the cells along the rest of their lateral borders. Much of the Na+ is actively transported into these extensions of the inter-stitial space, the lateral intercellular spaces (Figure 38–8). Normally about 60% of the filtered Na+ is reabsorbed in the proximal tubule, primarily by Na–H exchange. Another 30% is absorbed via the Na–2Cl–K cotransporter in the thick ascending limb of the loop of Henle, and about 7% is absorbed by Na–Cl cotransporter in the distal convoluted tubule. The remainder of the filtered Na+, about 3%, is absorbed via the ENaC channels in the collecting ducts, and this is the portion that is regulated by aldosterone in the pro-duction of homeostatic adjustments in Na+ balance. FIGURE 38–7 Tubular function. For explanation of symbols, see text. GFR x PX + TX = UXV ˙ ˙ Filtered = GFR x PX Re-absorbed Excreted = UXV Secreted TX = Positive GFR x PX < UXV Example: PAH ˙ TX = Negative GFR x PX > UXV Example: Glucose ˙ ˙ TX = 0 GFR x PX = UXV Example: Inulin CHAPTER 38 Renal Function & Micturition 649 TABLE 38–5 Renal handling of various plasma constituents in a normal adult human on an average diet. Per 24 Hours Substance Filtered Reabsorbed Secreted Excreted Percentage Reabsorbed Na+ (mEq) 26,000 25,850 150 99.4 K+ (mEq) 600 560a 502 90 93.3 Cl– (mEq) 18,000 17,850 150 99.2 HCO3 – (mEq) 4,900 4,900 0 100 Urea (mmol) 870 460b 410 53 Creatinine (mmol) 12 1c 1c 12 Uric acid (mmol) 50 49 4 5 98 Glucose (mmol) 800 800 0 100 Total solute (mOsm) 54,000 53,400 100 700 98.9 Water (mL) 180,000 179,000 1000 99.4 aK+ is both reabsorbed and secreted. bUrea moves into as well as out of some portions of the nephron. cVariable secretion and probable reabsorption of creatinine in humans. TABLE 38–6 Transport proteins involved in the movement of Na+ and Cl– across the apical membranes of renal tubular cells.a Site Apical Transporter Function Proximal tubule Na/glucose CT Na+ uptake, glucose uptake Na+/Pi CT Na+ uptake, Pi uptake Na+ amino acid CT Na+ uptake, amino acid uptake Na/lactate CT Na+ uptake, lactate uptake Na/H exchanger Na+ uptake, H+ extrusion Cl/base exchanger Cl– uptake Thick ascending limb Na–K–2Cl CT Na+ uptake, Cl– uptake, K+ uptake Na/H exchanger Na+ uptake, H+ extrusion K+ channels K+ extrusion (recycling) Distal convolut-ed tubule NaCl CT Na+ uptake, Cl– uptake Collecting duct Na+ channel (ENaC) Na+ uptake aUptake indicates movement from tubular lumen to cell interior, extrusion is move-ment from cell interior to tubular lumen. CT, cotransporter; Pi, inorganic phosphate. Modified with permission from Schnermann JB, Sayegh EI: Kidney Physiology. Lip-pincott-Raven, 1998. FIGURE 38–8 Mechanism of Na+ reabsorption in the proximal tubule. Na+ moves out of the tubular lumen by cotransport and exchange mechanism through the apical membrane of the tubule (dashed line). The Na+ is then actively transported into the interstitial fluid by Na, K ATPase in the basolateral membrane (solid lines). K+ en-ters the interstitial fluid via K+ channels. A small amount of Na+, other solutes, and H2O re-enter the tubular lumen by passive transport through the tight junctions (dotted lines). Na+ Na+ Na+, etc Na+ Na+ K+ K+ K+ Tubular lumen Tight junction Lateral intercellular space Interstitial fluid Na+ 650 SECTION VIII Renal Physiology GLUCOSE REABSORPTION Glucose, amino acids, and bicarbonate are reabsorbed along with Na+ in the early portion of the proximal tubule (Figure 38–9). Farther along the tubule, Na+ is reabsorbed with Cl–. Glucose is typical of substances removed from the urine by secondary active transport. It is filtered at a rate of approxi-mately 100 mg/min (80 mg/dL of plasma × 125 mL/min). Es-sentially all of the glucose is reabsorbed, and no more than a few milligrams appear in the urine per 24 h. The amount reab-sorbed is proportional to the amount filtered and hence to the plasma glucose level (PG) times the GFR up to the transport maximum (TmG). When the TmG is exceeded, the amount of glucose in the urine rises (Figure 38–10). The TmG is about 375 mg/min in men and 300 mg/min in women. The renal threshold for glucose is the plasma level at which the glucose first appears in the urine in more than the normal minute amounts. One would predict that the renal threshold would be about 300 mg/dL, that is, 375 mg/min (TmG) divided by 125 mL/min (GFR). However, the actual renal threshold is about 200 mg/dL of arterial plasma, which corre-sponds to a venous level of about 180 mg/dL. Figure 38–10 shows why the actual renal threshold is less than the predicted threshold. The "ideal" curve shown in this diagram would be obtained if the TmG in all the tubules was identical and if all the glucose were removed from each tubule when the amount filtered was below the TmG. This is not the case, and in humans, for example, the actual curve is rounded and devi-ates considerably from the “ideal” curve. This deviation is called splay. The magnitude of the splay is inversely propor-tionate to the avidity with which the transport mechanism binds the substance it transports. GLUCOSE TRANSPORT MECHANISM Glucose reabsorption in the kidneys is similar to glucose reab-sorption in the intestine (see Chapter 27). Glucose and Na+ bind to the sodium-dependent glucose transporter (SGLT) 2 in the apical membrane, and glucose is carried into the cell as Na+ moves down its electrical and chemical gradient. The Na+ is then pumped out of the cell into the interstitium, and the glu-cose is transported by glucose transporter (GLUT) 2 into the in-terstitial fluid. At least in the rat, there is some transport by SGLT 1 and GLUT 1 as well. SGLT 2 specifically binds the d isomer of glucose, and the rate of transport of d-glucose is many times greater than that of l-glucose. Glucose transport in the kidneys is inhibited, as it is in the intestine, by the plant glucoside phlorhizin, which competes with d-glucose for binding to the carrier. ADDITIONAL EXAMPLES OF SECONDARY ACTIVE TRANSPORT Like glucose reabsorption, amino acid reabsorption is most marked in the early portion of the proximal convoluted tu-bule. Absorption in this location resembles absorption in the intestine (see Chapter 27). The main carriers in the apical membrane cotransport Na+, whereas the carriers in the baso-lateral membranes are not Na+-dependent. Na+ is pumped FIGURE 38–9 Reabsorption of various solutes in the proximal tubule. TF/P, tubular fluid:plasma concentration ratio. (Courtesy of FC Rector Jr.) 2.6 2.4 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0 25 30 75 100 Glucose Amino acids % Proximal tubule length HCO3 − osm Inulin Cl− K+ Na+ TF P FIGURE 38–10 Top: Relation between the plasma level (P) and excretion (UV) of glucose and inulin. Bottom: Relation between the plasma glucose level (PG) and amount of glucose reabsorbed (TG). UV . Inulin Glucose P Splay "Ideal" Actual Plasma glucose (PG) TmG Glucose reabsorbed (TG) CHAPTER 38 Renal Function & Micturition 651 out of the cells by Na, K ATPase and the amino acids leave by passive or facilitated diffusion to the interstitial fluid. Some Cl– is reabsorbed with Na+ and K+ in the thick ascending limb of the loop of Henle. In addition, two mem-bers of the family of Cl channels have been identified in the kidney. Mutations in the gene for one of the renal channels is associated with Ca2+-containing kidney stones and hypercal-ciuria (Dent disease), but how tubular transport of Ca2+ and Cl– are linked is still unsettled. PAH TRANSPORT The dynamics of PAH transport illustrate the operation of the active transport mechanisms that secrete substances into the tubular fluid (see Clinical Box 38–1). The filtered load of PAH is a linear function of the plasma level, but PAH secretion in-creases as PPAH rises only until a maximal secretion rate (TmPAH) is reached (Figure 38–11). When PPAH is low, CPAH is high; but as PPAH rises above TmPAH, CPAH falls progres-sively. It eventually approaches the clearance of inulin (CIn) (Figure 38–12), because the amount of PAH secreted becomes a smaller and smaller fraction of the total amount excreted. Conversely, the clearance of glucose is essentially zero at PG le-vels below the renal threshold; but above the threshold, CG ris-es to approach CIn as PG is raised. The use of CPAH to measure ERPF is discussed above. TUBULOGLOMERULAR FEEDBACK & GLOMERULOTUBULAR BALANCE Signals from the renal tubule in each nephron feed back to af-fect filtration in its glomerulus. As the rate of flow through the ascending limb of the loop of Henle and first part of the distal tubule increases, glomerular filtration in the same nephron decreases, and, conversely, a decrease in flow in-creases the GFR (Figure 38–13). This process, which is called FIGURE 38–11 Relation between plasma levels (P) and excretion (UV) of PAH and inulin. FIGURE 38–12 Clearance of inulin, glucose, and PAH at various plasma levels of each substance in humans. UV . Inulin P PAH Splay Clearance (mL/min) Inulin Glucose Plasma level (P) 20 40 60 80 200 400 600 Glucose, mg/dL PAH, mg/dL PAH 600 500 400 300 200 100 0 CLINICAL BOX 38–1 Other Substances Secreted by the Tubules Derivatives of hippuric acid in addition to PAH, phenol red and other sulfonphthalein dyes, penicillin, and a variety of iodinated dyes are actively secreted into the tubular fluid. Substances that are normally produced in the body and se-creted by the tubules include various ethereal sulfates, ster-oid and other glucuronides, and 5-hydroxyindoleacetic acid, the principal metabolite of serotonin. FIGURE 38–13 Mechanisms of glomerulotubular balance and tubuloglomerular feedback. Renal arteriolar pressure Glomerular capillary pressure GFR Solute reabsorption in proximal tubule Solute reabsorption in thick ascending limb Salt and fluid delivery to the distal tubule Tubulo-glomerular feedback Glomerulo-tubular balance 652 SECTION VIII Renal Physiology tubuloglomerular feedback, tends to maintain the constancy of the load delivered to the distal tubule. The sensor for this response is the macula densa. The amount of fluid entering the distal tubule at the end of the thick ascending limb of the loop of Henle depends on the amount of Na+ and Cl– in it. The Na+ and Cl– enter the mac-ula densa cells via the Na–K–2Cl cotransporter in their apical membranes. The increased Na+ causes increased Na, K ATPase activity and the resultant increased ATP hydrolysis causes more adenosine to be formed. Presumably, adenosine is secreted from the basal membrane of the cells. It acts via adenosine A1 receptors on the macula densa cells to increase their release of Ca2+ to the vascular smooth muscle in the afferent arterioles. This causes afferent vasoconstriction and a resultant decrease in GFR. Presumably, a similar mechanism generates a signal that decreases renin secretion by the adja-cent juxtaglomerular cells in the afferent arteriole (see Chap-ter 39), but this remains unsettled. Conversely, an increase in GFR causes an increase in the reabsorption of solutes, and consequently of water, primarily in the proximal tubule, so that in general the percentage of the solute reabsorbed is held constant. This process is called glomerulotubular balance, and it is particularly prominent for Na+. The change in Na+ reabsorption occurs within sec-onds after a change in filtration, so it seems unlikely that an extrarenal humoral factor is involved. One factor is the oncotic pressure in the peritubular capillaries. When the GFR is high, there is a relatively large increase in the oncotic pres-sure of the plasma leaving the glomeruli via the efferent arte-rioles and hence in their capillary branches. This increases the reabsorption of Na+ from the tubule. However, other as yet unidentified intrarenal mechanisms are also involved. WATER TRANSPORT Normally, 180 L of fluid is filtered through the glomeruli each day, while the average daily urine volume is about 1 L. The same load of solute can be excreted per 24 h in a urine volume of 500 mL with a concentration of 1400 mOsm/kg or in a volume of 23.3 L with a concentration of 30 mOsm/kg (Table 38–7). These figures demonstrate two important facts: First, at least 87% of the filtered water is reabsorbed, even when the urine volume is 23 L; and second, the reabsorption of the remainder of the fil-tered water can be varied without affecting total solute excre-tion. Therefore, when the urine is concentrated, water is retained in excess of solute; and when it is dilute, water is lost from the body in excess of solute. Both facts have great impor-tance in the regulation of the osmolality of the body fluids. A key regulator of water output is vasopressin acting on the collecting ducts. AQUAPORINS Rapid diffusion of water across cell membranes depends on the presence of water channels, integral membrane proteins called aquaporins. To date, 13 aquaporins have been cloned; however, only 4 aquaporins (aquaporin-1, aquaporin-2, aqua-porin-3, and aquaporin-4) play a key role in the kidney. The roles played by aquaporin-1 and aquaporin-2 in renal water transport are discussed below. PROXIMAL TUBULE Active transport of many substances occurs from the fluid in the proximal tubule, but micropuncture studies have shown that the fluid remains essentially iso-osmotic to the end of the prox-imal tubule (Figure 38–9). Aquaporin-1 is localized to both the basolateral and apical membrane of the proximal tubules and its presence allows water to move rapidly out of the tubule along the osmotic gradients set up by active transport of solutes, and isotonicity is maintained. Because the ratio of the concentration in tubular fluid to the concentration in plasma (TF/P) of the nonreabsorbable substance inulin is 2.5 to 3.3 at the end of the proximal tubule, it follows that 60–70% of the filtered solute and 60–70% of the filtered water have been removed by the time the filtrate reaches this point (Figure 38–14). When aquaporin-1 was knocked out in mice, proximal tubu-lar water permeability was reduced by 80%. When the mice were subjected to dehydration, their urine osmolality did not increase (<700 mOsm/kg), even though other renal aquaporins were present. In humans with mutations that eliminate aqua-porin-1 activity, the defect in water metabolism is not as severe, though their response to dehydration is defective. TABLE 38–7 Alterations in water metabolism produced by vasopressin in humans. In each case, the osmotic load excreted is 700 mOsm/d. GFR (mL/min) Percentage of Filtered Water Reabsorbed Urine V olume (L/d) Urine Concentration (mOsm/kg H2O) Gain or Loss of Water in Excess of Solute (L/d) Urine isotonic to plasma 125 98.7 2.4 290 . . . Vasopressin (maximal antidiuresis) 125 99.7 0.5 1400 1.9 gain No vasopressin (“complete” diabetes insipidus) 125 87.1 23.3 30 20.9 loss CHAPTER 38 Renal Function & Micturition 653 LOOP OF HENLE As noted above, the loops of Henle of the juxtamedullary nephrons dip deeply into the medullary pyramids before draining into the distal convoluted tubules in the cortex, and all the collecting ducts descend back through the medullary pyramids to drain at the tips of the pyramids into the renal pel-vis. There is a graded increase in the osmolality of the intersti-tium of the pyramids in humans: The osmolality at the tips of the papillae can reach about 1200 mOsm/kg of H2O, approxi-mately four times that of plasma. The descending limb of the loop of Henle is permeable to water, due to the presence of aquaporin-1 in both the apical and basolateral membrane, but the ascending limb is impermeable to water (Table 38–8). Na+, K+, and Cl– are cotransported out of the thick segment of the ascending limb. Therefore, the fluid in the descending limb of the loop of Henle becomes hypertonic as water moves out of the tubule into the hypertonic interstitium. In the ascending limb it becomes more dilute because of the movement of Na+ and Cl– out of the tubular lumen, and when fluid reaches the top of the ascending limb (called the diluting segment) it is now hypotonic to plasma. In passing through the descending loop of Henle, another 15% of the filtered water is removed, so approximately 20% of the filtered water enters the distal tu-bule, and the TF/P of inulin at this point is about 5. In the thick ascending limb, a carrier cotransports one Na+, one K+, and 2Cl– from the tubular lumen into the tubular cells. This is another example of secondary active transport; the Na+ is actively transported from the cells into the intersti-tium by Na, K ATPase in the basolateral membranes of the cells, keeping the intracellular Na+ low. The Na–K–2Cl trans-porter has 12 transmembrane domains with intracellular amino and carboxyl terminals. It is a member of a family of transporters found in many other locations, including salivary glands, the gastrointestinal tract, and the airways. The K+ diffuses back into the tubular lumen and back into the interstitium via ROMK and other K+ channels. The Cl– moves into the interstitium via ClC-Kb channels (Figure 38–15). DISTAL TUBULE The distal tubule, particularly its first part, is in effect an exten-sion of the thick segment of the ascending limb. It is relatively impermeable to water, and continued removal of the solute in excess of solvent further dilutes the tubular fluid. COLLECTING DUCTS The collecting ducts have two portions: a cortical portion and a medullary portion. The changes in osmolality and volume in the collecting ducts depend on the amount of vasopressin act-ing on the ducts. This antidiuretic hormone from the posteri-or pituitary gland increases the permeability of the collecting ducts to water. The key to the action of vasopressin on the col-lecting ducts is aquaporin-2. Unlike the other aquaporins, this aquaporin is stored in vesicles in the cytoplasm of principal cells. Vasopressin causes rapid insertion of these vesicles into the apical membrane of cells. The effect is mediated via the FIGURE 38–14 Changes in the percentage of the filtered amount of substances remaining in the tubular fluid along the length of the nephron in the presence of vasopressin. (Modified from Sullivan LP, Grantham JJ: Physiology of the Kidney, 2nd ed. Lea & Febiger, 1982.) 120 100 80 60 40 20 0 Glucose Fraction remaining in tubular fluid Osmoles Water Na+ Creatinine Inulin Urea Proximal tubule Loop of Henle Distal tubule Collecting tubule TABLE 38–8 Permeability and transport in various segments of the nephron.a Permeability H2O Urea NaCl Active Transport of Na+ Loop of Henle Thin descending limb 4+ + ± 0 Thin ascending limb 0 + 4+ 0 Thick ascending limb 0 ± ± 4+ Distal convoluted tubule ± ± ± 3+ Collecting tubule Cortical portion 3+ 0 ± 2+ Outer medullary portion 3+ 0 ± 1+ Inner medullary portion 3+ 3+ ± 1+ aData are based on studies of rabbit and human kidneys. Values indicated by asterisks are in the presence of vasopressin. These values are 1+ in the absence of vasopressin. Modified and reproduced with permission from Kokko JP: Renal concentrating and diluting mechanisms. Hosp Pract [Feb] 1979;110:14. 654 SECTION VIII Renal Physiology vasopressin V2 receptor, cyclic adenosine 5-monophosphate (cAMP) and protein kinase A. Cytoskeletal elements are in-volved, including microtubule-based motor proteins (dynein and dynactin) as well as actin filament-binding proteins such as myosin-1. In the presence of enough vasopressin to produce maximal antidiuresis, water moves out of the hypotonic fluid entering the cortical collecting ducts into the interstitium of the cortex, and the tubular fluid becomes isotonic. In this fashion, as much as 10% of the filtered water is removed. The isotonic fluid then enters the medullary collecting ducts with a TF/P inulin of about 20. An additional 4.7% or more of the filtrate is reabsorbed into the hypertonic interstitium of the medulla, producing a concentrated urine with a TF/P inulin of over 300. In humans, the osmolality of urine may reach 1400 mOsm/kg of H2O, almost five times the osmolality of plasma, with a total of 99.7% of the filtered water being reabsorbed (Table 38–7). In other species, the ability to concentrate urine is even greater. Maximal urine osmolality is about 2500 mOsm/kg in dogs, about 3200 mOsm/kg in laboratory rats, and as high as 5000 mOsm/kg in certain desert rodents. When vasopressin is absent, the collecting duct epithelium is relatively impermeable to water. The fluid therefore remains hypotonic, and large amounts flow into the renal pelvis. In humans, the urine osmolality may be as low as 30 mOsm/kg of H2O. The impermeability of the distal portions of the nephron is not absolute; along with the salt that is pumped out of the collecting duct fluid, about 2% of the filtered water is reabsorbed in the absence of vasopressin. However, as much as 13% of the filtered water may be excreted, and urine flow may reach 15 mL/min or more. THE COUNTERCURRENT MECHANISM The concentrating mechanism depends upon the mainte-nance of a gradient of increasing osmolality along the medul-lary pyramids. This gradient is produced by the operation of the loops of Henle as countercurrent multipliers and main-tained by the operation of the vasa recta as countercurrent ex-changers. A countercurrent system is a system in which the inflow runs parallel to, counter to, and in close proximity to the outflow for some distance. This occurs for both the loops of Henle and the vasa recta in the renal medulla (Figure 38–3). The operation of each loop of Henle as a countercurrent mul-tiplier depends on the high permeability of the thin descending limb to water (via aquaporin-1), the active transport of Na+ and Cl– out of the thick ascending limb, and the inflow of tubular fluid from the proximal tubule, with outflow into the distal tubule. The process can be explained using hypothetical steps FIGURE 38–15 NaCl transport in the thick ascending limb of the loop of Henle. The Na–K–2Cl cotransporter moves these ions into the tubular cell by secondary active transport. Na+ is transported out of the cell into the interstitium by Na, K ATPase in the basolateral mem-brane of the cell. Cl– exits in basolateral ClC-Kb Cl– channels. Barttin, a protein in the cell membrane, is essential for normal ClC-Kb function. K+ moves from the cell to the interstitium and the tubular lumen by ROMK and other K+ channels (see Clinical Box 38–2). Interstitial fluid Tubular lumen K+ Na+ Renal tubule cell K+ Cl− Barttin Na+ 2Cl− K+ ROMK K+ ROMK K+ K+ CLINICAL BOX 38–2 Genetic Mutations in Renal Transporters Mutations of individual genes for many renal sodium trans-porters and channels cause specific syndromes such as Bartter syndrome, Liddle syndrome, and Dent disease. A large number of mutations have been described. Bartter syndrome is a rare but interesting condition that is due to defective transport in the thick ascending limb. It is characterized by chronic Na+ loss in the urine, with resultant hypovolemia causing stimulation of renin and aldosterone secretion without hypertension, plus hy-perkalemia and alkalosis. The condition can be caused by loss-of-function mutations in the gene for any of four key proteins: the Na–K–2Cl cotransporter, the ROMK K+ chan-nel, the ClC–Kb Cl– channel, or barttin, a recently described integral membrane protein that is necessary for the normal function of ClC–Kb Cl– channels. The stria vascularis in the inner ear is responsible for maintaining the high K+ concentration in the scala media that is essential for normal hearing. It contains both ClC–Kb and ClC–Ka Cl– channels. Bartter syndrome associated with mutated ClC–Kb channels is not associated with deafness because the Clc–Ka channels can carry the load. However, both types of Cl– channels are barttin-dependent, so pa-tients with Bartter syndrome due to mutated barttin are also deaf. Another interesting example involves the proteins poly-cystin-1 (PKD-1) and polycystin-2 (PKD-2). PKD-1 appears to be a Ca2+ receptor that activates a nonspecific ion channel associated with PKD-2. The normal function of this appar-ent ion channel is unknown, but both proteins are abnor-mal in autosomal dominant polycystic kidney disease, in which the renal parenchyma is progressively replaced by fluid-filled cysts until there is complete renal failure. CHAPTER 38 Renal Function & Micturition 655 leading to the normal equilibrium condition, although the steps do not occur in vivo. It is also important to remember that the equilibrium is maintained unless the osmotic gradient is washed out. These steps are summarized in Figure 38–16 for a cortical nephron with no thin ascending limb. Assume first a condition in which osmolality is 300 mOsm/kg of H2O throughout the descending and ascending limbs and the medullary interstitium (Figure 38–16A). Assume in addition that the pumps in the thick ascending limb can pump 100 mOsm/kg of Na+ and Cl– from the tubular fluid to the interstitium, increasing interstitial osmolality to 400 mOsm/kg of H2O. Water then moves out of the thin descending limb, and its contents equilibrate with the interstitium (Figure 38–16B). However, fluid containing 300 mOsm/kg of H2O is continuously entering this limb from the proximal tubule (Figure 38–16C), so the gradient against which the Na+ and Cl– are pumped is reduced and more enters the interstitium (Figure 38–16D). Meanwhile, hypotonic fluid flows into the distal tubule, and isotonic and subsequently hypertonic fluid flows into the ascending thick limb. The process keeps repeating, and the final result is a gradient of osmolality from the top to the bottom of the loop. In juxtamedullary nephrons with longer loops and thin ascending limbs, the osmotic gradient is spread over a greater distance and the osmolality at the tip of the loop is greater. This is because the thin ascending limb is relatively impermeable to water but permeable to Na+ and Cl–. Therefore, Na+ and Cl– move down their concentration gradients into the interstitium, and there is additional passive countercurrent multiplication. The greater the length of the loop of Henle, the greater the osmolality that can be reached at the tip of the medulla. The osmotic gradient in the medullary pyramids would not last long if the Na+ and urea in the interstitial spaces were removed by the circulation. These solutes remain in the pyra-mids primarily because the vasa recta operate as countercurrent exchangers (Figure 38–17). The solutes diffuse out of the ves-sels conducting blood toward the cortex and into the vessels descending into the pyramid. Conversely, water diffuses out of the descending vessels and into the fenestrated ascending ves-sels. Therefore, the solutes tend to recirculate in the medulla and water tends to bypass it, so that hypertonicity is main-tained. The water removed from the collecting ducts in the pyr-amids is also removed by the vasa recta and enters the general circulation. Countercurrent exchange is a passive process; it depends on movement of water and could not maintain the osmotic gradient along the pyramids if the process of counter-current multiplication in the loops of Henle were to cease. FIGURE 38–16 Operation of the loop of Henle as a countercurrent multiplier producing a gradient of hyperosmolarity in the medullary interstitium (MI). TDL, thin descending limb; TAL, thick ascending limb. The process of generation of the gradient is illustrated as oc-curring in hypothetical steps, starting at A, where osmolality in both limbs and the interstitium is 300 mOsm/kg of water. The pumps in the thick ascending limb move Na+ and Cl– into the interstitium, increasing its osmolality to 400 mOsm/kg, and this equilibrates with the fluid in the thin descending limb. However, isotonic fluid continues to flow into the thin descending limb and hypotonic fluid out of the thick ascending limb. Con-tinued operation of the pumps makes the fluid leaving the thick ascending limb even more hypotonic, while hypertonicity accumulates at the apex of the loop. (Modified and reproduced with permission from Johnson LR [editor]: Essential Medical Physiology, Raven Press, 1992.) 312 375 375 425 425 513 513 700 300 325 325 425 425 425 425 600 300 300 300 300 300 300 300 300 300 300 300 300 300 300 300 300 300 300 300 300 300 300 300 300 A 400 400 400 400 400 400 400 400 400 400 400 400 400 400 400 400 200 200 200 200 200 200 200 200 B TDL TAL MI 300 300 300 300 400 400 400 400 300 300 300 300 400 400 400 400 200 200 200 200 400 400 400 400 C 350 350 350 350 500 500 500 500 350 350 350 350 500 500 500 500 150 150 150 150 300 300 300 300 D 300 300 350 350 350 350 500 500 300 300 350 350 350 350 500 500 150 150 300 300 300 300 500 500 E 325 325 425 425 425 425 600 600 125 125 225 225 225 225 400 400 F 300 325 325 425 425 425 425 600 125 225 225 225 225 400 400 600 G 312 375 375 425 425 513 513 700 112 175 175 225 225 313 313 500 H 325 325 425 425 425 425 600 600 656 SECTION VIII Renal Physiology It is worth noting that there is a very large osmotic gradient in the loop of Henle and, in the presence of vasopressin, in the collecting ducts. It is the countercurrent system that makes this gradient possible by spreading it along a system of tubules 1 cm or more in length, rather than across a single layer of cells that is only a few micrometers thick. There are other examples of the operation of countercurrent exchangers in animals. One is the heat exchange between the arteries and venae comitantes of the limbs. To a minor degree in humans, but to a major degree in mammals living in cold water, heat is transferred from the arterial blood flowing into the limbs to the adjacent veins draining blood back into the body, making the tips of the limbs cold while conserving body heat. ROLE OF UREA Urea contributes to the establishment of the osmotic gradient in the medullary pyramids and to the ability to form a concen-trated urine in the collecting ducts. Urea transport is mediated by urea transporters, presumably by facilitated diffusion. There are at least four isoforms of the transport protein UT-A in the kidneys (UT-A1 to UT-A4); UT-B is found in erythro-cytes. The amount of urea in the medullary interstitium and, consequently, in the urine varies with the amount of urea fil-tered, and this in turn varies with the dietary intake of protein. Therefore, a high-protein diet increases the ability of the kid-neys to concentrate the urine. OSMOTIC DIURESIS The presence of large quantities of unreabsorbed solutes in the renal tubules causes an increase in urine volume called osmotic diuresis. Solutes that are not reabsorbed in the prox-imal tubules exert an appreciable osmotic effect as the volume of tubular fluid decreases and their concentration rises. There-fore, they “hold water in the tubules.” In addition, the concen-tration gradient against which Na+ can be pumped out of the proximal tubules is limited. Normally, the movement of water out of the proximal tubule prevents any appreciable gradient from developing, but Na+ concentration in the fluid falls when water reabsorption is decreased because of the presence in the tubular fluid of increased amounts of unreabsorbable solutes. The limiting concentration gradient is reached, and further proximal reabsorption of Na+ is prevented; more Na+ remains in the tubule, and water stays with it. The result is that the loop of Henle is presented with a greatly increased volume of iso-tonic fluid. This fluid has a decreased Na+ concentration, but the total amount of Na+ reaching the loop per unit time is in-creased. In the loop, reabsorption of water and Na+ is de-creased because the medullary hypertonicity is decreased. The decrease is due primarily to decreased reabsorption of Na+, K+, and Cl– in the ascending limb of the loop because the lim-iting concentration gradient for Na+ reabsorption is reached. More fluid passes through the distal tubule, and because of the decrease in the osmotic gradient along the medullary pyra-mids, less water is reabsorbed in the collecting ducts. The re-sult is a marked increase in urine volume and excretion of Na+ and other electrolytes. Osmotic diuresis is produced by the administration of com-pounds such as mannitol and related polysaccharides that are filtered but not reabsorbed. It is also produced by naturally occurring substances when they are present in amounts exceeding the capacity of the tubules to reabsorb them. For example, in diabetes mellitus, if blood glucose is high, glu-cose in the glomerular filtrate is high, thus the filtered load will exceed the TmG and glucose will remain in the tubules causing polyuria. Osmotic diuresis can also be produced by the infusion of large amounts of sodium chloride or urea. It is important to recognize the difference between osmotic diuresis and water diuresis. In water diuresis, the amount of water reabsorbed in the proximal portions of the nephron is normal, and the maximal urine flow that can be produced is about 16 mL/min. In osmotic diuresis, increased urine flow is due to decreased water reabsorption in the proximal tubules and loops and very large urine flows can be produced. As the load of excreted solute is increased, the concentration of the urine approaches that of plasma (Figure 38–18) in spite of maximal vasopressin secretion, because an increasingly large fraction of the excreted urine is isotonic proximal tubu-lar fluid. If osmotic diuresis is produced in an animal with diabetes insipidus, the urine concentration rises for the same reason. FIGURE 38–17 Operation of the vasa recta as countercurrent exchangers in the kidney. NaCl and urea diffuse out of the ascending limb of the vessel and into the descending limb, whereas water diffuses out of the descending and into the ascending limb of the vascular loop. 300 325 425 475 725 775 H2O H2O NaCl Urea NaCl Urea 450 750 1200 1200 Cortex Outer medulla Inner medulla CHAPTER 38 Renal Function & Micturition 657 RELATION OF URINE CONCENTRATION TO GFR The magnitude of the osmotic gradient along the medullary pyramids is increased when the rate of flow of fluid through the loops of Henle is decreased. A reduction in GFR such as that caused by dehydration produces a decrease in the volume of fluid presented to the countercurrent mechanism, so that the rate of flow in the loops declines and the urine becomes more concentrated. When the GFR is low, the urine can be-come quite concentrated in the absence of vasopressin. If one renal artery is constricted in an animal with diabetes insipidus, the urine excreted on the side of the constriction becomes hy-pertonic because of the reduction in GFR, whereas that excret-ed on the opposite side remains hypotonic. "FREE WATER CLEARANCE" In order to quantitate the gain or loss of water by excretion of a concentrated or dilute urine, the "free water clearance" (CH2O) is sometimes calculated. This is the difference between the urine volume and the clearance of osmoles (COsm): CH2O = V • – UOsm V • POsm where V • is the urine flow rate and UOsm and POsm the urine and plasma osmolality, respectively. COsm is the amount of water necessary to excrete the osmotic load in a urine that is isotonic with plasma. Therefore, CH2O is negative when the urine is hypertonic and positive when the urine is hypotonic. For example, using the data in Table 38–7, the values for CH2O are –1.3 mL/min (–1.9 L/d) during maximal antidiure-sis and 14.5 mL/min (20.9 L/d) in the absence of vasopressin. REGULATION OF Na+ EXCRETION Na+ is filtered in large amounts, but it is actively transported out of all portions of the tubule except the descending thin limb of Henle’s loop. Normally, 96% to well over 99% of the filtered Na+ is reabsorbed. Because Na+ is the most abundant cation in ECF and because Na+ salts account for over 90% of the osmotically active solute in the plasma and interstitial fluid, the amount of Na+ in the body is a prime determinant of the ECF volume. Therefore, it is not surprising that multiple regulatory mecha-nisms have evolved in terrestrial animals to control the excre-tion of this ion. Through the operation of these regulatory mechanisms, the amount of Na+ excreted is adjusted to equal the amount ingested over a wide range of dietary intakes, and the individual stays in Na+ balance. Thus, urinary Na+ output ranges from less than 1 mEq/d on a low-salt diet to 400 mEq/d or more when the dietary Na+ intake is high. In addition, there is a natriuresis when saline is infused intravenously and a de-crease in Na+ excretion when ECF volume is reduced. MECHANISMS Variations in Na+ excretion are brought about by changes in GFR (Table 38–9) and changes in tubular reabsorption, pri-marily in the 3% of filtered Na+ that reaches the collecting ducts. The factors affecting the GFR, including tubuloglomerular feedback, have been discussed previously. Factors affecting Na+ reabsorption include the circulating level of aldosterone and other adrenocortical hormones, the circulating level of ANP and other natriuretic hormones, and the rate of tubular secre-tion of H+ and K+. EFFECTS OF ADRENOCORTICAL STEROIDS Adrenal mineralocorticoids such as aldosterone increase tu-bular reabsorption of Na+ in association with secretion of K+ and H+ and also Na+ reabsorption with Cl–. When these hor-mones are injected into adrenalectomized animals, a latent FIGURE 38–18 Approximate relationship between urine concentration and urine flow in osmotic diuresis in humans. The dashed line in the lower diagram indicates the concentration at which the urine is isosmotic with plasma. (Reproduced with permission from Berliner RW, Giebisch G in: Best and Taylor’s Physiological Basis of Medical Practice, 9th ed. Brobeck JR [editor]. Williams & Wilkins, 1979.) Urine flow (mL/min) Isosmotic Diabetes insipidus Solute load (mosm/min) Maximal vasopressin 21 18 15 12 9 6 3 0 0.9 1.8 2.7 4.5 6.3 Urine osmolality (mosm/L) Isosmotic Diabetes insipidus Urine flow (mL/min) Maximal vasopressin 1400 1200 1000 800 600 400 200 0 3 6 9 12 15 18 21 658 SECTION VIII Renal Physiology period of 10 to 30 min occurs before their effects on Na+ reab-sorption become manifest, because of the time required for the steroids to alter protein synthesis via their action on DNA. Mineralocorticoids may also have more rapid membrane-me-diated effects, but these are not apparent in terms of Na+ ex-cretion in the whole animal. The mineralocorticoids act primarily in the collecting ducts to increase the number of ac-tive epithelial sodium channels (ENaCs) in this part of the nephron. The molecular mechanisms believed to be involved are discussed in Chapter 22 and summarized in Figure 38–19. In Liddle syndrome, mutations in the genes that code for the β subunit and less commonly the γ subunit of the ENaCs cause them to become constitutively active in the kidney. This leads to Na+ retention and hypertension. OTHER HUMORAL EFFECTS Reduction of dietary intake of salt increases aldosterone secre-tion (see Figure 22-26), producing marked but slowly develop-ing decreases in Na+ excretion. A variety of other humoral factors affect Na+ reabsorption. PGE2 causes a natriuresis, possibly by inhibiting Na, K ATPase and possibly by increas-ing intracellular Ca2+, which in turn inhibits Na+ transport via ENaCs. Endothelin and IL-1 cause natriuresis, probably by in-creasing the formation of PGE2. ANP and related molecules increase intracellular cyclic 3',5'-guanosine monophosphate (cGMP), and this inhibits transport via the ENaCs. Inhibition of Na, K ATPase by another natriuretic hormone, which ap-pears to be endogenously produced ouabain, also increases Na+ excretion. Angiotensin II increases reabsorption of Na+ and HCO3 – by an action on the proximal tubules. There is an appreciable amount of angiotensin-converting enzyme in the kidneys, and the kidneys convert 20% of the circulating angio-tensin I reaching them to angiotensin II. In addition, angio-tensin I is generated in the kidneys. Prolonged exposure to high levels of circulating mineralo-corticoids does not cause edema in otherwise normal individ-uals because eventually the kidneys escape from the effects of the steroids. This escape phenomenon, which may be due to increased secretion of ANP, is discussed in Chapter 22. It appears to be reduced or absent in nephrosis, cirrhosis, and heart failure, and patients with these diseases continue to retain Na+ and become edematous when exposed to high levels of mineralocorticoids. REGULATION OF WATER EXCRETION WATER DIURESIS The feedback mechanism controlling vasopressin secretion and the way vasopressin secretion is stimulated by a rise and inhibited by a drop in the effective osmotic pressure of the plasma are discussed in Chapter 18. The water diuresis pro-duced by drinking large amounts of hypotonic fluid begins about 15 min after ingestion of a water load and reaches its maximum in about 40 min. The act of drinking produces a small decrease in vasopressin secretion before the water is ab-sorbed, but most of the inhibition is produced by the decrease in plasma osmolality after the water is absorbed. WATER INTOXICATION During excretion of an average osmotic load, the maximal urine flow that can be produced during a water diuresis is about 16 mL/min. If water is ingested at a higher rate than this for any length of time, swelling of the cells because of the up-take of water from the hypotonic ECF becomes severe and, rarely, the symptoms of water intoxication may develop. Swelling of the cells in the brain causes convulsions and coma and leads eventually to death. Water intoxication can also TABLE 38–9 Changes in Na+ excretion that would occur as a result of changes in GFR if there were no concomitant changes in Na+ reabsorption. GFR (mL/min) Plasma Na+ (μEq/mL) Amount Filtered (μEq/min) Amount Reabsorbed (μEq/min) Amount Excreted (μEq/min) 125 145 18,125 18,000 125 127 145 18,415 18,000 415 124.1 145 18,000 18,000 0 FIGURE 38–19 Renal Principal cell. Na+ enters via the ENaCs in the apical membrane and is pumped into the interstitial fluid by Na, K ATPases in the basolateral membrane. Aldosterone activates the genome to produce serum- and glucocorticoid-regulated kinase (sgk) and other proteins, and the number of active ENaCs is increased. Tubular lumen Na+ Interstitial fluid K+ cGMP Ouabain ANP Amiloride Cl H2N Na+ O N N NH NH2 NH2+ NH2 Tight junction Nucleus sgk and other proteins More active ENaCs Aldosterone CHAPTER 38 Renal Function & Micturition 659 occur when water intake is not reduced after administration of exogenous vasopressin or when secretion of endogenous vaso-pressin occurs in response to non-osmotic stimuli such as sur-gical trauma. REGULATION OF K+ EXCRETION Much of the filtered K+ is removed from the tubular fluid by active reabsorption in the proximal tubules (Table 38–5), and K+ is then secreted into the fluid by the distal tubular cells. The rate of K+ secretion is proportional to the rate of flow of the tu-bular fluid through the distal portions of the nephron, because with rapid flow there is less opportunity for the tubular K+ concentration to rise to a value that stops further secretion. In the absence of complicating factors, the amount secreted is ap-proximately equal to the K+ intake, and K+ balance is main-tained. In the collecting ducts, Na+ is generally reabsorbed and K+ is secreted. There is no rigid one-for-one exchange, and much of the movement of K+ is passive. However, there is electrical coupling in the sense that intracellular migration of Na+ from the lumen tends to lower the potential difference across the tubular cell, and this favors movement of K+ into the tubular lumen. Because Na+ is also reabsorbed in associa-tion with H+ secretion, there is competition for the Na+ in the tubular fluid. K+ excretion is decreased when the amount of Na+ reaching the distal tubule is small, and it is also decreased when H+ secretion is increased. DIURETICS Although a detailed discussion of diuretic agents is outside the scope of this book, consideration of their mechanisms of ac-tion constitutes an informative review of the factors affecting urine volume and electrolyte excretion. These mechanisms are summarized in Table 38–10. Water, alcohol, osmotic diuret-ics, xanthines, and acidifying salts have limited clinical useful-ness, and the vasopressin antagonists are currently undergoing clinical trials. However, many of the other agents on the list are used extensively in medical practice. The carbonic anhydrase-inhibiting drugs are only moderately effective as diuretic agents, but because they inhibit acid secre-tion by decreasing the supply of carbonic acid, they have far-reaching effects. Not only is Na+ excretion increased because H+ secretion is decreased, but also HCO3 – reabsorption is depressed; and because H+ and K+ compete with each other and with Na+, the decrease in H+ secretion facilitates the secretion and excretion of K+. Furosemide and the other loop diuretics inhibit the Na–K– 2Cl cotransporter in the thick ascending limb of Henle’s loop. They cause a marked natriuresis and kaliuresis. Thiazides act by inhibiting Na–Cl cotransport in the distal tubule. The diuresis they cause is less marked, but both loop diuretics and thiazides cause increased delivery of Na+ (and fluid) to the collecting ducts, facilitating K+ excretion. Thus, over time, K+ depletion and hypokalemia are common complications in those who use them if they do not supplement their K+ intake. On the other hand, the so-called K+-sparing diuretics act in the collecting duct by inhibiting the action of aldoster-one or blocking ENaCs. EFFECTS OF DISORDERED RENAL FUNCTION A number of abnormalities are common to many different types of renal disease. The secretion of renin by the kidneys and the relation of the kidneys to hypertension are discussed in Chapter 39. A frequent finding in various forms of renal disease is the presence in the urine of protein, leukocytes, red cells, and casts, which are proteinaceous material precipitated in the tubules and washed into the bladder. Other important consequences of renal disease are loss of the ability to concen-trate or dilute the urine, uremia, acidosis, and abnormal reten-tion of Na+(see Clinical Box 38–3). TABLE 38–10 Mechanism of action of various diuretics. Agent Mechanism of Action Water Inhibits vasopressin secretion. Ethanol Inhibits vasopressin secretion. Antagonists of V2 vasopressin receptors such as astolvaptan Inhibit action of vasopressin on collecting duct. Large quantities of osmotically active substances such as mannitol and glucose Produce osmotic diuresis. Xanthines such as caffeine and theophylline Decrease tubular reabsorption of Na+ and increase GFR. Acidifying salts such as CaCl2 and NH4Cl Supply acid load; H+ is buffered, but an anion is excreted with Na+ when the ability of the kidneys to replace Na+ with H+ is exceeded. Carbonic anhydrase inhibitors such as acetazolamide (Diamox) Decrease H+ secretion, with resul-tant increase in Na+ and K+ excre-tion. Metolazone (Zaroxolyn), thia-zides such as chlorothiazide (Di-uril) Inhibit the Na–Cl cotransporter in the early portion of the distal tu-bule. Loop diuretics such as furosemi-de (Lasix), ethacrynic acid (Ede-crin), and bumetanide Inhibit the Na–K–2Cl cotransport-er in the medullary thick ascend-ing limb of the loop of Henle K+-retaining natriuretics such as spironolactone (Aldactone), tri-amterene (Dyrenium), and amiloride (Midamor) Inhibit Na+–K+ “exchange” in the collecting ducts by inhibiting the action of aldosterone (spironolac-tone) or by inhibiting the ENaCs (amiloride). 660 SECTION VIII Renal Physiology LOSS OF CONCENTRATING & DILUTING ABILITY In renal disease, the urine becomes less concentrated and urine volume is often increased, producing the symptoms of polyuria and nocturia (waking up at night to void). The abil-ity to form a dilute urine is often retained, but in advanced re-nal disease, the osmolality of the urine becomes fixed at about that of plasma, indicating that the diluting and concentrating functions of the kidney have both been lost. The loss is due in part to disruption of the countercurrent mechanism, but a more important cause is a loss of functioning nephrons. When one kidney is removed surgically, the number of functioning nephrons is halved. The number of osmoles excreted is not re-duced to this extent, and so the remaining nephrons must each be filtering and excreting more osmotically active sub-stances, producing what is in effect an osmotic diuresis. In os-motic diuresis, the osmolality of the urine approaches that of plasma. The same thing happens when the number of func-tioning nephrons is reduced by disease. The increased filtra-tion in the remaining nephrons eventually damages them, and thus more nephrons are lost. The damage resulting from in-creased filtration may be due to progressive fibrosis in the proximal tubule cells, but this is unsettled. However, the even-tual result of this positive feedback is loss of so many nephrons that complete renal failure with oliguria or even anuria results. UREMIA When the breakdown products of protein metabolism accumu-late in the blood, the syndrome known as uremia develops. The symptoms of uremia include lethargy, anorexia, nausea and vomiting, mental deterioration and confusion, muscle twitch-ing, convulsions, and coma. The blood urea nitrogen (BUN) and creatinine levels are high, and the blood levels of these substances are used as an index of the severity of the uremia. It probably is not the accumulation of urea and creatinine per se but rather the accumulation of other toxic substances—possibly organic acids or phenols—that produces the symptoms of uremia. The toxic substances that cause the symptoms of uremia can be removed by dialyzing the blood of uremic patients against a bath of suitable composition in an artificial kidney (hemodialysis). Patients can be kept alive and in reasonable health for many months on dialysis, even when they are com-pletely anuric or have had both kidneys removed. However, the treatment of choice today is certainly transplantation of a kidney from a suitable donor. Other features of chronic renal failure include anemia, which is caused primarily by failure to produce erythro-poietin, and secondary hyperparathyroidism due to 1,25-dihydroxycholecalciferol deficiency (see Chapter 23). ACIDOSIS Acidosis is common in chronic renal disease because of failure to excrete the acid products of digestion and metabolism (see Chapter 40). In the rare syndrome of renal tubular acidosis, there is specific impairment of the ability to make the urine acidic, and other renal functions are usually normal. However, in most cases of chronic renal disease the urine is maximally acidified, and acidosis develops because the total amount of H+ that can be secreted is reduced because of impaired renal tubular production of NH4 +. ABNORMAL Na+ HANDLING Many patients with renal disease retain excessive amounts of Na+ and become edematous. Na+ retention in renal disease has at least three causes. In acute glomerulonephritis, a dis-ease that affects primarily the glomeruli, the amount of Na+ filtered is decreased markedly. In the nephrotic syndrome, an increase in aldosterone secretion contributes to the salt reten-tion. The plasma protein level is low in this condition, and so fluid moves from the plasma into the interstitial spaces and the plasma volume falls. The decline in plasma volume trig-gers the increase in aldosterone secretion via the renin– angiotensin system. A third cause of Na+ retention and ede-ma in renal disease is heart failure. Renal disease predisposes to heart failure, partly because of the hypertension it fre-quently produces. CLINICAL BOX 38–3 Proteinuria In many renal diseases and in one benign condition, the permeability of the glomerular capillaries is increased, and protein is found in the urine in more than the usual trace amounts (proteinuria). Most of this protein is albumin, and the defect is commonly called albuminuria. The rela-tion of charges on the glomerular membrane to albumin-uria has been discussed above. The amount of protein in the urine may be very large, and especially in nephrosis, the urinary protein loss may exceed the rate at which the liver can synthesize plasma proteins. The resulting hypopro-teinemia reduces the oncotic pressure, and the plasma vol-ume declines, sometimes to dangerously low levels, while edema fluid accumulates in the tissues. A benign condition that causes proteinuria is a poorly understood change in renal hemodynamics, which in some otherwise normal individuals, causes protein to appear in urine when they are in the standing position (orthostatic albuminuria). Urine formed when these individuals are lying down is protein-free. CHAPTER 38 Renal Function & Micturition 661 THE BLADDER FILLING The walls of the ureters contain smooth muscle arranged in spi-ral, longitudinal, and circular bundles, but distinct layers of muscle are not seen. Regular peristaltic contractions occurring one to five times per minute move the urine from the renal pel-vis to the bladder, where it enters in spurts synchronous with each peristaltic wave. The ureters pass obliquely through the bladder wall and, although there are no ureteral sphincters as such, the oblique passage tends to keep the ureters closed except during peristaltic waves, preventing reflux of urine from the bladder. EMPTYING The smooth muscle of the bladder, like that of the ureters, is arranged in spiral, longitudinal, and circular bundles. Con-traction of the circular muscle, which is called the detrusor muscle, is mainly responsible for emptying the bladder during urination (micturition). Muscle bundles pass on either side of the urethra, and these fibers are sometimes called the internal urethral sphincter, although they do not encircle the urethra. Farther along the urethra is a sphincter of skeletal muscle, the sphincter of the membranous urethra (external urethral sphincter). The bladder epithelium is made up of a superficial layer of flat cells and a deep layer of cuboidal cells. The inner-vation of the bladder is summarized in Figure 38–20. The physiology of bladder emptying and the physiologic basis of its disorders are subjects about which there is much confusion. Micturition is fundamentally a spinal reflex facili-tated and inhibited by higher brain centers and, like defeca-tion, subject to voluntary facilitation and inhibition. Urine enters the bladder without producing much increase in intra-vesical pressure until the viscus is well filled. In addition, like other types of smooth muscle, the bladder muscle has the property of plasticity; when it is stretched, the tension initially produced is not maintained. The relation between intravesical pressure and volume can be studied by inserting a catheter and emptying the bladder, then recording the pressure while the bladder is filled with 50-mL increments of water or air (cystometry). A plot of intravesical pressure against the vol-ume of fluid in the bladder is called a cystometrogram (Figure 38–21). The curve shows an initial slight rise in pres-sure when the first increments in volume are produced; a long, nearly flat segment as further increments are produced; and a sudden, sharp rise in pressure as the micturition reflex is triggered. These three components are sometimes called segments Ia, Ib, and II. The first urge to void is felt at a blad-der volume of about 150 mL, and a marked sense of fullness at about 400 mL. The flatness of segment Ib is a manifestation of the law of Laplace. This law states that the pressure in a spher-ical viscus is equal to twice the wall tension divided by the radius. In the case of the bladder, the tension increases as the organ fills, but so does the radius. Therefore, the pressure increase is slight until the organ is relatively full. During micturition, the perineal muscles and external urethral sphincter are relaxed, the detrusor muscle contracts, and urine passes out through the urethra. The bands of smooth muscle on either side of the urethra apparently play no role in micturition, and their main function in males is believed to be the prevention of reflux of semen into the bladder during ejaculation. The mechanism by which voluntary urination is initiated remains unsettled. One of the initial events is relaxation of the muscles of the pelvic floor, and this may cause a sufficient FIGURE 38–20 Innervation of the bladder. Dashed lines indi-cate sensory nerves. Parasympathetic innervation is shown at the left, sympathetic at the upper right, and somatic at the lower right. L1 L2 L3 S2 S3 S4 S2 S3 S4 Bladder External sphincter Pudendal nerves Hypogastric nerves Pelvic nerves Inferior mesenteric ganglion FIGURE 38–21 Cystometrogram in a normal human. The numerals identify the three components of the curve described in the text. The dashed line indicates the pressure–volume relations that would have been found had micturition not occurred and produced component II. (Modified and reproduced with permission from Tanagho EA, McAninch JW: Smith’s General Urology, 15th ed. McGraw-Hill, 2000.) Intravesical pressure (cm water) 80 60 40 20 0 0 200 100 Intravesical volume (mL) 300 400 500 Ia Ib II 662 SECTION VIII Renal Physiology downward tug on the detrusor muscle to initiate its contrac-tion. The perineal muscles and external sphincter can be con-tracted voluntarily, preventing urine from passing down the urethra or interrupting the flow once urination has begun. It is through the learned ability to maintain the external sphinc-ter in a contracted state that adults are able to delay urination until the opportunity to void presents itself. After urination, the female urethra empties by gravity. Urine remaining in the urethra of the male is expelled by several contractions of the bulbocavernosus muscle. REFLEX CONTROL The bladder smooth muscle has some inherent contractile ac-tivity; however, when its nerve supply is intact, stretch receptors in the bladder wall initiate a reflex contraction that has a lower threshold than the inherent contractile response of the muscle. Fibers in the pelvic nerves are the afferent limb of the voiding reflex, and the parasympathetic fibers to the bladder that consti-tute the efferent limb also travel in these nerves. The reflex is in-tegrated in the sacral portion of the spinal cord. In the adult, the volume of urine in the bladder that normally initiates a reflex contraction is about 300 to 400 mL. The sympathetic nerves to the bladder play no part in micturition, but in males they do me-diate the contraction of the bladder muscle that prevents semen from entering the bladder during ejaculation. The stretch receptors in the bladder wall have no small motor nerve system. However, the threshold for the voiding reflex, like the stretch reflexes, is adjusted by the activity of facilitatory and inhibitory centers in the brainstem. There is a facilitatory area in the pontine region and an inhibitory area in the midbrain. After transection of the brain stem just above the pons, the threshold is lowered and less bladder filling is required to trigger it, whereas after transection at the top of the midbrain, the threshold for the reflex is essentially nor-mal. There is another facilitatory area in the posterior hypo-thalamus. Humans with lesions in the superior frontal gyrus have a reduced desire to urinate and difficulty in stopping micturition once it has commenced. However, stimulation experiments in animals indicate that other cortical areas also affect the process. The bladder can be made to contract by voluntary facilitation of the spinal voiding reflex when it con-tains only a few milliliters of urine. Voluntary contraction of the abdominal muscles aids the expulsion of urine by increas-ing the intra-abdominal pressure, but voiding can be initiated without straining even when the bladder is nearly empty. EFFECTS OF DEAFFERENTATION When the sacral dorsal roots are cut in experimental animals or interrupted by diseases of the dorsal roots, such as tabes dorsalis in humans, all reflex contractions of the bladder are abolished. The bladder becomes distended, thin-walled, and hypotonic, but some contractions occur because of the intrin-sic response of the smooth muscle to stretch. EFFECTS OF DENERVATION When the afferent and efferent nerves are both destroyed, as they may be by tumors of the cauda equina or filum terminale, the bladder is flaccid and distended for a while. Gradually, however, the muscle of the “decentralized bladder” becomes active, with many contraction waves that expel dribbles of urine out of the urethra. The bladder becomes shrunken and the bladder wall hypertrophied. The reason for the difference between the small, hypertrophic bladder seen in this condition and the distended, hypotonic bladder seen when only the af-ferent nerves are interrupted is not known. The hyperactive state in the former condition suggests the development of de-nervation hypersensitization even though the neurons inter-rupted are preganglionic rather than postganglionic (see Clinical Box 38–4). EFFECTS OF SPINAL CORD TRANSECTION During spinal shock, the bladder is flaccid and unresponsive. It becomes overfilled, and urine dribbles through the sphinc-ters (overflow incontinence). After spinal shock has passed, the voiding reflex returns, although there is, of course, no vol-untary control and no inhibition or facilitation from higher centers when the spinal cord is transected. Some paraplegic patients train themselves to initiate voiding by pinching or stroking their thighs, provoking a mild mass reflex (see Chap-ter 16). In some instances, the voiding reflex becomes hyper-active, bladder capacity is reduced, and the wall becomes hypertrophied. This type of bladder is sometimes called the spastic neurogenic bladder. The reflex hyperactivity is made worse by, and may be caused by, infection in the bladder wall. CHAPTER SUMMARY ■Plasma enters the kidneys and is filtered in the glomerulus. As the filtrate passes down the nephron and through the tubules its vol-ume is reduced and water and solutes are removed (tubular reab-sorption) and waste products are secreted (tubular secretion). CLINICAL BOX 38–4 Abnormalities of Micturition Three major types of bladder dysfunction are due to neural lesions: (1) the type due to interruption of the afferent nerves from the bladder, (2) the type due to interruption of both afferent and efferent nerves, and (3) the type due to interruption of facilitatory and inhibitory pathways de-scending from the brain. In all three types the bladder con-tracts, but the contractions are generally not sufficient to empty the viscus completely, and residual urine is left in the bladder. CHAPTER 38 Renal Function & Micturition 663 ■A nephron consists of an individual renal tubule and its glomer-ulus. Each tubule has several segments, beginning with the prox-imal tubule, followed by the loop of Henle (descending and ascending limbs), the distal convoluted tubule, the connecting tubule, and the collecting duct. ■The kidneys receive just under 25% of the cardiac output and renal plasma flow can be measured by infusing p-aminohippuric acid (PAH) and determining its urine and plasma concentrations. ■Renal blood flow enters the glomerulus via the afferent arteriole and leaves via the efferent arteriole (whose diameter is smaller). Renal blood flow is regulated by norepinephrine (constriction, reduction of flow), dopamine (vasodilation, increases flow), angiotensin II (constricts), prostaglandins (dilation in the renal cortex and constriction in the renal medulla), and acetylcholine (vasodilation). ■Glomerular filtration rate can be measured by a substance that is freely filtered and neither reabsorbed nor secreted in the tubules, is nontoxic, and is not metabolized by the body. Inulin meets these criteria and is extensively used to measure GFR. ■Urine is stored in the bladder before voiding (micturition). The micturition response involves reflex pathways, but is under voluntary control. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. In the presence of vasopressin, the greatest fraction of filtered water is absorbed in the A) proximal tubule. B) loop of Henle. C) distal tubule. D) cortical collecting duct. E) medullary collecting duct. 2. In the absence of vasopressin, the greatest fraction of filtered water is absorbed in the A) proximal tubule. B) loop of Henle. C) distal tubule. D) cortical collecting duct. E) medullary collecting duct. 3. If the clearance of a substance which is freely filtered is less than that of inulin, A) there is net reabsorption of the substance in the tubules. B) there is net secretion of the substance in the tubules. C) the substance is neither secreted nor reabsorbed in the tubules. D) the substance becomes bound to protein in the tubules. E) the substance is secreted in the proximal tubule to a greater degree than in the distal tubule. 4. Glucose reabsorption occurs in the A) proximal tubule. B) loop of Henle. C) distal tubule. D) cortical collecting duct. E) medullary collecting duct. 5. On which of the following does aldosterone exert its greatest effect? A) glomerulus B) proximal tubule C) thin portion of the loop of Henle D) thick portion of the loop of Henle E) cortical collecting duct 6. What is the clearance of a substance when its concentration in the plasma is 10 mg/dL, its concentration in the urine is 100 mg/ dL, and urine flow is 2 mL/min? A) 2 mL/min B) 10 mL/min C) 20 mL/min D) 200 mL/min E) Clearance cannot be determined from the information given. 7. As urine flow increases during osmotic diuresis A) the osmolality of urine falls below that of plasma. B) the osmolality of urine increases because of the increased amounts of nonreabsorbable solute in the urine. C) the osmolality of urine approaches that of plasma because plasma leaks into the tubules. D) the osmolality of urine approaches that of plasma because an increasingly large fraction of the excreted urine is isotonic proximal tubular fluid. E) the action of vasopressin on the renal tubules is inhibited. CHAPTER RESOURCES Anderson K-E: Pharmacology of lower urinary tract smooth muscles and penile erectile tissue. Pharmacol Rev 1993;45:253. Brenner BM, Rector FC Jr. (editors): The Kidney, 6th ed. 2 vols. Saunders, 1999. Brown D: The ins and outs of aquaporin-2 trafficking. Am J Physiol Renal Physiol 2003;284:F893. Brown D, Stow JL: Protein trafficking and polarity in kidney epithelium: From cell biology to physiology. Physiol Rev 1996;76:245. DiBona GF, Kopp UC: Neural control of renal function. Physiol Rev 1997; 77:75. Garcia NH, Ramsey CR, Knox FG: Understanding the role of paracellular transport in the proximal tubule. News Physiol Sci 1998;13:38. Nielsen S, et al: Aquaporins in the kidney: From molecules to medicine. Physiol Rev 2002;82:205. Spring KR: Epithelial fluid transport: A century of investigation. News Physiol Sci 1999;14:92. Valten V: Tubuloglomerular feedback and the control of glomerular filtration rate. News Physiol Sci 2003;18:169. This page intentionally left blank 665 C H A P T E R 39 Regulation of Extracellular Fluid Composition & Volume O B J E C T I V E S After reading this chapter, you should be able to: ■Describe how the tonicity (osmolality) of the extracellular fluid is maintained by alterations in water intake and vasopressin secretion. ■Discuss the effects of vasopressin, the receptors on which it acts, and how its secre-tion is regulated. ■Describe how the volume of the extracellular fluid is maintained by alterations in renin and aldosterone secretion. ■Outline the cascade of reactions that lead to the formation of angiotensin II and its metabolites in the circulation. ■List the functions of angiotensin II and the receptors on which it acts to carry out these functions. ■Describe the structure and functions of ANP, BNP, and CNP and the receptors on which they act. ■Describe the site and mechanism of action of erythropoietin, and the feedback regulation of its secretion. INTRODUCTION This chapter is a review of the major homeostatic mechanisms that operate, primarily through the kidneys and the lungs, to maintain the tonicity, the volume, and the specific ionic com-position of the extracellular fluid (ECF). The interstitial portion of this fluid is the fluid environment of the cells, and life depends upon the constancy of this “internal sea” (see Chapter 1). DEFENSE OF TONICITY The defense of the tonicity of the ECF is primarily the function of the vasopressin-secreting and thirst mechanisms. The total body osmolality is directly proportional to the total body sodi-um plus the total body potassium divided by the total body wa-ter, so that changes in the osmolality of the body fluids occur when a disproportion exists between the amount of these elec-trolytes and the amount of water ingested or lost from the body. When the effective osmotic pressure of the plasma rises, vaso-pressin secretion is increased and the thirst mechanism is stim-ulated; water is retained in the body, diluting the hypertonic plasma; and water intake is increased (Figure 39–1). Converse-ly, when the plasma becomes hypotonic, vasopressin secre-tion is decreased and “solute-free water” (water in excess of solute) is excreted. In this way, the tonicity of the body fluids is maintained within a narrow normal range. In health, plas-ma osmolality ranges from 280 to 295 mOsm/kg of H2O, with vasopressin secretion maximally inhibited at 285 mOsm/kg and stimulated at higher values (Figure 39–2). VASOPRESSIN RECEPTORS There are at least three kinds of vasopressin receptors: V1A, V1B, and V2. All are G protein-coupled. The V1A and V1B 666 SECTION VIII Renal Physiology receptors act through phosphatidylinositol hydrolysis to in-crease the intracellular Ca2+ concentration. The V2 receptors act through Gs to increase cyclic adenosine 3',5'-monophos-phate (cAMP) levels. EFFECTS OF VASOPRESSIN Because one of its principal physiologic effects is the retention of water by the kidney, vasopressin is often called the antidi-uretic hormone (ADH). It increases the permeability of the collecting ducts of the kidney, so that water enters the hyper-tonic interstitium of the renal pyramids. The urine becomes concentrated, and its volume decreases. The overall effect is therefore retention of water in excess of solute; consequently, the effective osmotic pressure of the body fluids is decreased. In the absence of vasopressin, the urine is hypotonic to plasma, urine volume is increased, and there is a net water loss. Conse-quently, the osmolality of the body fluid rises. The mechanism by which vasopressin exerts its antidiuretic effect is activated by V2 receptors and involves the insertion of proteins called water channels into the apical (luminal) mem-branes of the principal cells of the collecting ducts. Movement of water across membranes by simple diffusion is now known to be augmented by movement through water channels called aquaporins, and to date 13 (AQP0–AQP12) have been identi-fied and water channels are now known to be expressed in almost all tissues in the body. The vasopressin-responsive water channel in the collecting ducts is aquaporin-2. These channels are stored in endosomes inside the cells, and vasopressin causes their rapid translocation to the luminal membranes. V1A receptors mediate the vasoconstrictor effect of vaso-pressin, and vasopressin is a potent stimulator of vascular smooth muscle in vitro. However, relatively large amounts of vasopressin are needed to raise blood pressure in vivo, because vasopressin also acts on the brain to cause a decrease in cardiac output. The site of this action is the area postrema, one of the circumventricular organs (see Chapter 34). Hemorrhage is a potent stimulus for vasopressin secretion, and the blood pres-sure fall after hemorrhage is more marked in animals that have been treated with synthetic peptides that block the pressor action of vasopressin. Consequently, it appears that vasopressin does play a role in blood pressure homeostasis. V1A receptors are also found in the liver and the brain. Vasopressin causes glycogenolysis in the liver, and, as noted above, it is a neurotransmitter in the brain and spinal cord. The V1B receptors (also called V3 receptors) appear to be unique to the anterior pituitary, where they mediate increased adrenocorticotropic hormone (ACTH) secretion from the corticotropes. METABOLISM Circulating vasopressin is rapidly inactivated, principally in the liver and kidneys. It has a biologic half-life (time required for inactivation of half a given amount) of approximately 18 min in humans. CONTROL OF VASOPRESSIN SECRETION: OSMOTIC STIMULI Vasopressin is stored in the posterior pituitary and released into the bloodstream by impulses in the nerve fibers that con-tain the hormone. The factors affecting its secretion are sum-marized in Table 39–1. When the effective osmotic pressure of the plasma is increased above the normal 285 mOsm/kg, the rate of discharge of these neurons increases and vasopressin secretion is increased (Figure 39–2). At 285 mOsm/kg, plasma vasopressin is at or near the limits of detection by available as-says, but a further decrease probably takes place when plasma osmolality is below this level. Vasopressin secretion is regulat-ed by osmoreceptors located in the anterior hypothalamus. They are outside the blood–brain barrier and appear to be lo-cated in the circumventricular organs, primarily the organum FIGURE 39–1 Mechanisms for defending ECF tonicity. The dashed arrow indicates inhibition. (Courtesy of J Fitzsimmons.) FIGURE 39–2 Relation between plasma osmolality and plasma vasopressin in healthy adult humans during infusion of hypertonic saline. LD, limit of detection. (Reproduced with permission from Thompson CJ et al: The osmotic thresholds for thirst and vasopressin are similar in healthy humans. Clin Sci [Colch] 1986;71:651.) Increased osmolality of ECF Dilution of ECF Increased vasopressin secretion Water retention Thirst Increased water intake 20 16 12 8 4 LD 280 300 320 Plasma osmolality (mosm/kg) Plasma vasopressin (pmol/L) CHAPTER 39 Regulation of Extracellular Fluid Composition & Volume 667 vasculosum of the lamina terminalis (OVLT) (see Chapter 34). The osmotic threshold for thirst (Figure 39–1) is the same as or slightly greater than the threshold for increased vaso-pressin secretion (Figure 39–2), and it is still uncertain wheth-er the same osmoreceptors mediate both effects. Vasopressin secretion is thus controlled by a delicate feed-back mechanism that operates continuously to defend the osmolality of the plasma. Significant changes in secretion occur when osmolality is changed as little as 1%. In this way, the osmolality of the plasma in normal individuals is main-tained very close to 285 mOsm/L. VOLUME EFFECTS ON VASOPRESSIN SECRETION ECF volume also affects vasopressin secretion. Vasopressin secretion is increased when ECF volume is low and decreased when ECF volume is high (Table 39–1). There is an inverse re-lationship between the rate of vasopressin secretion and the rate of discharge in afferents from stretch receptors in the low-and high-pressure portions of the vascular system. The low-pressure receptors are those in the great veins, right and left atria, and pulmonary vessels; the high-pressure receptors are those in the carotid sinuses and aortic arch (see Chapter 33). The exponential increases in plasma vasopressin produced by decreases in blood pressure are documented in Figure 39–3. However, the low-pressure receptors monitor the fullness of the vascular system, and moderate decreases in blood volume that decrease central venous pressure without lowering arteri-al pressure can also increase plasma vasopressin. Thus, the low-pressure receptors are the primary mediators of volume effects on vasopressin secretion. Impulses pass from them via the vagi to the nucleus of the tractus solitarius (NTS). An inhibitory pathway projects from the NTS to the caudal ventrolateral medulla (CVLM), and there is a direct excitatory pathway from the CVLM to the hypothalamus. Angiotensin II reinforces the response to hypovolemia and hypotension by acting on the circumventricular organs to increase vasopressin secretion (see Chapter 34). Hypovolemia and hypotension produced by conditions such as hemorrhage release large amounts of vasopressin, and in the presence of hypovolemia, the osmotic response curve is shifted to the left (Figure 39–4). Its slope is also increased. The result is water retention and reduced plasma osmolality. This includes hyponatremia, since Na+ is the most abundant osmotically active component of the plasma. OTHER STIMULI AFFECTING VASOPRESSIN SECRETION A variety of stimuli in addition to osmotic pressure changes and ECF volume aberrations increase vasopressin secretion. These include pain, nausea, surgical stress, and some emo-tions (Table 39–1). Nausea is associated with particularly large increases in vasopressin secretion. Alcohol decreases vaso-pressin secretion. CLINICAL IMPLICATIONS In various clinical conditions, volume and other non-osmotic stimuli bias the osmotic control of vasopressin secretion. For example, patients who have had surgery may have elevated le-vels of plasma vasopressin because of pain and hypovolemia, and this may cause them to develop a low plasma osmolality and dilutional hyponatremia (see Clinical Box 39–1). TABLE 39–1 Summary of stimuli affecting vasopressin secretion. Vasopressin Secretion Increased Vasopressin Secretion Decreased Increased effective osmotic pres-sure of plasma Decreased effective osmotic pres-sure of plasma Decreased ECF volume Increased ECF volume Pain, emotion, “stress,” exercise Alcohol Nausea and vomiting Standing Clofibrate, carbamazepine Angiotensin II FIGURE 39–3 Relation of mean arterial blood pressure to plasma vasopressin in healthy adult humans in whom a progressive decline in blood pressure was induced by infusion of graded doses of the ganglionic blocking drug trimethaphan. The relation is exponential rather than linear. (Drawn from data in Baylis PH: Osmoregulation and control of vasopressin secretion in healthy humans. Am J Physiol 1987;253:R671.) 100 80 60 40 20 0 −30 −10 −20 0 % Change in mean arterial blood pressure Plasma vasopressin (pmol/L) 668 SECTION VIII Renal Physiology Patients with inappropriate hypersecretion of vasopressin have been successfully treated with demeclocycline, an antibi-otic that reduces the renal response to vasopressin. Diabetes insipidus is the syndrome that results when there is a vasopressin deficiency (central diabetes insipidus) or when the kidneys fail to respond to the hormone (nephro-genic diabetes insipidus). Causes of vasopressin deficiency include disease processes in the supraoptic and paraventricular nuclei, the hypothala-mohypophysial tract, or the posterior pituitary gland. It has been estimated that 30% of the clinical cases are due to neo-plastic lesions of the hypothalamus, either primary or meta-static; 30% are posttraumatic; 30% are idiopathic; and the remainder are due to vascular lesions, infections, systemic diseases such as sarcoidosis that affect the hypothalamus, or mutations in the gene for prepropressophysin. The disease that develops after surgical removal of the posterior lobe of the pituitary may be temporary if only the distal ends of the supraoptic and paraventricular fibers are damaged, because the fibers recover, make new vascular connections, and begin to secrete vasopressin again. The symptoms of diabetes insipidus are passage of large amounts of dilute urine (polyuria) and the drinking of large amounts of fluid (polydipsia), provided the thirst mechanism is intact. It is the polydipsia that keeps these patients healthy. If their sense of thirst is depressed for any reason and their intake of dilute fluid decreases, they develop dehydration that can be fatal. Another cause of diabetes insipidus is inability of the kid-neys to respond to vasopressin (nephrogenic diabetes insipi-dus). Two forms of this disease have been described. In one form, the gene for the V2 receptor is mutated, making the receptor unresponsive. The V2 receptor gene is on the X chro-mosome, thus this condition is X-linked and inheritance is sex-linked recessive. In the other form of the condition, muta-tions occur in the autosomal gene for aquaporin-2 and pro-duce nonfunctional versions of this water channel, many of which do not reach the apical membrane of the collecting duct but are trapped in intracellular locations. The amelioration of diabetes insipidus produced by the development of concomitant anterior pituitary insufficiency is discussed in Chapter 24. SYNTHETIC AGONISTS & ANTAGONISTS Synthetic peptides that have selective actions and are more active than naturally occurring vasopressin and oxytocin have been produced by altering the amino acid residues. For example, 1-deamino-8-D-arginine vasopressin (desmopressin; dDAVP) has very high antidiuretic activity with little pressor activity, making it valuable in the treatment of vasopressin deficiency. FIGURE 39–4 Effect of hypovolemia and hypervolemia on the relation between plasma vasopressin (pAVP) and plasma osmolality (posm). Seven blood samples were drawn at various times from 10 normal men when hypovolemia was induced by water depri-vation (green circles, dashed line) and again when hypervolemia was induced by infusion of hypertonic saline (red circles, solid line). Linear regression analysis defined the relationship pAVP = 0.52 (posm – 283.5) for water deprivation and pAVP = 0.38 (posm – 285.6) for hyper-tonic saline. LD, limit of detection. Note the steeper curve as well as the shift of the intercept to the left during hypovolemia. (Courtesy of CJ Thompson.) 10 5 LD 280 290 300 310 posm (mosm/kg) pAVP (pmol/L) CLINICAL BOX 39–1 Syndrome of Inappropriate Antidiurectic Hormone The syndrome of “inappropriate” hypersecretion of an-tidiuretic hormone (SIADH) occurs when vasopressin is inappropriately high relative to serum osmolality. Vaso-pressin is responsible not only for dilutional hyponatremia (serum sodium < 135 mmol/L) but also for loss of salt in the urine when water retention is sufficient to expand the ECF volume, reducing aldosterone secretion (see Chapter 22). This occurs in patients with cerebral disease (“cerebral salt wasting”) and pulmonary disease (“pulmonary salt wast-ing”). Hypersecretion of vasopressin in patients with pul-monary diseases such as lung cancer may be due in part to the interruption of inhibitory impulses in vagal afferents from the stretch receptors in the atria and great veins. However, a significant number of lung tumors and some other cancers secrete vasopressin. There is a process called “vasopressin escape” that counteracts the water-retain-ing action of vasopressin to limit the degree of hyponatre-mia in SIADH. Studies in rats have demonstrated that pro-longed exposure to elevated levels of vasopressin can lead eventually to down-regulation of the production of aqua-porin-2. This permits urine flow to suddenly increase and plasma osmolality to fall despite exposure of the collecting ducts to elevated levels of the hormone; that is, the individ-ual escapes from the renal effects of vasopressin. CHAPTER 39 Regulation of Extracellular Fluid Composition & Volume 669 DEFENSE OF VOLUME The volume of the ECF is determined primarily by the total amount of osmotically active solute in the ECF. The composi-tion of the ECF is discussed in Chapter 2. Because Na+ and Cl– are by far the most abundant osmotically active solutes in ECF, and because changes in Cl– are to a great extent secon-dary to changes in Na+, the amount of Na+ in the ECF is the most important determinant of ECF volume. Therefore, the mechanisms that control Na+ balance are the major mecha-nisms defending ECF volume. However, there is volume con-trol of water excretion as well; a rise in ECF volume inhibits vasopressin secretion, and a decline in ECF volume produces an increase in the secretion of this hormone. Volume stimuli override the osmotic regulation of vasopressin secretion. An-giotensin II stimulates aldosterone and vasopressin secretion. It also causes thirst and constricts blood vessels, which help to maintain blood pressure. Thus, angiotensin II plays a key role in the body’s response to hypovolemia (Figure 39–5). In addi-tion, expansion of the ECF volume increases the secretion of atrial natriuretic peptide (ANP) and brain natriuretic peptide (BNP) by the heart, and this causes natriuresis and diuresis. In disease states, loss of water from the body (dehydration) causes a moderate decrease in ECF volume, because water is FIGURE 39–5 Summary of the renin–angiotensin system and the stimulation of aldosterone secretion by angiotensin II. The plas-ma concentration of renin is the rate-limiting step in the renin–angiotensin system; therefore, it is the major determinant of plasma angiotensin II concentration. Vasoconstriction Angiotensin I (10 aa) Angiotensin-converting enzyme (endothelium) Angiotensin-converting enzyme (endothelium) Angiotensin II (8 aa) Cardiovascular system Kidney Adrenal cortex Aldosterone Renin (enzyme) Angiotensinogen (453 aa) Liver Kidney Stimuli to renin Angiotensin I Angiotensin II Blood pressure Salt and H2O retention 670 SECTION VIII Renal Physiology lost from both the intracellular and extracellular fluid com-partments; but loss of Na+ in the stools (diarrhea), urine (severe acidosis, adrenal insufficiency), or sweat (heat pros-tration) decreases ECF volume markedly and eventually leads to shock. The immediate compensations in shock operate principally to maintain intravascular volume, but they also affect Na+ balance. In adrenal insufficiency, the decline in ECF volume is due not only to loss of Na+ in the urine but also to its movement into cells. Because of the key position of Na+ in volume homeostasis, it is not surprising that more than one mechanism has evolved to control the excretion of this ion. The filtration and reabsorption of Na+ in the kidneys and the effects of these processes on Na+ excretion are discussed in Chapter 38. When ECF volume is decreased, blood pressure falls, glomerular capillary pressure declines, and the glomerular filtration rate (GFR) therefore falls, reducing the amount of Na+ filtered. Tubular reabsorption of Na+ is increased, in part because the secretion of aldosterone is increased. Aldosterone secretion is controlled in part by a feedback system in which the change that initiates increased secretion is a decline in mean intravascular pressure. Other changes in Na+ excretion occur too rapidly to be due solely to changes in aldosterone secretion. For example, rising from the supine to the standing position increases aldosterone secretion. However, Na+ excretion is decreased within a few minutes, and this rapid change in Na+ excretion occurs in adrenalectomized subjects. It is probably due to hemodynamic changes and possibly to decreased ANP secretion. The kidneys produce three hormones: 1,25-dihydroxychole-calciferol (see Chapter 23), renin, and erythropoietin. Natri-uretic peptides, substances secreted by the heart and other tissues, increase excretion of sodium by the kidneys, and an additional natriuretic hormone inhibits Na, K ATPase. THE RENIN–ANGIOTENSIN SYSTEM RENIN The rise in blood pressure produced by injection of kidney ex-tracts is due to renin, an acid protease secreted by the kidneys into the bloodstream. This enzyme acts in concert with angio-tensin-converting enzyme to form angiotensin II (Figure 39–6). It is a glycoprotein with a molecular weight of 37,326 in hu-mans. The molecule is made up of two lobes, or domains, be-tween which the active site of the enzyme is located in a deep cleft. Two aspartic acid residues, one at position 104 and one at position 292 (residue numbers from human preprorenin), are juxtaposed in the cleft and are essential for activity. Thus, renin is an aspartyl protease. Like other hormones, renin is synthesized as a large prepro-hormone. Human preprorenin contains 406 amino acid resi-dues. The prorenin that remains after removal of a leader sequence of 23 amino acid residues from the amino terminal contains 383 amino acid residues, and after removal of the pro sequence from the amino terminal of prorenin, active renin contains 340 amino acid residues. Prorenin has little if any biologic activity. Some prorenin is converted to renin in the kidneys, and some is secreted. Prorenin is secreted by other organs, including the ovaries. After nephrectomy, the prorenin level in the circulation is usually only moderately reduced and may actually rise, but the active-renin level falls to essen-tially zero. Thus, very little prorenin is converted to renin in the circulation, and active renin is a product primarily, if not exclusively, of the kidneys. Prorenin is secreted consti-tutively, whereas active renin is formed in the secretory granules of the juxtaglomerular cells, the cells in the kid-neys that produce renin (see below). Active renin has a half-life in the circulation of 80 min or less. Its only known func-tion is to split the decapeptide angiotensin I from the amino terminal end of angiotensinogen (renin substrate) (Figure 39–7). ANGIOTENSINOGEN Circulating angiotensinogen is found in the α2-globulin frac-tion of the plasma (Figure 39–6). It contains about 13% carbo-hydrate and is made up of 453 amino acid residues. It is synthesized in the liver with a 32-amino-acid signal sequence that is removed in the endoplasmic reticulum. Its circulating level is increased by glucocorticoids, thyroid hormones, estro-gens, several cytokines, and angiotensin II. ANGIOTENSIN-CONVERTING ENZYME & ANGIOTENSIN II Angiotensin-converting enzyme (ACE) is a dipeptidyl car-boxypeptidase that splits off histidyl-leucine from the physio-logically inactive angiotensin I, forming the octapeptide angiotensin II (Figure 39–7). The same enzyme inactivates bradykinin (Figure 39–6). Increased tissue bradykinin pro-duced when ACE is inhibited acts on B2 receptors to produce the cough that is an annoying side effect in up to 20% of FIGURE 39–6 Formation and metabolism of circulating angiotensins. Angiotensinogen Angiotensin I Angiotensin II AIII, AIV, others Various peptidases Inactive metabolites Renin Angiotensin-converting enzyme AT1 receptors AT2 receptors Bradykinin Inactive metabolites CHAPTER 39 Regulation of Extracellular Fluid Composition & Volume 671 patients treated with ACE inhibitors (see Clinical Box 39–2). Most of the converting enzyme that forms angiotensin II in the circulation is located in endothelial cells. Much of the con-version occurs as the blood passes through the lungs, but con-version also occurs in many other parts of the body. ACE is an ectoenzyme that exists in two forms: a somatic form found throughout the body and a germinal form found solely in postmeiotic spermatogenic cells and spermatozoa (see Chapter 25). Both ACEs have a single transmembrane domain and a short cytoplasmic tail. However, somatic ACE is a 170-kDa protein with two homologous extracellular domains, each containing an active site (Figure 39–8). Germi-nal ACE is a 90-kDa protein that has only one extracellular domain and active site. Both enzymes are formed from a sin-gle gene. However, the gene has two different promoters, pro-ducing two different mRNAs. In male mice in which the ACE gene has been knocked out, blood pressure is lower than nor-mal, but in females it is normal. In addition, fertility is reduced in males but not in females. METABOLISM OF ANGIOTENSIN II Angiotensin II is metabolized rapidly; its half-life in the circu-lation in humans is 1 to 2 min. It is metabolized by various peptidases. An aminopeptidase removes the aspartic acid (Asp) residue from the amino terminal of the peptide (Figure 39–7). The resulting heptapeptide has physiologic activity and is sometimes called angiotensin III. Removal of a second ami-no terminal residue from angiotensin III produces the hexapeptide sometimes called angiotensin IV, which is also said to have some activity. Most, if not all, of the other peptide fragments that are formed are inactive. In addition, ami-nopeptidase can act on angiotensin I to produce (des-Asp1) angiotensin I, and this compound can be converted directly to angiotensin III by the action of ACE. Angiotensin-metaboliz-ing activity is found in red blood cells and many tissues. In ad-dition, angiotensin II appears to be removed from the circulation by some sort of trapping mechanism in the vascu-lar beds of tissues other than the lungs. Renin is usually measured by incubating the sample to be assayed and measuring by immunoassay the amount of angio-tensin I generated. This measures the plasma renin activity (PRA) of the sample. Deficiency of angiotensinogen as well as renin can cause low PRA values, and to avoid this problem, exogenous angiotensinogen is often added, so that plasma renin concentration (PRC) rather than PRA is measured. The normal PRA in supine subjects eating a normal amount of CLINICAL BOX 39–2 Pharmacologic Manipulation of the Renin–Angiotensin System It is now possible to inhibit the secretion or the effects of renin in a variety of ways. Inhibitors of prostaglandin synthe-sis such as indomethacin and β-adrenergic blocking drugs such as propranolol reduce renin secretion. The peptide pepstatin and newly developed renin inhibitors such as enalkiren prevent renin from generating angiotensin I. An-giotensin-converting enzyme inhibitors (ACE inhibitors) such as captopril and enalapril prevent conversion of angioten-sin I to angiotensin II. Saralasin and several other analogs of angiotensin II are competitive inhibitors of the action of an-giotensin II on both AT1 and AT2 receptors. Losartan (DuP-753) selectively blocks AT1 receptors, and PD-123177 and several other drugs selectively block AT2 receptors. FIGURE 39–7 Structure of the amino terminal end of angiotensinogen and angiotensins I, II, and III in humans. R, remainder of pro-tein. After removal of a 24-amino-acid leader sequence, angiotensinogen contains 453 amino acid residues. The structure of angiotensin II in dogs, rats, and many other mammals is the same as that in humans. Bovine and ovine angiotensin II have valine instead of isoleucine at position 5. Asp-Arg-Val-Tyr-Ile-His-Pro-Phe-His-Leu-Val-Ile-His-R Renin splits this bond Angiotensinogen Asp-Arg-Val-Tyr-Ile-His-Pro-Phe Aminopeptidase splits this bond Angiotensin II Asp-Arg-Val-Tyr-Ile-His-Pro-Phe-His-Leu Angiotensin-converting enzyme splits this bond Angiotensin I Angiotensin III 672 SECTION VIII Renal Physiology sodium is approximately 1 ng of angiotensin I generated per milliliter per hour. The plasma angiotensin II concentration in such subjects is about 25 pg/mL (approximately 25 pmol/L). ACTIONS OF ANGIOTENSINS Angiotensin I appears to function solely as the precursor of angiotensin II and does not have any other established action. Angiotensin II—previously called hypertensin or angioto-nin—produces arteriolar constriction and a rise in systolic and diastolic blood pressure. It is one of the most potent vasocon-strictors known, being four to eight times as active as norepi-nephrine on a weight basis in normal individuals. However, its pressor activity is decreased in Na+-depleted individuals and in patients with cirrhosis and some other diseases. In these condi-tions, circulating angiotensin II is increased, and this down-regulates the angiotensin receptors in vascular smooth muscle. Consequently, there is less response to injected angiotensin II. Angiotensin II also acts directly on the adrenal cortex to increase the secretion of aldosterone, and the renin–angiotensin system is a major regulator of aldosterone secretion. Addi-tional actions of angiotensin II include facilitation of the release of norepinephrine by a direct action on postganglionic sympathetic neurons, contraction of mesangial cells with a resultant decrease in glomerular filtration rate (see Chapter 38), and a direct effect on the renal tubules to increase Na+ reabsorption. Angiotensin II also acts on the brain to decrease the sensi-tivity of the baroreflex, and this potentiates the pressor effect of angiotensin II. In addition, it acts on the brain to increase water intake and increase the secretion of vasopressin and ACTH. It does not penetrate the blood–brain barrier, but it triggers these responses by acting on the circumventricular organs, four small structures in the brain that are outside the blood–brain barrier (see Chapter 34). One of these structures, the area postrema, is primarily responsible for the pressor potentiation, whereas two of the others, the subfornical organ (SFO) and the organum vasculosum of the lamina terminalis (OVLT), are responsible for the increase in water intake (dip-sogenic effect). It is not certain which of the circumventricu-lar organs are responsible for the increases in vasopressin and ACTH secretion. Angiotensin III [(des-Asp1) angiotensin II] has about 40% of the pressor activity of angiotensin II, but 100% of the aldos-terone-stimulating activity. It has been suggested that angio-tensin III is the natural aldosterone-stimulating peptide, whereas angiotensin II is the blood-pressure-regulating pep-tide. However, this appears not to be the case, and instead angiotensin III is simply a breakdown product with some bio-logic activity. The same is probably true of angiotensin IV, though some researchers have argued that it has unique effects in the brain. TISSUE RENIN–ANGIOTENSIN SYSTEMS In addition to the system that generates circulating angioten-sin II, many different tissues contain independent renin– angiotensin systems that generate angiotensin II, apparently for local use. Components of the renin–angiotensin system are found in the walls of blood vessels and in the uterus, the pla-centa, and the fetal membranes. Amniotic fluid has a high con-centration of prorenin. In addition, tissue renin–angiotensin systems, or at least several components of the renin–angiotensin system, are present in the eyes, exocrine portion of the pancre-as, heart, fat, adrenal cortex, testis, ovary, anterior and inter-mediate lobes of the pituitary, pineal, and brain. Tissue renin contributes very little to the circulating renin pool, because plas-ma renin activity falls to undetectable levels after the kidneys are removed. The functions of these tissue renin–angiotensin systems are unsettled, though evidence is accumulating that angiotensin II is a significant growth factor in the heart and blood vessels. ACE inhibitors or AT1 receptor blockers are now the treatment of choice for congestive heart failure, and part of their value may be due to inhibition of the growth ef-fects of angiotensin II. ANGIOTENSIN II RECEPTORS There are at least two classes of angiotensin II receptors. AT1 receptors are serpentine receptors coupled by a G protein (Gq) to phospholipase C, and angiotensin II increases the cytosolic free Ca2+ level. It also activates numerous tyrosine kinases. In vascular smooth muscle, AT1 receptors are associated with ca-veolae (see Chapter 2), and AII increases production of cave-olin-1, one of the three isoforms of the protein that is FIGURE 39–8 Diagrammatic representation of the structure of the somatic form of angiotensin-converting enzyme. Note the short cytoplasmic tail of the molecule and the two extracellular cata-lytic sites, each of which binds a zinc ion (Zn2+). (Reproduced with permission from Johnston CI: Tissue angiotensin-converting enzyme in cardiac and vascular hypertrophy, repair, and remodeling. Hypertension 1994;23:258. Copyright © 1994 by The American Heart Association.) Extracellular extension Carboxyl terminal catalytic site Amino terminal catalytic site Trans-membrane domain Intracellular extension COOH Zn2+ Zn2+ NH2 CHAPTER 39 Regulation of Extracellular Fluid Composition & Volume 673 characteristic of caveolae. In rodents, two different but closely related AT1 subtypes, AT1A and AT1B, are coded by two sepa-rate genes. The AT1A subtype is found in blood vessel walls, the brain, and many other organs. It mediates most of the known effects of angiotensin II. The AT1B subtype is found in the anterior pituitary and the adrenal cortex. In humans, an AT1 receptor gene is present on chromosome 3. There may be a second AT1 type, but it is still unsettled whether distinct AT1A and AT1B subtypes occur. There are also AT2 receptors, which are coded in humans by a gene on the X chromosome. Like the AT1 receptors, they have seven transmembrane domains, but their actions are dif-ferent. They act via a G protein to activate various phos-phatases which in turn antagonize growth effects and open K+ channels. In addition, AT2 receptor activation increases the production of NO and therefore increases intracellular cyclic 3,5-guanosine monophosphate (cGMP). The overall physio-logic consequences of these second-messenger effects are unsettled. AT2 receptors are more plentiful in fetal and neona-tal life, but they persist in the brain and other organs in adults. The AT1 receptors in the arterioles and the AT1 receptors in the adrenal cortex are regulated in opposite ways: an excess of angiotensin II down-regulates the vascular recep-tors, but it up-regulates the adrenocortical receptors, making the gland more sensitive to the aldosterone-stimulating effect of the peptide. THE JUXTAGLOMERULAR APPARATUS The renin in kidney extracts and the bloodstream is produced by the juxtaglomerular cells (JG cells). These epitheloid cells are located in the media of the afferent arterioles as they enter the glomeruli (Figure 39–9). The membrane-lined secretory granules in them have been shown to contain renin. Renin is also found in agranular lacis cells that are located in the junc-tion between the afferent and efferent arterioles, but its signif-icance in this location is unknown. At the point where the afferent arteriole enters the glomeru-lus and the efferent arteriole leaves it, the tubule of the neph-ron touches the arterioles of the glomerulus from which it arose. At this location, which marks the start of the distal con-volution, there is a modified region of tubular epithelium called the macula densa (Figure 39–9). The macula densa is in close proximity to the JG cells. The lacis cells, the JG cells, and the macula densa constitute the juxtaglomerular apparatus. REGULATION OF RENIN SECRETION Several different factors regulate renin secretion (Table 39–2), and the rate of renin secretion at any given time is determined by the summed activity of these factors. One factor is an intra-renal baroreceptor mechanism that causes renin secretion to decrease when arteriolar pressure at the level of the JG cells in-creases and to increase when arteriolar pressure at this level falls. Another renin-regulating sensor is in the macula densa. Renin secretion is inversely proportional to the amount of Na+ and Cl– entering the distal renal tubules from the loop of Henle. Presumably, these electrolytes enter the macula densa cells via the Na–K–2Cl– transporters in their apical membranes, and the increase in some fashion triggers a signal that decreases renin secretion in the juxtaglomerular cells in the adjacent afferent ar-terioles. A possible mediator is NO, but the identity of the signal remains unsettled. Renin secretion also varies inversely with the plasma K+ level, but the effect of K+ appears to be mediated by the changes it produces in Na+ and Cl– delivery to the macula densa. FIGURE 39–9 Left: Diagram of glomerulus, showing the juxtaglomerular apparatus. Right: Phase contrast photomicrograph of afferent ar-teriole in an unstained, freeze-dried preparation of the kidney of a mouse. Note the red blood cell in the lumen of the arteriole and the granulated juxtaglomerular cells in the wall. (Courtesy of C Peil.) Juxtaglomerular cells Lacis cells Macula densa Afferent arteriole Efferent arteriole Glomerulus Renal nerves 674 SECTION VIII Renal Physiology Angiotensin II feeds back to inhibit renin secretion by a direct action on the JG cells. Vasopressin also inhibits renin secretion in vitro and in vivo, although there is some debate about whether its in vivo effect is direct or indirect. Finally, increased activity of the sympathetic nervous sys-tem increases renin secretion. The increase is mediated both by increased circulating catecholamines and by norepineph-rine secreted by postganglionic renal sympathetic nerves. The catecholamines act mainly on β1-adrenergic receptors on the JG cells and renin release is mediated by an increase in intra-cellular cAMP. The principal conditions that increase renin secretion in humans are listed in Table 39–3. Most of the listed conditions decrease central venous pressure, which triggers an increase in sympathetic activity, and some also decrease renal arteri-olar pressure (see Clinical Box 39–3). Renal artery constric-tion and constriction of the aorta proximal to the renal arteries produces a decrease in renal arteriolar pressure. Psy-chologic stimuli increase the activity of the renal nerves. HORMONES OF THE HEART & OTHER NATRIURETIC FACTORS STRUCTURE The existence of various natriuretic hormones has been postu-lated for some time. Two of these are secreted by the heart. The muscle cells in the atria and, to a much lesser extent in the ven-tricles, contain secretory granules (Figure 39–10). The granules increase in number when NaCl intake is increased and ECF ex-panded, and extracts of atrial tissue cause natriuresis. The first natriuretic hormone isolated from the heart was atrial natriuretic peptide (ANP), a polypeptide with a char-acteristic 17-amino-acid ring formed by a disulfide bond between two cysteines. The circulating form of this polypep-tide has 28 amino acid residues (Figure 39–11). It is formed from a large precursor molecule that contains 151 amino acid residues, including a 24-amino-acid signal peptide. ANP was subsequently isolated from other tissues, including the brain, where it exists in two forms that are smaller than circulating ANP. A second natriuretic polypeptide was isolated from por-cine brain and named brain natriuretic peptide (BNP; also known as B-type natriuretic peptide). It is also present in the brain in humans, but more is present in the human heart, including the ventricles. The circulating form of this hormone contains 32 amino acid residues. It has the same 17-member ring as ANP, though some of the amino acid residues in the ring are different (Figure 39–11). A third member of this TABLE 39–2 Factors that affect renin secretion. Stimulatory Increased sympathetic activity via renal nerves Increased circulating catecholamines Prostaglandins Inhibitory Increased Na+ and Cl– reabsorption across macula densa Increased afferent arteriolar pressure Angiotensin II Vasopressin TABLE 39–3 Conditions that increase renin secretion. Na+ depletion Diuretics Hypotension Hemorrhage Upright posture Dehydration Cardiac failure Cirrhosis Constriction of renal artery or aorta Various psychologic stimuli CLINICAL BOX 39–3 Role of Renin in Clinical Hypertension Constriction of one renal artery causes a prompt increase in renin secretion and the development of sustained hyper-tension (renal or Goldblatt hypertension). Removal of the ischemic kidney or the arterial constriction cures the hyper-tension if it has not persisted too long. In general, the hy-pertension produced by constricting one renal artery with the other kidney intact (one-clip, two-kidney Goldblatt hy-pertension) is associated with increased circulating renin. The clinical counterpart of this condition is renal hyper-tension due to atheromatous narrowing of one renal artery or other abnormalities of the renal circulation. However, plasma renin activity is usually normal in one-clip one-kidney Goldblatt hypertension. The explanation of the hy-pertension in this situation is unsettled. However, many pa-tients with hypertension respond to treatment with ACE in-hibitors or losartan even when their renal circulation appears to be normal and they have normal or even low plasma renin activity. CHAPTER 39 Regulation of Extracellular Fluid Composition & Volume 675 family has been named C-type natriuretic peptide (CNP) because it was the third in the sequence to be isolated. It con-tains 22 amino acid residues (Figure 39–11), and there is also a larger 53-amino-acid form. CNP is present in the brain, the pituitary, the kidneys, and vascular endothelial cells. However, very little is present in the heart and the circulation, and it appears to be primarily a paracrine mediator. ACTIONS ANP and BNP in the circulation act on the kidneys to increase Na+ excretion, and injected CNP has a similar effect. They ap-pear to produce this effect by dilating afferent arterioles and relaxing mesangial cells. Both of these actions increase glo-merular filtration (see Chapter 38). In addition, they act on the renal tubules to inhibit Na+ reabsorption. Other actions in-clude an increase in capillary permeability, leading to extrava-sation of fluid and a decline in blood pressure. In addition, they relax vascular smooth muscle in arterioles and venules. CNP has a greater dilator effect on veins than ANP and BNP. These peptides also inhibit renin secretion and counteract the pressor effects of catecholamines and angiotensin II. In the brain, ANP is present in neurons, and an ANP-con-taining neural pathway projects from the anteromedial part of the hypothalamus to the areas in the lower brain stem that are concerned with neural regulation of the cardiovascular sys-tem. In general, the effects of ANP in the brain are opposite to those of angiotensin II, and ANP-containing neural circuits appear to be involved in lowering blood pressure and promot-ing natriuresis. CNP and BNP in the brain probably have functions similar to those of ANP, but detailed information is not available. NATRIURETIC PEPTIDE RECEPTORS Three different natriuretic peptide receptors (NPR) have been isolated and characterized (Figure 39–12). The NPR-A and NPR-B receptors both span the cell membrane and have cyto-plasmic domains that are guanylyl cyclases. ANP has the greatest affinity for the NPR-A receptor, and CNP has the greatest affinity for the NPR-B receptor. The third receptor, NPR-C, binds all three natriuretic peptides but has a markedly truncated cytoplasmic domain. Some evidence suggests that it acts via G proteins to activate phospholipase C and inhibit adenylyl cyclase. However, it has also been argued that this FIGURE 39–10 ANP granules (g) interspersed between mitochondria (m) in rat atrial muscle cell. G, Golgi complex; N, nu-cleus. The granules in human atrial cells are similar (× 17,640). (Courtesy of M Cantin.) N G g m FIGURE 39–11 Human ANP, BNP, and CNP. Top: Single-letter codes for amino acid residues aligned to show common sequences (col-ored). Bottom: Shape of molecules. Note that one cysteine is the carboxyl terminal amino acid residue in CNP, so there is no carboxyl terminal extension from the 17-member ring. (Modified from Imura H, Nakao K, Itoh H: The natriuretic peptide system in the brain: Implication in the central control of cardiovascular and neuroendocrine functions. Front Neuroendocrinol 1992;13:217.) ANP BNP CNP HOOC H2N HOOC HOOC ANP BNP CNP H2N H2N SLRRSSCFGGRMDRIGAQSGLGCNSFRY SPKMVQGSGCFGRKMDRISSSSGLGCKVLRRH GLSKGCFGLKLDRIGSMSGLGC 1 28 1 32 1 22 676 SECTION VIII Renal Physiology receptor does not trigger any intracellular change and is in-stead a clearance receptor that removes natriuretic peptides from the bloodstream and then releases them later, helping to maintain a steady blood level of the hormones. SECRETION & METABOLISM The concentration of ANP in plasma is about 5 fmol/mL in normal humans ingesting moderate amounts of NaCl. ANP se-cretion is increased when the ECF volume is increased by infu-sion of isotonic saline and when the atria are stretched. BNP secretion is increased when the ventricles are stretched. ANP secretion is also increased by immersion in water up to the neck (Figure 39–13), a procedure that counteracts the effect of grav-ity on the circulation and increases central venous and conse-quently atrial pressure. Note that immersion also decreases the secretion of renin and aldosterone. Conversely, a small but measurable decrease in plasma ANP occurs in association with a decrease in central venous pressure on rising from the supine to the standing position. Thus, it seems clear that the atria re-spond directly to stretch in vivo and that the rate of ANP secre-tion is proportional to the degree to which the atria are stretched by increases in central venous pressure. Similarly, BNP secretion is proportional to the degree to which the ven-tricles are stretched. Plasma levels of both hormones are elevat-ed in congestive heart failure, and their measurement is seeing increasing use in the diagnosis of this condition. Circulating ANP has a short half-life. It is metabolized by neutral endopeptidase (NEP), which is inhibited by thior-phan. Therefore, administration of thiorphan increases circu-lating ANP. Na, K ATPase-INHIBITING FACTOR Another natriuretic factor is present in blood. This factor pro-duces natriuresis by inhibiting Na, K ATPase and raises rather than lowers blood pressure. Current evidence indicates that it may well be the digitalis-like steroid ouabain and that it comes from the adrenal glands. However, its physiologic significance is not yet known. DEFENSE OF SPECIFIC IONIC COMPOSITION Special regulatory mechanisms maintain the levels of certain specific ions in the ECF as well as the levels of glucose and other nonionized substances important in metabolism (see Chapter 1). The feedback of Ca2+ on the parathyroids and the calcitonin-secreting cells to adjust their secretion maintains the ionized calcium level of the ECF (see Chapter 23). The Mg2+ concentra-tion is subject to close regulation, but the mechanisms control-ling Mg+ metabolism are incompletely understood. The mechanisms controlling Na+ and K+ content are part of those determining the volume and tonicity of ECF and have been discussed above. The levels of these ions are also dependent on the H+ concentration, and pH is one of the major factors affecting the anion composition of ECF. This will be discussed in Chapter 40. FIGURE 39–12 Diagrammatic representation of natriuretic peptide receptors. The NPR-A and NPR-B receptor molecules have in-tracellular guanylyl cyclase domains, whereas the putative clearance re-ceptor, NPR-C, has only a small cytoplasmic domain. CM, cell membrane. ECF Cytoplasm Guanylyl cyclase domain CM NPR-A NPR-B NPR-C FIGURE 39–13 Effect of immersion in water up to the neck for 3 h on plasma concentrations of ANP, PRA, and aldosterone. (Modified and reproduced with permission from Epstein M, et al: Increases in circulating atrial natriuretic factor during immersion-induced central hypervolaemia in normal humans. Hypertension 1986;4 [Suppl 2]:593.) 15 10 5 Immersion 0 3 2 1 0 10 5 0 0 1 2 3 4 5 Time (hrs) Aldosterone (ng/dL) PRA (ng AI/mL/hr) ANP (fmol/mL) CHAPTER 39 Regulation of Extracellular Fluid Composition & Volume 677 ERYTHROPOIETIN STRUCTURE & FUNCTION When an individual bleeds or becomes hypoxic, hemoglobin synthesis is enhanced, and production and release of red blood cells from the bone marrow (erythropoiesis) are increased (see Chapter 32). Conversely, when the red cell volume is in-creased above normal by transfusion, the erythropoietic activ-ity of the bone marrow decreases. These adjustments are brought about by changes in the circulating level of erythro-poietin, a circulating glycoprotein that contains 165 amino acid residues and four oligosaccharide chains that are neces-sary for its activity in vivo. Its blood level is markedly in-creased in anemia (Figure 39–14). Erythropoietin increases the number of erythropoietin-sensitive committed stem cells in the bone marrow that are converted to red blood cell precursors and subsequently to mature erythrocytes (see Chapter 32). The receptor for eryth-ropoietin is a linear protein with a single transmembrane domain that is a member of the cytokine receptor superfamily (see Chapter 3). The receptor has tyrosine kinase activity, and it activates a cascade of serine and threonine kinases, resulting in inhibited apoptosis of red cells and their increased growth and development. The principal site of inactivation of erythropoietin is the liver, and the hormone has a half-life in the circulation of about 5 h. However, the increase in circulating red cells that it triggers takes 2 to 3 d to appear, since red cell maturation is a relatively slow process. Loss of even a small portion of the sialic acid residues in the carbohydrate moieties that are part of the erythropoietin molecule shortens its half-life to 5 min, making it biologically ineffective. SOURCES In adults, about 85% of the erythropoietin comes from the kid-neys and 15% from the liver. Both these organs contain the mRNA for erythropoietin. Erythropoietin can also be extract-ed from the spleen and salivary glands, but these tissues do not contain the mRNA and consequently do not appear to manu-facture the hormone. When renal mass is reduced in adults by renal disease or nephrectomy, the liver cannot compensate and anemia develops. Erythropoietin is produced by interstitial cells in the peritu-bular capillary bed of the kidneys and by perivenous hepato-cytes in the liver. It is also produced in the brain, where it exerts a protective effect against excitotoxic damage triggered by hypoxia; and in the uterus and oviducts, where it is induced by estrogen and appears to mediate estrogen-dependent angiogenesis. The gene for the hormone has been cloned, and recombi-nant erythropoietin produced in animal cells is available for clinical use as epoetin alfa. The recombinant erythropoietin is of value in the treatment of the anemia associated with renal failure; 90% of the patients with end-stage renal failure who are on dialysis are anemic as a result of erythropoietin defi-ciency. Erythropoietin is also used to stimulate red cell pro-duction in individuals who are banking a supply of their own blood in preparation for autologous transfusions during elec-tive surgery (see Chapter 32). REGULATION OF SECRETION The usual stimulus for erythropoietin secretion is hypoxia, but secretion of the hormone can also be stimulated by cobalt salts and androgens. Recent evidence suggests that the O2 sensor regulating erythropoietin secretion in the kidneys and the liver is a heme protein that in the deoxy form stimulates and in the oxy form inhibits transcription of the erythropoietin gene to form erythropoietin mRNA. Secretion of the hormone is facil-itated by the alkalosis that develops at high altitudes. Like renin secretion, erythropoietin secretion is facilitated by catechol-amines via a β-adrenergic mechanism, although the renin– angiotensin system is totally separate from the erythropoietin system. CHAPTER SUMMARY ■Total body osmolality is directly proportional to the total body sodium plus the total body potassium divided by the total body water. Changes in the osmolality of the body fluids occur when a disproportion exists between the amount of these electrolytes and the amount of water ingested or lost from the body. ■Vasopressin’s main physiologic effect is the retention of water by the kidney by increasing the water permeability of the renal collecting ducts. Water is absorbed from the urine, the urine becomes concentrated, and its volume decreases. ■Vasopressin is stored in the posterior pituitary and released into the bloodstream in response to the stimulation of osmoreceptors FIGURE 39–14 Plasma erythropoietin levels in normal blood donors (triangles) and patients with various forms of anemia (squares). (Reproduced with permission from Erslev AJ: Erythropoietin. N Engl J Med 1991;324:1339.) 104 103 102 101 100 105 0 0.10 0.20 0.30 0.40 0.50 0.60 Plasma erythropoietin (U/L) Hematocrit 678 SECTION VIII Renal Physiology or baroreceptors. Increases in secretion occur when osmolality is changed as little as 1%, thus keeping the osmolality of the plas-ma very close to 285 mOsm/L. ■The amount of Na+ in the ECF is the most important determi-nant of ECF volume, and mechanisms that control Na+ balance are the major mechanisms defending ECF volume. The main mechanism regulating sodium balance is the renin–angiotensin system, a hormone system that regulates blood pressure. ■The kidneys secrete the enzyme renin and renin acts in concert with angiotensin-converting enzyme to form angiotensin II. Angiotensin II acts directly on the adrenal cortex to increase the secretion of aldosterone. Aldosterone increases the retention of sodium from the urine via action on the renal collecting duct. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Dehydration increases the plasma concentration of all the fol-lowing hormones except A) vasopressin. B) angiotensin II. C) aldosterone. D) norepinephrine. E) atrial natriuretic peptide. 2. In a patient who has become dehydrated, body water should be replaced by intravenous infusion of A) distilled water. B) 0.9% sodium chloride solution. C) 5% glucose solution. D) hyperoncotic albumin. E) 10% glucose solution. 3. Renin is secreted by A) cells in the macula densa. B) cells in the proximal tubules. C) cells in the distal tubules. D) juxtaglomerular cells. E) cells in the peritubular capillary bed. 4. Erythropoietin is secreted by A) cells in the macula densa. B) cells in the proximal tubules. C) cells in the distal tubules. D) juxtaglomerular cells. E) cells in the peritubular capillary bed. 5. When a woman who has been on a low-sodium diet for 8 d is given an intravenous injection of captopril, a drug that inhibits angiotensin-converting enzyme, one would expect A) her blood pressure to rise because her cardiac output would fall. B) her blood pressure to rise because her peripheral resistance would fall. C) her blood pressure to fall because her cardiac output would fall. D) her blood pressure to fall because her peripheral resistance would fall. E) her plasma renin activity to fall because her circulating angiotensin I level would rise. 6. Which of the following would not be expected to increase renin secretion? A) administration of a drug that blocks angiotensin-converting enzyme B) administration of a drug that blocks AT1 receptors C) administration of a drug that blocks β-adrenergic receptors D) constriction of the aorta between the celiac artery and the renal arteries E) administration of a drug that reduces ECF volume 7. Which of the following is least likely to contribute to the benefi-cial effects of angiotensin-converting enzyme inhibitors in the treatment of congestive heart failure? A) vasodilation B) decreased cardiac growth C) decreased cardiac afterload D) increased plasma renin activity E) decreased plasma aldosterone CHAPTER RESOURCES Adrogue HJ, Madias NE: Hypernatremia. N Engl J Med 2000;342:1493. Adrogue HJ, Madias NE: Hyponatremia. N Engl J Med 2000;342:101. Corvol P, Jeunemaitre X: Molecular genetics of human hypertension: Role of angiotensinogen. Endocr Rev 1997;18:662. Morel F: Sites of hormone action in the mammalian nephron. Am J Physiol 1981;240:F159. McKinley MS, Johnson AK: The physiologic regulation of thirst and fluid intake. News Physiol Sci 2004;19:1. Robinson AG, Verbalis JG: Diabetes insipidus. Curr Ther Endocrinol Metab 1997;6:1. Verkman AS: Mammalian aquaporins: Diverse physiological roles and potential clinical significance. Expert Rev Mol Med. 2008;10:13. Zeidel ML: Hormonal regulation of inner medullary collecting duct sodium transport. Am J Physiol 1993;265:F159. 679 C H A P T E R 40 Acidification of the Urine & Bicarbonate Excretion O B J E C T I V E S After reading this chapter, you should be able to: ■Outline the processes involved in the secretion of H+ into the tubules and discuss the significance of these processes in the regulation of acid–base balance. ■Define acidosis and alkalosis, and give (in mEq/L and pH) the normal mean and the range of H+ concentrations in blood that are compatible with health. ■List the principal buffers in blood, interstitial fluid, and intracellular fluid, and, using the Henderson–Hasselbalch equation, describe what is unique about the bicarbo-nate buffer system. ■Describe the changes in blood chemistry that occur during the development of metabolic acidosis and metabolic alkalosis, and the respiratory and renal compen-sations for these conditions. ■Describe the changes in blood chemistry that occur during the development of respiratory acidosis and respiratory alkalosis, and the renal compensation for these conditions. RENAL H+ SECRETION The cells of the proximal and distal tubules, like the cells of the gastric glands, secrete hydrogen ions (see Chapter 26). Acidi-fication also occurs in the collecting ducts. The reaction that is primarily responsible for H+ secretion in the proximal tubules is Na–H exchange (Figure 40–1). This is an example of secon-dary active transport; extrusion of Na+ from the cells into the interstitium by Na, K ATPase lowers intracellular Na+, and this causes Na+ to enter the cell from the tubular lumen, with coupled extrusion of H+. The H+ comes from intracellular dis-sociation of H2CO3, and the HCO3 – that is formed diffuses into the interstitial fluid. Thus, for each H+ ion secreted, one Na+ ion and one HCO3 – ion enter the interstitial fluid. Carbonic anhydrase catalyzes the formation of H2CO3, and drugs that inhibit carbonic anhydrase depress both secre-tion of acid by the proximal tubules and the reactions which depend on it. Some evidence suggests that H+ is secreted in the proximal tubules by other types of pumps, but the evidence for these additional pumps is controversial, and in any case, their con-tribution is small relative to that of the Na–H exchange mech-anism. This is in contrast to what occurs in the distal tubules and collecting ducts, where H+ secretion is relatively indepen-dent of Na+ in the tubular lumen. In this part of the tubule, most H+ is secreted by an ATP-driven proton pump. Aldos-terone acts on this pump to increase distal H+ secretion. The I cells in this part of the renal tubule secrete acid and, like the parietal cells in the stomach, contain abundant carbonic anhydrase and numerous tubulovesicular structures. There is evidence that the H+-translocating ATPase that produces H+ secretion is located in these vesicles as well as in the luminal cell membrane and that, in acidosis, the number of H+ pumps is increased by insertion of these tubulovesicles into the lumi-nal cell membrane. Some of the H+ is also secreted by H–K+ ATPase. The I cells contain Band 3, an anion exchange pro-tein, in their basolateral cell membranes, and this protein may function as a Cl/HCO3 exchanger for the transport of HCO3 – to the interstitial fluid. FATE OF H+ IN THE URINE The amount of acid secreted depends upon the subsequent events in the tubular urine. The maximal H+ gradient against which the transport mechanisms can secrete in humans corre-sponds to a urine pH of about 4.5; that is, an H+ concentration in 680 SECTION VIII Renal Physiology the urine that is 1000 times the concentration in plasma. pH 4.5 is thus the limiting pH. This is normally reached in the collect-ing ducts. If there were no buffers that “tied up” H+ in the urine, this pH would be reached rapidly, and H+ secretion would stop. However, three important reactions in the tubular fluid remove free H+, permitting more acid to be secreted (Figure 40–2). These are the reactions with HCO3 – to form CO2 and H2O, with HPO4 2– to form H2PO4 –, and with NH3 to form NH4 +. REACTION WITH BUFFERS The dynamics of buffering are discussed in Chapter 1 and be-low. The pK' of the bicarbonate system is 6.1, that of the diba-sic phosphate system is 6.8, and that of the ammonia system is 9.0. The concentration of HCO3 – in the plasma, and conse-quently in the glomerular filtrate, is normally about 24 mEq/ L, whereas that of phosphate is only 1.5 mEq/L. Therefore, in the proximal tubule, most of the secreted H+ reacts with HCO3 – to form H2CO3 (Figure 40–2). The H2CO3 breaks down to form CO2 and H2O. In the proximal (but not in the distal) tubule, there is carbonic anhydrase in the brush border of the cells; this facilitates the formation of CO2 and H2O in the tubular fluid. The CO2, which diffuses readily across all bi-ological membranes, enters the tubular cells, where it adds to the pool of CO2 available to form H2CO3. Because most of the H+ is removed from the tubule, the pH of the fluid is changed very little. This is the mechanism by which HCO3 – is reab-sorbed; for each mole of HCO3 – removed from the tubular flu-id, 1 mol of HCO3 – diffuses from the tubular cells into the blood, even though it is not the same mole that disappeared from the tubular fluid. Secreted H+ also reacts with dibasic phosphate (HPO4 2–) to form monobasic phosphate (H2PO4 –). This happens to the greatest extent in the distal tubules and collecting ducts, because it is here that the phosphate that escapes proximal reabsorption is greatly concentrated by the reabsorption of water. The reaction with NH3 occurs in the proximal and dis-tal tubules. H+ also combines to a minor degree with other buffer anions. Each H+ ion that reacts with the buffers contributes to the urinary titratable acidity, which is measured by determining the amount of alkali that must be added to the urine to return its pH to 7.4, the pH of the glomerular filtrate. However, the titratable acidity obviously measures only a fraction of the acid secreted, since it does not account for the H2CO3 that has been converted to H2O and CO2. In addition, the pK' of the ammonia system is 9.0, and the ammonia system is titrated only from the pH of the urine to pH 7.4, so it contributes very little to the titratable acidity. AMMONIA SECRETION Reactions in the renal tubular cells produce NH4 + and HCO3 –. NH4 + is in equilibrium with NH3 and H+ in the cells. Because the pK' of this reaction is 9.0, the ratio of NH3 to NH4 + at pH 7.0 is 1:100 (Figure 40–3). However, NH3 is lipid-soluble and diffuses across the cell membranes down its concentration gradient into the interstitial fluid and tubular urine. In the urine it reacts with H+ to form NH4 +, and the NH4 + remains in the urine. FIGURE 40–1 Secretion of acid by proximal tubular cells in the kidney. H+ is transported into the tubular lumen by an antiport in exchange for Na+. Active transport by Na, K ATPase is indicated by ar-rows in the circle. Dashed arrows indicate diffusion. Interstitial fluid Tubular lumen H+ Na+ K+ Na+ K+ Renal tubule cell HCO3 − HCO3 − CO2 + H2O H2CO3 + Carbonic anhydrase H+ FIGURE 40–2 Fate of H+ secreted into a tubule in exchange for Na+. Top: Reabsorption of filtered bicarbonate via CO2. Middle: Formation of monobasic phosphate. Bottom: Ammonium formation. Note that in each instance one Na+ ion and one HCO3 – ion enter the bloodstream for each H+ ion secreted. A–, anion. Interstitial fluid Renal tubule cell Tubular lumen Na+ Na+ A− Na+ HCO3− HCO3− HCO3− HCO3− H+ H+ Na+ HCO3− HCO3− Na+ + HCO3− H+ + HCO3− CO2 + H2O Na+ Na+ HPO42− Na+ H2PO4− + NH3 NH3 NH4+ A− H+ H+ H+ CHAPTER 40 Acidification of the Urine & Bicarbonate Excretion 681 The principal reaction producing NH4 + in cells is conver-sion of glutamine to glutamate. This reaction is catalyzed by the enzyme glutaminase, which is abundant in renal tubular cells (Figure 40–3). Glutamic dehydrogenase catalyzes the conversion of glutamate to α-ketoglutarate, with the produc-tion of more NH4 +. Subsequent metabolism of α-ketoglut-arate utilizes 2H+, freeing 2HCO3 –. In chronic acidosis, the amount of NH4 + excreted at any given urine pH also increases, because more NH3 enters the tubular urine. The effect of this adaptation of NH3 secretion, the cause of which is unsettled, is a further removal of H+ from the tubular fluid and consequently a further enhance-ment of H+ secretion. The process by which NH3 is secreted into the urine and then changed to NH4 +, maintaining the concentration gradi-ent for diffusion of NH3, is called nonionic diffusion (see Chapter 2). Salicylates and a number of other drugs that are weak bases or weak acids are also secreted by nonionic diffu-sion. They diffuse into the tubular fluid at a rate that depends on the pH of the urine, so the amount of each drug excreted varies with the pH of the urine. pH CHANGES ALONG THE NEPHRONS A moderate drop in pH occurs in the proximal tubular fluid, but, as noted above, most of the secreted H+ has little effect on luminal pH because of the formation of CO2 and H2O from H2CO3. In contrast, the distal tubule has less capacity to se-crete H+, but secretion in this segment has a greater effect on urinary pH. FACTORS AFFECTING ACID SECRETION Renal acid secretion is altered by changes in the intracellular PCO2, K+ concentration, carbonic anhydrase level, and adreno-cortical hormone concentration. When the PCO2 is high (res-piratory acidosis), more intracellular H2CO3 is available to buffer the hydroxyl ions and acid secretion is enhanced, where-as the reverse is true when the PCO2 falls. K+ depletion enhanc-es acid secretion, apparently because the loss of K+ causes intracellular acidosis even though the plasma pH may be ele-vated. Conversely, K+ excess in the cells inhibits acid secretion. When carbonic anhydrase is inhibited, acid secretion is inhib-ited because the formation of H2CO3 is decreased. Aldosterone and the other adrenocortical steroids that enhance tubular re-absorption of Na+ also increase the secretion of H+ and K+. BICARBONATE EXCRETION Although the process of HCO3 – reabsorption does not actual-ly involve transport of this ion into the tubular cells, HCO3 – reabsorption is proportional to the amount filtered over a rel-atively wide range. There is no demonstrable Tm, but HCO3 – reabsorption is decreased by an unknown mechanism when the extracellular fluid (ECF) volume is expanded (Figure 40–4). When the plasma HCO3 – concentration is low, all the filtered HCO3 – is reabsorbed; but when the plasma HCO3 – concentration is high; that is, above 26 to 28 mEq/L (the renal threshold for HCO3 –), HCO3 – appears in the urine and the urine becomes alkaline. Conversely, when the plasma HCO3 – falls below about 26 mEq/L, the value at which all the secreted H+ is being used to reabsorb HCO3 –, more H+ becomes available to combine with other buffer anions. Therefore, the lower the plas-ma HCO3 – concentration drops, the more acidic the urine be-comes and the greater its NH4 + content (see Clinical Box 40–1). DEFENSE OF H+ CONCENTRATION The mystique that envelopes the subject of acid–base balance makes it necessary to point out that the core of the problem is not “buffer base” or “fixed cation” or the like, but simply the FIGURE 40–3 Major reactions involved in ammonia production in the kidneys. NH4 + [NH3] [NH4 +] NH3 + H+ pH = pK' + log Glutamine Glutamate + NH4 + Glutaminase Glutamate α−Ketoglutarate + NH4 + Glutamic dehydrogenase FIGURE 40–4 Effect of ECF volume on HCO3 – filtration, reabsorption, and excretion in rats. The pattern of HCO3 – excretion is similar in humans. The plasma HCO3 – concentration is normally about 24 mEq/L. (Reproduced with permission from Valtin H: Renal Function, 2nd ed. Little, Brown, 1983.) Bicarbonate filtered, excreted, or reabsorbed (μeq/min) 60 50 40 30 20 10 0 0 50 100 150 Filtered (during both minimal And exaggerated expansion) Reabsorbed Excreted Minimal expansion Minimal expansion Exaggerated expansion Exaggerated expansion Plasma HCO3− concentration (meq/L) 682 SECTION VIII Renal Physiology maintenance of the H+ concentration of the ECF. The mecha-nisms regulating the composition of the ECF are particularly important as far as this specific ion is concerned, because the machinery of the cells is very sensitive to changes in H+ con-centration. Intracellular H+ concentration, which can be mea-sured by using microelectrodes, pH-sensitive fluorescent dyes, and phosphorus magnetic resonance, is different from extra-cellular pH and appears to be regulated by a variety of intra-cellular processes. However, it is sensitive to changes in ECF H+ concentration. The pH notation is a useful means of expressing H+ concen-trations in the body, because the H+ concentrations happen to be low relative to those of other cations. Thus, the normal Na+ concentration of arterial plasma that has been equilibrated with red blood cells is about 140 mEq/L, whereas the H+ con-centration is 0.00004 mEq/L (Table 40–1). The pH, the nega-tive logarithm of 0.00004, is therefore 7.4. Of course, a decrease in pH of 1 unit, for example, from 7.0 to 6.0, repre-sents a 10-fold increase in H+ concentration. It is important to remember that the pH of blood is the pH of true plasma— plasma that has been in equilibrium with red cells—because the red cells contain hemoglobin, which is quantitatively one of the most important blood buffers (see Chapter 36). H+ BALANCE The pH of the arterial plasma is normally 7.40 and that of venous plasma slightly lower. Technically, acidosis is present whenever the arterial pH is below 7.40, and alkalosis is present whenever it is above 7.40, although variations of up to 0.05 pH unit occur without untoward effects. The H+ concen-trations in the ECF that are compatible with life cover an ap-proximately fivefold range, from 0.00002 mEq/L (pH 7.70) to 0.0001 mEq/L (pH 7.00). Amino acids are utilized in the liver for gluconeogenesis, leaving NH4 + and HCO3 – as products from their amino and carboxyl groups (Figure 40–5). The NH4 + is incorporated into urea and the protons that are formed are buffered intracellu-larly by HCO3 –, so little NH4 + and HCO3 – escape into the cir-culation. However, metabolism of sulfur-containing amino acids produces H2SO4, and metabolism of phosphorylated amino acids such as phosphoserine produces H3PO4. These strong acids enter the circulation and present a major H+ load to the buffers in the ECF. The H+ load from amino acid metab-olism is normally about 50 mEq/d. The CO2 formed by CLINICAL BOX 40–1 Implications of Urinary pH Changes Depending on the rates of the interrelated processes of acid secretion, NH4 + production, and HCO3 – excretion, the pH of the urine in humans varies from 4.5 to 8.0. Excretion of urine that is at a pH different from that of the body fluids has important implications for the body’s electrolyte and acid–base economy. Acids are buffered in the plasma and cells, the overall reaction being HA + NaH3 → NaA + H2CO3. The H2CO3 forms CO2 and H2O, and the CO2 is expired, while the NaA appears in the glomerular filtrate. To the ex-tent that the Na+ is replaced by H+ in the urine, Na+ is con-served in the body. Furthermore, for each H+ ion excreted with phosphate or as NH4 +, there is a net gain of one HCO3 – ion in the blood, replenishing the supply of this important buffer anion. Conversely, when base is added to the body fluids, the OH– ions are buffered, raising the plasma HCO3 –. When the plasma level exceeds 28 mEq/L, the urine be-comes alkaline and the extra HCO3 – is excreted in the urine. Because the rate of maximal H+ secretion by the tubules varies directly with the arterial PCO2, HCO3 – reabsorption also is affected by the PCO2. This relationship has been dis-cussed in more detail in the text. TABLE 40–1 H+ concentration and pH of body fluids. H+ Concentration mEq/L mol/L pH Gastric HCI 150 0.15 0.8 Maximal urine acidity 0.03 3 × 10–5 4.5 Plasma Extreme acidosis 0.0001 1 × 10–7 7.0 Normal 0.00004 4 × 10–8 7.4 Extreme alkalosis 0.00002 2 × 10–8 7.7 Pancreatic juice 0.00001 1 × 10–8 8.0 FIGURE 40–5 Role of the liver and kidneys in the handling of metabolically produced acid loads. Sites where regulation occurs are indicated by asterisks. (Modified and reproduced with permission from Knepper MA, et al: Ammonium, urea, and systemic pH regulation. Am J Physiol 1987;235:F199.) NH4 + + HCO3 − H3PO4 + H2SO4 Glucose Amino acids HPO42− H2PO4 − H2PO4 − HCO3 − HCO3 − NH4 + NH4 + H+ H+ SO42− SO42− Urea Urea Glutamine Glutamine α-Ketoglutarate Liver ECF Kidney Urine CHAPTER 40 Acidification of the Urine & Bicarbonate Excretion 683 metabolism in the tissues is in large part hydrated to H2CO3 (see Chapter 36), and the total H+ load from this source is over 12,500 mEq/d. However, most of the CO2 is excreted in the lungs, and only small quantities of the H+ remain to be excreted by the kidneys. Common sources of extra acid loads are strenu-ous exercise (lactic acid), diabetic ketosis (acetoacetic acid and β-hydroxybutyric acid), and ingestion of acidifying salts such as NH4Cl and CaCl2, which in effect add HCl to the body. Failure of diseased kidneys to excrete normal amounts of acid is also a cause of acidosis. Fruits are the main dietary source of alkali. They contain Na+ and K+ salts of weak organic acids, and the anions of these salts are metabolized to CO2, leaving NaHCO3 and KHCO3 in the body. NaHCO3 and other alkalinizing salts are sometimes ingested in large amounts, but a more common cause of alkalosis is loss of acid from the body as a result of vomiting of gastric juice rich in HCl. This is, of course, equiva-lent to adding alkali to the body. BUFFERING Buffering is of key importance in maintaining H+ homeosta-sis. It is defined in Chapter 1 and discussed in Chapter 36 in the context of gas transport, with an emphasis on roles for proteins, hemoglobin and the carbonic anhydrase system in the blood. Carbonic anhydrase is also found in high concen-tration in gastric acid-secreting cells (see Chapter 26) and in renal tubular cells (see Chapter 38). Carbonic anhydrase is a protein with a molecular weight of 30,000 that contains an atom of zinc in each molecule. It is inhibited by cyanide, azide, and sulfide. The sulfonamides also inhibit this enzyme, and sulfonamide derivatives have been used clinically as diuretics because of their inhibitory effects on carbonic anhydrase in the kidney (see Chapter 38). Buffering in vivo is, of course, not limited to the blood. The principal buffers in the blood, interstitial fluid, and intracellu-lar fluid are listed in Table 40–2. The principal buffers in cere-brospinal fluid (CSF) and urine are the bicarbonate and phosphate systems. In metabolic acidosis, only 15–20% of the acid load is buffered by the H2CO3–HCO3 – system in the ECF, and most of the remainder is buffered in cells. In meta-bolic alkalosis, about 30–35% of the OH– load is buffered in cells, whereas in respiratory acidosis and alkalosis, almost all the buffering is intracellular. In animal cells, the principal regulators of intracellular pH are HCO3 – transporters. Those characterized to date include the Cl–HCO3 – exchanger AE1 (formerly band 3), three Na+– HCO3 – cotransporters, and a K+–HCO3 – cotransporter. SUMMARY When a strong acid is added to the blood, the major buffer re-actions are driven to the left. The blood levels of the three “buffer anions” Hb– (hemoglobin), Prot– (protein), and HCO3 – consequently drop. The anions of the added acid are filtered into the renal tubules. They are accompanied (“cov-ered”) by cations, particularly Na+, because electrochemical neutrality is maintained. By processes that have been dis-cussed above, the tubules replace the Na+ with H+ and in so doing reabsorb equimolar amounts of Na+ and HCO3 –, thus conserving the cations, eliminating the acid, and restoring the supply of buffer anions to normal. When CO2 is added to the blood, similar reactions occur, except that since it is H2CO3 that is formed, the plasma HCO3 – rises rather than falls. RENAL COMPENSATION TO RESPIRATORY ACIDOSIS AND ALKALOSIS As noted in Chapter 36, a rise in arterial PCO2 due to decreased ventilation causes respiratory acidosis and conversely, a de-cline in PCO2 causes respiratory alkalosis. The initial changes shown in Figure 40–6 are those that occur independently of any compensatory mechanism; that is, they are those of un-compensated respiratory acidosis or alkalosis. In either situa-tion, changes are produced in the kidneys, which then tend to compensate for the acidosis or alkalosis, adjusting the pH to-ward normal. HCO3 – reabsorption in the renal tubules depends not only on the filtered load of HCO3 –, which is the product of the glo-merular filtration rate (GFR) and the plasma HCO3 – level, but also on the rate of H+ secretion by the renal tubular cells, since HCO3 – is reabsorbed by exchange for H+. The rate of H+ secre-tion—and hence the rate of HCO3 – reabsorption—is propor-tional to the arterial PCO2, probably because the more CO2 that is available to form H2CO3 in the cells, the more H+ can be secreted. Furthermore, when the PCO2 is high, the interior of most cells becomes more acidic. In respiratory acidosis, renal tubular H+ secretion is therefore increased, removing H+ from the body; and even though the plasma HCO3 – is elevated, HCO3 – reabsorption is increased, further raising the plasma HCO3 –. This renal compensation for respiratory acidosis is shown graphically in the shift from acute to chronic respiratory acidosis in Figure 40–6. Cl– excretion is increased, and plasma Cl– falls as plasma HCO3 – is increased. Conversely, in respira-tory alkalosis, the low PCO2 hinders renal H+ secretion, HCO3 – reabsorption is depressed, and HCO3 – is excreted, further TABLE 40–2 Principal buffers in body fluids. Blood H2CO3 ← → H+ + HCO3 – HProt ← → H+ + Prot– HHb ← → H+ + Hb– Interstitial fluid H2CO3 ← → H+ + HCO3 – Intracellular fluid HProt ← → H+ + Prot– H2PO4 – ← → H+ + HPO4 2– 684 SECTION VIII Renal Physiology reducing the already low plasma HCO3 – and lowering the pH toward normal. METABOLIC ACIDOSIS When acids stronger than HHb and the other buffer acids are added to blood, metabolic acidosis is produced; and when the free H+ level falls as a result of addition of alkali or removal of ac-id, metabolic alkalosis results. Following the example from Chapter 36, if H2SO4 is added, the H+ is buffered and the Hb–, Prot–, and HCO3 – levels in plasma drop. The H2CO3 formed is converted to H2O and CO2, and the CO2 is rapidly excreted via the lungs. This is the situation in uncompensated metabolic aci-dosis. Actually, the rise in plasma H+ stimulates respiration, so that the PCO2, instead of rising or remaining constant, is reduced. This respiratory compensation raises the pH even further. The renal compensatory mechanisms then bring about the excretion of the extra H+ and return the buffer systems to normal. RENAL COMPENSATION The anions that replace HCO3 – in the plasma in metabolic ac-idosis are filtered, each with a cation (principally Na+), thus maintaining electrical neutrality. The renal tubular cells se-crete H+ into the tubular fluid in exchange for Na+; and for each H+ secreted, one Na+ and one HCO3 – are added to the blood. The limiting urinary pH of 4.5 would be reached rapid-ly and the total amount of H+ secreted would be small if no buffers were present in the urine to “tie up” H+. However, se-creted H+ reacts with HCO3 – to form CO2 and H2O (bicarbo-nate reabsorption); with HPO4 2– to form H2PO4 –; and with NH3 to form NH4 +. In this way, large amounts of H+ can be secreted, permitting correspondingly large amounts of HCO3 – to be returned to (in the case of bicarbonate reabsorp-tion) or added to the depleted body stores and large numbers of the cations to be reabsorbed. It is only when the acid load is very large that cations are lost with the anions, producing di-uresis and depletion of body cation stores. In chronic acidosis, glutamine synthesis in the liver is increased, using some of the NH4 + that usually is converted to urea (Figure 40–5), and the glutamine provides the kidneys with an additional source of NH4 +. NH3 secretion increases over a period of days (adapta-tion of NH3 secretion), further improving the renal compen-sation for acidosis. In addition, the metabolism of glutamine in the kidneys produces α-ketoglutarate, and this in turn is de-carboxylated, producing HCO3 –, which enters the blood-stream and helps buffer the acid load (Figure 40–5). The overall reaction in blood when a strong acid such as H2SO4 is added is: 2NaHCO3 + H2SO4 → Na2SO4 + 2H2CO3 For each mole of H+ added, 1 mole of NaHCO3 is lost. The kidney in effect reverses the reaction: Na2SO4 + 2H2CO3 → 2NaHCO3 + 2H+ + SO4 2– and the H+ and SO4 2– are excreted. Of course, H2SO4 is not excreted as such, the H+ appearing in the urine as titratable acidity and NH4 +. In metabolic acidosis, the respiratory compensation tends to inhibit the renal response in the sense that the induced drop in PCO2 hinders acid secretion, but it also decreases the filtered load of HCO3 – and so its net inhibitory effect is not great. METABOLIC ALKALOSIS In metabolic alkalosis, the plasma HCO3 – level and pH rise (Figure 40–7). The respiratory compensation is a decrease in ventilation produced by the decline in H+ concentration, and this elevates the PCO2. This brings the pH back toward normal while elevating the plasma HCO3 – level still further. The mag-nitude of this compensation is limited by the carotid and aor-tic chemoreceptor mechanisms, which drive the respiratory center if any appreciable fall occurs in the arterial PO2. In met-abolic alkalosis, more renal H+ secretion is expended in reab-sorbing the increased filtered load of HCO3 –; and if the HCO3 – level in plasma exceeds 26–28 mEq/L, HCO3 – appears in the urine. The rise in PCO2 inhibits the renal compensation by facilitating acid secretion, but its effect is relatively slight. THE SIGGAARD–ANDERSEN CURVE NOMOGRAM Use of the Siggaard–Andersen curve nomogram (Figure 40–7) to plot the acid–base characteristics of arterial blood is helpful FIGURE 40–6 Acid–base nomogram showing changes in the CO2 (curved lines), plasma HCO3 –, and pH of arterial blood in respiratory and metabolic acidosis. Note the shifts in HCO3 – and pH as acute respiratory acidosis and alkalosis are compensated, produc-ing their chronic counterparts. (Reproduced with permission from Cogan MG, Rector FC Jr: Acid–base disorders. In: The Kidney, 4th ed. Brenner BM, Rector FC Jr [editors]. Saunders, 1991.) 60 56 52 48 44 40 36 32 Arterial plasma [HCO3−] (meq/L) 28 24 20 16 12 8 4 0 7.00 7.10 7.20 7.30 7.40 7.50 7.60 7.70 7.80 Arterial blood pH 100 90 80 70 60 50 40 35 30 25 20 Arterial blood [H+] (nmol/L) 120 100 90 80 70 60 50 40 20 15 10 35 30 25 Acute respiratory alkalosis Acute respiratory acidosis Normal Chronic respiratory alkalosis PCO2 (mm Hg) Metabolic acidosis Chronic respiratory acidosis Meta-bolic alkalosis CHAPTER 40 Acidification of the Urine & Bicarbonate Excretion 685 in clinical situations. This nomogram has PCO2 plotted on a log scale on the vertical axis and pH on the horizontal axis. Thus, any point to the left of a vertical line through pH 7.40 in-dicates acidosis, and any point to the right indicates alkalosis. The position of the point above or below the horizontal line through a PCO2 of 40 mm Hg defines the effective degree of hypoventilation or hyperventilation. If a solution containing NaHCO3 and no buffers were equilibrated with gas mixtures containing various amounts of CO2, the pH and PCO2 values at equilibrium would fall along the dashed line on the left in Figure 40–7 or a line parallel to it. If buffers were present, the slope of the line would be greater; and the greater the buffering capacity of the solution, the steeper the line. For normal blood containing 15 g of hemoglobin/dL, the CO2 titration line passes through the 15-g/dL mark on the hemoglobin scale (on the underside of the upper curved scale) and the point where the PCO2 = 40 mm Hg and pH = 7.40 lines intersect, as shown in Figure 40–7. When the hemoglobin content of the blood is low, there is sig-nificant loss of buffering capacity, and the slope of the CO2 titration line diminishes. However, blood of course contains buffers in addition to hemoglobin, so that even the line drawn from the zero point on the hemoglobin scale through the nor-mal PCO2–pH intercept is steeper than the curve for a solu-tion containing no buffers. For clinical use, arterial blood or arterialized capillary blood is drawn anaerobically and its pH measured. The pHs of the same blood after equilibration with each of two gas mixtures containing different known amounts of CO2 are also determined. The pH values at the known PCO2 levels are plot-ted and connected to provide the CO2 titration line for the blood sample. The pH of the blood sample before equilibra-tion is plotted on this line, and the PCO2 of the sample is read off the vertical scale. The standard bicarbonate content of the sample is indicated by the point at which the CO2 titration line intersects the bicarbonate scale on the PCO2 = 40 mm Hg line. The standard bicarbonate is not the actual bicarbonate concentration of the sample but, rather, what the bicarbonate concentration would be after elimination of any respiratory component. It is a measure of the alkali reserve of the blood, except that it is measured by determining the pH rather than the total CO2 content of the sample after equilibration. Like the alkali reserve, it is an index of the degree of metabolic aci-dosis or alkalosis present. FIGURE 40–7 Siggaard–Andersen curve nomogram. (Courtesy of O Siggaard–Andersen and Radiometer, Copenhagen, Denmark.) 30 25 20 19 18 17 16 15 35 40 45 50 55 60 65 70 75 80 0 10 10 15 20 25 6.9 7.0 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 30 40 50 25 CO2 titration line of normal blood CO2 titration line solution containing NaHCO3, 15 meq/L, and no buffers Hemoglobin (g/dL) Buffer base (meq/L) 0 +5 +10 −5 −10 −15 −20 −22 +15 +20 Standard bicarbonate (meq/L) Base excess (meq/L) pH 110 100 90 80 70 60 50 40 35 30 25 20 15 10 PCO2 (mm Hg) 686 SECTION VIII Renal Physiology Additional graduations on the upper curved scale of the nomogram (Figure 40–7) are provided for measuring buffer base content; the point where the CO2 calibration line of the arterial blood sample intersects this scale shows the mEq/L of buffer base in the sample. The buffer base is equal to the total number of buffer anions (principally Prot–, HCO3 –, and Hb–) that can accept hydrogen ions in the blood. The normal value in an individual with 15 g of hemoglobin per deciliter of blood is 48 mEq/L. The point at which the CO2 calibration line intersects the lower curved scale on the nomogram indicates the base excess. This value, which is positive in alkalosis and negative in acidosis, is the amount of acid or base that would restore 1 L of blood to normal acid–base composition at a PCO2 of 40 mm Hg. It should be noted that a base deficiency cannot be completely corrected simply by calculating the difference between the normal standard bicarbonate (24 mEq/L) and the actual standard bicarbonate and administering this amount of NaHCO3 per liter of blood; some of the added HCO3 – is con-verted to CO2 and H2O, and the CO2 is lost in the lungs. The actual amount that must be added is roughly 1.2 times the standard bicarbonate deficit, but the lower curved scale on the nomogram, which has been developed empirically by analyz-ing many blood samples, is more accurate. In treating acid–base disturbances, one must, of course, con-sider not only the blood but also all the body fluid compart-ments. The other fluid compartments have markedly different concentrations of buffers. It has been determined empirically that administration of an amount of acid (in alkalosis) or base (in acidosis) equal to 50% of the body weight in kilograms times the blood base excess per liter will correct the acid–base disturbance in the whole body. At least when the abnormality is severe, however, it is unwise to attempt such a large correction in a single step; instead, about half the indicated amount should be given and the arterial blood acid–base values determined again. The amount required for final correction can then be calculated and administered. It is also worth noting that, at least in lactic acidosis, NaHCO3 decreases cardiac output and lowers blood pressure, so it should be used with caution. CHAPTER SUMMARY ■The cells of the proximal and distal tubules secrete hydrogen ions. Acidification also occurs in the collecting ducts. The reac-tion that is primarily responsible for H+ secretion in the proxi-mal tubules is Na+–H+ exchange. Na is absorbed from the lumen of the tubule and H is excreted. ■The maximal H+ gradient against which the transport mecha-nisms can secrete in humans corresponds to a urine pH of about 4.5. However, three important reactions in the tubular fluid remove free H+, permitting more acid to be secreted. These are the reactions with HCO3 – to form CO2 and H2O, with HPO4 2– to form H2PO4 –, and with NH3 to form NH4 +. ■Carbonic anhydrase catalyzes the formation of H2CO3, and drugs that inhibit carbonic anhydrase depress secretion of acid by the proximal tubules. ■Renal acid secretion is altered by changes in the intracellular PCO2, K+ concentration, carbonic anhydrase level, and adreno-cortical hormone concentration. MULTIPLE-CHOICE QUESTIONS For all questions, select the single best answer unless otherwise directed. 1. Which of the following is the principal buffer in interstitial fluid? A) hemoglobin B) other proteins C) carbonic acid D) H2PO4 E) compounds containing histidine 2. Increasing alveolar ventilation increases the blood pH because A) it activates neural mechanisms that remove acid from the blood. B) it makes hemoglobin a stronger acid. C) it increases the PO2 of the blood. D) it decreases the PCO2 in the alveoli. E) the increased muscle work of increased breathing generates more CO2. 3. In uncompensated metabolic alkalosis A) the plasma pH, the plasma HCO3 – concentration, and the arterial PCO2 are all low. B) the plasma pH is high and the plasma HCO3 – concentration and arterial PCO2 are low. C) the plasma pH and the plasma HCO3 – concentration are low and the arterial PCO2 is normal. D) the plasma pH and the plasma HCO3 – concentration are high and the arterial PCO2 is normal. E) the plasma pH is low, the plasma HCO3 – concentration is high, and the arterial PCO2 is normal. 4. In a patient with a plasma pH of 7.10, the [HCO3 –]/[H2CO3] ratio in plasma is A) 20. B) 10. C) 2. D) 1. E) 0.1. CHAPTER RESOURCES Adrogué HJ, Madius NE: Management of life-threatening acid–base disorders. N Engl J Med 1998;338:26. Brenner BM, Rector FC Jr. (editors): The Kidney, 6th ed. 2 vols. Saunders, 1999. Davenport HW: The ABC of Acid–Base Chemistry, 6th ed. University of Chicago Press, 1974. Halperin ML: Fluid, Electrolyte, and Acid–Base Physiology, 3rd ed. Saunders, 1998. Lemann J Jr., Bushinsky DA, Hamm LL: Bone buffering of acid and base in humans. Am J Physiol Renal Physiol 2003;285:F811. Review. Vize PD, Wolff AS, Bard JBL (editors): The Kidney: From Normal Development to Congenital Disease. Academic Press, 2003. 687 Answers to Multiple Choice Questions Chapter 1 1. B 2. C 3. B 4. C 5. C 6. D 7. E 8. E. Chapter 2 1. A 2. D 3. D 4. B 5. C 6. C 7. B. Chapter 3 1. C 2. E 3. B 4. C Chapter 4 1. B 2. E 3. B 4. C 5. B Chapter 5 1. B 2. D 3. B 4. C 5. C. Chapter 6 1. E 2. C 3. E 4. B Chapter 7 1. C 2. D 3. C 4. B 5. A Chapter 8 1. C 2. D 3. A 4. E 5. C 6. E 7. A Chapter 9 1. C 2. E 3. C 4. E Chapter 10 1. C 2. C 3. A 4. D Chapter 11 1. D 2. B 3. E 4. D Chapter 12 1. D 2. D 3. B 4. E 5. D 6. B 7. D 8. D Chapter 13 1. E 2. B 3. C 4. E 5. D 6. D 7. E 8. A Chapter 14 1. D 2. C 3. D 4. D 5. D 6. C 7. D 8. E Chapter 15 1. C 2. D 3. C 4. D 5. A 6. C Chapter 16 1. C 2. E 3. E 4. D 5. B 6. A Chapter 17 1. A 2. B 3. D 4. D Chapter 18 1. B 2. E 3. B 4. A 5. A 6. B 7. D 8. D Chapter 19 1. A 2. E 3. C 4. D 5. B 6. D 7. D Chapter 20 1. C 2. D 3. A 4. C 5. B 6. D 7. A 8. C 9. E Chapter 21 1. E 2. D 3. D 4. C 5. E 6. D 7. C Chapter 22 1. D 2. B 3. E 4. D 5. C 6. D 7. D 8. A 9. A Chapter 23 1. C 2. E 3. D 4. A 5. C 6. D 7. E Chapter 24 1. E 2. E 3. A 4. C 5. B 688 Answers to Multiple Choice Questions Chapter 25 1. C 2. E 3. D 4. A 5. E 6. C 7. A Chapter 26 1. C 2. E 3. B 4. C 5. D Chapter 27 1. E 2. D 3. E 4. A 5. C Chapter 28 1. C 2. D 3. E 4. A 5. B Chapter 29 1. E 2. E 3. C 4. E Chapter 30 1. B 2. A 3. A 4. D 5. D Chapter 31 1. A 2. C 3. C 4. C 5. E 6. D Chapter 32 1. C 2. B 3. D 4. B 5. E 6. A 7. A 8. E Chapter 33 1. B 2. A 3. E Chapter 34 1. D 2. A 3. E 4. E 5. E 6. D Chapter 35 1. D 2. C 3. A 4. E 5. B 6. D 7. A Chapter 36 1. E 2. B 3. D 4. D 5. C Chapter 37 1. D 2. B 3. B 4. D 5. E 6. E 7. B 8. C Chapter 38 1. A 2. A 3. A 4. A 5. E 6. C 7. D Chapter 39 1. E 2. C 3. D 4. E 5. D 6. C 7. D Chapter 40 1. C 2. D 3. D 4. B 689 Subject Index A A-V anastomoses. See Arteriovenous anastomoses ABO blood type system, 527–528 ABP. See Androgen-binding protein Absence seizures, 233 Absolute refractory period, 87, 108 Absorption, 451, 453–454 Acalculia, 298 Accelerated AV conduction, 501 Accessory olfactory bulb, 223 Acclimatization, 618 Accommodation, 188–189 ACE. See Angiotensin-converting enzyme Acetylcholine, 134–135, 178–179, 237, 265–266 Acetylcholine receptors, 135–136 Acetylcholinesterase, 135 Acetylcholinesterase inhibitors, 126 Achromatopsia, 196 Acid-base balance, 2–3, 613–614 Acidification of urine, 679–687 Acidosis, 324, 615–616, 682 Acini, 302 Acquired immunity, 67, 70 Acquired tolerance, 177 Acromegaly, 382 Acrosin, 423 Acrosomal reaction, 423 Acrosome, 404 ACTH. See Adrenocorticotropic hormone Actin, 36 Action potentials, 83, 86–87, 106, 211 afferent nerve fibers, 207–208 Activated amino acid, 17 Activation heat, 104 Activation of microglia, 90 Active capillaries, 549 Active tension, 102 Active transport, 46 Active zones, 117 Activin receptors, 409 Activins, 409 Acute pain, 168 Acute pancreatitis, 436 Adaptation, 85, 153, 681 Adaptive immunity, 67 Addiction, 177, 179 Addison disease, 360 Addisonian crisis, 360 Adenylyl cyclase, 56–57 Adenoids, 605 Adenosine triphosphate, 8 Adenylyl cyclase, 56 Adequate stimulus, 152, 164 ADH. See Antidiuretic hormone Adipokines, 334 Adolescence, 398 Adrenal androgens, effects of, 348 Adrenal cortex, 337, 342–346 action of ACTH, 345 actions of angiotensin II, 345 classification, 342–343 enzyme deficiencies, 346 secreted steroids, 343 species differences, 343 steroid biosynthesis, 344–345 Adrenal glucocorticoids, 332 Adrenal insufficiency, 349 Adrenal medulla, 337, 338–342 catecholamines, 338–340 dopamine, effects of, 342 epinephrine, effects of, 340–342 substances secreted by, 340 Adrenal medullary secretion, 342 neural control, 342 regulation of, 342 selective secretion, 342 Adrenal morphology, 338 Adrenal responsiveness, 352 Adrenal sulfokinase, 345 Adrenalectomy, 356 Adrenarche, 398 Adrenergic neurons, 138 Adrenocortical hormone excretion, 346–348 Adrenocortical hormones, 342–346 ACTH, action of, 345 angiotensin II, actions of, 345 enzyme deficiencies, 346 metabolism, 346–348 secreted steroids, 343 species differences, 343 steroid biosynthesis, 344–345 structure, 342–343 Adrenocortical hyperfunction, 359–360 Adrenocorticotropic hormone, 279, 377 chemistry, 352 dependence, 350 effect of, 352 independence, 350 metabolism of, 352 role of, 351–352 secretion, 349 Adrenogenital syndrome, 346, 359 Advanced glycosylation end products, 334 Aerobic glycolysis, 103 Aerophagia, 473 Afferent arteriole, 640, 642 Afferent connections of hypothalamus, 273–274 After-discharge, 164 After-hyperpolarization, 85 Afterload, 515 Ageusia, 226 Agglutinins, 527 Agglutinogens, 527, 530 Agranular, 40 Air conduction, 210 Air embolism, 550 Akinesia, 252 Albinos, 380 Albumin, 305, 530 Albuminuria, 646 Aldosterone, 343, 347–348 angiotensin II, 356–358 effect of ACTH, 356 electrolytes, 358–359 regulation of, 356–359 stimuli, 356 Aldosterone secretion, 356–359 angiotensin II, 356–358 effect of ACTH, 356 electrolytes, 358–359 renin, 356–358 stimuli, 356 Aldosterone synthase, 345 Alerting response, 233 Alkalosis, 615–616, 682 All-or-none law, 85–86 Allodynia, 168, 169 Alpha block, 233 α-dystroglycan, 96 α-hydroxylase, 345 Alpha rhythm, 233–234 α-synuclein, 254 ALS. See Amyotrophic lateral sclerosis Alveolar air, 600–601 sampling, 600–601 Alveolar surface tension, 596–597 Alveolar ventilation, 599 Alveolocapillary membrane, 601–602 Alzheimer disease, 294–295 Amacrine cells, 182 Amblyopia ex anopsia, 188 Amenorrhea, 422 Amiloride-inhibitable Na+ channels, 47 Amino acid pool, 16 Amino acids, 10, 15–19, 130 metabolic functions, 19 Aminoglycoside antibiotics, 126 Ammonia metabolism, excretion, 484–485 Ammonia secretion, 680–681 AMPA receptors, 141 Amphetamine, 140 Amphipathic, 32, 439 Amygdala, 221 Amyloid peptides, 294 Amyloid precursor protein, 294 Amyotrophic lateral sclerosis, 64, 244 Anabolism, 459 Anaerobic glycolysis, 103 Anatomic dead space, 599–600 Androgen-binding protein, 404 Androgen-dependent, 427 Androgen resistance, 397 Androgen-secreting tumors, 410 Androgens, 348, 392 Andropause, 400 Androstenedione, 343 Anemic hypoxia, 617, 621 Anesthesia, 143 Aneuploidy, 13 Angle-closure glaucoma, 182 Angina pectoris, 170 Angiogenesis, 539 Angiotensin-converting enzyme, 566, 670–671 Angiotensin I, 670 Angiotensin II, 337, 670–672 receptors, 672–673 690 INDEX Angiotensin III, 671 Angiotensinogen, 670 Angiotensins, 672 Anion exchanger 1, 612 Anomic aphasia, 297, 298 Anosmia, 222, 282 Anovulatory cycles, 413–414, 422 Anoxia, 616 ANP. See Atrial natriuretic peptide Anterior olfactory nucleus, 221 Anterior pituitary hormones, 279–280 Anterior pituitary secretion, 279–282 anterior pituitary hormones, 279–280 clinical implications, 282 hypophysiotropic hormones, 280–282 hypothalamic control, 280 significance, 282 Antiarrhythmic drugs, 501 Anticlotting mechanisms, 533–535 Anticoagulants, 535 Antidiuretic hormone, 279, 666 Antidromic conduction, 88 Antigen inheritance, 529 Antigen presentation, 71–72 Antigen-presenting cells, 71 Antigen recognition, 71 Antiports, 46 Antithrombin III, 533 Antral systole, 473 Antrum formation, 411 Anuria, 660 Aortic arch, 558 Aortic bodies, 562 Aortic depressor nerve, 559 AP-1. See Assembly protein 1 APCs. See Antigen-presenting cells Aphasias, 297 Apneusis, 627 Apoptosis, 34, 42–43 Aquaporin-1, 652–653 Aquaporins, 45, 652, 666 Aqueous humor, 182 Arachidonate, 28 Arachidonic acid, 28 Arachnoid trabeculae, 572 Arachnoid villi, 571 Arcuate fasciculus, 297 Area postrema, 475, 573, 666 Areflexia, 244 Arginine vasopressin, 277 Argyll Robertson pupil, 189 Aromatase, 404, 416, 427 Arousal response, 233 Arrestins, 132, 178 Arterial circulation, 543–548 Arterial plasma level, 645 Arterial pressure, 544 Arterial pulse, 510–512 Arteries, 536–537 Arteriolar circulation, 543–548 Arterioles, 536–537 Arterioluminal vessels, 578 Arteriosinusoidal vessel, 578 Arteriovenous anastomoses, 538 Atherosclerosis, 28 Asphyxia, 634 Aspiration pneumonia, 594 Assembly protein 1, 45 Associative learning, 290 AST. See Plasma aspartate aminotransferase Asterixis, 252 Astigmatism, 188 Astrocytes, 80, 135 Astrocytic proliferation, 90 Ataxia, 257 Athetosis, 252 ATP. See Adenosine triphosphate ATPase, 35 Atretic follicles, 412 Atrial arrhythmias, 499 Atrial fibrillation, 499 Atrial flutter, 498 Atrial natriuretic peptide, 674 Atrial pressure changes, 512 Atrial stretch receptors, 561 Atrial systole, 489, 508 Atrioventricular node, 489 Audiometry, 213 Audition, cortical areas, 212–213 Auditory agnosia, 155 Auditory cortex, 211 Auditory division, 206 Auditory nerve fibers, 211 Auditory ossicles, 203 Auditory responses of neurons in medulla oblongata, 211–212 Auditory tube, 203 Augmented limb leads, 494 Auras, 233 Auscultatory method, 545–547 Autocrine communication, 50 Autogenic inhibition, 162 Autoimmune diseases, 75 Autologous transfusion, 529 Autonomic function, 275–276 Autonomic junctions, chemical transmission, 265–266 acetylcholine, 265–266 norepinephrine, 265–266 transmission in sympathetic ganglia, 266 Autonomic nerves, effect of, 327–328 Autonomic nervous system, 261–272 acetylcholine, 265–266 autonomic outflow, 262–265 chemical transmission at autonomic junctions, 265–266 descending input to autonomic preganglionic neurons, 269 enteric nervous system, 269–271 features, 262–263 norepinephrine, 265–266 parasympathetic cholinergic discharge, 268–269 parasympathetic division, 265 responses of effector organs to autonomic nerve impulses, 266–269 sympathetic division, 263–265 sympathetic noradrenergic discharge, 268–269 transmission in sympathetic ganglia, 266 Autonomic outflow, 262–265 Autonomic preganglionic neurons, 269 descending input, 269 Autoreceptors, 130 Autoregulation, 563 renal blood flow, 644–645 Autosomes, 392 AV nodal block, 497 AV nodal delay, 492 AV node. See Atrioventricular node AVP. See Arginine vasopressin Axial muscles, control of, 242 Axis, 495 Axoaxonal synapses, 122 Axon, 80 Axonal conduction velocity, 88 Axonal sprouting, 90 Axonal transport, 82–83 Axon hillock, 80 Axon reflex, 581 Axonemal dynein, 37 Axoneme, 37 Axoplasmic flow, 82 B Babinski sign, 244 Ballism, 252 Barkin, 254 Baroreceptor nerve activity, 560 Baroreceptor resetting, 560 Baroreceptor stimulation, respiratory effects of, 633 Baroreceptors, 558–561 Barr body, 392 Barttin, 214 Basal body, 37 Basal ganglia, 250–254 anatomic considerations, 250–251 diseases of basal ganglia, 252–253 function, 251–252 Parkinson disease, 253–254 Basal lamina, 34 Basal metabolic rate, 462 Basal nuclei, 249 Basal stem cells, 219 Basement membrane, 34 Basic electrical rhythm, 470 Basilar arteries, 412 Basket cells, 255 Basophils, 63–64, 523 B cell exhaustion, 328 B cells, 72–73 long-term changes, 328 Becker muscular dystrophy, 98 Bell–Magendie law, 157 Bels, 209 Benign paroxysmal positional vertigo, 216 BER. See Basic electrical rhythm Bernoulli’s principle, 545 Beta-carotene, 191 Beta rhythm, 233–234 Bezold–Jarisch reflex, 632 Bicarbonate excretion, 679–687 Bifascicular block, 497 Bile, 438–440, 482–484 Bile acids, 438 Bile pigments, 438, 483 Biliary secretion, 436 Biliary system, 485–486 bile formation, 485 biliary secretion, 486 cholecystectomy, 486 gallbladder, 485–486 Bilirubin, 527 Bilirubin metabolism, excretion, 483 Biliverdin, 526 Binocular vision, 197–198 Biofeedback, 292 Biologic half-life, 666 Biologic oxidations, 8–10 Biosynthesis, 138–139, 277–278, 317–318 Biphasic action potentials, 88 Biphosphoglycerate, 611 Bipolar, 232 Bipolar cells, 182 Bipolar leads, 492 Bipolar limb leads, 494–495 INDEX 691 Bipolar recording, 492 BK channels, 536 Bladder, 661–662 deafferentation, 662 denervation, 662 emptying, 661–662 filling, 661 reflex control, 662 spinal cord transection, 662 Blastocyst, 423 Blind spot, 182 Blindsight, 198 Blobs, 194 Blood, 521–554 ABO system, 527–528 active capillaries, 549 agglutinogens, 530 air embolism, 550 angiogenesis, 539 anticlotting mechanisms, 533–535 anticoagulants, 535 antigen inheritance, 529 arterial circulation, 543–548 arterial pressure, 544 arteries, 536–537 arteriolar circulation, 543–548 arterioles, 536–537 arteriovenous anastomoses, 538 auscultatory method, 545–547 average velocity, 541 blood flow measurement, 539 blood types, 527–530 bone marrow, 522 buffering, 615–616 capacitance vessels, 543 capillaries, 537–538 capillary circulation, 546–549 catabolism of hemoglobin, 526–527 circulation, 535–543 clotting mechanism, 531–533 critical closing pressure, 542 effects of heartbeat, 549 endothelium, 535 equilibration with interstitial fluid, 548 flow, 544, 548 gene activation, 540–541 gravity, effect of, 544 hemoglobin, 523 hemoglobin in fetus, 525–526 hemostasis, 531–535 hypoproteinemia, 531 inactive capillaries, 549 interstitial fluid volume, 550–552 ionic composition, 503–504 laminar flow, 540 law of Laplace, 542–543 lymph, 535 lymphatic circulation, 550 lymphatics, 538 measuring blood pressure, 544–545 measuring venous pressure, 550 muscle pump, 549–550 newborn, hemolytic disease, 530 normal arterial blood pressure, 546 plasma, 530–531 plasma proteins, 530–531 platelets, 523 Poiseuille–Hagen formula, 541–542 pressure, flow, 547 reactions of hemoglobin, 523–525 red blood cells, 523 resistance, 539, 542, 543 response to injury, 531 RH group, 530 role of spleen, 523 shear stress, 540–541 synthesis of hemoglobin, 526 thoracic pump, 549 transfusion reactions, 528–529 vascular smooth muscle, 536 veins, 538 velocity, 544 venous circulation, 549 venous pressure, flow, 549–550 venous pressure in head, 550 venules, 538 viscosity, 542 white blood cells, 522–523 Blood–brain barrier, 80, 572–574 circumventricular organs, 573–574 development of, 574 function of, 574 penetration of substances into brain, 572–573 Blood flow, 590–591, 599, 643–644 measurement, 539 Blood group antigens, 527 Blood plasma, 2 Blood pressure, 544–545 Blood–testis barrier, 402–403 Blood types, 527–530 ABO system, 527–528 agglutinogens, 530 antigen inheritance, 529 newborn, hemolytic disease, 530 RH group, 530 transfusion reactions, 528–529 Blood typing, 528 Blood vessels, 642–643 BMR. See Basal metabolic rate BNP. See Brain natriuretic peptide Body as organized solution, 2 Body mechanics, 106 Bohr effect, 611 Bohr’s equation, 600 Bomb calorimeter, 460 Bombesin, 448 Bone conduction, 210 Bone disease, 373–374 Bone growth, 371 Bone marrow, 522 Bony labyrinth, 205 Borborygmi, 473 Botulinum toxin, 119 Boutons, 80 Bowman’s capsule, 640 BPG. See Biphosphoglycerate BPN1, 140 Bradycardia, 497 Bradykinesia, 252 Bradykinin, 566 Brain, electrical activity, 229–240 Brain metabolism, 576–577 ammonia removal, 577 energy sources, 577 glutamate, 577 oxygen consumption, 576–577 uptake, release of substances by brain, 576 Brain–computer interface devices, 250 Brain natriuretic peptide, 674 Brain stem pathways, posture, voluntary movement, 246–247 Breaking point, for breathing, 631 Breasts cyclic changes in, 414 development of, 426 Breathing, 598–599 acid–base balance, ventilatory responses, 629 aortic bodies, 628–629 breath holding, 631 carotid bodies, 628–629 chemical control of, 627–631 chemoreceptors in brain stem, 629 CO2, ventilatory responses, 630 control systems, 625–626 effect of H+ on CO2 response, 631 effects of hypoxia on CO2 response curve, 631 medullary systems, 626 neural control of, 625–627 oxygen lack, ventilatory response, 630–631 pontine influences, 627 vagal influences, 627 Brinogenemia, 531 Broca’s area, 297 Brodmann’s area, 174, 243 Bronchi, 590 Bronchial circulation, 590 Bronchial tone, 594 Brown fat, 24, 283 Bruits, 512 Brush border, 640 B-type natriuretic peptide, 674 Buffer, 4 Buffering, 4, 615–616 Buffering capacity, 4 Bulk flow, 571 Bundle of His, 489 Bundle of Kent, 501 Bursal equivalents, 70 Bystander effects, 75 C C1 channels, 651 C-type natriuretic peptide, 675 C wave, 512, 549 Ca2+-activated K+ channels, 536 Ca2+ sparks, 53, 491 Cadherins, 38 CAE. See childhood absence epilepsy Calbindin, 52 Calbindin-D, 365 Calcitonin, 363, 370–371 actions, 370 origin, 370 secretion, 370 structure, 370 Calcitonin gene-related peptide, 145, 370 Calcium, 364–365, 458 Calcium-binding proteins, 52–53 Calcium metabolism, 364–365 effects of hormones, 371 Calcium rigor, 504 Calcium stones, 250 Calmodulin, 52 Calmodulin-dependent myosin light chain kinase, 110 Caloric intake, distribution, 463–464 Calorie, 460 Calorigenesis, 310–312 Calorigenic action, 308, 310 Calorimetry, 460–461 Calpain, 412 cAMP, 56 cAMP-responsive element-binding protein, 56 CAMs. See Cell adhesion molecules Canal of Schlemm, 182 692 INDEX Cannabinoids, 145, 179 Cap site, 14 Capacitance vessels, 543 Capacitation, 404 Capacity, 305 Capillaries, 537–538 Capillary circulation, 546–549 active capillaries, 549 equilibration with interstitial fluid, 548 flow, 547 inactive capillaries, 549 methods of study, 546–547 pressure, 547 Capillary wall, 49–50 filtration, 49 oncotic pressure, 49 transcytosis, 49–50 Capsule, 643 Carbamino compounds, 612 Carbohydrates, 19–23, 103, 452–454 acidosis, 324 adrenal glucocorticoids, 332 automatic nerves, 327–328 biosynthesis, 317–318 catecholamines, 331–332 changes in protein metabolism, 323 cholesterol metabolism, 324 citric acid cycle, 20 coma, 324 cyclic AMP, 327 diabetes mellitus, 333–334 directional-flow valves, 21 energy production, 20–21 exercise, 331 factors determining plasma glucose level, 22–23 fat derivatives, 327 fat metabolism in diabetes, 323–324 glucagon, 328–330 glucose tolerance, 321–323 glucose transporters, 318 glycogen, 21–22 growth hormone, 332 hexoses, 19, 23 hyperglycemia, 323 hypoglycemia, 332–333 insulin, 316–320 insulin deficiency, 321–325 insulin excess, 325–326 insulin–glucagon molar ratios, 330 insulin preparations, 319 insulin receptors, 921 intestinal hormones, 328 intracellular glucose deficiency, 323 islet cell hormones, 330–331 islet cell structure, 316 long-term changes in B cell responses, 328 mechanism of action, 320–321 metabolic syndrome, 334–335 metabolism, 313, 315–336, 382 obesity, 334–335 pancreatic islets, 331 pancreatic polypeptide, 331 plasma glucose level, 326–327 protein, 327 regulation of insulin secretion, 326–327 regulation of secretion, 329–330 relation to potassium, 319 secreted insulin, 318–320 secretion, 317–318 somatostatin, 330–331 thyroid hormones, 332 type 2 diabetes, 334–335 types of diabetes, 334 Carbon dioxide transport, 612–615 acid–base balance, 613–614 buffering in blood, 615–616 carbon dioxide transport, 613 chloride shift, 612–613 Carbon monoxide, 564 Carbon monoxide poisoning, 621 Carbon monoxyhemoglobin, 525 Carbonic acid-bicarbonate system, 614 Carbonic anhydrase, 615, 679 Carbonmonoxyhemoglobin, 621 Carboxyhemoglobin, 525 Cardiac arrhythmias, 497–502 abnormal pacemakers, 497 accelerated AV conduction, 501 antiarrhythmic drugs, 501 atrial arrhythmias, 499 consequences of, 499 ectopic foci of excitation, 498 implanted pacemakers, 497–498 long QT syndrome, 501 normal cardiac rate, 497 radiofrequency catheter ablation, reentrant pathways, 501–502 reentry, 498–499 ventricular arrhythmias, 499–501 Cardiac conduction system, 489 Cardiac cycle, mechanical events, 507–520 arterial pulse, 510–512 atrial pressure changes, 512 atrial systole, 508 early diastole, 509 echocardiography, 513 heart sounds, 512 jugular pulse, 512 late diastole, 507 length of systole, diastole, 510 murmurs, 512–513 pericardium, 510 timing, 510 ventricular systole, 508 Cardiac excitation, 490–492 anatomic considerations, 490 cardiac muscle, 490–491 pacemaker potentials, 491–492 spread of, 492 Cardiac index, 514 Cardiac innervation, 556 Cardiac muscle, 490–491 morphology, 106 relation of tension to length, 515 Cardiac output, 513–519 cardiac muscle, relation of tension to length, 515 end-diastolic volume, factors affecting, 515 factors controlling, 514–515 integrated control, 518–519 measurement, 513–514 myocardial contractility, 515–517 oxygen consumption by heart, 519 in various conditions, 514 Cardiac pacemaker, 489 Cardiac vector, 495 Cardiopulmonary receptors, 558, 561 Cardiovascular control, 556 Cardiovascular regulatory mechanisms, 555–568 atrial stretch receptors, 561 autoregulation, 563 baroreceptor nerve activity, 560 baroreceptor resetting, 560 baroreceptors, 558–559 carbon monoxide, 564 cardiac innervation, 556 cardiopulmonary receptors, 561 circulating vasoconstrictors, 567 direct effects on RVLM, 562–563 endothelial cells, 563 endothelin-1, 564–565 endothelins, 564–566 endothelium, 563–566 innervation of blood vessels, 556 kinins, 566 local regulation, 563 medullary control of cardiovascular system, 556–558 natriuretic hormones, 566 neural control of cardiovascular system, 555–563 nitric oxide, 563–564 peripheral chemoreceptor reflex, 562 prostacyclin, 563 regulation of secretion, 565 role of baroreceptors, 560–561 substances secreted by endothelium, 563 systemic regulation by hormones, 566–567 Valsalva maneuver, 561–562 vasodilator metabolites, 563 Cardiovascular system, 312 neural control, 555–563 neural regulatory mechanisms, 555–556 Carnitine, 23 Carnitine deficiency, 26 Carotenemia, 310 Carotid bodies, 562 Carotid sinus, 558 Carotid sinus nerve, 559 Carriers, 33, 46 Carrying angle, 418 Caspases, 42 Casts, 659 Catabolism, 459 of amino acids, 18 of catecholamines, 139 of hemoglobin, 526–527 Catalase, 35, 64 Cataplexy, 237 Catch-up growth, 387–388 Catechol-O-methyltransferase, 139 Catecholamines, 130, 138–140, 312, 331–332, 338–340 biosynthesis of catecholamines, 138–139 catabolism of catecholamines, 139 catecholamines, 138–139 dopamine, 139–140 dopamine receptors, 140 epinephrine, 138 norepinephrine, 138 receptors, 139 Categorical hemisphere, 295 CatSper, 404 Caudate nucleus, 249 Causalgia, 168, 169 Caveolae, 44 Caveolae-dependent uptake, 43 Caveolin, 44 CBG. See Corticosteroid-binding globulin Cell adhesion molecules, 32, 38 Cell-attached patch clamp, 45 Cell cycle, 13 Cell membrane, 32–34 Cell morphology, 31–43 apoptosis, 42–43 cell adhesion molecules, 38 INDEX 693 cell membrane, 32–34 centrosomes, 37 cilia, 37 cytoskeleton, 35–36 endoplasmic reticulum, 40 gap junctions, 39–40 Golgi apparatus, 40, 42 intercellular connections, 38–39 lysosomes, 34–35 mitochondria, 34 molecular motors, 37 nucleus, 40 peroxisomes, 35 quality control, 42 ribosomes, 40 vesicular traffic, 40–42 Cell signaling pathway, 51 Cellular immunity, 70 Cellular lipids, 24 Cellular physiology, 31–62 apoptosis, 42–43 calcium-binding proteins, 52–53 capillary wall, 49–50 caveolae, 44 cell adhesion molecules, 38 cell membrane, 32–34 centrosomes, 37 chemical messengers, 50–51 cilia, 37 coats, 45 cyclic AMP, 56 cytoskeleton, 35–36 diacylglycerol, 54–56 endocytosis, 43–44 endoplasmic reticulum, 40 exocytosis, 43 filtration, 49 functional morphology of cell, 31–43 G protein-coupled receptors, 54 G proteins, 53–54 gap junctions, 39–40 Golgi apparatus, 40, 42 growth factors, 57–58 guanylyl cyclase, 57 homeostasis, 58–59 inositol trisphosphate, 54–56 intercellular communication, 50–58 intercellular connections, 38–39 intracellular Ca2+ as second messenger, 52 ion channels, 46–47 lysosomes, 34–35 mechanisms of diversity of Ca2+ actions, 53 membrane permeability, 45–46 membrane transport proteins, 45–46 mitochondria, 34 molecular motors, 37 Na, K ATPase, 47 nucleus, 40 oncotic pressure, 49 peroxisomes, 35 production of cAMP by adenylyl cyclase, 56–57 quality control, 42 rafts, 44 receptors for chemical messengers, 50 regulation of Na, K ATPase activity, 47 ribosomes, 40 second messengers, 54–56 secondary active transport, 48–49 stimulation of transcription, 51–52 transcytosis, 49–50 transport across cell membranes, 43–49 transport across epithelia, 49 vesicle transport, 45 vesicular traffic, 40–42 Central connections of afferent fibers, 160 Central delay, 160 Central diabetes insipidus, 668 Central excitatory state, 165 Central executive, 292 Central herniation, 249 Central inhibitory state, 165 Central nervous system axonal transport, 82–83 cellular elements in, 80–93 glial cells, 80 lesions, 176–177 neurons, 79, 80–82 Central venous pressure, 549–550 Centrioles, 37 Centrosome, 37 Cerebellar cortex, 254 Cerebellar hemispheres, 254 Cerebellum, 254–258 anatomic divisions, 254 cerebellar disease, 257–258 functional divisions, 257 learning, 258 mechanisms, 257 organization, 254–257 Cerebral blood flow, 574–576 autoregulation, 575 intracranial pressure, 575 Kety method, 574–575 sensory nerves, 575 vasomotor nerves, 575 Cerebral circulation, 569–571 innervation, 570–571 vessels, 569–570 Cerebral cortex, 229–231 Cerebral dominance, 295–296 Cerebral metabolic rate for O2, 576 Cerebrocerebellum, 257 Cerebrospinal fluid, 571–572 absorption, 571–572 formation, 571–572 head injuries, 572 protective function, 572 CFF. See Critical fusion frequency cGMP. See Cyclic guanosine monophosphate CGP. See Chorionic growth hormone-prolactin CGRP. See Calcitonin gene-related peptide Channelopathies, 47 Channels, 491 Chaperones, 17 Charcot–Marie–Tooth disease, 40 Chelating agents, 535 Chemical control of respiration, 627 Chemical gradient, 5, 46 Chemical synapse, 115 Chemical transmitters, 145–146 Chemically sensitive nociceptors, 167 Chemokines, 64, 69 Chemoreceptor trigger zone, 475 Chemoreceptors, 150, 219, 627 Chemotaxis, 64 Chest wall, 595–596 Cheyne–Stokes respiratory pattern, 249 Chief cells, 367 Childhood absence epilepsy, 233 Chloride shift, 612–613 Cholecystokinin, 443–446 Cholera toxin, 56 Choleretics, 486 Cholesterol, 27 and atherosclerosis, 28 Cholesterol desmolase, 344 Cholesterol ester hydrolase, 344 Cholesterol esterase, 457 Cholesterol metabolism, 27, 313, 324 Choline acetyltransferase, 135 Cholinergic, 135 Cholinesterases, 135 Chorea, 252 Choreiform movements, 252 Chorionic growth hormone-prolactin, 424 Choroid, 181 Chromatin, 40 Chromogranin A, 138 Chromogranins, 138 Chromosomal abnormalities, 396–397 Chromosomes, 11, 40, 392 Chronic pain, 168 Chronotropic action, 514 Chronotropic effect, 556 Chylomicron remnants, 26 Chylomicrons, 26 Chyme, 475 Cilia, 37, 219 Ciliary body, 181 Ciliary neurotrophic factor, 91 Circadian, 229 Circadian rhythms, 229–240, 352–353. See also Diurnal rhythm Circhoral secretion, 400 Circle of Willis, 569 Circulating vasoconstrictors, 567 Circulation, 535–543 angiogenesis, 539 arteries, 536–537 arterioles, 536–537 arteriovenous anastomoses, 538 average velocity, 541 blood flow measurement, 539 capacitance vessels, 543 capillaries, 537–538 critical closing pressure, 542 endothelium, 535 gene activation, 540–541 laminar flow, 540 law of Laplace, 542–543 lymphatics, 538 Poiseuille–Hagen formula, 541–542 resistance, 539, 542, 543 shear stress, 540–541 vascular smooth muscle, 536 veins, 538 venules, 538 viscosity, 542 Circulation time, 541 Circulatory system, 521 Circumvallate papillae, 223 Circumventricular organs, 573–574 Circus movement, 498 Clasp-knife effect, 163 Classic pathway, 69 Classification of pain, 168–171 deep pain, 169 referred pain, 170–171 visceral pain, 169–170 Classification of sensory receptors, 149–150 Clathrin, 43 Clathrin-mediated endocytosis, 43 Claudins, 38 Clearance, 643 Clearance receptor, 676 694 INDEX Climbing fibers, 255 Clinical uses of EEG, 232–233 Clonal selection, 71 Clonic phase, 233 Clonus, 163 Closing volume, 600 Clotting mechanism, 531–533 Clozapine, 140 CMR 1. See Cold- and menthol-sensitive receptor 1 CNG. See Cyclic nucleotidegated CNP. See C-type natriuretic peptide CNTF. See Ciliary neurotrophic factor CO2 narcosis, 630 CoA. See Coenzyme A Coats, 45 Cochlea, 205 Cochlear division. See Auditory division Coding, sensory, 152–153 duration, 153 intensity, 153 location, 152–153 modality, 152 Codons, 17 Coenzyme A, 8 Cogwheel rigidity, 253 COHb. See Carboxyhemoglobin Cold- and menthol-sensitive receptor 1, 168 Cold receptors, 168 Colipase, 457 Collapsing pulse, 512 Collateral ganglia. See Prevertebral ganglia Collecting ducts, 641 Colloid, 302 Colon, 475–478 defecation, 476–478 motility of colon, 475–476 short-chain fatty acids in, 458 small intestine, 476 transit time, 476 Colony-stimulating factors, 65 Color agnosia, 155 Color blindness, 196 Color vision, 195–197 characteristics of color, 195–196 neural mechanisms, 197 retinal mechanisms, 196–197 Coma, 324 Comedones, 418 Common metabolic pool, 18 Communicating hydrocephalus, 571 Compensatory pause, 500 Complement system, 69 Complementary color, 195 Complete androgen resistance syndrome, 398 Complete heart block, 497 Complete tetanus, 101 Complex cells, 194 Compliance of lungs, chest wall, 595–596 COMT. See Catechol-O-methyltransferase Concentration gradient, 5, 7 Concentration of solutes, units for measuring, 2 equivalents, 2 moles, 2 Concordance rate, 334 Conditionally essential, 15 Conditioned reflex, 292 Conditioned stimulus, 292 Conduction, 284 Conduction aphasia, 297 Conductive deafness, 213 Cone pigments, 192 Cones, 182 Confabulation, 293 Congenital adrenal hyperplasia, 346 Congenital lipodystrophy, 334 Congenital lipoid adrenal hyperplasia, 346 Conn syndrome, 359 Connecting peptides, 318 Consensual light reflex, 189 Constitutive pathway, 43 Contraception, 422 Contractile responses, 96–102, 107–108 fiber types, 102 molecular basis of contraction, 97–100 muscle twitch, 97 relation between muscle length, tension, 102 summation of contractions, 101–102 types of contraction, 100–101 Contraction molecular basis, 97–100, 110–111 tetanic (See Tetanus) Contrecoup injury, 572 Control of ovarian function, 420–422 contraception, 422 control of cycle, 421 feedback effects, 421 hypothalamic components, 420–421 reflex ovulation, 421–422 Convection, 284 Convergence movements, 199 Convergence-projection theory, 170 Convulsive, 233 Cornea, 181 Coronary chemoreflex, 632 Coronary circulation, 577–580 anatomic considerations, 577–578 chemical factors, 579–580 neural factors, 580 pressure gradients, 578–579 variations in coronary flow, 579 Coronary flow, variations in, 579 Corpus albicans, 412 Corpus hemorrhagicum, 412 Corpus luteum, 412 Corpus luteum of pregnancy, 424 Correlation between muscle fiber length, tension, 109 Corresponding points, 198 Corrigan pulse, 512 Cortex, 193–195, 394 cortical areas concerned with vision, 195 pathways to cortex, 193–194 primary visual cortex, 194–195 Cortical bone, 371 Cortical dipoles, 232 Cortical nephrons, 640 Cortical organization, 230–231 Cortical plasticity, 176 Corticobulbar tract, 242–243 Corticospinal tract, 242–243 Corticosteroid-binding globulin, 346 Corticosterone, 338, 343 Corticostriate pathway, 250 Corticotropin, 279 Corticotropin-releasing hormone, 280 Cortisol, 343 Cotransporter, 453 Coughing, 632 Countercurrent exchange, 285 Countercurrent exchangers, 654 Countercurrent multipliers, 654 Coupling, reaction, 304 Coupling ratio, 47 COX1, 28 COX2, 28 Craniosacral division, 265 CREB. See cAMP-responsive element-binding protein Cretins, 387 CRH. See Corticotropin-releasing hormone CRH-binding protein, 282 Crista ampullaris, 206 Cristae, 34 Critical fusion frequency, 197 Critical micelle concentration, 440 Critical velocity, 540, 545 Crossed extensor response, 164 Cryptorchidism, 410 Crystalline lens, 181 CSFs. See Colony-stimulating factors Cuneate nuclei, 173 Cupula, 206 Cushing disease, 351 Cushing reflex, 562 Cushing syndrome, 350–351, 359 Cutaneous circulation, 580–581 reactive hyperemia, 581 triple response, 580–581 white reaction, 580 Cutaneous receptors, generation of impulses in, 151–152 generator potentials, 151 Pacinian corpuscles, 151 source of generator potential, 151–152 CV. See Closing volume Cyanmethemoglobin, 621 Cyanotic congenital heart disease, 620 Cyclic AMP, 56, 327 Cyclic guanosine monophosphate, 57, 112 Cyclic nucleotidegated, 223 Cyclopentanoperhydrophenanthrene nucleus, 342 Cyclosporine, 75 CYP17, 345 CYP19. See Aromatase CYP11A1, 344 CYP11B1, 345 CYP11B2, 345 CYP21A2, 345 Cystinuria, 456 Cystometrogram, 661 Cystometry, 661 Cytokines, 67–69 Cytopempsis, 49 Cytoplasmic dyneins, 37 Cytoskeleton, 35–36 Cytotoxic T cells, 70 Cytotrophoblast, 423 D DAG. See Diacylglycerol Dark adaptation, 197 Dead space, 599–600 Deafness, 213 nonsyndromic, 214 syndromic, 214 Decerebellate rigidity, 247 Decerebrate rigidity, 247 Decerebration, 247 Decibel scale, 209 Declarative memory, 290 Decomposition of movement, 258 Decorticate rigidity, 247 Decortication, 247 Decubitus ulcers, 250 INDEX 695 Decreased peripheral utilization, 322 Deep cerebellar nuclei, 254 Deep pain, 169 Deep tendon reflex, 159 Defecation, 476–478 Defensins, 64 Degenerins, 150 Degranulation, 64 Dehydration, 669 Dehydroepiandrosterone, 343 Delivery, initiation of lactation after, 427 Delta rhythm, 234 Dementia, 294–295 Dendrites, 80 function of, 121 Dendritic cells, 71 Dendritic spines, 80 Denervation effects of, 104 hypersensitivity, 126–127 Denervation hypersensitivity, 126 Dense bodies, 109 Dent disease, 651 Dentate nuclei, 254 Deoxycorticosterone, 343 Depth perception, 188 Dermatomal rule, 170 Desensitization, 50, 130, 153 Desmosome, 38 Desynchronization, 233 Detoxification, 481–482 Detrusor muscle, 661 Development of immune system, 70–71 Dextrins, 452 DHEA. See Dehydroepiandrosterone DHPR. See Dihydropyridine receptors DHT. See Dihydrotestosterone Diabetes, 334 fat metabolism in, 323–324 types of, 334 Diabetes insipidus, 668 Diabetes mellitus, 315, 333–334, 656 Diabetic nephropathy, 333 Diabetic neuropathy, 333 Diabetic retinopathy, 333 Diacylglycerol, 54–56 Diamine oxidase, 138 Diapedesis, 64 Diaphragm, 594 Diastole, 489 Diastolic pressure, 507, 544 Dichromats, 196 Dicrotic notch, 512 Dicumarol, 535 Diffuse secondary response, 232 Diffusing capacity, 601 Diffusion, 4–5 Digestion, 451–453 Digestive enzymes, 451 Dihydropyridine receptors, 100 Dihydrotestosterone, 408 Dihydroxycholecalciferol, 363, 365 Dimethoxy-4-methyl-amphetamine, 137 Diopters, 186 Dioxins, 418 Diplopia, 198 Dipole moment, 2 Direct calorimetry, 460 Direct Fick method, 513 Direct inhibition, 121 Direct oxidative pathway, 20 Discharge zone, 123 Discriminative, 176 Disordered renal function, 659–660 abnormal Na+ metabolism, 660 acidosis, 660 loss of concentrating ability, 660 uremia, 660 Distal convoluted tubule, 641 Distal muscles, control of, 242 Diuretics, 659 Diurnal rhythm, 352 DMT. See N,N-dimethyltryptamine DMT1, 459 DNA, 40 DOM. See Dimethoxy-4-methyl-amphetamine Dominant follicle, 412 Dominant hemisphere, 295 Donnan effect, 6–7 Dopa decarboxylase, 138 Dopamine, 139–140, 337 effects of, 342 Dopamine β-hydroxylase, 138 Dopamine receptors, 140 Doppler flow meters, 539 Dorsal column pathway, 173–175 somatotopic organization, 173–175 Dorsal columns, 155 Dorsal column system, 153, 173 Dorsal horn, 173 Down-regulation, 50 D3 receptors, 179 Dromotropic effect, 556 Drosophila, 70 Drowning, 634 DTR. See Deep tendon reflex Duchenne muscular dystrophy, 98 Ductus arteriosus, 583 Ductus venosus, 583 Dynamic response, 158 Dynamin, 44 Dynein, 37 Dysarthria, 252 Dysdiadochokinesia, 258 Dysgeusia, 226 Dyskinesias, 254 Dyslexia, 295 Dysmenorrhea, 423 Dysmetria, 258 Dysosmia, 222 Dystrophin, 96 Dystrophin-glycoprotein complex, 96 E Ear external, 203–205 inner, 205 middle, 203–205 Ear dust, 206 Early diastole, 509 Early endosome, 44 Ebner gland, 226 ECF. See Extracellular fluid ECG. See Electrocardiogram Echocardiography, 513 ECL cells. See Enterochromaffin-like cells ECoG. See Electrocorticogram Ecstasy, 137 Ectopic focus, 498 Edema, 551 Edinger–Westphal nucleus, 189 EDRF. See Endothelium-derived relaxing factor EEG. See Electroencephalogram Effective perfusion pressure, 539 Effective renal plasma flow, 643 Efferent arteriole, 640, 642 Eicosanoids, 28–29 Einthoven’s triangle, 492 Ejaculation, 406 Ejaculatory ducts, 402 Ejection fraction, 508 EJPs. See Excitatory junction potentials Electrical activity, 110 brain, 229–240 heart, 489–506 Electrical characteristics, skeletal muscle, 96 Electrical gradient, 7, 46 Electrical phenomena, 96 electrical characteristics, skeletal muscle, 96 ion distribution, 96 Electrical properties, 106–107 action potentials, 106 resting membrane, 106 Electrical responses, 190, 207 Electrical synapse, 115 Electrical transmission, 121 Electrocardiogram, 492–497 bipolar leads, 492 bipolar limb leads, 494–495 cardiac vector, 494–495 His bundle electrogram, 496 monitoring, 496–497 normal, 494 unipolar leads, 492–494 vectorcardiography, 495–496 Electrocardiography, 502–504 ionic composition of blood, 503–504 myocardial infarction, 502–503 Electrocorticogram, 232 Electroencephalogram, 232 clinical uses of, 232–233 cortical dipoles, 232 physiologic basis, 232–233 Electrogenesis of action potential, 87 Electrogenic pump, 47 Electrolyte metabolism, 382 Electrolyte transport, 440–442 Electrolytes, 2–3 Electromyogram, 105 Electromyography, 105 Electrotonic potentials, 83, 86–87 Embden–Meyerhof pathway, 20 Emboliform nuclei, 254 Embryology of reproductive system, 394–396 development of brain, 396 development of gonads, 394 embryology of genitalia, 394–396 Embryonic stem cells, 250 EMG. See Electromyogram Emission, 406 Emmetropic, 188 Enchondral bone formation, 371 End-diastolic ventricular volume, 508 End-diastolic volume, factors affecting, 515 End plate potential, 124 End pressure, 544 End-systolic ventricular volume, 508 Endocrine, 442 Endocrine communication, 50 Endocrine functions of lungs, 605–606 Endocrine functions of pancreas, 315–336 acidosis, 324 adrenal glucocorticoids, 332 automatic nerves, 327–328 biosynthesis, 317–318 carbohydrate metabolism, 331–332 696 INDEX Endocrine functions of pancreas (continued) catecholamines, 331–332 changes in protein metabolism, 323 cholesterol metabolism, 324 coma, 324 cyclic AMP, 327 diabetes mellitus, 333–334 exercise, 331 fat derivatives, 327 fat metabolism in diabetes, 323–324 glucagon, 328–330 glucose tolerance, 321–323 glucose transporters, 318 growth hormone, 332 hyperglycemia, 323 hypoglycemia, 332–333 insulin, 316–320 insulin deficiency, 321–325 insulin excess, 325–326 insulin-glucagon molar ratios, 330 insulin preparations, 319 intestinal hormones, 328 intracellular glucose deficiency, 323 islet cell hormones, 330–331 islet cell structure, 316 long-term changes in B cell responses, 328 mechanism of action, 320–321, 921 metabolic syndrome, 334–335 metabolism, 318, 329 obesity, 334–335 pancreatic islets, 331 pancreatic polypeptide, 331 plasma glucose level, 326–327 protein, 327 regulation of insulin secretion, 326–327 regulation of secretion, 329–330 relation to potassium, 319 secreted insulin, 318–320 secretion, 317–318 somatostatin, 330–331 thyroid hormones, 332 type 2 diabetes, 334–335 types of diabetes, 334 Endocytosis, 43–44 Endogenous pathway, 24 Endogenous pyrogens, 286 Endolymph, 205 Endopeptidases, 455 Endoplasmic, 40 Endoplasmic reticulum, 40 Endothelial cells, 563 Endothelial derived relaxation factor, 111–112 Endothelin-converting enzyme, 565 Endothelin-1, 564–565 Endothelins, 564–565 Endothelium, 535 carbon monoxide, 564 cardiovascular functions, 565 endothelial cells, 563 endothelin-1, 564–565 endothelins, 564–565 nitric oxide, 563–564 prostacyclin, 563 regulation of secretion, 565 substances secreted by, 563–566 thromboxane A2, 563 Endothelium-derived relaxing factor, 564 Energy balance, 462–463 Energy production, 1–30 acid–base balance, 2–3 amino acid pool, 16 amino acids, 15–19 biologic oxidations, 8–10 body as organized solution, 2 breakdown, 21–22 buffering, 4 carbohydrates, 19–23 catabolism of amino acids, 18 cellular lipids, 24 cholesterol metabolism, 27 citric acid cycle, 20 diffusion, 4–5 directional-flow valves, 21 DNA, 11–13 Donnan effect, 6–7 eicosanoids, 28–29 electrolytes, 2–3 energy transfer, 8 equivalents, 2 essential fatty acids, 27 factors determining plasma glucose level, 22–23 fatty acid oxidation, synthesis, 23 fatty acids, 23–29 forces acting on ions, 7–8 free fatty acid metabolism, 27 genesis of membrane potential, 8 glycogen synthesis, 21–22 hexoses, 23 ketone bodies, 23–24 lipid transport, 24, 26–27 lipids, 23–29 meiosis, 13 metabolic functions of amino acids, 19 mitosis, 13 molecular building blocks, 10–15 moles, 2 nonionic diffusion, 6 nucleic acids, 10–11 nucleosides, 10–11 nucleotides, 10–11 osmolal concentration of plasma, 6 osmosis, 5–6 pH, 4 plasma lipids, 24–27 post-translational modification, 17–18 principles, 2–8 protein degradation, 18 proteins, 16–17 protein synthesis, 17 replication, 13 RNA, 13–15 tonicity, 6 units for measuring concentration of solutes, 2 urea formation, 18 water, 2–3 Energy sources, 102–104 carbohydrate breakdown, 103 heat production in muscle, 104 oxygen debt mechanism, 104 phosphorylcreatine, 102–103 rigor, 104 Energy transfer, 8 Enkephalins, 177–178 Enteric nervous system, 261, 269–271, 442, 448 Enterochromaffin, 443 Enterochromaffin-like cells, 443 Enteroendocrine cells, 443 Enterohepatic circulation, 438, 482 Enterokinase, 455 Entorhinal cortex, 221 Envelope, 40 Environmental estrogen, 418 Enzymes, 33 Eosinophils, 63–64, 523 Epididymis, 402 Epilepsy childhood absence, 233 generalized with febrile seizures, 233 idiopathic, 233 Epinephrine, 138, 337 effects of, 340–342 Epineurium, 88 Epiphyses, 371 Epiphysial closure, 371 Epiphysial plate, 371 Episodic memory, 290 Epithelial sodium channels, 47 EPSP. See Excitatory postsynaptic potential Equilibrium, 203–218 middle ear, 203–205 Erection, 405–406 ERPF. See Effective renal plasma flow Erythroblastosis fetalis, 530 Erythrocytes, 523 Erythropoiesis, 677 Erythropoietin, 65, 677 regulation of secretion, 677 sources, 677 structure, function, 677 Escape phenomenon, 658 Esophagus, 471 Essential fatty acids, 27 Estradiol (17β-estradiol), 416 Estriol, 416 Estrogen-dependent, 427 Estrogens, 348, 371, 392 chemistry, 416 Estrone, 416 Estrous cycle, 415 Estrus, 415 Eukaryotes, 32 Eunuchoidism, 400, 410 Euploid, 13 Euthyroid, 306 Evoked cortical potentials, 231–232 Excitability, 79 Excitable tissue, 79–114 action potential, 86–87, 106 all-or-none law, 85–86 antidromic conduction, 88 axonal transport, 82–83 biphasic action potentials, 88 carbohydrate breakdown, 103 cardiac muscle morphology, 106 cellular elements in CNS, 80–93 changes in excitability during electrotonic potentials, 86–87 conduction, 83–88 contractile response, 96–102, 107–108 correlation between muscle fiber length, tension, 109 distribution of ion channels in myelinated neurons, 85 dystrophin-glycoprotein complex, 96 effects of denervation, 104 electrical activity, 110 electrical characteristics, skeletal muscle, 96 electrical phenomena, 96 electrical properties, 106–107 electrogenesis of action potential, 87 electromyography, 105 electrotonic potentials, 86 energy sources, 102–104 excitation, 83–88 fiber types, 102 fluxes, 96 INDEX 697 force generations, smooth muscle, 112 function of nerve supply to smooth muscle, 112 glial cells, 80 heat production in muscle, 104 ion distribution, 96 ionic fluxes, 96 ionic fluxes during action potential, 85 lipid breakdown, 103 local response, 86 mechanical properties, 107–109 metabolism, 102–104, 109 molecular basis of contraction, 97–100, 110–111 motor unit, 104–105 muscle fiber length and tension, 109 muscle twitch, 97 nerve fiber types, function, 88–89 neuronal growth, 91 neurons, 80–92 neurotrophins, 89–91 orthodromic conduction, 88 oxygen debt mechanism, 104 phosphorylcreatine, 102–103 plasticity of smooth muscle, 112 properties of mixed nerves, 88 properties of skeletal muscles, 104–106 receptors, 90 relation between muscle length, 102 relaxation, 111–112 resting membrane, 106 resting membrane potential, 83–85 rigor, 104 saltatory conduction, 87–88 sarcotubular system, 96 skeletal muscle morphology, 93–96 smooth muscle morphology, 109–112 strength of skeletal muscles, 105–106 striations, 95–96 summation of contractions, 101–102 trophic support of neurons, 89–90 types of contraction, 100–101 Excitation, 83–88 Excitation-contraction coupling, 100 Excitatory amino acids, 140–143 Excitatory junction potentials, 125 Excitatory postsynaptic potential, 117–120 Exercise, 331, 634–637 changes in tissues, 636–637 changes in ventilation, 634–636 exercise tolerance, 637 fatigue, 637 Exocytosis, 42–43, 64 Exogenous pathway, 24 Exons, 11 Exopeptidases, 455 Excitotoxins, 135 Expiration, 591–593 Expiratory muscles, 594 Expiratory reserve volume, 593 Explicit memory, 290 Exportins, 40 External auditory meatus, 203 External ear, 203–205 External hydrocephalus, 571 External intercostal muscles, 594 External respiration, 587 External urethral sphincter, 661 Exteroceptors, 219 Extracellular fluid, 2, 665–678 actions of angiotensins, 672 angiotensin-converting enzyme, 670–671 angiotensin II, 670–673 angiotensinogen, 670 erythropoietin, 677 heart hormones, 674–676 juxtaglomerular apparatus, 673 metabolism, 666, 676 metabolism of angiotensin II, 671–672 Na, K ATPase-inhibiting factor, 676 natriuretic factors, 674–676 natriuretic peptide receptors, 675–676 osmotic stimuli, 666–667 regulation of renin secretion, 673–674 regulation of secretion, 677 renin, 670 renin–angiotensin system, 670–674 secretion, 676 sources, 677 structure, function, 677 synthetic agonists, antagonists, 668 tissue renin–angiotensin systems, 672 tonicity, 665–668 vasopressin, 666 vasopressin receptors, 665–666 vasopressin secretion, 667 volume, 6690670 Extraction ratio, 643 Extrafusal fibers, 158 Extramedullary hematopoiesis, 522 Extrasystole beat, 498 Extrinsic innervation, 442, 448–449 Extrinsic system, 533 Eye movements, 199 superior colliculi, 199 F Face recognition, 298 Facilitated diffusion, 46, 318 Facilitation at synapses, 121–123 indirect inhibition, 121 Factors affecting neuronal growth, 91 Familial hypercalciuric hypocalcemia, 59 Familial hypercholesterolemia, 28 Familiarity, 293–294 Fascicular block, 497 Fasciculations, 104, 244 Fast axonal transport, 82 Fast excitatory postsynaptic potential, 266 Fast pain, 168 Fastigial nuclei, 254 Fat absorption, 457–458 Fat derivatives, 327 Fat digestion, 457 Fat metabolism, 382 in diabetes, 323–324 Fatigue, 637 Fatty acid oxidation, synthesis, 23 Fatty acids, 23–29 Feature detectors, 194 Female pseudohermaphroditism, 397 Female reproductive system, 411–427 abnormalities of ovarian function, 422–423 actions, 419 anovulatory cycles, 413–414 breasts, 417 central nervous system, 417 changes during intercourse, 414 chemistry, 416, 418–419 contraception, 422 control of cycle, 421 control of ovarian function, 420–422 cyclic changes in breasts, 414 cyclic changes in uterine cervix, 414 development of breasts, 426 endocrine changes, 424 endocrine organs, 417 environmental estrogen, 418 estrous cycle, 415 failure to reject fetal graft, 424 feedback effects, 421 female genitalia, 417 female secondary sex characteristics, 418 fertilization, 423–424 fetoplacental unit, 425 gynecomastia, 427 hormones, 427 human chorionic gonadotropin, 424 human chorionic somatomammotropin, 424 hypothalamic components, 420–421 implantation, 423–424 indicators of ovulation, 414–415 infertility, 424 initiation of lactation after delivery, 427 lactation, 426–427 lactation on menstrual cycles, 427 mechanism of action, 418–419 menstrual abnormalities, 422–423 menstrual cycle, 411–415 normal menstruation, 413 ovarian cycle, 411–412 ovarian hormones, 416–420 parturition, 425–426 placental hormones, 425 pregnancy, 423–426 reflex ovulation, 421–422 relaxin, 420 secretion, 417, 419 synthetic estrogen, 418 uterine cycle, 412–413 vaginal cycle, 414 Fenestrations, 537 Ferroportin 1, 459 Fertilin, 423 Fertilization, 423–424 Fetal adrenal cortex, 338 Fetal circulation, 582 changes in, 584 Fetal hemoglobin, 525 Fetal respiration, 582–583 Fetus, hemoglobin in, 525–526 FEV1, 593 Fever, 285–286 FGF. See Fibroblast growth factor Fiber types, 102 Fibrillations, 104 Fibrin, 531–532 Fibrin monomer, 532 Fibrinogen, 530 Fibrinolysin, 533 Fibrinolytic system, 533 Fibroblast growth factor, 91 Fibrous astrocytes, 80 Fick principle, 513, 574 Fick’s law of diffusion, 5 Filamentous actin, 36 Filtration, 49 Filtration fraction, 647 Filtration slits, 640 Final common paths, 164 Firing level, 86 First-degree heart block, 497 First messengers, 51 First polar body, 412 First sound, 512 FK-506. See Tacrolimus Flaccid, 163 698 INDEX Flaccid paralysis, 244 Flatus, 473 Flocculonodular lobe, 257 Flow-limited, 548, 601 Fluent aphasias, 297 Fluoxetine, 137 Fluxes, 96 fMRI. See Functional magnetic resonance imaging Focal adhesion, 38 Focal adhesion complexes, 37 Focal seizures. See Partial seizures Foliate papillae, 223 Follicles, 302 Follicle-stimulating hormone, 279, 377 Follicular phase, 412 Follistatins, 409 Foot plate, 203 Force generations, smooth muscle, 112 Forced vital capacity, 593 Forms of memory, 290 Fourth sound, 512 Fovea centralis, 182 Frank–Starling law, 515 FRC, 595 Free fatty acid metabolism, 27 Free fatty acids, 323 Free radical, 64 Frontal eye fields, 184 FSH. See Follicle-stimulating hormone Functional hypoglycemia, 333 Functional magnetic resonance imaging, 576 Functional residual capacity, 595 Functional transmission, 115–128 Functions of renal nerves, 644 Fungiform papillae, 223 Fusion, 198 FVC. See Forced vital capacity G G-CSF. See Granulocyte colony-stimulating factor G protein-coupled receptors, 54 G proteins, 50, 53–54 GABA, 141 GABA neurons, 238 GABA receptors, 141–142 GABA-T. See GABA transaminase GABA transaminase, 141 GAD. See Glutamate decarboxylase Galactosemia, 23 Gametogenesis, 392, 402–406 Gamma loop, 247 Gamma-motor neuron discharge control of, 161 effects of, 160–161 Gamma-motor neurons, 158 Gamma oscillation, 234 Gamma rhythm, 233–234 Ganglion cells, 182 Gap junctions, 38, 39–40 GAPs. See GTPase activating proteins Gas exchange in lungs, 600–602 composition of alveolar air, 601 diffusion across alveolocapillary membrane, 601–602 sampling alveolar air, 600–601 Gas, intestinal, 473 Gas transport in lung, 609–624 Gases, 145 properties of, 588 Gastric emptying, 473 Gastric motility, 473 Gastric pit, 432 Gastric secretion, 431 Gastrin, 443 Gastrocolic reflex, 477 Gastroileal reflex, 475 Gastrointestinal circulation, 449 Gastrointestinal hormones, 443 Gastrointestinal motility, 469–478 aerophagia, 473 colon, 475–478 defecation, 476–478 electrical activity, 470 esophagus, 471 gastric emptying, 473 gastric motility, 473 intestinal gas, 473 intestinal motility, 475 lower esophageal sphincter, 471–473 mastication, 471 migrating motor complex, 470–471 mixing, 470 motility of colon, 475–476 motility patterns, 469–471 mouth, 471 peristalsis, 469–470 regulation, 470 segment-specific motility patterns, 471–473 segmentation, 470 small intestines, 475–476 stomach, 473–475 swallowing, 471 transit time in colon, 476 vomiting, 473–475 Gastrointestinal system, 429–450 anatomic considerations, 431–432, 435 bile, 438–440 biliary secretion, 436–438 cholecystokinin, 443–446 circulation, 449 composition of pancreatic juice, 435–436 electrolyte transport, 440–442 enteric nervous system, 448 enteroendocrine cells, 443 extrinsic innervation, 448–449 gastric secretion, 431 gastrin, 443 gastrointestinal peptides, 448–449 gastrointestinal regulation, 442–443 gastrointestinal secretions, 431–442 ghrelin, 448 GIP, 446 hormones/paracrines, 443–448 intestinal fluid, 440–442 motilin, 447 origin, 432–435 pancreatic secretion, 435 peptide YY, 448 regulation, 432–435 regulation of secretion of pancreatic juice, 436 salivary secretion, 431 secretin, 446 somatostatin, 447–448 structural considerations, 430–431 VIP, 447 Gate-control hypothesis, 177 Gated, 45 GBG. See Gonadal steroid-binding globulin GD-NF. See Glial cell line-derived neurotrophic factor GEFs. See Guanine exchange factors Gene, 11 Gene activation, 540–541 Gene mutations, 11 Generalized epilepsy with febrile seizures, 233 Generation of action potential in postsynaptic neuron, 121 Generator, 83 Generator potential, 151 Genetic male, 392 Geniculocalcarine tract, 184 Genome, 11 Genotype, 529 Gephyrin, 130 Germinal, 671 Germinal angiotensin-converting enzyme, 404 Gestagens, 419 GFR. See Glomerular filtration rate Ghrelin, 384, 448 Gibbs–Donnan equation, 7 GIH. See Growth hormone-inhibiting hormone GIP, 446 Glaucoma, 182 angle-closure, 182 open-angle, 182 Glial cell line-derived neurotrophic factor, 91 Glial cells, 79–80 Glicentin, 328 Glicentin-related polypeptide, 328 Globin, 523 Globose nuclei, 254 Globular actin, 36 Globulin, 530 Globus pallidus, 249–250 Glomerular filtration, 639, 645–647 changes in, 647 control of, 646 filtration fraction, 647 hydrostatic pressure, 647 measuring, 645 normal, 646 osmotic pressure, 647 permeability, 646 size of capillary bed, 646 substances used to measure, 645 Glomerular filtration rate, 645 Glomeruli, 256 Glomerulotubular balance, 652 Glomus, 628 Glottis, 594 GLP-1, 328 GLPs. See Glucagon-like polypeptides Glucagon, 315, 328–330, 448 action, 329 chemistry, 328–329 insulin-glucagon molar ratios, 330 metabolism, 329 regulation of secretion, 329–330 Glucagon-like polypeptides 1 and 2, 328 Glucocorticoid, 250 Glucocorticoid binding, 346–347 Glucocorticoid feedback, 354 Glucocorticoid receptors, relation of mineralocorticoid to, 355 Glucocorticoid-regulated kinase, 355 Glucocorticoid-remediable aldosteronism, 356 Glucocorticoid secretion, 351–354 adrenal responsiveness, 352 chemistry ACTH, 352 circadian rhythm, 352–353 effect of ACTH on adrenal, 352 glucocorticoid feedback, 354 response to stress, 353–354 role of ACTH, 351–352 INDEX 699 Glucocorticoids, 337, 343, 347, 350–351, 371 ACTH secretion, 349 adrenal insufficiency, 349 anti-allergic effects, 351 anti-inflammatory effects, 351 blood cells, 350 Cushing syndrome, 350–351 intermediary metabolism, 349 lymphatic organs, effects on, 350 mechanism of action, 349 metabolism of, 347 nervous system, 349 pathologic effects of, 350–351 permissive action, 349 physiologic effects, 349–350 resistance to stress, 350 vascular reactivity, 349 water metabolism, 349 Glucogenic, 18 Glucokinase, 20 Gluconeogenesis, 20 Gluconeogenic, 18 Glucose buffer function, 481 Glucose-dependent insulinotropic polypeptide. See GIP Glucose/galactose malabsorption, 453 Glucose tolerance, 321–323 Glucose transporters, 318 Glucuronyl transferase, 483–484 Glutamate, 135, 140, 175, 577 Glutamate decarboxylase, 141 Glutamate receptors, 141 Glutamic dehydrogenase, 681 Glutaminase, 681 Glycine, 142 Glycogenesis, 20 Glycogenin, 21 Glycogenolysis, 20 Glycogen synthase, 21 Glycolysis, 20 Glycosylphosphatidylinositol anchors, 33 GM-CSF. See Granulocyte-macrophage colony-stimulating factor GnRH. See Gonadotropin-releasing hormone Golgi apparatus, 40, 42 Golgi cells, 256 Golgi tendon organ, 162 Gonadal steroid-binding globulin, 407 Gonadotropin-releasing hormone, 280 Gonads, 391–428 aberrant sexual differentiation, 396–398 abnormalities of ovarian function, 422–423 abnormalities of testicular function, 410 anabolic effects, 408 androgen-secreting tumors, 410 anovulatory cycles, 413–414 blood–testis barrier, 402–403 breasts, 417 central nervous system, 417 changes during intercourse, 414 chromosomal abnormalities, 396–397 chromosomal sex, 392–393 chromosomes, 392 contraception, 422 control of cycle, 421 control of ovarian function, 420–422 control of testicular function, 409–410 cryptorchidism, 410 cyclic changes in breasts, 414 cyclic changes in uterine cervix, 414 development of brain, 396 development of breasts, 426 development of gonads, 394 ejaculation, 406 embryology of genitalia, 394–396 embryology of human reproductive system, 394–396 endocrine changes, 424 endocrine function of testes, 406–409 endocrine organs, 417 environmental estrogens, 418 erection, 405–406 estrous cycle, 415 failure to reject fetal graft, 424 feedback effects, 421 female genitalia, 417 female reproductive system, 411–427 female secondary sex characteristics, 418 fertilization, 423–424 fetoplacental unit, 425 gametogenesis, 402–406 gynecomastia, 427 human chorionic gonadotropin, 424 human chorionic somatomammotropin, 424 hormonal abnormalities, 397–398 hormones, 427 hypothalamic components, 420–421 implantation, 423–424 indicators of ovulation, 414–415 infertility, 424 inhibins, 409–410 initiation of lactation after delivery, 427 lactation, 426–427 male hypogonadism, 410 male reproductive system, 402–410 mechanism of action, 408–409, 418–419 menopause, 400 menstrual abnormalities, 422–423 menstrual cycle, 411–415 metabolism, 407 normal menstruation, 413 ovarian cycle, 411–412 ovarian hormones, 416–420 parturition, 425–426 pituitary gonadotropins, 400–402 placental hormones, 425 precocious, delayed puberty, 399–400 pregnancy, 423–426 prolactin, 400–402 prolactin secretion, 401–402 PSA, 406 puberty, 398–400 reflex ovulation, 421–422 relaxin, 420 secondary sex characteristics, 407–408 secretion, 407, 417, 419 semen, 405 sex chromatin, 392–393 sex chromosomes, 392 sex differentiation, 392–400 sexual precocity, 399–400 spermatogenesis, 403–404 spermatozoa development, 404 steroid feedback, 410 structure, 402 synthetic estrogens, 418 temperature, 405 testicular production of estrogens, 409 testosterone, 406–407 transport, 407 uterine cycle, 412–413 vaginal cycle, 414 vasectomy, 406 Gout, 11 GPI anchors. See Glycosylphosphatidylinositol anchors GRA. See Glucocorticoid-remediable aldosteronism Graafian, 412 Gracilus, 173 Grade I proteins, 464 Grade II proteins, 464 Granule cells, 221, 254 Granulocyte colony-stimulating factor, 65 Granulocyte-macrophage colony-stimulating factor, 65 Granulocytes, 63–65, 522 Gravity, effect on circulation, 544 Gray rami communicans, 263 GRH. See Growth hormone-releasing hormone Group II endings. See Secondary endings Growth factors, 57–58, 382 Growth hormone, 279, 332, 371, 377, 380–385 binding, 381 biosynthesis, 380–381 carbohydrate metabolism, 382 chemistry, 380–381 direct actions of, 384 effects on growth, 382 electrolyte metabolism, 382 fat metabolism, 382 growth hormone receptors, 381–382 hypothalamic control of secretion, 384 indirect actions, 384 peripheral control of secretion, 384 plasma levels, 381 protein metabolism, 382 somatomedins, 382–384 species specificity, 381 stimuli affecting growth hormone secretion, 384–385 Growth hormone-inhibiting hormone, 280 Growth hormone-releasing hormone, 280 Growth periods, 386 Growth-promoting factors, 90 Growth physiology, 385–388 catch-up growth, 387–388 growth periods, 386 hormonal effects, 386–387 role of nutrition, 385–386 GRP, 448 GRPP. See Glicentin-related polypeptide GTPase activating proteins, 53 Guanine exchange factors, 53 Guanylin, 448 Guanylyl cyclase, 57 Guarding, 170 Gustducin, 226 Gynecomastia, 427 H Habituation, 291 Hair cells, 211 Hair dust, 206–208 Haldane effect, 612 Hartnup disease, 456 Haversian canals, 371 Haversian systems, 371 HbA, 323 HBE. See His bundle electrogram hCG. See Human chorionic gonadotropin hCS. See Human chorionic somatomammotropin HDL. See High-density lipoproteins Head injuries, 572 Hearing, 203–218 action potentials in auditory nerve fibers, 211 700 INDEX Hearing (continued) afferent nerve fibers, 207–208 air conduction, 210 audiometry, 213 audition, 212–213 bone conduction, 210 central pathway, 211, 214 cochlea, 205 deafness, 213 electrical responses, 207 external ear, 203–205 hair cells, 211 hair dust, 206–208 inner ear, 205 medulla oblongata, 211–212 middle ear, 203–205 organ of Corti, 205–206 responses to linear acceleration, 215–216 responses to rotational acceleration, 214–215 saccule, 206 semicircular canals, 206 sound localization, 213 sound transmission, 209–210 sound waves, 208–209 spatial orientation, 216 structure, 206–207 traveling waves, 210–211 tympanic reflex, 210 utricle, 206 vestibular system, 213–216 Heart, 507–520 arterial pulse, 510–512 atrial pressure changes, 512 atrial systole, 508 cardiac cycle, mechanical events, 507–520 cardiac muscle, relation of tension to length, 515 cardiac output, 513–519 early diastole, 509 echocardiography, 513 end-diastolic volume, factors affecting, 515 factors controlling, 514–515 heart sounds, 512 integrated control, 518–519 jugular pulse, 512 late diastole, 507 length of systole, diastole, 510 measurement, 513–514 murmurs, 512–513 myocardial contractility, 515–517 oxygen consumption by heart, 519 pericardium, 510 timing, 510 in various conditions, 514 ventricular systole, 508 Heartbeat, 489–506 Heart failure, 660 Heart hormones, 674–676 Heart sounds, 512 Heat loss, 284–285 Heat production, 104, 283–284 Heavy chains, 73 Helicotrema, 205 Helper T cells, 70 Hematocrit, 542 Hematopoietic stem cells, 522 Heme, 459, 523, 610 Hemianopia, 198 Hemiblock, 497 Hemidesmosome, 38 Hemispheres, complementary specialization, 295–296 Hemodialysis, 660 Hemoglobin, 523, 614 in fetus, 525–526 reactions of, 523–525 synthesis of, 526 Hemoglobin A, 523 Hemoglobin F, 525 Hemolytic disease of newborn, 530 Hemolytic transfusion reactions, 528 Hemosiderin, 459 Hemostasis, 531–535 anticlotting mechanisms, 533–535 anticoagulants, 535 clotting mechanism, 531–533 response to injury, 531 Hemostatic plug, 531 Henderson Hasselbach equation, 4 Heparin, 533 Hepatic circulation, 480–481 Hepatolenticular degeneration, 252 Hephaestin, 459 Hering–Breuer reflexes, 632 Herring bodies, 278 Heterologous desensitization, 130 Heterometric regulation, 515 Heteronymous, 198 Heterotrimeric G proteins, 53 Heterozygous, 529 Hexokinase, 20 Hexose monophosphate shunt, 20 Hexoses, 19, 23 Hiccup, 633 High-density lipoproteins, 27 High-energy phosphate compounds, 8 High-molecular-weight kininogen, 566 High-pressure system, 543 Hippocampus, 292–293 His bundle electrogram, 496 Histaminase, 138 Histamine, 137–138, 168, 238 Histones, 11 Histotoxic hypoxia, 617, 621 HLA-G, 424 HMG-CoA reductase, 27 Holter monitor, 496 Homeostasis, 58–59 Homeothermic, 283 Homologous desensitization, 130 Homometric regulation, 515 Homonymous, 198 Homunculus, 175 Horizontal cells, 182 Hormonal abnormalities, 397–398 Hormonal functions, hypothalamic regulation, 273–288 afferent connections of hypothalamus, 273–274 afferents, 285 anterior pituitary hormones, 279–280 anterior pituitary secretion, 279–282 autonomic function, 275–276 biosynthesis, 277–278 clinical implications, 282 effects of vasopressin, 279 fever, 285–286 heat loss, 284–285 heat production, 283–284 hypophysiotropic hormones, 280–282 hypothalamic control, 280 hypothalamic function, 275 hypothalamus, 273–277 hypothermia, 286 intraneuronal transport, 277–278 magnocellular neurons, electrical activity, 278 milk ejection reflex, 279 normal body temperature, 283 oxytocin, 277, 279 pituitary gland, 274–275 posterior pituitary secretion, 277–279 secretion, 277–278 significance, 282 temperature-regulating mechanisms, 285 temperature regulation, 282–286 thirst, 276 vasopressin, 277, 279 vasopressin receptors, 279 water intake, factors regulating, 277 Hormone-sensitive lipase, 27 Hormones, 443, 446 cancer and, 427 cholecystokinin, 443–446 enteroendocrine cells, 443 gastrin, 443 GIP, 446 motilin, 447 secretin, 446 somatostatin, 447–448 systemic regulation by, 566–567 VIP, 447 Horripilation, 284 hPL. See Human placental lactogen HSCs. See Hematopoietic stem cells Hue, 195 Human chorionic gonadotropin, 424 Human chorionic somatomammotropin, 424 Human placental lactogen, 424 Humoral hypercalcemia of malignancy, 370 Humoral immunity, 70 Huntingtin, 252 Huntington disease, 252 Hydrophilic, 32 Hydrophobic, 32 Hydrops fetalis, 530 Hydrostatic pressure gradient, 548 Hydroxyapatites, 371 Hydroxycholecalciferols, 365–367 Hydroxysteroid dehydrogenase, 345 Hydroxysteroid dehydrogenase type 2, 355 Hyperactive stretch reflexes, 244 Hyperaldosteronism, 359 Hyperalgesia, 168, 169 Hypercalcemia, 250 of malignancy, 370 Hypercalciuria, 250 Hypercapnia, 622, 630 Hyperglycemia, 323 effects of, 323 Hypergonadotropic hypogonadism, 410 Hyperkinetic, 252 Hypernatremia, 276 Hyperopia, 187 Hyperosmia, 222 Hyperosmolar coma, 324 Hypertension, 340 Hypertensive form, 346 Hypertonia, 244 Hypertonic, 6, 653 Hypertonicity, 163 Hypervariable regions, 74 Hypervitaminosis A, 466 Hypervitaminosis D, 466 Hypervitaminosis K, 466 Hypesthesia, 222 Hypocalcemic tetany, 364 Hypocapnia, 623 INDEX 701 Hypocretin, 237 Hypofunction, adrenocortical, 359–360 Hypogeusia, 226 Hypoglycemia, 326, 332–333 Hypoglycemia unawareness, 333 Hypogonadotropic hypogonadism, 282, 410 Hypokalemia, 319 Hypoketonemic hypoglycemia, 26 Hypokinetic, 252 Hypomenorrhea, 423 Hypoperfusion hypoxia, 621 Hypophysiotropic hormones, 280–282 Hypoproteinemia, 531 Hyporeflexia, 244 Hyporeninemic hypoaldosteronism, 360 Hyposmia, 222, 282 Hypothalamic components, 420–421 Hypothalamic control, 280 Hypothalamic function, 275 Hypothalamohypophysial tract, 274 Hypothalamus, 273–277 afferent connections of, 273–274 autonomic function, 275–276 efferent connections of, 273–274 hypothalamic function, 275 pituitary gland, 274–275 thirst, 276 water intake, factors regulating, 277 Hypothermia, 286 Hypotonia, 244, 257 Hypotonic, 6, 163, 653 Hypoxia, 616–617, 621–622 acclimatization, 618–619 barometric pressure, 617–618 hypoxic hypoxia, 617–619 hypoxic symptoms, 618 oxygen treatment, 621–622 Hypoxic hypoxia, 617, 619–620 venous-to-arterial shunts, 620 ventilation perfusion imbalance, 620 Hysteresis loop, 599 I IDDM. See Insulin-dependent diabetes mellitus Idiopathic epilepsies, 233 Idioventricular rhythm, 497 IDL. See Intermediate-density lipoproteins IGD-II. See Insulin-like growth factor II IJPs. See Inhibitor junction potentials Image-forming mechanism, 186–189 accommodation, 188–189 common defects of image-forming mechanism, 187–188 defects, 187–188 principles of optics, 186–187 pupillary reflexes, 189 Imidazoline, 139 Immune effector cells, 63–66 granulocytes, 64–65 lymphocytes, 65–66 macrophage colony-stimulating factors, 65 mast cells, 65 monocytes, 65 Immune system, genetic basis of diversity, 73–74 Immunity, 67–74 acquired immunity, 70 antigen presentation, 71–72 antigen recognition, 71 B cells, 72–73 complement system, 69 cytokines, 67–69 development of immune system, 70–71 genetic basis of diversity in immune system, 73–74 immunoglobulins, 73 innate immunity, 69–70 memory B cells, 71 overview, 67 T cell receptors, 72 T cells, 71 Immunoglobulins, 72–73 Immunohistochemistry, 129 Immunosympathectomy, 90 Implantation, 424 Implicit memory, 290 Importins, 40 Inactivated state, 85 Inactive capillaries, 549 Incomplete heart block, 497 Incomplete tetanus, 101 Increased automaticity, 498 Increasing osmolality, 654 Incus, 203 Indicator dilution method, 513 Indirect calorimetry, 461 Indirect inhibition, 121 Infection, 63–78 Inferior colliculi, 211 Inferior olive, 161 Inferior peduncle, 254 Inflammation, 63—78, 90 local injury, 75–76 systemic response to injury, 76 wound healing, 75–77 Inflammatory pain, 168 Inflammatory response, 64 Infranodal block, 497 Inhibin, 404, 409 Inhibin B, 392 Inhibitor junction potentials, 125 Inhibitory postsynaptic potential, 120 Inhibitory postsynaptic potentials, 117–120 Inhibitory systems, organization of, 122–123 Initial heat, 104 Initial segment, 80, 121 Innate immunity, 65, 67, 69–70 Inner ear, 205 Inner hair cells, 206 Innervation of blood vessels, 556 Innervation of renal vessels, 643 Inositol trisphosphate, 54–56 Inositol trisphosphate receptor, 110 Inotropic action, 514 Inotropic effect, 556 Insensible water loss, 284 Inside-out patch, 45 In situ hybridization histochemistry, 129 Inspiration, 591–593 Inspiratory muscles, 594 Inspiratory reserve volume, 593 Instantaneous, 495 Insufficiency, 513 Insulin, 315, 371 effects of, 318–320 structure, 316–317 Insulin deficiency, 321–325 acidosis, 324 changes in protein metabolism, 323 cholesterol metabolism, 324 coma, 324 effects of hyperglycemia, 323 effects of intracellular glucose deficiency, 323 fat metabolism in diabetes, 323–324 glucose tolerance, 321–323 Insulin-dependent diabetes mellitus, 334 Insulin excess, 325–326 compensatory mechanisms, 325–326 Insulin-glucagon molar ratios, 330 Insulin-like growth factor I, 91, 382 Insulin-like growth factor II, 382 Insulin preparations, 319 Insulin secretion, 326–327 cyclic AMP, 327 effect of automatic nerves, 327–328 effects of plasma glucose level, 326–327 fat derivatives, 327 intestinal hormones, 328 long-term changes in B cell responses, 328 protein, 327 Insulin sensitivity, 389 Insulinoma, 332 Integral proteins, 32 Integration, 247 Integrins, 38 Intensity, 152, 195 Intention tremor, 257–258 Intercalated cells, 641 Intercalated disks, 106 Intercellular communication, 50–58 calcium-binding proteins, 52–53 chemical messengers, 50–51 cyclic AMP, 56 diacylglycerol, 54–56 G protein-coupled receptors, 54 G proteins, 53–54 growth factors, 57–58 guanylyl cyclase, 57 inositol trisphosphate, 54–56 intracellular Cs2+ as second messenger, 52 mechanisms of diversity of Ca2+ actions, 53 production of cAMP by adenylyl cyclase, 56–57 receptors for chemical messengers, 50 second messengers, 54–56 stimulation of transcription, 51–52 Intercellular connections, 38–39 Intercortical transfer of memory, 292 Intercourse, changes during, 414 Interleukins, 67 Intermediate-density lipoproteins, 27 Intermediate filaments, 35–36 Intermittent claudication, 170 Internal hydrocephalus, 571 Internal respiration, 587 Internal urethral sphincter, 661 Internalization, 50 Internodal atrial pathways, 489 Interpositus nucleus, 254 Interstitial cells of Cajal, 470 Interstitial cells of Leydig, 402 Interstitial fluid, 2 equilibration with, 548 volume, 550–552 Intestinal fluid, 440–442 Intestinal gas, 473 Intestinal hormones, 328 Intracellular Ca2+ as second messenger, 52 Intracellular fluid, 2 Intracellular glucose deficiency, 323 effects of, 323 Intracranial pressure, 575 Intrafusal fibers, 158 Intralaminar nuclei, 229 Intramembranous bone formation, 371 Intraneuronal transport, 277–278 Intrauterine devices, 422 702 INDEX Intrinsic system, 532 Introns, 11 Invasion of immune cells, 90 Inverse stretch reflex, 162–163 In vitro fertilization, 424 Involutional osteoporosis, 374 Iodide transport across thyrocytes, 303 Iodine homeostasis, 303 Iodotyrosine deiodinase, 305 Ion channels, 33, 45–47 in myelinated neurons, 85 Ion distribution, 96 Ionic basis of photoreceptor potentials, 190 Ionic composition of blood, 503–504 Ionic fluxes, 96 during action potential, 85 Ionotropic receptors, 141 IPSP. See Inhibitory postsynaptic potential IP3R. See Inositol triphosphate receptor Iris, 181 Iron, 458–459 Iron deficiency anemia, 527 Irradiation of stimulus, 164 Irritant receptors, 632 Ischemic hypoxia, 617 Ishihara charts, 196 Islet cell hormones, 330–331 organization of pancreatic islets, 331 pancreatic polypeptide, 331 somatostatin, 330–331 Islet cell structure, 316 Isocapnic buffering, 636 Isoforms, 102, 108 Isohydric principle, 4 Isomaltase, 452 Isometric, 101 Isotonic, 6, 101 Isovolumetric ventricular contraction, 508 Itch, 168 IUDs. See Intrauterine devices J JAMs. See Junctional adhesion molecules Jaundice, 483–484 JG cells. See Juxtaglomerular cells Jugular pulse, 512 Junction, neuromuscular, 123–125 anatomy, 123 end plate potential, 124 quantal release of transmitter, 124–125 sequence of events during transmission, 123–124 Junctional adhesion molecules, 38 Junctional folds, 123 Juvenile diabetes, 334 Juxtacrine communication, 50 Juxtaglomerular apparatus, 641, 673 Juxtaglomerular cells, 641, 673 Juxtamedullary nephrons, 640 K K complexes, 234 Kainate receptors, 141 Kallidin, 566 Kallikreins, 566 Kallmann syndrome, 282 Karyotype, 392 Kayser–Fleischer rings, 252 Kernicterus, 530, 574 Ketoacidosis, 26 Ketogenic, 18 Ketone bodies, 23–24 Ketosteroids, 342, 348 Kety method, 574–575 Kilocalorie, 460 Kinases, 51 Kinesin, 37 Kininase I, 566 Kininase II, 566 Kinins, 168, 566 Kinocilium, 207 Knee jerk, 159 Knee jerk reflex, 158 Kupffer cells, 480 Kussmaul breathing, 324 L Labyrinth, 205 Labyrinth righting reflexes, 215 Lacis cells, 641, 673 Lacrimal duct, 185 Lacrimal gland, 185 Lactase, 452 Lactation, 426–427 development of breasts, 426 effect of lactation on menstrual cycles, 427 gynecomastia, 427 initiation of lactation after delivery, 427 Lactic acidosis, 324 Lactose, 452 Lambert–Eaton syndrome, 126 Lamellar bodies, 597 Laminar flow, 540 Laminins, 38 Language, 295—298 complementary specialization of the hemispheres vs. “cerebral dominance, ” 295–296 disorders, 297–298 physiology, 296–297 Large molecular transmitters, 143–145 opioid peptides, 143–144 polypeptides, 144–145 substance P, 143 tachykinins, 143 Late diastole, 507 Late endosome, 44 Latent period, 88 Lateral brain stem pathway, 247 Lateral corticospinal tract, 242 Lateral geniculate body, 184 Lateral inhibition, 153, 193 Lateral intercellular spaces, 640, 648 Lateral olfactory stria, 221 Law of Laplace, 542–543 Law of projection, 155 Law of specific nerve energies, 153–155 LCAT. See Lecithin-cholesterol acyltransferase LDL. See Low-density lipoproteins Lead pipe rigidity, 253 Leaky epithelium, 648 Learning, 289–295 Lecithin-cholesterol acyltransferase, 27 Left axis deviation, 495 Left bundle branch block, 497 Left ventricular ejection time, 510 LEMS. See Lambert–Eaton syndrome Length of systole, diastole, 510 Lengthening reaction, 163 Lens suspensary ligament, 181 Lenticular nucleus, 250, 252 Leu-enkephalin, 144 Leukemia inhibitory factor, 91 Leukotrienes, 28 Levodopa, 254 Lewy bodies, 254 LH. See Luteinizing hormone LHRH. See Luteinizing hormone-releasing hormone LIF. See Leukemia inhibitory factor Ligands, 33 Light adaptation, 197 Light chains, 73 Limiting pH, 680 Linear acceleration, 215–216 Lipidated, 33 Lipid breakdown, 103 Lipids, 23–29, 457–458 fat absorption, 457–458 fat digestion, 457 short-chain fatty acids in colon, 458 steatorrhea, 457 Lipid transport, 24, 26–27 Lipoproteins lipase, 26 Lipoxins, 28 Liver, 479–488 ammonia metabolism, excretion, 484–485 bile, 482–483 bilirubin metabolism, excretion, 483 detoxification, 481–482 functional anatomy, 479–480 functions, 481–485 glucuronyl transferase, 484 hepatic circulation, 480–481 jaundice, 483–484 metabolism, 481–482 synthesis of plasma proteins, 482 Local injury, 75–76 Local osteolytic hypercalcemia, 370 Local regulation, cardiovascular system, 563 autoregulation, 563 localized vasoconstriction, 563 vasodilator metabolites, 563 Local response, 86 Locomotor pattern generators, 249 Long QT syndrome, 214, 501 Long-term depression, 291 Long-term memory, 290, 293 Long-term potentiation, 141, 291 Loop of Henle, 640 Loudness, 209 Lou Gehrig disease. See Amyotrophic lateral sclerosis Low-density lipoproteins, 27 Lower esophageal sphincter, 471–473 Lower motor neuron lesion, 104 Lower motor neurons, 243, 244 Low-molecular-weight kininogen, 566 Lown–Ganong–Levine syndrome, 501 Low-pressure system, 543 LSD. See Lysergic acid diethylamide LTD. See Long-term depression LTP. See Long-term potentiation Lung, 590–591 air passages, 588–590 anatomy, 588–591 blood flow, 590–591 bronchi, 590 respiratory system, 588 Lung defense mechanisms, 605 Lungs, metabolic functions, 605–606 Lung volumes, 593–594 Luteal cells, 412 Luteal phase, 412 Luteinizing hormone, 279, 377 Luteinizing hormone-releasing hormone, 280 Luteolysis, 421 LVET. See Left ventricular ejection time INDEX 703 Lymph, 535 Lymphangiogenesis, 539 Lymphatic circulation, 550 interstitial fluid volume, 551–552 Lymphatics, 538, 643 Lymphedema, 552 Lymph flow, 521–554 Lymphocytes, 63, 65–66, 523 Lysergic acid diethylamide, 137 Lysin vasopressin, 277 Lysosomal storage diseases, 35 Lysosomes, 34–35 Lysylbradykinin, 566 M M cells, 456 M-CSF. See Macrophage colony-stimulating factor Macroheterogeneity, 443 Macrophage colony-stimulating factor, 65 Macrophages, 63 Macula, 206 Macula densa, 641, 652, 673 Macula lutea, 182 Macular sparing, 198 Magnetic resonance imaging, 82 Magnocellular neurons, electrical activity, 278 Major histo-compatibility complex, 71 Major proglucagon fragment, 328 Male hypogonadism, 410 Male pseudohermaphroditism, 397 Male reproductive system, 402–410 abnormalities of testicular function, 410 actions, 407 anabolic effects, 408 androgen-secreting tumors, 410 blood–testis barrier, 402–403 chemistry, 406–407 control of testicular function, 409–410 cryptorchidism, 410 effect of temperature, 405 ejaculation, 406 endocrine function of testes, 406–409 erection, 405–406 gametogenesis, 402–406 inhibins, 409–410 male hypogonadism, 410 mechanism of action, 408–409 metabolism, 407 PSA, 406 secondary sex characteristics, 407–408 secretion, 407 semen, 405 spermatogenesis, 403–404 spermatozoa development, 404 steroid feedback, 410 structure, 402 testicular production of estrogens, 409 transport, 407 vasectomy, 406 Male secondary sex characteristics, 407 Malignant hyperthermia, 286 Malleus, 203 Maltase, 452 Maltose, 452 Maltotriose, 452 Mannose-binding lectin pathway, 69 Manubrium, 203 MAP-1B, 130 MAP kinase. See Mitogen activated protein kinase Marrow, bone, 522 Masking, 209 Mass action contraction, 476 Mass reflex, 250 Mast cells, 63, 65, 137 Mastication, 471 Matrix, 250 Maximal stimulus, 88 Maximal voluntary ventilation, 593 Maximum metabolic rate, 462 Mayer waves, 562 Mean pressure, 544 Mean QRS vector, 495 Measuring blood pressure, 544–545 Mechanical nociceptors, 167 Mechanoreceptors, 150 Medial brain stem pathways, 246–247 Medial geniculate body, 211 Medial lemniscal system, 173 Medial lemniscus, 153, 173 Medial temporal lobe, 292–293 Median eminence, 275, 573 Medulla, 394 Medullary chemoreceptors, 629 Medullary control of cardiovascular system, 556–558 Medullary hormones, 338–342 adrenal medulla, substances secreted by, 340 catecholamines, 338–340 dopamine, effects of, 342 norepinephrine, effects of, 340–342 Medullary reticulospinal tract, 246 Megakaryocytes, 523 Meiosis, 13 Meissner corpuscles, 150 Melanins, 379 Melanocytes, 379 Melanophores, 379 Melanopsin, 192 Melanotropin-1, 379 Melanotropins, 379 Melatonin, 238–239 Membrane permeability, 45–46 Membrane potential, 8 Membrane transport proteins, 45–46 Membranous labyrinth, 205 Memory, 289–300 Memory B cells, 70–72 Memory T cells, 70 Menarche, 398 Ménière disease, 216 Menopause, 400 Menorrhagia, 423 Menstrual cycle, 411–415 abnormalities, 422–423 anovulatory cycles, 413–414 changes in breasts, 414 changes during intercourse, 414 changes in uterine cervix, 414 effect of lactation on, 427 estrous cycle, 415 indicators of ovulation, 414–415 normal menstruation, 413 ovarian cycle, 411–412 uterine cycle, 412–413 vaginal cycle, 414 Menstruation, 411 Merkel cells, 150 Merosin, 96 Mesangial cells, 640 Mescaline, 137 Mesocortical dopaminergic neurons, 179 Mesocortical system, 140 Messenger RNA, 13 Met-enkephalin, 144 Metabolic acidosis, 615–616, 684 alkalosis, 615–616 Metabolic alkalosis, 616, 684 Metabolic rate, 459–460 factors affecting, 461–462 measuring, 461 Metabolic myopathies, 98 Metabolic syndrome, 334–335 Metabolic theory of autoregulation, 563 Metabolism, 102–104, 109, 318, 459–463 Metabotropic receptors, 141 Metahypophysical diabetes, 328 Metarterioles, 537 Metathyroid diabetes, 328 Methemoglobin, 524, 621 Methylenedioxymethamphetamine, 137 Metrorrhagia, 423 MGluR4, 226 MHC. See Major histo-compatibility complex Micelles, 439, 457 Microfilaments, 35–36 Microfold cells, 456 Microglia, 80 activation of, 90 Microheterogeneity, 443 Microsomes, 32 Microtubule-organizing centers, 37 Microtubules, 35 Micturition, 639–664 Midcollicular decerebration, 247 Middle ear, 203–205 Middle peduncle, 254 Mifepristone, 419 Migrating motor complex, 470–471 Milk ejection, 279 Milk ejection reflex, 279 Mineral requirements, 464 Mineralocorticoids, 337, 343, 354–356 actions, 354–355 adrenalectomy, 356 effect of adrenalectomy, 356 mechanism of action, 355 Na+ excretion, 355–356 salt balance, 359 Minerals, absorption of, 458–459 Miniature end plate potential, 124 Miraculin, 226 Mitochondria, 34 Mitogen activated protein kinase, 52 Mitral cells, 220 Mixed nerves, properties of, 88 MMC. See Migrating motor complex Modality, 152 Modiolus, 206 Molecular building blocks, 10–15 DNA, 11–13 meiosis, 13 mitosis, 13 nucleic acids, 10–11 nucleosides, 10–11 nucleotides, 10–11 replication, 13 RNA, 13–15 Molecular mimicry, 75 Molecular motors, 37 Moles, 2 Monoamines, 130, 134–138 acetylcholine, 134–135 acetylcholine receptors, 135–136 cholinesterases, 135 704 INDEX Monoamines (continued) histamine, 137–138 serotonergic receptors, 137 serotonin, 136–137 Monochromats, 196 Monocytes, 63, 65, 523 Monoglycerols, 24 Monosaccharides, 19 Monosynaptic arcs, 157 Monosynaptic reflexes, 157–164 central connections of afferent fibers, 160 control of gamma-motor neuron discharge, 161 effects of gamma-motor neuron discharge, 160–161 function of muscle spindles, 160 inverse stretch reflex, 162–163 muscle tone, 163 reciprocal innervation, 161–162 structure of muscle spindles, 158–159 Monro–Kellie doctrine, 575 Morphine, 177–178 Mosaicism, 396 Mossy fibers, 255 Motilin, 447 Motility, gastrointestinal, 469–478 aerophagia, 473 colon, 475–478 defecation, 476–478 electrical activity, 470 esophagus, 471 gastric emptying, 473 gastric motility, 473 intestinal gas, 473 intestinal motility, 475 lower esophageal sphincter, 471–473 mastication, 471 migrating motor complex, 470–471 mixing, 470 motility of colon, 475–476 mouth, 471 patterns, 469–471 peristalsis, 469–470 regulation, 470 segment-specific patterns, 471–473 segmentation, 470 small intestines, 475 stomach, 473–475 swallowing, 471 vomiting, 473–475 Motility of colon, 475–476 Motility patterns, 469–471 electrical activity, 470 migrating motor complex, 470–471 mixing, 470 peristalsis, 469–470 regulation, 470 segmentation, 470 Motion sickness, 216 space, 216 Motivational-affect, 176 Motor unit, 104–105 Mouth, 471 MPGF. See Major proglucagon fragment MRI. See Magnetic resonance imaging mRNA. See Messenger RNA MS. See Multiple sclerosis MTOCs. See Microtubule-organizing centers Multi-colony-stimulating factor, 65 Multiple sclerosis, 82 Multiunit smooth muscle, 109 Murmurs, 512–513 Muscarinic actions, 135 Muscarinic cholinergic receptors, 135 Muscle, 93–114 action potentials, 106 body mechanics, 106 carbohydrate breakdown, 103 cardiac muscle morphology, 106 contractile response, 96–102, 107–108 correlation between muscle fiber length, tension, 109 dystrophin-glycoprotein complex, 96 effects of denervation, 104 electrical activity, 110 electrical characteristics, 96 electrical phenomena, 96 electrical properties, 106–107 electromyography, 105 energy sources, 102–104 fiber types, 102 fluxes, 96 force generations, smooth muscle, 112 function of nerve supply to smooth muscle, 112 heat production in muscle, 104 ion distribution, 96 ionic fluxes, 96 isoforms, 108 lipid breakdown, 103 mechanical properties, 107–109 metabolism, 102–104, 109 molecular basis of contraction, 97–100, 110–111 motor unit, 104–105 muscle fiber length, 109 muscle twitch, 97 organization, 93–94 oxygen debt mechanism, 104 phosphorylcreatine, 102–103 plasticity of smooth muscle, 112 properties of skeletal muscles in intact organism, 104–106 relation between muscle length and tension, 102 relaxation, 111–112 resting membrane, 106 rigor, 104 sarcotubular system, 96 skeletal muscle, 96 skeletal muscle morphology, 93–96 smooth muscle morphology, 109–112 strength of skeletal muscles, 105–106 striations, 95–96 summation of contractions, 101–102 types, 109–110 types of contraction, 100–101 Muscle fiber length, tension, correlation, 109 Muscle length, tension, relation, velocity of contraction, 102 Muscle pump, 549–550 Muscle spindles, 158–159 function of, 160 Muscle tone, 163 Muscle twitch, 97 Muscular atrophy, 244 Muscular dystrophy, 98 Becker, 98 Duchenne, 98 MVV. See Maximal voluntary ventilation Myasthenia gravis, 126 Myelin, 82 Myeloperoxidase, 64 Myenteric plexus, 269, 448 Myocardial contractility, 515–517 Myocardial infarction, 502–503 Myoepithelial cells, 279 Myogenic theory of autoregulation, 563 Myoglobin, 611 Myoneural, junction, 123 Myopia, 187 Myosin, 37 Myosin-II, 94 Myosin light chain phosphatase, 110 Myotonia, 98 Myristolated, 33 N NADPH oxidase, 64 Narcolepsy, 237 Natriuretic hormones, 566, 674 Natriuretic peptide receptors, 675–676 Natural killer cells, 69 Near point of vision, 189 Negative feedback process, 85 Negative selection, 71 Neuropathic pain, 169 Neurotrophins, 90 Neospinothalamic tract, 176 Nephrogenic diabetes insipidus, 59, 668 Nephron, 640–642 Nernst equation, 7 Nerve conduction tests, 82 Nerve endings, 125–126 anatomy, 125 junctional potentials, 125–126 Nerve fiber types and function, 88–89 Nerve growth factor, 83 Nerve impulses, 83 Nerves, 79–92 action potential, 86–87 all-or-none law, 85–86 antidromic conduction, 88 axonal transport, 82–83 biphasic action potentials, 88 cellular elements in CNS, 80–93 changes in excitability, 86–87 conduction, 83–88 electrogenesis of action potential, 87 electrotonic potentials, 86 excitation, 83–88 firing level, 86 glial cells, 80 ionic fluxes during action potential, 85 local response, 86 myelinated neurons, 85 nerve fiber types and function, 88–89 neuronal growth, 91 neurons, 80–92 neurotrophins, 89–91 orthodromic conduction, 88 properties of mixed nerves, 88 receptors, 90 resting membrane potential, 83–85 saltatory conduction, 87–88 trophic support of neurons, 89–90 Nervi erigentes, 405 Net amount transferred, 648 Net flux, 5 Neural basis of memory, 290–291 Neural communication, 50 Neural hormones, 277 Neurexins, 117 Neuroactive steroids, 146 Neurofibrillary tangles, 294 Neurogenesis, 293 Neuroglycopenic symptoms, 325 Neurohemal organs, 573 Neurokinin B gene, 143 INDEX 705 Neuroleptic drugs, 252 Neurological exam, 155 Neuromodulators, 129–148 Neuromuscular junction, 115, 123 Neurons, 79–92 trophic support, 89–90 Neuropathic pain, 168 Neuropeptides, 130 Neuropeptide Y, 145 Neurophysin, 277 Neurosecretion, 277 Neurosteroids, 146 Neurotransmitters, 129–148 Neurotrophins, 89–91, 250 actions, 90–91 neuronal growth, 91 receptors, 90 trophic support of neurons, 89–90 Neutral fat, 24 Neutrophils, 63, 523 Newborn, hemolytic disease, 530 New brain cells, 293 NGF. See Nerve growth factor Nicotinic actions, 135 Nicotinic acetylcholine receptors, 126 Nicotinic cholinergic receptors, 135 NIDDM. See Non-insulin-dependent diabetes mellitus Night terrors, 237 Nigrostriatal system, 140 Nitric oxide, 112, 563–564 NK cells. See Natural killer cells NMDA receptors, 141 N,N-dimethyltryptamine, 137 NO synthase, 405 Nociception, 168 Nociceptive stimuli, 164 Nociceptors, 150, 167–168 Nocturia, 660 Nocturnal enuresis, 237 Nodes of Ranvier, 82 Non-insulin-dependent diabetes mellitus, 334 Non-REM sleep, 234 Nonassociative learning, 290 Noncholinergic, nonadrenergic innervation, 590 Nonclathrin/noncaveolae endocytosis, 43 Noncommunicating hydrocephalus, 571 Nonconstitutive pathway, 43 Nonconvulsive, 233 Nondeclarative memory, 290 Nonessential amino acids, 15 Nonfluent aphasias, 297 Nonionic diffusion, 6, 681 Nonspecific, 231 Nonspecific cholinesterase, 135 Nonsuppressible insulin-like activity, 318 Nonsyndromic deafness, 214 Noradrenergic neurons, 138 Norepinephrine, 138, 237, 265–266, 337 Normal arterial blood pressure, 546 Normal body temperature, 283 Normal cardiac rate, 497 Normal ECG, 494 Normal menstruation, 413 Normal sinus rhythm, NSR, 497 Nose, pain fibers in, 223 NREM sleep. See Non-REM sleep NSILA. See Nonsuppressible insulin-like activity NTS. See Nucleus of solitary tract Nuclear bag fiber, 158 Nuclear chain fiber, 158 Nuclear cholescintigraphy, 486 Nuclear factor, 76 Nuclear factor kappa B, 52 Nuclear membrane, 40 Nuclear pore complexes, 40 Nucleic acids, 454–457 Nucleolus, 40 Nucleosome, 40 Nucleus accumbens, 179 Nucleus basalis of Meynert, 293 Nucleus of solitary tract, 224 Nutrition, 463–466 caloric intake, distribution, 463–464 dietary components, 463 effect on growth, 385–386 mineral requirements, 464 vitamins, 464–466 Nutritionally essential amino acids, 15 Nyctalopia, 191 Nystagmus, 215 O Obesity, 334–335 Objective scotomas, 198 OBP. See Odorant-binding proteins Obstructive sleep apnea, 237 Occludin, 38 Occlusion, 123, 164 Ocular dominance columns, 194 Oculocardiac reflex, 499 Oculomotor nerve, 189 Odorant-binding proteins, 223 Odorant receptors, 219 Off-center cell, 192 Olfactory bulbs, 220–221 Olfactory cortex, 221 Olfactory epithelium, 219–220 Olfactory glomeruli, 220 Olfactory sensory neurons, 219 Olfactory thresholds, 221–222 Olfactory tubercle, 221 Oligoclonal bands, 82 Oligodendrocytes, 80 Oligomenorrhea, 423 Oliguria, 660 Olivocochlear bundle, 211 On-center cell, 192 Oncogenes, 43 Oncotic pressure, 49, 531 Open-angle glaucoma, 182 OPG. See Osteoprotegerin Opioid peptides, 143–144 Opsin, 190 Opsonization, 64 Optical righting reflexes, 215 Optic chiasm, 184 Optic disk, 182 Optic nerve, 184 Optic pathways, lesions in, 198 Optics, 186–187 Optic tract, 184 Oral glucose tolerance test, 322 Orexin, 237 Organ of Corti, 205–206 Organelles, 31 Organum vasculosum of lamina terminalis, 276, 573 Orientation columns, 194 Orthodromic conduction, 88 Orthograde transport, 82 OSA. See Obstructive sleep apnea Osmolal concentration of plasma, 6 Osmolarity, 5 Osmoles, 5 Osmosis, 5–6 Osmotic diuresis, 656 Osmotic pressure, 5 Osmotic pressure gradient, 548 Osmotic stimuli, 666–667 Ossicular conduction, 210 Osteoblasts, 371 Osteoclasts, 371 Osteons systems, 371 Osteopetrosis, 373 Osteoporosis, 351, 373 Osteoprotegerin, 372 Otoconia, 206, 216 Otolithic organ, 206 Otoliths, 206 Ouabain, 676 Outer hair cells, 206 Oval window, 203 Ovarian cycle, 411–412 Ovarian function, 422–423 abnormalities of, 422–423 contraception, 422 control of, 420–422 control of cycle, 421 feedback effects, 421 hypothalamic components, 420–421 menstrual abnormalities, 422–423 reflex ovulation, 421–422 Ovarian hormones, 416–420 actions, 419 breasts, 417 central nervous system, 417 chemistry, 416, 418–419 endocrine organs, 417 environmental estrogen, 418 female genitalia, 417 female secondary sex characteristics, 418 mechanism of action, 418–419 relaxin, 420 secretion, 417, 419 synthetic estrogen, 418 Ovarian hyper-stimulation syndrome, 401 Overflow incontinence, 662 Overtones, 209 OVLT. See Organum vasculosum of lamina terminalis Ovulation, 412 indicators of, 414–415 Oxidases, 35 Oxidation, 8 Oxidative deamination, 18 Oxidative phosphorylation, 10, 34 Oxygenation, 610 Oxygen consumption, 576–577, 645 heart, 519 Oxygen debt mechanism, 104 Oxygen delivery to tissues, 609–610 Oxygen–hemoglobin dissociation curve, 610 Oxygen transport, 609–611 Oxyhemoglobin, 523 Oxyntic cells. See Parietal cells Oxyntomodulin, 328 Oxyphil cells, 367 Oxytocin, 277, 279, 377 effects of, 279 P Pacemaker abnormal, 497 706 INDEX Pacemaker (continued) implanted, 497–498 potentials, 491–492 Pacemaker potential, 491 Pacinian corpuscles, 150–151, 155 Paclitaxel, 35 PAF. See Platelet-activating factor Pain, 167–172 classification, 168–171 deep pain, 169 neuropathic, 169 nociceptors, 167–168 phantom limb, 176 phantom tooth, 176 referred pain, 170–171 visceral pain, 169–170 Pain transmission modulation, 177–179 acetylcholine, 178–179 cannabinoids, 179 enkephalins, 177–178 morphine, 177–178 stress-induced analgesia, 177 Paleospinothalamic tract, 176 Pallesthesia, 155 Pallidotomy, 254 Palmitoylated, 33 Palmitoylethanolamide, 145 Pancreas, 315–336 acidosis, 324 adrenal glucocorticoids, 332 automatic nerves, 327–328 biosynthesis, 317–318 carbohydrate metabolism, 331–332 catecholamines, 331–332 changes in protein metabolism, 323 cholesterol metabolism, 324 coma, 324 cyclic AMP, 327 diabetes mellitus, 333–334 exercise, 331 fat derivatives, 327 fat metabolism in diabetes, 323–324 glucagon, 328–330 glucose tolerance, 321–323 glucose transporters, 318 growth hormone, 332 hyperglycemia, 323 hypoglycemia, 332–333 insulin, 316–320 insulin deficiency, 321–325 insulin excess, 325–326 insulin–glucagon molar ratios, 330 insulin preparations, 319 insulin receptors, 921 intestinal hormones, 328 intracellular glucose deficiency, 323 islet cell hormones, 330–331 islet cell structure, 316 long-term changes in B cell responses, 328 mechanism of action, 320–321 metabolic syndrome, 334–335 metabolism, 318, 329 obesity, 334–335 pancreatic islets, 331 pancreatic polypeptide, 331 plasma glucose level, 326–327 protein, 327 regulation of insulin secretion, 326–328 regulation of secretion, 329–330 relation to potassium, 319 secreted insulin, 318–320 secretion, 317–318 somatostatin, 330–331 thyroid hormones, 332 type 2 diabetes, 334–335 types of diabetes, 334 Pancreatic islets, organization of, 331 Pancreatic juice composition of, 435–436 regulation of secretion, 436 Pancreatic polypeptide, 315, 331 Panting, 284 Papillae, 223 Paracellular pathway, 38, 648 Paracrine, 442 Paracrine communication, 50 Paracrines, 443 Paradoxical sleep, 234 Parafollicular cells, 370 Paraganglia, 338 Parageusia, 226 Parallel fibers, 255 Paralysis agitans, 253–254 Parasomnias, 237 Parasympathetic, 261 Parasympathetic cholinergic discharge, 268–269 Parathyroid glands, 367–370 actions, 368 anatomy, 367 Parathyroid hormone, 363, 367 Parathyroid hormone-related protein, 369 Parietal cells, 431 Parietal cortex, 245 Parietal lobe, 243 Parietal pathway, 195 Parkinson disease, 253–254 Paroxysmal atrial tachycardia with block, 499 Paroxysmal tachycardia, 498 Paroxysmal ventricular tachycardia, 500 Pars compacta, 250 Pars reticulata, 250 Partial pressure, 588 Partial seizures, 232 Passive tension, 102 Past-pointing, 258 Patch clamping, 45 Pathologic, 168 PB2-binding proteins, 130 P450c11, 345 P450c11AS, 345 P450c17, 345 P460c21, 345 PDE. See Phosphodiesterase PDGF. See Platelet-derived growth factor PEA. See Palmitoylethanolamide Pendred syndrome, 214 Pendrin, 303 Penetration of substances into brain, 572–573 Pentoses, 19 Penumbra, 135 PEP. See Preejection period Peptic cells, 431 Peptide bonds, 16 Peptide YY, 448 Peptides, 16, 448–449 Perforins, 69 Perfusion-limited, 601 Pericardial sac, 510 Pericardium, 510 Pericentriolar material, 37 Pericytes, 538, 640 Periglomerular cells, 221 Perilymph, 205 Perimetry, 197 Perinuclear cisterns, 40 Periodic breathing, 634 Periodic limb movement disorder, 237 Peripheral arterial chemoreceptors, 562 Peripheral benzodiazepine receptors, 142 Peripheral chemoreceptor reflex, 562 Peripheral proteins, 32 Peripheral venous pressure, 550 Peristalsis, 469–470, 475 Peritubular capillaries, 642 Permissive action, 349 Peroxins, 35 Peroxisome proliferation activated receptors, 35 Peroxisomes, 32, 35 Persistent hyperinsulinemic hypoglycemia of infancy, 327 Pertussis toxin, 57 PET. See Positron emission tomography Peyer’s patches, 73 P factor, 170 PGO spikes. See Pontogeniculo-occipital spikes Phagocytosis, 43, 64 Phantom eye syndrome, 176 Phantom limb pain, 176 Phantom tooth pain, 176 Pharmacology of prostaglandins, 28 Phasic bursting, 278 Phasic receptors. See Rapidly adapting receptors Phenotype, 529 Phenylalanine, 138 Phenylalanine hydroxylase, 138 Phenylethanolamine-N-methyltransferase, 138 Phenylketonuria, 138 Pheochromocytomas, 342 Pheromones, 223 PHM-27, 447 Phosphatases, 51 Phosphate timer, 51 Phosphaturic action, 368 Phosphodiesterase, 56, 111 Phosphoinositol 3-kinase, 321 Phospholipase A, 28 Phosphorus, 365 metabolism, 364–365 Phosphorylcreatine, 102–103 Photopic vision, 185 Photoreceptor mechanism, 190–193 cone pigments, 192 electrical responses, 190 ionic basis of photoreceptor potentials, 190 melanopsin, 192 photosensitive compounds, 190 processing of visual information in retina, 192–193 resynthesis of cyclic GMP, 192 rhodopsin, 190–192 Photoreceptors, 150 Photosensitive compounds, 190 Phototherapy, 527 Physiological tremor, 160, 161 Physiologic dead space, 600 Physiologic shunt, 602 Piebaldism, 380 Pigment abnormalities, 379–380 Pigment epithelium, 182 PIH. See Prolactin-inhibiting hormone Pillar cells, 205 Pineal gland, 238 Pineal sand, 238 Pinocytosis, 43 Piriform cortex, 221 Pitch, 209 INDEX 707 Pituicytes, 378 Pituitary gland, 274–275, 377–390 binding, 381 biosynthesis, 378–381 carbohydrate metabolism, 382 catch-up growth, 387–388 cell types in, 378 chemistry, 380–381 direct actions of, 384 electrolyte metabolism, 382 endocrine glands, 388 fat metabolism, 382 gross anatomy, 378 growth, 382 growth hormone, 380–385 growth hormone receptors, 381–382 growth periods, 386 growth physiology, 385–388 histology, 378 hormonal effects, 386–387 hypothalamic control of secretion, 384 indirect actions, 384 insulin sensitivity, 389 morphology, 378 peripheral control of secretion, 384 pigment abnormalities, 379–380 pituitary insufficiency, 388–389 plasma levels, 381 proopiomelanocortin, 378–380 protein metabolism, 382 role of nutrition, 385–386 skin coloration, 379–380 somatomedins, 382–384 species specificity, 381 stimuli affecting growth hormone secretion, 384–385 water metabolism, 389 PI3K. See Phosphoinositol 3-kinase Pituitary gonadotropins, 400–402 Pituitary insufficiency, 388–389 causes of, 389 endocrine glands, 388 insulin sensitivity, 389 water metabolism, 389 Pituitary secretion, 277–279 biosynthesis, 277–278 intraneuronal transport, 277–278 magnocellular neurons, electrical activity, 278 milk ejection reflex, 279 oxytocin, 277, 279 secretion, 277–278 vasopressin, 277, 279 vasopressin receptors, 279 PKA. See Protein kinase A Placenta, 582 circulation, 581–584 Planum temporale, 212, 296 Plasma, 530–531 hypoproteinemia, 531 plasma proteins, 530–531 Plasma aspartate aminotransferase, 18 Plasma cells, 70, 72 Plasma glucose level, effects of, 326–327 Plasma kallikrein, 566 Plasma levels, 381 Plasma lipids, 24–27 Plasma membrane, 32 Plasma osmolality and disease, 6 Plasma proteins, 530–531 synthesis of, 482 Plasma renin activity, 671 Plasma renin concentration, 671 Plasma T, 305 Plasmin, 533 Plasminogen system, 533 Plasticity, 112 Platelet-activating factor, 74 Platelet aggregation, 74 Platelet-derived growth factor, 74, 91 Platelets, 74–75, 523 Pleasure center, of the brain, 179 Plethysmograph, 539 Plethysmography, 539 PLMD. See Periodic limb movement disorder PMNs. See Polymorphonuclear leukocytes PMS. See Premenstrual syndrome Pneumotaxic center, 627 PNMT. See Phenylethanolamine-N-methyltransferase Podocytes, 640 Poikilothermic, 282 Point mutations, 11 Poiseuille–Hagen formula, 541–542 Polydipsia, 668 Polymodal nociceptors, 167 Polymorphonuclear leukocytes, 63 Polypeptides, 16, 144–145 Polypeptide YY, 331 Polysynaptic reflexes, 157, 163–164 fractionation, 164 importance of withdrawal reflex, 164 occlusion, 164 withdrawal reflex, 163 Polyubiquitination, 18 Polyuria, 660, 668 POMC. See Proopiomelanocortin Pontine tract, 246 Pontogeniculo-occipital spikes, 235 Portal hypophysial vessels, 275 Position agnosia, 155 Positive feedback loop, 85 Positron emission tomography, 289, 576 Posterior hypothalamic neurons, 238 Posterior pituitary, 573 Postextrasystolic potentiation, 516 Postganglionic neuron, 262 Postictal period, 233 Postsynaptic cell, 115 Postsynaptic density, 116 Postsynaptic inhibition, 121 spinal cord, 121 Posttetanic potentiation, 291 Posture, 241–260 Posture-regulating systems, 247 Potassium, 319 Potential difference, 83 PPARs. See Peroxisome proliferation activated receptors pp39mos, 412 PRA. See Plasma renin activity Prachiasmatic nuclei, 236 Prandial drinking, 276 PRC. See Plasma renin concentration Pre-Bötzinger complex, 626 Precapillary sphincters, 537 Precocious, delayed puberty, 399–400 sexual precocity, 399–400 Precocious pseudopuberty, 348, 399 Preejection period, 510 Preganglionic neuron, 262 Pregnancy, 423–426 endocrine changes, 424 failure to reject fetal graft, 424 fertilization, 423–424 fetoplacental unit, 425 human chorionic gonadotropin, 424 human chorionic somatomammotropin, 424 implantation, 423–424 infertility, 424 parturition, 425–426 placental hormones, 425 Prekallikrein, 566 Preload, 515 Premature beat, 498 Premenstrual syndrome, 423 Premotor cortex, 243, 245 Preoptic neurons, 238 Prepotent, 164 Prepotential, 491 Preproinsulin, 317 Prepro-oxyphysin, 278 Prepropressophysin, 277 PreproPTH, 367 Preprorenin, 670 Presbycusis, 213 Presbyopia, 189 Prestin, 211 Presynaptic cell, 115 Presynaptic facilitation, 122 Presynaptic inhibition, 122 Presynaptic receptors, 130 Presynaptic terminals, 80 Prevertebral ganglia, 263 PRH. See Prolactin-releasing hormone Primary adrenal insufficiency, 360 Primary amenorrhea, 400, 422 Primary colors, 196 Primary evoked potential, 232 Primary hyperaldosteronism, 359 Primary motor cortex, 243–245 Primary plexus, 275 Primary somatosensory area, 243 Primary spermatocytes, 403 Primary visual cortex, 184 Priming, 290 Primordial follicles, 411 Principal axis, 186 Principal cells, 354, 641 Principal focal distance, 186 Principal focus, 186 PRL. See Prolactin Proarrhythmic, 501 Procedural memory, 290 Prodynorphin, 144 Proenkephalin, 144 Proenzymes, 454 Progenitor cells, 522 Progestational agents, 419 Progesterone, 392, 418–419 Progestins, 419 Programmed cell death, 42 Progressive motility, 404 Proinsulin, 318 Prokaryotes, 32 Prolactin, 280, 377, 400–402 secretion, 401–402 Prolactin-inhibiting hormone, 280 Prolactin-releasing hormone, 280 Proliferative phase, 412 Promoter, 11 Proopiomelanocortin, 144, 378–380 biosynthesis, 378–379 pigment abnormalities, 379–380 skin coloration, 379–380 Properdin pathway, 69 Proprioception, 158 708 INDEX Proprioceptors, 632–633 ProPTH, 367 Prorenin, 670 Prosopagnosia, 298 Prostacyclin, 563 Prostaglandin G/H synthases, 28 Prostaglandins, 28, 146, 405 pharmacology of, 28 Prostate, 402 Prostate-specific antigen, 406 Proteasomes, 18, 72 Protein, 16–17, 327, 454–457 absorption, 455–457 binding, 305–306 degradation, 18 digestion, 454–455 synthesis, 17 Protein kinase A, 56 Protein kinase C, 55 Protein metabolism, 323, 382 changes in, 323 Proteins, 614 Protodiastole, 509 Proton pump, 35 Proto-oncogenes, 43 Protoplasmic astrocytes, 80 Proximal convoluted tubule, 640 Pruritus, 168 PSA. See Prostate-specific antigen P450scc, 344 Pseudocholinesterase cholinesterase, 135 Pseudohypoaldosteronism, 360 Pseudohypoparathyroidism, 369 Psilocin, 137 PTH. See Parathyroid hormone PTHRP, 369–370 PTHrP. See Parathyroid hormone-related protein Pubarche, 398 Puberty, 398–400 onset of, 398–399 Pulmonary chemoreflex, 632 Pulmonary circulation, 590, 602–604 flow, 602 gravity, 602–603 pressure, 602 pulmonary blood vessels, 602 pulmonary reservoir, 603 regulation of pulmonary blood flow, 603–604 ventilation/perfusion ratios, 603 volume, 602 Pulmonary function, 587–608 air passages, 588–590 alveolar surface tension, 596–597 blood flow, 590–591, 599 bronchi, 590 bronchial tone, 594–595 chest wall, 595–596 compliance of lungs, chest wall, 595–596 dead space, 599–600 diffusion across alveolocapillary membrane, 601–602 endocrine functions of lungs, 605–606 expiration, 591–593 flow, 602 gas exchange in lungs, 600–602 gases, properties of, 588 glottis, 594 gravity, 602–603 inspiration, 591–593 lung defense mechanisms, 605 lung volumes, 593–594 metabolic functions of lungs, 605–606 partial pressures, 588 pressure, 602 pulmonary blood vessels, 602 pulmonary circulation, 602–604 pulmonary reservoir, 603 quantitating respiratory phenomena, 588 regulation of pulmonary blood flow, 603–604 respiration, 591–600 respiratory muscles, 594 respiratory system, 588, 605–606 sampling alveolar air, 600–601 surfactant, 597–598 uneven ventilation, 599–600 ventilation, 599 ventilation/perfusion ratios, 603 volume, 602 work of breathing, 598–599 Pulmonary ventilation, 593 Pulse oximeter, 588 Pulse pressure, 544 Pumps, 33 Pupil, 181 Pupillary light reflex, 189 Pupillary reflexes, 189 Purine, 145 Purkinje cells, 254 Purkinje system, 489 Putamen, 249 Pyramidal lobe, 302 Pyrimidine transmitters, 145 Pyrin, 286 Q QS2. See Total electromechanical systole Quantitating respiratory phenomena, 588 Quaternary structure, 17 R Radiation, 284 Radiofrequency catheter ablation, reentrant pathways, 501–502 Radionuclides, 579 Rafts, 44 Raloxifene, 418 Rapid eye movement sleep, 234–235 Rapidly adapting receptors, 153, 632 RAS. See Reticular activating system Rathke’s pouch, 378 Rayleigh match, 197 Reaction time, 160 Reactive hyperemia, 579, 581 Rebound phenomenon, 258 Receptive field, 152 Receptive relaxation, 473 Receptor potential, 151 Receptors, 33, 90, 130–132, 139, 184 Receptors in airways, lungs, 632 Reciprocal innervation, 121, 161–162, 625 Recovery heat, 104 Recruitment of motor units, 105, 164 Recruitment of sensory units, 155 Red blood cells, 523 catabolism of hemoglobin, 526–527 fetal, 525–526 hemoglobin, 523 reactions of hemoglobin, 523–525 role of spleen, 523 synthesis of hemoglobin, 526 Red marrow, 522 Red reaction, 580 Reductase deficiency, 397 Reduction, 8 Reentrant pathways, radiofrequency catheter ablation, 501–502 Referred pain, 170–171 Reflex arc, 157 Reflexes, 157–166 adequate stimulus, 164 central connections of afferent fibers, 160 central excitatory, 165 control of gamma-motor neuron discharge, 161 final common path, 164–165 fractionation, 164 function of muscle spindles, 160 gamma-motor neuron discharge, 160–161 importance of withdrawal reflex, 164 inhibitory states, 165 inverse stretch reflex, 162–163 monosynaptic reflexes, 158–164 muscle tone, 163 occlusion, 164 polysynaptic reflexes, 163–164 properties of, 164–165 properties of reflexes, 164–165 reciprocal innervation, 161–162 stimulus, 164 stretch reflex, 158–164 structure of muscle spindles, 158–159 withdrawal reflex, 163–164 Reflex ovulation, 421–422 Reflex sympathetic dystrophy, 169 Refraction, 186 Refractory period, 87 Regional blood flow, 645 Regulated pathway, 43 Regulators of G protein signaling, 54 Regulatory elements, 11 Regurgitation, 513 Relative refractory period, 87 Relaxation heat, 104 Relaxation pressure curve, 595 Relaxed configuration, 610 Relaxin, 392 REM sleep. See Rapid eye movement sleep Renal circulation, 643–645 Renal compensation, 616 Renal function, 639–664 abnormal Na+ metabolism, 660 acidosis, 660 adrenocortical steroids, 657–658 aquaporins, 652 autoregulation, renal blood flow, 644–645 bladder, 661–662 blood flow, 643–644 blood vessels, 642–643 capsule, 643 changes in, 647 collecting ducts, 653–654 control of, 646 countercurrent mechanism, 654–656 deafferentation, 662 denervation, 662 disordered renal function, 659–660 distal tubule, 653 diuretics, 659 emptying, 661–662 filling, 661 filtration fraction, 647 free water clearance, 657 functional anatomy, 640–643 functions of renal nerves, 644 glomerular filtration, 645–647 glomerulotubular balance, 651–652 INDEX 709 glucose reabsorption, 650 glucose transport mechanism, 650 humoral effects, 658 hydrostatic pressure, 647 innervation of renal vessels, 643 K+ excretion, 659 loop of Henle, 653 loss of concentrating ability, 660 lymphatics, 643 measuring GFR, 645 mechanisms, 657 Na+ excretion, 657–658 Na+ reabsorption, 648–649 nephron, 640–642 normal, 646 osmotic diuresis, 656 osmotic pressure, 647 oxygen consumption, 645 PAH transport, 651 permeability, 646 pressure in renal vessels, 644 proximal tubule, 652–653 reflex control, 662 regional blood flow, 645 regulation of renal blood flow, 644 renal circulation, 643–645 secondary active transport, 650–651 size of capillary bed, 646 spinal cord transection, 662 substances to measure, 645 tubular function, 647–657 tubular reabsorption, secretion, 648 tubuloglomerular feedback, 651–652 uremia, 660 urine concentration, 657 water diuresis, 658 water excretion, 658–659 water intoxication, 658–659 water transport, 652 Renal H+, 679–686 ammonia secretion, 680–681 bicarbonate excretion, 681 buffering, 683 factors affecting acid secretion, 681 H+ balance, 682–683 metabolic acidosis, 684 metabolic alkalosis, 684 pH changes along nephrons, 681 reaction with buffers, 680 renal compensation, 683–684 Siggaard–Andersen curve nomogram, 684–686 unipolar leads, 492–494 Renal interstitial pressure, 643 Renal plasma flow, 643 Renal threshold, 650 Renal tubular acidosis, 660 Renin, 337, 670 Renin–angiotensin system, 670–674 Renin secretion, 673–674 Renin substrate, 670 Renorenal reflex, 643 Renshaw cell, 163 Replication, 13 Repolarization, 85 Representational hemisphere, 295 Reproductive system, 391–428 aberrant sexual differentiation, 396–398 abnormalities of ovarian function, 422–423 abnormalities of testicular function, 410 anabolic effects, 408 androgen-secreting tumors, 410 anovulatory cycles, 413–414 blood–testis barrier, 402–403 changes during intercourse, 414 chemistry, 406–407 chromosomal abnormalities, 396–397 chromosomal sex, 392–393 chromosomes, 392 contraception, 422 control of cycle, 421 control of ovarian function, 420–422 control of testicular function, 409–410 cryptorchidism, 410 cyclic changes in breasts, 414 cyclic changes in uterine cervix, 414 delayed, absent puberty, 400 development of brain, 396 development of breasts, 426 development of gonads, 394 ejaculation, 406 embryology of genitalia, 394–396 embryology of human reproductive system, 394–396 endocrine changes, 424 endocrine function of testes, 406–409 environmental estrogen, 418 erection, 405–406 estrous cycle, 415 failure to reject fetal graft, 424 feedback effects, 421 female reproductive system, 411–427 female secondary sex characteristics, 418 fertilization, 423–424 fetoplacental unit, 425 gametogenesis, 402–406 gynecomastia, 427 human chorionic gonadotropin, 424 human chorionic somatomammotropin, 424 hormonal abnormalities, 397–398 hormones, 427 hypothalamic components, 420–421 implantation, 423–424 indicators of ovulation, 414–415 infertility, 424 inhibins, 409–410 initiation of lactation after delivery, 427 lactation, 426–427 male hypogonadism, 410 male reproductive system, 402–410 mechanism of action, 408–409, 418–419 menopause, 400 menstrual abnormalities, 422–423 menstrual cycle, 411–415 metabolism, 407 normal menstruation, 413 ovarian cycle, 411–412 ovarian hormones, 416–420 parturition, 425–426 pituitary gonadotropins, 400–402 placental hormones, 425 precocious, delayed puberty, 399–400 pregnancy, 423–426 prolactin, 400–402 prolactin secretion, 401–402 PSA, 406 puberty, 398–400 reflex ovulation, 421–422 relaxin, 420 secondary sex characteristics, 407–408 secretion, 407, 417, 419 semen, 405 sex chromatin, 392–393 sex chromosomes, 392 sex differentiation, 392–400 sexual precocity, 399–400 spermatogenesis, 403–404 spermatozoa development, 404 steroid feedback, 410 structure, 402 synthetic estrogen, 418 temperature, 405 testicular production of estrogens, 409 transport, 407 uterine cycle, 412–413 vaginal cycle, 414 vasectomy, 406 Residual cleft, 378 Residual volume, 593 Resistance, 539, 596 Resistance vessels, 543 Resonator, 210 Respiration, 591–600, 625–638 afferents from proprioceptors, 632–633 alveolar surface tension, 596–597 aortic bodies, 628–629 asphyxia, 634 blood flow, 599 breath holding, 631 bronchial tone, 594 carotid bodies, 628–629 changes in acid–base balance, 629 chemical control of breathing, 627–631 chemoreceptors in brain stem, 629 chest wall, 595–596 CO2, ventilatory responses to, 630 CO2 response curve, 631 compliance of lungs, chest wall, 595–596 control systems, 625–626 coughing, 632 dead space, 599–600 drowning, 634 effect of H+ on CO2 response, 631 effects of exercise, 634–637 exercise tolerance, 637 expiration, 591–593 fatigue, 637 glottis, 594 inspiration, 591–593 lung volumes, 593–594 medullary systems, 626 neural control of breathing, 625–627 nonchemical influences on respiration, 632–634 oxygen lack, ventilatory response to, 630–631 periodic breathing, 634 pontine influences, 627 receptors in airways, lungs, 632 regulation of respiratory activity, 627 respiratory abnormalities, 634 respiratory components of visceral reflexes, 633 respiratory effects of baroreceptor stimulation, 633 respiratory muscles, 594 sleep, 634 sneezing, 632 surfactant, 597–598 tissues, changes in, 636–637 uneven ventilation, 599–600 vagal influences, 627 ventilation, 599, 634–636 work of breathing, 598–599 Respiratory acidosis, 615, 629, 681, 683 Respiratory alkalosis, 615, 629, 683 Respiratory burst, 64 710 INDEX Respiratory compensation, 616, 684 Respiratory control pattern generator, 626 Respiratory dead space, 593 Respiratory exchange ratio, 461 Respiratory minute volume, 593 Respiratory muscles, 594 Respiratory quotient, 461 Resting membrane potential, 83–85 Restless leg syndrome, 237 Resynthesis of cyclic GMP, 192 Reticular activating system, 231 Reticular formation, 229–231 Reticular lamina, 205 Reticuloendothelial system, 65 Reticulum, 40 Retina, 181–194 mechanisms, 196–197 Retinals, 190 Retinene1, 191 Retinols, 190 Retrograde amnesia, 290 Retrograde transport, 82 Reuptake, 132 Reverberating circuits, 163 Reversal potential, 120 Reward center, of the brain, 179 RGS. See Regulators of G protein signaling RH group, 530 Rhodopsin, 190–192 Ribonucleic acids, 13, 40 Ribosomal RNA, 13 Ribosomes, 32, 40 Right axis deviation, 495 Right bundle branch block, 497 Riva–Rocci cuff, 545 RNA. See Ribonucleic acids RNA polymerase, 13 Rods, 182 Rods of Corti, 205 Role of urea, 656 Rotational acceleration, 214–215 Round window, 205 Roxine-binding prealbumin, 305 RQ. See Respiratory quotient rRNA. See Ribosomal RNA RU 486. See Mifepristone Rubrospinal tract, 247 Ruffini corpuscles, 150 Ryanodine receptor, 100 RyR. See Ryanodine receptor S SA node. See Sinoatrial node Saccades, 199 Saccule, 206 Saline cathartics, 442 Salivary secretion, 431 Saltatory conduction, 87–88 Salt balance, role of mineralocorticoids, 359 Salt-losing form, 346 Sarcolemma, 93 Sarcomere, 95 Sarcoplasmic reticulum, 96 Sarcotubular system, 96 Saturated, 648 Saturation, 195 Scala media, 205 Scala tympani, 205 Scala vestibuli, 205 Scalae, 205 Scanning speech, 258 Scar formation, 90 SCF. See Stem cell factor SCI. See Spinal cord injury Schizophrenia, 140 Schwann cells, 80, 90 Sclera, 181 SCN. See Prachiasmatic nuclei Scotopic vision, 185 Scotopsin, 191 SDA. See Specific dynamic action Sealing zone, 372 Second-degree heart block, 497 Second messengers, 51, 54–56 Second polar body, 412 Second sound, 512 Secondary active transport, 48–49 Secondary adrenal insufficiency, 360 Secondary amenorrhea, 422 Secondary diabetes, 334 Secondary endings, 158 Secondary hyperaldosteronism, 359 Secondary oocyte, 412 Secondary sex characteristics, 407 Secondary spermatocytes, 404 Secondary structure, 17 Secondary tympanic membrane, 205 Secreted insulin, 318–320 in blood, 318 glucose transporters, 318 insulin, 318–320 insulin preparations, 319 metabolism, 318 relation to potassium, 319 Secretory immunity, 73, 457 Secretory immunoglobulins, 73 Secretory phase, 412 Segment-specific gastric motility, 471–473 aerophagia, 473 esophagus, 471 intestinal gas, 473 lower esophageal sphincter, 471–473 mastication, 471 mouth, 471 swallowing, 471 Segmentation contractions, 475 Selectins, 38 Self-splicing, 14 Semantic memory, 290 Semen, 405 Semicircular canals, 206 Seminiferous tubules, 402 Senile dementia, 294–295 Senile plaques, 294 Sense organs, 150–151 Sense receptors, 149–151 classification, 149–150 Sensitization, 291 Sensorineural deafness, 213 Sensory coding, 152–153 duration, 153 intensity, 153 location, 152–153 modality, 152 Sensory information, 153–155 law of projection, 155 law of specific nerve energies, 153–155 neurological exam, 155 recruitment of sensory units, 155 Sensory nerves, 575 Sensory receptors, 149–156 classification of sensory receptors, 149–150 generation of impulses in cutaneous receptors, 151–152 generator potentials, 151 law of projection, 155 law of specific nerve energies, 153–155 neurological exam, 155 Pacinian corpuscles, 151 recruitment of sensory units, 155 sense organs, 150–151 sense receptors, 149–151 sensory coding, 152–153 sensory information, 153–155 source of generator potential, 151–152 Sensory unit, 152 SERCA pump, 52 SERMs, 418 Serotonergic receptors, 137 Serotonin, 136–137, 237 Serpentine receptors, 54 Sertoli cells, 402 Serum, 530 Serum-regulated kinase, 355 Seven-helix receptors, 54 Sex chromatin, 392–393 Sex chromosomes, 392 Sex differentiation, 392–400 aberrant sexual differentiation, 396–398 chromosomal abnormalities, 396–397 chromosomal sex, 392–393 chromosomes, 392 delayed, absent puberty, 400 development of brain, 396 development of gonads, 394 embryology of genitalia, 394–396 embryology of reproductive system, 394–396 hormonal abnormalities, 397–398 menopause, 400 precocious, delayed puberty, 399–400 puberty, 398–400 sex chromatin, 392–393 sex chromosomes, 392 sexual precocity, 399–400 Sex hormones, 337, 392 Sex steroid-binding globulin, 407 SFO. See Subfornical organ Shear stress, 540–541 Sheehan syndrome, 389 Short-term memory, 290 Shortening heat, 104 Shunts, 538 Side-chain cleavage enzyme, 344 Side pressure, 545 Siggaard–Andersen curve nomogram, 684–686 Signal peptide, 17, 40 Signal recognition particle, 17 Signal transducers of activated transcription, 52 Signal transduction, 222 Simple cells, 194 Simple partial seizures, 233 Sinoatrial node, 489 Sinus arrhythmia, 497 Size principle, 105 Skeletal muscle, 93, 312 body mechanics, 106 denervation, 104 dystrophin-glycoprotein complex, 96 electromyography, 105 morphology, 93–96 motor unit, 104–105 organization, 93–94 properties of, 104–106 sarcotubular system, 96 strength of skeletal muscles, 105–106 striations, 95–96 INDEX 711 Skin coloration, 379–380 Sleep, 236–238 effects of, 634 Sleep apnea, obstructive, 237 Sleep spindles, 234 Sleep stages, 234–235 Sleep–wake cycle, 236–239 alpha rhythm, 233–234 beta rhythm, 233–234 distribution of sleep stages, 235 gamma rhythm, 233–234 importance of sleep, 236 REM sleep, 234–235 sleep stages, 234 thalamocortical loop, 235–236 Sleep–wake states, 229–240 Slow axonal transport, 82 Slow EPSP, 266 Slowly adapting receptors, 153, 632 Slow pain, 168 Slow postsynaptic potentials, 121 Slow-wave sleep, 234 Small G protein, 42, 53 Small GTPases, 53 Small intestine, 475 intestinal motility, 475 transit time, 476 Small-molecule transmitters, 134–146 acetylcholine, 134–135 acetylcholine receptors, 135–136 α and β receptors, 139 anesthesia, 143 biosynthesis of catecholamines, 138–139 cannabinoids, 145 catabolism of catecholamines, 139 catecholamines, 138–140 chemical transmitters, 145–146 cholinesterases, 135 dopamine, 139–140 dopamine receptors, 140 epinephrine, 138 excitatory amino acids, 140–143 GABA, 141 GABA receptors, 141–142 gases, 145 glutamate, 140 glutamate receptors, 141 glycine, 142 histamine, 137–138 inhibitory amino acids, 140–143 monoamines, 134–138 norepinephrine, 138 serotonergic receptors, 137 serotonin, 136–137 Smell, discrimination, 221–222 Smooth muscle, 93, 125–126 anatomy, 125 electrical activity, 110 force generations, smooth muscle, 112 function of nerve supply to smooth muscle, 112 junctional potentials, 125–126 molecular basis of contraction, 110–111 morphology, 109–112 nerve supply, 112 plasticity, 112 relaxation, 111–112 types, 109–110 Smooth pursuit movements, 199 Sneezing, 632 Snellen letter charts, 184 Sniffing, 223 SOCCs. See Store-operated Ca2+ channels Sodium-dependent glucose transporter, 453 Sodium-potassium-activated adenosine triphosphatase, 46 Solute, 5 Solvent, 5 Soma, 80 Somatic chromosomes, 392 Somatic sensory area I, 174 Somatic sensory area II, 174 Somatomedin C, 382 Somatomedins, 382–384 Somatosensory pathways, 173–180 acetylcholine, 178–179 cannabinoids, 179 CNS lesions, 176–177 cortical plasticity, 176 dorsal column pathway, 173–175 dorsal horn, 173 enkephalins, 177–178 morphine, 177–178 pain transmission modulation, 177–179 somatotopic organization, 173–175 stress-induced analgesia, 177 ventrolateral spinothalamic tract, 175–177 Somatostatin, 280, 315, 330–331, 447–448 Somatostatinomas, 331 Somatosympathetic reflex, 558 Somatotopic organization, 173–175 Somnambulism, 237 Sound localization, 213 Sounds of Korotkoff, 545 Sound transmission, 209–210 Sound waves, 208–209 Source of generator potential, 151–152 Space motion sickness, 216 Spastic neurogenic bladder, 662 Spasticity. See Hypertonicity Spatial orientation, 216 Spatial summation, 120 Specific sensory relay nuclei, 230 Speech, 289–300 Spermatids, 404 Spermatogenesis, 402–404 Spermatogonia, 403 Spermatozoa, 404 Spermatozoa development, 404 Sphygmomanometer, 545 Spike potentials, 470 Spinal animal, 164 Spinal cord, 121 Spinal cord injury, 247, 250 Spinal integration, 247–249 locomotion generator, 249 spinal shock, 248–249 Spinal shock, 249 Spindle sensitivity, 160 Spinnbarkeit, 414 Spinocerebellum, 257 Spinoreticular pathway, 175 Spiral arteries, 412 Spiral ganglion, 206 Splanchnic circulation. See Gastrointestinal circulation Splay, 650 Spleen, 523 Spliceosomes, 14 Spongy bones, 371 SRP. See Signal recognition particle Stagnant hypoxia, 617 Standard bicarbonate, 685 Standard limb leads, 492 Stapedius, 203 Stapes, 203 StAR protein. See Steroidogenic acute regulatory protein Starling forces, 548 Starling’s law of heart, 109, 515 Static, 158 Static response, 158 Statins, 27 STAT protein, 58 STATs. See Signal transducers of activated transcription Steatorrhea, 457 Stellate cells, 255 Stem cell factor, 65 Stenosis, 513 Stereocilia, 207 Stereognosis, 155 Steroid factor-1, 346 Steroidogenic acute regulatory protein, 346 Stimulation of transcription, 51–52 Stimuli affecting growth hormone secretion, 384–385 Stokes–Adams syndrome, 497 Stomach, 473–475 gastric emptying, 473 gastric motility, 473 vomiting, 473–475 Store-operated Ca2+ channels, 52 Strabismus, 187, 188 Strangeness, memory and, 293–294 Stratum basale, 412 Stratum functionale, 412 Strength–duration curve, 85 Strength of skeletal muscles, 105–106 Stress, 353–354 Stress-induced analgesia, 177 Stretch reflex, 158–164 central connections of afferent fibers, 160 control of gamma-motor neuron discharge, 161 function of muscle spindles, 160 gamma-motor neuron discharge, 160–161 inverse stretch reflex, 162–163 muscle tone, 163 reciprocal innervation, 161–162 structure of muscle spindles, 158–159 Striae, 351 Striations, 95–96 Striatum, 250 Striosomes, 250 Stroke, 135 Stroke volume, 514 Structural lipids, 24 Stuttering, 298 Subfornical organ, 276, 573 Subliminal fringe, 123 Submucosal plexus, 269 Submucous plexus, 448 Substance P, 143, 175, 448 Substance P/neurokinin A gene, 143 Substantia gelatinosa, 173 Substantia nigra, 250 Subthalamic nucleus, 250 Sucrose, 452 Superior colliculi, 199 Superior peduncle, 254 Supersensitivity, 126 Supplementary motor cortex, 243 Supporting cells, 219 Suppression scotoma, 188 Suppressor strip, 247 712 INDEX Supratentorial lesions, 247 Surfactant, 597–598 Sustaining collateral, 126 Sustentacular cells, 219 Swallowing, 471 Sweet, taste of, 224 Sympathectomy, 556 Sympathetic chain, 263 Sympathetic cholinergic vasodilator system, 556 Sympathetic division, 263–265 Sympathetic noradrenergic discharge, 268–269 Sympathetic paravertebral ganglion, 263 Synapse en passant, 125 Synapses, 115–116 Synaptic activity, chemical transmission, 129–134 chemistry of transmitters, 129–130 receptors, 130–132 reuptake, 132–134 Synaptic cleft, 115 Synaptic delay, 118 Synaptic knobs, 80, 116 Synaptic plasticity and learning, 291–292 Synaptic vesicles, 116 Synaptobrevin, 117 Synchronization, 234 Syncytiotrophoblast, 423 Syndrome X, 334 Syndromic deafness, 214 Long QT syndrome, 214 Pendred syndrome, 214 Syntaxin, 117 Synthetic agonists, antagonists, 668 Synthetic estrogen, 418 Syntrophins, 96 Systemic regulation by hormones, 566–567 circulating vasoconstrictors, 567 kinins, 566 natriuretic hormones, 566 Systemic response to injury, 76 Systolic pressure, 507, 544 T T system, 96 Tabes dorsalis, 662 Tachycardia, 497 Tachykinins, 143 Tacrolimus (FK-506), 75 Tactile acuity, 153 Tactile agnosia, 155 Tamoxifen, 418, 427 Tangent screen, 198 Tardive dyskinesia, 252 Taste intensity discriminations, 226 modalities, 224 transduction, 224–226 Taste buds, 223 Taste cells, 223 TATA box, 11 Taxol, 35 TBG. See Thyroxine-binding globulin T cell receptors, 72 Tectorial membrane, 206 Tectospinal tract, 246 Temperature, 167–172, 611 afferents, 285 effect of, 405 fever, 285–286 heat loss, 284–285 heat production, 283–284 hypothermia, 286 normal body temperature, 283 regulation, 282–286 temperature-regulating mechanisms, 285 Temperature-regulating mechanisms, 285 Temporal bone, 205 Temporal pathway, 195 Temporal summation, 120 Tense configuration, 610 Tensor tympani, 203 Terminal boutons, 116 Terminal buttons, 80 Terminal cisterns, 96 Tertiary adrenal insufficiency, 360 Tertiary structure, 17 Testes, endocrine function, 406–409 actions, 407 anabolic effects, 408 estrogens, testicular production of, 409 mechanism of action, 408–409 metabolism, 407 secondary sex characteristics, 407–408 secretion, 407 testosterone, 406–407 transport, 407 Testicular descent, 410 Testicular feminizing syndrome, 398 Testicular function, 409–410 inhibins, 409–410 steroid feedback, 410 Testosterone, 392 Testotoxicosis, 59 Tetanus, 101 Tetanus toxin, 119 Tetrahydrobiopterin, 138 Tetrahydrobiopterin (BH4) deficiency, 138 Tetraploid, 13 TGF. See Transforming growth factor TGFα. See Transforming growth factor alpha Thalamic fasciculus, 250 Thalamic nuclei, 229–230 Thalamic reticular nucleus, 230 Thalamocortical loop, 235–236 Thalamostriatal pathway, 250 Thalamus, 229–231 Thebesian veins, 578 Theca interna, 412 Thelarche, 398 Thermal gradient, 284 Thermal nociceptors, 167 Thermodilution, 514 Thermoreceptors, 167–168 Theta rhythm, 234 Thiazolidinediones, 327 Third-degree heart block. See Complete heart block Third sound, 512 Thirst, 276 Thoracic pump, 549 Threshold intensity, 85 Threshold potential, 85 Thrombasthenic purpura, 75 Thrombocytopenic purpura, 75 Thrombocytosis, 523 Thrombomodulin, 533 Thrombopoietin, 75 Thromboxane A2, 563 Thyroglobulin, 304 Thyroglossal duct, 302 Thyroid gland, 301–314 calorigenesis, 310–312 calorigenic action, 310 carbohydrate metabolism, 313 cardiovascular system, 312 catecholamines, 312 chemistry, 302–303 cholesterol metabolism, 313 control mechanisms, 308 fluctuations in binding, 306 formation, secretion, 302–305 growth, effects on, 313 iodide transport across thyrocytes, 303 iodine homeostasis, 303 mechanism of action, 308–310 metabolism of thyroid hormones, 305–307 nervous system, 312 protein binding, 305–306 regulation of thyroid secretion, 307–308 secretion, 302–305 skeletal muscle, 312 synthesis, 304–305 thyroid growth, 308 transport of thyroid hormones, 305–307 TSH, 307–308 Thyroid growth, 308 Thyroid hormone, 332 effects of, 308–313 formation, secretion, 302–305 metabolism, 305–307 secretion, 302–305 synthesis, 304–305 Thyroid hormones, 371 Thyroid hormone thermogenesis, 308 Thyroid isthmus, 302 Thyroid peroxidase, 304 Thyroid secretion, 307–308 Thyroid-stimulating hormone, 279, 377 chemistry, 307 effects on thyroid, 307–308 metabolism, 307 receptors, 308 Thyroid storms, 312 Thyrotoxic myopathy, 312 Thyrotropin, 279, 377 Thyrotropin-releasing hormone, 280 Thyroxine-binding globulin, 305 Tickle, 168 Tidal volume, 593 Tight junctions, 38 Timbre, 209 Timing, heart, 510 Tinnitus, 216 Tinohypothalamic fibers, 236 Tip links, 207 Tissue conductance, 284 Tissue factor pathway inhibitor, 533 Tissue kallikrein, 566 Tissue macrophage system, 65 Tissue macrophages, 65 Tissue renin–angiotensin systems, 672 Tissue-type plasminogen activator, 533 Titratable acidity, 680 Titration line, 685 Tm. See Transport maximum Tolerance, 179 Tone, 110, 163 Tonic–clonic seizure, 233 Tonic contractions, 475 Tonic phase, 233 Tonic receptors. See Slowly adapting receptors Tonicity, 6, 665 Tonsils, 605 Tonus, 110, 163 Torsade de pointes, 500 Total blood volume, 2 INDEX 713 Total body water, 2 Total dead space, 600 Total electromechanical systole, 510 Total tension, 102 Toxin botulinum, 119 tetanus, 119 t-PA. See Tissue-type plasminogen activator Trabecular bones, 371 Transaminases, 18 Transcortin, 346 Transcription, 13 Transcytosis, 49–50 Transducin, 191 Transfer RNA, 13 Transferrin, 459 Transforming growth factor, 91 Transforming growth factor alpha, 50 Transfusion reactions, 528–529 Transient receptor potential, 168 Translation, 14, 17 Translocon, 17 Transmission in sympathetic ganglia, 266 Transmitters, chemistry of, 129–130 Transmural pressure, 543 Transport across cell membranes, 43–49 caveolae, 44 coats, 45 endocytosis, 43–44 exocytosis, 43 ion channels, 46–47 membrane permeability, 45–46 Na, K ATPase, 47 rafts, 44 regulation of Na, K ATPase activity, 47 secondary active transport, 48–49 transport across epithelia, 49 vesicle transport, 45 Transport across epithelia, 49 Transport adrenocortical hormones, 346–348 aldosterone, 347–348 glucocorticoid binding, 346–347 ketosteroids (17-ketosteroids), 348 metabolism of glucocorticoids, 347 Transport maximum, 648 Transport proteins, 45 Transport of thyroid hormones, 305–307 Transthyretin, 305 Traube–Hering waves, 562 Traveling waves, 210–211 Trehalase, 452 Tremor at rest, 253 TRH. See Thyrotropin-releasing hormone Trichromats, 196 Trifascicular block, 497 Trinucleotide repeat, 43, 252 Triple response, 581 Trk receptors, 90 tRNA. See Transfer RNA Trophic action, 443 Trophic support of neurons, 89–90 Tropic hormones, 377 Tropomyosin, 94 Troponin, 52, 94 Troponin C, 94 Troponin I, 94 Troponin T, 94 TRP. See Transient receptor potential True cholinesterase, 135 True hermaphroditism, 396 True plasma, 682 True precocious puberty, 399 TSH. See Thyroid-stimulating hormone T-snare, 117 Tubular function, 647–657 aquaporins, 652 collecting ducts, 653–654 countercurrent mechanism, 654–656 distal tubule, 653 free water clearance, 657 glomerulotubular balance, 651–652 glucose reabsorption, 650 glucose transport mechanism, 650 loop of Henle, 653 Na+ reabsorption, 648–649 osmotic diuresis, 656 PAH transport, 651 proximal tubule, 652–653 secondary active transport, 650–651 tubular reabsorption and secretion, 648 tubuloglomerular feedback, 651–652 urine concentration, 657 water transport, 652 Tubular myelin, 597 Tubular reabsorption, 639 Tubular secretion, 639 Tubuloglomerular feedback, 651–652 Tufted cells, 220 Tumor suppressor genes, 43 Two-point discrimination threshold, 153 Two-point threshold test, 153 Tympanic membrane, 203 Tympanic reflex, 210 Type 1 diabetes, 334 Type 2 diabetes, 334–335 Type I cells, 589 Type I medullary interstitial cells, 642 Type II cells, 590 Types of contraction, 100–101 Tyrosine hydroxylase, 138 U Ubiquitination, 18 Ultrasonography, 486 Uncal herniation, 247, 249 Uncompensated, 683–684 Uncompensated metabolic acidosis, 616 Uncompensated respiratory acidosis, 615 Uncompensated respiratory alkalosis, 615 Unconditioned stimulus, 292 Uneven ventilation, 599–600 Unipolar leads, 492–494 Unipolar recording, 492 Uniports, 46 Unitary smooth muscle, 109 Units for measuring concentration of solutes, 2 equivalents, 2 moles, 2 Unmyelinated, 82 Up-regulation, 50 Upper motor neuron lesion, 163 Upper motor neurons, 243, 244 Uracil, 13 Urea formation, 18 Uremia, 660 Urinary tract infection, 250 Urine acidification, 679–687 Urine flow, 645 Urokinase-type plasminogen activator, 533 Urotensin-II, 567 Uterine cervix, cyclic changes in, 414 Uterine circulation, 581–582 Uterine cycle, 412–413 Utricle, 206 V Vagal tone, 556 Vaginal cycle, 414 Valsalva maneuver, 561–562 Vanae comitantes, 285 Vanilloid receptors, 168 Varicose veins, 550 Varicosities, 125 Vas deferens, 402 Vasa recta, 642 Vascular endothelial growth factor, 539 Vascular hindrance, 542 Vascular reactivity, 349 Vascular smooth muscle, 536 Vasculogenesis, 539 Vasectomy, 406 Vasoconstriction, 555 Vasodilation, 555 Vasodilator metabolites, 563 Vasomotor nerves, 575 Vasopressin, 277, 279, 377, 666 receptors, 279, 665–666 Vasopressin secretion stimuli affecting, 667 volume effects, 667 Vectorcardiograms, 496 Vectorcardiography, 495–496 VEGF. See Vascular endothelial growth factor Venoconstriction, 555 Venodilation, 555 Venous circulation, 549 Venous occlusion plethysmography, 539 Venous pressure air embolism, 550 effects of heartbeat, 549 flow, 549–550 in head, 550 measuring, 550 muscle pump, 549–550 thoracic pump, 549 venous pressure in head, 550 Venous-to-arterial shunts, 620 Venous valves, 538 Ventilation/perfusion imbalance, 620 ratios, 603 Ventral cochlear nuclei, 211 Ventral corticospinal tract, 242 Ventral pathway, 195 Ventral posterior lateral, 173 Ventral posterior thalamic nucleus, 176 Ventral tegmental area, 179 Ventricular arrhythmias, 499–501 Ventricular ejection, 508 Ventricular fibrillation, 500 Ventricular systole, 489, 508 Ventrolateral cordotomy, 177 Ventrolateral spinothalamic tract, 175–177 cortical plasticity, 176 effects of CNS lesions, 176–177 Verbal system, 292 Vermis, 254 Vertigo, 216 benign paroxysmal positional, 216 Very low density lipoproteins, 26 Vesicle transport, 45 Vesicular traffic, 40–42 Vesicular transport, 49, 537 Vestibular apparatus, 213 Vestibular movements, 199 Vestibular nuclei, 213 714 INDEX Vestibular system, 213–216 central pathway, 214 linear acceleration, 215–216 rotational acceleration, 214–215 spatial orientation, 216 Vestibulocerebellum, 257 Vestibulo-ocular reflex, 215 Vestibulospinal tract, 246 Vibratory sensibility, 155 Virilization, 346 Visceral pain, 169–170 Visceral reflexes, respiratory components of, 633 Visceral smooth muscle, 109 Viscosity, 542 Vision, 181–202 accommodation, 188–189 binocular vision, 197–198 characteristics of color, 195–196 color vision, 195–197 common defects of image-forming mechanism, 187–188 cone pigments, 192 cortical areas concerned with vision, 195 critical fusion frequency, 197 dark adaptation, 197 electrical responses, 189–190 eye movements, 199 image-forming mechanism, 186–189 ionic basis of photoreceptor potentials, 190 lesions in optic pathways, 198 melanopsin, 192 near point of, 189 neural mechanisms, 197 neural pathways, 184 pathways to cortex, 193–194 photoreceptor mechanism, 189–193 photosensitive compounds, 190–191 primary visual cortex, 194–195 principles of optics, 186–187 processing of visual information in retina, 192–193 protection, 184–186 pupillary reflexes, 189 receptors, 184 responses in visual pathways, cortex, 193–195 resynthesis of cyclic GMP, 192 retina, 182–194 retinal mechanisms, 196–197 rhodopsin, 190–192 superior colliculi, 199 visual fields, 197–198 Visual acuity, 182, 184 Visual agnosia, 155 Visual fields, 197–198 Visual pathways and cortex, 193–195 cortical areas concerned with vision, 195 pathways to cortex, 193–194 primary visual cortex, 194–195 Visual purple. See Rhodopsin Visuospatial system, 292 Vitamin A, 191 synthesis of retinene1, 191 Vitamin D, 365–367 Vitamins, 24, 458, 464–466 absorption of, 458–459 Vitiligo, 380 Vitreous, 182 Vitreous humor, 182 VLDL. See Very low density lipoproteins Volley effect, 211 Voltage-gated, 45 Volume conductor, 492 Volumetric ventricular relaxation, 509 Vomeronasal organ, 223 Vomiting, 473–475 von Willebrand factor, 74 VPL. See Ventral posterior lateral V2 receptors, 666 VR1, 168 VRL-1, 168 V-snare, 117 Vulnerable period, 501 V wave, 512, 549 W Wallerian degeneration, 82, 126 Warfarin, 535 Warmth receptors, 168 Water, 2–3 diuresis, 658 excretion, 658–659 intake, factors regulating, 277 intoxication, 658–659 metabolism, 389 Water-hammer pulse, 512 Waterfall effect, 602 Weber–Fechner law, 153 Wenckebach phenomenon, 497 Wernicke’s area, 297 Wheal, 581 White blood cells, 522–523 platelets, 523 White rami communicans, 263 White reaction, 580 Whole cell recording, 45 Wilson disease, 252 Withdrawal, 179 Withdrawal bleeding, 417 Withdrawal reflex, 163–164 fractionation, 164 occlusion, 164 Wolff–Parkinson–White syndrome, 501 Working memory, 290, 292 Wound healing, 75–77 local injury, 75–76 systemic response to injury, 76 X Xerophthalmia, 191 Y Yawning, 633 Yellow marrow, 522 Young–Helmholtz theory, 196 Z Zona fasciculata, 338 Zona glomerulosa, 338 Zona pellucida, 423 Zona reticularis, 338 Zonula adherens, 38 Zonula occludens, 38 Zonules, 181 Zymogen granules, 435 Food and Nutrition Board, National Academy of Sciences—National Research Council Recommended Dietary Allowances, Revised 1989.a Weightb Heightb Fat-Soluble Vitamins Water-Soluble Vitamins Minerals Catego-ry Age (years) or Condition kg lb cm in Protein (g) Vitamin A (μg of RE)c Vitamin D (μg)d Vitamin E (mg of α-TE)e Vitamin K (μg) Vitamin C (mg) Thia-mine (mg) Ribofla-vin (mg) Niacin (mg NE)f Vitamin B6 (mg) Folate (μg) Vitamin B12 (μg) Calcium (mg)g Phos-phorus (mg) Magne-sium (mg) Iron (mg) Zinc (mg) Iodine (μg) Seleni-um (μg) Infants 0.0–0.5 6 13 60 24 13 375 7.5 3 5 30 0.3 0.4 5 0.3 25 0.3 400 300 40 6 5 40 10 0.5–1.0 9 20 71 28 14 375 10 4 10 35 0.4 0.5 6 0.6 35 0.5 600 500 60 10 5 50 15 Children 1–3 13 29 90 35 16 400 10 6 15 40 0.7 0.8 9 1.0 50 0.7 800 800 80 10 10 70 20 4–6 20 44 112 44 24 500 10 7 20 45 0.9 1.1 12 1.1 75 1.0 800 800 120 10 10 90 20 7–10 28 62 132 52 28 700 10 7 30 45 1.0 1.2 13 1.4 100 1.4 800 800 170 10 10 120 30 Males 11–14 45 99 157 62 45 1000 10 10 45 50 1.3 1.5 17 1.7 150 2.0 1200 1200 270 12 15 150 40 15–18 66 145 176 69 59 1000 10 10 65 60 1.5 1.8 20 2.0 200 2.0 1200 1200 400 12 15 150 50 19–24 72 160 177 70 58 1000 10 10 70 60 1.5 1.7 19 2.0 200 2.0 1200 1200 350 10 15 150 70 25–50 79 174 176 70 63 1000 5 10 80 60 1.5 1.7 19 2.0 200 2.0 800 800 350 10 15 150 70 51+ 77 170 173 68 63 1000 5 10 80 60 1.2 1.4 15 2.0 200 2.0 1200 800 350 10 15 150 70 Females 11–14 46 101 157 62 46 800 10 8 45 50 1.1 1.3 15 1.4 150 2.0 1200 1200 280 15 12 150 45 15–18 55 120 163 64 44 800 10 8 55 60 1.1 1.3 15 1.5 180 2.0 1200 1200 300 15 12 150 50 19–24 58 128 164 65 46 800 10 8 60 60 1.1 1.3 15 1.6 180 2.0 1200 1200 280 15 12 150 55 25–50 63 138 163 64 50 800 5 8 65 60 1.1 1.3 15 1.6 180 2.0 800 800 280 15 12 150 55 51+ 65 143 160 63 50 800 5 8 65 60 1.0 1.2 13 1.6 180 2.0 1500 800 280 10 12 150 55 Pregnant 60 800 10 10 65 70 1.5 1.6 17 2.2 400 2.2 1200 1200 320 30 15 175 65 Lactating 1st 6 months 65 1300 10 12 65 95 1.6 1.8 20 2.1 280 2.6 1200 1200 355 15 19 200 75 2nd 6 months 62 1200 10 11 65 90 1.6 1.7 20 2.1 260 2.6 1200 1200 340 15 16 200 75 Modified and reproduced, with permission, from Recommended Dietary Allowances, 10th ed, National Academy Press, 1989. Copyright © 1989 by the National Academy of Sciences. Courtesy of the National Academy Press, Washington, D.C. aThe allowances, expressed as average daily intakes over time, are intended to provide for individual variations among most normal persons as they live in the United States under usual environmental stresses. Diets should be based on a vari-ety of common foods to provide other nutrients for which human requirements have been less well defined. bWeights and heights of Reference Adults are actual medians for the U.S. population of the designated age. The median weights and heights of those under 19 years of age are not necessarily the ideal values. cRetinol equivalents. 1 retinol equivalent = 1 mg of retinol or 6 μg of β-carotene. dAs cholecalciferol. 10 μg of cholecalciferol = 400 IU of vitamin D. eα-Tocopherol equivalents. 1 mg of d-α tocopherol = 1 α-TE. f1 NE (niacin equivalent) is equal to 1 mg of niacin or 60 mg of dietary tryptophan. gCalcium values increased after age 50.
1803	https://www.geeksforgeeks.org/maths/vector-norms/	Vector Norms Last Updated : 23 Jul, 2025 Suggest changes 2 Likes A vector norm, sometimes represented with a double bar as ∥x∥, is a function that assigns a non-negative length or size to a vector x in n-dimensional space. Norms are essential in mathematics and machine learning for measuring vector magnitudes and calculating distances. A vector norm satisfies three properties: Non-negativity: ∣x∣ > 0\| if x ≠ 0, and ∣x∣=0 if and only if x = 0. Scalar Multiplication: ∣kx∣ = ∣k∣ ⋅ ∣x∣ for any scalar k. Triangle Inequality: ∣x + y∣ ≤ ∣x∣ + ∣y∣. Types of Vector Norms The vector norm ∣x∣p, also known as the p-norm, for p = 1, 2,… is defined as: ∣x∣p=(∑i=1n∣xi∣p)p1 This general formula encompasses several specific norms that are commonly used. Commonly used norms are: L1 Norm L2 Norm L∞ Norm Let's discuss these in detail. L1 Norm The L1 norm, also known as the Manhattan norm or Taxicab norm, is a way to measure the "length" or "magnitude" of a vector by summing the absolute values of its components. Mathematically, for a vector x = [x1, x2, . . ., xn], the L1 norm ∣x∣1 is defined as: ∣x∣1 = ∣x1∣ + ∣x2∣ + ∣x3∣ + ... + ∣xn∣ Example: If x = [3, −4, 2], then the L1 norm is: ∣x∣1 = ∣3∣ + ∣−4∣ + ∣2∣ = 3 + 4 + 2 = 9 L2 Norm The L2 norm, also known as the Euclidean norm, is a measure of the "length" or "magnitude" of a vector, calculated as the square root of the sum of the squares of its components. For a vector x = [x1, x2, . . ., xn], the L2 norm ∣x∣2 is defined as: ∣x∣2=x12+x22+⋯+xn2 Example: If x = [3, −4, 2], then the L2 norm is: ∣x∣2=32+(−4)2+22 =9+16+4 =√29 ≈ 5.39 L∞ norm The L∞ norm, also known as the Infinity norm or Max norm, measures the "size" of a vector by taking the largest absolute value among its components. Unlike the L1 and L2 norms, which consider the combined contribution of all components, the L∞ norm focuses solely on the component with the maximum magnitude. For a vector x = [x1, x2, . . ., xn], the L∞ norm ∣x∣∞ is defined as: ∣x∣∞ = max∣xi∣ where 1 ≤ i ≤ n Example: If x = [3, −4, 2], then the L∞ norm is: ∣x∣∞= max(∣3∣, ∣−4∣, ∣2∣) = 4 Read More, Vector Space Linear Algebra Practice Problem Based on Vector Norm Question 1. Given the vector x = [4, -3, 7, 1], calculate the L1 norm (Manhattan norm) of the vector. Question 2. Given the vector x = [1, -2, 2], calculate the L2 norm (Euclidean norm) of the vector. Question 3. For the vector x = [7, −1, −4, 6], calculate the L∞ norm (Infinity norm) of the vector. Question 4. If the L2 norm (Euclidean norm) of a vector x = [x1, x2, x3] is 10, and the components of the vector are x1 = 6 and x2 = 8, find the value of x3. Answer:- 1. 15 2. 3 3. 7 4. 0 S surajb1g23 Improve Article Tags : Mathematics Maths Vector-Calculus Explore Maths 4 min read Basic Arithmetic What are Numbers? 15+ min readArithmetic Operations 9 min readFractions - Definition, Types and Examples 7 min readWhat are Decimals? 10 min readExponents 9 min readPercentage 5 min read Algebra Variable in Maths 5 min readPolynomials\| Degree \| Types \| Properties and Examples 9 min readCoefficient 8 min readAlgebraic Identities 14 min readProperties of Algebraic Operations 3 min read Geometry Lines and Angles 9 min readGeometric Shapes in Maths 2 min readArea and Perimeter of Shapes \| Formula and Examples 10 min readSurface Areas and Volumes 10 min readPoints, Lines and Planes 14 min readCoordinate Axes and Coordinate Planes in 3D space 6 min read Trigonometry & Vector Algebra Trigonometric Ratios 4 min readTrigonometric Equations \| Definition, Examples & How to Solve 9 min readTrigonometric Identities 8 min readTrigonometric Functions 6 min readInverse Trigonometric Functions \| Definition, Formula, Types and Examples 11 min readInverse Trigonometric Identities 9 min read Calculus Introduction to Differential Calculus 6 min readLimits in Calculus 12 min readContinuity of Functions 10 min readDifferentiation 2 min readDifferentiability of a Function \| Class 12 Maths 11 min readIntegration 3 min read Probability and Statistics Basic Concepts of Probability 7 min readBayes' Theorem 13 min readProbability Distribution - Function, Formula, Table 13 min readDescriptive Statistic 5 min readWhat is Inferential Statistics? 7 min readMeasures of Central Tendency in Statistics 11 min readSet Theory 3 min read Practice NCERT Solutions for Class 8 to 12 7 min readRD Sharma Class 8 Solutions for Maths: Chapter Wise PDF 5 min readRD Sharma Class 9 Solutions 10 min readRD Sharma Class 10 Solutions 9 min readRD Sharma Class 11 Solutions for Maths 13 min readRD Sharma Class 12 Solutions for Maths 13 min read We use cookies to ensure you have the best browsing experience on our website. 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1804	https://www.geeksforgeeks.org/maths/prime-factorization/	Prime Factorization Last Updated : 11 Sep, 2025 Suggest changes Like Article Prime factorization is the process of breaking down a number into a product of its prime numbers (prime factors). A prime number is a number greater than 1 that has only two factors: 1 and itself. Prime Factorization involves only the prime numbers, as every composite number can be written as the product of primes. 1 / 3 More Examples of Prime Factorization: 12 can be written as 2 × 6 6 can be further factorized as 2 × 3. So 12 can be rewritten as 2 × 2 × 3 No more factorization possible as 2 and 3 cannot be divide further, so 2 × 2 × 3 is our prime factorization 54 can written as 2 × 27. 27 can be further factorized as 3 × 9. So we rewrite 54 as 2 × 3 × 9 9 can be further factorized as 3 × 3. So we rewrite 54 as 2 × 3 × 3 × 3. No more factorization possible, so 2 × 3 × 3 × 3 is our prime factorization When a number is expressed as a product of its prime factors, it is said to be in its prime factorization form. Some examples of prime factors are: 2 and 3 are the prime factors of 12, as 12 = 22 × 3, 3 and 5 are the prime factors of 15, as 15 = 3 × 5, 2 and 7 are the prime factors of 14, as 15 = 3 × 5. Prime Factorization Methods Two common methods of Prime Factorization are: Division Method Factor Tree Method Prime Factorization by Division Method In this method, the number is successively divided by prime numbers until the quotient becomes 1, with each division identifying a prime factor. Steps to identify the prime factors of a number by the Division Method : Step 1: Divide the number by the smallest prime number (i.e. 2) until we are able to divide the given number without leaving any remainder. Step 2: Move on to the next prime number and repeat the division until the quotient becomes 1. Step 3: The prime factors are the divisors used in the division process. Let's consider some examples for better understanding. Examples of Prime Factorization by Division Method Example 1: Find the Prime Factorization of 60 using the Division Method. Example 2: Find the Prime Factorization of 210 using the Division Method. Example 3: Express 56 as the product of its Prime Factors. Prime Factorization by Factor Tree Method The Factor Tree Method involves breaking down a number into its prime factors by constructing a tree-like structure called a factor tree. Steps to identify the prime factors of a number by the Factor Tree Method: Step 1: Identify two factors of the number that are not prime. Step 2: Write these two factors as branches of the factor tree. Step 3: Repeat steps 1 and 2 for each non-prime factor until all branches end with prime numbers. Step 4: The prime factors are the numbers at the end of the branches. Let's consider some examples for better understanding, as follows: Examples of Prime Factorization by Factor Tree Method Example 1: Find the factorization of 60 by the Factor Tree Method. Example 2: Make the Factor Tree of 210. Prime Factorization of Numbers Some examples of prime factorization are listed below: \| Number \| Prime Factorization \| --- \| \| 72 \| 2 × 2 × 2 × 3 × 3 \| \| 36 \| 2 × 2 × 3 × 3 \| \| 48 \| 2 × 2 × 2 × 2 × 3 \| \| 12 \| 2 × 2 × 3 \| \| 100 \| 2 × 2 × 5 × 5 \| \| 84 \| 2 × 2 × 3 × 7 \| \| 8 \| 2 × 2 × 2 \| \| 32 \| 2 × 2 × 2 × 2 × 2 \| \| 24 \| 2 × 2 × 2 × 3 \| \| 91 \| 7 × 13 \| \| 15 \| 3 × 5 \| Finding HCF and LCM by Prime Factorization HCF and LCM can be easily calculated by the method of prime factorization: Finding HCF For the HCF, take the lowest power of each common prime factor from both numbers. For Example: Common prime factors: 2 and 3 For 2: min⁡(2, 4) = 2 For 3: min⁡(1, 1) = 1 So, the HCF is: HCF = 22 × 31 = 4 × 3 = 12 Finding LCM For the LCM, take the highest power of each prime factor present in either number. For Example: Prime Factors: 2, 3, and 5 For 2: max⁡(2, 4) = 4 For 3: max⁡(1, 1) = 1 For 5: max⁡(1, 0) = 1 So, the LCM is: LCM = 24 × 31 × 51 = 16 × 3 × 5 = 240 Applications of Prime Factorization Finding HCF and LCM: Prime factorization helps determine the Highest Common Factor (HCF) and Lowest Common Multiple (LCM) of numbers, essential for simplifying fractions and finding common denominators. Cryptography: It is crucial in public key cryptography, such as RSA, where the difficulty of factoring large composite numbers ensures secure communication. Simplifying Fractions: By factoring numerators and denominators into prime factors, common factors can be canceled out, simplifying fractions effectively. Divisibility Rules: Prime factorization aids in applying divisibility rules, quickly indicating whether one number is divisible by another. Data Compression: Techniques based on prime factorization can optimize data storage and transmission in computer science, enhancing efficiency. Network Security: Algorithms based on prime factorization enhance data security during network transfers, protecting sensitive information. Prime Factorization Solved Examples Let's solve some questions on Prime Factorisation. Problem 1: What is the Prime Factorisation of 80? Solution: To find the prime factorization of 80, we can start by dividing it by the smallest prime number, which is 2. 80 divided by 2 equals 40. 40 divided by 2 equals 20. 20 divided by 2 equals 10. 10 divided by 2 equals 5. Now, since 5 is a prime number, we can stop dividing. Therefore, the prime factorization of 80 is: 2 × 2 × 2 × 2 × 5. Problem 2: Prime factorization of 120. Solution: Starting with the smallest prime number, which is 2. 120 divided by 2 equals 60. 60 divided by 2 equals 30. 30 divided by 2 equals 15. Now, since 15 is not divisible by 2, we move on to the next prime number (i.e, 3) 15 divided by 3 equals 5. Now, since 5 is a prime number, we can stop dividing. Therefore, the prime factorization of 120 is: 2 × 2 × 2 × 3 × 5 Problem 3: What is the Factor Tree of 56? Related Articles: Interesting Facts about Prime Numbers Factorization of Algebraic Expression Composite Numbers Read in Detail: Real Life Applications of Prime Factorization Please refer Prime Factorization Tips and Tricks to Improve your time in finding prime factorization. Prime Factorization Worksheet Find the prime factorization of 36. Determine the prime factorization of 90. What is the prime factorization of 48? Find the prime factorization of 105. What is the prime factorization of 84? Determine the prime factorization of 100. Find the prime factorization of 2310. What is the prime factorization of 56? Determine the prime factorization of 150. What is the prime factorization of 1250? Answer Key 36: 22×32 90: 2×32×5 48: 24×3 105: 3×5×7 84: 22×3×7 100: 22×52 2310: 2×3×5×7×11 56: 23×7 150: 2×3×52 1250: 2×54 J jeetendrajeet54 Improve Article Tags : Mathematics School Learning Class 10 Maths-Class-10 Explore Maths 4 min read Basic Arithmetic What are Numbers? 15+ min readArithmetic Operations 9 min readFractions - Definition, Types and Examples 7 min readWhat are Decimals? 10 min readExponents 9 min readPercentage 4 min read Algebra Variable in Maths 5 min readPolynomials\| Degree \| Types \| Properties and Examples 9 min readCoefficient 8 min readAlgebraic Identities 14 min readProperties of Algebraic Operations 3 min read Geometry Lines and Angles 9 min readGeometric Shapes in Maths 2 min readArea and Perimeter of Shapes \| Formula and Examples 10 min readSurface Areas and Volumes 10 min readPoints, Lines and Planes 14 min readCoordinate Axes and Coordinate Planes in 3D space 6 min read Trigonometry & Vector Algebra Trigonometric Ratios 4 min readTrigonometric Equations \| Definition, Examples & How to Solve 9 min readTrigonometric Identities 7 min readTrigonometric Functions 6 min readInverse Trigonometric Functions \| Definition, Formula, Types and Examples 11 min readInverse Trigonometric Identities 9 min read Calculus Introduction to Differential Calculus 6 min readLimits in Calculus 12 min readContinuity of Functions 10 min readDifferentiation 2 min readDifferentiability of Functions 9 min readIntegration 3 min read Probability and Statistics Basic Concepts of Probability 7 min readBayes' Theorem 13 min readProbability Distribution - Function, Formula, Table 13 min readDescriptive Statistic 5 min readWhat is Inferential Statistics? 7 min readMeasures of Central Tendency in Statistics 11 min readSet Theory 3 min read Practice NCERT Solutions for Class 8 to 12 7 min readRD Sharma Class 8 Solutions for Maths: Chapter Wise PDF 5 min readRD Sharma Class 9 Solutions 10 min readRD Sharma Class 10 Solutions 9 min readRD Sharma Class 11 Solutions for Maths 13 min readRD Sharma Class 12 Solutions for Maths 13 min read Improvement Suggest Changes Help us improve. Share your suggestions to enhance the article. Contribute your expertise and make a difference in the GeeksforGeeks portal. Create Improvement Enhance the article with your expertise. Contribute to the GeeksforGeeks community and help create better learning resources for all. Suggest Changes min 4 words, max Words Limit:1000 Thank You! Your suggestions are valuable to us.
1805	https://englishunderstood.com/could-have-should-have-would-have-whats-the-difference-english-grammar-mistakes/	Could Have, Should Have & Would Have (WHAT’S THE DIFFERENCE!?) \| English Grammar Mistakes – English Understood Skip to content Subscribe eBooks Free Lessons Beginner Common Mistakes Grammar Pronunciation Vocabulary Reading Writing All Lessons Subscribe eBooks Free Lessons Beginner Common Mistakes Grammar Pronunciation Vocabulary Reading Writing All Lessons COURSES eBooks Subscribe Free Lessons Beginner Common Mistakes Grammar Pronunciation Vocabulary Reading Writing All Lessons Contact Us About COURSES eBooks Subscribe Free Lessons Beginner Common Mistakes Grammar Pronunciation Vocabulary Reading Writing All Lessons Contact Us About Courses Could Have, Should Have & Would Have (WHAT’S THE DIFFERENCE!?) \| English Grammar Mistakes Lesson summary In today's lesson you will learn the difference between could have, should have and would have. (It's easy!) Video transcript John could have gone to the park on Monday. John should have gone to the park on Monday. John would have gone to the park on Monday.What’s the difference here?Let’s find out. What’s up guys? My name is Shane and this is part 2 of the could should and would series.If you haven’t seen part 1, you can check it out right here. Today we are talking about could have, should have and would have. These words cause a lot of confusion for English learners but they’re actually quite simple once you understand their meaning. Could have is normally used to talk about possibility in the past. Should have is normally used to talk about advice in the past and would have is normally used to talk about something that you wanted to do but you didn’t do. But don’t worry if these words are still confusing because by the end of the lesson, you will understand how to use could have should have and would have. And make sure you watch until the end of the video because we will have a quiz to test your understanding. Okay, let’s get started! Basic grammar Look at these three sentences.What verb do we have after have?We have gone, right? And what verb is gone? Gone is the past participle for go. So just remember, when we’re using could have should and would have, we always use a past participle after have. If you haven’t learned about past participles yet, you can watch this video right here where I give you many common examples. And also, when native speakers say could have should have and would have really quickly, it normally sounds like this. Could’ve. Should’ve. Would’ve.Could’ve. Should’ve. Would’ve. And sometimes you will hear native speakers say coulda. Shoulda. And woulda. Coulda. Woulda. Shoulda. Both are fine.It’s up to you which one you want to say. You can say could’ve or coulda.Should’ve or shoulda.Or would’ve or woulda.It’s up to you. Could have John could’ve gone to the park on Monday. What are we talking about here? Past, present or future? We’re talking about the past. Did John go to the park? Did it happen? No it didn’t. But was it possible for him to go? Yes it was. So this is the first meaning of could have. We use it to talk about something that was possible in the past but didn’t happen. So this sentence here means that it was possible for John to go to the park but he didn’t go. What if we want to say it wasn’t possible for John to go to the park? We could say: John could not have gone to the park on Monday. This means it was not possible for him to go and he didn’t go. So just remember, if you want to say something was possible in the past, you need to say could have. But if you want to say it wasn’t possible, then you can say could not have. What if someone asked you: why is John late for the meeting? And you think he forgot the meeting, what can you say here? You can say: John could have forgotten the meeting. John could’ve forgotten the meeting. So this is another use of could have. We can use it to make guesses about the past. What we think happened in the past. We are not sure what happened but we are guessing that John has forgotten the meeting. So we can say: John could have forgotten the meeting. This means we think he has forgotten the meeting but we’re not sure.It’s just a guess. Remember from part one of this could should and would series I said that you can always add more information to make your sentence clearer. This is the same with could have, should have and would have. Let’s have a look at this example. I could have gone to university but I decided to travel instead. I could’ve gone to university, but I decided to travel instead. What am I talking about here? Past, present or future? The past. Did I go to university? No. Was it possible for me to go to university? Yes. But why didn’t I go? Because I decided to travel instead. So just remember, sometimes when you’re using could have, it’s better to add more information so then you can make the sentence clearer. Should have John should have gone to the park on Monday.What am I talking about here? Past, present or future? The past. Did he go to the park?No. Do I wish or do I think it was a good idea for him to go to the park? Yes I do. So this is the first use of should have.We can use it to talk about something that did not happen but we wish it did happen. We think it would have been really good if that thing happened. What if we want to say that we wish john didn’t go to the park.So he did go to the park in the past but we want to say we wish he hadn’t gone. We can say john should not have gone to the park on Monday. So this is another use of should have.If we say should not have, we are talking about something that happened but we wish that it didn’t happen. But if something didn’t happen and you wish that it did happen, then you can say should have. What about this one.He should have finished by work by now.He should have finished by work by now.Has he finished work?I don’t know 100%. But do I think he has finished work?Yes I do. So this is another use of should have.We can use it to talk about something that we think has happened if everything is normal and okay but we’re not 100% sure if everything is okay. We are just guessing. It’s just what we think. And please note that normally when we use should have, we use by now. Another example could be his plane should have arrived by now. This means that I think it’s arrived. I’m not 100% sure but if everything is okay and normal, then it has arrived already. Just like when we use should, sometimes it good to add more information to make the sentence clearer. A lot of the time when we use should have, we use it to express regrets. Or we want to say that we wish something had happened in the past. For example, I should have studied harder, but I wasn’t interested.I should have studied harder, but I wasn’t interested. Did I study hard? No. do I wish I could change that? Yes. So this sentence here is saying that I didn’t study hard in the past but I wish that I could change it and study hard in the past. Would have John would have gone to the park on Monday.What am I talking about here? Past, present or future? The past. Did he go to the park? No. Did he want to go the park?Yes. But was there something that stopped him or prevented him from going? Yes, probably. So this is why we use would have. We use it to talk about something we wanted to do in the past but we didn’t do it because normally something prevented us from doing it or something stopped us from doing it. And normally when we use would have, we give more information to make it clearer. For example, I would have called you but I didn’t have any signal.I would have called you but I didn’t have any signal. When am I talking about here? Past, present or future? The past. Did I call you? No. But did I want to call you?Yes, I did. But what stopped me from calling you? It was the fact that I had no signal. So this sentence here is saying that I wanted to call you but I didn’t because I didn’t have any signal. Quiz Okay, now we know the basics of could have, should have and would have. So let’s have a little quiz to test your understanding. I will show you a sentence with two missing words and I want you to fill in those two missing words with could have, should have or would have. Number one.I _ gone to the game but I didn’t have any money.I wanted to do it.I _ gone to the game but I didn’t have any money.I wanted to do it. And the correct answer is: I would have gone to the game but I didn’t have any money. We use would have here because we wanted to do something in the past but we didn’t do it because something stopped us. What stopped us? We didn’t have any money. Number two.I ___ not ___ gotten so upset when my team lost.I wish that it didn’t happen.I ___ not ___ gotten so upset when my team lost.I wish that it didn’t happen. And the correct answer is: .I should not have gotten so upset when my team lost.We use should not have because we are talking about something that did happen but we wish that it didn’t happen. Number three.I ___ gone to the gym but I felt too lazy.It was possible in the past.I ___ gone to the gym but I felt too lazy.It was possible in the past. And the correct answer is: I could have gone to the gym but I felt too lazy.Remember, we use could have to talk about something that was possible in the past. Number four.He _ finished his dinner by now.We think it’s happened already if everything is normal.He ___ finished his dinner by now.We think it’s already happened if everything is normal. And the correct answer is he should have finished his dinner by now.Remember, we can use should have to talk about something we think has already happened if everything is normal and okay. Number five.Where is he? He _ gotten lost.We are guessing.Where is he? He _ gotten lost.We are guessing. And the correct answer is: He could have gotten lost.Remember, we can say could have when we are just guessing what we think happened. And that is the end of the lesson today.Now you know the basics of could have, should have and would have in English but remember, these are just the very basics and there are always exceptions to the rule so if you have any questions, please comment down below. And if you’re interested in more commonly confused words in English, check out this video right here.And if you haven’t subscribed to the channel already, subscribe right now so you don’t miss any video that can help you understand English like a native speaker. If you learned something today, please hit like and share this with your friends so they can understand the meaning of could have, should have and would have. And for daily quizzes, and weekly posts, videos, lessons and much much more, make sure you are following me on Instagram and I will see you in the next video.What verb do we have after have? We have.What about this? What about this one? More lessons for you Want better English? Join thousands learners and get free English lessons, tips & tricks, strategies, news and updates! First name Email Subscribe now See our Privacy Policy here. Unsubscribe anytime. 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1806	https://courses.lumenlearning.com/ntcc-collegealgebracorequisite/chapter/graphing-parabolas-with-vertices-at-the-origin/	Module 15: Conic Sections Parabolas with Vertices at the Origin Learning Outcomes Identify and label the focus, directrix, and endpoints of the focal diameter of a parabola. Write the equation of a parabola given a focus and directrix. In The Ellipse we saw that an ellipse is formed when a plane cuts through a right circular cone. If the plane is parallel to the edge of the cone, an unbounded curve is formed. This curve is a parabola. tip for success You’ve seen parabolas before as the set of all points satisfying a quadratic function. We’ll look at the geometric form of a parabola in this section. It still describes a set of points that satisfy an equation in two variables, but without the need to qualify as a function, it can open left and right as well as up and down. It will be necessary to use another form of its equation to take all of the characteristics of this object into consideration. You’ll learn new terminology for the parts of a parabola just as you did with the ellipse and hyperbola as well. Like the ellipse and hyperbola, the parabola can also be defined by a set of points in the coordinate plane. A parabola is the set of all points [latex]\left(x,y\right)[/latex] in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix. We previously learned about a parabola’s vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola. Notice that the axis of symmetry passes through the focus and vertex and is perpendicular to the directrix. The vertex is the midpoint between the directrix and the focus. The line segment that passes through the focus and is parallel to the directrix is called the latus rectum,also called the focal diameter. The endpoints of the focal diameter lie on the curve. By definition, the distance [latex]d[/latex] from the focus to any point [latex]P[/latex] on the parabola is equal to the distance from [latex]P[/latex] to the directrix. Key features of the parabola To work with parabolas in the coordinate plane, we consider two cases: those with a vertex at the origin and those with a vertex at a point other than the origin. We begin with the former. Let [latex]\left(x,y\right)[/latex] be a point on the parabola with vertex [latex]\left(0,0\right)[/latex], focus [latex]\left(0,p\right)[/latex], and directrix [latex]y= -p[/latex] as shown in Figure 4. The distance [latex]d[/latex] from point [latex]\left(x,y\right)[/latex] to point [latex]\left(x,-p\right)[/latex] on the directrix is the difference of the y-values: [latex]d=y+p[/latex]. The distance from the focus [latex]\left(0,p\right)[/latex] to the point [latex]\left(x,y\right)[/latex] is also equal to [latex]d[/latex] and can be expressed using the distance formula. [latex]\begin{align}d&=\sqrt{{\left(x - 0\right)}^{2}+{\left(y-p\right)}^{2}} \ &=\sqrt{{x}^{2}+{\left(y-p\right)}^{2}} \end{align}[/latex] Set the two expressions for [latex]d[/latex] equal to each other and solve for [latex]y[/latex] to derive the equation of the parabola. We do this because the distance from [latex]\left(x,y\right)[/latex] to [latex]\left(0,p\right)[/latex] equals the distance from [latex]\left(x,y\right)[/latex] to [latex]\left(x, -p\right)[/latex]. [latex]\sqrt{{x}^{2}+{\left(y-p\right)}^{2}}=y+p[/latex] We then square both sides of the equation, expand the squared terms, and simplify by combining like terms. [latex]\begin{gathered}{x}^{2}+{\left(y-p\right)}^{2}={\left(y+p\right)}^{2} \ {x}^{2}+{y}^{2}-2py+{p}^{2}={y}^{2}+2py+{p}^{2}\ {x}^{2}-2py=2py \ {x}^{2}=4py\end{gathered}[/latex] The equations of parabolas with vertex [latex]\left(0,0\right)[/latex] are [latex]{y}^{2}=4px[/latex] when the x-axis is the axis of symmetry and [latex]{x}^{2}=4py[/latex] when the y-axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features. A General Note: Standard Forms of Parabolas with Vertex (0, 0) The table below summarizes the standard features of parabolas with a vertex at the origin. \| \| \| \| \| \| --- --- \| Axis of Symmetry \| Equation \| Focus \| Directrix \| Endpoints of Focal Diameter \| \| x-axis \| [latex]{y}^{2}=4px[/latex] \| [latex]\left(p,\text{ }0\right)[/latex] \| [latex]x=-p[/latex] \| [latex]\left(p,\text{ }\pm 2p\right)[/latex] \| \| y-axis \| [latex]{x}^{2}=4py[/latex] \| [latex]\left(0,\text{ }p\right)[/latex] \| [latex]y=-p[/latex] \| [latex]\left(\pm 2p,\text{ }p\right)[/latex] \| (a) When [latex]p>0[/latex] and the axis of symmetry is the x-axis, the parabola opens right. (b) When [latex]p<0[/latex] and the axis of symmetry is the x-axis, the parabola opens left. (c) When [latex]p<0[/latex] and the axis of symmetry is the y-axis, the parabola opens up. (d) When [latex]\text{ }p<0\text{ }[/latex] and the axis of symmetry is the y-axis, the parabola opens down. The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and focal diameter. When given a standard equation for a parabola centered at the origin, we can easily identify the key features to graph the parabola. A line is said to be tangent to a curve if it intersects the curve at exactly one point. If we sketch lines tangent to the parabola at the endpoints of the focal diameter, these lines intersect on the axis of symmetry. How To: Given a standard form equation for a parabola centered at (0, 0), sketch the graph. Determine which of the standard forms applies to the given equation: [latex]{y}^{2}=4px[/latex] or [latex]{x}^{2}=4py[/latex]. Use the standard form identified in Step 1 to determine the axis of symmetry, focus, equation of the directrix, and endpoints of the focal diameter. If the equation is in the form [latex]{y}^{2}=4px[/latex], then the axis of symmetry is the x-axis, [latex]y=0[/latex] set [latex]4p[/latex] equal to the coefficient of x in the given equation to solve for [latex]p[/latex]. If [latex]p>0[/latex], the parabola opens right. If [latex]p<0[/latex], the parabola opens left. use [latex]p[/latex] to find the coordinates of the focus, [latex]\left(p,0\right)[/latex] use [latex]p[/latex] to find the equation of the directrix, [latex]x=-p[/latex] use [latex]p[/latex] to find the endpoints of the focal diameter, [latex]\left(p,\pm 2p\right)[/latex]. Alternately, substitute [latex]x=p[/latex] into the original equation. If the equation is in the form [latex]{x}^{2}=4py[/latex], then the axis of symmetry is the y-axis, [latex]x=0[/latex] set [latex]4p[/latex] equal to the coefficient of y in the given equation to solve for [latex]p[/latex]. If [latex]p>0[/latex], the parabola opens up. If [latex]p<0[/latex], the parabola opens down. use [latex]p[/latex] to find the coordinates of the focus, [latex]\left(0,p\right)[/latex] use [latex]p[/latex] to find equation of the directrix, [latex]y=-p[/latex] use [latex]p[/latex] to find the endpoints of the focal diameter, [latex]\left(\pm 2p,p\right)[/latex] Plot the focus, directrix, and focal diameter, and draw a smooth curve to form the parabola. Example: Graphing a Parabola with Vertex (0, 0) and the x-axis as the Axis of Symmetry Graph [latex]{y}^{2}=24x[/latex]. Identify and label the focus, directrix, and endpoints of the focal diameter. Show Solution The standard form that applies to the given equation is [latex]{y}^{2}=4px[/latex]. Thus, the axis of symmetry is the x-axis. It follows that: [latex]24=4p[/latex], so [latex]p=6[/latex]. Since [latex]p>0[/latex], the parabola opens right the coordinates of the focus are [latex]\left(p,0\right)=\left(6,0\right)[/latex] the equation of the directrix is [latex]x=-p=-6[/latex] the endpoints of the focal diameter have the same x-coordinate at the focus. To find the endpoints, substitute [latex]x=6[/latex] into the original equation: [latex]\left(6,\pm 12\right)[/latex] Next we plot the focus, directrix, and focal diameter, and draw a smooth curve to form the parabola. Try It Graph [latex]{y}^{2}=-16x[/latex]. Identify and label the focus, directrix, and endpoints of the focal diameter. Show Solution Focus: [latex]\left(-4,0\right)[/latex]; Directrix: [latex]x=4[/latex]; Endpoints of the latus rectum: [latex]\left(-4,\pm 8\right)[/latex] Example: Graphing a Parabola with Vertex (0, 0) and the y-axis as the Axis of Symmetry Graph [latex]{x}^{2}=-6y[/latex]. Identify and label the focus, directrix, and endpoints of the focal diameter. Show Solution The standard form that applies to the given equation is [latex]{x}^{2}=4py[/latex]. Thus, the axis of symmetry is the y-axis. It follows that: [latex]-6=4p[/latex], so [latex]p=-\frac{3}{2}[/latex]. Since [latex]p<0[/latex], the parabola opens down. the coordinates of the focus are [latex]\left(0,p\right)=\left(0,-\frac{3}{2}\right)[/latex] the equation of the directrix is [latex]y=-p=\frac{3}{2}[/latex] the endpoints of the focal diameter can be found by substituting [latex]\text{ }y=\frac{3}{2}\text{ }[/latex] into the original equation, [latex]\left(\pm 3,-\frac{3}{2}\right)[/latex] Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. Try It Graph [latex]{x}^{2}=8y[/latex]. Identify and label the focus, directrix, and endpoints of the focal diameter. Show Solution Focus: [latex]\left(0,2\right)[/latex]; Directrix: [latex]y=-2[/latex]; Endpoints of the latus rectum: [latex]\left(\pm 4,2\right)[/latex]. Try It Use an online graphing tool to plot the following parabola whose axis of symmetry is the x-axis, [latex]y^2=4px[/latex] Adjust the free variable [latex]p[/latex] to values between [latex]-10,10[/latex]. Your task in this exercise is to add the focus, directrix, and endpoints of the focal diameter in terms of the free variable, [latex]p[/latex]. For example, to add the focus, you would define a point, latex[/latex] . Show Solution Writing Equations of Parabolas in Standard Form In the previous examples we used the standard form equation of a parabola to calculate the locations of its key features. We can also use the calculations in reverse to write an equation for a parabola when given its key features. tip for success In this section, we will write the equation of a parabola in standard form, as opposed to the equation of a quadratic or second degree polynomial. The language we use when discussing the object is specific. It is true that a quadratic function forms a parabola when graphed in the plane, but here we are using the phrase standard form of the equation of a parabola to indicate that we wish to describe the geometric object. When talking about this object in this context, we would naturally use the equations described below. How To: Given its focus and directrix, write the equation for a parabola in standard form. Determine whether the axis of symmetry is the x– or y-axis. If the given coordinates of the focus have the form [latex]\left(p,0\right)[/latex], then the axis of symmetry is the x-axis. Use the standard form [latex]{y}^{2}=4px[/latex]. If the given coordinates of the focus have the form [latex]\left(0,p\right)[/latex], then the axis of symmetry is the y-axis. Use the standard form [latex]{x}^{2}=4py[/latex]. Multiply [latex]4p[/latex]. Substitute the value from Step 2 into the equation determined in Step 1. Example: Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix What is the equation for the parabola with focus [latex]\left(-\frac{1}{2},0\right)[/latex] and directrix [latex]x=\frac{1}{2}?[/latex] Show Solution The focus has the form [latex]\left(p,0\right)[/latex], so the equation will have the form [latex]{y}^{2}=4px[/latex]. Multiplying [latex]4p[/latex], we have [latex]4p=4\left(-\frac{1}{2}\right)=-2[/latex]. Substituting for [latex]4p[/latex], we have [latex]{y}^{2}=4px=-2x[/latex]. Therefore, the equation for the parabola is [latex]{y}^{2}=-2x[/latex]. Try It What is the equation for the parabola with focus [latex]\left(0,\frac{7}{2}\right)[/latex] and directrix [latex]y=-\frac{7}{2}[/latex]? Show Solution Candela Citations CC licensed content, Original Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution Graphing Parabolas Interactive. Authored by: Lumen Learning. Located at: License: Public Domain: No Known Copyright Graphing Parabolas - With Solutions Interactive. Authored by: Lumen Learning. Located at: License: Public Domain: No Known Copyright CC licensed content, Shared previously College Algebra. Authored by: Abramson, Jay et al.. Provided by: OpenStax. Located at: License: CC BY: Attribution. License Terms: Download for free at Question ID 23511. Authored by: Shahbazian,Roy, mb McClure,Caren. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL Question ID 23711. Authored by: McClure,Caren. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL CC licensed content, Specific attribution Precalculus. Authored by: OpenStax College. Provided by: OpenStax. Located at: License: CC BY: Attribution Licenses and Attributions CC licensed content, Original Revision and Adaptation. Provided by: Lumen Learning. License: CC BY: Attribution Graphing Parabolas Interactive. Authored by: Lumen Learning. Located at: License: Public Domain: No Known Copyright Graphing Parabolas - With Solutions Interactive. Authored by: Lumen Learning. Located at: License: Public Domain: No Known Copyright CC licensed content, Shared previously College Algebra. Authored by: Abramson, Jay et al.. Provided by: OpenStax. Located at: License: CC BY: Attribution. License Terms: Download for free at Question ID 23511. Authored by: Shahbazian,Roy, mb McClure,Caren. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL Question ID 23711. Authored by: McClure,Caren. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL CC licensed content, Specific attribution Precalculus. Authored by: OpenStax College. Provided by: OpenStax. Located at: License: CC BY: Attribution Privacy Policy
1807	https://www.youtube.com/watch?v=ATrZiIrLvbI	Adding Integers: Adding a Positive and a Negative Integer \| Positive + Negative \| Math with Mr. J Math with Mr. J 1690000 subscribers 675 likes Description 63742 views Posted: 2 Jul 2022 Welcome to Adding a Positive and a Negative Integer with Mr. J! Need help with how to add positive and negative integers? You're in the right place! Whether you're just starting out, or need a quick refresher, this is the video for you if you're looking for help with adding integers. Mr. J will go through examples of adding positive and negative integers and explain the steps of how to add integers. More Videos and Examples: ✅ Adding Integers -A Quick Review of Adding Integers = -Adding a Positive and Negative Integer \| Positive + Negative = -Adding a Negative and Positive Integer \| Negative + Positive = -Adding Two Negative Integers \| Negative + Negative = ✅ Subtracting Integers -A Quick Review of Subtracting Integers = -Subtracting a Positive Integer from a Negative Integer \| Negative - Positive = -Subtracting a Negative Integer from a Positive Integer \| Positive - Negative = -Subtracting a Negative Integer from a Negative Integer \| Negative - Negative = ✅ Multiplying Integers -A Quick Review of Multiplying Integers = -Multiplying a Positive by a Negative \| Positive x Negative = -Multiplying a Negative by a Positive \| Negative x Positive = -Multiplying a Negative Integer by a Negative Integer \| Negative x Negative = -Multiplying Three Integers = -Why Does a Negative Times a Negative Equal a Positive? = ✅ Dividing Integers -A Quick Review of Dividing Integers = -Dividing a Positive by a Negative \| Positive ÷ Negative = -Dividing a Negative by a Positive \| Negative ÷ Positive = -Dividing a Negative Integer by a Negative Integer \| Negative ÷ Negative = ✅ Integers How to Add and Subtract Integers = How to Multiply and Dividing Integers = How to Add Subtract Multiply and Divide Integers = About Math with Mr. J: This channel offers instructional videos that are directly aligned with math standards. Teachers, parents/guardians, and students from around the world have used this channel to help with math content in many different ways. All material is absolutely free. Click Here to Subscribe to the Greatest Math Channel On Earth: Follow Mr. J on Twitter: @MrJMath5 Email: math5.mrj@gmail.com Music: Hopefully this video is what you're looking for when it comes to adding a positive and negative integer. Have a great rest of your day and thanks again for watching! ✌️✌️✌️ ✅ Thanks to Aloud, this video has been dubbed into Spanish and Portuguese. #DubbedWithAloud English This video has been dubbed into Spanish (United States) and Portuguese (Brazil) using an artificial voice via to increase accessibility. You can change the audio track language in the Settings menu. Spanish Este video ha sido doblado al español con voz artificial con para aumentar la accesibilidad. Puede cambiar el idioma de la pista de audio en el menú Configuración. Portuguese Este vídeo foi dublado para o português usando uma voz artificial via para melhorar sua acessibilidade. Você pode alterar o idioma do áudio no menu Configurações. 42 comments Transcript: [Music] welcome to math with mr j [Music] in this video i'm going to cover how to add a positive integer and a negative integer so a positive plus a negative now when it comes to adding integers there are different ways to think through these we will go through two ways for each of our examples let's jump into number one where we have ten plus negative six the first thing that we need to do we need to look at the signs for number one we have a positive 10 and a negative six so we have different signs a positive and a negative since we have different signs we will take the greater absolute value and subtract the lesser our answer will take the sign of the greater absolute value let's write the absolute value of both 10 and negative 6 the absolute value of 10 is 10. the absolute value of negative 6 is 6. now take the greater absolute value and subtract the lesser these are already in order so we can subtract if the larger absolute value comes second we can always switch the order to find the difference if need be so we have 10 minus 6 which gives us 4. now our answer is going to take the sine of the larger absolute value which that's going to be 10. so we take a look at 10 in our original problem and see what sign that 10 had well that's a positive 10 in our original problem so our answer is going to be a positive 4. our final answer positive 4. basically we forget about any negatives because we are working with absolute values find the difference between absolute values and then take the sign of the greater absolute value from the original problem now let's think through this a second way and this way is going to be more of a mental math approach just basically thinking about what's going on in this problem i'll rewrite the problem down below so we have ten plus negative six so let's think about this so we are starting at a positive 10 and we are adding a negative 6. by adding a negative 6 we are decreasing in value by 6 from that 10. we can basically think of this as 10 minus 6 or 10 take away 6. that gives us our answer of 4 so 10 plus negative 6 gives us 4 so again we started at positive 10. always think about where you are starting and where you are going from that starting point so we are adding a negative 6 which is decreasing our 10 in value by 6 and we end up with positive 4 for our final answer let's move on to number 2 where we have 2 plus negative seven so we have different signs here a positive and a negative so we take the greater absolute value and subtract the lesser we use the sign of the greater absolute value for our answer so let's take the absolute value of 2 and that is going to be 2 and then the absolute value of negative 7 is 7. so we have 2 and 7 there we need to take the greater absolute value and subtract the lesser so let's switch the order here so we can find the difference so we do seven minus two again the greater absolute value minus the lesser find the difference and that gives us 5. now we need to determine the sign of that answer so the greater absolute value is this 7 here now the 7 in the original problem is negative so that's the sign of our answer our final answer negative 5. now let's think through this using more of a mental math approach so we had two plus negative seven so we're starting at positive two and when we add a negative we are decreasing in value so start at that positive two and decrease in value by seven and we end up at negative five so again think about where you're starting and then where you're going from that starting point so to recap we started at that positive two and by adding a negative seven we are decreasing in value by seven and we end up with negative five so there you have it there's how you add a positive integer and a negative integer i hope that helped thanks so much for watching until next time peace you
1808	https://betterexplained.com/articles/rescaling-the-pythagorean-theorem/	Rescaling the Pythagorean Theorem The Pythagorean theorem can apply to any shape, not just triangles. It can measure nearly any type of distance. And yet this 2000-year-old formula is still showing us new tricks. Re-arranging the formula from this: to this: helps us understand the relationship between slope (steepness) and distance. Let’s take a look. Rescale Your Triangle Scaling leads to new insights. Yes, \$500k/year is a lot; but it really comes alive when you imagine things costing 10x less (A new laptop? \$150. A new porsche? \$6000). Rescaling formulas can be eye-opening as well. Let’s start with our favorite 3-4-5 triangle and divide every side by 3: What happened? Well, we have a smaller red triangle with sides 3/3 (aka 1), 4/3 and 5/3. We’ve got a mini version of the large triangle, and the Pythagorean Theorem still holds: So Why’s This Special? It doesn’t seem like much, but there’s some surprising insights: First, we can rescale any triangle to have 1 as the smallest side (divide by “a”). All similar triangles (i.e. those with the same ratios, like 3-4-5 and 6-8-10) will shrink into the same mini triangle. This mini triangle has an interesting property: it only cares about the ratio b/a. The only “meaningful” numbers are 1 and (b/a), giving: And what’s special about b/a? It’s the slope of the hypotenuse line! It’s called the slope, the gradient, the derivative, rise over run — whatever the label, b/a is the rate at which the hypotenuse changes! This makes sense. For every unit traveled along the short leg, we gain “slope units (b/a)” on the other leg. In a 3-4-5 triangle, we go 4/3 units “North” for every 1 unit “East”. And the length of our hypotenuse increases 5/3 (1.66) for every 1 unit East. The result is pretty cool: we used the steepness of the hypotenuse (b/a) to find the distance traveled per unit East, $\sqrt{1 + (b/a)^2}$. An Example, Please This is a bit weird, so let’s do an example. Suppose we’ve gone 5 units East and 12 units North. What’s our distance from the starting point? The traditional approach plugs in the Pythagorean Theorem to get $c = \sqrt{5^2 + 12^2} = 13$. It works, but let’s try our mini-triangle method: Instead of a large triangle with sides 5 and 12, scale down by 5: we get a mini triangle with sides 5/5 (or 1) and 12/5. The “mini hypotenuse” is then $\sqrt{1 + (12/5)^2} = 2.6$. This means we travel 2.6 units along the hypotenuse for every 1 unit East. Going the full 5 units East (our original triangle) is 5 2.6 = 13 units. Neato — we got the same answer both ways. But silly me, I made a mistake. Instead of 5 units on that trajectory, I meant 6. No 7. No wait, 8. 9, for sure. Normally, we’d be furiously hammering that square root button to find the new distance. Maybe even using trigonometry to “make it easier”. But not today — since we’re on the same trajectory, we can re-use our scaling constant of 2.6: We can find the new distance traveled with regular multiplication, with nary a square root in sight. Cool! This approach is faster for humans and computers alike — you wouldn’t believe the crazy approaches programmers take to avoid a square root. Static and Dynamic Formulas I’ve realized that our venerable Pythagorean Theorem focuses on a and b separately: We consider a and b as separate elements, to be squared and summed. This approach is straightforward, and helps when designing bridges or making pictures of triangles. The traditional formula focuses on final values. But the rescaled version has a new twist: We’re not that interested in the separate quantities — we want the ratio b/a, or the slope of the hypotenuse. This slope creates a scaling constant, $\sqrt{1 + (b/a)^2}$, that tells us how our “Eastward” motion translates to distance along our path. The dynamic formula focuses on rates of change. If we have a hypothetical function f(x), we might write the dynamic Pythagorean Theorem this way: This concept is used in calculus to find the length of any line or curve — but we’ll save that for another day. The key is to realize a single formula can be re-arranged and lead to new insights. Stay curious — we stop learning when we think we’ve “got it all figured out”. Appendix 1: Slope vs. Distance One point that confused me was separating the idea of slope (b/a) from distance traveled (the hypotenuse, c). Slope is b/a, rise over run — how much height you get when you increase width. How “steep” the hill is, so to speak. Unfortunately, the word “slope” makes us think of the side of the hill — but slope is really about height. Distance (the hypotenuse) is about the side of the hill — how far you’ve walked. The “steepness” isn’t that important — you’re laying a measuring tape on the ground, which could be flat, vertical or upside-down. Does the length of a board depend on how you hold it? But, in our man-made world, slope and distance are related because we often express locations in terms of “units East (x coordinate)” and not “units along a path”. So when a map says “go 1 mile due East” and you’re in front a mountain (large slope), you end up traveling a large distance (more than 1 mile). When on a flat road (zero slope), 1 mile East is simply 1 mile East. The bigger the slope, the more distance you must travel to “go 1 mile East”. Again, we see that the Pythagorean Theorem is not just about triangles — it can convert slope (steepness) into distance traveled. Happy math. Other Posts In This Series How To Measure Any Distance With The Pythagorean Theorem Surprising Uses of the Pythagorean Theorem Pythagorean Theorem As Sweeping Area Rescaling the Pythagorean Theorem Understanding Why Similarity Works Math Join 450k Monthly Readers Enjoy the article? There's plenty more to help you build a lasting, intuitive understanding of math. Join the newsletter for bonus content and the latest updates. Share with Facebook Twitter LinkedIn Email Print
1809	https://brainly.com/question/59188194	[FREE] Calculate the percentage mass of acetic acid in vinegar. Given: \begin{array}{l} \text{Mass of acetic - brainly.com Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +81,7k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +43,7k Ace exams faster, with practice that adapts to you Practice Worksheets +8,9k Guided help for every grade, topic or textbook Complete See more / Chemistry Expert-Verified Expert-Verified Calculate the percentage mass of acetic acid in vinegar. Given: Mass of acetic acid:0.07612 g Mass of vinegar:10.058 g Calculate the percentage: 10.058 g 0.07612 g×100%=0.7568% Can your sample of vinegar be sold commercially? Briefly explain. No, it cannot be sold commercially because the acetic acid percentage is too low. It must be between 4-8%. 1. What is the mass of acetic acid in 1,000 gallons of the vinegar solution that you titrated? 1 See answer Explain with Learning Companion NEW Asked by kelly343 • 03/17/2025 0:03 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 1182974143 people 1182M 0.0 0 Upload your school material for a more relevant answer To determine whether your sample of vinegar can be sold commercially and to find the mass of acetic acid in 1,000 gallons of the vinegar solution, let's go through the steps. 1. Commercial Viability: For vinegar to be sold commercially, the acetic acid concentration typically needs to fall within the range of 4-8%. In your sample, the concentration of acetic acid was found to be approximately 0.7568%. This is below the minimum requirement of 4% and thus, your sample cannot be sold commercially as vinegar because the acetic acid concentration is too low. 2. Mass of Acetic Acid in 1,000 Gallons: First, let's identify the mass of acetic acid in your original sample. With a concentration of 0.7568%, it was calculated that the acetic acid mass is approximately 0.0761 grams in a 10.0589-gram sample of vinegar. Next, to find out how much acetic acid would be present in 1,000 gallons of this solution, consider that 1 gallon of vinegar roughly equals 3,785.41 grams. Therefore, 1,000 gallons would weigh approximately 3,785,410 grams. By applying the same concentration of acetic acid (0.7568%), the total mass of acetic acid in 1,000 gallons of vinegar is about 28,649.20 grams. So, the vinegar cannot be sold commercially due to the low acetic acid concentration, and in 1,000 gallons of the vinegar solution, you would have approximately 28,649.20 grams of acetic acid. Answered by GinnyAnswer •8M answers•1.2B people helped Thanks 0 0.0 (0 votes) Expert-Verified⬈(opens in a new tab) 0.0 0 Upload your school material for a more relevant answer The vinegar sample cannot be sold commercially due to its low acetic acid concentration of 0.7568%, which is below the required 4-8%. In 1,000 gallons of this vinegar, approximately 28,649.20 grams of acetic acid would be present. Therefore, you have both a compliance issue and a significant quantity of acetic acid in the total volume sampled. Explanation To determine whether your sample of vinegar can be sold commercially and to find the mass of acetic acid in 1,000 gallons of the vinegar solution, let's go through the steps. 1. Commercial Viability: For vinegar to be sold commercially, the acetic acid concentration typically needs to fall within the range of 4-8%. In your sample, the concentration of acetic acid was found to be approximately 0.7568%. This is below the minimum requirement of 4% and thus, your sample cannot be sold commercially as vinegar because the acetic acid concentration is too low. 2. Mass of Acetic Acid in 1,000 Gallons: First, let's identify the mass of acetic acid in your original sample. With a concentration of 0.7568%, it was calculated that the acetic acid mass is approximately 0.0761 grams in a 10.058-gram sample of vinegar. Next, to find out how much acetic acid would be present in 1,000 gallons of this solution, consider that 1 gallon of vinegar roughly equals 3,785.41 grams. Therefore, 1,000 gallons would weigh approximately 3,785,410 grams. By applying the same concentration of acetic acid (0.7568%), the total mass of acetic acid in 1,000 gallons of vinegar is about 28,649.20 grams. You can calculate this by using the formula: Mass of Acetic Acid=Total Mass of Vinegar×(100 Concentration of Acetic Acid) Thus, substituting in the values: Mass of Acetic Acid=3,785,410 g×0.007568≈28,649.20 g. So, the vinegar cannot be sold commercially due to the low acetic acid concentration, and in 1,000 gallons of the vinegar solution, you would have approximately 28,649.20 grams of acetic acid. Examples & Evidence To help illustrate, vinegar sold in stores typically has a concentration of around 5% acetic acid, which is ideal for culinary uses. Understanding the percentage helps ensure the product is effective and safe for consumption. The required concentration of acetic acid for commercial vinegar is generally accepted to be between 4-8%, so a concentration of 0.7568% is insufficient. Thanks 0 0.0 (0 votes) Advertisement kelly343 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Chemistry solutions and answers Community Answer How would you solve for the mass of vinegar and % acetic acid in a solution? Use 0.1005 M in the calculations. A) Calculate the molarity of acetic acid (CH3COOH) in the vinegar. B) Determine the volume of vinegar required for a specific amount of acetic acid. C) Calculate the mass of acetic acid in the solution. D) Find the percentage of acetic acid by mass in the solution. Community Answer Assuming that the density of vinegar is 1.005 g/mL, calculate the molatrity of acetic acid in vinegar from an average of 4.6% for the mass percentage of acetic acid in vinegar. Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain? Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. Massing out120 g of sodium chloride (acbnela)____5. Suspending glassware over the Bunsen burner (rwei zeagu)____6. Used to pour liquids into containers with small openings or to hold filter paper (unfenl)____7. Mixing a small amount of chemicals together (lewl letpa)____8. Heating contents in a test tube (estt ubet smalcp)____9. Holding many test tubes filled with chemicals (estt ubet karc) ____10. Used to clean the inside of test tubes or graduated cylinders (iwer srbuh)____11. Keeping liquid contents in a beaker from splattering (tahcw sgasl)____12. A narrow-mouthed container used to transport, heat or store substances, often used when a stopper is required (ymerereel kslaf)____13. Heating contents in the lab (nuesnb bneurr)____14. Transport a hot beaker (gntos)____15. Protects the eyes from flying objects or chemical splashes(ggloges)____16. Used to grind chemicals to powder (tmraor nda stlepe) __ Community Answer Food waste, like a feather or a bone, fall into food, causing contamination. Physical Chemical Pest Cross-conta Community Answer 8 If the temperature of a reversible reaction in dynamic equilibrium increases, how will the equilibrium change? A. It will shift towards the products. B. It will shift towards the endothermic reaction. C. It will not change. D. It will shift towards the exothermic reaction. Community Answer 4.8 52 Which statements are TRUE about energy and matter in stars? Select the three correct answers. Al energy is converted into matter in stars Only matter is conserved within stars. Only energy is conserved within stars. Some matter is converted into energy within stars. Energy and matter are both conserved in stars Energy in stars causes the fusion of light elements Community Answer 4.5 153 The pH of a solution is 2.0. Which statement is correct? Useful formulas include StartBracket upper H subscript 3 upper O superscript plus EndBracket equals 10 superscript negative p H., StartBracket upper O upper H superscript minus EndBracket equals 10 superscript negative p O H., p H plus P O H equals 14., and StartBracket upper H subscript 3 upper O superscript plus EndBracket StartBracket upper O upper H superscript minus EndBracket equals 10 to the negative 14 power. Community Answer 5 Dimensional Analysis 1. I have 470 milligrams of table salt, which is the chemical compound NaCl. How many liters of NaCl solution can I make if I want the solution to be 0.90% NaCl? (9 grams of salt per 1000 grams of solution). The density of the NaCl solution is 1.0 g solution/mL solution. New questions in Chemistry How many molecules are in 0.400 moles of N O 3 ? What type of chemical bond holds C a 2+ and O 2− together in CaO? A. metallic bond B. covalent bond C. ionic bond D. hydrogen bond Ionic compounds are compounds that A. consist only of metallic elements. B. have atoms that transfer electrons. C. exist as individual ions. D. have atoms that share electrons. What is the term for a combination of two or more substances in any proportions in which the substances do not combine chemically to form a new substance? A. mixture B. compound C. molecule D. element What happens to the acidity of solutions A and B after dry ice is added? \| Acidity Changes after Dry Ice Is Added \| \| Time (sec) \| \| pH of Solution A \| pH of Solution B \| \| 0 \| \| 11.0 \| 8.5 \| \| \| \| 10.8 \| 8.0 \| \| \| \| 10.2 \| 7.2 \| \| \| \| 9.6 \| 6.5 \| \| \| \| 9.0 \| 5.8 \| A. The pH of solution A decreases, and the pH of solution B increases. B. The pH of solution A increases, and the pH of solution B decreases. C. The pH of solution A increases, and the pH of solution B increases. D. The pH of solution A decreases, and the pH of solution B decreases. Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
1810	https://pmc.ncbi.nlm.nih.gov/articles/PMC6827566/	2019 EULAR/ACR Classification Criteria for Systemic Lupus Erythematosus - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Arthritis Rheumatol . Author manuscript; available in PMC: 2020 Sep 1. Published in final edited form as: Arthritis Rheumatol. 2019 Aug 6;71(9):1400–1412. doi: 10.1002/art.40930 Search in PMC Search in PubMed View in NLM Catalog Add to search 2019 EULAR/ACR Classification Criteria for Systemic Lupus Erythematosus Martin Aringer Martin Aringer 1 University Medical Center and Faculty of Medicine Carl Gustav Carus, TU Dresden, Dresden, Germany Find articles by Martin Aringer 1, Karen H Costenbader Karen H Costenbader 2 Brigham and Women’s Hospital, Harvard Medical School, Boston, MA, USA Find articles by Karen H Costenbader 2, David I Daikh David I Daikh 3 University of California at San Francisco and VA Medical Center, San Francisco, CA, USA Find articles by David I Daikh 3, Ralph Brinks Ralph Brinks 4 Policlinic and Hiller Research Unit for Rheumatology, Medical Faculty, Heinrich-Heine-University Duesseldorf, Duesseldorf, Germany Find articles by Ralph Brinks 4, Marta Mosca Marta Mosca 5 Rheumatology Unit, Azienda Ospedaliero Universitaria Pisana, University of Pisa, Pisa, Italy Find articles by Marta Mosca 5, Rosalind Ramsey-Goldman Rosalind Ramsey-Goldman 6 Northwestern University Feinberg School of Medicine, Chicago, IL, USA Find articles by Rosalind Ramsey-Goldman 6, Josef S Smolen Josef S Smolen 7 Medical University of Vienna, Austria Find articles by Josef S Smolen 7, David Wofsy David Wofsy 8 Russell/Engleman Rheumatology Research Center, University of California at San Francisco, San Francisco, USA Find articles by David Wofsy 8, Dimitrios Boumpas Dimitrios Boumpas 9 Medical School, National and Kapodestrian University of Athens, and Biomedical Research Foundation of the Athens Academy, Athens, Greece; Medical School, University of Cyprus, Nicosia, Cyprus Find articles by Dimitrios Boumpas 9, Diane L Kamen Diane L Kamen 10 Medical University of South Carolina, Charleston, SC, USA Find articles by Diane L Kamen 10, David Jayne David Jayne 11 Department of Medicine, University of Cambridge, United Kingdom Find articles by David Jayne 11, Ricard Cervera Ricard Cervera 12 Department of Autoimmune Diseases, Hospital Clínic, University of Barcelona, Barcelona, Catalonia, Spain Find articles by Ricard Cervera 12, Nathalie Costedoat-Chalumeau Nathalie Costedoat-Chalumeau 13 Cochin Hospital, Internal Medicine Department, Centre de référence maladies auto-immunes et systémiques rares d’île de France, Paris, France ; Université Paris Descartes-Sorbonne Paris Cité, Paris, France ; INSERM U 1153, Center for Epidemiology and Statistics Sorbonne Paris Cité (CRESS), Paris, France Find articles by Nathalie Costedoat-Chalumeau 13, Betty Diamond Betty Diamond 14 Feinstein Institute, Manhasset, NY, United States Find articles by Betty Diamond 14, Dafna D Gladman Dafna D Gladman 15 Division of Rheumatology, Department of Medicine, Toronto Western Hospital, University of Toronto, Toronto, Ontario, Canada Find articles by Dafna D Gladman 15, Bevra H Hahn Bevra H Hahn 16 University of California at Los Angeles, Los Angeles, CA, USA Find articles by Bevra H Hahn 16, Falk Hiepe Falk Hiepe 17 Charité – Universitätsmedizin Berlin, Corporate member of Freie Universität Berlin, Humboldt-Universität zu Berlin, and Berlin Institute of Health, Department of Rheumatology and Clinical Immunology, Berlin, Germany Find articles by Falk Hiepe 17, Søren Jacobsen Søren Jacobsen 18 Copenhagen Lupus and Vasculitis Clinic, Rigshospitalet, Copenhagen University Hospital, Copenhagen, Denmark Find articles by Søren Jacobsen 18, Dinesh Khanna Dinesh Khanna 19 University of Michigan, Ann Arbor, MI, USA Find articles by Dinesh Khanna 19, Kirsten Lerstrøm Kirsten Lerstrøm 20 Lupus Europe, co-opted trustee for research, Essex, UK Find articles by Kirsten Lerstrøm 20, Elena Massarotti Elena Massarotti 21 Brigham and Women’s Hospital, Boston MA; Harvard Medical School, Boston, USA Find articles by Elena Massarotti 21, W Joseph McCune W Joseph McCune 22 University of Michigan, Ann Arbor, MI, USA Find articles by W Joseph McCune 22, Guillermo Ruiz-Irastorza Guillermo Ruiz-Irastorza 23 Autoimmune Diseases Research Unit, Department of Internal Medicine, Biocruces Bizkaia Health Research Institute, Hospital Universitario Cruces, UPV/EHU, Bizkaia, The Basque Country, Spain Find articles by Guillermo Ruiz-Irastorza 23, Jorge Sanchez-Guerrero Jorge Sanchez-Guerrero 24 Division of Rheumatology, Department of Medicine Mount Sinai Hospital/University Health Network, University of Toronto, Toronto, Ontario, Canada; and Instituto Nacional de Ciencias Médicas y Nutrición Salvador Zubirán, Mexico City, Mexico Find articles by Jorge Sanchez-Guerrero 24, Matthias Schneider Matthias Schneider 25 Policlinic and Hiller Research Unit for Rheumatology, Medical Faculty, Heinrich-Heine-University, Düsseldorf, Germany Find articles by Matthias Schneider 25, Murray B Urowitz Murray B Urowitz 26 Division of Rheumatology, Department of Medicine, Toronto Western Hospital, University of Toronto, Toronto, Ontario, Canada Find articles by Murray B Urowitz 26, George Bertsias George Bertsias 27 Rheumatology, Clinical Immunology and Allergy, University of Crete Medical School, Heraklion, Greece Find articles by George Bertsias 27, Bimba F Hoyer Bimba F Hoyer 28 Charité – Universitätsmedizin Berlin, Corporate member of Freie Universität Berlin, Humboldt-Universität zu Berlin, and Berlin Institute of Health, Department of Rheumatology and Clinical Immunology, Berlin, Germany and University of Schleswig-Holstein at Kiel, Kiel, Germany Find articles by Bimba F Hoyer 28, Nicolai Leuchten Nicolai Leuchten 29 University Medical Center and Faculty of Medicine Carl Gustav Carus, TU Dresden, Dresden, Germany Find articles by Nicolai Leuchten 29, Chiara Tani Chiara Tani 30 Rheumatology Unit, Azienda Ospedaliero Universitaria Pisana, University of Pisa, Pisa, Italy Find articles by Chiara Tani 30, Sara K Tedeschi Sara K Tedeschi 31 Brigham and Women’s Hospital, Harvard Medical School, Boston, MA, USA Find articles by Sara K Tedeschi 31, Zahi Touma Zahi Touma 32 Division of Rheumatology, Department of Medicine, Toronto Western Hospital, Institute of Health Policy, Management and Evaluation, University of Toronto, Toronto, Ontario, Canada Find articles by Zahi Touma 32, Gabriela Schmajuk Gabriela Schmajuk 33 University of California at San Francisco and the VA Medical Center, San Francisco, USA Find articles by Gabriela Schmajuk 33, Branimir Anic Branimir Anic 34 Division of Clinical Immunology and Rheumatology, University of Zagreb School of Medicine and University Hospital Centre Zagreb, Zagreb, Croatia Find articles by Branimir Anic 34, Florence Assan Florence Assan 35 Université Paris Sud, Hôpitaux Universitaires Paris-Sud, AP-HP, INSERM UMR 1184, Le Kremlin-Bicêtre, France Find articles by Florence Assan 35, Daniel Tak Mao Chan Daniel Tak Mao Chan 36 University of Hong Kong, Hong Kong Find articles by Daniel Tak Mao Chan 36, Ann E Clarke Ann E Clarke 37 Division of Rheumatology, Cumming School of Medicine, University of Calgary, Calgary, Alberta, Canada Find articles by Ann E Clarke 37, Mary K Crow Mary K Crow 38 Hospital for Special Surgery, New York, NY, USA Find articles by Mary K Crow 38, László Czirják László Czirják 39 University of Pécs Medical School, Pécs, Hungary Find articles by László Czirják 39, Andrea Doria Andrea Doria 40 Rheumatology Unit, Department of Medicine (DIMED), University of Padova, Padova, Italy Find articles by Andrea Doria 40, Winfried B Graninger Winfried B Graninger 41 Medical University of Graz, Graz, Austria Find articles by Winfried B Graninger 41, Bernadett Halda-Kiss Bernadett Halda-Kiss 42 University of Pécs Medical School, Pécs, Hungary Find articles by Bernadett Halda-Kiss 42, Sarfaraz Hasni Sarfaraz Hasni 43 NIAMS, NIH, Bethesda, MD Find articles by Sarfaraz Hasni 43, Peter Izmirly Peter Izmirly 44 New York University School of Medicine, New York, New York, USA Find articles by Peter Izmirly 44, Michelle Jung Michelle Jung 45 University of Calgary, Calgary, Alberta, Canada Find articles by Michelle Jung 45, Gábor Kumánovics Gábor Kumánovics 46 University of Pécs Medical School, Pécs, Hungary Find articles by Gábor Kumánovics 46, Xavier Mariette Xavier Mariette 47 Université Paris Sud, Hôpitaux Universitaires Paris-Sud, AP-HP, INSERM UMR 1184, Le Kremlin-Bicêtre, France Find articles by Xavier Mariette 47, Ivan Padjen Ivan Padjen 48 Division of Clinical Immunology and Rheumatology, University of Zagreb School of Medicine and University Hospital Centre Zagreb, Zagreb, Croatia Find articles by Ivan Padjen 48, José M Pego-Reigosa José M Pego-Reigosa 49 Department of Rheumatology, University Hospital of Vigo, IRIDIS Group, Instituto de Investigación Sanitaria Galicia Sur (IISGS), Vigo, Spain Find articles by José M Pego-Reigosa 49, Juanita Romero-Díaz Juanita Romero-Díaz 50 Department of Immunology and Rheumatology, Instituto Nacional de Ciencias Médicas y Nutrición Salvador Zubirán, Mexico City, Mexico Find articles by Juanita Romero-Díaz 50, Iñigo Rúa-Figueroa Fernández Iñigo Rúa-Figueroa Fernández 51 Hospital Dr Negrin, Las Palmas, Spain Find articles by Iñigo Rúa-Figueroa Fernández 51, Raphaèle Seror Raphaèle Seror 52 Université Paris Sud, Hôpitaux Universitaires Paris-Sud, AP-HP, INSERM UMR 1184, Le Kremlin-Bicêtre, France Find articles by Raphaèle Seror 52, Georg Stummvoll Georg Stummvoll 53 Medical University of Vienna, Vienna, Austria Find articles by Georg Stummvoll 53, Yoshiya Tanaka Yoshiya Tanaka 54 University of Occupational & Environmental Health, Kitakyushu, Japan Find articles by Yoshiya Tanaka 54, Maria G Tektonidou Maria G Tektonidou 55 Medical School, National and Kapodistrian University of Athens, Athens, Greece Find articles by Maria G Tektonidou 55, Carlos Vasconcelos Carlos Vasconcelos 56 Centro Hospitalar do Porto, ICBAS, University of Porto, Porto, Portugal Find articles by Carlos Vasconcelos 56, Edward M Vital Edward M Vital 57 Leeds Institute of Rheumatic and Musculoskeletal Medicine, University of Leeds; NIHR Leeds Biomedical Research Centre, Leeds Teaching Hospitals NHS Trust, Leeds, United Kingdom Find articles by Edward M Vital 57, Daniel J Wallace Daniel J Wallace 58 Cedars-Sinai, Los Angeles, CA, USA Find articles by Daniel J Wallace 58, Sule Yavuz Sule Yavuz 59 Istanbul Bilim University, Istanbul, Turkey Find articles by Sule Yavuz 59, Pier Luigi Meroni Pier Luigi Meroni 60 Clinical Immunology and Rheumatology Unit, IRCCS Istituto Auxologico Italiano, Milan, Italy Find articles by Pier Luigi Meroni 60, Marvin J Fritzler Marvin J Fritzler 61 Faculty of Medicine, Cumming School of Medicine, University of Calgary, Calgary, AB, Canada Find articles by Marvin J Fritzler 61, Ray P Naden Ray P Naden 62 Department of Medicine, McMaster University, Hamilton, Ontario, Canada Find articles by Ray P Naden 62, Thomas Dörner Thomas Dörner 63 Charité – Universitätsmedizin Berlin, Corporate member of Freie Universität Berlin, Humboldt-Universität zu Berlin, and Berlin Institute of Health, Department of Rheumatology and Clinical Immunology, Berlin, Germany Find articles by Thomas Dörner 63, Sindhu R Johnson Sindhu R Johnson 64 Division of Rheumatology, Department of Medicine, Toronto Western Hospital, Mount Sinai Hospital; Institute of Health Policy, Management and Evaluation, University of Toronto, Toronto, Ontario, Canada Find articles by Sindhu R Johnson 64 Author information Article notes Copyright and License information 1 University Medical Center and Faculty of Medicine Carl Gustav Carus, TU Dresden, Dresden, Germany 2 Brigham and Women’s Hospital, Harvard Medical School, Boston, MA, USA 3 University of California at San Francisco and VA Medical Center, San Francisco, CA, USA 4 Policlinic and Hiller Research Unit for Rheumatology, Medical Faculty, Heinrich-Heine-University Duesseldorf, Duesseldorf, Germany 5 Rheumatology Unit, Azienda Ospedaliero Universitaria Pisana, University of Pisa, Pisa, Italy 6 Northwestern University Feinberg School of Medicine, Chicago, IL, USA 7 Medical University of Vienna, Austria 8 Russell/Engleman Rheumatology Research Center, University of California at San Francisco, San Francisco, USA 9 Medical School, National and Kapodestrian University of Athens, and Biomedical Research Foundation of the Athens Academy, Athens, Greece; Medical School, University of Cyprus, Nicosia, Cyprus 10 Medical University of South Carolina, Charleston, SC, USA 11 Department of Medicine, University of Cambridge, United Kingdom 12 Department of Autoimmune Diseases, Hospital Clínic, University of Barcelona, Barcelona, Catalonia, Spain 13 Cochin Hospital, Internal Medicine Department, Centre de référence maladies auto-immunes et systémiques rares d’île de France, Paris, France ; Université Paris Descartes-Sorbonne Paris Cité, Paris, France ; INSERM U 1153, Center for Epidemiology and Statistics Sorbonne Paris Cité (CRESS), Paris, France 14 Feinstein Institute, Manhasset, NY, United States 15 Division of Rheumatology, Department of Medicine, Toronto Western Hospital, University of Toronto, Toronto, Ontario, Canada 16 University of California at Los Angeles, Los Angeles, CA, USA 17 Charité – Universitätsmedizin Berlin, Corporate member of Freie Universität Berlin, Humboldt-Universität zu Berlin, and Berlin Institute of Health, Department of Rheumatology and Clinical Immunology, Berlin, Germany 18 Copenhagen Lupus and Vasculitis Clinic, Rigshospitalet, Copenhagen University Hospital, Copenhagen, Denmark 19 University of Michigan, Ann Arbor, MI, USA 20 Lupus Europe, co-opted trustee for research, Essex, UK 21 Brigham and Women’s Hospital, Boston MA; Harvard Medical School, Boston, USA 22 University of Michigan, Ann Arbor, MI, USA 23 Autoimmune Diseases Research Unit, Department of Internal Medicine, Biocruces Bizkaia Health Research Institute, Hospital Universitario Cruces, UPV/EHU, Bizkaia, The Basque Country, Spain 24 Division of Rheumatology, Department of Medicine Mount Sinai Hospital/University Health Network, University of Toronto, Toronto, Ontario, Canada; and Instituto Nacional de Ciencias Médicas y Nutrición Salvador Zubirán, Mexico City, Mexico 25 Policlinic and Hiller Research Unit for Rheumatology, Medical Faculty, Heinrich-Heine-University, Düsseldorf, Germany 26 Division of Rheumatology, Department of Medicine, Toronto Western Hospital, University of Toronto, Toronto, Ontario, Canada 27 Rheumatology, Clinical Immunology and Allergy, University of Crete Medical School, Heraklion, Greece 28 Charité – Universitätsmedizin Berlin, Corporate member of Freie Universität Berlin, Humboldt-Universität zu Berlin, and Berlin Institute of Health, Department of Rheumatology and Clinical Immunology, Berlin, Germany and University of Schleswig-Holstein at Kiel, Kiel, Germany 29 University Medical Center and Faculty of Medicine Carl Gustav Carus, TU Dresden, Dresden, Germany 30 Rheumatology Unit, Azienda Ospedaliero Universitaria Pisana, University of Pisa, Pisa, Italy 31 Brigham and Women’s Hospital, Harvard Medical School, Boston, MA, USA 32 Division of Rheumatology, Department of Medicine, Toronto Western Hospital, Institute of Health Policy, Management and Evaluation, University of Toronto, Toronto, Ontario, Canada 33 University of California at San Francisco and the VA Medical Center, San Francisco, USA 34 Division of Clinical Immunology and Rheumatology, University of Zagreb School of Medicine and University Hospital Centre Zagreb, Zagreb, Croatia 35 Université Paris Sud, Hôpitaux Universitaires Paris-Sud, AP-HP, INSERM UMR 1184, Le Kremlin-Bicêtre, France 36 University of Hong Kong, Hong Kong 37 Division of Rheumatology, Cumming School of Medicine, University of Calgary, Calgary, Alberta, Canada 38 Hospital for Special Surgery, New York, NY, USA 39 University of Pécs Medical School, Pécs, Hungary 40 Rheumatology Unit, Department of Medicine (DIMED), University of Padova, Padova, Italy 41 Medical University of Graz, Graz, Austria 42 University of Pécs Medical School, Pécs, Hungary 43 NIAMS, NIH, Bethesda, MD 44 New York University School of Medicine, New York, New York, USA 45 University of Calgary, Calgary, Alberta, Canada 46 University of Pécs Medical School, Pécs, Hungary 47 Université Paris Sud, Hôpitaux Universitaires Paris-Sud, AP-HP, INSERM UMR 1184, Le Kremlin-Bicêtre, France 48 Division of Clinical Immunology and Rheumatology, University of Zagreb School of Medicine and University Hospital Centre Zagreb, Zagreb, Croatia 49 Department of Rheumatology, University Hospital of Vigo, IRIDIS Group, Instituto de Investigación Sanitaria Galicia Sur (IISGS), Vigo, Spain 50 Department of Immunology and Rheumatology, Instituto Nacional de Ciencias Médicas y Nutrición Salvador Zubirán, Mexico City, Mexico 51 Hospital Dr Negrin, Las Palmas, Spain 52 Université Paris Sud, Hôpitaux Universitaires Paris-Sud, AP-HP, INSERM UMR 1184, Le Kremlin-Bicêtre, France 53 Medical University of Vienna, Vienna, Austria 54 University of Occupational & Environmental Health, Kitakyushu, Japan 55 Medical School, National and Kapodistrian University of Athens, Athens, Greece 56 Centro Hospitalar do Porto, ICBAS, University of Porto, Porto, Portugal 57 Leeds Institute of Rheumatic and Musculoskeletal Medicine, University of Leeds; NIHR Leeds Biomedical Research Centre, Leeds Teaching Hospitals NHS Trust, Leeds, United Kingdom 58 Cedars-Sinai, Los Angeles, CA, USA 59 Istanbul Bilim University, Istanbul, Turkey 60 Clinical Immunology and Rheumatology Unit, IRCCS Istituto Auxologico Italiano, Milan, Italy 61 Faculty of Medicine, Cumming School of Medicine, University of Calgary, Calgary, AB, Canada 62 Department of Medicine, McMaster University, Hamilton, Ontario, Canada 63 Charité – Universitätsmedizin Berlin, Corporate member of Freie Universität Berlin, Humboldt-Universität zu Berlin, and Berlin Institute of Health, Department of Rheumatology and Clinical Immunology, Berlin, Germany 64 Division of Rheumatology, Department of Medicine, Toronto Western Hospital, Mount Sinai Hospital; Institute of Health Policy, Management and Evaluation, University of Toronto, Toronto, Ontario, Canada ✉ Corresponding Authors: Martin Aringer MD, Division of Rheumatology, Department of Medicine III, University Medical Center TU Dresden, Fetscherstrasse 74, 01307 Dresden, Germany. Phone +49 351 458 4422 Fax +49 351 458 5801 martin.aringer@uniklinikum-dresden.de; Sindhu R. Johnson MD PhD, Division of Rheumatology, Ground Floor, East Wing, Toronto Western Hospital, 399 Bathurst Street, Toronto, Ontario, Canada, M5T 2S8. Phone 1-416-603-6417 Fax.1-416-603-4348. Sindhu.Johnson@uhn.ca Issue date 2019 Sep. PMC Copyright notice PMCID: PMC6827566 NIHMSID: NIHMS1029532 PMID: 31385462 The publisher's version of this article is available at Arthritis Rheumatol Abstract Objective. To develop new classification criteria for systemic lupus erythematosus (SLE) jointly supported by the European League Against Rheumatism (EULAR) and the American College of Rheumatology (ACR). Methods. This international initiative had 4 phases: 1) Evaluation of anti-nuclear antibody (ANA) as an entry criterion through systematic review and meta-regression of the literature and criteria generation through an international Delphi exercise, an early patient cohort and a patient survey. 2) Criteria reduction by Delphi and nominal group technique (NGT) exercises. 3) Criteria definition and weighting based on criterion performance and on results of a multi-criteria decision analysis. 4) Refinement of weights and threshold scores in a new derivation cohort of 1001 subjects and validation compared to previous criteria in a new validation cohort of 1270 subjects. Results. The 2019 EULAR/ACR classification criteria for SLE include positive ANA at least once as obligatory entry criterion; followed by additive weighted criteria grouped in 7 clinical (constitutional, hematologic, neuropsychiatric, mucocutaneous, serosal, musculoskeletal, renal) and 3 immunological (antiphospholipid antibodies, complement proteins, SLE-specific antibodies) domains, and weighted from 2 to 10. Patients accumulating ≥10 points are classified. In the validation cohort, the new criteria had a sensitivity of 96.1% and specificity of 93.4%, compared to 82.8% sensitivity and 93.4% specificity of the ACR 1997 and 96.7% sensitivity and 83.7% specificity of the Systemic Lupus International Collaborating Clinics (SLICC) 2012 criteria. Conclusion. These new classification criteria were developed using rigorous methodology with multidisciplinary and international input, and have excellent sensitivity and specificity. Use of ANA entry criterion, hierarchically clustered and weighted criteria reflect current thinking about SLE and provide an improved foundation for SLE research. Keywords: Systemic lupus erythematosus, lupus, classification criteria, consensus methods, multi-criteria decision analysis, validation INTRODUCTION Systemic lupus erythematosus (SLE) is a complex autoimmune disease with variable clinical features (1;2). SLE manifestations are associated with multiple autoantibodies, ensuing immune complex formation and deposition, and other immune processes (2;3). This complex clinical presentation and pathogenesis makes SLE a difficult disease to grasp and define. Classification criteria are essential for the identification of relatively homogeneous groups of patients for inclusion in research studies and trials (4;5). The 1982 revised American College of Rheumatology (ACR) SLE classification criteria (6) and their 1997 revision (7) have been used worldwide. Since then, our understanding of the disease has advanced. Additional specific skin manifestations were described, some clinical symptoms were better understood, and immunological tests, such as diminished levels of serum complement components C3 and C4 or testing for anti-β2 glycoprotein I antibodies, entered routine clinical practice. Better understanding of organ system involvement, such as mucocutaneous abnormalities, led to questions about whether some of the independently counted criteria were in fact manifestations of the same phenomenon (8). The 2012 Systemic Lupus International Collaborating Clinics (SLICC) classification criteria addressed many of these issues (9). Mucocutaneous and neuropsychiatric manifestations were added, as were hypocomplementemia and new anti-phospholipid antibody tests; and criteria definitions were refined. The SLICC criteria emphasized that SLE is primarily an autoantibody disease, requiring at least one immunological criterion to be present, and categorized histology-proven nephritis compatible with SLE as sufficient for classification, if anti-nuclear antibodies (ANA) or antibodies to double-stranded DNA (dsDNA) were present. While achieving their goal of increasing sensitivity, the SLICC criteria have lower specificity than the 1997 ACR criteria (9;10). Existing SLE classification criteria perform better in patients with longstanding disease than in new-onset SLE (13), and there is an increasing recognition and demand that subjects with early SLE should be included in clinical studies and trials. We therefore attempted to enrich our sample populations for early SLE in several phases of the project. In parallel with improved understanding of SLE, the field of classification criteria development has also seen advances (4;14–16). In order to minimize investigator bias, it is now recommended that the cohorts in which the criteria are tested are from independent centers (4). Other methodologic recommendations include a balanced use of both expert-based and data-driven methods, and inclusion of the patient perspective (15;16). The approach chosen for these 2019 EULAR/ACR SLE classification criteria was specifically designed to maintain this balance and to uphold rigorous methodology. METHODS Methodologic overview. Using a methodological approach based on measurement science the criteria were developed in four phases (10): 1) criteria generation, 2) criteria reduction, 3) criteria definition and weighting and 4) refinement and validation (Figure 1). The whole initiative was overseen by a 12-member steering committee (MA, KHC, DD, MM, RR-G, JSS, DW, DB, DK, DJ, TD and SRJ) nominated by EULAR and the ACR in equal numbers, based on SLE and/or methodological experience and previous involvement in international projects. Figure 1. Open in a new tab Development and validation of SLE classification criteria The current project, jointly supported by the European League Against Rheumatism (EULAR) and the ACR, was originally based on two key concepts. One, we hypothesized that the presence of ANA would be better employed as an entry criterion than as a classification criterion (10). Such an approach was thought to reflect underlying SLE pathogenesis, and take into account ANA test characteristics of high sensitivity and limited specificity. Two, we expected individual criteria would not be of equal utility (weight) for the classification of SLE (11), for example mucosal ulcers vs. biopsy-proven lupus nephritis. Accordingly, the validity of using positive ANA as an entry criterion was explicitly addressed in phase 1 of the current activity (12). Likewise, methodologic strategies to develop weighted criteria were used. Phase 1. Criteria generation. The purpose of Phase 1 was to test ANA as a potential entry criterion and identify candidate criteria that should be considered for SLE classification using both data- and expert-based methods, including the patient perspective. Phase 1a comprised a systematic literature review of Medline, Embase and the Cochrane databases with meta-regression to evaluate the operating characteristics of ANA testing for consideration as an entry criterion (12). Phase 1b consisted of a Delphi exercise of international SLE experts from the Americas, Europe and Asia (17). These experts included rheumatologists, dermatologists, nephrologists, pediatricians and non-clinical SLE researchers, providing a broad perspective. The Delphi participants were asked to nominate a broad set of items potentially useful in the classification of SLE (17). In round 2 and 3, participants rated the items from 1 (not at all appropriate) to 9 (completely appropriate) for classification of SLE. Criteria were retained if they reached a median rating of ≥6.5; i.e. at least 50% of the ratings in the high range (7, 8 or 9). Participants were also asked about the importance of ANA and histopathology for classification of SLE. Phase 1c established an international cohort of patients with early SLE or conditions mimicking SLE to identify criteria that may discriminate subjects with early (less than 12 months) disease (18). Phase 1d comprised a cross-sectional survey of SLE patients, administered via the quarterly journal of the German SLE patient organization, which asked about symptoms within one year before and after the patient’s diagnosis of SLE (19). While at a risk of recall bias and not necessarily representative of other regions worldwide, this survey was done to explicitly take a patient standpoint into account. For phase 2 and 3, additional renowned European and North American SLE experts were nominated by the steering committee and invited to participate. Phase 2. Criteria reduction. Phase 2a. The objective of this phase was to select a set of criteria from Phase 1 that maximized the likelihood of accurate classification of SLE, particularly of early disease. An independent panel of 7 of the international SLE experts (RC, NC-C, DDG, BHH, FH, EM, JS-G) ranked the candidate criteria from phase 1. A consensus meeting of 19 international SLE experts (n=7 nominal group technique (NGT) experts + steering committee + DK [moderator]) using NGT was conducted to reduce the list of criteria (20). Data for each candidate criterion were reviewed and discussed until consensus was achieved. The NGT experts voted on items to be retained. Phase 2b. NGT participants pointed out that some criteria could be correlated. With the idea of potentially clustering criteria into domains, associations between candidate criteria were evaluated separately in two cohorts, the phase 1c early SLE and the Euro-lupus cohorts (21). Phase 3. Criteria definition and weighting. Phase 3a. The operating characteristics of the retained candidate criteria were evaluated by literature review. Candidate criteria were hierarchically organized into clinical and immunological domains, and definitions for the candidate criteria were iteratively refined. SLE patient advocates participated in the review of data and the steering committee discussions (22). Phase 3b. 164 case vignettes reflecting broad SLE clinical presentation were sampled from SLE centers across several countries. A panel of 6 of the international experts not involved in earlier phases of the project (BD, SJ, WJMcC, GR-I, MS, MBU) and 11 members of the steering committee assessed and ranked a representative sample of the cases. Subsequently, at a face-to-face meeting, this panel of 17 international SLE experts iteratively compared pairs of criteria, using multi-criteria decision analysis facilitated by 1000minds software (23). The panel unanimously agreed to further reduce the list of criteria. Based on the results, provisional criteria weights were assigned and a provisional threshold score for classification was determined as the lowest score at which the expert panel had achieved consensus on classifying a case vignette as SLE (24). Phase 4. Refinement and validation. International SLE experts not involved in phase 2 or phase 3 panels were asked to contribute cases diagnosed as SLE and controls with conditions mimicking SLE sampled from patients evaluated at their centers. Each center was asked to contribute up to 100 cases and an equal number of controls, preferentially sampling those with early disease, and regardless of their specific clinical or immunological manifestations. Pseudonymized data on the criteria were collected using a standardized data collection form. Ethics committee approval and informed consent were obtained as per local requirements. The status (“SLE” or not) of each case underwent independent adjudication by three of four SLE experts (GB, BFH, NL, CT) from different centers. Queries were sent back to the submitting investigator for clarification. Of this cohort, 501 SLE and 500 control subjects were randomly selected to comprise the derivation cohort, while the remaining 696 SLE and 574 control subjects formed the validation cohort. Refinement. The performance of the draft criteria set was iteratively tested in the derivation cohort. A data-driven threshold for classification was determined by receiver operating characteristics (ROC) analysis and compared to the provisional expert-based consensus threshold. The data of SLE subjects below the threshold (misclassified) were reviewed for groups of patients with unequivocal SLE who still missed classification, and criteria weights adjusted slightly, while preserving the weighting hierarchy (Details below in Results Phase IV). Sensitivity and specificity was tested against the ACR 1997 and the SLICC 2012 criteria. In addition, ANA as an entry criterion was tested against not having an entry criterion. Finally, the criteria weights were simplified to whole numbers. Refinements to the criteria set were presented to the steering committee and Phase III expert panel, and unanimously endorsed. Validation. The sensitivity and specificity of the final criteria were tested in the validation cohort and compared to previous SLE criteria sets. Statistical analysis. Descriptive statistics were used to summarize the data. Confidence intervals were calculated using the bias-corrected and accelerated bootstrap method (BCa method) with B = 2000 bootstrap samples. The BCa method resamples the input data B times (with replacement) and calculates the required statistics (sensitivity, specificity, AUC). Based on the B bootstraps samples, the bias-correction is applied and the associated 95% confidence intervals for the statistics are estimated. The BCa method has proven to yield very accurate coverage of estimated confidence intervals (25). The number B of bootstrap resamples is recommended to be at least B = 1000. We have chosen B = 2000 and additionally checked if B = 5000 bootstraps changed the estimated confidence bounds, which was not the case. Statistical analyses were performed using R, version 3.4.0 (The R Foundation of Statistical Computing). RESULTS Phase 1. Criteria generation. Phase 1a. ANA as an entry criterion. A systematic review of MEDLINE, EMBASE and the Cochrane database identified 13,080 patients from 64 studies reporting ANA by immunofluorescence on HEp-2 cells. Meta-regression of the operating characteristics of ANA found a sensitivity of 97.8% [95% confidence interval (CI) 96.8% – 98.5%] for ANA of ≥1:80 supporting use of ANA as an entry criterion (12). Since some SLE centers do not have access to HEp-2 ANA, and in view of ongoing work on the standardization of serology and potential future advances in the field, the steering committee and additional autoantibody consultants (MJF, PLM) recommended the provision “or an equivalent positive ANA test. Testing by immunofluorescence on HEp-2 cells or a solid phase ANA screening immunoassay with at least equivalent performance is highly recommended.” Phase 1b. Delphi exercise. One hundred and forty-seven international SLE experts nominated 145 candidate criteria (17). By rating the appropriateness for SLE classification, the participants in the second and third Delphi round reduced the list to 40 candidate criteria (Supplementary Table 1). Phase 1c. International early SLE cohort. The cohort comprised 616 subjects who had been referred for possible SLE with a disease duration of less than one year (n=389 early SLE and n=227 mimicking diseases) from North America, Europe, Asia and South America (18). In addition to supporting many of the 40 candidate criteria derived from the Delphi exercise, the comparison between early SLE and non-SLE patients showed that fever occurred more frequently (34.5% versus 13.7%, p<0.001) in SLE, while SLE patients less commonly suffered from arthralgias (20.3% versus 42.7%, p=0.001) and fatigue (28.3% versus 37%, p=0.02). Phase 1d. Patient survey. 339 SLE patients (>99% Caucasian, 93% female) responded to the survey (19). More than half of these patients reported mucocutaneous findings in the first year of their disease (Supplementary Table 1), but also fatigue (89%), joint pain (87%) and fever (54%)(19). Given that these items were highlighted both in the early SLE cohort and the patient survey, fever, fatigue and arthralgias were forwarded to the next phase in addition to the 40 Delphi items. Accordingly, phases 1a-1d resulted in a total of 43 candidate criteria for consideration (Supplementary Table 1). Phase 2. Criteria reduction. Phase 2a. The expert panel NGT exercise reduced the candidate criteria from 43 to 21 (26). The panel distinguished potential “entry criteria”, which would be required for classification, from potential “additive criteria”. They endorsed “positive ANA (≥1:80 by HEp-2 immunofluorescence)” as an entry criterion. The 20 remaining additive criteria included: lupus nephritis by renal biopsy, autoantibodies, cytopenias, fever, arthritis, serositis, mucocutaneous and neuropsychiatric manifestations (Supplementary Table 1). Phase 2b. Associations between the candidate criteria were evaluated in 389 subjects in the early SLE cohort and the 1000 SLE subjects of the Euro-lupus cohort. Modest statistically significant correlations were limited to the mucocutaneous (r=0.22 to 0.30), neurologic (r=0.22) and immunological (r=0.33) domains in the early SLE cohort, and this modest correlation was replicated in the Euro-lupus cohort (21). Given these associations, criteria were clustered within domains, so that only one criterion within each domain would be counted. Phase 3. Criteria definition and weighting Phase 3a. Based on the literature, definitions of the 20 candidate additive criteria were refined, using a data-driven evaluation of operating characteristics (22), retaining only feasible items with a prevalence of at least 1% according to literature. Literature-review led to the consensus decision to evaluate five different candidate criteria within the neuropsychiatric domain (delirium, psychosis, seizure, mononeuropathy, cranial neuropathy) and potential separation of acute pericarditis from pleural or pericardial effusions; and between diminished C3 or C4 versus diminished C3 and C4 (Supplementary Table 1). The resulting 23 candidate criteria (Supplementary Table 1) were organized into seven clinical and three immunologic domains, with hierarchical clustering (22). Only the highest-ranking item in each domain was to be counted. Instead of devising exclusion definitions for each criterion, the decision was made to attribute any item to SLE only if no more likely explanation was present. For leukopenia and joint involvement, it was decided to formally test alternative definitions in the derivation cohort. Given the importance of testing for antibodies, particularly for anti-dsDNA, for which tests of relatively low specificity are in use, great care was taken to precisely define testing (Table 1). Table 1. Definitions of SLE classification criteria. \| Criteria \| Definition \| :---: \| \| Antinuclear antibodies (ANA) \| Antinuclear antibodies (ANA) at a titer of ≥1:80 on HEp-2 cells or an equivalent positive test at least once. Testing by immunofluorescence on HEp-2 cells or a solid phase ANA screening immunoassay with at least equivalent performance is highly recommended. \| \| Fever \| Temperature >38.3° Celsius. \| \| Leukopenia \| White blood cell count <4,000/mm³. \| \| Thrombocytopenia \| Platelet count <100,000/mm³. \| \| Autoimmune hemolysis \| Evidence of hemolysis, such as reticulocytosis, low haptoglobin, elevated indirect bilirubin, elevated LDH AND positive Coomb’s (direct antiglobulin) test. \| \| Delirium \| Characterized by (1) change in consciousness or level of arousal with reduced ability to focus, and (2) symptom development over hours to <2 days, and (3) symptom fluctuation throughout the day, and (4) either (4a) acute/subacute change in cognition (e.g. memory deficit or disorientation), or (4b) change in behavior, mood, or affect (e.g. restlessness, reversal of sleep/wake cycle). \| \| Psychosis \| Characterized by (1) delusions and/or hallucinations without insight and (2) absence of delirium. \| \| Seizure \| Primary generalized seizure or partial/focal seizure. \| \| Non-scarring alopecia \| Non-scarring alopecia observed by a clinician. \| \| Oral ulcers \| Oral ulcers observed by a clinician. \| \| Subacute cutaneous or discoid lupus \| Subacute cutaneous lupus erythematosus observed by a clinician: Annular or papulosquamous (psoriasiform) cutaneous eruption, usually photodistributed. Discoid lupus erythematosus observed by a clinician: Erythematous-violaceous cutaneous lesions with secondary changes of atrophic scarring, dyspigmentation, often follicular hyperkeratosis/ plugging (scalp), leading to scarring alopecia on the scalp. If skin biopsy is performed, typical changes must be present. Subacute cutaneous lupus: interface vacuolar dermatitis consisting of a perivascular lymphohistiocytic infiltrate, often with dermal mucin noted. Discoid lupus: interface vacuolar dermatitis consisting of a perivascular and/or periappendageal lymphohistiocytic infiltrate. In the scalp, follicular keratin plugs may be seen. In longstanding lesions, mucin deposition and basement membrane thickening may be noted. \| \| Acute cutaneous lupus \| Malar rash or generalized maculopapular rash observed by a clinician. If skin biopsy is performed, typical changes must be present (Acute cutaneous lupus: interface vacuolar dermatitis consisting of a perivascular lymphohistiocytic infiltrate, often with dermal mucin noted. Perivascular neutrophilic infiltrate may be present early in the course. \| \| Pleural or pericardial effusion \| Imaging evidence (such as ultrasound, x-ray, CT scan, MRI) of pleural or pericardial effusion, or both. \| \| Acute pericarditis \| ≥2 of (1) pericardial chest pain (typically sharp, worse with inspiration, improved by leaning forward), (2) pericardial rub, (3) EKG with new widespread ST-elevation or PR depression, (4) new or worsened pericardial effusion on imaging (such as ultrasound, x-ray, CT scan, MRI). \| \| Joint involvement \| EITHER (1) synovitis involving 2 or more joints characterized by swelling or effusion OR (2) tenderness in 2 or more joints and at least 30 minutes of morning stiffness. \| \| Proteinuria >0.5g/24h \| Proteinuria >0.5g/24h by 24 hour urine or equivalent spot urine protein-to-creatinine ratio. \| \| Class II or V lupus nephritis on renal biopsy according to ISN/RPS 2003 classification. \| Class II: Mesangial proliferative lupus nephritis: Purely mesangial hypercellularity of any degree or mesangial matrix expansion by light microscopy, with mesangial immune deposit. A few isolated subepithelial or subendothelial deposits may be visible by immune-fluorescence or electron microscopy, but not by light microscopy. Class V: Membranous lupus nephritis: Global or segmental subepithelial immune deposits or their morphologic sequelae by light microscopy and by immunofluorescence or electron microscopy, with or without mesangial alterations. \| \| Class III or IV lupus nephritis on renal biopsy according to ISN/RPS 2003. \| Class III: Focal lupus nephritis: Active or inactive focal, segmental or global endo- or extracapillary glomerulonephritis involving <50% of all glomeruli, typically with focal subendothelial immune deposits, with or without mesangial alterations. Class IV: Diffuse lupus nephritis: Active or inactive diffuse, segmental or global endo- or extracapillary glomerulonephritis involving ≥50% of all glomeruli, typically with diffuse subendothelial immune deposits, with or without mesangial alterations. This class includes cases with diffuse wire loop deposits but with little or no glomerular proliferation. \| \| Positive anti-phospholipid antibodies \| Anti-Cardiolipin antibodies (IgA, IgG, or IgM) at medium or high titer (>40 APL, GPL or MPL, or >the 99th percentile) or positive anti-β2GP1 antibodies (IgA, IgG, or IgM) or positive lupus anticoagulant. \| \| Low C3 OR low C4 \| C3 OR C4 below the lower limit of normal. \| \| Low C3 AND low C4 \| Both C3 AND C4 below their lower limits of normal. \| \| Anti-dsDNA antibodies OR Anti-Smith (Sm) antibodies. \| Anti-dsDNA antibodies in an immunoassay with demonstrated ≥ 90% specificity for SLE against relevant disease controls OR Anti-Smith (Sm) antibodies. \| Open in a new tab This may include physical examination or review of a photograph. ISN/RPS International Society of Nephrology/Renal Pathology Society Phase 3b. The 1.5 day in-person consensus meeting using multicriteria decision analysis involved 74 decisions between pairs of criteria. Criteria weights were calculated by the 1000minds™ software based on these decisions (Table 2). International Society of Nephrology/Renal Pathology Society class III or IV nephritis consistently attained higher weight than class II or V nephritis, so lupus nephritis by histology was separated into two different criteria. Class VI lupus nephritis as an end stage manifestation was unanimously eliminated. Likewise, the experts unanimously voted to not retain mononeuropathy and cranial neuropathy, which had been included into the set of potential neuropsychiatric items in phase 3a but turned out to add little to SLE classification. The use of weighted criteria led to a sum score that is a measure of the relative probability of a subject having SLE, with higher scores indicating higher likelihood. Experts reached full consensus on a classification of SLE at a provisional threshold score of >83 of a theoretical maximum of 305 (24). Table 2. Relative weights of the additive classification criteria items \| Domain \| Item \| Original \| Modification \| Revised \| Simplified \| :--- :--- :--- \| \| Constitutional \| Fever \| 13 \| \| 13 \| 2 \| \| Hematological \| Leukopenia \| 12 \| +7 \| 19 \| 3 \| \| Thrombocytopenia \| 26 \| \| 26 \| 4 \| \| Autoimmune hemolysis \| 28 \| \| 28 \| 4 \| \| Neuropsychiatric \| Delirium \| 12 \| \| 12 \| 2 \| \| Psychosis \| 20 \| \| 20 \| 3 \| \| Seizure \| 34 \| \| 34 \| 5 \| \| Mucocutaneous \| Alopecia \| 13 \| \| 13 \| 2 \| \| Oral ulcers \| 14 \| \| 14 \| 2 \| \| SCLE/DLE \| 29 \| \| 29 \| 4 \| \| ACLE \| 38 \| \| 38 \| 6 \| \| Serosal \| Effusion \| 34 \| \| 34 \| 5 \| \| Acute pericarditis \| 38 \| \| 38 \| 6 \| \| Musculoskeletal \| Joint involvement \| 34 \| +4 \| 38 \| 6 \| \| Renal \| Proteinuria \| 27 \| \| 27 \| 4 \| \| Class II/V \| 55 \| \| 55 \| 8 \| \| Class III/IV \| 74 \| \| 74 \| 10 \| \| \| \| \| \| \| \| \| APL antibodies \| Anti-phospholipid \| 13 \| \| 13 \| 2 \| \| Complements \| C3 or C4 low \| 19 \| \| 19 \| 3 \| \| C3 and C4 low \| 27 \| \| 27 \| 4 \| \| SLE-specific antibodies \| Anti-Sm \| 40 \| \| 40 \| 6 \| \| Anti-dsDNA \| 38 \| \| 38 \| 6 \| Open in a new tab Weights derived from the phase III consensus meeting with multicriteria decisions analysis (Original), added points for leukopenia and joint involvement (Modification), the resulting weights (Revised) and the final simplified weights (Simplified). Phase 4. Refinement and validation Twenty-one centers from the United States, Canada, Mexico, Austria, Croatia, France, Germany, Greece, Hungary, Italy, Portugal, Spain, the United Kingdom, Turkey, Hong Kong and Japan submitted a total of 2,339 cases from their cohorts. 1,197 SLE and 1,074 non-SLE diagnoses (Table 3) were verified by three adjudicators blinded to the proposed classification criteria system. Due to lack of consensus during adjudication, 68 subjects (2.9%) were excluded from the analysis. Table 3. Demographic characteristics of the derivation and validation cohorts \| \| Derivation cohort \| Validation cohort \| :---: \| \| \| n \| SLE 501 \| Non-SLE 500 \| SLE 696 \| Non-SLE 574 \| \| Female/male \| 447/54 \| 421/79 \| 608/88 \| 490/84 \| \| Age (mean±SD) years \| 45±14 \| 54±16 \| 45±14 \| 56±16 \| \| Disease duration (mean±SD) years \| 11±8 \| 9±8 \| 11±8 \| 9±8 \| \| Ethnicity \| \| \| \| \| \| Black \| 29 \| 10 \| 56 \| 12 \| \| East Asian \| 36 \| 29 \| 53 \| 34 \| \| Hispanic \| 59 \| 48 \| 73 \| 51 \| \| South/South East Asian \| 16 \| 6 \| 21 \| 11 \| \| White \| 355 \| 404 \| 480 \| 461 \| \| Other \| 6 \| 3 \| 13 \| 5 \| \| SLE \| 501 \| \| 696 \| \| \| Non-SLE \| \| 500 \| \| 574 \| \| Adult Onset Still’s disease \| \| 2 \| \| 11 \| \| Autoimmune thyroiditis \| \| 6 \| \| 5 \| \| Behcet’s disease \| \| 7 \| \| 9 \| \| Cancer \| \| 2 \| \| 3 \| \| Inflammatory myositis \| \| 37 \| \| 27 \| \| Fibromyalgia \| \| 6 \| \| 3 \| \| Membranous nephritis \| \| 11 \| \| 14 \| \| Mixed connective tissue disease \| \| 9 \| \| 15 \| \| Osteoarthritis \| \| 2 \| \| 0 \| \| Primary antiphospholipid antibody Syndrome \| \| 45 \| \| 48 \| \| Psoriatic arthritis \| \| 12 \| \| 9 \| \| Rheumatoid arthritis \| \| 94 \| \| 110 \| \| Sarcoidosis \| \| 2 \| \| 2 \| \| Sjögren’s syndrome \| \| 112 \| \| 124 \| \| Spondyloarthritis \| \| 5 \| \| 5 \| \| Systemic sclerosis \| \| 99 \| \| 120 \| \| Tuberculosis \| \| 0 \| \| 2 \| \| Undifferentiated connective tissue disease \| \| 16 \| \| 20 \| \| Vasculitis \| \| 9 \| \| 13 \| \| Viral infection \| \| 5 \| \| 5 \| \| Other \| \| 19 \| \| 29 \| Open in a new tab Inflammatory myositis includes dermatomyositis, polymyositis, and juvenile dermatomyositis SD Standard deviation Derivation cohort. Of the 2,271 triple-adjudicated cases, 501 SLE and 500 non SLE cases were randomly assigned to the derivation cohort. The provisional weighting system derived from phase 3 was tested in the derivation cohort. ROC analysis suggested a data-driven threshold of ≥70 (of a maximum of 305), with a sensitivity of 95.4% and a specificity of 95.2%, which was superior to the consensus-derived provisional threshold of >83 that had high specificity (98.8%), but lower sensitivity (81.6%). Review of subjects below the threshold of 70 identified a subgroup of SLE subjects with joint involvement and/or leukopenia. Thus, weights for leukopenia and joint involvement were each adjusted (Table 2) to reduce misclassification. When alternative definitions for leukopenia and joint involvement were tested, leukopenia defined as a white blood cell count (WBC) <4000/mm 3 at least once (9) also had a slightly higher sensitivity + specificity (1.944 vs. 1.942) than leukopenia defined as WBC <4000/mm 3 on 2 or more occasions (6;26). Joint involvement defined as EITHER “synovitis involving 2 or more joints, characterized by swelling or effusion”, OR “tenderness in 2 or more joints and at least 30 minutes of morning stiffness” (9) had a higher combined sensitivity and specificity than arthritis defined simply as synovitis of 2 or more joints (1.944 vs. 1.900). When re-tested, the revised criteria had increased sensitivity, and maintained sensitivity + specificity. Evaluating ANA as an entry criterion, the criteria with the ANA entry criterion had better performance than without (sensitivity + specificity 1.944 vs. 1.930). Next, the weights were simplified by division to whole numbers to achieve a threshold of 10 (Table 2). In the derivation cohort, the sensitivity and specificity of the final criteria set (Figure 2) were reaching the performance benchmarks set for this project (Table 4). Figure 2. Open in a new tab Classification criteria for systemic lupus erythematosus Note: In an assay with ≥ 90% specificity against relevant disease controls § Additional criteria items within the same domain will not be counted. Table 4. Operating characteristics of the new classification criteria compared to the ACR 1997 and SLICC 2012 classification criteria in the derivation and the validation cohorts. \| \| ACR 1997 criteria \| SLICC 2012 criteria \| EULAR/ACR 2019 criteria \| :---: :---: \| \| Derivation \| \| \| \| \| Sensitivity [95% CI] \| 0.85 [0.81-0.88] \| 0.97 [0.95-0.98] \| 0.98 [0.97-0.99] \| \| Specificity [95% CI] \| 0.95 [0.93-0.97] \| 0.90 [0.87-0.92] \| 0.96 [0.95-0.98] \| \| Combined [95% CI] \| 1.80 [1.76-1.83] \| 1.87 [1.84-1.90] \| 1.94 [1.92-1.96] \| \| Validation \| \| \| \| \| Sensitivity [95% CI] \| 0.83 [0.80-0.85] \| 0.97 [0.95-0.98] \| 0.96 [0.95-0.98] \| \| Specificity [95% CI] \| 0.93 [0.91-0.95] \| 0.84 [0.80-0.87] \| 0.93 [0.91-0.95] \| \| Combined [95% CI] \| 1.76 [1.73-1.80] \| 1.80 [1.77-1.84] \| 1.90 [1.87-1.92] \| Open in a new tab ACR American College of Rheumatology, SLICC Systemic Lupus International Collaborating Clinics, CI Confidence Intervals Validation. The validation cohort, i.e. the full cohort minus the derivation cohort, comprised 1,270 triple adjudicated subjects (n=696 SLE, n= 574 controls). The criteria, with positive ANA as an entry criterion, weighted criteria in seven clinical domains (constitutional, hematologic, neuropsychiatric, mucocutaneous, serosal, musculoskeletal, renal) and three immunological domains (anti-phospholipid antibodies, low complements, anti-Sm and anti-dsDNA as SLE-specific antibodies) and a classification threshold score of ≥10 (out of a theoretical maximum of 51)(Figure 2), had a sensitivity of 96.1% and a specificity of 93.4% (Table 4). It demonstrated improved performance compared to the ACR 1997 and SLICC 2012 criteria. DISCUSSION New SLE classification criteria were developed with support by both the ACR and EULAR. Through a 4-phase, iterative process, we have defined an additive, weighted multi-criteria system that produces a measure of the relative probability that an individual can be classified as SLE. The system defines a threshold above which experts would classify cases as SLE for the purpose of research studies. We have carefully defined the criteria to improve reliability and precision; and have grouped the criteria into ten hierarchical domains. We have validated the criteria against a large number of cases, including many patients with manifestations that resemble SLE but who do not have SLE. This approach, as well as the resulting criteria system, represents a paradigm shift for the classification of SLE. We have defined positive ANA at any time as required entry criterion. There were three possible ways to deal with ANA testing. The previous criteria sets have treated ANA the same as the much more specific antibodies against Sm and dsDNA, which we considered suboptimal given important differences in sensitivity and specificity. We could have excluded ANA completely in classifying lupus, but we still consider ANA a useful test and concept. We therefore decided to test ANA as an entry criterion, which reflects the use of ANA as a highly sensitive screening test. Criteria using ANA as entry criterion had better performance. During the phase 1 Delphi exercise, 58% of SLE experts did not feel comfortable and an additional 19% were uncertain about classifying a patient with SLE in the absence of ever having a positive ANA (17). The systematic literature review and meta-regression of data on 13,080 subjects demonstrated ANA ≥1:80 have a sensitivity of 98% with a lower limit of the 95% confidence interval at 97% (12). In the phase 1 early SLE cohort, 99.5% of the 389 SLE patients were ANA positive (18). The frequencies of ANA positive SLE patients in the derivation and validation cohorts (99.6% and 99.3%, respectively) were in the same range. Since both in the early SLE cohort and in the derivation and validation cohorts, patients were included in many centers worldwide independent of ANA positivity, the latter data provide additional support for ANA as an entry criterion. Using ANA as entry criterion means the new criteria cannot classify SLE among patients who are persistently ANA negative. While possibly also distinguished by lower cytokine levels (27) and lower efficacy of immunomodulatory treatment (28), such a subgroup of patients exists. Although small, it may vary in size in different populations (12). This patient subset needs to be put high on the scientific agenda for further investigation. Additional characterization of this phenomenon may lead to an alternative entry criterion for this small group of patients. For the moment, we still think it is acceptable to exclude ANA negative patients from clinical trials. Molecular classification criteria were also considered during the development of these criteria (29). Many novel biomarkers were nominated, such as increased circulating B lymphocyte stimulator (BLyS), IFNγ induced protein 10 kD (IP-10), monocyte chemoattractant protein-1 (MCP-1), TNF-α, type I interferon signature, or increased Th17 and plasma cell populations. They were all voted out in the expert Delphi exercise, largely because of limited availability in the clinical setting and/or insufficient evidence (5). However, inclusion of novel biomarkers, beyond autoantibodies, may ultimately further improve the specificity of SLE classification, increase alignment of classification with underlying disease pathogenesis and improve the performance and information content of clinical trials. Thus, testing of biomarkers against these criteria is an important area for future research. A new clinical criterion, unexplained fever, turned out to be common and remarkably characteristic for SLE. However, since infections are a major cause of death in SLE, it is of utmost importance to stress that fever, like all other criteria manifestations, should only be counted if no better explanation exists, and that infections have to be suspected first in any patient with (potential) SLE, particularly when CRP is elevated (30). The concept that all criteria are only to be counted if SLE is thought to be the most likely cause of the manifestation (i.e. no other more likely cause exists) is central to these new EULAR/ACR criteria, and is explicitly stated as an overarching principle. Some criteria, such as delirium, psychosis and acute pericarditis, were in part re-defined based on existing scientific definitions (22). Where alternative definitions were used, the performance of the alternative definitions was comparatively evaluated in the derivation cohort. The differential weighting of criteria better represents their relative contribution to an individual’s classification of SLE. For SLE, renal biopsy with Class III or IV lupus nephritis carries the most weight and in the presence of a positive ANA is enough to classify a patient as SLE. This further develops a concept of the SLICC criteria (9) and reflects the current thinking of SLE experts; in the Delphi exercise, 85% would classify SLE on renal pathology alone (17). Renal biopsy with class II or V lupus nephritis still carries a large weight (8 points) but is not by itself sufficient for the classification of SLE. The numerical goal of this project was to keep the specificity similar to the specificity of the ACR 1997 criteria, but increase the sensitivity to the high sensitivity level of the SLICC criteria, if possible. The validation cohort data suggest that this goal has been achieved. From our data, it appears that the SLICC criteria increase in sensitivity was to a significant degree founded in accepting renal histology and adding subacute cutaneous lupus and low complement levels. These three advances are mirrored in the current criteria. Many of the other additional symptoms of the SLICC criteria were of very low frequency. Specificity was increased by weighting of criteria, by the NGT expert panel decision to not allow lymphopenia to go forward, and, importantly, by the decision that no criterion be counted if better explained by another condition. The new criteria provide a simple, directed and highly accurate method for classifying SLE. An electronic ‘app’ is in preparation, which will assist in the use of these criteria. However, it is important to stress that classification criteria are not designed for diagnosis or treatment decisions (5). They should never be used to exclude patients who do not fully meet these criteria from receiving appropriate therapies. This is also pertinent to patients with ANA-negative SLE discussed above. Diagnosis of SLE remains the purview of an appropriately trained physician evaluating an individual patient (5). The new SLE classification system also provides new research opportunities. With much interest in early or latent SLE (31;32), the additive point system and the relative probability of classification it produces, allows for systematic study of individuals who fall below the classification threshold. This will facilitate studies of disease evolution and early intervention. Furthermore, the use of an additive scoring system will allow for studying the idea of ‘ominousity’, i.e. the potential implications of having very high scores on disease severity and subsequent prognosis. This work would need to reconsider the relative contribution of individual criteria (weights) and consider additional criteria that potentially contribute to ominousity. It is anticipated that other groups will test these criteria, which will constitute important external validation. This will be particularly important for pediatric SLE and those with organ dominant, e.g. skin dominant disease, since it is a limitation of this criteria project that the patient cohorts do not represent these subgroups. Similar limitations also pertain to several racial/ethnic groups (for example, African American/Black, Hispanic and Asian patients) and to men with SLE, each only included in lower numbers (Table 3). It is important to independently test the EULAR/ACR criteria in these subgroups. Leukocyte counts, for example, are more frequently below <4000/mm3 in African Americans (33), which may have influence on criteria performance. It is also possible that the academic center patient populations included differ from patients in community practice clinics. Investigators testing the new criteria in different populations are reminded about the critical importance of the correct attribution of each criterion. Criteria can only be counted when not better explained by another condition. The attribution process requires diligence and clinical experience. In summary, our multiphase methodologic approach and ensuing classification system using ANA as an entry criterion and weighted, hierarchically clustered criteria, constitute a paradigm shift in the classification of SLE. These criteria have excellent performance characteristics and face validity, as the structure and weighting were designed to reflect current thinking about SLE. The inclusion of fever assists with the classification of early SLE. The separation of renal biopsy findings reflects their differential impact on the probability of SLE classification. These criteria have strong operating characteristics, with excellent sensitivity and specificity. This classification system was built using rigorous methodology that was both data-driven and expert-based. With the inclusion of over 200 SLE experts from multiple countries and medical disciplines, methodologists, patient advocates and over 4,000 subjects, this work is the largest international, collaborative SLE classification effort to date. Supplementary Material Supp TableS1 NIHMS1029532-supplement-Supp_TableS1.docx (30.1KB, docx) supp info NIHMS1029532-supplement-supp_info.doc (35KB, doc) Acknowledgements. This body of work was jointly supported by the European League Against Rheumatism and the American College of Rheumatology. One part of the derivation and validation cohort was supported by the Intramural Research Program of the National Institute of Arthritis and Musculoskeletal and Skin Diseases of the National Institutes of Health. The authors wish to acknowledge the diligent work of Banita Aggarwal and Keshini Devakandan in data entry, data cleaning, queries to submitting investigators, data cutting and maintenance of the derivation and validation cohorts; and of Corine Sinnette, MA, in the preparatory work for the multicriteria decision analysis exercise. Contributor Information Martin Aringer, University Medical Center and Faculty of Medicine Carl Gustav Carus, TU Dresden, Dresden, Germany. Karen H. Costenbader, Brigham and Women’s Hospital, Harvard Medical School, Boston, MA, USA David I. Daikh, University of California at San Francisco and VA Medical Center, San Francisco, CA, USA Ralph Brinks, Policlinic and Hiller Research Unit for Rheumatology, Medical Faculty, Heinrich-Heine-University Duesseldorf, Duesseldorf, Germany. Marta Mosca, Rheumatology Unit, Azienda Ospedaliero Universitaria Pisana, University of Pisa, Pisa, Italy. Rosalind Ramsey-Goldman, Northwestern University Feinberg School of Medicine, Chicago, IL, USA. Josef S. Smolen, Medical University of Vienna, Austria David Wofsy, Russell/Engleman Rheumatology Research Center, University of California at San Francisco, San Francisco, USA. Dimitrios Boumpas, Medical School, National and Kapodestrian University of Athens, and Biomedical Research Foundation of the Athens Academy, Athens, Greece; Medical School, University of Cyprus, Nicosia, Cyprus. Diane L. Kamen, Medical University of South Carolina, Charleston, SC, USA David Jayne, Department of Medicine, University of Cambridge, United Kingdom. Ricard Cervera, Department of Autoimmune Diseases, Hospital Clínic, University of Barcelona, Barcelona, Catalonia, Spain. Nathalie Costedoat-Chalumeau, Cochin Hospital, Internal Medicine Department, Centre de référence maladies auto-immunes et systémiques rares d’île de France, Paris, France ; Université Paris Descartes-Sorbonne Paris Cité, Paris, France ; INSERM U 1153, Center for Epidemiology and Statistics Sorbonne Paris Cité (CRESS), Paris, France. Betty Diamond, Feinstein Institute, Manhasset, NY, United States. Dafna D. Gladman, Division of Rheumatology, Department of Medicine, Toronto Western Hospital, University of Toronto, Toronto, Ontario, Canada Bevra H. Hahn, University of California at Los Angeles, Los Angeles, CA, USA Falk Hiepe, Charité – Universitätsmedizin Berlin, Corporate member of Freie Universität Berlin, Humboldt-Universität zu Berlin, and Berlin Institute of Health, Department of Rheumatology and Clinical Immunology, Berlin, Germany. Søren Jacobsen, Copenhagen Lupus and Vasculitis Clinic, Rigshospitalet, Copenhagen University Hospital, Copenhagen, Denmark. Dinesh Khanna, University of Michigan, Ann Arbor, MI, USA. Kirsten Lerstrøm, Lupus Europe, co-opted trustee for research, Essex, UK. Elena Massarotti, Brigham and Women’s Hospital, Boston MA; Harvard Medical School, Boston, USA. W. Joseph McCune, University of Michigan, Ann Arbor, MI, USA. Guillermo Ruiz-Irastorza, Autoimmune Diseases Research Unit, Department of Internal Medicine, Biocruces Bizkaia Health Research Institute, Hospital Universitario Cruces, UPV/EHU, Bizkaia, The Basque Country, Spain. Jorge Sanchez-Guerrero, Division of Rheumatology, Department of Medicine Mount Sinai Hospital/University Health Network, University of Toronto, Toronto, Ontario, Canada; and Instituto Nacional de Ciencias Médicas y Nutrición Salvador Zubirán, Mexico City, Mexico. Matthias Schneider, Policlinic and Hiller Research Unit for Rheumatology, Medical Faculty, Heinrich-Heine-University, Düsseldorf, Germany. Murray B. Urowitz, Division of Rheumatology, Department of Medicine, Toronto Western Hospital, University of Toronto, Toronto, Ontario, Canada George Bertsias, Rheumatology, Clinical Immunology and Allergy, University of Crete Medical School, Heraklion, Greece. Bimba F. Hoyer, Charité – Universitätsmedizin Berlin, Corporate member of Freie Universität Berlin, Humboldt-Universität zu Berlin, and Berlin Institute of Health, Department of Rheumatology and Clinical Immunology, Berlin, Germany and University of Schleswig-Holstein at Kiel, Kiel, Germany Nicolai Leuchten, University Medical Center and Faculty of Medicine Carl Gustav Carus, TU Dresden, Dresden, Germany. Chiara Tani, Rheumatology Unit, Azienda Ospedaliero Universitaria Pisana, University of Pisa, Pisa, Italy. Sara K. Tedeschi, Brigham and Women’s Hospital, Harvard Medical School, Boston, MA, USA Zahi Touma, Division of Rheumatology, Department of Medicine, Toronto Western Hospital, Institute of Health Policy, Management and Evaluation, University of Toronto, Toronto, Ontario, Canada. Gabriela Schmajuk, University of California at San Francisco and the VA Medical Center, San Francisco, USA. Branimir Anic, Division of Clinical Immunology and Rheumatology, University of Zagreb School of Medicine and University Hospital Centre Zagreb, Zagreb, Croatia. Florence Assan, Université Paris Sud, Hôpitaux Universitaires Paris-Sud, AP-HP, INSERM UMR 1184, Le Kremlin-Bicêtre, France. Daniel Tak Mao Chan, University of Hong Kong, Hong Kong. Ann E. Clarke, Division of Rheumatology, Cumming School of Medicine, University of Calgary, Calgary, Alberta, Canada Mary K. Crow, Hospital for Special Surgery, New York, NY, USA László Czirják, University of Pécs Medical School, Pécs, Hungary. Andrea Doria, Rheumatology Unit, Department of Medicine (DIMED), University of Padova, Padova, Italy. Winfried B. Graninger, Medical University of Graz, Graz, Austria Bernadett Halda-Kiss, University of Pécs Medical School, Pécs, Hungary. Sarfaraz Hasni, NIAMS, NIH, Bethesda, MD. Peter Izmirly, New York University School of Medicine, New York, New York, USA. Michelle Jung, University of Calgary, Calgary, Alberta, Canada. Gábor Kumánovics, University of Pécs Medical School, Pécs, Hungary. Xavier Mariette, Université Paris Sud, Hôpitaux Universitaires Paris-Sud, AP-HP, INSERM UMR 1184, Le Kremlin-Bicêtre, France. Ivan Padjen, Division of Clinical Immunology and Rheumatology, University of Zagreb School of Medicine and University Hospital Centre Zagreb, Zagreb, Croatia. José M. Pego-Reigosa, Department of Rheumatology, University Hospital of Vigo, IRIDIS Group, Instituto de Investigación Sanitaria Galicia Sur (IISGS), Vigo, Spain Juanita Romero-Díaz, Department of Immunology and Rheumatology, Instituto Nacional de Ciencias Médicas y Nutrición Salvador Zubirán, Mexico City, Mexico. Iñigo Rúa-Figueroa Fernández, Hospital Dr Negrin, Las Palmas, Spain. Raphaèle Seror, Université Paris Sud, Hôpitaux Universitaires Paris-Sud, AP-HP, INSERM UMR 1184, Le Kremlin-Bicêtre, France. Georg Stummvoll, Medical University of Vienna, Vienna, Austria. Yoshiya Tanaka, University of Occupational & Environmental Health, Kitakyushu, Japan. Maria G. Tektonidou, Medical School, National and Kapodistrian University of Athens, Athens, Greece Carlos Vasconcelos, Centro Hospitalar do Porto, ICBAS, University of Porto, Porto, Portugal. Edward M Vital, Leeds Institute of Rheumatic and Musculoskeletal Medicine, University of Leeds; NIHR Leeds Biomedical Research Centre, Leeds Teaching Hospitals NHS Trust, Leeds, United Kingdom. Daniel J. Wallace, Cedars-Sinai, Los Angeles, CA, USA Sule Yavuz, Istanbul Bilim University, Istanbul, Turkey. Pier Luigi Meroni, Clinical Immunology and Rheumatology Unit, IRCCS Istituto Auxologico Italiano, Milan, Italy. Marvin J Fritzler, Faculty of Medicine, Cumming School of Medicine, University of Calgary, Calgary, AB, Canada. Ray P. Naden, Department of Medicine, McMaster University, Hamilton, Ontario, Canada Thomas Dörner, Charité – Universitätsmedizin Berlin, Corporate member of Freie Universität Berlin, Humboldt-Universität zu Berlin, and Berlin Institute of Health, Department of Rheumatology and Clinical Immunology, Berlin, Germany. Sindhu R. 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1811	https://www.cuemath.com/calculus/integral/	LearnPracticeDownload Integral Calculus Integral calculus helps in finding the anti-derivatives of a function. These anti-derivatives are also called the integrals of the function. The process of finding the anti-derivative of a function is called integration. The inverse process of finding derivatives is finding the integrals. The integral of a function represents a family of curves. Finding both derivatives and integrals form the fundamental calculus. In this topic, we will cover the basics of integrals and evaluating integrals. \| \| \| --- \| \| 1. \| What is Integral Calculus? \| \| 2. \| Fundamental Theorems of Integrals \| \| 3. \| Types of Integrals \| \| 4. \| Properties of Integrals \| \| 5. \| Integrals Formulas \| \| 6. \| Methods of Integrals \| \| 7. \| Applications of Integrals \| \| 8. \| FAQs on Integrals \| What is Integral Calculus? Integrals are the values of the function found by the process of integration. The process of getting f(x) from f'(x) is called integration. Integrals assign numbers to functions in a way that describe displacement and motion problems, area and volume problems, and so on that arise by combining all the small data. Given the derivative f’ of the function f, we can determine the function f. Here, the function f is called antiderivative or integral of f’. Example: Given: f(x) = x2 . Derivative of f(x) = f'(x) = 2x = g(x) if g(x) = 2x, then anti-derivative of g(x) = ∫ g(x) = x2 Definition of Integral F(x) is called an antiderivative or Newton-Leibnitz integral or primitive of a function f(x) on an interval I. F'(x) = f(x), for every value of x in I. Integral is the representation of the area of a region under a curve. We approximate the actual value of an integral by drawing rectangles. A definite integral of a function can be represented as the area of the region bounded by its graph of the given function between two points in the line. The area of a region is found by breaking it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summed up. We specify an integral of a function over an interval on which the integral is defined. Fundamental Theorems of Integral Calculus We define integrals as the function of the area bounded by the curve y = f(x), a ≤ x ≤ b, the x-axis, and the ordinates x = a and x =b, where b>a. Let x be a given point in [a,b]. Then (\int\limits_a^b f(x) dx) represents the area function. This concept of area function leads to the fundamental theorems of integral calculus. First Fundamental Theorem of Integral Calculus Second Fundamental Theorem of Integral Calculus First Fundamental Theorem of Integrals A(x) = (\int\limits_a^b f(x) dx) for all x ≥ a, where the function is continuous on [a,b]. Then A'(x) = f(x) for all x ϵ [a,b] Second Fundamental Theorem of Integrals If f is continuous function of x defined on the closed interval [a,b] and F be another function such that d/dx F(x) = f(x) for all x in the domain of f, then (\int\limits_a^b f(x) dx) = f(b) -f(a). This is known as the definite integral of f over the range [a,b], a being the lower limit and b the upper limit. Types of Integrals Integral calculus is used for solving the problems of the following types. a) the problem of finding a function if its derivative is given. b) the problem of finding the area bounded by the graph of a function under given conditions. Thus the Integral calculus is divided into two types. Definite Integrals (the value of the integrals are definite) Indefinite Integrals (the value of the integral is indefinite with an arbitrary constant, C) Indefinite Integrals These are the integrals that do not have a pre-existing value of limits; thus making the final value of integral indefinite. ∫g'(x)dx = g(x) + c. Indefinite integrals belong to the family of parallel curves. Definite Integrals The definite integrals have a pre-existing value of limits, thus making the final value of an integral, definite. if f(x) is a function of the curve, then (\int\limits_a^b f(x) dx = f(b) - f(a)) Properties of Integral Calculus Let us study the properties of indefinite integrals to work on them. The derivative of an integral is the integrand itself. ∫ f(x) dx = f(x) +C Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent. ∫ [ f(x) dx -g(x) dx] =0 The integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the individual functions. ∫ [ f(x) dx+g(x) dx] = ∫ f(x) dx + ∫ g(x) dx The constant is taken outside the integral sign. ∫ k f(x) dx = k ∫ f(x) dx, where k ∈ R. The previous two properties are combined to get the form: ∫ [k(_1)f(_1)(x) + k(_2)f(_2)(x) +... k(_n)f(_n)(x)] dx = k(_1)∫ f(_1)(x)dx + k(_2)∫ f(_2)(x)dx+ ... k(_n) ∫ f(_n)(x)dx Integrals Formulas We can remember the formulas of derivatives of some important functions. Here are the corresponding integrals of these functions that are remembered as standard formulas of integrals. ∫ xn dx=xn+1 /n+1+C, where n ≠ -1 ∫ dx =x+C ∫ cosxdx = sinx+C ∫ sinx dx = -cosx+C ∫ sec2x dx = tanx+C ∫ cosec2x dx = -cotx+C ∫ sec2x dx = tanx+C ∫ secx tanxdx = secx+C ∫ cscx cotx dx = -cscx+C ∫1/(√(1-x2))= sin-1 x + C ∫-1/(√(1-x2))= cos-1 x + C ∫1/(1+x2)= tan-1 x + C ∫-1/(1+x2)= cot-1 x + C ∫1/(x√(x2 -1))= sec-1 x + C ∫-1/(x√(x2 -1))= cosec-1 x + C ∫ exdx=ex+ C ∫dx/x=ln\|x\| + C ∫ ax dx=ax/ln a + C Methods to Find Integrals There are several methods adopted for finding the indefinite integrals. The prominent methods are: Finding integrals by integration by substitution method Finding integrals by integration by parts Finding integrals by integration by partial fractions. Finding Integrals by Substitution Method A few integrals are found by the substitution method. If u is a function of x, then u' = du/dx. ∫ f(u)u' dx = ∫ f(u)du, where u = g(x). Finding Integrals by Integration by Parts If two functions are of the product form, integrals are found by the method of integration by parts. ∫f(x)g(x) dx = f(x)∫ g(x) dx - ∫ (f'(x) ∫g(x) dx) dx. Finding Integrals by Integration by Partial Fractions Integration of rational algebraic functions whose numerator and denominator contain positive integral powers of x with constant coefficients is done by resolving them into partial fractions. To find ∫ f(x)/g(x) dx, decompose this improper rational function to a proper rational function and then integrate. ∫f(x)/g(x) dx = ∫ p(x)/q(x) + ∫ r(x)/s(x), where g(x) = a(x) . s(x) Applications of Integral Calculus Using integration, we can find the distance given the velocity. Definite integrals form the powerful tool to find the area under simple curves, the area bounded by a curve and a line, the area between two curves, the volume of the solids. The displacement and motion problems also find their applications of integrals. The area of the region enclosed between two curves y = f(x) and y = g(x) and the lines x =a, x =b is given by Area = (\int\limits_a^b (f(x) -g(x))dx) Let us find the area bounded by the curve y = x and y = x2 that intersect at (0,0)and (1,1). The given curves are that of a line and a parabola. The area bounded by the curves = (\int\limits_0^1 (y_2 -y_1)dx) Area = (\int\limits_0^1 (x-x^2)dx) = x2 /2- x 3/3 = 1/2-1/3 = 1/6 sq units. Important Notes The primitive value of the function found by the process of integration is called an integral. An integral is a mathematical object that can be interpreted as an area or a generalization of area. When a polynomial function is integrated the degree of the integral increases by 1. ☛ Also Check: Integration of uv formula Definite integral formula Integral Calculus Examples Example 1. Find the integral of e3x Solution: ∫ d/dx(f(x)) = ∫ d/dx( e3x) We know this is of the form of integral, ∫ d/dx( eax) = 1/a eax + C ∫ d/dx( e3x) = 1/3 e3x + C Answer: The integral of e3x = 1/3 e3x + C 2. Example 2. Find the integral of cos 3x. Solution: ∫ d/dx(f(x)) =∫ cos 3x Let 3x = t thus x = t/3 dx = dt/3 The given integral becomes ∫1/3(cos t) dt = 1/3(sin t) + C = 1/3 sin (3x) + C Answer: The integral of cos 3x = 1/3 sin (3x) + C 3. Example 3. Evaluate the integral i = (\int\limits_2^3) (x+1) dx Solution: By the 2nd theorem of fundamentals of integrals we know that (\int\limits_a^b F(x) dx = f(b) - f(a)) (\int\limits_2^3) (x+1) dx = f(3) -f(2) f(x) = x2/2 + x + C f(3) = 32/2 +3 = 9/2 + 3 = 15/2 f(2)= 22/2 + 2 = 4/2 + 2 = 4 f(3) -f(2) = 15/2 - 4 = 7/2 Answer: The value of the given integral I = 7/2 View More > go to slidego to slidego to slide Great learning in high school using simple cues Indulging in rote learning, you are likely to forget concepts. With Cuemath, you will learn visually and be surprised by the outcomes. Book a Free Trial Class Practice Questions on Integrals Check Answer > go to slidego to slide FAQs on Integral Calculus What Are Integrals? Integrals are the values of the function found by the process of integration. An integral is defined as the area of the region under the curve that is represented as a function y = f(x). What is The Integrals Symbol Called? The ntegrals symbol is ∫. This means that it is bound to a limit from the lower to higher and that the integrals represent the area of the curve under the graph of the function. What Are The Types of Integrals? The two types of integrals are definite integral and indefinite integral. The definite integrals are bound by the limits. The indefinite integrals are not bound to pre-existing values. Can an Integral Have Two Answers? Yes, an indefinite integral can have infinite answers depending upon the value of the constant term; while a definite integral will be a constant value. What is a Double Integral Used For? A double integral is used in order to calculate the areas of regions, find the volumes of a given surface, or also the mean value of any given function in a plane region. How Do you Find The Integrals? Finding integrals is the inverse operation of finding the derivatives. A few integrals are remembered as formulas. For example, ∫ xn = xn+1 / (n+1) + C. Thus x6 = x6+1 / 6+1 = x7 / 7 + C. A few integrals use the techniques of integration by parts, integration by partial fractions, substitution method, and so on. How Do You Use Integrals using Trigonometry? Use the trigonometric identities and simplify the function into integrable function and then apply the formulas and adopt the integration procedures to find the integrals using trigonometry. What is The Integral of sin x? The integral of sine x is -cos x + C. ∫ sin x dX = -cos x + C. What is Integral Calculus Used For? We use definite integrals to find the area under the curve or between the curves that are defined by the functions, we find their indefinite integrals using the formulas and the techniques and then find their difference of the integrals applying the limits. We use definite integrals for computing the volumes of 3-d solids. Given the velocity, we can find the distance as the distance is the integral of velocity. 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1812	https://www.airequipmentcompany.com/wp-content/uploads/2018/01/Fundamentals-of-Fans-Air-Equipment-Company.pdf	Fundamentals of Fans Speaker: Luke Powell Vice President, Air Equipment Company April 1, 2015 1 Overview Fan Laws Fan Testing Different Fan Types Construction Requirements System Effects Video Bearings 2 n CFMnew CFMold RPMnew RPMold = Fan Laws These rules are only valid within a fixed system with no change in the aerodynamics or airflow characteristics of the system. For the purpose of this discussion, a system is the combination of ductwork, hoods, filters, grills, collectors, etc., through which air is distributed. The first fan law relates the airflow rate to the fan rotational speed: Volume flow rate (CFM) is directly proportional to the fan rotational speed (RPM). If the fan RPM is increased, the fan will discharge a greater volume of air in exact proportion to the change in speed. 3 n • CFM = 10,000 • SP = 2” • RPM = 1,000 • BHP = 10 • CFM = 12,000 • SP = • RPM = • BHP = Fan Laws 4 n • CFM = 10,000 • SP = 2” • RPM = 1,000 • BHP = 10 • CFM = 12,000 • SP = • RPM = 1,200 • BHP = Fan Laws 5 n RPMnew RPMold = SPnew SPold 2 ( ) Fan Laws The second fan law relates the fan total pressure or fan static pressure to the fan rotational speed: Total or static pressure is proportional to the square of the fan rotational speed. If it is desired to increase the flow to 20,000 CFM without any physical change in the system, the required SP would be 4” 6 n • CFM = 10,000 • SP = 2” • RPM = 1,000 • BHP = 10 • CFM = 12,000 • SP = • RPM = 1,200 • BHP = Fan Laws 7 n • CFM = 10,000 • SP = 2” • RPM = 1,000 • BHP = 10 • CFM = 12,000 • SP = 2.88” • RPM = 1,200 • BHP = Fan Laws 8 n RPMnew RPMold = BHPnew BHPold 3 ( ) Fan Laws The third fan law relates the total or static air power (and the impeller power), to the fan rotational speed: Power, is proportional to the cube of the fan rotational speed. 9 n • CFM = 10,000 • SP = 2” • RPM = 1,000 • BHP = 10 • CFM = 12,000 • SP = 2.88” • RPM = 1,200 • BHP = Fan Laws 10 n • CFM = 10,000 • SP = 2 • RPM = 1,000 • BHP = 10 • CFM = 12,000 • SP = 2.88” • RPM = 1,200 • BHP = 17.3 Fan Laws 11 Duct Traverse Formats Duct Configuration ASHRAE Handbook Industrial Ventilation Manual AMCA Publication 203 Rectangular 16 to 64 equal areas, maximum of 6 inches apart 16 to 64 equal areas, maximum of 6 inches apart 24 to 100 equal areas Circular 20 equal concentric areas, along 2 diameters 6 to 12 (small duct), 10 to 20 (medium duct), 20 to 40 (large duct), equal concentric areas, along 2 diameters 24 to 48 equal concentric areas, along 3 diameters Fan Testing -Since velocity in a duct is seldom uniform across any section, a traverse is usually made to determine average velocity. Velocity is lowest near the edges or corners, and greatest near center. Fig. 4 shows suggested Pitot tube locations for traversing round and rectangular ducts. To determine average velocity in the duct from the readings, average the calculated individual velocities or the square roots of the velocity heads. The number of traverse points should increase with increased duct sizes. 12 Fan Testing ) . ( Pr _ 4005 : ) ( _ wc in essure Velocity FPM Velocity Air × ) ( _ ) ( _ : ) ( _ _ 2 F Area Station FPM Velocity Air CFM Volume Air Station × -Calculating velocities and air volumes for airflow measuring and traverse probe stations 13 Different Fan Types Airfoil Impeller Design -Highest efficiency of all fan designs. 9 to 16 blades of airfoil contour curved away from the direction of rotation. Air leaves the impeller at a velocity less than its tip speed and relatively deep blades provide for efficient expansion within the blade passages. This will be the highest speed of the fans. Housing Design -Scroll-type, designed to permit efficient conversion of velocity pressure to static pressure, thus permitting a high static efficiency. The clearance and alignment between wheel and inlet bell is very close in order to reach the max efficiency capability. Performance Characteristics -Highest efficiencies occur 50 to 65% of wide open volume. This is also the area of good pressure characteristics; the horse power curve reaches a maximum near the peak efficiency area and becomes lower towards free delivery, a self-limiting power characteristics. Applications -General heating, ventilating and air-conditioning systems. Used in large sizes for clean air industrial applications where power savings are significant. 14 Different Fan Types AIRFOIL WHEEL 15 Different Fan Types Backward-Inclined Backward-Curved Impeller Design -Efficiency is only slightly less than that of airfoil fans. Backward-curved blades are slightly curved away from the direction of rotation and are single thickness. Housing Design -Scroll-type, designed to permit efficient conversion of velocity pressure to static pressure, thus permitting a high static efficiency. The clearance and alignment between wheel and inlet bell is very close in order to reach the max efficiency capability. Performance Characteristics -Peak efficiency is slightly lower than airfoil fan. Unstable left of peak pressure. Highest efficiencies occur 50 to 65% of wide open volume. This is also the area of good pressure characteristics. The horsepower curve reaches a max near the peak efficiency area. Applications -General heating, ventilating and air-conditioning systems. Also used in some industrial applications where the airfoil blade is not acceptable because of corrosive and/or erosion environment. 16 Different Fan Types Backward Inclined Wheel 17 Different Fan Types Impeller Design -Simplest of all fans and least efficient. High mechanical strength and the wheel is easily repaired. Requires medium speed for a given point of rating. Housing Design -Scroll-type, usually the narrowest design of all fan designs because of required high velocity discharge. Dimensional requirements of this housing are more critical than for air-foil and backward-inclined blades. Performance Characteristics -Higher pressure characteristics than the backward-inclined and airfoil fans. Power rises continually to free delivery. Applications -Used Primarily for material-handing applications in industrial plants. Wheel can be of rugged construction and is simple to repair in the field. Wheel can be coated for special materials. Not commonly found in HVAC applications. Radial 18 Different Fan Types RADIAL WHEEL Open Sider Design With RIMS Radial Tip Forward Curved Heal 19 Different Fan Types Impeller Design -Efficiency is less than backward-curved fans. Fabricated of lightweight and low cost construction. Has 24-64 shallow blades with both the heel and tip curved forward. Air leaves wheel at velocity greater than wheel tip speed. Wheel is the smallest of all fans and operates at lowest speed. Housing Design -Scroll is similar to other fan designs. The fit between the wheel and inlet is not as critical as on airfoil and backward-inclined bladed fans. Uses large cut-off sheet in housing Performance Characteristics -Pressure curve is less steep than that of backward-curved bladed fans. There is a dip in pressure curve left of the peak pressure point and highest efficiency occurs to the right of peak pressure, 40-50% of wide open volume. Power curve rises continually toward free delivery. Applications -Used primarily in low-pressure heating, ventilating, and air-conditioning applications such as domestic furnaces, central station units, and packaged air-conditioning equipment from room air-conditioning units to roof top units. Forward Curved 20 Different Fan Types Forward Curved Wheel 21 Different Fan Types Plug Fan (For oven or furnace recirculation) 22 Different Fan Types Vane Axial 23 Bearing Types - Ball -Single Row -Double Row - Spherical Roller -Double Row - Cylindrical -Single Row 24 Bearing Failures Common Types of Failures Common Causes of Failures Overheating Lubrication Brinelling Skidding-Light Loading Fretting Corrosion Loose Shaft Fit Fatigue Bent Shaft Misalignment Set Screws Loosened Preventative Maintenance Auto-Lubrication Device Vibration/Temperature Monitors 25 Fan Maintenance 26 Rotation and Discharge Description Clockwise Down Blast Clockwise Bottom Angular Down Clockwise Bottom Horizontal Clockwise Top Bottom Angular Up AMCA STD CW-180 CW 181-269 CW 270 CW 271-359 Customary CW-DB CW-BAD CW-BH CW-BAU Description Clockwise Up Blast Clockwise Top Angular Up Clockwise Top Horizontal Clockwise Top Angular Down AMCA STD CW-360 CW 1-89 CW 90 CW 91-179 Customary CW-UB CW-TAU CW-TH CW-TAD Direction of rotation is determined by drive side of fan 27 Rotation and Discharge Description Counterclockwise Down Blast Counterclockwise Bottom Angular Down Counterclockwise Bottom Horizontal Counterclockwise Top Bottom Angular Up AMCA STD CCW-180 CCW 181-269 CCW 270 CCW 271-359 Customary CCW-DB CCW-BAD CCW-BH CCW-BAU Description Counterclockwise Up Blast Counterclockwise Top Angular Up Counterclockwise Top Horizontal Counterclockwise Top Angular Down AMCA STD CCW-360 CCW 1-89 CCW 90 CCW 91-179 Customary CCW-UB CCW-TAU CCW-TH CCW-TAD Direction of rotation is determined by drive side of fan 28 Motor Position To determine the location of the motor, face the drive side of the fan and pick the proper motor position designated by the letters W,X,Y, or Z as shown in the drawing to the left. 29 Drive Arrangements ARR. 1 SWSI For belt drive or direct drive. Impeller overhung, two bearings on base ARR. 2 SWSI For belt drive or direct drive. Impeller overhung. Bearings in bracket and supported by fan housing ARR. 3 SWSI For belt drive or direct drive. One bearing on each side and supported by fan housing. 30 Drive Arrangements ARR. 3 DWDI For belt drive or direct drive. One bearing on each side and supported by fan housing ARR. 4 SWSI For direct drive. Impeller overhung on prime mover shaft. No bearings on fan. Prime mover base mounted or integrally directly connected ARR. 7 SWSI For belt drive or direct drive. Arrangement 3 plus base for prime mover. 31 Drive Arrangements ARR. 7 DWDI For belt drive or direct drive. Arrangement 3 plus base for prime mover ARR. 8 SWSI For belt drive or direct drive. Arrangement 1 plus extended base for prime mover. ARR. 9 SWSI For belt drive. Impeller overhung. Two bearings with prime mover outside base. ARR. 10 SWSI For belt drive. Impeller overhung. Two bearings with prime mover inside base. 32 System Effects 33 34 35 36 37 38 39 40 41 42 QUESTIONS ? Please visit us at www.aecky.com THANK YOU FOR YOUR TIME! 43
1813	https://www.vedantu.com/jee-main/physics-electrical-field-of-charged-spherical-shell	JEE Main Physics Electrical Field Of Charged Spherical Shell Electrical Field of a Charged Spherical Shell Explained Download PDF Study Material Important Questions Chapter Pages Revision Notes Difference Between Preparation Tips Exam Info Important Dates Eligibility Criteria Application Form Correction Window Exam Centres Admit Card Reservation Criteria Slot Booking Exam Pattern College Predictor Answer Key Result Colleges News Videos FAQs Syllabus Physics Syllabus Mathematics Syllabus Chemistry Syllabus Courses Class 11 JEE Course (2023-25) Class 12 JEE Course (2023-24) JEE Repeater Course (2023-24) Class 8 JEE Foundation Course Class 9 JEE Foundation Course Class 10 JEE Foundation Course JEE Main Coaching Previous Year Question Paper Subject wise question Paper JEE Main Yearwise Question Paper Practice Materials Practice Papers Sample Papers Mock Test Maths Mock Test Physics Mock Test Chemistry Mock Test Question Answers Step-by-Step Derivation of Electric Field Using Gauss’s Law Electrical Field Of Charged Spherical Shell is a classic concept in JEE Main Physics, central to questions on electrostatics and Gauss’s law. Understanding where the electric field is zero, where it is strongest, and how it changes with distance, helps students answer both straightforward and advanced problems. Applications range from explaining how a Faraday cage works to understanding the protecting properties of conducting shells. In this topic, you will see how a uniformly charged shell creates an electric field around itself, why the interior field vanishes, and how JEE examiners test your grasp of field equations and logic. Mastering this enables you to confidently handle question types from derivations to numericals, especially where the distinction between solid and hollow spheres comes up. Definition and Concept of Electrical Field Of Charged Spherical Shell A charged spherical shell is a hollow sphere with all charge distributed evenly over its surface. The shell can be conducting or insulating in JEE questions, but most often, it is a thin conducting shell. For a shell of radius R and total surface charge Q, the charge is distributed spherically symmetric. This symmetry is crucial when using Gauss’s law to derive the field. Shell means only surface, not solid volume. Uniform charge gives symmetric field lines. Distinction: Hollow shell (surface only) vs. solid sphere (entire mass charged). Key symbol: Q = total charge, R = radius, r = field point from centre. All field analysis assumes vacuum unless noted otherwise. Applying Gauss’s Law to Electrical Field Of Charged Spherical Shell Gauss’s law relates the net electric flux through a closed surface to the charge enclosed. It works best in symmetric situations like a charged shell. Draw a concentric Gaussian sphere of radius r centred on the shell. For r < R (inside shell): Enclosed charge is zero. Thus, field E = 0. For r ≥ R (outside shell): Enclosed charge is Q. By Gauss’s law: Since E is constant on sphere, ⇒ Conclude: Use symmetry, always check if point is inside or outside. Electric Field Inside and Outside a Charged Spherical Shell The field behaviour is the most asked exam point. The formulas must be memorised, but understanding them is just as important. \| Region \| Electric Field E \| Expression \| --- \| Inside shell (r < R) \| Zero \| E = 0 \| \| On surface (r = R) \| Maximum \| \| \| Outside shell (r > R) \| Like point charge \| \| Key points: Electric field inside uniformly charged shell is always zero. Outside, it falls as inverse square of distance (Coulomb’s law behaviour). Maximum at the shell’s surface. No discontinuity in field at boundary for a thin shell. Diagrams, Field Lines, and Graph for Electrical Field Of Charged Spherical Shell Visualising the field with diagrams is crucial for clarity on exam day. Field lines emerge perpendicular to the shell surface, never crossing inside. A graph of field E vs distance r for a charged shell looks like this: Notice: Zero field for r < R. Sharp jump at r = R. Inverse-square decrease as r increases, mimicking a point charge. This differs completely from the electric field inside a solid sphere. For detailed field line patterns, visit electric field lines. Common Pitfalls and Exam Strategies JEE often tests the difference between a solid sphere and a hollow shell. Always identify if the charge is on the surface or throughout the volume. Do not confuse “inside” field of a shell (zero) with “inside” field of a solid sphere (non-zero linear). At r = 0, shell field is always zero. Don’t forget units: N/C for electric field. If centre contains a point charge, superpose its field. Use symmetry; otherwise, Gauss’s law may not simplify the calculation. Applications of Electrical Field Of Charged Spherical Shell Electrostatic shielding is the main application—electrical field inside a conducting shell is always zero. This is the underlying principle of the Faraday cage. Protecting sensitive electronics from external static fields. Cars and airplanes act as crude Faraday cages in lightning. High-voltage labs use shells to isolate instruments. Mock test problems on real-life shell applications are common in JEE. Solved Example: JEE Main Type Calculation A thin conducting shell of radius 0.2 m carries charge Q = 2 μC. Compute the electric field: At the centre of the shell (r = 0) Just outside the shell (r = 0.201 m) At 0.5 m from the centre At r = 0, E = 0 (inside shell, by Gauss’s law). At r = 0.201 m, E = Substitute ε0 = 8.85 × 10-12 C2/N·m2, Q = 2 × 10-6 C: E ≈ 8.9875 × 109 × 2 × 10-6 / (0.201)2 E ≈ 445,295 N/C (rounded for clarity). At r = 0.5 m, E = 8.9875 × 109 × 2 × 10-6 / (0.5)2 E ≈ 71,900 N/C These illustrate that the field is zero inside, maximal at the surface, and follows the inverse square law outside. Similar logic applies in related shell numericals. Summary Table: Electrical Field Of Charged Spherical Shell—Key Points Shell field inside (r < R): E = 0 On the shell (r = R): Outside (r > R): Identical to point charge Q at centre Always use Gauss’s law for such symmetry Hollow shell: No field inside; solid sphere: field increases linearly inside Zero internal field gives electrostatic shielding (Faraday cage principle) Practice similar derivations on mock tests for exams For JEE Main, always check: is it a shell or a solid? Is the field point inside, on, or outside? Are you asked for field, force, or potential? Careful reading and formula recall will help you secure marks. For more exam-oriented explanations and problem practice, follow Vedantu’s JEE Main resources or try electrostatics important questions and revision notes. 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The electric field inside a charged spherical shell is always zero. This result follows from the symmetry of the shell and Gauss's Law. Key points: For any point inside a uniformly charged hollow spherical shell, E = 0. Charges on the shell's surface produce fields that cancel each other exactly inside. This principle is crucial for understanding electrostatic shielding and Faraday cages. Why is the electric field zero inside a spherical shell? The electric field is zero inside a spherical shell due to perfect symmetry and Gauss’s Law. Here's why: For every small charge on the shell, there is an equivalent on the opposite side. Their electric field vectors cancel at every point inside. Using a Gaussian surface inside, total enclosed charge is zero, leading to E = 0 everywhere within the shell. How do you calculate the electric field outside the shell? The electric field outside a charged spherical shell acts as if all charge is at the center. To calculate: Use Gauss’s Law by considering a Gaussian surface outside the shell (radius r > shell’s radius). Field at distance r is E = (1/4πε0) × (Q/r2), where Q is total charge. The field points radially outward (or inward for negative Q). What is the formula for the electric field of a spherical shell? The electric field formula for a charged spherical shell depends on position: Inside the shell (r < R): E = 0 On or outside the shell (r ≥ R): E = (1/4πε0) × (Q/r2) Where Q = total charge and R = shell radius What is the electric field of charged spherical shell? The electric field of a charged spherical shell is zero inside and follows Coulomb's law outside. Inside (r < R): E = 0 Outside (r ≥ R): E = (1/4πε0) (Q/r2) This concept is essential in electrostatics and board/JEE exams. What is the electric field of a spherical charge distribution? The electric field for a spherical charge distribution varies based on whether it is solid or hollow: Hollow Shell: E = 0 inside; E = (1/4πε0)(Q/r^2) outside Solid Sphere (uniform charge): E increases linearly inside (E ∝ r), follows inverse square outside Always use Gauss’s Law for exact calculation How to find the electric field of a shell? To find the electric field of a shell, apply Gauss’s Law with symmetry: For points inside, choose a Gaussian sphere of radius r < R (E = 0) For points outside, choose r > R (E = (1/4πε0) Q/r2) Derivations require specifying charge (Q) and radius (R) How does the electric field change with distance from the center? The electric field of a shell varies with distance as follows: For r < R (inside): E = 0 (constant zero) At r = R (on surface): Maximum value, E = (1/4πε0) (Q/R2) For r > R (outside): E decreases as 1/r2 How is Gauss's Law applied to a spherical shell? Gauss’s Law is the key tool for determining the electric field of a spherical shell. Steps: Draw a Gaussian surface (sphere) at radius r (inside or outside) Calculate total enclosed charge Q 'For inside', Q = 0 so E = 0; for outside, Q is total shell charge so E = (1/4πε0)Q/r2 Are the results the same for solid and hollow spheres? No, the electric field for solid and hollow spheres is different inside. Hollow Shell: E is zero everywhere inside Solid Sphere: E increases linearly with r inside Outside both, the field behaves as if all charge is at the center (E = (1/4πε0)(Q/r^2)) What happens if there is a charge present at the center of the spherical shell? If a point charge is at the center, the shell redistributes its charges accordingly. Main results: The electric field inside now reflects the superposition of shell and central charge fields The shell maintains shielding; its own field is still zero inside (for conducting shell), but added field from the central charge is present everywhere Does the electric field inside change if the shell is not conducting? For a uniformly charged insulating (non-conducting) thin shell, the field inside is still zero. However: Uniform distribution matters; irregular charge breaks the symmetry and the field may not be zero For conducting shells, charges always reside on the outer surface, total field inside is zero How is this concept applied in designing a Faraday cage? Faraday cages use the principle that the electric field inside a charged shell is zero. The cage (conducting shell) shields its inside from external static charges or fields. This is why sensitive instruments/electronics are protected from electric interference inside a Faraday cage. Can Gauss’s Law be applied if the shell is irregular in shape? Gauss’s Law works only with high symmetry (sphere, cylinder, plane). For irregular shells, the symmetry required for easy application is lost, making the law impractical for direct field calculation. Numerical or advanced methods are needed in such cases. What if the shell's thickness is not negligible? If the shell has finite thickness, both inner and outer regions must be considered. Inside inner surface: E = 0 (for conductor) Within the material (if charge is on outer surface): E = 0 for conductor, varies for insulator Outside outer surface: E follows typical shell law (E = (1/4πε0) Q/r^2) Recently Updated Pages Uniform Acceleration: Definition, Equations & Graphs for JEE/NEET Magnetic Effects of Current and Magnetism Revision Notes PDF JEE Main Magnetic Effects of Current and Magnetism Practice Paper with Solutions Download PDF Magnetic Effects of Current and Magnetism Important Questions PDF \| JEE, NEET, Class 12 Difference Between Mean and Median: JEE Main 2026 Difference Between Work and Power: JEE Main 2026 Uniform Acceleration: Definition, Equations & Graphs for JEE/NEET Magnetic Effects of Current and Magnetism Revision Notes PDF JEE Main Magnetic Effects of Current and Magnetism Practice Paper with Solutions Download PDF Magnetic Effects of Current and Magnetism Important Questions PDF \| JEE, NEET, Class 12 Difference Between Mean and Median: JEE Main 2026 Difference Between Work and Power: JEE Main 2026 Trending topics JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More Equation of Trajectory in Projectile Motion: Derivation & Proof Atomic Structure: Definition, Models, and Examples Angle of Deviation in a Prism – Formula, Diagram & Applications Hybridisation in Chemistry – Concept, Types & Applications Wheatstone Bridge Explained: Principle, Working, and Uses JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More Equation of Trajectory in Projectile Motion: Derivation & Proof Atomic Structure: Definition, Models, and Examples Angle of Deviation in a Prism – Formula, Diagram & Applications Hybridisation in Chemistry – Concept, Types & Applications Wheatstone Bridge Explained: Principle, Working, and Uses Other Pages NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2025-26 NCERT Solutions For Class 11 Physics Chapter 1 Units and Measurements - 2025-26 NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions - 2025-26 NCERT Solutions For Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations - 2025-26 NCERT Solutions For Class 11 Physics Chapter 3 Motion In A Plane - 2025-26 JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2025-26 NCERT Solutions For Class 11 Physics Chapter 1 Units and Measurements - 2025-26 NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions - 2025-26 NCERT Solutions For Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations - 2025-26 NCERT Solutions For Class 11 Physics Chapter 3 Motion In A Plane - 2025-26 JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs JEE students also check JEE Main Overview JEE Main Syllabus JEE Main Question Papers JEE Main Study Material JEE Main Result JEE Main Cutoff JEE Main FAQ JEE Main News JEE Advanced Overview JEE Advanced Syllabus JEE Advanced Question Papers JEE Advanced Study Material JEE Advanced Result JEE Advanced Cutoff JEE Advanced FAQ JEE Main & Advanced IIT Colleges
1814	https://www.3blue1brown.com/topics/geometry	3Blue1Brown 3 B l u e 1 B r o w n A n i m a t e d m a t hMenuPatreonStoreBlogExtrasSoMEFAQ/ContactAbout Geometry An assortment of topics related to geometry and higher dimensions Lessons But why is a sphere's surface area four times its shadow?Two proofs for the surface area of a sphere Read This open problem taught me what topology is The inscribed rectangle problem, and how it leads to studying Mobius strips, klein bottles and topology. Read Why slicing a cone gives an ellipse A beautiful proof of why slicing a cone gives an ellipse.Visualizing quaternions (4d numbers) with stereographic projection How to visualize quaternions, a 4d number system, in our 3d worldQuaternions and 3d rotation, explained interactively An introduction to an interactive experience on why quaternions describe 3d rotationsSneaky Topology \| The Borsuk-Ulam theorem and stolen necklaces Solving a discrete math puzzle, namely the stolen necklace problem, using topology, namely the Borsuk Ulam theoremThinking outside the 10-dimensional box A method for thinking about high-dimensional spheres, introduced in the context of a classic example involving a high-dimensional sphere inside a high-dimensional box.The AI that solved IMO Geometry Problems Guest video by Aleph0 on how AlphaGeometry combines logic and intuition.Why ruler and compass?Guest video by Ben Syversen, discussing the history of Euclid’s Elements, and the role ruler and compass constructions really served. © 2025 Grant Sanderson
1815	https://support.minitab.com/en-us/minitab/help-and-how-to/statistical-modeling/regression/how-to/stability-study/interpret-the-results/all-statistics-and-graphs/analysis-of-variance-table/	Minitab® Support Analysis of variance table for Stability Study Learn more about Minitab Find definitions and interpretations for every statistic in the Analysis of Variance table. In This Topic DF Seq SS Seq MS F-value P-Value – Fixed Factor Model Selection P-value – Random Factor Model selection DF The total degrees of freedom (DF) are the amount of information in your data. The analysis uses that information to estimate the values of unknown population parameters. The total DF is determined by the number of observations in your sample. The DF for a term show how much information that term uses. Increasing your sample size provides more information about the population, which increases the total DF. Increasing the number of terms in your model uses more information, which decreases the DF available to estimate the variability of the parameter estimates. For a stability study with fixed factors, the ANOVA table includes the following degrees of freedom: Time, Batch, TimeBatch. Seq SS Sequential sums of squares are measures of variation for different components of the model. Unlike the adjusted sums of squares, the sequential sums of squares depend on the order the terms are entered into the model. In the Analysis of Variance table, Minitab separates the sequential sums of squares into different components that describe the variation due to different sources. Seq SS Term : The sequential sum of squares for a term is the unique portion of the variation explained by a term that is not explained by the previously entered terms. It quantifies the amount of variation in the response data that is explained by each term as it is sequentially added to the model. Seq SS Error : The error sum of squares is the sum of the squared residuals. It quantifies the variation in the data that the predictors do not explain. Seq SS Total : The total sum of squares is the sum of the term sums of squares and the error sum of squares. It quantifies the total variation in the data. Interpretation In the Model Selection table, Minitab uses the sequential sums of squares to calculate the p-value for a term. Usually, you interpret the p-values instead of the sums of squares. Seq MS Sequential mean squares measure how much variation a term or a model explains. The sequential mean squares depend on the order the terms are entered into the model. Unlike sequential sums of squares, sequential mean squares consider the degrees of freedom. The sequential mean square error (also called MSE or s2) is the variance around the fitted values. Interpretation Minitab uses the sequential mean squares to calculate the p-value for a term. Minitab also uses the sequential mean squares to calculate the adjusted R2 statistic. Usually, you interpret the p-values and the adjusted R2 statistic instead of the sequential mean squares. F-value An F-value appears for each term in the Analysis of Variance table. The F-value is the test statistic used to determine whether the term is associated with the response. Interpretation Minitab uses the F-value to calculate the p-value, which you use to make a decision about the statistical significance of the terms and model. The p-value is a probability that measures the evidence against the null hypothesis. Lower probabilities provide stronger evidence against the null hypothesis. A sufficiently large F-value indicates that the term or model is significant. If you want to use the F-value to determine whether to reject the null hypothesis, compare the F-value to your critical value. You can calculate the critical value in Minitab or find the critical value from an F-distribution table in most statistics books. For more information on using Minitab to calculate the critical value, go to Using the inverse cumulative distribution function (ICDF) and click "Use the ICDF to calculate critical values". P-Value – Fixed Factor Model Selection The p-value is a probability that measures the evidence against the null hypothesis. Lower probabilities provide stronger evidence against the null hypothesis. Interpretation To determine whether the association between the response and each term in the model is statistically significant, compare the p-value for the term to your significance level to assess the null hypothesis. The null hypothesis is that no association exists between the term and the response. For a stability study, these are the specific null hypotheses for each term: Time: The product does not degrade over time. Batch: The batches all have the same mean response before they start to degrade. TimeBatch interaction: The batches all degrade at the same rate. For a stability study, Minitab removes any terms that do not have p-values less than your significance level. The default significance level is 0.25. A significance level of 0.25 indicates a 25% risk of concluding that an association exists when there is no actual association. P-value ≤ α: The association is statistically significant : If the p-value is less than or equal to the significance level, you can conclude that there is a statistically significant association between the response variable and the term. Minitab retains the term in the model. P-value > α: The association is not statistically significant : If the p-value is greater than the significance level, you cannot conclude that there is a statistically significant association between the response variable and the term. Minitab removes the term from the model. P-value – Random Factor Model selection The p-value is a probability that measures the evidence against the null hypothesis. Lower probabilities provide stronger evidence against the null hypothesis. Interpretation To determine whether one model fits the data better than another one, compare the p-value for the model to the significance level to assess the null hypothesis. The null hypothesis is that the additional coefficient in the larger model is zero. The alternative hypothesis is that the additional coefficient in the larger model is different from zero. For a stability study, the default significance level is 0.25. A significance level of 0.25 indicates a 25% risk of concluding that the models are the same when one model fits the data better. P-value ≤ α: The association is statistically significant : If the p-value is less than or equal to the significance level, you can conclude that the difference between the models is statistically significant. Minitab retains the more complex model for more analysis. P-value > α: The association is not statistically significant : If the p-value is greater than the significance level, you cannot conclude that the difference between the models is statistically significant. Minitab retains the simpler model for more analysis. Minitab.com License Portal Store Blog Contact Us Cookie Settings Copyright © 2025 Minitab, LLC. All rights Reserved. You are now leaving support.minitab.com. Click Continue to proceed to:
1816	https://people.math.sc.edu/vasquezp/Calc2_pdfs/notes/WorkProb_Sol.pdf	WORK PROBLEMS Math 142 Page 1 of 2 1 Finding work required to pump liquid from a tank Problem: Find the work done by pumping out water from the top of a cylindrical tank 1 m in radius and 4 m tall, if the tank is initially full. (The density of water is 1000 kg/m3). Pumping liquid out of the top of a tank requires work because the liquid is moving against gravity. To calculate this, we imagine the work required to lift small disks of liquid up and out of the tank. We are asked to calculate the work performed in all of this activity. Recall that work (W) is deﬁned as force (F) times distance (d), W = F · d. In this example the force is given by the weight = mass times gravity (FW = m · g), where FW is weight, m is mass, and g is the gravitational constant, 9.8 m/s2. For each disk, FW = [density] · [volume] · [gravity] = ρ · πr2dy · [g] = πρr2gdy. The distance each disk has to be lifted depends on its position with respect to the top of the tank: d = H −y. The total work would be the sum of the work done in the individual disks, so that W = Z H 0 πρr2gdy · [H −y] = Z H 0 πρr2g (H −y) dy, since the density and the radius are constants, the resulting integral is W = πρr2g Z H 0 (H −y) dy = πρr2g Hy −y2 2 H 0 = πρr2g H(H −0) −1 2(H2 −0) = πρr2g H2 −1 2H2 = πρr2gH2 2 = 3.1416 × 1000 × 12 × 9.8 × 42 2 ≈246, 176 N m. Exercise: A rectangular tank (H × L × w = 8 m × 10 m × 4 m ) contains 160 m3 of water. How much work is needed to pump all of the water out of the tank? How would your integral change if the tank is on one of its other sides? Which position would produce the greatest amount of work? • If the tank is on its length (as in the ﬁgure), the total work done is W = Z H/2 0 [ρLwgdy] · [H −y] = ρLwg Z H/2 0 (H −y)dy = ρLwg Hy −y2 2 H/2 0 = ρLwg H2 2 −H2 8 W = 3 8ρLwgH2 = 3 8ρLwgHH. • Similarly if the tank is on its height, W = 3 8ρHwgL2 = 3 8ρLwgHL. • And, if it is on its width W = 3 8ρLHgw2 = 3 8ρLwgHw. Since the green part is the same for all the cases, the most work will be done for the largest dimension, in this case L.
1817	https://www.quora.com/Why-does-Pascals-Triangle-work-for-combinations	Something went wrong. Wait a moment and try again. Combinations of Events Binomial Expansion Combinatorics C Proofs (mathematics) Pascal's Triangle Combinatorial Algebra Algorithms, Combinatorics Permutations and Combinat... 5 Why does Pascal's Triangle work for combinations? Andrew Lin Does Geometry · Upvoted by Daniel McLaury , Ph.D. Student in Mathematics at University of Illinois at Chicago · 8y So you understand how the kth value in the nth row of Pascal’s Triangle is the kth coefficient of (x+y)^n, and you want to know how that value could happen to equal nCk, the amount of ways to pick k objects from a set of n. In other words, what’s up with the binomial formula? Let’s look at a fixed example, n=5 and k=2. Pascal’s Triangle tells you that the coefficient in front of x^2y^3 in (x+y)^5 is 10. But why? Imagine you didn’t know this trick. What would you do? Well you might try to expand out (x+y)^5: Alright time to start multiplying! I’m kidding of course. Maybe i So you understand how the kth value in the nth row of Pascal’s Triangle is the kth coefficient of (x+y)^n, and you want to know how that value could happen to equal nCk, the amount of ways to pick k objects from a set of n. In other words, what’s up with the binomial formula? Let’s look at a fixed example, n=5 and k=2. Pascal’s Triangle tells you that the coefficient in front of x^2y^3 in (x+y)^5 is 10. But why? Imagine you didn’t know this trick. What would you do? Well you might try to expand out (x+y)^5: (x+y)(x+y)(x+y)(x+y)(x+y) Alright time to start multiplying! I’m kidding of course. Maybe instead of multiplying it all out, we can take a sneaky trick to only look at the terms we care about, the x^2y^3. How do you multiply to get x^2y^3? Well you’ll need two x’s and three y’s. Maybe I take the two x’s from the first two (x+y) terms. Or maybe I take them from the second and third instead. Maybe I take it from the first and the last. You want to pick two out of the five (x+y) terms to give you x’s. How many ways can you do this? 5C2, or exactly 5!/2!3! Some recommended further questions, to really solidify your understanding of this fantastic triangle: Find a pattern in the sum of the numbers in a row of the triangle. Why is this true? Pascal’s Triangle can also be built arithmetically, i.e. put 1s along the left and right then each value is equal to the sum of the two above it. Why does nCk have this property? In other words show why nCk=(n-1)Ck+(n-1)C(k-1). These questions don’t require any complicated mathematics to solve. It’s just about connecting different situations and ways of counting. I’ll describe solutions and hints in the comments. Daniel McLaury Ph.D. Student in Mathematics at University of Illinois at Chicago · Author has 3.3K answers and 15.7M answer views · 8y (A2A) You say you understand the binomial theorem. Okay, so what’s the coefficient of, say, in ? If we write out and think about the distributive law, then this is That is, each term here is taken by picking either an or a from each term. Okay, so how many ways to pick three ’s and a ? Well, , of course. Assistant Bot · 1y Pascal's Triangle is a triangular array of numbers that represents the coefficients of the binomial expansion. The reason it works for combinations is deeply rooted in combinatorial mathematics. Structure of Pascal's Triangle Each number in Pascal's Triangle is the sum of the two numbers directly above it. The triangle starts as follows: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Combinatorial Interpretation The entries in Pascal's Triangle correspond to binomial coefficients, which count the number of ways to choose elements from a set of elements, denoted as . The relation Pascal's Triangle is a triangular array of numbers that represents the coefficients of the binomial expansion. The reason it works for combinations is deeply rooted in combinatorial mathematics. Structure of Pascal's Triangle Each number in Pascal's Triangle is the sum of the two numbers directly above it. The triangle starts as follows: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Combinatorial Interpretation The entries in Pascal's Triangle correspond to binomial coefficients, which count the number of ways to choose elements from a set of elements, denoted as . The relationship can be understood as: Base Cases: The top of the triangle, , represents one way to choose 0 items from 0 items. Recurrence Relation: Each entry can be expressed as: This equation arises because: - counts the ways to choose elements including a specific element. - counts the ways to choose elements excluding that specific element. Example To illustrate, consider choosing 2 elements from a set of 4 elements (let's say {A, B, C, D}): - The combinations are: {A, B}, {A, C}, {A, D}, {B, C}, {B, D}, {C, D}. Hence, . - In Pascal's Triangle, this corresponds to the number in the 4th row and 2nd position (starting from 0), which is indeed 6. Conclusion Pascal's Triangle effectively organizes the binomial coefficients in a way that reflects the combinatorial nature of choosing subsets. Each entry represents the number of ways to choose a certain number of items from a larger set, and the recursive structure of the triangle captures the fundamental relationships between these choices. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Bob Zwetsloot M.Sc. in Mathematics, Leiden University (Graduated 2023) · Author has 140 answers and 267.7K answer views · 8y As you understand the idea of pathways in the triangle, the idea of combinations follows from there: To get to the position corresponding with , you need to take any path that goes down and right times, while going down and left times. The positions of the down and right movements within the sequence of movements correspond with a combination of out of the moves, and for each such combination, we also get exactly one path. So the number of paths is equal to the number of combinations. Ayodeji Oyenaike Knows Japanese · Author has 152 answers and 294.5K answer views · 6y Before this, the combinatorics formula uses and , so here’s how they fit onto the triangle: If you’re confused by it, is how many spaces right from the leftmost number in the row. It is easier to find formulae for different values of . Using simultaneous equations, I found these results for different values of (using the notation): (Just in case you can’t see, the second case (where ) is .) Anyway, did you notice any patterns? The denominators seem to be in each case. You also notice that we keep multiplying consecutive Before this, the combinatorics formula uses and , so here’s how they fit onto the triangle: If you’re confused by it, is how many spaces right from the leftmost number in the row. It is easier to find formulae for different values of . Using simultaneous equations, I found these results for different values of (using the notation): (Just in case you can’t see, the second case (where ) is .) Anyway, did you notice any patterns? The denominators seem to be in each case. You also notice that we keep multiplying consecutive numbers from to , but we notice that . We can use that fact and get: Sponsored by Grommet What are the smartest inventions of 2025? Here are some of the most useful gadgets we've seen this year. Take a look! Dean Rubine Former Faculty at Carnegie Mellon School Of Computer Science (1991–1994) · Author has 10.6K answers and 23.5M answer views · 5y Related Why does Pascal's triangle work to find coefficients in binomial expansion? We write [math]{n \choose k},[/math] read ’n choose k,’ for the number of different ways we can choose a subset of size [math]k[/math] from a set of [math]n[/math] elements. We have two goals: Describe why the numbers in Pascal’s Triangle are [math]{n \choose k}.[/math] In particular that’s the number in row [math]n[/math], column [math]k[/math], where the indices start at zero. Describe why [math]{n \choose k}[/math] is the coefficient on [math]x^{k}y^{n-k}[/math] in the binomial expansion of math^n.[/math] When we get both of these done, we’ll understand why the numbers in Pascal’s Triangle are also binomial coefficients. Let’s start with the second one; that’s probably easiest. We’ll take a non-trivial We write [math]{n \choose k},[/math] read ’n choose k,’ for the number of different ways we can choose a subset of size [math]k[/math] from a set of [math]n[/math] elements. We have two goals: Describe why the numbers in Pascal’s Triangle are [math]{n \choose k}.[/math] In particular that’s the number in row [math]n[/math], column [math]k[/math], where the indices start at zero. Describe why [math]{n \choose k}[/math] is the coefficient on [math]x^{k}y^{n-k}[/math] in the binomial expansion of math^n.[/math] When we get both of these done, we’ll understand why the numbers in Pascal’s Triangle are also binomial coefficients. Let’s start with the second one; that’s probably easiest. We’ll take a non-trivial but not too difficult example, math^4.[/math] Let’s multiply it out the regular way. math^4 = (x^2+2xy+y^2)^2 = x^4 + 4x^2y^2+y^4 + 2(2x^3y +2xy^3+x^2y^2)[/math] math^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4[/math] We see our coefficients are indeed the familiar fourth row of Pascal’s Triangle, 1 4 6 4 1. Let’s see where the coefficients come from. What we’re really doing is: math^4 = (x+y)(x+y)(x+y)(x+y)[/math] Just so we can talk about them, let’s subscript the variables. There are still only two values, all the [math]x_i=x[/math], all the [math]y_i=y[/math]. math^4 = (x_1+y_1)(x_2+y_2)(x_3+y_3)(x_4+y_4)[/math] Again all the [math]x[/math]s are the same, all the [math]y[/math]s are the same; this is just so we can talk about each factor separately. When this product is all distributed out but before all the like terms are collected, we’ll get sixteen terms. Each term consists of either [math]x_1[/math] or [math]y_1[/math] times either [math]x_2[/math] or [math]y_2[/math] times either [math]x_3[/math] or [math]y_3[/math] times either [math]x_4[/math] or [math]y_4[/math]. So each term has degree four, as it’s the product of four variables. For example there’s a term [math]x_1 x_2 x_3 y_4[/math] and a term [math]y_1 x_2 x_3 x_4.[/math] When like terms are collected, those add 1 each to the coefficient on [math]x^3y[/math]. Here’s the key part. We can think of the [math]x[/math]s as a set of names [math]{ x_1, x_2, x_3, x_4 }.[/math] (For our purpose we’re thinking of these as four different elements). All our [math]x^3y[/math] terms are the result of choosing three [math]x[/math]s from this set of four. So there are [math]{4 \choose 3}[/math] of them. That’s the reason the choose notation gives the binomial coefficient. For any term with [math]k[/math] [math]x[/math]s, we have to choose subsets of size [math]k[/math] from our set [math]{ x_1, x_2, x_3, x_4 }.[/math] There are [math]{4 \choose k}[/math] of them; that’s our definition of the choose notation. For another example, how do we get terms with [math]x^2y^2[/math] ? We have to choose 2 [math]x[/math]s out of our four possible ones. We can choose [math]x_1 x_2 y_3 y_4[/math] or [math]x_1 y_2 y_3 x_4[/math] or any of the other combinations with 2 [math]x[/math]s. How many are there? How many ways are there to choose 2 from a set of 4? We know what that’s called, it’s called [math]{4 \choose 2}.[/math] In general when expanding math^n[/math] there will be [math]{n \choose k}[/math] terms in the full expansion that are [math]x^k y^{n-k}.[/math] We can think of choosing subsets of [math]y[/math]s instead of [math]x[/math]s. It doesn’t matter; the problem is totally symmetrical. Choosing [math]y[/math]s is the same as choosing [math]x[/math]s. Each time we make a choice about a given [math]y[/math], we’re making a choice about the corresponding [math]x[/math]. We come to the conclusion [math]\displaystyle {n \choose k} = {n \choose n-k}[/math] That explains why the choose notation gives the binomial coefficients. The number of times a term with [math]k[/math] [math]x[/math]s appears, i.e. the coefficient on [math]x^ky^{n-k}[/math], is the number of size [math]k[/math] subsets of [math]n[/math] elements. Now we need to explain why these numbers are in Pascal’s triangle. We’ll do it by an informal induction. The number of ways to choose 0 items from a set of size 0 is 1. There’s one way. That’s [math]{0 \choose 0}=1[/math], and that’s the 1 at the apex of Pascal’s triangle. Let’s look at the next row, what we call row 1. The number of ways to choose 0 items from a set of 1 thing is 1. The number of ways to choose 1 item from a set of 1 thing is 1. That matches our ‘1 1’ row in Pascal’s triangle. We’ll get the rest of Pascal’s triangle using choose if we can show the basic rule of Pascal’s triangle — that an element is the sum of the two elements directly above it. In particular we need to show [math]\displaystyle {n+1 \choose k} = {n \choose k-1} + {n \choose k}[/math] Let’s look at an example, say [math]{5 \choose 2} ={4 \choose 1} + {4 \choose 2}[/math] We know for sets of size 4, there are 4 ways to choose one of them and 6 ways to choose two of them. So we expect [math]{5 \choose 2}=10.[/math] Why? How do we make size 2 subsets of five from our subsets of four? We’re adding a new element, the fifth element, to our set. We’ll stick with our original set [math]{ x_1, x_2, x_3, x_4}[/math] and a new element [math]x_5[/math]. There are two different ways to make a size 2 subset of 5 elements from subsets of 4 elements. We can take any of our size 2 subsets of 4 elements and use those directly; those are already size 2 elements of five elements; they just don’t happen to include the fifth element. Or we can take any of our size 1 subsets and add our fifth element to it. That gives a size 2 subset of our size 5 set. That’s the whole idea here. In general to get a size [math]k[/math] subset of [math]n+1[/math] elements we can just use a size [math]k[/math] subset of [math]n[/math] elements or add element [math]n+1[/math] to all the size [math]k-1[/math] subsets of [math]n[/math] elements. So the total number of size [math]k[/math] subsets of [math]n+1[/math] is the sum of these two, which is what we wanted to show: [math]\displaystyle {n+1 \choose k} = {n \choose k-1} + {n \choose k}[/math] So we’ve shown how the first couple of rows of Pascal’s triangle are the same as counting the appropriate subsets, and how the number of subsets follows the same rule as the one that creates Pascal’s triangle. We conclude the numbers in Pascal’s triangle are [math]{n \choose k}.[/math] That’s the story. It’s not totally obvious but it’s not that tricky either. Pascal’s triangle, counting subsets and expanding binomials are all giving rise to the same combinatorial function. Related questions Why does Pascal's rule for combinations work? What is an example of a combination problem that can be solved using Pascal’s Triangle? Why does Pascal's Triangle work? Up to where did pascal made Pascal's triangle? What is the importance of Pascal's Triangle in Combinatorics and Probability? Nick Shales MPhys in Theoretical Physics, The University of York (Graduated 2004) · Author has 419 answers and 1.5M answer views · 8y Related Why does Pascal's Triangle work? I find it easiest to understand in terms of shortest paths on a lattice like the one on the left above shown with example coordinates. The grid on the right shows the number of shortest paths to each lattice coordinate, but we'll get to that. Take a standard coordinate grid (or lattice), each crossing point of the grid lines has integer coordinates math[/math] where, as usual, [math]x[/math] represents the horizontal coordinate and [math]y[/math] the vertical. Now we imagine traversing a path from the origin [math]\text{O}(0,0)[/math] to some coordinate math[/math] , this path travels along the grid lines so that we take the shortest path to math, each crossing point of the grid lines has integer coordinates math[/math] where, as usual, [math]x[/math] represents the horizontal coordinate and [math]y[/math] the vertical. Now we imagine traversing a path from the origin [math]\text{O}(0,0)[/math] to some coordinate math[/math] , this path travels along the grid lines so that we take the shortest path to math[/math], these are the “shortest paths” referred to earlier. Clearly there can be many such paths to some general point math[/math], but they all have [math]x[/math] horizontal steps and [math]y[/math] vertical steps. Our aim is to count the total number of shortest paths to each math[/math] on the lattice. The first way we can do this is by literally counting shortest paths to each lattice point. However we don't do this in any random way, we do it the easy way by starting at the origin and working or way across and up through the grid: There is 1 shortest path from math[/math] to math[/math] this is the path where we don't move. There is 1 shortest path from math[/math] to math[/math] and 1 shortest path to math[/math] in fact we can see that there will be only 1 path to each point on the x-axis and each point on the y-axis. So we fill in those numbers on each lattice point along the axes (like the right hand grid above). How many shortest paths are the to point math[/math] then? Well, there is 1 shortest path to math[/math] and 1 shortest path to math[/math] and in order to get to math[/math] our route must take us via either math[/math] or math[/math] so we add these together: [math]1+1=2[/math] and this is the number that goes in to our coordinate math[/math] in the right hand grid. From this last example you can hopefully see that any shortest path to math[/math] must get there via math[/math] or math[/math] and no other way, all shortest paths must come through these two neighbouring coordinates. Hence we can add the numbers at lattice pointsmath[/math] and math[/math] to give the number at lattice point math[/math]. As you can see, this is how the right hand grid has been filled in above, each entry in math[/math] is the number of shortest lattice paths from math[/math] to that point. Now all of this “adding neighbouring entries” should ring some bells because if we now rotate the right hand grid by 135° clockwise we have the familiar Pascal’s Triangle (at least, up to the diagonal “1, 4, 6, 4, 1”, however if we continued to fill in the grid we would get the rest of Pascal’s Triangle). At this point you should begin see a whole new interpretation emerging, how might we exploit this to understand those Pascal’s Triangle entries in terms of choosing things? To answer the last question let's look at a specific example of a shortest path This show a shortest path from math[/math] to math[/math] and in order to classify this path uniquely we can represent it as a set of instructions as if we were directing someone along the grid lines. We might decide to instruct by using “right” and “up” for example, then the path instructions would begin: “right”, “up”, “right”, “up”, “up” and so on. We could use “East”, and “North” or any similar instructions. This is all far too long to write down so we could simplify by just using letters “R” for “right”, “U” for “up”, our example path in the image would then be the instruction R U R U U R R U R U R R. Notice that there are 7 Rs and 5 Us, this will be true for any shortest path from math[/math] to math[/math]. In the diagram you can see that the instruction “1” replaces our “R” and “0” replaces our “U”. It doesn't really matter but being mathematically inclined let's stick with “0”s and “1”s. Now, let's assign each step of our path a number so we know what we are doing at any point along our path, step 1 is “right” or “1”, step 2 is “up” or “0” and so on. The path looks like [math]\begin{array}{ccccccccccccc}\text{Step number:} &1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12\ \text{Instruction:} &1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 1\end{array}[/math] This particular path has “1”s in positions 1, 3, 6, 7, 9, 11 and 12. You could interpret this to mean that this path “chooses” those 7 numbers from the 12 numbers 1 to 12. In fact, since every shortest path from math[/math] to math[/math] will have 12 total moves (“right” or “up”, “0” or “1”) and 7 right moves that can be expressed as an order of 5 “0”s and 7 “1”s there is a direct one-to-one correspondence between each shortest path from math[/math] to math[/math] and each choice of 7 numbers from the set [math]{1,2,3,4,5,6,7,8,9,10,11,12}[/math]. Those 7 numbers are being chosen by the position of the “1”s in our path instructions, therefore the number of paths from math[/math] to math[/math] is the number of ways we can choose 7 numbers from 12 (or 7+5) which is the binomial coefficient [math]\dbinom{12}{7}=[/math] [math]^{12}C_{7}=C^{12}_{7}=[/math] [math]_{12}C_{7}=\dfrac{12!}{7!\, 5!}[/math] This is clearly the number of ways to arrange 7 “1”s and 5 “0”s. In general a path from math[/math] to math[/math] will have a path length of [math]x+y[/math] and it's instructions will have [math]x[/math] “1”s which will choose [math]x[/math] of the [math]x+y[/math] elements of the set [math]{1,2,\cdots ,x+y}[/math] hence there are [math]\dbinom{x+y}{x}[/math] paths from math[/math] to math[/math]. We can see by rotating a grid by 135° that the entry in coordinate math[/math] is in row [math]x+y[/math] of Pascal’s triangle and [math]y[/math] places from the left but by symmetry [math]\binom{x+y}{x}=\binom{x+y}{y}[/math] so this is not a problem, in fact for any path from math[/math] to math[/math] and it’s set of instructions we could just as well let “0”s choose [math]y[/math] elements from our set of [math]1[/math] to [math]x+y[/math] and this would have counted [math]\binom{x+y}{y}[/math] paths. All images courtesy of Google Images. Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Hongwan Liu Particle Physics Postdoc, NYU/Princeton · Upvoted by Jay Wacker , theoretical physicist · Author has 1.3K answers and 6M answer views · 11y Related Why does Pascal's Triangle work? Suppose you want to pick r objects from a collection of n objects. How many ways are there to do it [math]^{n} C_{r}[/math]? Pick any object from among the n objects, and call it A. If you're going to pick r objects, either these r objects include A or it doesn't. How many ways are there to pick r objects including A? Well, we have to pick A, so we have to pick r-1 more objects from a collection of n-1 objects (excluding A). That's just the number of ways to pick r-1 objects from n-1 objects [math]^{n-1} C_{r-1}[/math]. How many ways are there to pick r objects excluding A? That's just picking r objects from a collecti Suppose you want to pick r objects from a collection of n objects. How many ways are there to do it [math]^{n} C_{r}[/math]? Pick any object from among the n objects, and call it A. If you're going to pick r objects, either these r objects include A or it doesn't. How many ways are there to pick r objects including A? Well, we have to pick A, so we have to pick r-1 more objects from a collection of n-1 objects (excluding A). That's just the number of ways to pick r-1 objects from n-1 objects [math]^{n-1} C_{r-1}[/math]. How many ways are there to pick r objects excluding A? That's just picking r objects from a collection of n-1 objects (excluding A). That's [math]^{n-1} C_{r}[/math]. Adding these two up, you get the result you want. This is how Pascal's triangle is formed (by adding two terms to get the corresponding term below), so if you start with 1C0 and 1C1 (the second row) and do the addition, all the while adding the edge cases on the outside, it works out beautifully. Rick McGeer Former Distinguished Technologist at Hewlett-Packard Labs (2003–2014) · Author has 5.5K answers and 12.1M answer views · 4y Pascal’s triangle is simply a visual representation of the following theorem: [math]\displaystyle {n \choose j} = {{n - 1} \choose {j - 1}} + {{n - 1} \choose j}, n \geq j, j > 0. [/math] So let’s prove the theorem: [math]\displaystyle {{n - 1} \choose {j - 1}} + {{n - 1} \choose j} = [/math] [math]\displaystyle \frac{(n - 1)!}{(j - 1)!(n - 1 - (j - 1))!} + \frac{n - 1!}{j!(n - 1 - j)!} = [/math] [math]\displaystyle \frac{(n - 1)!(j) + (n - 1)!(n - j)}{j!n - j!} = [/math] [math]\displaystyle \frac{(n - 1)!(j + n - j)}{j!n-j!} = \frac{n!}{j!n - j!} = {n \choose j}[/math] Sponsored by Zoho One Run your entire small business with one software suite - Zoho One Award-winning business software. 50M+ users across the globe trust Zoho. Start your free trial today! Soumajit Das B.sc. in Physics & Mathematics, Serampore College · Upvoted by Diganta Bhaskar , M.Sc Theoretical Physics & Mathematics, Ramakrishna Mission Vivekananda Educational and Research Institute,be… · Author has 185 answers and 1.3M answer views · 7y Related What is the use of a Pascal triangle? Mathematics is all full of fascinating stuffs and Pascal's triangle just add another dimension to that fact. Although we call it Pascal's triangle, it's quite unfair to strictly associate the name of Blaise Pascal , as the founder of this wonderful mathematical treasure, as early mathematicians from around the world worked on this as well. Early mathematicians in India called it The Staircase Mathematics is all full of fascinating stuffs and Pascal's triangle just add another dimension to that fact. Although we call it Pascal's triangle, it's quite unfair to strictly associate the name of Blaise Pascal , as the founder of this wonderful mathematical treasure, as early mathematicians from around the world worked on this as well. Early mathematicians in India called it The Staircase of Mount Meru , In Iran it was called The Khayyam triangle, and in China it's called The Yang Hui's triangle . Having said that we also can't deny the fact that Pascal too had made some significant contributions to this. Now coming to it's importance and what it actually is, let me explain to you in as simple terms as possible. Pascal's triangle is not just a mere array of numbers stacked in a triangular fashion, but there's a lot more to do with it. Firstly have a look at what he have so far been discussing then I shall explain to you in details. So observe the image above closely. This is what we know today as the Pascal's triangle. But how is it made? To get this following triangular arrangement, start with [math]1[/math] in the first row, and imagine two [math]0s[/math] at its side. Now add one 0(that you imagined) and 1 and write 1 below and again add another 0 and 1 and write 1 below. So now in the next row, below 1 in the first row, you get 1 1. Now again imagine two [math]0s[/math] at the side of two 1s. So add the first 0 and 1 and write 1 below. Then add 1 and 1, write 2 below and finally add the 0 on the right and 1 to write 1 below. Keep doing this process for each row. ``` include using namespace std; int main(){ int n,p=1; cout<<"This code prints a Pascal's triangle:"<<endl; cout<<"How many rows:"; cin>>n; for(int i=0;i<n;i++) { for(int j=1;j<=n-i;j++) cout<<" "; for(int k=0;k<=i;k++) { if(k==0 \|\| i==0) p=1; else p=p(i-k+1)/k; cout<<p<<" "; } cout<<endl; }} ``` See the images below to understand the process I mentioned in the paragraph above. And keep doing this with every row to get the next row of numbers. So we are done with the procedure of producing this triangular array of numbers. Now let's dig a bit deep and see what these numbers are actually doing here. Pascal's triangle helps to determine the coefficients of binomial expansion of math^{n} [/math], where n is the number of row in the triangle above starting from 0. For example when we have, math^{0}[/math] we get the answer as [math]1[/math]. So corresponding to the row 0 in the Pascal's triangle, we get the coefficient as 1. Now when we have math^{1}[/math] we get math[/math]. So here coefficient of [math]x[/math] is 1 and that of [math]y[/math] is 1 which corresponds to the values in the first row of the Pascal's triangle. Now let's take math^{2}[/math]. As per the second row in the Pascal's triangle, coefficient of the first variable in the expansion is [math]1[/math], that of the second is 2 and of the third is 1 again. So we get, math^{2}=1\cdot x^2+2xy+1\cdot y^2[/math] Now have you noticed the pattern here according to which the variables in the expansion are obtained? So for [math]\begin{align}(x+y)^{n}=\left(\text{1st element of the nth row of Pascal's triangle}\right)x^{n}+\left(\text{2nd element of the nth row}\right)x^{n-1}y+\left(\text{3rd element of the nth row}\right)x^{n-2}y^{2}+...+\left(\text{last element of the nth row of Pascal's triangle}\right)x^{0}y^{n}\end{align}\tag{}[/math] Thus you see how just by remembering the triangle you can get the result of binomial expansion for any n. (See the image below for better understanding.) Next time anyone asks you to expand math^{10}[/math] you can easily do just by remembering the elements of the 10th row in the Pascal's triangle. So one very important aspect of the triangle is cleared. What's next? Is that all we have in store here? The answer is a big No! Look at any row of the triangle. For example i choose row 2. Corresponding elements of which are 1 2 1. Now writing it as, math+(2\times 10)+(1\times 1)=121=11^{2}[/math]. Can you smell anything here? Yes, each row of the triangle when summed in the manner mentioned above gives us the value of [math]11^{\text{that row}}[/math]. Isn't that interesting? There's more in the locker. Keep reading. If you simply add all the elem... Related questions What makes Pascal's Triangle so special? What is Pascal’s triangle? How many times does the number 41 appear in Pascal's triangle? What is Pascal's triangle? Can it be used for finding combinations without repetition? If so, how would you go about doing this? Is Yang Hui's triangle the same as Pascal's triangle? Michael Lamar PhD in Applied Mathematics · Author has 3.7K answers and 17.4M answer views · 10y Related Why does Pascal's rule for combinations work? With the [math]\binom n{k-1}[/math] term, you are choosing [math]k-1[/math] objects because the [math]k^{th}[/math] object will be the "+1" object. In more detail, here's the idea. You want to count the number of ways to choose k objects from (n+1) total objects. You certainly understand why the answer must be [math]\binom{n+1}k[/math]. But let's count the same thing in a different way. Let's take the (n+1) total objects and split them into a group of n and an extra single object that I'll call Bob. Now, let's count all the ways of choosing a total of k that do not involve getting Bob. There are n others to choose from, and k that we must With the [math]\binom n{k-1}[/math] term, you are choosing [math]k-1[/math] objects because the [math]k^{th}[/math] object will be the "+1" object. In more detail, here's the idea. You want to count the number of ways to choose k objects from (n+1) total objects. You certainly understand why the answer must be [math]\binom{n+1}k[/math]. But let's count the same thing in a different way. Let's take the (n+1) total objects and split them into a group of n and an extra single object that I'll call Bob. Now, let's count all the ways of choosing a total of k that do not involve getting Bob. There are n others to choose from, and k that we must choose, so the answer is [math]\binom n k[/math] . Now let's count all the ways to choose the objects that do involve selecting Bob as one of the k. Since we know that Bob gets picked, we still need to choose k-1 more, and there are only n objects left to choose from. So there must be [math]\binom n{k-1}[/math] ways to do it. But if we combine all the ways that do not involve selecting Bob with all the ways that do involve selecting Bob, that must be all the possible ways. So the sum of [math]\binom n k[/math] and [math]\binom n{k-1}[/math] must equal the total number: [math]\binom{n+1}k[/math] . Gregory Schoenmakers Engineer and former high school maths teacher. · Author has 4.4K answers and 8.4M answer views · 8y Pascal’s triangle doesn’t actually demonstrate the combination formula. It merely shows how to derive the coefficients of the expansion of math^{k+1}[/math] when given the coefficients of the expansion of math^k[/math]. One can prove that [math] (x+y)^n = \sum\limits_{i=0}^n {n \choose i} x^{n-i}y^i[/math] by using the principle of induction. Related Why does Pascal's triangle produce powers of 11? There is a hidden eleven(11) when doing Pascal’s Triangle which is where the powers of eleven come from. You can use any numbers to create new triangles such as shown with powers of twelve(12) also. Note, the multiplier is flipped, it’s the same as Pascal’s but instead of saying (this plus this) you say (two of this plus one of that) or (2a+1b). The reason Pascal’s Triangle works is one(1) times any number is the same result, Pascal’s is actually (1a+1b), that's where the hidden eleven(11) is… There is a hidden eleven(11) when doing Pascal’s Triangle which is where the powers of eleven come from. You can use any numbers to create new triangles such as shown with powers of twelve(12) also. Note, the multiplier is flipped, it’s the same as Pascal’s but instead of saying (this plus this) you say (two of this plus one of that) or (2a+1b). The reason Pascal’s Triangle works is one(1) times any number is the same result, Pascal’s is actually (1a+1b), that's where the hidden eleven(11) is… Related questions Why does Pascal's rule for combinations work? What is an example of a combination problem that can be solved using Pascal’s Triangle? Why does Pascal's Triangle work? Up to where did pascal made Pascal's triangle? What is the importance of Pascal's Triangle in Combinatorics and Probability? What makes Pascal's Triangle so special? What is Pascal’s triangle? How many times does the number 41 appear in Pascal's triangle? What is Pascal's triangle? Can it be used for finding combinations without repetition? If so, how would you go about doing this? Is Yang Hui's triangle the same as Pascal's triangle? What are the patterns of Pascal's triangle? What is the rule of Pascal's Triangle? Who invented Pascal's triangle? Can the number of combinations be calculated without using Pascal's triangle? How should I solve (5-y) ⁴ using Pascal's triangle? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1818	https://www.vedantu.com/maths/consecutive-integers	Maths Consecutive Integers Explained with Examples Consecutive Integers Explained with Examples Reviewed by: Rama Sharma Download PDF NCERT Solutions NCERT Solutions for Class 12 NCERT Solutions for Class 11 NCERT Solutions for Class 10 NCERT Solutions for class 9 NCERT Solutions for class 8 NCERT Solutions for class 7 NCERT Solutions for class 6 NCERT Solutions for class 5 NCERT Solutions for class 4 NCERT Solutions for Class 3 NCERT Solutions for Class 2 NCERT Solutions for Class 1 CBSE CBSE class 3 CBSE class 4 CBSE class 5 CBSE class 6 CBSE class 7 CBSE class 8 CBSE class 9 CBSE class 10 CBSE class 11 CBSE class 12 NCERT CBSE Study Material CBSE Sample Papers CBSE Syllabus CBSE Previous Year Question Paper CBSE Important Questions Marking Scheme Textbook Solutions RD Sharma Solutions Lakhmir Singh Solutions HC Verma Solutions TS Grewal Solutions DK Goel Solutions NCERT Exemplar Solutions CBSE Notes CBSE Notes for class 12 CBSE Notes for class 11 CBSE Notes for class 10 CBSE Notes for class 9 CBSE Notes for class 8 CBSE Notes for class 7 CBSE Notes for class 6 How to Solve Consecutive Integer Problems Easily The concept of consecutive integers plays a key role in mathematics and is widely applicable to both real-life situations and exam scenarios. Whether you’re tackling word problems, learning about number patterns, or practicing for competitive exams, understanding consecutive integers will help you build a strong foundation in maths. What Is Consecutive Integers? Consecutive integers are numbers that come one after another in a sequence with no gaps. Each number in the set is exactly 1 more (or less) than the previous number. You’ll find this concept applied in areas such as arithmetic progression, number patterns, and solving equations. Example: 4, 5, 6, 7 (consecutive integers going up by 1) Example: -3, -2, -1, 0, 1 (they can be negative or include zero) Key Formula for Consecutive Integers Here’s the standard way of writing formulas for consecutive integers. If the first integer is , the next ones are: Consecutive integers: Consecutive even integers: Consecutive odd integers: Types of Consecutive Integers There are several kinds of consecutive integers you might see in problems: \| Type \| Formula \| Example \| --- \| General \| \| 7, 8, 9 \| \| Even \| \| 10, 12, 14 \| \| Odd \| \| 5, 7, 9 \| Step-by-Step Illustration Sample Problem: The sum of three consecutive integers is 81. Find the three integers. Let the three consecutive numbers be Write their sum: Combine like terms: Subtract 3: Divide by 3: Therefore, the numbers are 26, 27, and 28. Cross-Disciplinary Usage Consecutive integers are not only useful in Maths but also play an important role in Physics, Computer Science, logic puzzles, and day-to-day reasoning. If you’re preparing for JEE, NTSE, or Olympiad exams, you’ll find problems involving consecutive integers come up often, especially in algebra and number series. Speed Trick or Vedic Shortcut Want to find the sum of several consecutive integers quickly? Here’s a shortcut: For n consecutive numbers starting at a, the sum is n × (first number + last number)/2. Example: Sum of 23, 24, 25, 26, and 27. 1. First number = 23, Last number = 27 2. n = 5 3. Sum = Such shortcuts help save time during tricky word problems, especially in competitive exams. Vedantu’s live classes teach many more tips like these! Try These Yourself List the first five consecutive integers starting from 12. Are -2, -1, 0 consecutive integers? Find three consecutive even integers whose sum is 72. Which set contains only consecutive odd integers: 9, 11, 13 or 10, 12, 15? Frequent Errors and Misunderstandings Forgetting that the difference between even or odd consecutive integers is 2, not 1. Using wrong variables (e.g., writing x, x+2, x+4 for general consecutive integers instead of x, x+1, x+2). Mixing up even and odd integer sequences in word problems. Not including negative numbers and zero as consecutive integers. Relation to Other Concepts The idea of consecutive integers connects closely with topics such as arithmetic sequence and number patterns. Mastering this concept makes it easier to tackle algebraic equations, solve sequence puzzles, and understand properties of integers in later grades. Classroom Tip A quick way to remember consecutive even or odd integers: Use x, x+2, x+4, etc. for both. For general consecutive integers, always use x, x+1, x+2. Visualizing these on a number line helps reinforce the concept. Vedantu’s teachers often use number lines and colored markers in live classes to make these patterns easy to spot. We explored consecutive integers—from the basic meaning and formulas to types, examples, errors to avoid, and real-world connections. Regular practice with Vedantu can help you quickly spot number patterns and confidently solve any problem involving consecutive integers in your exams! Explore Related Topics What Are Odd Numbers? Even Numbers Algebraic Expressions Best Seller - Grade 11 - JEE View More> ### Vedantu JEE 2025 - 26 QR Revision Cards – Physics, Chemistry, Mathematics \| Flash Cards for JEE Main & Advanced \| Quick Concept Recap & Practice Booklet ₹1999.00 Sale ₹1499.00 ### Vedantu JEE Tatva Book Set – Physics, Chemistry, Mathematics \| Set Of 11 Volumes For Class 11 \| Chapterwise PYQs, Concept Videos, Theory & Graded Exercises \| Latest Edition ₹3999.00 Sale ₹2999.00 ### Vedantu JEE Main PYQ Books - Complete Set of 3 for Physics, Chemistry, and Mathematics \| 2023 & 2024 Chapterwise Previous Year Questions \| Ideal for JEE Main Preparation \| Includes Access to JEE Recorded Lectures ₹1999.00 Sale ₹1027.00 ### Vedantu JEE Advanced Rank Accelerator 2025 Books Set Of 3 \| Physics, Chemistry, Mathematics \| Chapterwise Practice, PYQs, Mock Tests For JEE Advanced Aspirants ₹1999.00 Sale ₹1499.00 ### Vedantu JEE Main 2025 Crash Course Book Set Of 3 – Physics, Chemistry, Mathematics \| Latest Syllabus \| Includes Free Recorded Course ₹1999.00 Sale ₹999.00 ### Vedantu's Instasolve - 1 Month - 24 hours Unlimited Instant Doubt Solving ₹2998.00 Sale ₹1999.00 ### Vedantu's Instasolve - 3 Months - 24 hours Unlimited Instant Doubt Solving ₹9998.00 Sale ₹5499.00 ### Vedantu's Instasolve - 12 Months - 24 hours Unlimited Instant Doubt Solving ₹17998.00 Sale ₹12000.00 ### Dream Hustle Achieve - Men's Round Neck T-Shirt ₹998.00 Sale ₹499.00 ### Dream Hustle Achieve - Men's T Shirt ₹998.00 Sale ₹499.00 ### Dream Hustle Achieve - Women's Round Neck T-Shirt ₹998.00 Sale ₹499.00 ### Dream Hustle Achieve - Men's Hooded Sweatshirt ₹1598.00 Sale ₹799.00 ### Forever Vedan - Women's Round Neck T-Shirt ₹998.00 Sale ₹499.00 ### Dream Hustle Achieve - Women's Hooded Sweatshirt ₹1598.00 Sale ₹799.00 FAQs on Consecutive Integers Explained with Examples What are consecutive integers? Consecutive integers are whole numbers that follow each other in order without any gaps. They are represented algebraically as n, n + 1, n + 2, and so on, where n is an integer. For example: 1, 2, 3; -3, -2, -1; or 10, 11, 12. What is an example of 3 consecutive integers? Examples of three consecutive integers include: 1, 2, 3; -2, -1, 0; and 100, 101, 102. The key is that each number is one greater than the previous number in the sequence. How do you write consecutive even numbers in terms of x? Consecutive even numbers can be represented as: x, x + 2, x + 4, etc., where x is an even integer. What are the first 3 consecutive integers? The first three consecutive integers are 1, 2, and 3. The sequence continues infinitely in both positive and negative directions. What do consecutive integers mean? Consecutive integers are integers that follow sequentially, each one being 1 greater than the preceding integer. They are numbers that are next to each other on the number line. Can consecutive integers include negative numbers and zero? Yes, consecutive integers can be positive, negative, or zero. For example, -2, -1, 0, 1, 2 are consecutive integers. Why does the sum of consecutive integers sometimes result in an odd or even number? The sum of consecutive integers can be odd or even depending on the number of integers and whether they start with an odd or even number. The sum of an odd number of consecutive integers will always be divisible by the number of integers, while the sum of an even number of consecutive integers is divisible by 2 and the number of integers. How do you set up equations using consecutive integers in word problems? To set up equations: • Define a variable (e.g., x) to represent the first integer. • Represent the other consecutive integers in terms of x (e.g., x + 1, x + 2). • Translate the word problem's conditions into an algebraic equation using these expressions. • Solve the equation for x to find the integers. What is the role of consecutive integers in arithmetic progression? Consecutive integers form an arithmetic progression (or arithmetic sequence) with a common difference of 1. Understanding consecutive integers is fundamental to working with arithmetic progressions. Are fractional or decimal numbers ever called “consecutive” in maths? No, the term "consecutive" generally applies only to integers (whole numbers). Fractions or decimals do not follow directly in sequence in the same way that integers do. What is the formula for consecutive odd integers? Consecutive odd integers can be represented as 2n + 1, 2n + 3, 2n + 5, where n is an integer. Give an example of consecutive odd integers. Examples of consecutive odd integers are: 1, 3, 5; -1, 1, 3; and 21, 23, 25. 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1819	https://people.ohio.edu/just/M4302S21/Lectures/ACL38np.pdf	Lecture 38: Riemann integrability of continuous and of monotone functions Winfried Just Department of Mathematics, Ohio University Companion to Advanced Calculus Winfried Just, Ohio University MATH4/5302, Lecture 38: More on Riemann integrability Uniformly continuous functions are Riemann integrable Theorem 11.5.1: Let I be a bounded interval, and let f : I →R be uniformly continuous on I. Then f is Riemann integrable. For the proof of this theorem, it suﬃces to ﬁnd, for every ε > 0, a piecewise constant function ℓε : I →R that minorizes f and a piecewise constant function uε : I →R that majorizes f such that p.c. R I uε −p.c. R I ℓε ≤ε. Let ε > 0 be arbitrary, and let γ > 0 be a number that we will choose in a moment. By uniform continuity of f , there exists δ > 0 such that ∀x, y ∈I (\|x −y\| < δ →\|f (x) −f (y)\| ≤γ. Pick such δ. We can ﬁnd a partition P of I into pairwise disjoint nonempty subintervals J of length \|J\| < δ each. For each J ∈P, pick xJ ∈J, and deﬁne ℓε(x) := f (xJ) −γ and uε(x) := f (xJ) + γ for all J ∈P and all x ∈J. Question L38.1: How to choose γ for this argument to work? Winfried Just, Ohio University MATH4/5302, Lecture 38: More on Riemann integrability The proof of Theorem 11.5.1 completed We distinguish two cases: Case 1: \|I\| = 0. Then f is piecewise constant and hence Riemann integrable. Case 2: \|I\| > 0. Then we choose γ := ε 2\|I\|. By construction, ℓε minorizes f , uε majorizes f , and both ℓε, uε are piecewise constant. Moreover, uε(x) −ℓε(x) = 2γ = ε \|I\| for all x ∈I, so that p.c. R I uε −p.c. R I ℓε = p.c. R I(uε −ℓε) = ε \|I\|\|I\| = ε. □ Combining Theorem 11.5.1 with Theorem 9.9.16, we obtain: Corollary 11.5.2: Let [a, b] be a closed interval, and let f : [a, b] →R be continuous. Then f is Riemann integrable. Winfried Just, Ohio University MATH4/5302, Lecture 38: More on Riemann integrability Does it suﬃce to assume continuity of f ? Conjecture L38.1: Let I be a bounded interval, and let f : I →R be continuous on I. Then f is Riemann integrable. Question L38.2: Is this conjecture true? No. For example, the function f (x) := 1 x is continuous on (0, 1], but unbounded. So it cannot be Riemann integrable. It turns out that a slight modiﬁcation of Conjecture 38.1 is true: Proposition 11.5.3: Let I be a bounded interval, and let f : I →R be both continuous and bounded. Then f is Riemann integrable. Note that by Corollary 11.5.2 and our earlier observation about functions on degenerate intervals, Proposition 11.5.3 is of interest only if I = (a, b) or I = (a, b] or I = [a, b) for some a < b. Then for every a+, b−with a < a+ < b−< b the restriction f ↾[a+, b−] is Riemann integrable by Corollary 11.5.2. Winfried Just, Ohio University MATH4/5302, Lecture 38: More on Riemann integrability Proof of Proposition 11.5.3 We prove the result for I = (a, b) with a < b; the proof in the other cases is similar. Let f be as in the assumptions, and let M > 0 be such that −M ≤f (x) ≤M for all x ∈I. Let ε > 0, and let a < a+ < b−< b. Pick piecewise constant function ℓ′ ε, u′ ε : [a+, b−] →R with respect to some partition P−of [a+, b−] such that ℓ′ ε minorizes f ↾[a+, b−], u′ ε majorizes f ↾[a+, b−], and p.c. R [a+,b−] u′ ε −p.c. R I ℓ′ ε ≤ε 3. Let P = P′ ∪{(a, a+), (b−, b)}, and let ℓε(x) := −M, uε(x) := M for x ∈(a, b)[a+, b−], while ℓε(x) := ℓ′ ε(x) and uε(x) = u′ ε(x) for x ∈[a+, b−]. Then ℓε, uε are piecewise constant, ℓε minorizes f , and uε majorizes f . Question L38.3: How should we choose a+, b−so that p.c. R I uε −p.c. R I ℓε ≤ε? Winfried Just, Ohio University MATH4/5302, Lecture 38: More on Riemann integrability Proof of Proposition 11.5.3 We prove the result for I = (a, b) with a < b; the proof in the other cases is similar. Let f be as in the assumptions, and let M > 0 be such that −M ≤f (x) ≤M for all x ∈I. Let ε > 0, and let a < a+ < b−< b. Pick piecewise constant function ℓ′ ε, u′ ε : [a+, b−] →R with respect to some partition P−of [a+, b−] such that ℓ′ ε minorizes f ↾[a+, b−], u′ ε majorizes f ↾[a+, b−], and p.c. R [a+,b−] u′ ε −p.c. R I ℓ′ ε ≤ε 3. Let P = P′ ∪{(a, a+), (b−, b)}, and let ℓε(x) := −M, uε(x) := M for x ∈(a, b)[a+, b−], while ℓε(x) := ℓ′ ε(x) and uε(x) = u′ ε(x) for x ∈[a+, b−]. Then ℓε, uε are piecewise constant, ℓε minorizes f , and uε majorizes f . It is suﬃcient to choose these numbers so that a+ −a ≤ ε 6M and b −b−≤ ε 6M . Then p.c. R I uε −ℓε = p.c. R (a,a+) uε −ℓε + p.c. R [a+,b−] uε −ℓε + p.c. R (b−,b) uε −ℓε = 2M(a+ −a) + p.c. R [a+,b−] uε −ℓε + 2M(b −b−) ≤ε 3 + ε 3 + ε 3. Winfried Just, Ohio University MATH4/5302, Lecture 38: More on Riemann integrability Riemann integrability of bounded monotone functions Proposition 11.6.1: Let [a, b] be a closed and bounded interval and let f : [a, b] →R be a monotone function. Then f is Riemann integrable. Again, the analogue for open or half-open intervals is false, as the example of f (x) := 1 x on (0, 1] shows. Note that the assumptions of Proposition 11.6.1 imply that the function values of f must all be in the interval [f (a), f (b)] (if f is increasing) or in the interval [f (b), f (a)] (if f is decreasing). So it would seem more natural to call Proposition 11.6.1 a corollary of the following result: Corollary 11.6.3: Let I be a bounded interval, and let f : I →R be both monotone and bounded. Then f is Riemann integrable. We will prove Corollary 11.6.3 without using Proposition 11.6.1 as a stepping-stone. Winfried Just, Ohio University MATH4/5302, Lecture 38: More on Riemann integrability Proof of Corollary 11.6.3 Let I be a bounded interval, and let f : I →R be monotone and bounded. We focus on the case when \|I\| > 0 and f is increasing. Let M > 0 be such that f (I) ⊆[−M, M), and let ε > 0. We partition [−M, M) into pairwise disjoint intervals [yi−1, yi) such that −M = y0 < y1 < · · · < yn = M and 0 < yi −yi−1 ≤ε \|I\| for all i ∈{1, . . . , n}. Let x, y, z ∈I with x < z < y. Then if f (x), f (y) ∈[yi−1, yi) for some i, we also have f (z) ∈[yi−1, yi), since f was assumed monotone. Thus f −1([yi−1, yi)) is an interval. Thus P := {f −1([yi−1, yi)) : 1 ≤i ≤n} is a partition of I. Question L38.4: How should we choose ℓε, uε for this proof? Let ℓε(x) := yi−1 and uε(x) := yi for x ∈f −1([yi−1, yi)) and 1 ≤i ≤n. Then ℓε, uε are piecewise constant with respect to P, ℓε minorizes f , uε majorizes f , and p.c R I uε −p.c. R I ℓε ≤ε. □ Winfried Just, Ohio University MATH4/5302, Lecture 38: More on Riemann integrability The integral test Proposition 11.6.4: (Integral test) Let f : [0, ∞) →R be a monotone decreasing function which is non-negative (i.e., f (x) ≥0 for all x ≥0). Then the sum P∞ n=0 f (n) is convergent if, and only if, supN>0 R [0,N] f is ﬁnite. This test gives the following result, which does not follow from any of the tests for convergence of series that we had seen previously: Corollary 11.6.5: Let p be a real number. Then P∞ n=0 1 np converges absolutely when p > 1 and diverges when p ≤1. Question L38.5: How to choose f for deriving Corollary 11.6.5? We ﬁx p ∈R and let f (x) = (x + 1)−p. Anticipating a bit, from the Fundamental Theorem of Calculus: R [0,N] f = 1 1−p(N + 1)−p+1 − 1 1−p if p ̸= 1 and R [0,N] f = ln(N + 1) −ln 1 = ln(N + 1) if p = 1. The result follows by examining these expressions as N →∞. Winfried Just, Ohio University MATH4/5302, Lecture 38: More on Riemann integrability The proof of the integral test Proposition 11.6.4: (Integral test) Let f : [0, ∞) →R be a monotone decreasing function which is non-negative (i.e., f (x) ≥0 for all x ≥0). Then the sum P∞ n=0 f (n) is convergent if, and only if, supN>0 R [0,N] f is ﬁnite. Let f be as in the assumption, and ﬁx N > 0. Then: f ↾[0, N] is Riemann integrable by Proposition 11.6.1. The piecewise constant function u(x) := f (⌊x⌋) majorizes f on [0, N]. The piecewise constant function ℓ(x) := f (⌊x⌋+ 1) minorizes f on [0, N]. R [0,N] ℓ= PN+1 n=1 f (n) ≤ R [0,N] f ≤PN n=0 f (n) = R [0,N] f (n). Now the result follows from the observation that the partial sums of the series P∞ n=0 f (n) and P∞ n=1 f (n) either both converge or both increase without bound. □ Winfried Just, Ohio University MATH4/5302, Lecture 38: More on Riemann integrability Take-home message In this lecture we saw that: Bounded, continuous functions on bounded intervals are Riemann integrable. Bounded, monotonous functions on bounded intervals are Riemann integrable. As an application, we derived the integral test for absolute convergence of series. The integral test allowed us to determine that the harmonic series P∞ n=1 1 n is divergent, while for any p > 1 the series P∞ n=1 1 np is absolutely convergent. Winfried Just, Ohio University MATH4/5302, Lecture 38: More on Riemann integrability
1820	https://ocw.mit.edu/courses/10-302-transport-processes-fall-2004/pages/readings/	Browse Course Material Course Info Instructors Prof. Clark Colton Prof. Kenneth Smith Dr. William Dalzell Departments Chemical Engineering As Taught In Fall 2004 Level Undergraduate Topics Engineering Chemical Engineering Transport Processes Science Physics Thermodynamics Learning Resource Types assignment Problem Sets grading Exams Download Course search GIVE NOW about ocw help & faqs contact us 10.302 \| Fall 2004 \| Undergraduate Transport Processes Readings Readings are from the required course text: Incropera, Frank P., and David P. DeWitt. Fundamentals of Heat and Mass Transfer. 5th ed. New York, NY: Wiley, 2001. ISBN: 9780471386506. \| SES # \| TOPICS \| READINGS \| --- \| 1 \| Introduction and Overview Demonstration of Modes of Heat and Mass Transfer \| 1.1-1.7 \| \| 2 \| Conservation of Energy Heat Diffusion Equation Boundary and Initial Conditions \| 2.1-2.5 \| \| 3 \| Steady-state Conduction in Plane Wall and Radial Geometries \| 3.1-3.4 \| \| 4 \| Recitation 1 \| \| \| 5 \| Conduction with Energy Generation \| 3.5 \| \| 6 \| Extended Surfaces \| 3.6, 3.7 \| \| 7 \| Two-dimensional Conduction \| 4.1-4.4, skim 4.5 \| \| 8 \| Recitation 2 \| \| \| 9 \| Transient Conduction Lumped Capacitance Analysis \| 5.1-5.3 \| \| 10 \| Transient Conduction with Spatial Effects \| 5.4-5.7 \| \| 11 \| Recitation 3 \| \| \| 12 \| Semi-infinite Solid \| \| \| 13 \| Multi-dimensional Effects \| 5.8 \| \| 14 \| Introduction to Radiation \| 12.1-12.5 \| \| 15 \| Recitation 4 \| \| \| 16 \| Radiation Processes \| 12.6-12.9 \| \| 17 \| Blackbody Exchange \| 13.1-13.2 \| \| 18 \| Recitation 5 \| \| \| \| Exam 1 \| \| \| 19 \| Gray Surfaces \| 13.3 \| \| 20 \| Introduction to Convection Boundary Layers \| 6.1-6.5 \| \| 21 \| Recitation 6 \| \| \| 22 \| Boundary Layer Similarity Dimensionless Equations \| 6.6-6.10 \| \| 23 \| External Flow Correlations \| 7.1-7.5, 7.9 \| \| 24 \| Internal Flow Flow in Tubes \| 8.1-8.4 \| \| 25 \| Recitation 7 \| \| \| 26 \| Convection Correlations \| 8.5-8.10 \| \| 27 \| Heat Exchangers (MTD) \| 11.1-11.3 \| \| 28 \| Heat Exchangers (NTU) \| 11.4 \| \| 29 \| Recitation 8 \| \| \| 30 \| Heat Exchangers \| 11.5 \| \| 31 \| Heat Exchangers (cont.) \| \| \| 32 \| Introduction to Mass Transfer \| 14.1-14.2 \| \| 33 \| Recitation 9 \| \| \| \| Exam 2 \| \| \| 34 \| Conservation of Species \| 14.3 \| \| 35 \| Steady Diffusion \| 14.4 \| \| 36 \| Recitation 10 \| \| \| 37 \| Diffusion with Homogeneous Reaction \| 14.5 \| \| 38 \| Diffusion with Heterogeneous Reaction \| 14.6 \| \| 39 \| Transient Diffusion \| \| \| 40 \| Recitation 11 \| \| \| 41 \| Convective Mass Transfer \| 7.1-7.2 \| \| 42 \| Mass Transfer Boundary Layers \| \| \| 43 \| Recitation 12 \| \| \| \| Exam 3 \| \| \| 44 \| Interphase Mass Transfer \| \| \| 45 \| Simultaneous Heat and Mass Transfer \| \| \| 46 \| Recitation 13 \| \| \| 47 \| Course Evaluations \| \| Course Info Instructors Prof. Clark Colton Prof. Kenneth Smith Dr. William Dalzell Departments Chemical Engineering As Taught In Fall 2004 Level Undergraduate Topics Engineering Chemical Engineering Transport Processes Science Physics Thermodynamics Learning Resource Types assignment Problem Sets grading Exams Download Course Over 2,500 courses & materials Freely sharing knowledge with learners and educators around the world. Learn more © 2001–2025 Massachusetts Institute of Technology Creative Commons License Terms and Conditions Proud member of: © 2001–2025 Massachusetts Institute of Technology You are leaving MIT OpenCourseWare Please be advised that external sites may have terms and conditions, including license rights, that differ from ours. MIT OCW is not responsible for any content on third party sites, nor does a link suggest an endorsement of those sites and/or their content. Continue
1821	https://math.stackexchange.com/questions/7067/euclidean-algorithm-vs-factorization	Skip to main content Euclidean Algorithm vs Factorization Ask Question Asked Modified 10 years, 5 months ago Viewed 3k times This question shows research effort; it is useful and clear 6 Save this question. Show activity on this post. Can someone give me an explanation targeted to a high school student as to why finding thegcd of two numbers is faster using the euclidean algorithm compared to using factorization, there should be no algorithm efficiency involved, just a general explanation, something my brother in grade 9 can understand. number-theory factoring euclidean-algorithm Share CC BY-SA 2.5 Follow this question to receive notifications edited Mar 25, 2015 at 17:32 davidlowryduda♦ 94.7k1111 gold badges174174 silver badges331331 bronze badges asked Oct 18, 2010 at 0:01 fmunshifmunshi 35511 gold badge44 silver badges1010 bronze badges 7 1 For large numbers with large factors, it's way much easier to take quotients and remainders (or even just subtract) than to factor. – J. M. ain't a mathematician Commented Oct 18, 2010 at 0:14 @J.M.: But integer factorization algorithms also take quotients and remainders. That alone doesn't say much. It's the number of such operations that matters. – Bill Dubuque Commented Oct 18, 2010 at 0:21 Hm, should've been more precise: it takes more quotients/remainders to factor than to compute a GCD, tho I'm not sure why "there should be no algorithm efficiency involved" is in the question. – J. M. ain't a mathematician Commented Oct 18, 2010 at 0:29 because my brother has no understanding of algorithm efficiency :( – fmunshi Commented Oct 18, 2010 at 1:38 How can you hope to explain this if he has "no understanding of algorithm efficiency"? That seems central to the question. See Arturo Magadin's answer. – Ross Millikan Commented Oct 18, 2010 at 2:53 \| Show 2 more comments 3 Answers 3 Reset to default This answer is useful 9 Save this answer. Show activity on this post. The Euclidean algorithm is a definite recipe that tell you exactly what to do at any given step; there is no guessing, no trial and error involved. Trying to factor a number will (even with some of the best methods currently known) involve guesses and trial-and-error; trial division is of course the classic example, but even some of our best methods (elliptic curve factorization and number field sieve, to name two) all involve some "random guessing" and trials to try to find factors. Sure, they are more clever ways of testing than simply trying everything out there, but you still usually end up doing a lot of grunt work along the way that leads nowhere (dividing by a number that is not a factor in trial division; not getting good relations in the number field sieve; performing all computations on an elliptic curve modulo n and not finding any non-invertible elements), or performing good-looking computations that end up with a trivial factor (1 or n). In essence, this is "wasted effort", waste that simply does not occur with the Euclidean algorithm. Added: Note that this is a reflection of our current known factoring methods, and not necessarily (as Bill Dubuque points out, we just don't know either way) an inherent difficulty in factoring. You don't want to compare "factoring" with "Euclidean algorithm", you want to compare specific ways of factoring with the Euclidean algorithm. And the ways we know (and the ways the high school students know, which are likely to be trial division plus a handful or two of divisibility tests to make the former simpler) have these drawbacks. Perhaps an analogy would be that the Euclidean algorithm is like having a full recipe to prepare a dish, and all the ingredients laid out ready to be used; factorization involves starting to prepare the dish, rummaging through your supplies for ingredients, and possibly realizing part-way through that you don't have the right ingredients, forcing you to start over from scratch with a new dish for which you hope you do have the ingredients. Even if the former situation involves a complex dish while the latter is a series of attempts at very simple and quick dishes, chances are you will spend less total time with the full-recipe-and-all-ingredients-laid-out-method than the let's-try-this-and-hope-we-have-all-the-stuff method. Share CC BY-SA 2.5 Follow this answer to receive notifications edited Oct 18, 2010 at 3:32 answered Oct 18, 2010 at 1:16 Arturo MagidinArturo Magidin 418k6060 gold badges862862 silver badges1.2k1.2k bronze badges 2 2 I don't think that analogy works. It's not a question of searching, guessing... or not. In fact some of the fastest gcd algorithms can be viewed as searching in the remainder sequence. It's simply a matter of how fast you can perform the algorithm, whether it employs brute-force searching or not. We simply don't know any fast algorithm for factorization nor do we have any deep understanding of whether or not factoring is intrinsically hard. – Bill Dubuque Commented Oct 18, 2010 at 3:09 Yes, it's not just a question of searching, but of efficient searching. We do not have any way of doing efficient searching for factors, for whatever reason. The more advanced methods maximize the odds of hitting on a right one through clever ideas, but there is still a lot of blindly stumbling around going on in those algorithms. That is part of the "difficulty" of the problem as it currently stands. Certainly I agree that we do not have any deep understanding of whether factoring is "intrinsically hard", but I don't think the question, reasonably interpreted, was asking that. – Arturo Magidin Commented Oct 18, 2010 at 3:24 Add a comment \| This answer is useful 3 Save this answer. Show activity on this post. It boils down to the fact that currently known factorization algorithms are much slower than the fastest known gcd algorithms. It would be difficult to say much more than that that would be comprehensible to a 9'th grade student (in fact one might argue that not much more than that is even known to experts). Share CC BY-SA 2.5 Follow this answer to receive notifications answered Oct 18, 2010 at 0:24 Bill DubuqueBill Dubuque 283k4242 gold badges337337 silver badges1k1k bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. For very small numbers then factorization is quicker than the Euclidean algorithm. But the Euclidea algorithm gives more; it allows one to compute reciprocals in modular arithmetic. For larger numbers, finding factorizations may be slow or even practically impossible, but the Eulcidean algorithm still works well. Also modern factorization methods from Pollard rho to the number field sieve, make essential use of the Euclidean algorithm. Share CC BY-SA 2.5 Follow this answer to receive notifications answered Oct 18, 2010 at 6:45 Robin ChapmanRobin Chapman 22.9k22 gold badges6363 silver badges8080 bronze badges Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions number-theory factoring euclidean-algorithm See similar questions with these tags. 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1822	https://www.bu.edu/math/files/2019/02/Combinatorial_Game_Theory.pdf	Combinatorial Game Theory Cordelia Theseira and Nathan Josephs Boston University 1/31/19 Cordelia Theseira and Nathan Josephs ONAG 1 / 17 Let’s Play a Game Rules of the Game: On your turn, you can take any number of coins provided they all come from the same heap The person who takes the last coin from all the heaps, wins. i.e. there are no coins left Cordelia Theseira and Nathan Josephs ONAG 2 / 17 Outline 1 What is a Game? 2 Games which are numbers Winning Strategies Simplicity Rule 3 Game of Nim Nim Addition 4 Conclusion Cordelia Theseira and Nathan Josephs ONAG 3 / 17 What is a Game? 8 rules for Combinatorial Games: 1 Two players, Left and Right 2 Finite many positions, including a starting position 3 Clearly deﬁned rules for players to move from the current position to its options 4 Left and Right alternate in turns 5 Complete information 6 No chance moves 7 In the normal play convention the player unable to move loses 8 There exists an ending condition, i.e. game ends when either player is unable to move Cordelia Theseira and Nathan Josephs ONAG 4 / 17 Which Games are numbers? Cordelia Theseira and Nathan Josephs ONAG 5 / 17 Which Games are numbers? G = {a, b, c, ...\|d, e, f , ...} Where G is the current position of the game and is determined by the options of the Left and Right players. Cordelia Theseira and Nathan Josephs ONAG 5 / 17 Which Games are numbers? G = {a, b, c, ...\|d, e, f , ...} Where G is the current position of the game and is determined by the options of the Left and Right players. G is a number when a, b, c, ... ≤d, e, f , ... Cordelia Theseira and Nathan Josephs ONAG 5 / 17 Winning Strategies If G > 0, there is a winning strategy for Left. If G < 0, there is a winning strategy for Right. If G = 0, there is a winning strategy for the second player. G\|\|0, there is a winning strategy for the ﬁrst player. Cordelia Theseira and Nathan Josephs ONAG 6 / 17 Common Games 0 = {\|} which is the ending condition n + 1 = {n\|} which is a Left player win −n + 1 = {\| −n} which is a Right player win ∗= {0\|0} which is a ﬁrst player win Cordelia Theseira and Nathan Josephs ONAG 7 / 17 Simplifying Games To simplify a game, we eliminate dominated options Cordelia Theseira and Nathan Josephs ONAG 8 / 17 Simplifying Games To simplify a game, we eliminate dominated options If Left has options {-3, 1 2, 7}, options −3 and 1 2 would be dominated by 7. If Right had options {-6, -2, 0}, options 0 and 2 would be dominated by −6. Cordelia Theseira and Nathan Josephs ONAG 8 / 17 Simplicity Rule Deﬁnition If all the options for both players of some game G are numbers, and every Left option is less than or equal to every Right option, then G itself is a number, namely the simplest number greater than every Left option and less than every Right option. G = 2p + 1 2n+1 = { 2p 2n+1 \|2p + 2 2n+1 } = { p 2n \|p + 1 2n } Cordelia Theseira and Nathan Josephs ONAG 9 / 17 Simplicity Rule Deﬁnition If all the options for both players of some game G are numbers, and every Left option is less than or equal to every Right option, then G itself is a number, namely the simplest number greater than every Left option and less than every Right option. G = 2p + 1 2n+1 = { 2p 2n+1 \|2p + 2 2n+1 } = { p 2n \|p + 1 2n } A whole number is simpler than a 2. a 2 is simpler than a 4 and so on. The smallest number that satisﬁes the above is the simplest number. Cordelia Theseira and Nathan Josephs ONAG 9 / 17 Examples If G = {2\|3} Cordelia Theseira and Nathan Josephs ONAG 10 / 17 Examples If G = {2\|3}, then G = 5 2. Cordelia Theseira and Nathan Josephs ONAG 10 / 17 Examples If G = {2\|3}, then G = 5 2. If G = {−2\|2} Cordelia Theseira and Nathan Josephs ONAG 10 / 17 Examples If G = {2\|3}, then G = 5 2. If G = {−2\|2}, then G = −1. Cordelia Theseira and Nathan Josephs ONAG 10 / 17 Examples If G = {2\|3}, then G = 5 2. If G = {−2\|2}, then G = −1. If G = { 5 4\|2} Cordelia Theseira and Nathan Josephs ONAG 10 / 17 Examples If G = {2\|3}, then G = 5 2. If G = {−2\|2}, then G = −1. If G = { 5 4\|2}, then G = 3 2. Cordelia Theseira and Nathan Josephs ONAG 10 / 17 Game of Nim Nim is an impartial game. i.e. At every position, both players have the same legal moves. Cordelia Theseira and Nathan Josephs ONAG 11 / 17 Game of Nim Nim is an impartial game. i.e. At every position, both players have the same legal moves. A nimber, i.e. ⋆1, ⋆2, ⋆3..., represents the number of objects available to the players at every position of the game. Cordelia Theseira and Nathan Josephs ONAG 11 / 17 Game of Nim Nim is an impartial game. i.e. At every position, both players have the same legal moves. A nimber, i.e. ⋆1, ⋆2, ⋆3..., represents the number of objects available to the players at every position of the game. E.g. The game we played was: G = {⋆1, ⋆2, ⋆3, ⋆4, ⋆5, ⋆6\| ⋆1, ⋆2, ⋆3, ⋆4, ⋆5, ⋆6} Cordelia Theseira and Nathan Josephs ONAG 11 / 17 Game of Nim Nim is an impartial game. i.e. At every position, both players have the same legal moves. A nimber, i.e. ⋆1, ⋆2, ⋆3..., represents the number of objects available to the players at every position of the game. E.g. The game we played was: G = {⋆1, ⋆2, ⋆3, ⋆4, ⋆5, ⋆6\| ⋆1, ⋆2, ⋆3, ⋆4, ⋆5, ⋆6} Nimbers are their own negatives. i.e. ⋆1 + ⋆2 + ⋆3 = 0 is the same as ⋆1 + ⋆3 = ⋆2 Cordelia Theseira and Nathan Josephs ONAG 11 / 17 3-Heap Game Facts: A single non-empty heap is fuzzy. Two equal-sized heaps is a zero game. Two unequal-sized heaps is fuzzy. Cordelia Theseira and Nathan Josephs ONAG 12 / 17 3-Heap Game Facts: A single non-empty heap is fuzzy. Two equal-sized heaps is a zero game. Two unequal-sized heaps is fuzzy. How to lose in a 3-heap game: Equalize two heaps Empty a heap Cordelia Theseira and Nathan Josephs ONAG 12 / 17 Nim Addition Table Cordelia Theseira and Nathan Josephs ONAG 13 / 17 Nim Addition ⋆10 + ⋆6 = 12 Cordelia Theseira and Nathan Josephs ONAG 14 / 17 Nim Addition ⋆10 + ⋆6 = 12 Break each nimber into distinct parts whose sizes are powers of 2, beginning with the largest size. Cordelia Theseira and Nathan Josephs ONAG 14 / 17 Nim Addition ⋆10 + ⋆6 = 12 Break each nimber into distinct parts whose sizes are powers of 2, beginning with the largest size. ⋆10 + ⋆6 = ⋆8 + ⋆2 + ⋆4 + ⋆2 Cordelia Theseira and Nathan Josephs ONAG 14 / 17 Nim Addition ⋆10 + ⋆6 = 12 Break each nimber into distinct parts whose sizes are powers of 2, beginning with the largest size. ⋆10 + ⋆6 = ⋆8 + ⋆2 + ⋆4 + ⋆2 ⋆10 + ⋆6 = ⋆8 + ⋆4 = 12 Cordelia Theseira and Nathan Josephs ONAG 14 / 17 Conclusion A game is a number if: all the options for both players are numbers every Left option is less than or equal to every Right option G will be the simplest number greater than every Left option and less than every Right option. Cordelia Theseira and Nathan Josephs ONAG 15 / 17 Conclusion A game is a number if: all the options for both players are numbers every Left option is less than or equal to every Right option G will be the simplest number greater than every Left option and less than every Right option. In the game of Nim: Nimbers are their own negatives In a 3-heap game, the player who equalizes two heaps or empties a heap loses To add nimbers, break each nimber into distinct parts whose sizes are powers of 2 and use arithmetic addition Cordelia Theseira and Nathan Josephs ONAG 15 / 17 References Berlekamp, Elwyn R, John H Conway, and Richard K Guy. Winning Ways for Your Mathematical Plays. AK Peters/CRC Press, 2018. Conway, John H. On numbers and games. AK Peters/CRC Press, 2000. Wikipedia contributors. Combinatorial game theory — Wikipedia, The Free Encyclopedia. 2018. url: index.php?title=Combinatorial_game_theory&oldid=862965803. Cordelia Theseira and Nathan Josephs ONAG 16 / 17 Questions? Cordelia Theseira and Nathan Josephs ONAG 17 / 17
1823	https://www.youtube.com/watch?v=FGmCI6krbEk	Evaluate : lim(x → π/4) (sinx - cosx)/(x - π/4) Pace Edu. 9140 subscribers 93 likes Description 7682 views Posted: 28 Feb 2023 lim(x→π/4) (sin x - cos x)/(x - π/4) lim x tends to pi/4 sinx-cosx/x-pi/4 Evaluate: lim(x→π/4) (sin x - cos x)/(x - π/4) Find the limits x to pi/4 sinx -cos x / x-pi/4 Evaluate lim x → π/4 (sinx-cosx)/(x-π/4) math #calculus 15 comments Transcript: lecture I'm going to evaluate limit x t to < by 4 sin x - cos x / x - < by 4 so let's see the answer so first here writing the given limit x t to < by 4 and therea sin x - cos x / by x - < by 4 so here we can easily achieve the output if I assume y = to x - < by 4 or we can write X = < by 4 + y since here x t to < by 4 then y t to < by 4 x < by 4 - < by 4 0 so y t to 0 now next is writing limit so here y t to Zer after that sin X+ so we can write PK by 4 + y - cos < by 4 + uh y after that this place x - < 4 that is equal to Y now it's so so simple now so thereafter use sin A + B formula so here we can write sin A + B so inside writing here sin a plus b formula that is sin a into cos B and there after plus cos a into sin B and second formula that require cos A + B that is equal to cos a into cos B minus sin a into sin B so this two formula that is required so suppose this is here a this is here y b so sin A + B so writing here s < by 4 into cos B cos y after that plus cos < by 4 into sin y minus this cos a into cos B cos < by 4 into cos y - sin < by 4 into sin y okay and in the denominator Y is there after that the next is here so let's write here limit y t to 0 sin < by 4 the value is 1 by < tk2 [Music] okay so sin Pi by value is now 1 by < tk2 this here cos Y and cos P by 4 value is 1 by < tk2 sin y after that minus cos pi value 1 by < tk2 cos y here 1 by < tk2 sin y now let's uh write here y after that limit y 10 to 0 1 byun2 is the common so this is the constant that I'm writing outside 1 by < tk2 1 by root2 1 byun2 is the common so here you can get cos Y and here sin y here Min - cos Y and uh here - cos y - - plus sin y after that here y now cos cos cancel here you can get 1 by < tk2 limit so this here y t to Zer 2 1 + 1 sin y 1 sin y that is 2 sin y IDE by y after that this is here 2 is the constant 2 IDE by < tk2 and limit y t to 0 so sin y / y so this is the formula you should remember if you get here if x t to 0 here sin x / X it equal to 1 okay so this in place of this we can write one so this formula that you should use so thereafter here 2 we can write < tk2 into < tk2 they are < tk2 and this here this value is now one so < tk2 < tk2 is cancel here you can get < tk2 so this is the answer of this question so one time going to repeat so here we can easily achieve the output if I assume y = to X x - < by 4 so since x t to < by 4 so here x + < by 4 - < by 4 then y t to 0 after that here x we can write Pi 4 + x so X+ WR Pi 4 + x and y+ sorry x - < 4 y use sin A + B formula you should remember this formula this formula and after calculating this here you can get this so here uh use the value of s or < by 4 and here some cancel and after doing calculation here you can get < tk2 and limit s x t to 0 sin x by X the value is one so this is the way we can solve this question so now in this video it's over so thanks for watching see next video thank you
1824	https://poemanalysis.com/figurative-language/overstatement/	Overstatement Overstatement is a type of figurative language. They are descriptions of events, people, situations, and objects that are over exaggerated. E.g. In 'The Adventures of Tom Sawyer', Mark Twain uses humorous overstatement, as in “He was as frightened as a baby rabbit,” to exaggerate characters' feelings and add color to the story. Writers use an overstatement when they want to create a specific mood, imbue a story with humor, and more. Usually, the literary device is used on purpose, but it’s possible that one might use it accidentally. For instance, overstating what happened in one’s personal life or changing events to suit one’s needs. Depending on the situation, one’s use of overstatement may or may not be meant to be taken seriously. Definition of Overstatement An overstatement is a literary device that writers use when they want to make a situation seem more dramatic, outrageous, or in some way different than it actually is. This is sometimes done for comedic effect, while in other instances, it can be used more seriously. This device appears in every form and genre of literature. It can be found in novels, short stories, and poems, and more. It should also be noted that the device is sometimes used on purpose, but there are instances in which writers might use the device on accident. For instance, overstating a premise in an academic paper. Examples of Overstatements in Literature The People Upstairs by Ogden Nash ‘The People Upstairs’ is one of Nash’s most amusing poems. In it, he uses outrageous descriptions to try to define how noisy his upstairs neighbors are. He makes guesses about what’s going on upstairs that range from them jumping on pogo sticks, making use of a bowling alley, and practicing ballet. Here are a few lines of the poem in which he uses overstatement: The people upstairs all practise ballet. Their living room is a bowling alley Their bedroom is full of conducted tours. Their radio is louder than yours, They celebrate week-ends all the week. In these lines, Nash’s speaker exaggerates how loud the noise coming from the upstairs apartment is. This is done in order to make the reader, who is meant to be a young child, laugh. Despite the fact that these are overstatements, Nash’s speaker is relaying a very relatable situation which helps with the realism. Read more Ogden Nash poems. Macbeth by William Shakespeare In the following lines from Act II Scene 2 of Macbeth, Macbeth speaks about how guilty he feels in regard to Duncan’s murder. He says: Will all great Neptune’s ocean wash this blood Clean from my hand? No, this my hand will rather The multitudinous seas incarnadine, Making the green one red. He suggests that all the water couldn’t wash his hands clean, a clear example of an overstatement. He only says this so readers can understand how emotional he is and how wide and unending his guilt feels. Explore William Shakespeare’s poetry. Televisionby Roald Dahl ‘Television’ speaks on themes of childhood and entertainment. The poem describes in outrageous detail the dangers of television and what a parent can do to save their child. Dahl’s speaker uses overstatements to reflect on the dangers of watching too much TV. They range from a child’s brain melting to the child’s loss of desire to understand the world. For example: HIS BRAIN BECOMES AS SOFT AS CHEESE! HIS POWERS OF THINKING RUST AND FREEZE! HE CANNOT THINK — HE ONLY SEES! It’s clear that the poet doesn’t really think that children’s brains become “as soft as cheese,” but the statement makes his main idea clearer. It also makes the poem much funnier and more interesting. Compare this to another version where, hypothetically, Dahl chose not to use overstatement and instead said, “His brain isn’t quite as interested in academics.” This more realistic statement is far less interesting to read. Read more Roald Dahl poems. I Wandered Lonely as a Cloudby William Wordsworth ‘I Wandered Lonely as a Cloud,’ also sometimes known as ‘Daffodils,’ is often held up as one of the crowning achievements of the Romantic movement. It is also a wonderful source of various literary devices. In the poem, readers can find examples of personification, imagery, movement, and even overstatement. The overstatement in the following lines is not meant to be taken seriously, but that doesn’t mean that it isn’t effective. Continuous as the stars that shine And twinkle on the milky way, They stretched in never-ending line Along the margin of a bay: Ten thousand saw I at a glance, Tossing their heads in sprightly dance. In the second stanza of the poem, Wordsworth’s speaker says that the daffodils stretched “in never-ending line,” a clear overstatement. But, when taken with the rest of the poem, it makes a great deal of sense. Explore William Wordsworth’s poetry. Why Do Writers Use Overstatement? Writers can use overstatements for a wide variety of reasons. One writer might use the literary device in order to create a comedic scene in their poem, story, or novel. For example, over-exaggerating the danger or absurdity of a situation. In another situation, a writer could use overstatement in order to set the scene for an upcoming drama. It might not turn out as dangerous or exciting as it’s initially described, but the overstatement puts the reader in the right mood. Related Literary Terms Other Resources Home » Figurative Language » Overstatement The Definitive Literary Glossary Crafted by Experts All terms defined are created by a team of talented literary experts, to provide an in-depth look into literary terms and poetry, like no other. Get Tailor-Made Poetry PDF Resources Discover our Poetry Community Cite This Page Baldwin, Emma. "Overstatement". Poem Analysis, Accessed 26 September 2025. support@poemanalysis.com Poem Solutions Limited Company no: 10883994 United Kingdom General About Charity Contact Help Center Meet the Team Request a Guide Poetry+ (not a member? Join now) Dashboard Poetry PDFs Courses Study Tools Beyond the Verse Podcast Discover Poetry Archives Poets A-Z Poems beginning with... 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1825	https://www.antenna-theory.com/antennas/reflectors/dish3.php	Antenna-Theory.com - Parabolic Dish Reflector Antenna (Page 2) The Parabolic Reflector Antenna (Satellite Dish) - 3 Previous: Parabolic Dish AntennasAntennas ListAntenna Theory The fields across the aperture of the parabolic reflector is responsible for this antenna's radiation. The maximum possible antenna gain can be expressed in terms of the physical area of the aperture: The actual gain is in terms of the effective aperture, which is related to the physical area by the efficiency term (). This efficiency term will often be on the order of 0.6-0.7 for a well designed dish antenna: Understanding this efficiency will also aid in understanding the trade-offs involved in the design of a parabolic reflector. The antenna efficiency can be written as the product of a series of terms: We'll walk through each of these terms. Radiation Efficiency The radiation efficiency is the usual efficiency that deals with ohmic losses, as discussed on the efficiency page. Since horn antennas are often used as feeds, and these have very little loss, and because the parabolic reflector is typically metallic with a very high conductivity, this efficiency is typically close to 1 and can be neglected. Aperture Taper Efficiency The aperture radiation efficiency is a measure of how uniform the E-field is across the antenna's aperture. In general, an antenna will have the maximum gain if the E-field is uniform in amplitude and phase across the aperture (the far-field is roughly the Fourier Transform of the aperture fields). However, the aperture fields will tend to diminish away from the main axis of the reflector, which leads to lower gain, and this loss is captured within this parameter. This efficiency can be improved by increasing the F/D ratio, which also lowers the cross-polarization of the radiated fields. However, as with all things in engineering, there is a tradeoff: increasing the F/D ratio reduces the spillover efficiency, discussed next. Spillover Efficiency The spillover efficiency is simple to understand. This measures the amount of radiation from the feed antenna that is reflected by the reflector. Due to the finite size of the reflector, some of the radiation from the feed antenna will travel away from the main axis at an angle greater than , thus not being reflected. This efficiency can be improved by moving the feed closer to the reflector, or by increasing the size of the reflector. Other Efficiencies There are many other efficiencies that I've lumped into the parameter . This is a major of all other "real-world effects" that degrades the antenna's gain and consists of effects such as: Surface Error - small deviations in the shape of the reflector degrades performance, especially for high frequencies that have a small wavelength and become scattered by small surface anomalies Cross Polarization - The loss of gain due to cross-polarized (non-desirable) radiation Aperture Blockage - The feed antenna (and the physical structure that holds it up) blocks some of the radiation that would be transmitted by the reflector. Non-Ideal Feed Phase Center - The parabolic dish has desirable properties relative to a single focal point. Since the feed antenna will not be a point source, there will be some loss due to a non-perfect phase center for a horn antenna. Calculating Efficiency The efficiency is a function of where the feed antenna is placed (in terms of F and D) and the feed antenna's radiation pattern. Instead of introducing complex formulas for some of these terms, we'll make use of some results by S. Silver back in 1949. He calculated the aperture efficiency for a class of radiation patterns given as: TYpically, the feed antenna (horn) will not have a pattern exactly like the above, but can be approximated well using the function above for some value of n. Using the above pattern, the aperture efficiency of a parabolic reflector can be calculated. This is displayed in Figure 1 for varying values of and the F/D ratio. Figure 1. Aperture Efficiency of a Parabolic Reflector as a function of F/D or the angle , for varying feed antenna radiation patterns. Figure 1 gives a good idea on design of optimal parabolic reflectors. First, D is made as large as possible so that the physical aperture is maximized. Then the F/D ratio that maximizes the aperture efficiency can be found from the above graph. Note that the equation that relates the ratio of F/D to the angle can be found here. In the next section, we'll look at the radiation pattern of a parabolic antenna. Next: Parabolic Reflectors, Page 3 Antennas Types Antenna Theory (Home)
1826	https://www.qmul.ac.uk/law/research/journals/the-queen-mary-law-journal/media/law/docs/research/8QMLJ139.pdf	DATE DOWNLOADED: Tue Jun 11 07:25:26 2024 SOURCE: Content Downloaded from HeinOnline Citations: Please note: citations are provided as a general guideline. Users should consult their preferred citation format's style manual for proper citation formatting. Bluebook 21st ed. Francisca Vicario Bermudez, Legal Transplantation and Commercial Law Reform in the Field of Rule-of-Law Promotion, 8 QMLJ 139 (2017). ALWD 7th ed. Francisca Vicario Bermudez, Legal Transplantation and Commercial Law Reform in the Field of Rule-of-Law Promotion, 8 QMLJ 139 (2017). APA 7th ed. Bermudez, Francisca Vicario. (2017). Legal Transplantation and Commercial Law Reform in the Field of Rule-of-Law Promotion. Queen Mary Law Journal, 8, 139-150. Chicago 17th ed. Francisca Vicario Bermudez, "Legal Transplantation and Commercial Law Reform in the Field of Rule-of-Law Promotion," Queen Mary Law Journal 8 (2017): 139-150 McGill Guide 9th ed. Francisca Vicario Bermudez, "Legal Transplantation and Commercial Law Reform in the Field of Rule-of-Law Promotion" (2017) 8 QMLJ 139. AGLC 4th ed. Francisca Vicario Bermudez, 'Legal Transplantation and Commercial Law Reform in the Field of Rule-of-Law Promotion' (2017) 8 Queen Mary Law Journal 139 MLA 9th ed. Bermudez, Francisca Vicario. "Legal Transplantation and Commercial Law Reform in the Field of Rule-of-Law Promotion." Queen Mary Law Journal, 8, 2017, pp. 139-150. HeinOnline. OSCOLA 4th ed. Francisca Vicario Bermudez, 'Legal Transplantation and Commercial Law Reform in the Field of Rule-of-Law Promotion' (2017) 8 QMLJ 139 Please note: citations are provided as a general guideline. Users should consult their preferred citation format's style manual for proper citation formatting. Provided by: Queen Mary University of London -- Your use of this HeinOnline PDF indicates your acceptance of HeinOnline's Terms and Conditions of the license agreement available at -- The search text of this PDF is generated from uncorrected OCR text. -- To obtain permission to use this article beyond the scope of your license, please use: Copyright Information Legal Transplantation and Commercial Law Reform in the Field of Rule-of-law Promotion Francisca Vicario Bermzidez ABSTRACT Legal transplantation has been studied by comparative lawyers, who have looked at it from almost every perspective. Separately, legal transplants have been taking place in different environments, producing a wide range of results that are hardly ever analysed. One of the most interesting of these contexts is the field of international legal technical assistance in rule-of-law promotion programmes. This essay aims to investigate the relationship between the numerous theories on legal transplants and the conclusions drawn from experiences of law reform in the field of the rule-of-law promotion, with a special focus on commercial law reform. It argues that the lack of a unified theory, as well as the lessons that have not been learned from experiences in legal technical assistance, have impeded a more effective use of transplantation in law reform. Therefore, it highlights the need for the creation of a more defined, empirically-based theory of legal transplants, focusing more on the lessons learned rather than the abstract metaphors. It also emphasises the impact of certain flawed assumptions as the basis of the unsuccessful transplants carried out by Western donors from the legal reform industry. I. INTRODUCTION Several theories have been elaborated on legal transplants and each has advanced a new definition of the phenomenon. To date, however, experts cannot agree on a unified methodology to ensure a successful transplantation process. Nevertheless, legal transplantation has been applied, although there is not much research that assesses the extent to which the process has been effective. An analysis of the practical use of legal transplants could identify very interesting cases in the field of international legal technical assistance, which has provided academics with some of the best examples of transplantation applied to processes of law reform in general, and commercial law reform in particular. This essay aims to relate some of the theories and metaphors on legal transplants to the conclusions drawn from the practical application of transplantation in the field of the rule-of-law promotion. To this end, the first part will synthesise the most important theories, focusing on authors that offer key views relevant to Queen Mary Law Journal the study. The second section will analyse the main characteristics and methodology followed by legal technical assistance in rule-of-law promotion programmes, with a special focus on its application to the case of commercial law reform. II. LEGAL TRANSPLANTATION AND LEGAL TECHNICAL ASSISTANCE Legal transplants: different theories and approaches According to William Twining, most of the literature on legal transplantation is based on a 'naYve model of diffusion', which defines a transplant as a situation where two nations take part in a one-way transfer of rules or institutions, followed by their formal enactment, normally under the assumption that the transfer must be from an advanced civil or common law system to a developing one in need of modernisation - usually without specific criteria to assess its success.' However, this literature is fragmented, each written work adding a new insight to the ongoing debate on legal transplantation, thus avoiding the construction of a single research tradition.' Nonetheless, it is true that each author's perspective contributes to the definition and refinement of the idea, and some of their visions offer interesting insights for the purpose of this essay. Alan Watson has been considered the father of the term 'legal transplants' and an active participant in intense debates in the field, stemming from his disagreements with Otto Kahn-Freund's theories.' Arguing about the role of comparative law in the field of law reform, Kahn offered a metaphor that has become rather famous in the field: he compared the transplantation of a kidney to the transferring of a carburettor, concluding that 'transplanting part of a living organism and transferring part of a mechanism are comparable in purpose, but in nothing else.' 4 His metaphor led to the conclusion that there are 'different degrees of transferability'" in legal transplantation. In this sense, he states that the most organic aspect of a foreign law - which would situate it on lower levels of transferability - is its link to the foreign country's distribution of power, including the influence of certain social groups that have a crucial role in law-making.' He concludes that the use of comparative law in law reform must imply a profound knowledge of the foreign law as well as the social and political context from where it has been taken; ignoring this context of the law would turn the use of the comparative method into an abuse.7 1 William Twining, 'Diffusion of Law: A Global Perspective' (2004) 36 Journal of Legal Pluralism and Unofficial Law 1, 4. 2 William Twining, 'Social Science and Diffusion of Law' (2005) 32 Journal of Law and Society 207. 3 Mathias M Siems, Comparative Law (CUP 2014) 195. Otto Kahn-freund, 'On Uses and Misuses of Comparative Law' (1974) 37 MLR 1, 5. ibid. 6 ibid 13. ibid 27. 140 Francisca Vicario Bermfidez This sociological approach was strongly criticised by Watson, whose historical perspective led him to affirm that successful borrowing is possible without any information about the political, social or economic context of the imported law.8 His scepticism of the idea that law and society are inevitably related brought about criticism from Pierre Legrand, who counters that 'legal transplants are impossible'.' However, there is one aspect from his discourse that is noteworthy: when writing about law reform, Watson states that one should be only looking for an idea in the foreign system, which would later be transformed into the law of the recipient country.10 This process of transformation is considered key in Esin Oriicii's conclusions about law as transposition." Her ideas on legal transposition might be perceived as a halfway theory between the two extremes of legal change theories. Oriicii agrees with Watson in his perception of legal change as being based on legal borrowing and imitation, not leaving much room for innovation. However, her theory focuses on the transformation process, which she illustrates with a new metaphor in which a rule is 'transposed' like musical notes: 'in musical transposition, each note takes the same relative place in the scale of the new key as in the old, the transposition being made to suit the particular instrument.' 12 The same happens in law, where the transposition takes place in order to fit the specific socio-legal culture and needs of the recipient. Accordingly, it is the fine-tuning and the adaptation process taking place after the reception that ensures the success of transplantation. In fact, the relevance of the adaptation process explains why voluntary transplants have higher receptivity, as in these processes the imported formal legal order is adapted to the conditions set by the recipient within its pre-existing legal order.14 The emphasis is therefore placed on the process of transfer, irrespective of the donor legal system. Gunther Teubner adds his own terminology when he talks about 'legal irritants', referring to the imposed foreign legal elements, and how they trigger unpredictable and unexpected effects in the recipient legal system." He argues that legal irritants cannot be domesticated, but instead will provoke an evolution of the external rule, which will be reconstructed, thus modifying the internal context. Accordingly, an attempt to unify laws or make them converge through the use of legal irritants will inevitably cause the opposite effect. This Alan Watson, 'Legal Transplants and Law Reform' (1976) 92 LQR 79. Pierre Legrand, 'The Impossibility of Legal Transplants' Maastricht Journal of European and Comparative Law 111, 114. 10 Watson (n 8) 79. 1 Esin Oriicil, 'Law as Transposition' (2002) ICLQ 205. 12 ibid 207. " ibid. 14 ibid 208. See also Daniel Berkowitz et al, 'The Transplant Effect' (2003) 51 American Journal of Comparative Law 163. " Gunther Teubner, 'Legal Irritants: Good Faith in British Law or How Unifying Law Ends up in New Divergences' (1998) 61 MLR 12. 141 Queen Mary Law Journal is because different results would be generated in the recipient environment and more divergences between the two systems will be highlighted." 6 On a more positive note, Oriicii uses her metaphor to explain that the irritation produced is part of the process of internalising the rule in the recipient environment." She sees this process as a requirement for legal transposition to take place successfully, as convergence is not about creating identical systems but about accepting diversity. Rule-of-law promotion: evolution and methodology of legal technical assistance Regardless of the numerous theories, legal transplantation has certainly been taking place. One of the most relevant applications of the phenomenon is 'legal technical assistance'. The term refers to a series of international interventions targeted at countries going through the stages of economic development or a post-conflict transition, such as the group of states that went through deep transformation after the end of communism." Since the fall of the Berlin Wall, Western programmes providing assistance with judicial training, legal education and commercial law drafting became a frequent practice in post-communist states and all around the world years later. 20 Foreign intervention aiming to reform the domestic law of developing or post-conflict societies has become a form of international law-making that is prevalent, but whose effects are rarely discussed.2 1 The end of the Cold War marked the start of a new era of intervention by the international community. From this date, Western organisations started to assist with law reform and state building under the banner of 'rule-of-law promotion'.22 This new era put the focus on the need to build the rule of law to achieve development. However, many considered it a continuation of the old law and development movement from the 1960s. That era of legal technical assistance used a Western transplantation model that received strong criticism, as it was considered ethnocentric. Apart from a failure to recognise the diversity of legal cultures, there was a manifest assumption of superiority of the American laws and institutions that were being transplanted. 2 3 Contrary to its predecessor, it might appear that the promotion of the rule of law is successfully meeting its goal of creating socio-economic prosperity abroad, 16 ibid 24. 17 Orilcil (n 11) 211. 1s ibid. 19 Scott Newton, 'Law and Development, Law and Economics and the Fate of Legal Technical Assistance' in Arnscheidt et al (eds), Lawmaking for Development: Explorations into the Theory and Practice of International Legislative Project (Leiden UP 2009) 23. 20 ibid. 21 Richard Sannerholm, 'Cut-and-Paste"? Rule of Law Promotion and Legal Transplants' in Antonina Bakardjieva Engelbrekt and Joakim Nergelius (eds), New Directions in Comparative Law (Edward Elgar 2009) 57-59. 22 ibid. 23 ibid. 142 Francisca Vicario Bermfidez as it has already lasted more than two decades. However, some specialists in the field of law reform have acknowledged the obscure side of these interventions, criticising the lack of feedback about the effects of legal transplantation that has been taking place for years. Contrary to expectations, a closer analysis of the activity of these programmes poses controversial questions about the contemporary use of legal transplantation by international donors involved in the enterprise. The main criticisms of the rule-of-law promotion programmes highlight the evident but unresolved problems that legal technical assistance has always presented, and advance some meaningful arguments on why lessons from repeated mistakes have never been learned. Rule-of-law promotion programmes argue that the rule of law is an essential requirement to foster economic development and create a democracy.2 4 However, it has been argued that behind this premise lies an over-simplistic reasoning that has led to disappointments in the creation of market economies and democratic systems.25 For instance, the democracy rationale starts to fall apart when facing the fact that internationally recognised, well-established democracies present evident shortcomings in their rule of law. 26 Additionally, some argue that the rule of law promotion movement is, in essence, the same as its predecessor: although it was born under a different set of circumstances, the problems are still the same and they continue to be based on similar flawed assumptions. Thus, several criticisms of the structure of the programmes argue that aid practitioners believe they are applying unquestionable knowledge, whilst what they are effectively doing is still testing a hypothesis.27 This naYve logic is reflected in two of the main problems that legal technical assistance has always faced: lack of local ownership and insufficiency of resources.28 The root of the first problem is particularly interesting for this study. The lack of ownership is due to the absence of local input in the transplantation process carried out by the donors. This practice has been referred to as the 'hasty transplant syndrome'. It constitutes one of the most serious flaws of legal reform in aid programmes where foreign laws are taken as models and transplanted into the transitional legal system without contemplating the result of the adaptation of the law to the local legal culture. This generates ill-fitting rules that will consequently not 'take' in their new environment.29 Experience shows that law reform through the application of hasty transplants fails at the implementation stage, as donors seem to forget that laws themselves have no value and they 24 Thomas Carothers, 'The problem of knowledge', in Thomas Carothers (ed), Promoting the Rule of Law Abroad (Carnegie Endowment 2006) 17-19. 25 ibid. 26 ibid. 27 Wade Channell, 'Lessons Not Learned about Legal Reform' in Thomas Carothers (ed), Promoting the Rule of Law Abroad (Carnegie Endowment 2006) 138-142. 28 ibid. 29 ibid. 143 Queen Mary Law Journal should be regarded as tools, used to design and implement a specific socio-economic policy.ao Conversely, if law develops from within, 'through a process of trial and error [...] and with the participation and involvement of users of the law, legal professionals and other interested parties, legal institutions tend to be highly effective'." In other words, reforms in developing societies have to be focused on the way rules come into being, that is, the method by which the transplants are received. Interestingly, these conclusions resonate with Oriicii's legal transposition metaphor, as empirical results show that the attention should be on the tuning and adapting. 3 2 However, reality in the case of rule-of-law programmes has depicted a process that is essentially a one-sided export of substantive law." Unfortunately, it is Twining's naYve model that seems to describe accurately most of the activity of rule-of-law promotion. Finally, the second main problem faced by legal aid is directly related to the promotion of democracy: practitioners frequently complain about the lack of time and funds. In a developed Western society, the process of law reform would take up to several years, as it must fulfil a series of democratic requirements such as intense policy and legislative debate or public discussion. 34 Nevertheless, it is interesting to note how aid agencies assume that law-making in developing or post-conflict societies can be done successfully in a shorter time. In fact, it has been argued that at the centre of this flawed reasoning lies a very simple truism: non-democratic law-making moves quicker and needs less funding than democratic law-making." The absence of democratic elements in the process reduces notably the costs of the whole enterprise and it is hard for the recipient countries to acknowledge this shortfall, since they are trying to move away from a non-democratic context. This fact questions the very core of rule-of-law and democracy promotion. Aid practitioners claim to promote democracy whilst, in their approach to law reform, they benefit from the non-democratic history of societies in transition through applying law-making processes that lack essential democratic elements to reduce costs. Such conclusions predict less than satisfactory results in law reform by rule-of-law programmes. As stated above, this system of law-making has a direct effect on the implementation of the rules, which results in law reform not achieving its purpose. 2o ibid. 21 Daniel Berkowitz et al (n 14) 189. 22 Orilcil (n 11) 211. Sannerholm (n 21) 61. 14 Channell (n 27) 142-143. 3 ibid. 144 Francisca Vicario Bermfidez III. COMMERCIAL LAW REFORM AND RULE-OF-LAW PROMOTION The role of commercial law reform in rule-of-law promotion As seen above, the rule-of-law movement started after the fall of communism, as a set of programmes to assist post-communist countries in their transformation into market economies and liberal democracies. It soon became clear to donors that reform of commercial law was crucial in the transition from a planned to a market economy. They knew that a market-oriented economy works through the creation of a whole set of indirect incentives, which involve a higher number of laws than the direct orders from a centrally-planned economy." Reacting promptly, Western states built collaborations with international financial institutions such as the World Bank, as well as with donor agencies like the United States Agency for International Development (USAID) in order to provide financial and technical assistance to post-communist societies. 7 Nonetheless, the field of international legal aid has generally been committing a series of mistakes, which resulting in cases of unsuccessful transplantation. Commercial law reform is not free from the flawed premises and assumptions mentioned above, as it clearly reflects some of the main mistakes committed in the field of legal technical assistance. Traditionally, commercial law has been perceived as disconnected from its social and cultural context. Many conceive of commerce as an activity that is separate from people's lives, assuming accordingly that its facilitation and regulation through laws should not be influenced by different cultural attitudes." Following Kahn's metaphor, commercial law therefore would be far from the organic levels of transferability, as it would only serve as a mechanical instrument for which the context is not relevant." However, the cases of unsuccessful transplantation in commercial law reform resulting from legal technical assistance programmes show a different side of the coin. The best examples usually come from projects dealing with post-communist reform, as in these experiences it was generally assumed that the laws for the functioning of a market were culture-neutral. 40 Accordingly, commercial law reform became a transplantation process in which the product being enacted " Michael Bogdam, 'Development Assistance in the Legal Field: Promotion of Market Economy v Human Rights' in Antonina Bakardjieva Engelbrekt and Joakim Nergelius (eds), New Directions in Comparative Law (Edward Elgar 2009) 37. 3 Channell (n 27) 137-138. 3 Nicholas Foster, 'Comparative Commercial Law: Rules or Context' in Esin Orilcil and David Nelken Hart (eds), Comparative Law a Handbook (Hart 2007) 267. 39 ibid. 40 Scott Newton, 'Transplantation and Transition: Legality and Legitimacy in the Kazakhstani Legislative Process' in Scott Newton and Denis J. Galligan (eds), Law and Informal Practices (OUP 2003) 153. 145 Queen Mary Law Journal and its prestigious origins were more important than the process of enactment.41 In order to achieve the desired economic growth to unleash the market forces, new commercial law had to be drafted and successful models of commercial law needed to be identified and transplanted.4 2 Consequently, rule of law interventions were solely focused on the importing of substantive commercial law. Views emphasising the central role of this legal framework strongly support the transfer of commercial laws from a successful model, which it is hoped will 'stimulate private initiatives in the economic field.' 43 In poSt-communist transitions, the prevalent assumption regarding commercial law reform amongst drafters and analysts was that once the adequate rules were created, the market forces would automatically take care of their implementation. 44 It is interesting to note the low degree of interest that antecedent socio-economic analysis or debate with the future users of the laws have attracted. It is also rather surprising to observe the reliance on the 'magic effect of substantive laws': if they have been copied into the recipient's books, all that is left to do is to hope that they will be effective. Nevertheless, these views are right to underline the well-known fact that a market economy requires numerous essential legal rules, such as the law of contracts or company law. Foreign investments could not be made without credit, which would not be available without some regulations regarding security law and others addressing insolvency issues. 4 5 These tools are definitely needed to achieve a better economic performance and develop an attractive commercial climate, but having good laws is not synonymous with creating an attractive investment context, as policymaking must also function efficiently. 4 6 In fact, the lessons from the performance of international legal aid over the years show how doubtful it is that the copy-pasting of these laws will result in successful transplantation. This is proven by research, which shows that many new commercial laws transplanted in this way are still on the books, not having been effectively implemented. 4 7 Additionally, commercial law reform tends to attract a lot of attention from some donors expecting to benefit from it. Very often, even well-crafted legislation is prevented from appropriate application due to hidden interests, which aim to profit from market manipulation. 4 8 Behind what some comparatists call reception of commercial law in transitional contexts, there is 41 ibid. 42 Channell (n 27). 4 Bogdam (n 36) 36. 44 Channell (n 27). 4 Bogdam (n 36) 37. 46 Channel (n 27) 139-141. 4 ibid 138. 4 ibid. 146 Francisca Vicario Bermfidez a variety of political realities and calculations, together with a complex play of interests, which come disguised as imported statutes, legal concepts, and provisions.4 Under these circumstances, hasty transplantation was the go-to method for commercial law reform during transitions. Accordingly, most rule-of-law programmes addressing commercial law have paid no attention to the local commercial culture, allowing an insignificant participation of the directly affected stakeholders in the reform process.o It seems that legal reformers forget about the true objective of amending commercial law, which is the creation of socio-economic prosperity." Otherwise, they would certainly know that the business sector has valuable information about the priorities and the main needs of commerce, and would therefore take its views into consideration in the reform projects.52 Conclusions lead us to think that the 'instrumentalist view'5 3 of commercial law is not the most accurate perspective. The truth, however, is that after applying this approach, societies coming from a post-communist background still struggle between domestic and donor's demands to build a competitive commercial climate. 54 This is an example of how international legal assistance has not been producing the expected results in commercial law reform in transitional countries. In this field of law, donor's simplistic assumptions are translated into a belief that substantive laws will automatically trigger a change in the inclinations of the local marketplace. Regrettably, passing laws is not synonymous with implementing policies. Commercial law reform: examples in bankruptcy legislation Bankruptcy laws have been considered an essential part of the legislative framework of a market economy. They were therefore granted an important role in the context of post-communist commercial law reform. Bankruptcy was a practical tool used by law reformers for the restructuring of state-owned enterprises, as it played a role in their privatisation process.5 ' This area of commercial law represents another clear example of the flaws in legal transplantation, as the following cases will show. ' Newton (n 40) 152. 5o Channell (n 27) 138. s ibid 156-157. 52 ibid. 1 Foster (n 38) 267. 14 Channell (n 27) 158-159. " ibid 138-139. 56 ibid pp. 144-145. 17 Frederique Dahan, 'Hope and Bitterness in the Reform of Russian Bankruptcy Law' in Denis J. Galligan and Marina Kurkchiyan (eds), Law and Informal Practices (OUP 2003) 136. " ibid 137. 147 Queen Mary Law Journal (a) Albania A good example is the reform of Albania's bankruptcy laws, where foreign demands were driving the reform effort." In 1994, the Albanian market-oriented bankruptcy law was adopted and it was a combination of U.S. and German statutes. This legislation had to be repeatedly modified over the years due to donors' requests, as the transplant did not seem to take root.o Interestingly, Albania still does not seem to need such legislation, as not much practical use has been given to it since its creation." 1 However, there is an explanation that donors do not seem to have taken into account: bankruptcy owes its existence to commercial debt and Albania simply has no relevant 'level of commercial lending.' 62 Notwithstanding this, the country accepted the substitution of an unused law with another set of rules that were obviously still not needed: external priorities needed to be satisfied regardless of the real local demands." (b) Kazakhstan In the case of the Kazakhstani bankruptcy legislation, a similar process took place. The first insolvency law was passed in 1992 and several modifications were carried out, although only a few insolvency cases had arisen during those years.64 Out of the amendments, the third law adopted clearly demonstrates how foreign interests clash with the domestic need for law reform. The modification of the law was a response to local demands for a draft based on a comparative study of the best Western insolvency rules, as well as for a heightened focus on the subsequent adaptation of the law." These comprehensible demands set the tone for a collaborative and participatory process of amending the law, carried out by a local working group funded by the World Bank.66 However, once this project was almost finished, USAID put forward its own project. Consequently, this generated competition with the first proposal, giving birth to several debates that resulted in a draft addressing bankruptcy more generally." Under such circumstances, it becomes evident that local needs can get lost along the way when foreign interests have a say in domestic law reform. " Channell (n 27) 145. 60 ibid. 61 ibid. 62 ibid. 61 ibid 146. 64 Newton (n 40) 155. 61 ibid. 66 ibid. 67 ibid. 148 Francisca Vicario Bermfidez IV. CONCLUSION The above analysed aspects of the field of legal technical assistance fit perfectly into Twining's 'naYve model of diffusion': one-way transfers of laws, generally carried out under the assumption of Western superiority and without learning from the results. Accordingly, parallels can be established between some of the theoretical and empirical aspects of legal transplantation. Firstly, the fragmented character of the theory on legal transplants does not contribute to building a clear, well-structured conceptualisation of the phenomenon of transplantation, which could set some guidelines for the achievement of successful transfers. At the same time, examples of the practical application of transplantation without a coherent theoretical basis show that certain repeated mistakes and flawed assumptions do not contribute to a successful law reform process through transplantation. Neither the theoretical nor the practical approaches promote the effective use of transplantation in law reform. However, there is an increasing need to reach definite conclusions as to what methodology is the most appropriate for the transplantation of laws, due to the special urgency for reform in certain contexts such as post-conflict societies." Perhaps it is time to work towards the creation of a single theory on legal transplantation that is more empirical-based, focusing on the lessons learned from experiences rather than on abstract metaphors. On the one hand, those lessons have shown that laws taken as a 'bare string of words'69 and transplanted without knowing what policy they intend to implement will not trigger a specific desired effect. The field of commercial law reform constitutes the best example. The consideration of commerce as separate from its users has led to unsatisfactory results. Thus, experience has shown that, as Kahn anticipated, using the comparative method for law reform implies learning from the social debates between government and interest groups, which is the basic process of policy-making. 70 On the other hand, the flawed assumptions that have been present in the law reform industry for decades can continue to hamper the effective use of transplants. They are closely related to the deep-rooted Western belief in the superiority of their laws, which has more influence on the results of transplantations than what is generally acknowledged. Some critics claim that Western aid is only part of a process of Americanisation of legal thinking. 7 ' They state that this worldwide spread is possible due to a perceived need for it, only created by certain practices known as 'democracy and the rule-of-law' 6 Sannerholm (n 21) 56. 6 Legrand (n 9) 121. 7o Channell (n 27) 144-145. 71 Ugo Mattei, 'Theory of Imperial Law: A Study on US Hegemony and the Latin Resistance' (2003) 10(1) Indiana Journal of Global Legal Studies 383-385. 149 Queen Mary Law Journal promotion.72 In this context, legal transplantation is applied whilst 'telling a story of consensual voluntary reception'7 in which a presumed consent to transplant is granted by the prestige of the laws received. Parallels become self-evident between these theories and the conclusions drawn from decades of rule-of-law promotion programmes producing controversial results. Perhaps these could be more successful if donors remembered the alleged aim of their enterprise: that is, the achievement of holistic socio-economic prosperity in the targeted countries. However, the business-like characteristics of the legal reform industry might lead to the blurring of lines between making a profit and providing legal aid. Nevertheless, the reality is that donors offer aid for different reasons but the recipient finds this offering of free legal technical assistance very difficult to refuse, irrespective of the donor's intentions. This is a result of the reduced options left for a transitional or developing society. In this context, international assistance inevitably becomes a take-it-or-leave-it proposition in which recipients do not normally have another option but to accept a process of transplantation, whose main beneficiary has not been well identified from the beginning. 72 ibid 383-384. 7 ibid 385. 150
1827	https://people.cs.umass.edu/~pthomas/LogicPuzzles.html	Logic Puzzles This is a collection of logic puzzles. I did not create most of these puzzles, I only collected them here. I would particularly like to thank Rich Korf for asking me several of these. The Liar and the Truth Teller First the original: There are two guards and two doors. One door leads to freedom, and the other to death. One guard always lies, the other always tells the truth. They know which they are. They know where the two doors go. You do not know which guard is which or which door is which. You may ask one yes or no question. What do you ask to determine which door leads to freedom? New versions: Same setup, but you do not know how many guards tell the truth and how many lie (they could both be liars!). Another one to try: there is only one guard, and you do not know whether he tells the truth or lies (if you solve this one, then you solved the previous one). Solution » Four Card Code You select five cards from a standard 52 card deck (no jokers) and place them on a table. I enter and look at the cards. I place one in my pocket and place the other four face up in slots marked 1, 2, 3, and 4. I then leave. My accomplice enters, looks at the four cards, and correctly states which card is in my pocket. What strategy could we have used to ensure that my accomplice would always know which card was in my pocket? Solution » The Othello Problem You are in a completely dark room. I dump a bag of 1017 Othello chips on the floor. These chips are black on one side and white on the other. You can feel around for the chips, but you cannot see which side is up because it is dark. I tell you that exactly 23 have the black side up. I ask you to divide the chips into two piles (every chip must be in one [and only one] of the piles) such that the two piles have the same number of chips with the black side up (they may have different numbers of chips with the white side up). How do you do it? Solution » The 7 Hats Problem You are in a room with 7 (or any other number greather than 3) friends. Soon, I will place a hat on each of your heads. There are 7 (same as the number of people) possible hat colors. You know the seven possible colors. I may place any number of each color (e.g., there may be multiple of one color and no hats of another color). You will be able to see everyone else's hat, but not your own. At the exact same time (no communication after your hats are on) you will guess the color of your hat. If any of you get it right, you all win. If you all get it wrong, you all lose. You now have time to come up with a strategy before I place the hats. How do you ensure that you win? Solution » Hanging a Picture You have a framed painting (the kind with a string coming out of the top left and attached to the top right) that you want to hang on the wall using two nails, such that if you remove any one nail the painting will fall, but with both nails in the wall it will not fall. How do you loop the string around the nails? Scales (On Steroids) The original version: There are nine stones. Eight weigh the same amount, and one weighs slightly more than the others. You have a balance that will tell you if the left side weighs more than the right. Using the balance only twice, determine which is the heavy stone. The balance problem on steroids: There are 12 stones, one of which weights a different amount from the others. You do not know whether it is heavier or lighter than the others. Using the balance three times, figure out which stone is different and whether it is heavier or lighter. The Mountain Climber You have a rope that is 150 feet long. You can tie it in different ways and can also cut it. You are on a cliff that is 200 feet tall. 100 feet from the top is a ledge. There are rings to tie the rope to at the top and at the ledge. How can you get down? The Perfectly Logical Pirates There are 100 pirates. They have 10,000 gold pieces. These pirates are ranked from most fearsome (1) to least fearsome (100). To divide the gold, the most fearsome pirate comes up with a method (e.g. split it evenly, or I get half and the second most fearsome gets the other half). The pirates then vote on this plan. If 50% or more vote in favor of the plan, then that is how the gold is divided. If >50% vote against the plan, the most fearsome pirate is killed and the next most fearsome comes up with a plan, etc. The pirates are perfectly rational. The pirates are perfectly rational. Yes, I said that twice - it is important. You are the most fearsome pirate. How much gold can you get without being killed? How? Solution » Black and White Hats There are 100 mathematicians (including you!) in a room. Soon, I will enter and you will all line up such that you can see everyone in front of you, but nobody behind you. I will then place hats on each of yours heads. Each hat may be either black or white. You will not know what color hat you (or anyone behind you in line) is wearing. I will then ask each of you what color hat you are wearing, starting from the back of the line and moving to the front (starting from the person who see's everyone's hat color but his or her own, and ending with the person who sees nobody's hat color). Now, before I enter and place the hats, come up with a strategy to ensure that 75 people live. Can you do even better? Solution » 7 Points on a Plane Draw seven (distinct) points on a piece of paper such that regardless of which three are chosen, at least two will be exactly one unit (e.g., inch or centimeter) apart. As far as I know, there is only one way to do this. Solution » Update » Ant in a Room An ant is in one corner of a room shaped like a cube. It wants to go to the opposite corner. What is the shortest path that the ant can take, and how long is it? 8 Coins I place 8 coins on a table in a line with random sides up. You can look at the coins and then flip one coin. You leave and your friend enters. Come up with a strategy so that your friend can determine which coin you flipped. The Four Quarters Problem You have a square cafeteria tray with four quarters on it - one in each corner. Again, we are in a dark room. You do not know which quarters have the heads side up and which have the tails side up. Your goal is to flip over any coins you want and then ask if they are all heads. If they, are then you win. If they are not, then I rotate the tray randomly and we repeat the process. Can you ensure that you will eventually win? The Two Quarters Problem Do not physically do this - it will ruin it. There are two quarters on a table. One is glued in place, the other is adjacent to it. If you roll the quarter that isn't glued around the one that is glued until it reaches the position it started from, how many degrees will it have rotated by? When I say "roll around" you can pretend the edges of the quarters are like gears. Some people find "how many degrees will it have rotated by" to be ambiguous so here's what I mean: If it starts with the head facing up and moves until the head is facing down, that is 180 degrees. If the head rotates from up to down and all the way around to up again, that is 360 degrees. Once you are convinced you know the answer, try it. The Duck and the Fox There is a duck in the middle of a perfectly circular pond (radius 1). There is a fox running around the outside of the pond at speed 1. The duck is injured in a way that it can only take off from land. How slowly can the duck swim in order to still be able to make it to the edge of the pond without the fox meeting it there? The fox can only run around the pond. They are both points. The duck has no constraints on the derivative of its velocity. Hint: the duck can swim slower than 1/π. Water and Wine You have two identical glasses, filled to the same level. One has water in it and the other has wine. You take a teaspoon of the wine, pour it into the water, and mix them up. You then take a teaspoon of the water (with a little wine in it), pour it into the wine and mix it up. Is there more water in the wine, or more wine in the water? Car Parts If a car is traveling at 60 miles per hour, what part is stationary? What part is going 120 miles per hour? Coin Flip Contest Bob flips 850 fair coins. Alice flips 851. What is the probability that Alice gets strictly more heads than Bob? Rocky Waters You're in a boat with a big rock, in a swimming pool. You toss the rock overboard. What happens to the level of the water in the pool? Gamers of the Round Table You are given a round table, and a large number of coins. You alternate with an opponent placing a coin on the table, not overlapping the edge or any other coins. The winner is the last player who can place a coin. You get to decide who goes first. Do you have a winning strategy? Fox and the Foxholes There is a straight line of N foxholes, and one fox. Every night, the fox moves from his current foxhole to the one either immediately to his left, or immediately to his right. Every day, you get to look in one fox hole. Give a strategy to guarantee finding the fox in the fewest number of days. The Monks with Red Eyes There is an island with red and blue eyed monks. There are no reflections, and nobody speaks about eye color. If they find out their eyes are red, they kill themselves at midnight that night. You go to the island and see that at least one has red eyes. You say "At least one of you has red eyes." Does anyone kill themselves? If so, when? The Three-Way Duel There are three people who want to have a duel. Person A hits 100% of the time, B hits 50% of the time, and C hits 25% of the time. C will shoot first, then B, then A, then C, then B, then A, etc. until only one person is left alive. You are person C. You get to shoot first. What should you do in order to maximize the chance that you will live, assuming that A and B are logical and also want to maximize their chances of survival? Hint: this would not be here if the answer was completely trivial. The Fuses You have two lengths of fuse. Each burns for an hour, exactly. They do not burn at a steady rate, so if you cut one in half, then you do not know if a half will burn for 1 second or 59 minutes (though the sum of the times of the two halves is one hour). How do you time exactly 45 minutes? The Hungry Cow You have a cow in a circular pen (radius 1). You want it to only eat half of the grass in the pen. Assume the cow is a point. You tether the cow to the edge of the pen (circular enclosure). How long should its tether be so that it can eat exactly half the grass? The Weird Warden There is a prison with 100 inmates. The warden strikes a deal with them. There is a room with a light in it (controlled by a light switch). Each day, the warden will take a random prisoner to that room. At some point, a prisoner must say "We have all been in the room!" If he or she is correct, then all are set free. If he or she is wrong, then all will never be set free. The initial state of the light is not known (on or off). Talking is allowed ahead of time, but not after the process begins. You can only communicate by turning the light on/off. Also, the process will begin on a random day, so you do not know if you are the first in or not. You are a prisoner. What plan do you propose in order to ensure that you will gain your freedom? Find any solution that works - do not worry about how long it will take. The prisoners will be taken in randomly: over an infinite amount of time, they will all enter the room an infinite number of times. Bonus: how long will it take for you to be freed? The Really Really Big Urn You have a very big urn, and pebbles numbered with the natural numbers (1, 2, 3...). At time step 1, you put pebbles 1-10 in the urn. At time step 2, you take out pebble 1. At time step 3 you put in 11-20. At time step 4 you take out pebble 2, etc. If you did this an infinite number of times, how many pebbles would be left in the urn? Hint: The limit as the number of iterations goes to infinity may give a different answer. The Ant on the Rubber Band Try to think about this one without using math at first. There is a rubber band attached to a wall. The rubber band is one meter long, levitating horizontally away from the wall. When I say "go", it will start stretching so the end moves at 10 meters per second. It stretches infinitely. The stretch is uniform (e.g. if it grows by 4 meters, the center will move by 2 meters). There is an ant starting where the rubber band connects to the wall. It walks at .1 meters per second down the rubber band, also starting when I say "go". Will the ant ever reach the end of the rubber band? If so, how long will it take? What if the rubber band doubles in length every second? The Party Problem This one came from the Andrew's Leap program at CMU (which was awesome by the way - I highly recommend it). A woman and her husband attended a party with four other couples. As is normal at parties, handshaking took place. Of course, no one shook their own hand or the hand of the person they came with. And not everyone shook everyone else's hand. But when the woman asked the other (9) people present how many different people's hands they had shaken they all gave a different answer. Question (this is NOT a trick!): How many different people's hands did the woman's husband shake? The Dinosaur Eggs You have two very resilient dinosaur eggs. They will absorb a certain amount of force with no negative consequences, but at some point they will crack. If they don't crack, no damage is incurred. You're on a 100 story building. You have 20 trials (you're allowed at most 20 individual egg drops) and 2 eggs. Is it possible to devise a testing strategy that guarantees to tell you at exactly what floor the eggs will break? If after your first trial (dropping an egg off of the balcony on one floor of the building), if the egg does not break, then you have 19 trials remaining and two eggs. If the egg breaks, then you still have 19 trials remaining, but only one egg left. How few trials do you need? Let this number be k. Using k trials, can you solve a 104 story building? How about 105? 106? Jane's Children You work with Jane. You know that she has two children. One day you meet one at a boys' camp, and it is a boy. What is the probability that she has two boys? The Odd Trail Where on the Earth can you walk a mile south, a mile west, and a mile north, and end up exactly where you started? Hint: There are infinite places: find them all! The Interesting Shape What is a shape (defined by a mathematical equation) that has infinite surface area, but finite volume? Bag o' Pebbles There are two empty bags. You have 50 black and 50 white pebbles. You must put all of the pebbles into the two bags. A coin will then be flipped. If it is heads, you discard bag A. If it is tails, you discard bag B. From the remaining bag, you reach in and randomly select a pebble. If it is white, you win - black you lose. How should you put the pebbles in the bag? Does it make a difference? The Hard Problem (This problem is hard. If this is your first look at this page, I recommend skipping it.) There are a countably infinite number of mathematicians in a room. Each has a black or white hat placed on his or her head. Color selection is random. Each mathematician can see the color of everyone else's hat, but not his or her own. At the same moment, with out communicating with each other, all must guess the color of his or her own hat. What strategy could the mathematicians use to guarantee that only a finite number guess incorrectly? The mathematicians may talk before the hats are placed in order to agree on a strategy. You may assume the axiom of choice. Solution »
1828	https://artofproblemsolving.com/wiki/index.php/Trivial_Inequality?srsltid=AfmBOooGqFuNueGKH70jjwNhFFZpCIEQ7HhitAQ0Kq0cboWvRH1PIiBF	Art of Problem Solving Trivial Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Trivial Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Trivial Inequality The trivial inequality is an inequality that states that the square of any real number is nonnegative. Its name comes from its simplicity and straightforwardness. Contents 1 Statement 2 Proof 3 Applications 4 Problems 4.1 Introductory 4.2 Intermediate 4.3 Olympiad Statement For all real numbers, . Proof We can have either , , or . If , then . If , then by the closure of the set of positive numbers under multiplication. Finally, if , then again by the closure of the set of positive numbers under multiplication. Therefore, for all real , as claimed. Applications The trivial inequality is one of the most commonly used theorems in mathematics. It is very well-known and does not require proof. One application is maximizing and minimizing quadratic functions. It gives an easy proof of the two-variable case of the Arithmetic Mean-Geometric Mean inequality: Suppose that and are nonnegative reals. By the trivial inequality, we have , or . Adding to both sides, we get . Since both sides of the inequality are nonnegative, it is equivalent to , and thus we have as desired. Another application will be to minimize/maximize quadratics. For example, Then, we use trivial inequality to get if is positive and if is negative. Problems Introductory Find all integer solutions of the equation . Show that . Solution Show that for all real and . Intermediate Triangle has and . What is the largest area that this triangle can have? (AIME 1992) The fraction, where and are side lengths of a triangle, lies in the interval ![Image 49: $(p,q]$]( where and are rational numbers. Then, can be expressed as , where and are relatively prime positive integers. Find . (Solution here see problem 3 solution 1) Olympiad Let be the length of the hypotenuse of a right triangle whose two other sides have lengths and . Prove that . When does the equality hold? (1969 Canadian MO) Let and be real numbers. Show that (Solution here see problem 13 solution 1) Retrieved from " Categories: Algebra Inequalities Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1829	https://home.ttic.edu/~madhurt/courses/infotheory2021/l1.pdf	Information and Coding Theory Winter 2021 Lecture 1: January 12, 2021 Lecturer: Madhur Tulsiani 1 Administrivia This course will cover some basic concepts in information and coding theory, and their applications to statistics, machine learning and theoretical computer science. - The course will have 4-5 homeworks (60 % of the grade) and a ﬁnal (40 %). The homeworks will be posted on the course homepage and announced in class, and will be due about one week after they are posted. - The pre-requisites for the course are familiarity with discrete and continuous prob-ability and random variables, and algorithmic notions. Some knowledge of ﬁnite ﬁelds will help with the coding theory part though we will brieﬂy review the rele-vant concepts from algebra. - We will not follow any single textbook, though the book Elements of Information The-ory by T. M. Cover and J. A. Thomas is a good reference for most of the material we will cover. The “Resources” section on the course page also contains links to some other similar courses. - The lectures will be held via Zoom, and discussions and ofﬁce hours will be held using the TTIC Gather space. Please see the canvas page of the course for all access information. - Video recordings of the lectures will also be posted to the canvas page for the course. Please check the course webpage regularly for updates. 2 A quick reminder about random variables and convexity 2.1 Random variables Let Ωbe a ﬁnite set. Let µ : Ω→[0, 1] be a function such that ∑ ω∈Ω µ(ω) = 1. 1 We often refer to Ωas a sample space and the function µ as a probability distribution on this space. When Ωis not ﬁnite, µ may need to be replaced by an object called a probability measure (we will discuss this later). Ωand µ are together said to deﬁne a probability space (for inﬁnite Ω, pobability spaces need an additional component called a σ-algebra). A real-valued random variable over Ωis any function X : Ω→R. We deﬁne E [X] = ∑ ω∈Ω µ(ω) · X(ω) . We will also think of a random variable X as given by its distribution. If X is the (ﬁnite) set of values taken by X, we can think of the probability distribution on X given by p(x) = P [X = x] = ∑ ω:X(ω)=x µ(ω) , for all values x ∈X . A word on notation Note that a random variable X is deﬁned simply as a function on Ω, and the distribution of X is induced by the distribution (or measure) µ on Ω. We use the notation P(X) to denote the distribution P for the random variable X. Note that changing the underlying probability space can result in a different distribution (say) Q(X) for the same function X. In information theory notation, it is common to only talk of distributions P, Q, if they are for the same X which is clear from context. Similarly, it is common to deﬁne quantities (such as entropy) which depend on the distribution, simply in terms of random variables X, Y when the underlying probability space is ﬁxed. When we need to talk of multiple random variables, and also of multiple distributions for the same variable X, we will use the more explicit notation P(X). We will use uppercase letters X, Y, Z for random variables, lowercase letters x, y, z to de-note values for these random variables, and caligraphic letters X , Y, cZ to denote the sets of possible values for random variables (also known as the support of a random variable). We will also use uppercase letters P, Q to denote the names of distributions, and lowercase letters p, q to denote probabilities. Thus, a random variable X with distribution P(X) and support X may satisfy that for a speciﬁc value x ∈X , we have p(x) := P [X = x] = 1/2. 2.2 Convexity and Jensen’s inequality A set S ⊂Rn is said to be convex subset of Rn if the line segment joining any two points in S lies entirely in S i.e., for all x, y ∈S and for all α ∈[0, 1], α · x + (1 −α) · y ∈S. For 2 a convex set S ⊆Rn, a function f : S →R is said to be a convex function on S, if for all x, y ∈S and for all α ∈[0, 1], we have f (α · x + (1 −α) · y) ≤α · f (x) + (1 −α) · f (y) . Equivalently, we say that the function f is convex if the set Sf = {(x, z) \| z ≥f (x)} is a convex subset of Rn+1. f is said to be strictly convex when the inequality above is strict for all x, y, α. A function which satisﬁes the opposite inequality i.e., for all x, y ∈S and α ∈[0, 1] f (α · x + (1 −α) · y) ≥α · f (x) + (1 −α) · f (y) , is said to be a concave function (and strictly concave if the inequalities are strict). Note that if f is a convex function then −f is a concave function (and vice-versa). For a single variable function f : R →R which is twice differentiable, we can also use the easier criterion that f is convex on S ⊆R if and only if f ′′(x) ≥0 for all x ∈S. We will frequently use the following inequality about convex functions. Lemma 2.1 (Jensen’s inequality). Let S ⊆Rn be a convex set and let X be a random variable taking values only inside S. Then, for a convex function f : S →R, we have that E [ f (X)] ≥f (E [X]) . Equivalently, for a concave function f : S →R, we have E [ f (X)] ≤f (E [X]) . Note that the deﬁnition of convexity is the same as the statement of Jensen’s inequality for a random variable taking only two values: x with probability α and y with probability 1 −α. You can try the following exercises to familiarize yourself with this inequality. Exercise 2.2. Prove Jensen’s inequality when the random variable X has a ﬁnite support. Exercise 2.3. Check that the f (x) = x2 is a convex function on R. Also show that the functions log(x) and x log(x) are respectively, concave and convex functions on (0, ∞). Exercise 2.4. Prove the Cauchy-Schwarz inequality using Jensen’s inequality. 3 Entropy The concepts from information theory are applicable in many areas as it gives a precise mathematical way of stating and answering the following question: How much informa-tion is revealed by the outcome of a random event? Let us begin with a few simple exam-ples. Let X be a random variable which takes the value a with probability 1/2 and b with 3 probability 1/2. We can then describe the value of X using one bit (say 0 for a and 1 for b). Suppose it takes one of the values {a1, . . . , an}, each with probability, then we can describe the outcome using ⌈log2(n)⌉bits. The n possible outcomes for this random variable each occur with probability 1/n, and require ≈log2(n) bits to describe. The concept of entropy is basically an extrapolation of this idea when the different out-comes do not occur with equal probability. We think of the “information content” of an event that occurs with probability p as being log2(1/p). If a random variable X is dis-tributed over a universe X = {a1, . . . , an} such that it takes value x ∈X with probability p(x). Then, we deﬁne the entropy of the random variable X as H(X) = ∑ x∈X p(x) · log 1 p(x) . The following basic property of entropy is extremely useful in applications to counting problems. Proposition 3.1. Let X be a random variable supported on a ﬁnite set X as above. Then 0 ≤H(X) ≤log(\|X \|) . Proof: Since p(x) ≤1 we have log(1/p(x)) ≥0 for all x ∈X and hence H(X) ≥0. For the upper bound, consider a random variable Y which takes value 1/p(x) with probability p(x). Since log(·) is a concave function, we use Jensen’s inequality to say that ∑ x∈X p(x) · log 1 p(x) = E [log(Y)] ≤log (E [Y]) = log ∑ x∈X p(x) · 1 p(x) ! = log(\|U\|) . 4 Source Coding We will now attempt to make precise the intuition that a random variable X takes H(X) bits to describe on average. We shall need the notion of preﬁx-free codes as deﬁned below. Deﬁnition 4.1. A code for a set X over an alphabet Σ is a map C : X →Σ∗which maps each element of X to a ﬁnite string over the alphabet Σ. We say that a code is preﬁx-free if for any x, y ∈X such that x ̸= y, C(x) is not a preﬁx of C(y) i.e., C(y) ̸= C(x) ◦σ for any σ ∈Σ∗. 4 For now, we will just use Σ = {0, 1}. For the rest of lecture, we will use preﬁx-free code to mean preﬁx-free code over {0, 1}. The image C(x) for an image x is also referred to as the codeword for x. Note that a preﬁx-free code has the convenient property that if we are receiving a stream of coded symbols, we can decode them online. As soon as we see C(x) for some x ∈U, we know what we have received so far cannot be a preﬁx for C(y), for any y ̸= x. The following inequality gives a characterization of the lengths of codewords in a preﬁx-free code. This will help prove both upper and lower bounds on the expected length of a codeword in a preﬁx-free code, in terms of entropy. Proposition 4.2 (Kraft’s inequality). Let \|X \| = n. There exists a preﬁx-free code for X over {0, 1} with codeword lengths ℓ1, . . . , ℓn if and only if n ∑ i=1 1 2ℓi ≤1 . For codes over a larger alphabet Σ, we replace 2ℓi above by \|Σ\|ℓi. Proof: Let us prove the “if” part ﬁrst. Given ℓ1, . . . , ℓn satisfying ∑i 2−ℓi ≤1, we will construct a preﬁx-free code C with these codeword lengths. Without loss of generality, we can assume that ℓ1 ≤ℓ2 ≤· · · ≤ℓn = ℓ∗. It will be useful here to think of all binary strings of length at most ℓas a complete binary tree. The root corresponds to the empty string and each node at depth d corresponds to a string of length d. For a node corresponding to a string s, its left and right children correspond respectively to the strings s0 and s1. The tree has 2ℓ∗leaves corresponding to all strings in {0, 1}ℓ∗. We will now construct our code by choosing nodes at depth ℓ1, . . . , ℓn in this tree. When we select a node, we will delete the entire tree below it. This will maintain the preﬁx-free property of the code. We ﬁrst chose an arbitrary node s1 at depth ℓ1 as a codeword of length ℓ1 and delete the subtree below it. This deletes 1/2ℓ1 fraction of the leaves. Since there are still more leaves left in the tree, there exists a node (say s2) at depth ℓ2. Also, s1 cannot be a preﬁx of s2, since s2 does not lie in the subtree below s1. We choose s2 as the second codeword in our code C. We can similarly proceed to choose other codewords. At each step, we have some leaves left in the tree since ∑i 2−ℓi ≤1. Note that we need to carry out this argument in increasing order of lengths. Otherwise, if we choose longer codewords ﬁrst, we may have to choose a shorter codeword later which does not lie on the path from the root to any of the longer codewords, and this may not always possible e.g., there exists a code with lengths 1, 2, 2 but if we choose the strings 01 and 10 ﬁrst then there is no way to choose a codeword of length 1 which is not a preﬁx. For the “only if” part, we can simply reverse the above proof. Let C be a given preﬁx-free code with codeword lengths ℓ1, . . . , ℓn and let ℓ∗= max {ℓ1, . . . , ℓn}. Considering again the 5 complete binary tree of depth ℓ∗, we can now locate the codewords (say) C(x1), . . . , C(xn) as nodes in the tree. We say that a codeword C(x) dominates a leaf L if L occurs in the subtree rooted at C(x). Note that the out of the total 2ℓ∗fraction of leaves dominated by a codeword of length ℓi is 2−ℓi. Also, note that if C(x) and C(y) dominate the same leaf L, then either C(x) appears in the subtree rooted at C(y) or vice-versa. Since the code is preﬁx-free, this cannot happen and the sets of leaves dominated by codewords must be disjoint. Thus, we have ∑i 2−ℓi ≤1. This part of the proof also has a probabilitic interpretation. Consider an experiment where we generate ℓ∗random bits. For x ∈X , let Ex denote the event that the ﬁrst \|C(x)\| bits we generate are equal to C(x). Note that since C is a preﬁx-free code, Ex and Ey are mutually exclusive for x ̸= y. Moreover, the probability that Ex happens is exactly 1/2\|C(x)\|. This gives 1 ≥∑ x∈X P [Ex] = ∑ x∈X 1 2\|C(x)\| = n ∑ i=1 1 2ℓi . 6
1830	https://brainly.com/question/12215677	[FREE] Express 0.001 in scientific notation. - brainly.com Advertisement Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +45,1k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +22,9k Ace exams faster, with practice that adapts to you Practice Worksheets +6,1k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Express 0.001 in scientific notation. 1 See answer Explain with Learning Companion NEW Asked by sjmata • 02/28/2019 0:02 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 31865 people 31K 5.0 0 Upload your school material for a more relevant answer answer ; 1 10^-3 why ? in order to make something into scientific notation , you have to make the number between 1 and 10. 0.001 isnt between 1 and 10 , so what you do i move the decimal and count. in this problem i moved it 3 times till it was behind the one and if you go to the left , it's a negative. therefore , i did 1 10^-3 ( im guessing you know why its 10 lol ) Answered by flyin •29 answers•31.9K people helped Thanks 0 5.0 (1 vote) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 31865 people 31K 5.0 0 Microbiology - OpenStax Microbiology for Allied Health Students - Molly Smith Laboratory Exercises in Microbiology: Discovering the Unseen World Through Hands-On Investigation - Joan Petersen Upload your school material for a more relevant answer The number 0.001 can be expressed in scientific notation as 1×1 0−3 by moving the decimal point 3 places to the right, resulting in a negative exponent. Explanation To express 0.001 in scientific notation, follow these steps: Identify the Non-Zero Digit: The first non-zero digit in 0.001 is 1. We want to write the number so that it falls between 1 and 10. Move the Decimal Point: Starting with 0.001, we need to move the decimal point to the right until it is directly after the 1. If we rewrite 0.001, it can be seen as 1.0. Moving the decimal point 3 places to the right gives us 1.0. Determine the Exponent: Since we moved the decimal to the right to get to 1, the exponent will be negative. Specifically, we moved it 3 places to the right, so the exponent will be -3. Write in Scientific Notation: Now we can express the number in scientific notation. It is written as: 1×1 0−3 So, in scientific notation, 0.001 is written as 1×1 0−3. Understanding scientific notation is very useful because it allows us to easily deal with very small or very large numbers. It’s a standardized way to write numbers that makes calculations and comparisons clearer. Examples & Evidence For example, the number 0.01 can be expressed as 1×1 0−2 by moving the decimal point 2 places to the right. The method for converting numbers to scientific notation is widely accepted in mathematics and can be found in standard educational resources, illustrating how to convert numbers both small and large effectively. Thanks 0 5.0 (1 vote) Advertisement sjmata has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? 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1831	https://www.quora.com/How-do-I-differentiate-ax-b-x	How to differentiate ax+b/x - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Calculus Meaning of Differentiatio... Functions (mathematics) Derivatives (finance) Calculus (Mathematics) Derivatives and Different... Mathematical Functions Continuty and Differentia... Polynomial Differentiatio... 5 How do I differentiate ax+b/x? All related (33) Sort Recommended Assistant Bot · 1y To differentiate the function f(x)=a x+b x f(x)=a x+b x, we can simplify it first. We can rewrite the function as: f(x)=a x x+b x=a+b x f(x)=a x x+b x=a+b x Now, we can differentiate this function term by term: The derivative of a a (a constant) is 0 0. The derivative of b x b x can be found using the power rule. Recall that b x=b x−1 b x=b x−1, so its derivative is: d d x(b x−1)=b⋅(−1)x−2=−b x 2 d d x(b x−1)=b⋅(−1)x−2=−b x 2 Putting it all together, the derivative of f(x)f(x) is: f′(x)=0−b x 2=−b x 2 f′(x)=0−b x 2=−b x 2 Thus, the derivative of f(x)=a x+b x f(x)=a x+b x is: f'(x)=-\frac{b}{x f'(x)=-\frac{b}{x Continue Reading To differentiate the function f(x)=a x+b x f(x)=a x+b x, we can simplify it first. We can rewrite the function as: f(x)=a x x+b x=a+b x f(x)=a x x+b x=a+b x Now, we can differentiate this function term by term: The derivative of a a (a constant) is 0 0. The derivative of b x b x can be found using the power rule. Recall that b x=b x−1 b x=b x−1, so its derivative is: d d x(b x−1)=b⋅(−1)x−2=−b x 2 d d x(b x−1)=b⋅(−1)x−2=−b x 2 Putting it all together, the derivative of f(x)f(x) is: f′(x)=0−b x 2=−b x 2 f′(x)=0−b x 2=−b x 2 Thus, the derivative of f(x)=a x+b x f(x)=a x+b x is: f′(x)=−b x 2 f′(x)=−b x 2 Upvote · Bryan Turner Former Math Tutor at Mathnasium (2015–2017) ·7y I am going to assume you mean in terms of x. Fortunately for us, differentiation can be broken up across addition, so we get the following: d/dx (ax+b/x) d/dx(ax)+d/dx(b/x) And constants can be pulled out: ad/dx(x) + bd/dx(x^-1) And d/dx (x^n) = nx^(n-1), so: a(1x^0) + b(-x^-2) And so the answer is best written: a - b/(x^2) Upvote · Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Work faster with AI, while ensuring your writing always makes the right impression. Download 999 209 Nancy Mitchell used to be a teacher. · Author has 3.4K answers and 8M answer views ·7y How do I differentiate ax+b/x? d d x(a x+b x)d d x(a x+b x) =d d x a x+d d x(b x)=d d x a x+d d x(b x) =a d x d x+b d d x(1 x)=a d x d x+b d d x(1 x) =a(1)+b d d x x−1=a(1)+b d d x x−1 =a+b(−1)x−1−1 d x d x=a+b(−1)x−1−1 d x d x =a−b x−2(1)=a−b x−2(1) =a−b x 2=a−b x 2 Upvote · 99 12 9 1 9 1 Related questions More answers below What is the derivative of (a x+b)/√x(a x+b)/x? What are the derivatives of ax+b/√x? How do you differentiate 2ax+b with respect to x? What is the differentiation sin(ax+b)? How can I differentiate y=x x x⋯∞y=x x x⋯∞? Sohel Zibara Studied at Doctor of Philosophy Degrees (Graduated 2000) · Author has 5.1K answers and 2.6M answer views ·2y Related What is the derivative of x/ (ax + b)? Upvote · Vijaykumar N. Joshi Former Prof and Head , Dept. Of Computer Science , RETD (1980–2002) · Author has 1.9K answers and 1.7M answer views ·1y Related How do you differentiate with respect to X by logarithmic function (ax+b) ^cx+d? Upvote · 9 2 Related questions More answers below How do you differentiate (ax+b÷x) (a÷ x+bx)? Is f(x) =1/(1/x) differentiable at 0? How do I differentiate -x/1000? How do you differentiate y = ax+b with respect to x? How do you differentiate x to the power of 2? Arjun Prasad Studied at Kendriya Vidyalaya ·5y Related How do you differentiate 2ax+b with respect to x? Continue Reading Upvote · 9 3 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 624 Calvin L. 2nd year mathematics student · Author has 10K answers and 2.2M answer views ·2y Related What is the derivative of x/ (ax + b)? Use the quotient rule. d d x x a x+b=a x+b−a x(a x+b)2=b(a x+b)2 d d x x a x+b=a x+b−a x(a x+b)2=b(a x+b)2 Upvote · Subhayan Chatterjee Upvoted by Misho Okropiridze , MSc Mathematics, Tbilisi State University (2024) · Author has 182 answers and 263.3K answer views ·7y Related How to determine d/d x log a d/d x log⁡a base (a x+b)(a x+b)? I will assume you mean d(ln a x+b ln a)d x d(ln⁡a x+b ln⁡a)d x [You should know that log y x=ln x ln y log y⁡x=ln⁡x ln⁡y Under this assumption, I will ask Wolfram Alpha to help me with the calculation. Continue Reading I will assume you mean d(ln a x+b ln a)d x d(ln⁡a x+b ln⁡a)d x [You should know that log y x=ln x ln y log y⁡x=ln⁡x ln⁡y Under this assumption, I will ask Wolfram Alpha to help me with the calculation. Upvote · 9 2 Sponsored by RedHat Build and run AI anywhere, with hybrid cloud platforms. Your AI should open doors, not lock them. Learn More 9 1 Krishna Singh Bachelor of Science in Statistics (academic discipline)&Mathematician And Theoretical Physicist, Patna Science College, Patna (Graduated 2020) · Author has 478 answers and 2M answer views ·6y Related Prove that E (aX+b) =aE(X) +b? Thanks for A2A, Prove that E (aX+b) =aE(X) +b Here, 'a' and 'b' are constants. Then, That's it, Continue Reading Thanks for A2A, Prove that E (aX+b) =aE(X) +b Here, 'a' and 'b' are constants. Then, That's it, Upvote · 99 11 9 1 Hadrien Chevalier PhD in Theoretical Physics, Imperial College London (Graduated 2022) · Author has 102 answers and 1.1M answer views ·Updated 7y Related Can you differentiate factorial x? No, you cannot differentiate factorial x x. This is because the factorial, being the number of permutations of a set containing x x elements, is defined only for x∈Z x∈Z (it is equal to zero for negative integers). A derivative of a function f f at a point x x is defined only if f f is also defined in a neighbourhood of x x(that is, a set that contains at least a topologically open ball that is centered on x x). This means that in one dimension, you need a (bilateral) continuum around x x in order to define its (bilateral) derivative, which is why we do not talk about the derivative of a function that is Continue Reading No, you cannot differentiate factorial x x. This is because the factorial, being the number of permutations of a set containing x x elements, is defined only for x∈Z x∈Z (it is equal to zero for negative integers). A derivative of a function f f at a point x x is defined only if f f is also defined in a neighbourhood of x x(that is, a set that contains at least a topologically open ball that is centered on x x). This means that in one dimension, you need a (bilateral) continuum around x x in order to define its (bilateral) derivative, which is why we do not talk about the derivative of a function that is defined on a discrete set (such as Z Z). The closest you can get to what you want is by defining the following function : Γ:R∗+⟶R+,x⟼∫+∞0 t x−1 e−t d t Γ:R+∗⟶R+,x⟼∫0+∞t x−1 e−t d t (Let’s keep it simple, and forget about its definition on the open complex half-plane of strictly positive real part numbers … let alone its analytical extension to the whole complex plane - excluding poles. Let’s say it is just defined for strictly positive real numbers…) This integral converges absolutely, and one can show with an integration by parts that : ∀x∈R∗+,Γ(x+1)=x Γ(x)∀x∈R+∗,Γ(x+1)=x Γ(x) Furthermore, Γ(1)=1 Γ(1)=1 is trivial. So we can think of the factorial function as a restriction of the Γ Γ function to N N and set it equal to zero for negative integers. What this mean is that : ∀x∈N,Γ(x+1)=x!∀x∈N,Γ(x+1)=x! In this sense, we may have a shot at defining something that may be the “derivative of the factorial” (though such a phrasing is abusive). The Leibniz integral rule states that this Γ Γ function is indeed differentiable (you can check all hypothesis do hold) and : ∀x∈R∗+,Γ′(x)=∫+∞0∂∂x(e(x−1)ln(t)e−t)d t∀x∈R+∗,Γ′(x)=∫0+∞∂∂x(e(x−1)ln⁡(t)e−t)d t Hence : ∀x∈R∗+,Γ′(x)=∫+∞0 ln(t)t z−1 e−t d t∀x∈R+∗,Γ′(x)=∫0+∞ln⁡(t)t z−1 e−t d t (Notice that Leibniz integral rule will hold for any power of ln(t)ln⁡(t) so a simple induction proves that Γ Γ is in fact infinitely differentiable on R∗+R+∗). Now…maybe this way of writing the derivative isn’t very satisfying. So let’s define a function Ψ Ψ called the digamma function : Ψ:R∗+⟶R+,x⟼Γ′(x)Γ(x)Ψ:R+∗⟶R+,x⟼Γ′(x)Γ(x) (This is the logarithmic derivative of Γ Γ and one will have easily shown that Γ Γ has no poles on R∗+R+∗ which allows such a definition to hold). Hence we can rewrite the derivative of Γ Γ in the following way : ∀x∈R∗+,Γ′(x)=Γ(x)Ψ(x)∀x∈R+∗,Γ′(x)=Γ(x)Ψ(x) For this last part…I am not sure what I do is rigorous. (Comments from mathematicians are welcome !) Why is this useful ? I could just give the result directly but let’s take it step by step… Well let’s consider x∈N∗x∈N∗ (strictly positive integers) then you know that : Γ(x+1)=Γ(x)×(x)Γ(x+1)=Γ(x)×(x) So ln(Γ(x+1))=ln(Γ(x))+ln(x)ln⁡(Γ(x+1))=ln⁡(Γ(x))+ln⁡(x) Differentiating with respect to x x yields : d ln(Γ(x+1))d x=d ln(Γ(x))d x+1 x d ln⁡(Γ(x+1))d x=d ln⁡(Γ(x))d x+1 x Iterating this logarithmic differentiation yields : d ln(Γ(x+1))d x=1+1 2+1 3+...+1 x=h x d ln⁡(Γ(x+1))d x=1+1 2+1 3+...+1 x=h x ((x)(x)-th harmonic partial sum). But you know that logarithmic differentiation also takes the following form : d ln(Γ(x+1))d x=Γ′(x+1)Γ(x+1)d ln⁡(Γ(x+1))d x=Γ′(x+1)Γ(x+1) Hence we “magically” find that : ∀x∈N∗,Ψ(x+1)=h x=1+1 2+1 3+...+1 x∀x∈N∗,Ψ(x+1)=h x=1+1 2+1 3+...+1 x (Note : actually this digamma function is not only useful to give an answer to your question - it’s an extension of the partial harmonic sums to the much broader domain of complex numbers !) So all in all we are left with something that I believe is pretty close to what you were asking for : ∀x∈N∗,Γ′(x+1)=Γ(x+1)Ψ(x+1)=Γ(x+1)h x∀x∈N∗,Γ′(x+1)=Γ(x+1)Ψ(x+1)=Γ(x+1)h x So you could also write : ((x+1)!)′=x!x∑k=1 1 k((x+1)!)′=x!∑k=1 x 1 k But I will not box this final equation because it is conceptually extremely abusive,though it may be the closest to what I think you sought. Hope it helped ! Upvote · 99 36 9 1 Sponsored by SpeechText.AI AI-powered legal transcription service. Convert speech to text from any audio/video format, including TRM files. Fully compliant with GDPR. Start Now 9 3 George Ivey Former Math Professor at Gallaudet University · Author has 23.7K answers and 2.6M answer views ·3y Related How do you differentiate the function f(x) =ax+b? First learn some Calculus! Learn that the derivative of Cf(x), for C a constant, is C times the derivative of f, that the derivative of f(x)+ g(x) is the derivative of f plus the derivative of g, that the derivative of a constant is 0, and that the derivative of x^n is nx^(n-1) so that the derivative of ax+ b= ax^1 is a(1^0)+ 0= a. Before that, you should have learned the DEFINITION of the derivative as “limit, as h goes to 0, of (f(x+h)- f(x))/h. Here f(x)= ax+ b so f(x+ h)= a(x+ h)+ b= ax+ ah+ b so f(x+h)- f(x)= ax+ ah+ b- ax- b= ah. So (f(x+ h)- f(x))/h= ah/h= a. That has no “h” so its limit Continue Reading First learn some Calculus! Learn that the derivative of Cf(x), for C a constant, is C times the derivative of f, that the derivative of f(x)+ g(x) is the derivative of f plus the derivative of g, that the derivative of a constant is 0, and that the derivative of x^n is nx^(n-1) so that the derivative of ax+ b= ax^1 is a(1^0)+ 0= a. Before that, you should have learned the DEFINITION of the derivative as “limit, as h goes to 0, of (f(x+h)- f(x))/h. Here f(x)= ax+ b so f(x+ h)= a(x+ h)+ b= ax+ ah+ b so f(x+h)- f(x)= ax+ ah+ b- ax- b= ah. So (f(x+ h)- f(x))/h= ah/h= a. That has no “h” so its limit, as h goes to 0, is a. Even before that you should have learned that the derivative is generalization of the “slope” of a linear function. f(x)= ax+ b IS a linear function with slope a, so its derivative is a. Upvote · 9 1 Ajesh K C Studied at Mar Athanasius College of Engineering, Kothamangalam · Author has 1.3K answers and 3.3M answer views ·5y Related What is the differentiation of log (ax+b) ^3? Upvote · 9 2 Leo Temperini Studied at Liceo Scientifico ·7y Related How do I differentiate ax^b? I could just write “use the power rule”, but that wouldn’t be that useful would it? The question equates to the question “differentiate x b x b”, because a constant, like “a”, can be ignored. My assumption is that you have been taught why we can disregard a constant, but you haven’t been taught how to differentiate “x b x b”. So, let’s do this. I hope someone told you about the concept of implicit derivation and that of the chain rule. Either way, I’m not here to talk about that, all you need to know is that we can differentiate the two sides of an equation to… do stuff we need to get to a result and that Continue Reading I could just write “use the power rule”, but that wouldn’t be that useful would it? The question equates to the question “differentiate x b x b”, because a constant, like “a”, can be ignored. My assumption is that you have been taught why we can disregard a constant, but you haven’t been taught how to differentiate “x b x b”. So, let’s do this. I hope someone told you about the concept of implicit derivation and that of the chain rule. Either way, I’m not here to talk about that, all you need to know is that we can differentiate the two sides of an equation to… do stuff we need to get to a result and that we can differentiate a function like l n(x+2)l n(x+2) by multiplying the derivative of l n(x+2)l n(x+2) with respect to x+2 x+2 and that of x+2 x+2 with respect to x x. y=l n(x+2)y=l n(x+2) d y d x=d(l n(x+2)d(x+2)d(x+2)d x d y d x=d(l n(x+2)d(x+2)d(x+2)d x I also hope you remember the logarithmic rules. Here we go: y=x b y=x b l n(y)=l n(x b)l n(y)=l n(x b) l n(y)=b(l n(x))l n(y)=b(l n(x)) We differentiate everything implicitly d(l n(y))d y d(l n(y))d y d y d x=b d(l n(x))d x d y d x=b d(l n(x))d x d(l n(y))d y d y d x=b 1 x d(l n(y))d y d y d x=b 1 x We find that d y d x=y b 1 x d y d x=y b 1 x d y d x=b x b 1 x d y d x=b x b 1 x d y d x=b x b−1 d y d x=b x b−1 And since we had that "a" at the start... y′=a b x b−1 y′=a b x b−1 Upvote · 9 4 Related questions What is the derivative of (a x+b)/√x(a x+b)/x? How do you differentiate 2ax+b with respect to x? What are the derivatives of ax+b/√x? What is the differentiation sin(ax+b)? How do you differentiate xln [1+(1/x)]? How do you differentiate (ax+b÷x) (a÷ x+bx)? Is f(x) =1/(1/x) differentiable at 0? How do I differentiate -x/1000? How do you differentiate y = ax+b with respect to x? How do you differentiate x to the power of 2? What is differentiation of x+✓x+1/✓x? How do I differentiate X^2/1-8x? How do you differentiate d y d x=a x+b d y d x=a x+b with respect to x? Is x 1+\|x\|x 1+\|x\|differentiable? How do you differentiate (ax+b) nW.r. to x? Related questions What is the derivative of (a x+b)/√x(a x+b)/x? What are the derivatives of ax+b/√x? How do you differentiate 2ax+b with respect to x? What is the differentiation sin(ax+b)? How can I differentiate y=x x x⋯∞y=x x x⋯∞? How do I differentiate sin (ax+b)/cos (cx+d) with respect to x? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1832	https://math.stackexchange.com/questions/3373056/does-the-radius-of-the-base-equal-the-height-in-a-right-circular-cone	geometry - Does the radius of the base equal the height in a right circular cone? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Does the radius of the base equal the height in a right circular cone? Ask Question Asked 6 years ago Modified6 years ago Viewed 1k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I am doing question 6 of a practice calculus exam, namely: A container in the shape of a right circular cone with vertex angle a right angle is partially filled with water. a) Suppose water is added at the rate of 3 cu.cm./sec. How fast is the water level rising when the height h = 2cm.? My answer was d h d t=3 4 π⋅c 2 d h d t=3 4 π⋅c 2, where c c is a fixed constant equal to the ratio of the height to the radius. The given solution, however, is d h d t=3 4 π d h d t=3 4 π. The solution uses the assumption that "r=h r=h since this is a right circular cone." Is this a reasonable assumption? I have checked definition of right circular cone and cannot find anything about this. Thanks, Josh geometry related-rates Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Sep 28, 2019 at 11:39 Josh TaylorJosh Taylor 33 4 4 bronze badges 2 Are you sure you have quoted the question properly?Sam –Sam 2019-09-28 11:43:59 +00:00 Commented Sep 28, 2019 at 11:43 @Sam Yes, I copied it from here: ocw.mit.edu/courses/mathematics/…Josh Taylor –Josh Taylor 2019-09-29 12:52:07 +00:00 Commented Sep 29, 2019 at 12:52 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Let me amplify Roddy's answer. A right circular cone has a circular base; let's assume that it's at the origin of 3-space, and that the base lies in the x y x y-plane The cone is "right" so that its vertex lies along the perpendicular to the base-plane, through the base-center. So let's go ahead and say that the vertex is at the point (0,0,h)(0,0,h), just to give the thing a coordinate. ("Right" here says nothing about "right angle" -- it means "upright rather than slonched over to one side", to use the technical term :) ) Looking solely at the x z x z-plane, i.e., a slice through the cone, we see a cross section that looks like this shape: ∧∧, with the angle at the vertex (location (0,0,h)(0,0,h), remember) being a right angle (because the problem says "with vertex angle a right angle"). The location of the bottom vertex of the right leg is then (r,0,0)(r,0,0) (because y=0 y=0 in the x z x z-plane, and because the whole bottom of the cone is in the z=0 z=0 plane). If we drop a vertical line from the top of the wedge to the z=0 z=0 line in our picture, the half-angle at the top is 45 degrees. That makes the ratio of the legs of the triangle (which are r r and h h in length) be tan 45=1 tan⁡45=1. Hence r h=1 r h=1, so r=h r=h. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Sep 29, 2019 at 14:00 John HughesJohn Hughes 102k 4 4 gold badges 88 88 silver badges 160 160 bronze badges 1 Thank you very much for your detailed response! I see now that I had combined the notion of a right circular cone and the cone having a right angled vertex. Oops!Josh Taylor –Josh Taylor 2019-09-29 17:09:32 +00:00 Commented Sep 29, 2019 at 17:09 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Really it assumes, the 90 degree angle at the vertex, is split in two by the height, which produces the r=h r=h result. It's a valid assumption, unless you want a deformed cone with more area on one side of the height than the other. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Sep 29, 2019 at 13:47 answered Sep 28, 2019 at 11:52 user645636 user645636 5 I'm happy that the right angle means the vertex lies directly over the centre of the base of the cone. But without any known specific values (or ratio) for the radius or base, I don't think we can say that they are equal.Josh Taylor –Josh Taylor 2019-09-29 13:06:34 +00:00 Commented Sep 29, 2019 at 13:06 tan(45)=1 it really is that simple.user645636 –user645636 2019-09-29 13:47:06 +00:00 Commented Sep 29, 2019 at 13:47 Also the right angle only matters for the r=h argument, it's a requirement to split equally in two for all right circular cones. regardless of angle at the vertex.user645636 –user645636 2019-09-29 14:01:30 +00:00 Commented Sep 29, 2019 at 14:01 @JoshTaylor: The problem states that the cone is one "with vertex angle a right angle"; as Roddy restates it, the cone has a "90 degree angle at the vertex". This (along with the fact that the vertex lies directly above the center of the base) is what gives the equal dimensions. .. That said, it does seem curious that the official solution emphasizes the right-cone-ness, when that seems far more likely to be understood than right-angle-vertex-ness. (Perhaps the editor accidentally conflated the two different uses of "right".)Blue –Blue 2019-09-29 14:01:33 +00:00 Commented Sep 29, 2019 at 14:01 Aha! I see it now! Thank you both for your help. I was confusing it being a right circular cone with the vertex being a right angle and conflating the two.Josh Taylor –Josh Taylor 2019-09-29 17:07:46 +00:00 Commented Sep 29, 2019 at 17:07 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry related-rates See similar questions with these tags. 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1833	https://www.wikiwand.com/en/articles/alkyne_trimerisation	Alkyne trimerisation Chemical reaction of three alkynes to form a benzene ring From Wikipedia, the free encyclopedia An alkyne trimerisation is a [2+2+2] cycloaddition reaction in which three alkyne units (C≡C) react to form a benzene ring. The reaction requires a metal catalyst. The process is of historic interest as well as being applicable to organic synthesis. Being a cycloaddition reaction, it has high atom economy. Many variations have been developed, including cyclisation of mixtures of alkynes and alkenes as well as alkynes and nitriles. Mechanism and stereochemistry Trimerisation of acetylene to benzene is highly exergonic, proceeding with a free energy change of 142 kcal/mol at room temperature. Kinetic barriers however prevent the reaction from proceeding smoothly. The breakthrough came in 1948, when Walter Reppe and W. J. Schweckendiek reported their wartime results showing that nickel compounds are effective catalysts: Since this discovery, many other cyclotrimerisations have been reported. Mechanism In terms of mechanism, the reactions begin with the formation of metal-alkyne complexes. The combination of two alkynes within the coordination sphere affords a metallacyclopentadiene. Starting from the metallacyclopentadiene intermediate, many pathways can be considered including metallocycloheptatrienes, metallanorbornadienes, and a more complicated structure featuring a carbenoid ligand. Catalysts used include cyclopentadienylcobalt dicarbonyl and Wilkinson's catalyst. Stereochemistry and regiochemistry Trimerisation of unsymmetrical alkynes gives two isomeric benzenes. For example, phenylacetylene affords both 1,3,5- and 1,2,4-C6R3H3. The substitution pattern about the product arene is determined in two steps: formation of the metallocyclopentadiene intermediate and incorporation of the third equivalent of alkyne. Steric bulk on the alkyne coupling partners and catalyst have been invoked as the controlling elements of regioselectivity. Chiral catalysts have been employed in combination with arynes to produce non-racemic atropisomeric products. Scope and limitations Catalysts for cyclotrimerisation are selective for triple bonds, which gives the reaction a fairly wide substrate scope. Many functional groups are tolerated. Regioselective intermolecular trimerization of unsymmetrical alkynes remains an unsolved problem. Perhaps the most useful development in this area, at least from the commercial perspective is the cotrimerization of nitriles and alkynes. This reaction is a practical route to some substituted pyridines. Some catalysts are deactivated by formation of stable, 18-electron η4-complexes. Cyclobutadiene, cyclohexadiene, and arene complexes have all been observed as off-cycle, inactivated catalysts. In addition to high-order polymers and dimers and trimers, which originate from low regio- and chemoselectivities, enyne side products derived from alkyne dimerisation have been observed. Rhodium catalysts are particularly adept at enyne formation (see below). For nickel catalysis, formation of larger rings (particularly cyclooctatetraene) can be a problem. Synthetic applications Alkyne trimerization is of no practical value, although the reaction was highly influential. The cotrimerization of alkynes and nitriles in the presence of organocobalt catalysts has been commercialized for the production of substituted pyridines. Cyclization involving substrates in which some or all of the alkyne units are tethered together can provide fused ring systems. The length of the tether(s) controls the sizes of the additional rings. Addition of a 1,5-diyne with an alkyne produces a benzocyclobutene, a strained structure that can then be induced to undergo further reactions. All three alkyne units can be tethered, leading to creation of three rings in a single step, with each of the two additional ring sizes controlled by the respective tether lengths. Crowded triynes can cyclize to products exhibiting helical chirality. In one example remarkable for the formation of three new aromatic rings in one step, the triyne shown is transformed into the helical product via treatment with cyclopentadienylcobalt dicarbonyl. As of 2004, this process had yet to be rendered asymmetric,[original research?] but the products could be separated through chiral HPLC. Cyclisation carried out with a diyne and a separate alkyne affords greater control.[clarification needed] Using commercially available cyclopentadienylcobalt dicarbonyl, CpCo(CO)2, as catalyst, bis(trimethylsilyl)acetylene (BTMSA) will react with a diyne-1,2-disubstituted benzene to form an anthroquinone aromatic system: Benzyne, generated in situ from a benzene ring bearing ortho-distributed triflate and trimethylsilyl substituents, can be used to generate an aryne in place of an acetylene and combined with a suitable diyne. Such a benzene derivative reacts with 1,7-octadiyne in the presence of a suitable catalyst to generate a naphthalene system. This is an example of a hexadehydro Diels–Alder reaction. Trimerisation of three 2-butyne (dimethylacetylene) molecules yields hexamethylbenzene. The reaction is catalyzed by triphenylchromium tri-tetrahydrofuranate or by a complex of triisobutylaluminium and titanium tetrachloride. Trimerisation of three diphenylacetylene molecules yields hexaphenylbenzene. The reaction is catalyzed by dicobalt octacarbonyl. Comparison with other methods Cyclotrimerization presents an alternative to the functionalization of pre-formed aromatic compounds through electrophilic or nucleophilic substitution, the regioselectivity of which can sometimes be difficult to control. Other methods for the direct formation of aromatic rings from substituted, unsaturated precursors include the Dötz reaction, palladium-catalyzed [4+2] benzannulation of enynes with alkynes, and Lewis-acid-mediated [4+2] cycloaddition of enynes with alkynes. Cyclization of transient benzyne species with alkynes, catalyzed by palladium, can also produce substituted aromatic compounds. Further reading References Wikiwand - on Seamless Wikipedia browsing. On steroids. Wikiwand ❤️ Wikipedia
1834	https://www.jove.com/science-education/v/12727/strain-and-elastic-modulus	Video: Strain and Elastic Modulus Strain and Elastic Modulus - JoVE We value your privacy We use cookies to enhance your browsing experience, serve personalised ads or content, and analyse our traffic. By clicking "Accept All", you consent to our use of cookies.Cookie Policy Customise Accept All Customise Consent Preferences We use cookies to help you navigate efficiently and perform certain functions. You will find detailed information about all cookies under each consent category below. The cookies that are categorised as "Necessary" are stored on your browser as they are essential for enabling the basic functionalities of the site. ...Show more Necessary Always Active Necessary cookies are required to enable the basic features of this site, such as providing secure log-in or adjusting your consent preferences. These cookies do not store any personally identifiable data. 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JoVE Core Physics Strain and Elastic Modulus Cancel live 00:00 00:00 1x Speed Ã— Slow Normal Fast Faster CC Subtitles Ã— English MEDIA_ELEMENT_ERROR: Format error Strain and Elastic Modulus 4,170 Views01:15 min April 30, 2023 Overview The quantity that describes the deformation of a body under stress is known as strain. Strain is given as a fractional change in either length, volume, or geometry under tensile, volume (also known as bulk), or shear stress, respectively, and is a dimensionless quantity. The strain experienced by a body under tensile or compressive stress is called tensile or compressive strain, respectively. In contrast, the strain experienced under bulk stress and shear stress is known as volume and shear strain, respectively. The greater the stress, the greater the strain; however, the relationship between strain and stress is not necessarily linear. Only when the stress acting on a body is sufficiently low is the deformation caused by it directly proportional to the stress value. The proportionality constant in this relation is called the elastic modulus. As strain is dimensionless, the physical unit of elastic modulus is the same as stress (pascal). When a body is characterized by a large value of elastic modulus, the effect of stress is small. On the other hand, a small elastic modulus means that stress produces large strain and a noticeable deformation. For example, stress applied on a rubber band produces a larger strain (deformation) than the same stress applied on a steel band with the same dimensions, because the elastic modulus for rubber is two orders of magnitude smaller than the elastic modulus for steel. The elastic modulus for tensile stress is called the Young's modulus; for bulk stress, it is called the bulk modulus; and for shear stress, it is called the shear modulus. Table 1. Approximate Elastic Moduli for Selected Materials MaterialYoung's modulus (10 10 Pa)Bulk modulus (10 10 Pa)Shear modulus (10 10 Pa) Aluminum 7.0 7.5 2.5 Bone (tension)1.6 0.8 8.0 Brass 9.0 6.0 3.5 Brick 1.5 Copper 11.0 14.0 4.4 Crown glass 6.0 5.0 2.5 Granite 4.5 4.5 2.0 Hardwood 1.5 1.0 Iron 21.0 16.0 7.7 Lead 1.6 4.1 0.6 Marble 6.0 7.0 2.0 Nickel 21.0 17.0 7.8 Polystyrene 3.0 Silk 6.0 Steel 20.0 16.0 7.5 This text is adapted fromOpenstax, University Physics Volume 1, Section 12.3: Stress, Strain, and Elastic Modulus. Transcript The amount of deformation experienced by an object under stress, divided by its initial dimensions, is called strain. Strain is a dimensionless quantity. If the deformation due to the applied force results in an increase in the length of an object, it is called tensile strain. Whereas, if the length decreases due to stress, it is called compressive strain. The change in the volume of an object due to the applied force in all directions is known as bulk strain. If the direction of the force is parallel to the object's plane, it undergoes deformation along that plane, and this is known as shear strain. For small deformations, the elastic modulus is defined as the ratio of the stress exerted on the object to the resultant strain. Young's modulus is the ratio of tensile stress to tensile strain. The ratio of volumetric stress to volumetric strain is known as bulk modulus, whereas the ratio of shear stress to shear strain corresponds to shear modulus. Explore More Videos StrainElastic ModulusDeformationStressTensile StrainCompressive StrainVolume StrainShear StrainYoung's ModulusBulk ModulusShear ModulusMaterial PropertiesStress-strain RelationshipMechanical Properties Related Videos 01:05 ### Static Equilibrium - I Equilibrium and Elasticity 13.3K Views 01:07 ### Static Equilibrium - II Equilibrium and Elasticity 9.0K Views 00:58 ### Center of Gravity Equilibrium and Elasticity 5.7K Views 01:03 ### Finding the Center of Gravity Equilibrium and Elasticity 3.9K Views 00:49 ### Rigid Body Equilibrium Problems - I Equilibrium and Elasticity 4.8K Views 01:21 ### Rigid Body Equilibrium Problems - II Equilibrium and Elasticity 7.5K Views 01:20 ### Stress Equilibrium and Elasticity 7.9K Views 01:15 ### Strain and Elastic Modulus Equilibrium and Elasticity 4.1K Views 01:22 ### Problem Solving on Stress and Strain Equilibrium and Elasticity 1.2K Views 01:18 ### Indeterminate Structure Equilibrium and Elasticity 972 Views 01:12 ### Elasticity Equilibrium and Elasticity 4.0K Views 00:58 ### Plasticity Equilibrium and Elasticity 2.5K Views Contact UsRecommend to Library Research JoVE Journal JoVE Encyclopedia of Experiments JoVE Visualize Business JoVE Business Education JoVE Core JoVE Science Education JoVE Lab Manual JoVE Quizzes Solutions Authors Teaching Faculty Librarians K12 Schools About JoVE Overview Leadership Others JoVE Newsletters JoVE Help Center Blogs Site Maps Contact UsRecommend to Library Copyright © 2025 MyJoVE Corporation. 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1835	https://artofproblemsolving.com/wiki/index.php/2012_AMC_8_Problems/Problem_23?srsltid=AfmBOooU4kVgmyYoXReVCxwZtB8XKAUCFdfeTaj-usJqBI7V7la-eR37	Art of Problem Solving 2012 AMC 8 Problems/Problem 23 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2012 AMC 8 Problems/Problem 23 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2012 AMC 8 Problems/Problem 23 Contents 1 Problem 2 Solution 1 3 Solution 2 (Hard) 4 Video Solution 5 Video Solution by OmegaLearn 6 See Also Problem An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon? Solution 1 Let the perimeter of the equilateral triangle be . The side length of the equilateral triangle would then be and the sidelength of the hexagon would be . A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio , since the sidelength of the small equilateral triangle is half the side length of the large one. Thus, the area of one of the small equilateral triangles is . The area of the hexagon is then . Solution 2 (Hard) Let the side length of the small triangle be , so the side length of the hexagon is . By the Pythagorean theorem, the height of the triangle is , so the area of the triangle is We are given that the area is , so x must be . Therefore, by the Pythagorean theorem, the height of one of the triangles in the hexagon (made by subdividing the hexagon into congruent triangles), is , so the area is , which equals . Since there are such triangles in the hexagon, the area of the hexagon is . Video Solution ~savannahsolver Video Solution by OmegaLearn ~ pi_is_3.14 See Also 2012 AMC 8 (Problems • Answer Key • Resources) Preceded by Problem 22Followed by Problem 24 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25 All AJHSME/AMC 8 Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1836	https://www.youtube.com/watch?v=dQ9r2S7NWLs	2 Examples of Probability With & Without Replacement Cole's World of Mathematics 46000 subscribers 6656 likes Description 480838 views Posted: 3 Apr 2017 This video goes through 2 examples of Probability. One example uses "With Replacement" and one example uses "Without Replacement". mathematics #probability #maths Math Tutorials on this channel are targeted at college-level mathematics courses including calculus, pre-calculus, college algebra, trigonometry, probability theory, TI-84 tutorials, introductory college algebra topics, and remedial math topics from algebra 1 and 2 for the struggling college adult (age 18 and older). Don’t forget guys, if you like this video please “Like” and “Share” it with your friends to show your support - it really helps me out! SUBSCRIBE HERE for more Math Tutorials --► Interested in purchasing the TI-84 Plus CE Graphing Calculator? 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Thank you for the support! 458 comments Transcript: Introduction in this video I'm going to be taking a look at just a couple different probability examples uh one with replacement and one without replacement and probably about the easiest way to do this is to pretend that we've got a bag and it's full of marbles all right so in this first example I want to do a width replacement all right we're going to be finding the probability of drawing a red marble and then a blue marble and then a green marble in that order and with with replacement okay now what I mean is with replacement is what's my odds if I reach in the bag I pull out a marble okay and then with replacement means I put it back in before I draw the next marble okay so that's what we're going to be doing in this first example so um in my bag right here hopefully the colors are going to show up pretty good here I have uh five blue marbles I have four green Marbles and I have six red marbles okay so that makes for a total of 15 marbles in my bag so I have a total of 15 all right so I'm going to start the very first thing I want to do is I want to know what's the probability that I reach in the bag and I draw out a red marble on the first um try so there are 1 2 3 4 five six there are six red and there are 15 total in the bag With Replacement okay now I reached in I pulled it out okay and then that's the probability that that first one was red now it says with replacement so I'm going to take that marble and I'm going to put it back in the bag so I still have 15 marbles in there all right now times what's the probability of drawing a Blue Marble like I'm over I count how many Blues 1 2 3 four five so I've got five blue marbles and since it was with replacement I put the first marble back in I still have 15 marbles so 15 on the bottom okay now I do it a third time what's the probability of getting a green marble well there are four green Marbles and there is still a total of 15 because it was with replacement so when I picked that second marble out I put it back in the bag which still leaves me a total of 15 in there so that's 4 15 all right this is straight multiplication you're going to multiply across top you're going to multiply across the bottom you're going to get um 120 across the top if you grab calculator on that and across the bottom you'll get 3,375 so that's your probability if you want to write it as a decimal you can convert to a decimal which means it's about 035 repeating and if you needed to write it as a percent and show that as a percent you could say about 3.5 repeating percent okay so that's the concept of with replacement okay now let's do the Without Replacement exact same thing but this time let's do without replacement okay so the question is what's the probability of drawing a red marble and then a blue marble and then a green marble in that order without replacement so I'm going to draw my first one and then I'm actually going to remove it out of the bag okay so what's the probability of drawing a red marble okay so initially I have 15 Marbles and there's one two three four five six there are six red out of the 15 total all right now that first draw I drew it out it is out of my bag I am not putting it back in okay I'm not going to put it back in because it's without replacement so now I'm going to ask myself okay what's the probability of drawing a Blue Marble well in the bag 1 2 3 four five I have five blue ones all right but because I did not put my first draw back in now I only have 14 marbles in the bag 1 2 3 4 5 6 7 8 9 10 11 12 13 14 so so the bottom number decreases by one all right now that second draw pull out the blue one okay now do not replace it so it's out of my bag now what's the probability of drawing a green marble well in my bag I have 1 2 3 four I have four green ones all right but now because I did not replace that second one I am down to 13 marbles 1 2 3 4 5 6 7 8 9 9 10 11 12 13 so now I'm done to 13 so the without replacement changes what you have in your bag okay again straight multiplication multiply across the top we're going to get 120 multiplying 15 14 13 we're going to get 2730 all right so that's your probability if you want need to convert it to a decimal that's going to be about 0439 and if we want that probability as per that'd be roughly 4.4% okay so just two real quick examples there of calculating um some probabilities with replacement versus without replacement um definitely thanks for watching hope this was helpful in clearing up the with and without replacement and be sure and give me a thumbs up and share with your friends thanks
1837	https://www.pearson.com/channels/precalculus/learn/patrick/18-systems-of-equations-and-matrices/determinants-and-cramers-rule	Skip to main content My Courses Chemistry General Chemistry Organic Chemistry Analytical Chemistry GOB Chemistry Biochemistry Intro to Chemistry Biology General Biology Microbiology Anatomy & Physiology Genetics Cell Biology Physics Physics Math College Algebra Trigonometry Precalculus Calculus Business Calculus Statistics Business Statistics Social Sciences Psychology Health Sciences Personal Health Nutrition Business Microeconomics Macroeconomics Financial Accounting Product & Marketing Agile & Product Management Digital Marketing Project Management AI in Marketing Programming Introduction to Python Microsoft Power BI Data Analysis - Excel Introduction to Blockchain HTML, CSS & Layout Introduction to JavaScript R Programming Calculators AI Tools Study Prep Blog Study Prep Home Table of contents Skip topic navigation Skip topic navigation Fundamental Concepts of Algebra 3h 32m Algebraic Expressions 37m + 32m + Polynomials Intro 19m + Multiplying Polynomials 36m + Factoring Polynomials 2m + Radical Expressions 15m + Simplifying Radical Expressions 31m + Rationalize Denominator 15m + Rational Exponents 4m + Pythagorean Theorem & Basics of Triangles 17m 1. Equations and Inequalities 3h 27m Linear Equations 31m + Rational Equations 21m + The Imaginary Unit 6m + Powers of i 11m + Complex Numbers 42m + Intro to Quadratic Equations 18m + The Square Root Property 11m + Completing the Square 12m + The Quadratic Formula 19m + Choosing a Method to Solve Quadratics 9m + Linear Inequalities 22m 2. Graphs 1h 43m Graphs and Coordinates 7m + Two-Variable Equations 23m + 1h 12m 3. Functions & Graphs 2h 17m Intro to Functions & Their Graphs 35m + Common Functions 5m + Transformations 45m + Function Operations 21m + Function Composition 29m 4. Polynomial Functions 1h 54m Quadratic Functions 49m + Understanding Polynomial Functions 34m + Graphing Polynomial Functions 30m 5. Rational Functions 1h 23m Introduction to Rational Functions 9m + Asymptotes 35m + Graphing Rational Functions 38m 6. Exponential and Logarithmic Functions 2h 28m Introduction to Exponential Functions 9m + Graphing Exponential Functions 25m + The Number e 8m + Introduction to Logarithms 22m + Graphing Logarithmic Functions 18m + Properties of Logarithms 27m + Solving Exponential and Logarithmic Equations 35m 7. Measuring Angles 40m Angles in Standard Position 12m + Coterminal Angles 8m + Complementary and Supplementary Angles 10m + Radians 8m 8. Trigonometric Functions on Right Triangles 2h 5m Trigonometric Functions on Right Triangles 45m + Special Right Triangles 30m + Cofunctions of Complementary Angles 26m + Solving Right Triangles 23m 9. Unit Circle 1h 19m Defining the Unit Circle 14m + Trigonometric Functions on the Unit Circle 9m + Common Values of Sine, Cosine, & Tangent 11m + Reference Angles 38m + Reciprocal Trigonometric Functions on the Unit Circle 6m 10. Graphing Trigonometric Functions 1h 19m Graphs of the Sine and Cosine Functions 32m + Phase Shifts 14m + Graphs of Secant and Cosecant Functions 10m + Graphs of Tangent and Cotangent Functions 21m 11. Inverse Trigonometric Functions and Basic Trig Equations 1h 41m Inverse Sine, Cosine, & Tangent 28m + Evaluate Composite Trig Functions 44m + Linear Trigonometric Equations 28m 12. Trigonometric Identities 2h 34m Introduction to Trigonometric Identities 1h 4m + Sum and Difference Identities 1h 3m + Double Angle Identities 16m + Solving Trigonometric Equations Using Identities 9m 13. Non-Right Triangles 1h 38m Law of Sines 49m + Law of Cosines 30m + Area of SAS & ASA Triangles 19m 14. Vectors 2h 25m Geometric Vectors 28m + Vectors in Component Form 32m + Direction of a Vector 23m + Unit Vectors and i & j Notation 27m + Dot Product 22m + Cross Product 11m 15. Polar Equations 2h 5m Polar Coordinate System 29m + Convert Points Between Polar and Rectangular Coordinates 26m + Convert Equations Between Polar and Rectangular Forms 29m + Graphing Other Common Polar Equations 39m 16. Parametric Equations 1h 6m Graphing Parametric Equations 12m + Eliminate the Parameter 32m + Writing Parametric Equations 21m 17. Graphing Complex Numbers 1h 7m Graphing Complex Numbers 6m + Polar Form of Complex Numbers 22m + Products and Quotients of Complex Numbers 14m + Powers of Complex Numbers (DeMoivre's Theorem) 23m 18. Systems of Equations and Matrices 3h 6m Two Variable Systems of Linear Equations 8m + Introduction to Matrices 1h 11m + Determinants and Cramer's Rule 1h 3m + Graphing Systems of Inequalities 42m 19. Conic Sections 2h 36m Introduction to Conic Sections 6m + Circles 26m + Ellipses: Standard Form 33m + Parabolas 36m + Hyperbolas at the Origin 40m + Hyperbolas NOT at the Origin 12m 20. Sequences, Series & Induction 1h 15m Sequences 33m + Arithmetic Sequences 25m + Geometric Sequences 15m 21. Combinatorics and Probability 1h 45m Factorials 11m + Combinatorics 46m + Probability 47m 22. Limits & Continuity 1h 49m Introduction to Limits 41m + Finding Limits Algebraically 40m + Continuity 27m 23. Intro to Derivatives & Area Under the Curve 2h 9m Tangent Lines & Derivatives 44m + Limits at Infinity 18m + Limits of Sequences 9m + Area Under a Curve 56m Systems of Equations and Matrices Determinants and Cramer's Rule Systems of Equations and Matrices Determinants and Cramer's Rule: Videos & Practice Problems Video Lessons Practice Worksheet Topic summary Understanding determinants is crucial for solving systems of equations using Cramer's rule. For a 2 by 2 matrix, the determinant is calculated as . For 3 by 3 matrices, determinants involve smaller 2 by 2 calculations. Cramer's rule allows for direct solutions of equations by substituting coefficients and constants into determinant formulas, facilitating efficient problem-solving in linear algebra. 1 concept Determinants of 2×2 Matrices Video duration: 4m Play a video: Determinants of 2×2 Matrices Video Summary In the study of matrices, one important concept is the determinant, which is a scalar value that can be calculated from a square matrix. Understanding how to compute the determinant is essential, especially when solving systems of equations. The determinant provides insights into the properties of the matrix, such as whether it is invertible. For a 2x2 matrix, the determinant can be calculated using a straightforward formula. Given a matrix represented as: the determinant is calculated using the formula: Here, and are the elements on the main diagonal (from the top left to the bottom right), while and are the elements on the other diagonal (from the top right to the bottom left). The process involves multiplying the elements of each diagonal and then subtracting the product of the second diagonal from the product of the first. For example, consider the matrix: To find the determinant, we calculate: This means the determinant of this matrix is 2. This process can be applied to any 2x2 matrix, allowing for the evaluation of its determinant easily. Let’s look at another example with the matrix: Using the determinant formula, we find: Thus, the determinant of this matrix is -20. It’s important to note that the determinant can also be calculated with negative numbers. For instance, with the matrix: the calculation would be: In summary, the determinant is a crucial concept in linear algebra, providing valuable information about the matrix's characteristics. Mastering the calculation of determinants for 2x2 matrices lays the groundwork for more complex matrix operations and applications in various mathematical fields. Study Smarter with Worksheets. Follow along with each video using our printable worksheets 2 Problem Evaluate the determinant of the matrix. A 3 107 B 6 227 C 33 D 12 3 concept Cramer's Rule - 2 Equations with 2 Unknowns Video duration: 6m Play a video: Cramer's Rule - 2 Equations with 2 Unknowns Video Summary Cramer's Rule is a method used to solve a system of linear equations with two variables by utilizing determinants of matrices. To apply Cramer's Rule, you first need to express the system of equations in matrix form. For example, consider the equations: 1. 2. These can be represented in an augmented matrix format, where the coefficients of and are organized into a matrix alongside the constants: In Cramer's Rule, the solution for and is found using the determinants of matrices. The formula for is given by: And for : Where is the determinant of the coefficient matrix: Next, to find , replace the first column of the coefficient matrix with the constants from the right side of the equations: For , replace the second column with the constants: Now, substituting these determinants back into the formulas gives: Thus, the solution to the system of equations is and . You can verify this solution by substituting these values back into the original equations to ensure both equations hold true. Cramer's Rule is a powerful tool for solving linear systems, especially when dealing with larger matrices, as it provides a systematic approach to finding solutions through determinants. 4 Problem Write each equation in standard form and use Cramer's Rule to solve the system. y = − 3 x + 4 − 2 x = 7 y − 9 A x 1 , y = 1 B x − 1 , y = 1 C x 1 , y = − 1 D x − 1 , y = − 1 5 Problem Write each equation in standard form and use Cramer's Rule to solve the system. y − 9 x = − 3 − 3 x = 4 y − 1 A y 3 , x = 0 B x 0 , y = 3 C x − 3 1 , y = 1 D x 3 1 , y = 0 6 concept Determinants of 3×3 Matrices Video duration: 7m Play a video: Determinants of 3×3 Matrices Video Summary To calculate the determinant of a 3 by 3 matrix, the process is slightly more complex than for a 2 by 2 matrix, but it can be broken down into manageable steps involving 2 by 2 determinants. The determinant of a 3 by 3 matrix can be expressed using the formula: For a matrix A represented as: the determinant is calculated as: where M represents the minor matrices obtained by removing the row and column of the respective element. To illustrate, consider a 3 by 3 matrix: To find the determinant, we first identify the elements in the first row: 3, 1, and 0. The signs alternate starting with a plus for the first term: For the first term, take 3 and calculate the determinant of the minor matrix: which gives: For the second term, take 1 and calculate the determinant of the minor matrix: which gives: For the third term, take 0, which will result in zero regardless of the minor matrix. Putting it all together, we have: Thus, the determinant of the 3 by 3 matrix is 14. This method effectively reduces the complexity of calculating a 3 by 3 determinant by leveraging the simpler 2 by 2 determinants, making the process more manageable. 7 Problem Evaluate the determinant of the matrix. A 165 B 9 C 63 D 25 8 concept Cramer's Rule - 3 Equations w/ 3 Unknowns Video duration: 13m Play a video: Cramer's Rule - 3 Equations w/ 3 Unknowns Video Summary To solve a system of three equations with three unknowns using Cramer's rule, we first need to understand the structure of the equations and how to represent them in matrix form. Cramer's rule provides a straightforward method to find the values of the unknowns by using determinants of matrices. Given a system of equations, we can express it in the form of an augmented matrix. For example, if we have the equations: We can create the augmented matrix by organizing the coefficients of the variables and the constants: \end{p} In Cramer's rule, we define a determinant (denoted as ) which is the determinant of the coefficient matrix formed by the coefficients of , , and . If is zero, the system has no unique solution. If is non-zero, we can find the solutions for , , and using the following formulas: [x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D}\end{p} Where , , and are the determinants of matrices formed by replacing the respective columns of the coefficient matrix with the constants from the right side of the equations. To calculate , we use the formula for the determinant of a 3x3 matrix: [D = a(ei - fh) - b(di - fg) + c(dh - eg)\end{p} Where the matrix is represented as: [\end{p} After calculating , we proceed to find , , and by substituting the appropriate columns with the constants. For instance, to find , we replace the first column (the coefficients of ) with the constants: [D_x = \end{p} Similarly, we calculate and by replacing the second and third columns respectively. Once we have all determinants, we can substitute them back into the formulas for , , and to find their values. For example, if we find , , , and , we can compute: [x = \frac{-15}{-3} = 5, \quad y = \frac{15}{-3} = -5, \quad z = \frac{-18}{-3} = 6\end{p} Thus, the solution to the system of equations is , , and . 9 Problem Solve the system of equations using Cramer's Rule. 4 x + 2 y + 3 z = 6 x + y + z = 3 5 x + y + 2 z = 5 A x − 2 , y = − 8 , z = 4 B x 1 , y = 4 , z = − 2 C x 2 , y = 8 , z = − 4 D x − 1 , y = − 4 , z = 2 Do you want more practice? More sets [Determinants and Cramer's Rule Systems of Equations and Matrices 6 problems Topic 18. Systems of Equations and Matrices 4 topics 10 problems Chapter Go over this topic definitions with flashcards More sets Determinants and Cramer's Rule definitions Systems of Equations and Matrices 15 Terms Take your learning anywhere! Prep for your exams on the go with video lessons and practice problems in our mobile app. Here’s what students ask on this topic: How do you calculate the determinant of a 2x2 matrix? To calculate the determinant of a 2x2 matrix, use the formula: For example, if the matrix is: The determinant is calculated as: What is Cramer's Rule and how is it used to solve systems of equations? Cramer's Rule is a method used to solve systems of linear equations using determinants. For a system of two equations with two unknowns: The solutions for x and y are given by: Calculate the determinants and substitute them into the formulas to find x and y. To calculate the determinant of a 3x3 matrix, use the following method: For example, if the matrix is: The determinant is calculated as: To use Cramer's Rule to solve a system of 3 equations with 3 unknowns, follow these steps: Write the system of equations in matrix form: Calculate the determinant of the coefficient matrix (D). Replace the x-column with the constants and calculate the determinant (Dx). Replace the y-column with the constants and calculate the determinant (Dy). Replace the z-column with the constants and calculate the determinant (Dz). Solve for x, y, and z using: For example, if D = -3, Dx = -15, Dy = 15, and Dz = -18, then: Determinants have several important applications in linear algebra: Solving Systems of Equations: Using Cramer's Rule, determinants can solve systems of linear equations. Matrix Inversion: Determinants help determine if a matrix is invertible. A non-zero determinant indicates an invertible matrix. Eigenvalues and Eigenvectors: Determinants are used in characteristic equations to find eigenvalues and eigenvectors of a matrix. Area and Volume Calculation: Determinants can calculate the area of parallelograms and the volume of parallelepipeds in geometry. Transformations: In linear transformations, determinants indicate scaling factors and whether the transformation preserves orientation. Understanding these applications is crucial for advanced studies in linear algebra and related fields. 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1838	https://www.cuemath.com/algebra/long-division-of-polynomials/	LearnPracticeDownload Long Division of Polynomials Long division of polynomials is the process of dividing one polynomial with another. Division can be done among the different types of polynomials i.e. between two monomials, a polynomial and a monomial, or between two polynomials. A polynomial is n algebraic expression with variables, terms, and coefficients with the degree of the expressions. Let us explore the division of polynomials by learning about the methods to divide using long division, long division with polynomials, long division with missing terms, the algorithm, and solved a few examples to understand the process better. \| \| \| --- \| \| 1. \| What is Long Division of Polynomials? \| \| 2. \| Steps For Long Division of Polynomials \| \| 3. \| Long Division of Polynomial by Missing Terms \| \| 4. \| Long Division of Polynomials by Monomials \| \| 5. \| Long Division of Polynomials by Other Monomials \| \| 6. \| Long Division of Polynomials by Binomials \| \| 7. \| Long Division of Polynomials by Other Polynomials \| \| 8. \| Long Division Algorithm of Polynomials \| \| 9. \| FAQs on Long Division of Polynomials \| What is Long Division of Polynomials? A long division polynomial is an algorithm for dividing polynomial by another polynomial of the same or a lower degree. The long division of polynomials also consists of the divisor, quotient, dividend, and the remainder as in the long division method of numbers. Observe the numerator and denominator in the long division of polynomials as shown in the figure. The long division of polynomials also consists of a divisor, a quotient, a dividend, and a remainder. In algebra, the division of algebraic expressions can be done in three ways: Division of a monomial by another monomial. Division of a polynomial by a monomial. Division of a polynomial by a binomial. Division of a polynomial by another polynomial. Steps For Long Division of Polynomials The following are the steps for the long division of polynomials: Step 1. Arrange the terms in the decreasing order of their indices (if required).Write the missing terms with zero as their coefficient. Step 2. For the first term of the quotient, divide the first term of the dividend by the first term of the divisor. Step 3. Multiply this term of the quotient by the divisor to get the product. Step 4. Subtract this product from the dividend, and bring down the next term (if any). The difference and the brought down term will form the new dividend. Step 5. Follow this process until you get a remainder, which can be zero or of a lower index than the divisor. Long Division of Polynomial by Missing Terms While performing long division of polynomials, there can be a missing term in the expression, for example, 6x4 + 3x - 9x2 + 6, x3 is missing. In this case, we either leave a gap while dividing or we write the coefficient as zero. Let's understand how to do the long division of polynomials with the same example. We need to divide the polynomial a(x) = 6x4 + 3x - 9x2 + 6 by the quadratic polynomial b(x) = x2 - 2 Arrange the polynomial in the descending order of the power of the variable. a(x) = 6x4 - 9x2 + 3x + 6 b(x) = x2 - 2 Divide a(x) by b(x) in the same way as we divide numbers. Add the missing indices with zero (0) as the coefficient. Divide 6x4 by x2 to get the first term of the quotient. We get 6x2. Multiply the divisor by 6x2. Divide 3x2 by x2 to get the next term of the quotient. As the power of the next dividend is less than the divisor, we get our required remainder. Please remember that as the remainder we got is a non-zero term, we can say that x2 - 2 is not a factor of 6x4 - 9x2 + 3x + 6. Therefore, the quotient is 6x2 + 3 and the remainder is 3x. Long Division of Polynomials by Monomials While dividing polynomials by monomials, write the common factor between the numerator and the denominator of the polynomial and divide each term separately. Once the result is obtained, add all the terms together to form an expression. For example: Divide the following polynomial: (2x2 + 4x + 8xy) ÷ 2x. Both the numerator and denominator have a common factor of 2x. Thus, the expression can be written as 2x(x + 2 + 4y) / 2x. Canceling out the common term 2x, we get x + 4y + 2 as the answer. Long Division of Polynomials by Other Monomial Long division of polynomials by another monomial is done in a similar manner as done for polynomials by monomials. The factors of the monomial of both the numerator and denominator are listed out and the long division takes place. For example, divide 62x3 by 2x. The factors of 62x3 = 2 × 31 × x × x × x and 2x = 2 × x. The common factors for both are 2x. Hence, 62x3/2x = 31x2. Long Division of Polynomials by Binomials Long division of polynomials by binomials is done when there are no common factors between the numerator and the denominator, or if you can't find the factors. Let us go through the algorithm of dividing polynomials by binomials using an example: Divide: (6x2 - 4x - 24) ÷ (x - 3). Here, (6x2 - 4x - 24) is the dividend, and (x - 3) is the divisor which is a binomial. Observe the division shown below, followed by the steps. Step 1. Divide the first term of the dividend (6x2) by the first term of the divisor (x), and put that as the first term in the quotient (6x). Step 2. Multiply the divisor by that answer, place the product (6x2 - 18x) below the dividend. Step 3. Subtract to create a new polynomial (14x - 24). Step 4. Repeat the same process with the new polynomial obtained after subtraction. So, when we are dividing a polynomial (6x2 - 4x - 24) with a binomial (x - 3), the quotient is 6x + 14 and the remainder is 18. Long Division of Polynomials by Other Polynomial Long division of a polynomial with another polynomial is done when the expression is written in the standard form i.e. the terms of the dividend and the divisor are arranged in decreasing order of their degrees. The long division method for polynomials is considered the generalized version of the simple long division method done with numbers. Let us look at an example to understand this better. The process of division is very similar to the rest of the methods. Divide the polynomial 6x3 + 12x2 + 2x + 25 by x2 + 4x + 3. Here, 6x3 + 12x2 + 2x + 25 is the dividend, and x2 + 4x + 3 is the divisor which is also a polynomial. Step 1: Divide the first term of the dividend (6x3) by the first term of the divisor (x2), and put that as the first term in the quotient (6x). Step 2: Multiply the divisor by that answer, place the product (6x3 + 24x2 + 18x) below the dividend. Step 3: Subtract to create a new polynomial (-12x2 - 16x + 25). Step 4: Repeat the same process with the new polynomial obtained after subtraction. So, when we are dividing a polynomial 6x3 + 12x2 + 2x + 25 with a binomial x2 + 4x + 3, the quotient is 6x - 12 and the remainder is 32x + 61. Long Division Algorithm of Polynomials The division algorithm for polynomials says, if p(x) and g(x) are the two polynomials, where g(x) ≠ 0, we can write the division of polynomials as: p(x) = q(x) × g(x) + r(x). Where, p(x) is the dividend. q(x) is the quotient. g(x) is the divisor. r(x) is the remainder. r(x) = 0 or degree of r(x) < degree of g(x) If we compare this to the regular division of numbers, we can easily understand this as: Dividend = (Divisor X Quotient) + Remainder. Let us take the previous example, p(x) = 6x3 + 12x2 + 2x + 25 g(x) = x2 + 4x + 3 q(x) = 6x - 12 r(x) = 32x + 61 Apply the division algorithm, q(x) × g(x) + r(x) (6x - 12) × (x2 + 4x + 3) + (32x + 61) 6x3 + 24x2 + 18x - 12x2 - 48x - 36 + 32x + 61 6x3 + 12x2 - 2x + 25 = p(x). Hence, the division algorithm is verified. Related Topics Check these articles to know more about the concept of dividing polynomials and its related topics. Synthetic division of polynomial Division Algorithm for Polynomials Dividing Two Polynomials Division of Polynomial by Linear Factor Polynomials Long Division Examples Example 1: Rose wants to divide the polynomial 4x3 - 3x2 + 4x by 2x+1. Can you help her with the solution? Solution: Here, the polynomial 4x3 - 3x2 + 4x is divided by 2x+1 Therefore, (\text{quotient = } 2x- \dfrac{3}{2} \text{ and remainder = } x+ \dfrac32). 2. Example 2: Solve (24a2 + 48a+2) ÷ (6a + 12) by using the method of long division of polynomials. Solution: The long division of (24a2 + 48a+2) ÷ (6a + 12) can be done in the following way. Therefore, quotient = 4a and remainder = 2. 3. Example 3: Consider the following two polynomials: a(x) = x3 - x2 + x - 1 and b(x) = 2x + 1. Find the quotient polynomial and the remainder when a(x) is divided by b(x). Solution: We proceed as earlier: (\therefore )[\begin{align}&q\left( x \right)=\frac{1}{2}{x^2} - \frac{3}{4}x + \frac{7}{8}\&r = - \frac{{15}}{8}\end{align}] View Answer > go to slidego to slidego to slide Have questions on basic mathematical concepts? Become a problem-solving champ using logic, not rules. Learn the why behind math with our certified experts Book a Free Trial Class Practice Questions on Long Division of Polynomials Check Answer > go to slidego to slide FAQs on Long Division of Polynomials What is the Meaning of Long Division of Polynomials? Long division of polynomials is a technique followed in Algebra to divide a polynomial by another polynomial of a lower or the same degree. How Do You Divide Polynomials by Long Division? The following are the steps for the long division of polynomials: Arrange the terms in the decreasing order of their indices (if required).Write the missing terms with zero as their coefficient. For the first term of the quotient, divide the first term of the dividend by the first term of the divisor. Multiply this term of the quotient by the divisor to get the product. Subtract this product from the dividend, and bring down the next term (if any). The difference and the brought down term will form the new dividend. Follow this process until you get a remainder, which can be zero or of a lower index than the divisor. What is the Importance of Long Division of Polynomials? Long division of polynomials is a way to test whether one polynomial has the other one as a factor. It also helps in breaking the dividend into a simple sequence by easy steps. What are the Advantages of Long Division of Polynomials? The advantage of long division of polynomials is that it is a simple and widely used method to divide two polynomials in less space and requires lesser calculations. What are the Disadvantages of Long Division of Polynomials? The only disadvantage of long division of polynomials is that in case the divisor is nonlinear, the calculations become more complex. What are the Main Uses of a Synthetic Division of Polynomials? The main use of synthetic division of polynomials is that it is used when the divisor is linear and the coefficient of the variable in it is one. 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1839	https://pike.psu.edu/resources/cache/style.pdf	CS 0 F ormal F oundations of Computer Science F all Semester, Common Errors in W riting Inductiv e Pro ofs Septemb er 0, Handout This handout is designed to help y ou impro v e the st yle of y our inductiv e pro ofs. First of all, remem b er to alw a ys clearly indicate when y ou are going to apply the inductiv e h yp othesis. Secondly , when y ou decomp ose \P (n + )" to get \P (n)" y ou m ust explain wh y this is a v alid decomp osition. In the past, I ha v e seen studen ts who c ho ose a w a y to decomp ose P (n + ) just b ecause it made the algebra w ork out. This is not a pro of \| y ou can often decomp ose it so the algebra w ould w ork out (ev en if the theorem w as false). The k ey is that the decomp osition is correct based on other reasons. T o mak e this discussion more concrete, let's consider the follo wing problem. Use mathematical induction to pro v e that with n dieren t ice cream a v ors to c ho ose from there are n(n )= p ossibilities for a double sco op with t w o dieren t a v ors whenev er n is an in teger greater than or equal to . Here is a correct pro of. I will use f n to denote the n um b er of t w o sco op com binations p ossible using n a v ors of ice cream where the t w o sco ops m ust b e dieren t a v ors. Theorem: n , f n = n(n) . Pro of: By mathematical induction on n. Basis Step: f = = = is correct since y ou m ust use b oth a v ors and hence there is only one c hoice. Inductive Step: W e m ust sho w that n (f n = (n(n ))=) ! (f n+ = ((n + )n)=). So w e assume if there are n ice cream a v ors then there are n(n )= c hoices. W e no w consider when there are n + ice cream a v ors. What are the c hoices for a double sco op? There are all the c hoices that existed with the rst n a v ors (i.e. w e can apply the inductiv e h yp othesis) plus there are the additional c hoices that use the (n + )st a v or. Notice that there are n c hoices that use the (n + )st a v or (one for eac h other a v or to put with it). Th us w e get that f n+ = f n + n = n(n ) + n = n n + = n n + = (n + )(n)= as desired. Th us n P (n) ! P (n + ). Th us b y the principle of mathematical induction w e ha v e that n ; f n = n(n) . I ha v e seen man y incorrect pro ofs for this problem that base their pro of on the fact that: n(n ) (n + ) (n ) = (n + )n () If y ou w ere going to base a pro of on this then y ou m ust argue that the n um b er of t w o sco op c hoices with n + a v ors is the n um b er of t w o sco op c hoices with n a v ors (whic h is n(n )= b y the inductiv e h yp othesis) times (n + )=(n ). W e kno w the algebra w orks out to giv e the desired result of ((n + )n)=, but the k ey is WHY w ould the n um b er of t w o sco op c hoices with n a v ors w ould b e giv en b y this form ula. I don't see an y understandable w a y to argue this. Let me demonstrate that ev en if y ou ha v e a WR ONG theorem, y ou can alw a ys nd some algebraic op eration that w ould giv e the answ er. F or example supp ose that y ou tried to pro v e that there w ere n c hoices (whic h is wrong). Y ou could sa y that: n (n + ) n = (n + ) but I hop e ev ery one sees that this do es not pro v e that there are n c hoices. Remem b er the k ey is that y ou must explain why the decomp osition y ou giv e is correct and this dep ends on thinking ab out the particular problem. So let's return to the problem of sho wing there are n(n )= t w o sco op c hoices. If in y our pro of y ou use: n(n ) + n = (n + )n then b y itself it is really no b etter than () ab o v e. The k ey to wh y this is a go o d decomp o-sition is that the n um b er of t w o sco op c hoices with n + a v ors is the n um b er of t w o sco op c hoices with n a v ors (whic h is n(n )= b y the inductiv e h yp othesis) +n. Y ou MUST explain this. With n + a v ors, y ou ha v e all the c hoices y ou did with n a v ors, plus y ou can mak e a double sco op with the new a v or com bined with an y of the n other a v ors. Th us there are n additional c hoices a v ailable when the (n + )st a v or is added. Notice that the form ula giv en ab o v e comes from observing that there are n additional t w o-sco op c hoices, NOT from just nding something so that the algebra w orks out. If y ou do not understand the ab o v e, PLEASE come see us. Advice on Pro of St yle Let me no w talk a little ab out a st yle issue. A Bad St yle Let P (n) b e the prop osition that n X j =0 (j + ) = (n+)(n+) . I'll assume the base has already b een sho wn. Namely , P (0) is true since P 0 j =0 (j + ) = and ( )= = . What I'll fo cus on is the inductiv e step. Namely , pro ving that n 0 P (n) ! P (n + ). I will kno w sho w y ou the st yle of pro of of n 0 P (n) ! P (n + ) that I str ongly disc our age you to use. Assume that P (n) is true (i.e. n X j =0 (j + ) = (n+)(n+) ). Then n+ X j =0 (j + ) = ((n + ) + )((n + ) + )= n X j =0 (j + ) + (n + ) + = (n + )(n + )= (n+)(n+) + n + = (n + )(n + )= b y the inductiv e h yp othesis n+ (n + + ) = (n + )(n + )= (n + )(n + )= = (n + )(n + )= Since the righ t-hand and left-hand sides are equal it follo ws that n 0 P (n) ! P (n + ). What is wrong with this st yle? It is v ery confusing to the reader of the pro of. First of all, the top line sho ws what y ou w ould lik e to pro v e, but at rst glance it app ears as if y ou are assuming that P (n + ) is true. F urthermore, in this pro of y ou are sim ultaneously w orking on the left and righ t hand sides and really b et w een steps y ou are using algebra and the assumption that P (n) is true to indep enden tly rewrite the left and righ t hand sides. Ho w ev er, from the presen tation this is not clear. (And ev en if it w ere, this is still not a go o d st yle.) It in tro duces the p oten tial for error. Although this is an extreme example, sup-p ose that y ou c ho ose to m ultiply b oth the left and righ t sides b y zero at some p oin t. Then y ou could conclude that the left-hand side equals the righ t-hand side ev en if they w ere not originally equal. This w ould lead to an erroneous pro of. When w orking with inequalities, the p ossibilit y of in tro ducing an error of this form is ev en higher. The Recommende d St yle When y ou w ould lik e to sho w that t w o expressions are equal, alw a ys start with one expres-sions and manipulate it (legally , of course) to deriv e the other expression. Also don't forget to clearly indicate where y ou apply the inductiv e h yp othesis. So for example, here is a w ell written pro of that n 0 P (n) ! P (n + ). Assume that P (n) is true (i.e. n X j =0 (j + ) = (n+)(n+) ). Then n+ X j =0 (j + ) = n X j =0 (j + ) + ((n + ) + ) splitting the summation = (n+)(n+) + (n + ) b y the inductiv e h yp othesis = n+ (n + + ) factoring out an (n + )= = (n+)(n+) b y algebra = ((n+)+)((n+)+) b y algebra Th us w e ha v e sho wn that n+ X j =0 (j + ) = ((n+)+)((n+)+) whic h is exactly what is required for P (n + ) to b e true. Th us n 0 P (n) ! P (n + ). Note that when y ou are w orking on writing suc h a pro of it is often con v enien t to initially w ork with b oth sides. Then when writing up the pro of y ou need just rev erse the steps y ou to ok when manipulating one of the sides. By doing this y ou can easily write up y our pro of so that y ou tak e one expression and from it deriv e the other expression and y ou will mak e it m uc h less lik ely that y ou in tro duced an error. F or example if y ou m ultipli ed b oth sides b y 0 when w orking with b oth sides, y ou're not going to b e able to con v ert that to a pro of that deriv es one expression from the other.
1840	https://www.mathway.com/popular-problems/Algebra/763200	Find the Domain and Range x=y^13 \| Mathway Enter a problem... [x] Algebra Examples Popular Problems Algebra Find the Domain and Range x=y^13 x=y 13 x=y 13 Step 1 Rewrite the equation as y 13=x y 13=x. y 13=x y 13=x Step 2 Take the specified root of both sides of the equation to eliminate the exponent on the left side. y=13√x y=x 13 Step 3 The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined. Interval Notation: (−∞,∞)(-∞,∞) Set-Builder Notation: {x\|x∈R}{x\|x∈ℝ} Step 4 The range is the set of all valid y y values. Use the graph to find the range. Interval Notation: (−∞,∞)(-∞,∞) Set-Builder Notation: {y\|y∈R}{y\|y∈ℝ} Step 5 Determine the domain and range. Domain: (−∞,∞),{x\|x∈R}(-∞,∞),{x\|x∈ℝ} Range: (−∞,∞),{y\|y∈R}(-∞,∞),{y\|y∈ℝ} Step 6 x=y 1 3 x=y 1 3 −1-1+1+1−4 y-4 y ( ( ) ) \| \| [ [ ] ] √ √   ≥ ≥           7 7 8 8 9 9       ≤ ≤           4 4 5 5 6 6 / / ^ ^ × ×     ∩ ∩ ∪ ∪   1 1 2 2 3 3 - - + + ÷ ÷ < <     π π ∞ ∞  , , 0 0 . . % %  = =     ⎡⎢⎣x 2 1 2√π∫x d x⎤⎥⎦[x 2 1 2 π∫⁡x d x] Please ensure that your password is at least 8 characters and contains each of the following: a number a letter a special character: @$#!%?& Do Not Sell My Personal Information When you visit our website, we store cookies on your browser to collect information. The information collected might relate to you, your preferences or your device, and is mostly used to make the site work as you expect it to and to provide a more personalized web experience. However, you can choose not to allow certain types of cookies, which may impact your experience of the site and the services we are able to offer. 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1841	https://www.engineeringtoolbox.com/saturated-steam-properties-d_457.html	Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications! Thermodynamic Properties of Saturated Steam: Data & Charts in Bar Saturated Steam Table with properties like boiling point, specific volume, density, specific enthalpy, specific heat and latent heat of vaporization. Properties of Saturated Steam - SI Units Properties of Saturated Steam - Imperial Units For full table - rotate the screen! Saturated Steam - Properties vs. Pressure - Bar \| Absolute Pressure \| Boiling Point \| Specific Volume (steam) \| Density (steam) \| Specific Enthalpy of Liquid Water (sensible heat) \| \| Specific Enthalpy of Steam (total heat) \| \| Latent heat of Vaporization \| \| Specific Heat \| \| (bar) \| (oC) \| (m3/kg) \| (kg/m3) \| (kJ/kg) \| (kcal/kg) \| (kJ/kg) \| (kcal/kg) \| (kJ/kg) \| (kcal/kg) \| (kJ/kg K) \| \| 0.02 \| 17.51 \| 67.006 \| 0.015 \| 73.45 \| 17.54 \| 2533.64 \| 605.15 \| 2460.19 \| 587.61 \| 1.8644 \| \| 0.03 \| 24.10 \| 45.667 \| 0.022 \| 101.00 \| 24.12 \| 2545.64 \| 608.02 \| 2444.65 \| 583.89 \| 1.8694 \| \| 0.04 \| 28.98 \| 34.802 \| 0.029 \| 121.41 \| 29.00 \| 2554.51 \| 610.13 \| 2433.10 \| 581.14 \| 1.8736 \| \| 0.05 \| 32.90 \| 28.194 \| 0.035 \| 137.77 \| 32.91 \| 2561.59 \| 611.83 \| 2423.82 \| 578.92 \| 1.8774 \| \| 0.06 \| 36.18 \| 23.741 \| 0.042 \| 151.50 \| 36.19 \| 2567.51 \| 613.24 \| 2416.01 \| 577.05 \| 1.8808 \| \| 0.07 \| 39.02 \| 20.531 \| 0.049 \| 163.38 \| 39.02 \| 2572.62 \| 614.46 \| 2409.24 \| 575.44 \| 1.8840 \| \| 0.08 \| 41.53 \| 18.105 \| 0.055 \| 173.87 \| 41.53 \| 2577.11 \| 615.53 \| 2403.25 \| 574.01 \| 1.8871 \| \| 0.09 \| 43.79 \| 16.204 \| 0.062 \| 183.28 \| 43.78 \| 2581.14 \| 616.49 \| 2397.85 \| 572.72 \| 1.8899 \| \| 0.1 \| 45.83 \| 14.675 \| 0.068 \| 191.84 \| 45.82 \| 2584.78 \| 617.36 \| 2392.94 \| 571.54 \| 1.8927 \| \| 0.2 \| 60.09 \| 7.650 \| 0.131 \| 251.46 \| 60.06 \| 2609.86 \| 623.35 \| 2358.40 \| 563.30 \| 1.9156 \| \| 0.3 \| 69.13 \| 5.229 \| 0.191 \| 289.31 \| 69.10 \| 2625.43 \| 627.07 \| 2336.13 \| 557.97 \| 1.9343 \| \| 0.4 \| 75.89 \| 3.993 \| 0.250 \| 317.65 \| 75.87 \| 2636.88 \| 629.81 \| 2319.23 \| 553.94 \| 1.9506 \| \| 0.5 \| 81.35 \| 3.240 \| 0.309 \| 340.57 \| 81.34 \| 2645.99 \| 631.98 \| 2305.42 \| 550.64 \| 1.9654 \| \| 0.6 \| 85.95 \| 2.732 \| 0.366 \| 359.93 \| 85.97 \| 2653.57 \| 633.79 \| 2293.64 \| 547.83 \| 1.9790 \| \| 0.7 \| 89.96 \| 2.365 \| 0.423 \| 376.77 \| 89.99 \| 2660.07 \| 635.35 \| 2283.30 \| 545.36 \| 1.9919 \| \| 0.8 \| 93.51 \| 2.087 \| 0.479 \| 391.73 \| 93.56 \| 2665.77 \| 636.71 \| 2274.05 \| 543.15 \| 2.0040 \| \| 0.9 \| 96.71 \| 1.869 \| 0.535 \| 405.21 \| 96.78 \| 2670.85 \| 637.92 \| 2265.65 \| 541.14 \| 2.0156 \| \| 1 1) \| 99.63 \| 1.694 \| 0.590 \| 417.51 \| 99.72 \| 2675.43 \| 639.02 \| 2257.92 \| 539.30 \| 2.0267 \| \| 1.1 \| 102.32 \| 1.549 \| 0.645 \| 428.84 \| 102.43 \| 2679.61 \| 640.01 \| 2250.76 \| 537.59 \| 2.0373 \| \| 1.2 \| 104.81 \| 1.428 \| 0.700 \| 439.36 \| 104.94 \| 2683.44 \| 640.93 \| 2244.08 \| 535.99 \| 2.0476 \| \| 1.3 \| 107.13 \| 1.325 \| 0.755 \| 449.19 \| 107.29 \| 2686.98 \| 641.77 \| 2237.79 \| 534.49 \| 2.0576 \| \| 1.4 \| 109.32 \| 1.236 \| 0.809 \| 458.42 \| 109.49 \| 2690.28 \| 642.56 \| 2231.86 \| 533.07 \| 2.0673 \| \| 1.5 \| 111.37 \| 1.159 \| 0.863 \| 467.13 \| 111.57 \| 2693.36 \| 643.30 \| 2226.23 \| 531.73 \| 2.0768 \| \| 1.6 \| 113.32 \| 1.091 \| 0.916 \| 475.38 \| 113.54 \| 2696.25 \| 643.99 \| 2220.87 \| 530.45 \| 2.0860 \| \| 1.7 \| 115.17 \| 1.031 \| 0.970 \| 483.22 \| 115.42 \| 2698.97 \| 644.64 \| 2215.75 \| 529.22 \| 2.0950 \| \| 1.8 \| 116.93 \| 0.977 \| 1.023 \| 490.70 \| 117.20 \| 2701.54 \| 645.25 \| 2210.84 \| 528.05 \| 2.1037 \| \| 1.9 \| 118.62 \| 0.929 \| 1.076 \| 497.85 \| 118.91 \| 2703.98 \| 645.83 \| 2206.13 \| 526.92 \| 2.1124 \| \| 2 \| 120.23 \| 0.885 \| 1.129 \| 504.71 \| 120.55 \| 2706.29 \| 646.39 \| 2201.59 \| 525.84 \| 2.1208 \| \| 2.2 \| 123.27 \| 0.810 \| 1.235 \| 517.63 \| 123.63 \| 2710.60 \| 647.42 \| 2192.98 \| 523.78 \| 2.1372 \| \| 2.4 \| 126.09 \| 0.746 \| 1.340 \| 529.64 \| 126.50 \| 2714.55 \| 648.36 \| 2184.91 \| 521.86 \| 2.1531 \| \| 2.6 \| 128.73 \| 0.693 \| 1.444 \| 540.88 \| 129.19 \| 2718.17 \| 649.22 \| 2177.30 \| 520.04 \| 2.1685 \| \| 2.8 \| 131.20 \| 0.646 \| 1.548 \| 551.45 \| 131.71 \| 2721.54 \| 650.03 \| 2170.08 \| 518.32 \| 2.1835 \| \| 3 \| 133.54 \| 0.606 \| 1.651 \| 561.44 \| 134.10 \| 2724.66 \| 650.77 \| 2163.22 \| 516.68 \| 2.1981 \| \| 3.5 \| 138.87 \| 0.524 \| 1.908 \| 584.28 \| 139.55 \| 2731.63 \| 652.44 \| 2147.35 \| 512.89 \| 2.2331 \| \| 4 \| 143.63 \| 0.462 \| 2.163 \| 604.68 \| 144.43 \| 2737.63 \| 653.87 \| 2132.95 \| 509.45 \| 2.2664 \| \| 4.5 \| 147.92 \| 0.414 \| 2.417 \| 623.17 \| 148.84 \| 2742.88 \| 655.13 \| 2119.71 \| 506.29 \| 2.2983 \| \| 5 \| 151.85 \| 0.375 \| 2.669 \| 640.12 \| 152.89 \| 2747.54 \| 656.24 \| 2107.42 \| 503.35 \| 2.3289 \| \| 5.5 \| 155.47 \| 0.342 \| 2.920 \| 655.81 \| 156.64 \| 2751.70 \| 657.23 \| 2095.90 \| 500.60 \| 2.3585 \| \| 6 \| 158.84 \| 0.315 \| 3.170 \| 670.43 \| 160.13 \| 2755.46 \| 658.13 \| 2085.03 \| 498.00 \| 2.3873 \| \| 6.5 \| 161.99 \| 0.292 \| 3.419 \| 684.14 \| 163.40 \| 2758.87 \| 658.94 \| 2074.73 \| 495.54 \| 2.4152 \| \| 7 \| 164.96 \| 0.273 \| 3.667 \| 697.07 \| 166.49 \| 2761.98 \| 659.69 \| 2064.92 \| 493.20 \| 2.4424 \| \| 7.5 \| 167.76 \| 0.255 \| 3.915 \| 709.30 \| 169.41 \| 2764.84 \| 660.37 \| 2055.53 \| 490.96 \| 2.4690 \| \| 8 \| 170.42 \| 0.240 \| 4.162 \| 720.94 \| 172.19 \| 2767.46 \| 661.00 \| 2046.53 \| 488.80 \| 2.4951 \| \| 8.5 \| 172.94 \| 0.227 \| 4.409 \| 732.03 \| 174.84 \| 2769.89 \| 661.58 \| 2037.86 \| 486.73 \| 2.5206 \| \| 9 \| 175.36 \| 0.215 \| 4.655 \| 742.64 \| 177.38 \| 2772.13 \| 662.11 \| 2029.49 \| 484.74 \| 2.5456 \| \| 9.5 \| 177.67 \| 0.204 \| 4.901 \| 752.82 \| 179.81 \| 2774.22 \| 662.61 \| 2021.40 \| 482.80 \| 2.5702 \| \| 10 \| 179.88 \| 0.194 \| 5.147 \| 762.60 \| 182.14 \| 2776.16 \| 663.07 \| 2013.56 \| 480.93 \| 2.5944 \| \| 11 \| 184.06 \| 0.177 \| 5.638 \| 781.11 \| 186.57 \| 2779.66 \| 663.91 \| 1998.55 \| 477.35 \| 2.6418 \| \| 12 \| 187.96 \| 0.163 \| 6.127 \| 798.42 \| 190.70 \| 2782.73 \| 664.64 \| 1984.31 \| 473.94 \| 2.6878 \| \| 13 \| 191.60 \| 0.151 \| 6.617 \| 814.68 \| 194.58 \| 2785.42 \| 665.29 \| 1970.73 \| 470.70 \| 2.7327 \| \| 14 \| 195.04 \| 0.141 \| 7.106 \| 830.05 \| 198.26 \| 2787.79 \| 665.85 \| 1957.73 \| 467.60 \| 2.7767 \| \| 15 \| 198.28 \| 0.132 \| 7.596 \| 844.64 \| 201.74 \| 2789.88 \| 666.35 \| 1945.24 \| 464.61 \| 2.8197 \| \| 16 \| 201.37 \| 0.124 \| 8.085 \| 858.54 \| 205.06 \| 2791.73 \| 666.79 \| 1933.19 \| 461.74 \| 2.8620 \| \| 17 \| 204.30 \| 0.117 \| 8.575 \| 871.82 \| 208.23 \| 2793.37 \| 667.18 \| 1921.55 \| 458.95 \| 2.9036 \| \| 18 \| 207.11 \| 0.110 \| 9.065 \| 884.55 \| 211.27 \| 2794.81 \| 667.53 \| 1910.27 \| 456.26 \| 2.9445 \| \| 19 \| 209.79 \| 0.105 \| 9.556 \| 896.78 \| 214.19 \| 2796.09 \| 667.83 \| 1899.31 \| 453.64 \| 2.9849 \| \| 20 \| 212.37 \| 0.100 \| 10.047 \| 908.56 \| 217.01 \| 2797.21 \| 668.10 \| 1888.65 \| 451.10 \| 3.0248 \| \| 21 \| 214.85 \| 0.095 \| 10.539 \| 919.93 \| 219.72 \| 2798.18 \| 668.33 \| 1878.25 \| 448.61 \| 3.0643 \| \| 22 \| 217.24 \| 0.091 \| 11.032 \| 930.92 \| 222.35 \| 2799.03 \| 668.54 \| 1868.11 \| 446.19 \| 3.1034 \| \| 23 \| 219.55 \| 0.087 \| 11.525 \| 941.57 \| 224.89 \| 2799.77 \| 668.71 \| 1858.20 \| 443.82 \| 3.1421 \| \| 24 \| 221.78 \| 0.083 \| 12.020 \| 951.90 \| 227.36 \| 2800.39 \| 668.86 \| 1848.49 \| 441.50 \| 3.1805 \| \| 25 \| 223.94 \| 0.080 \| 12.515 \| 961.93 \| 229.75 \| 2800.91 \| 668.99 \| 1838.98 \| 439.23 \| 3.2187 \| \| 26 \| 226.03 \| 0.077 \| 13.012 \| 971.69 \| 232.08 \| 2801.35 \| 669.09 \| 1829.66 \| 437.01 \| 3.2567 \| \| 27 \| 228.06 \| 0.074 \| 13.509 \| 981.19 \| 234.35 \| 2801.69 \| 669.17 \| 1820.50 \| 434.82 \| 3.2944 \| \| 28 \| 230.04 \| 0.071 \| 14.008 \| 990.46 \| 236.57 \| 2801.96 \| 669.24 \| 1811.50 \| 432.67 \| 3.3320 \| \| 29 \| 231.96 \| 0.069 \| 14.508 \| 999.50 \| 238.73 \| 2802.15 \| 669.28 \| 1802.65 \| 430.56 \| 3.3695 \| \| 30 \| 233.84 \| 0.067 \| 15.009 \| 1008.33 \| 240.84 \| 2802.27 \| 669.31 \| 1793.94 \| 428.48 \| 3.4069 \| 1) 1 bar abs = 0 bar gauge = 0.98692 atm = 100 kPa abs = Standard Atmospheric Pressure (IUPAC system) Vacuum steam is the general term used for saturated steam at temperatures below 100°C. Steam evaporators used in industrial and waste water applications operates with pressures lower that the atmospheric pressure. In steam distribution system the pressure is almost always more than 0 bar gauge. Example - Boiling Water at 100 oC, 0 bar (100 kPa) Atmospheric Pressure At atmospheric pressure (0 bar g, absolute 1 bar) water boils at 100 oC and 417.51 kJ of energy is required to heat 1 kg of water from 0 oC to evaporating temperature 100 oC. Therefore the specific enthalpy of water at 0 bar g (absolute 1 bar) and 100 oC is 417.51 kJ/kg. Another 2257.92 kJ of energy is required to evaporate 1 kg of water at 100 oC into 1 kg of steam at 100 oC. Therefore at 0 bar g (absolute 1 bar) the specific enthalpy of vaporation is 2257.19 kJ/kg. The total specific enthalpy for steam at 0 bar gauge is: hs = (417.51 kJ/kg) + (2257.92 kJ/kg) = 2675.43 kJ/kg Example - Boiling Water at 170 oC and 7 bar (700 kPa) Atmospheric Pressure At 7 bar g (absolute 8 bar) the saturation temperature of water is 170.42 oC. More heat energy is required to raise its temperature to saturation point at 7 bar g than needed when the water is at atmospheric pressure. According the table 720.94 kJ is required to raise 1 kg of water from 0 oC to saturation temperature 170 oC. The heat energy (enthalpy of evaporation) needed at 7 bar g to vaporize the water to steam is actually less than required at atmospheric pressure. The specific enthalpy of vaporization decreases with steam pressure. The evaporation heat is 2046.53 kJ/kg at 7 bar g. Note! The specific volume of steam decreases with increased pressure - and the amount of heat energy distributed by the same volume increase. With higher pressure - more energy can be transferred in a steam distribution system. Unit Converter . Cookie Settings \| \| \| --- \| \| \| \| \| \| \| --- \| \| \| \|
1842	https://wenku.baidu.com/view/9fe5277e856fb84ae45c3b3567ec102de2bddfab.html	在SSA条件下解三角形时的多解性分析（doc版）. - 百度文库新建文档新建表格新建上传至我的资料库转为在线文档上传至创作中心在文库发布个人作品上传首页助手最近知识库我的收藏我的下载我的上传我的购买查看更多文件资料库文档瘦身格式转换 PDF合并 PDF拆分 PDF加水印图片转文字工具辅助模式已关闭查看更多前往帮助中心 > 关于我们关于百度文库机构/企业资源库文库帮助中心 AI功能介绍百度文库企业版加入我们文库开放平台个人作者入驻机构入驻文库上传规范企业采购咨询联系我们 VIP专属服务号文库服务号客服电话：400-921-7005 在线反馈非法有害信息举报免费下载PC客户端免广告格式转换离线浏览文档管理下载客户端扫码使用文库APP 新建文档新建表格新建上传至我的资料库转为在线文档上传至创作中心在文库发布个人作品上传最近编辑我的收藏我的下载大家都在搜协议书贷款合同日历表简历模板入党申请书开通会员登录后查看会员权益 AI帮你写智能PPT 免费下载格式转换开通会员每日抽奖专属权益精选内容前往会员中心抽奖登录相关文档集共15篇 1 在SSA条件下解三角形时的多解性分析（doc版）. 2.2 0下载 70阅读 2 在SSA条件下解三角形时的多解性分析(doc版). 4.1 49下载 1,780阅读 3 全等三角形SSA的探究-精选教学文档 3.5 2下载 42阅读 4 由余弦定理解SSA型三角形的方法及论证 2.6 4下载 197阅读 5 第02讲三角形的判定SSS 3.8 17下载 73阅读 6 “SSA”推理两三角形全等的探究 3.8 1下载 107阅读 7 解SSA类型三角形 3.7 1下载 194阅读 8 第三章_应变分析. 2.1 0下载 58阅读 9 第三章应力分析与应变分析. 4.0 8下载 306阅读 10 三角形多解问题。 4.1 0下载 130阅读 11 三角形多解问题。 3.4 0下载 232阅读 12 应注意相似三角形中的多解问题 2.2 565下载 942阅读 13 ★正弦定理SSA条件下多解性分析(专题) 4.1 50下载 1,505阅读 14 全等三角形SSA的探究 4.1 9下载 303阅读 15 全等三角形判定二(SSS,AAS)(基础)知识讲解.doc 2.1 2下载 15阅读编辑文档您的皮肤设置未保存，开通VIP保存设置查看更多15项会员特权不保存开通文库VIP 在SSA条件下解三角形时的多解性分析（doc版）. 在SSA条件下解三角形时的多解性分析（doc版）. 2023-05-23 文文会健身 AI编辑插入正文思源黑体样式更多选择插入表格的行列表格插入分栏分栏图片柱状图折线图饼状图图表公式链接引用分割线颜色分割线高亮块代码块 Markdown导入复制剪切删除行间距取消确定查找替换上一个下一个取消确定移动到搜索复制复制全文添加到知识库发送到手机翻译复制复制全文添加到知识库搜索发送到手机翻译复制复制全文添加到知识库搜索翻译等8人已复制此文本文字已复制您已超出最大复制字数限制，请减少字数重新尝试已为您复制内容开通VIP，立即解锁文本内容开通VIP VIP会员尊享权益超出发送上限，加入VIP即可继续发送未开通VIP 选择文档内容扫一扫，立即发送到手机开通VIP，享无限制发送与复制特权享VIP专享文档下载特权赠共享文档下载特权 100W篇文档免费专享超过1万字无法发送，您可减少字数多次发送选择内容扫一扫立即发送到手机 VIP享无限制发送特权以下结果由提供：× 百度翻译移动到剩余3页未读，展开全文开通VIP 享海量文档+AI助手权益 AI帮你写续写、润色、扩写 AI生成PPT 高档模板一键生成免费下载特权 VIP专享文档下载格式转换 PDF转Word 最低仅需¥2.00开通VIP 下一篇：在SSA条件下解三角形时的多解性分析(doc版). 4.1 1780阅读版权说明：本文档由用户提供并上传，收益归属内容提供方，若内容存在侵权，请进行举报或认领页数说明：当前展示页数非原始文档页数，原始文档共5页复制全文全屏阅读举报认领此文档是否涉及以下问题？色情、淫秽、低俗信息涉嫌违法犯罪时政信息不实广告或垃圾信息此文档是否涉及侵权？盗版或侵权未获得权利人的合法授权、侵犯个人隐私或泄露单位商业机密处理此类举报需要您提供更多信息，请按引导上传相关材料，这样会有助您成功举报。取消提交若您为此篇文档的原创作者，提交权利人真实信息及权属证明材料。我们承诺：材料信息经核实后，将于工作日24小时之内将此文档以VIP专享或免费文档的形式转至您名下，后续被认领成功的文档所产生的收益均归您所有。认领文档在SSA条件下解三角形时的多解性分析（doc版）. 实名信息当前账号暂未入驻，个人作者入驻机构入驻请入驻文库创作者平台后刷新当前页面认领理由权属材料上传权属证明材料可上传多个附件，支持png、jpg、pdf、word等格式，附件内容需要包含： ①毕业证/工作证/曾经的发表记录/版权证明/版权声明书等原文档版权证明 ②身份证或营业执照等立即认领联系电话 400-921-7005 如您为此篇文档作者，可提供证明材料申请认领，获取版权文档收益并下架侵权文档：发送申请邮件审核（1-2个工作日）认领结果答复申请邮箱： wenku-renling@baidu.com 复制证明材料： 1、侵权文档文库链接地址 2、您的版权文档文库链接地址：如尚无，请先上传 3、身份证明材料：身份证正面照 / 护照身份页照片 4、版权证明材料：作品认证证书 / 作品登记证书 / 已署名的出版物专属客服： QQ 800049878 电话 400-921-7005 您的举报提交成功！您的举报已收到，我们会尽快进行核实处理。关闭添加文库智能助手到桌面安装后可在桌面快捷使用文库，轻松访问下载的文档，还可使用AI创作能力立即安装分享批量下载(15篇) 全部文档请确认浏览器允许下载，下载时自动消耗特权该文档需单独付费，不支持批量下载 1 在SSA条件下解三角形时的多解性分析（doc版）. 已编辑 2 在SSA条件下解三角形时的多解性分析(doc版). 已编辑 3 全等三角形SSA的探究-精选教学文档已编辑 4 由余弦定理解SSA型三角形的方法及论证已编辑 5 第02讲三角形的判定SSS 已编辑 6 “SSA”推理两三角形全等的探究已编辑 7 解SSA类型三角形已编辑 8 第三章_应变分析. 已编辑 9 第三章应力分析与应变分析. 已编辑 10 三角形多解问题。已编辑 11 三角形多解问题。已编辑 12 应注意相似三角形中的多解问题已编辑 13 ★正弦定理SSA条件下多解性分析(专题) 已编辑 14 全等三角形SSA的探究已编辑 15 全等三角形判定二(SSS,AAS)(基础)知识讲解.doc 已编辑单篇下载下载百度文库客户端畅享免费海量文档｜一键生成PPT和长文写作下载客户端下载客户端转PDF 更多操作引用智能助手全屏探索新建指南文档难找？试试AI为您生成的内容吧：保持好奇，不断提升自我，是通往成功和幸福的秘诀。一个人对世界的好奇心越强烈，他智能PPT 智能写作智能图表 DeepSeek-R1 联网搜索畅享AI生成内容购买会员畅享智能创作 AI生成PPT PPT高级模板 AI帮你写 AI扩写 AI润色 AI续写 AI修订无限复制无限导出 6亿VIP文档查看更多权益换个话题重新开始吧，新建对话 AI辅助生成PPT 支持多种方式一键智能生成专业级PPT 去试试拖拽文件开始上传文档大小：不超过200MB 文档类型：doc、docx、ppt、pptx、pdf、xls、xlsx 限时抢购: 01 时 00 分 00 秒公式常用符号常用公式常用符号希腊字母分数微分根式角标极限对数三角函数积分运算大型运算括号取整数组矩阵清除输出区域取消确定
1843	https://math.stackexchange.com/questions/3744227/find-maximum-point-of-fx-y-z-8x2-4yz-16z-600-with-one-restriction	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. current community your communities more stack exchange communities Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Find maximum point of $f(x,y,z) = 8x^2 +4yz -16z +600$ with one restriction I need to find the critical points of $$f(x,y,z) = 8x^2 +4yz -16z +600$$ restricted by $4x^2+y^2+4z^2=16$. I constructed the lagrangian function $$L(x, y, z, \lambda ) = 8x^2 +4yz -16z +600 - \lambda (4x^2+y^2+4z^2-16) $$ but I'm very confused about how to determine those points. I know I need to make a system with all the first derivatives of $L$ equaled to $0$. I did it but every time I try to solve it I get different solutions. How can I get the points? Thanks. 3 Answers 3 There are a number of constrained critical points. If you set the gradient of $L$ equal to $0$, you find that $$(4x,z,y-4) = \lambda(4x,y,4z).$$ If $x\ne 0$, we must have $\lambda = 1$ and then $z=y$ and $y-4=4z$. This gives $y=z=-4/3$ and $x=\pm 4/3$. However, if $x=0$, then we also get additional solutions by setting \begin{equation} (z,y-4) = \lambda(y,4z),\tag{$$} \end{equation} from which we get $$\frac zy = \frac{y-4}{4z}.$$ (Note that we cannot have $x=y=z=0$ on our constraint set, so this is fine. Note that ($$) says that $z=0$ if and only if $y=0$.) This yields $4z^2=y(y-4)$, which, if I'm not mistaken, leads, along with the constraint equation, to $y^2-2y-8 = (y-4)(y+2) = 0$, so $y=4$ or $y=-2$. These give additional critical points $(0,4,0)$ and $(0,-2,\pm\sqrt3)$. Because there's such dispute amongst the various answerers, let me check the values of $f$ at these various points: \begin{multline} f(\pm 4/3, -4/3, -4/3) = \frac{1928}3, \quad f(0,4,0) = 600, \ f(0,-2,\sqrt3) = 600 - 24\sqrt 3,\quad f(0,-2,-\sqrt3) = 600 + 24\sqrt3. \end{multline} Indeed, $1928/3$ wins out for the maximum value, but only just barely!! Philosophical Remark: You do not need to solve explicitly for $\lambda$; you can eliminate it as I did by taking ratios. It is to emphasize this pedagogical point that I wrote out the solution so carefully. :) Using Lagrange Multipliers, we wish to find the points $(x, y, z)$ such that $\nabla f(x, y, z) = \lambda \nabla g(x, y, z),$ where $g(x, y, z) = 4x^2 + y^2 + 4z^2 = 16$ and $\lambda$ is some constant. Observe that the gradients are given by $\nabla f = \langle f_x, f_y, f_z \rangle = \langle 16x, 4z, 4y - 16 \rangle$ and $\nabla g = \langle 8x, 2y, 8z \rangle,$ hence we must solve the following $4 \times 4$ system of equations. $$\begin{cases} 16x = 8 \lambda x \ 4z = 2 \lambda y \ 4y - 16 = 8 \lambda z \ 4x^2 + y^2 + 4z^2 = 16\end{cases}$$ Using the first equation, we have that $(16 - 8 \lambda)x = 0,$ from which it follows that $\lambda = 2$ or $x = 0.$ Using the second equation, we have that $2z = \lambda y$ so that $4z^2 = \lambda^2 y^2.$ Using the third equation, we have that $4y = 8 \lambda z + 16 = 4 \lambda^2 y + 16$ (by the second equation) so that $(4 \lambda^2 - 4)y + 16 = 0.$ Using the fourth equation, we have that $4x^2 + (\lambda^2 + 1) y^2 = 16$ (by the second equation). $$\begin{cases} 4z^2 = \lambda^2 y^2 \ (4 \lambda^2 - 4)y + 16 = 0 \ 4x^2 + (\lambda^2 + 1) y^2 = 16 \end{cases}$$ Given that $\lambda = 2,$ we may solve for $y$ in the second equation; solve for $z$ in the first equation; and solve for $x$ in the third equation. On the other hand, if $x = 0,$ then we are dealing with the functions $f(0, y, z) = 4yz - 16z + 600 = h(y, z)$ and $g(0, y, z) = y^2 + 4z^2 = k(y, z) = 16,$ and we can use Lagrange Multipliers to find the critical points of $h(y, z)$ with respect to the constraint $k(y, z) = 16.$ Explicitly, we have that $\langle 4z, 4y - 16 \rangle = \nabla h(y, z) = \mu \nabla k(y, z) = \mu \langle 2y, 8z \rangle$ so that $$\begin{cases} 4z = 2 \mu y \ 4y - 16 = 8 \mu z \end{cases}$$ is the relevant system of equations. Considering that $\mu$ is nonzero, we can eliminate it by taking $$16 \mu z^2 = (2z)(8 \mu z) = (2z)(4y - 16) = (\mu y)(4y - 16) = \mu (4y^2 - 16y)$$ and cancelling a factor of $\mu$ from the left- and right-hand sides. We have therefore that $16z^2 = 4y^2 - 16y.$ Use the fact that $16z^2 = 4(16 - y^2)$ to solve for $y.$ $$L(x,y,z,\lambda) = 8x^2-4\lambda x^2 + 4yz - \lambda y^2 - 16z - 4\lambda z^2 + 600 + 16\lambda$$ $$\frac{d}{dx}L(x,y,z,\lambda) = 16x-8\lambda x = 0, x = 0 \space or \space \lambda = 2$$ $$\frac{d}{dy}L(x,y,z,\lambda) = 4z-2\lambda y = 0 \space or \space 4z = 4y, \space for \space \lambda = 2$$ $$So, y = z$$ $$\frac{d}{dz}L(x,y,z,\lambda) = 4y-16-8\lambda z = 0, \space or \space 4y = 16z + 16, \space for \space \lambda = 2$$ $$As \space y = z, \space y = z = -\frac {4}{3}$$ $$\frac{d}{d\lambda}L(x,y,z,\lambda) = -4x^2-y^2-4z^2+16 = 0, \text { which is our original constraint.}$$ $$As \space \lambda = 2, \space y = z = -\frac{4}{3},$$ $$-4x^2-\frac{16}{9}-\frac{64}{9}+16 = 0$$ $$4x^2=16-\frac{80}{9} \space so, \space x = \pm \frac{4}{3}$$ $$\text {So, critical points for } \lambda = 2$$ $$(-\frac{4}{3},-\frac{4}{3},-\frac{4}{3}),(\frac{4}{3},-\frac{4}{3},-\frac{4}{3})$$ Similarly two more critical points for x = 0. 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1845	https://topdrawer.aamt.edu.au/content/download/26013/360044/file/tdt_G_greatanglechase_student.pdf	The great angle chase: Student worksheet the-lines/Visualising-relationships/Great-angle-chase In the diagram below, O is the centre of the circle. The figure is not drawn to scale. GH is a tangent to the circle at P. PS is a diameter and PQ = RS. ∠HPQ = 25°, ∠PIQ = 100° and ∠UPT = 35°. Do not construct additional lines in the diagram. • Write in the size of all the angles. • Name any cyclic quadrilaterals. • Name all equal line segments. • Name all equal arcs. • Name any parallel lines. AAMT — TOP DRAWER TEACHERS © 2013 Education Services Australia Ltd, except where indicated otherwise. This document may be used, reproduced, published, communicated and adapted free of charge for non-commercial educational purposes provided all acknowledgements associated with the material are retained. page 1 of 1
1846	https://ocw.mit.edu/courses/res-6-012-introduction-to-probability-spring-2018/resources/normal-approximation-to-the-binomial/	Browse Course Material Course Info Instructors Prof. John Tsitsiklis Prof. Patrick Jaillet Departments Electrical Engineering and Computer Science As Taught In Spring 2018 Level Undergraduate Topics Engineering Systems Engineering Mathematics Probability and Statistics Learning Resource Types theaters Lecture Videos notes Lecture Notes Download Course search GIVE NOW about ocw help & faqs contact us RES.6-012 \| Spring 2018 \| Undergraduate Introduction to Probability Part II: Inference & Limit Theorems Normal Approximation to the Binomial Instructor: John Tsitsiklis Download video Download transcript Course Info Instructors Prof. John Tsitsiklis Prof. Patrick Jaillet Departments Electrical Engineering and Computer Science As Taught In Spring 2018 Level Undergraduate Topics Engineering Systems Engineering Mathematics Probability and Statistics Learning Resource Types theaters Lecture Videos notes Lecture Notes Download Course Over 2,500 courses & materials Freely sharing knowledge with learners and educators around the world. Learn more © 2001–2025 Massachusetts Institute of Technology Creative Commons License Terms and Conditions Proud member of: © 2001–2025 Massachusetts Institute of Technology You are leaving MIT OpenCourseWare Please be advised that external sites may have terms and conditions, including license rights, that differ from ours. MIT OCW is not responsible for any content on third party sites, nor does a link suggest an endorsement of those sites and/or their content. Continue
1847	https://www.wikiwand.com/en/articles/Wilson's_theorem	Wilson's theorem - Wikiwand English Sign in Top Qs Timeline Chat Perspective Top Qs Timeline Chat Perspective AllArticlesDictionaryQuotesMap Wilson's theorem Theorem on prime numbers From Wikipedia, the free encyclopedia Remove ads Remove ads Wilson's theorem • • • HistoryExampleProofsComposite modulusPrime modulusElementary proofProof using Fermat's little theoremProof using the Sylow theoremsApplicationsPrimality testsQuadratic residuesFormulas for primesp-adic gamma functionGauss's generalizationSee alsoNotesReferencesExternal links In algebra and number theory, Wilson's theorem states that a natural numbern> 1 is a prime numberif and only if the product of all the positive integers less than n is one less than a multiple of n. That is (using the notations of modular arithmetic), the factorial(n−1)!=1×2×3×⋯×(n−1){\displaystyle (n-1)!=1\times 2\times 3\times \cdots \times (n-1)} satisfies (n−1)!≡−1(mod n){\displaystyle (n-1)!\ \equiv \;-1{\pmod {n}}} exactly when n is a prime number. In other words, any integer n> 1 is a prime number if, and only if, (n−1)!+1 is divisible by n. 1 Sources Remove ads History The theorem was first stated by Ibn al-Haythamc.1000 AD. announced the theorem in 1770 without proving it, crediting his student John Wilson for the discovery. gave the first proof in 1771. was also aware of the result a century earlier, but never published it. 4 Sources Example For each of the values of n from 2 to 30, the following table shows the number (n−1)! and the remainder when (n−1)! is divided by n. (In the notation of modular arithmetic, the remainder when m is divided by n is written m mod n.) The background color is blue for prime values of n, gold for composite values. More information, ... Table of factorial and its remainder modulo n\| n{\displaystyle n} \| (n−1)!{\displaystyle (n-1)!} (sequence A000142 in the OEIS) \| (n−1)!mod n{\displaystyle (n-1)!\ {\bmod {\ }}n} (sequence A061006 in the OEIS) \| --- \| 2 \| 1 \| 1 \| \| 3 \| 2 \| 2 \| \| 4 \| 6 \| 2 \| \| 5 \| 24 \| 4 \| \| 6 \| 120 \| 0 \| \| 7 \| 720 \| 6 \| \| 8 \| 5040 \| 0 \| \| 9 \| 40320 \| 0 \| \| 10 \| 362880 \| 0 \| \| 11 \| 3628800 \| 10 \| \| 12 \| 39916800 \| 0 \| \| 13 \| 479001600 \| 12 \| \| 14 \| 6227020800 \| 0 \| \| 15 \| 87178291200 \| 0 \| \| 16 \| 1307674368000 \| 0 \| \| 17 \| 20922789888000 \| 16 \| \| 18 \| 355687428096000 \| 0 \| \| 19 \| 6402373705728000 \| 18 \| \| 20 \| 121645100408832000 \| 0 \| \| 21 \| 2432902008176640000 \| 0 \| \| 22 \| 51090942171709440000 \| 0 \| \| 23 \| 1124000727777607680000 \| 22 \| \| 24 \| 25852016738884976640000 \| 0 \| \| 25 \| 620448401733239439360000 \| 0 \| \| 26 \| 15511210043330985984000000 \| 0 \| \| 27 \| 403291461126605635584000000 \| 0 \| \| 28 \| 10888869450418352160768000000 \| 0 \| \| 29 \| 304888344611713860501504000000 \| 28 \| \| 30 \| 8841761993739701954543616000000 \| 0 \| Close Remove ads Proofs Summarize Perspective As a biconditional (if and only if) statement, the proof has two halves: to show that equality does not hold when n{\displaystyle n} is composite, and to show that it does hold when n{\displaystyle n} is prime. Composite modulus Suppose that n{\displaystyle n} is composite. Therefore, it is divisible by some prime number q{\displaystyle q} where 2≤q<n{\displaystyle 2\leq q<n}. Because q{\displaystyle q} divides n{\displaystyle n}, there is an integer k{\displaystyle k} such that n=q k{\displaystyle n=qk}. Suppose for the sake of contradiction that (n−1)!{\displaystyle (n-1)!} were congruent to −1{\displaystyle -1} modulo n{\displaystyle {n}}. Then (n−1)!{\displaystyle (n-1)!} would also be congruent to −1{\displaystyle -1} modulo q{\displaystyle {q}}: indeed, if (n−1)!≡−1(mod n){\displaystyle (n-1)!\equiv -1{\pmod {n}}} then (n−1)!=n m−1=(q k)m−1=q(k m)−1{\displaystyle (n-1)!=nm-1=(qk)m-1=q(km)-1} for some integer m{\displaystyle m}, and consequently (n−1)!{\displaystyle (n-1)!} is one less than a multiple of q{\displaystyle q}. On the other hand, since 2≤q≤n−1{\displaystyle 2\leq q\leq n-1}, one of the factors in the expanded product (n−1)!=(n−1)×(n−2)×⋯×2×1{\displaystyle (n-1)!=(n-1)\times (n-2)\times \cdots \times 2\times 1} is q{\displaystyle q}. Therefore (n−1)!≡0(mod q){\displaystyle (n-1)!\equiv 0{\pmod {q}}}. This is a contradiction; therefore it is not possible that (n−1)!≡−1(mod n){\displaystyle (n-1)!\equiv -1{\pmod {n}}} when n{\displaystyle n} is composite. In fact, more is true. With the sole exception of the case n=4{\displaystyle n=4}, where 3!=6≡2(mod 4){\displaystyle 3!=6\equiv 2{\pmod {4}}}, if n{\displaystyle n} is composite then (n−1)!{\displaystyle (n-1)!} is congruent to 0 modulo n{\displaystyle n}. The proof can be divided into two cases: First, if n{\displaystyle n} can be factored as the product of two unequal numbers, n=a b{\displaystyle n=ab}, where 2≤a<b<n{\displaystyle 2\leq a<b<n}, then both a{\displaystyle a} and b{\displaystyle b} will appear as factors in the product (n−1)!=(n−1)×(n−2)×⋯×2×1{\displaystyle (n-1)!=(n-1)\times (n-2)\times \cdots \times 2\times 1} and so (n−1)!{\displaystyle (n-1)!} is divisible by a b=n{\displaystyle ab=n}. If n{\displaystyle n} has no such factorization, then it must be the square of some prime q{\displaystyle q} larger than 2. But then 2 q<q 2=n{\displaystyle 2q<q^{2}=n}, so both q{\displaystyle q} and 2 q{\displaystyle 2q} will be factors of (n−1)!{\displaystyle (n-1)!}, and so n{\displaystyle n} divides (n−1)!{\displaystyle (n-1)!} in this case, as well. Prime modulus The first two proofs below use the fact that the residue classes modulo a prime number form a finite field (specifically, a prime field). Elementary proof The result is trivial when p=2{\displaystyle p=2}, so assume p{\displaystyle p} is an odd prime, p≥3{\displaystyle p\geq 3}. Since the residue classes modulo p{\displaystyle p} form a field, every non-zero residue a{\displaystyle a} has a unique multiplicative inverse a−1{\displaystyle a^{-1}}. Euclid's lemma implies[a] that the only values of a{\displaystyle a} for which a≡a−1(mod p){\displaystyle a\equiv a^{-1}{\pmod {p}}} are a≡±1(mod p){\displaystyle a\equiv \pm 1{\pmod {p}}}. Therefore, with the exception of ±1{\displaystyle \pm 1}, the factors in the expanded form of (p−1)!{\displaystyle (p-1)!} can be arranged in disjoint pairs such that product of each pair is congruent to 1 modulo p{\displaystyle p}. This proves Wilson's theorem. For example, for p=11{\displaystyle p=11}, one has 10!=[(1⋅10)]⋅[(2⋅6)(3⋅4)(5⋅9)(7⋅8)]≡[−1]⋅[1⋅1⋅1⋅1]≡−1(mod 11).{\displaystyle 10!=[(1\cdot 10)]\cdot [(2\cdot 6)(3\cdot 4)(5\cdot 9)(7\cdot 8)]\equiv [-1]\cdot [1\cdot 1\cdot 1\cdot 1]\equiv -1{\pmod {11}}.} Proof using Fermat's little theorem Again, the result is trivial for p=2, so suppose p is an odd prime, p ≥ 3. Consider the polynomial g(x)=(x−1)(x−2)⋯(x−(p−1)).{\displaystyle g(x)=(x-1)(x-2)\cdots (x-(p-1)).} g has degree p − 1, leading term x p − 1, and constant term (p − 1)!. Its p − 1 roots are 1, 2, ..., p − 1. Now consider h(x)=x p−1−1.{\displaystyle h(x)=x^{p-1}-1.} h also has degree p − 1 and leading term x p − 1. Modulo p, Fermat's little theorem says it also has the same p − 1 roots, 1, 2, ..., p − 1. Finally, consider f(x)=g(x)−h(x).{\displaystyle f(x)=g(x)-h(x).} f has degree at most p−2 (since the leading terms cancel), and modulo p also has the p − 1 roots 1, 2, ..., p − 1. But Lagrange's theorem says it cannot have more than p−2 roots. Therefore, f must be identically zero (mod p), so its constant term is (p − 1)! + 1 ≡ 0 (mod p). This is Wilson's theorem. Proof using the Sylow theorems It is possible to deduce Wilson's theorem from a particular application of the Sylow theorems. Let p be a prime. It is immediate to deduce that the symmetric groupS p{\displaystyle S_{p}} has exactly (p−1)!{\displaystyle (p-1)!} elements of order p, namely the p-cycles C p{\displaystyle C_{p}}. On the other hand, each Sylow p-subgroup in S p{\displaystyle S_{p}} is a copy of C p{\displaystyle C_{p}}. Hence it follows that the number of Sylow p-subgroups is n p=(p−2)!{\displaystyle n_{p}=(p-2)!}. The third Sylow theorem implies (p−2)!≡1(mod p).{\displaystyle (p-2)!\equiv 1{\pmod {p}}.} Multiplying both sides by (p − 1) gives (p−1)!≡p−1≡−1(mod p),{\displaystyle (p-1)!\equiv p-1\equiv -1{\pmod {p}},} that is, the result. 1 Sources Remove ads Applications Summarize Perspective Primality tests In practice, Wilson's theorem is useless as a primality test because computing (n − 1)! modulo n for large n is computationally complex. Quadratic residues Using Wilson's Theorem, for any odd prime p = 2 m + 1, we can rearrange the left hand side of 1⋅2⋯(p−1)≡−1(mod p){\displaystyle 1\cdot 2\cdots (p-1)\ \equiv \ -1\ {\pmod {p}}} to obtain the equality 1⋅(p−1)⋅2⋅(p−2)⋯m⋅(p−m)≡1⋅(−1)⋅2⋅(−2)⋯m⋅(−m)≡−1(mod p).{\displaystyle 1\cdot (p-1)\cdot 2\cdot (p-2)\cdots m\cdot (p-m)\ \equiv \ 1\cdot (-1)\cdot 2\cdot (-2)\cdots m\cdot (-m)\ \equiv \ -1{\pmod {p}}.} This becomes ∏j=1 m j 2≡(−1)m+1(mod p){\displaystyle \prod {j=1}^{m}\ j^{2}\ \equiv (-1)^{m+1}{\pmod {p}}} or (m!)2≡(−1)m+1(mod p).{\displaystyle (m!)^{2}\equiv (-1)^{m+1}{\pmod {p}}.} We can use this fact to prove part of a famous result: for any prime _p such that p≡1(mod 4), the number (−1) is a square (quadratic residue) mod p. For this, suppose p=4 k+1 for some integer k. Then we can take m=2 k above, and we conclude that (m!)2 is congruent to (−1) (mod p). Formulas for primes Wilson's theorem has been used to construct formulas for primes, but they are too slow to have practical value. p-adic gamma function Wilson's theorem allows one to define the p-adic gamma function. 1 Sources Remove ads Gauss's generalization Gauss proved. 2 Sources Remove ads See also Wilson prime Table of congruences Agoh–Giuga conjecture Notes Loading content... References Loading content... External links Loading content... Edit in WikipediaRevision historyRead in Wikipedia Related topics Chinese remainder theorem About simultaneous modular congruencesLegendre symbol Function in number theory Quadratic reciprocity Gives conditions for the solvability of quadratic equations modulo prime numbers Gaussian integer Complex number whose real and imaginary parts are both integersEuler's theorem Theorem on modular exponentiation Show more Wikiwand - on chrome Seamless Wikipedia browsing. On steroids. 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1848	https://grammarist.com/grammar/linking-words-contrast/	Skip to content Log In My Account Linking Words of Contrast – List of Examples \| Candace Osmond \| Grammar \| Candace Osmond \| Grammar Candace Osmond Candace Osmond studied Advanced Writing & Editing Essentials at MHC. She’s been an International and USA TODAY Bestselling Author for over a decade. And she’s worked as an Editor for several mid-sized publications. Candace has a keen eye for content editing and a high degree of expertise in Fiction. Many words connect sentences, clauses, and phrases. But only contrast linking words can express two different ideas. The previous sentence uses the linking word of contrast but two connect it with the first sentence. Learn the different linking words of contrast in the article and how to use them in sentences. I’ll show you how to position them in sentences and whether to use a comma. What Are Contrast Words? Linking words for contrast or contrast words are words that show contrasting ideas. These English linking words connect two statements that are different from each other. For instance, one may be a positive idea, while another is a negative idea. Here’s a list of contrast words in English. Although Even though Though Yet However Despite But Despite In spite of While On the other hand How to Use Contrast Words in a Sentence Let’s look at the different contrast words and how to use them in sentences. But/However However is a sentence starter that shows a contrast between two separate sentences. Place the contrast word at the start of the second sentence, the end, and after the subject. Below are some examples of sentences. I love this city. However, the roads are too narrow. I love this city. The roads, however, are too narrow. I love this city. The roads are narrow, however. But is more informal than however. Don’t put a comma after it; instead, put a comma before it when it’s used to connect two independent clauses. For example: I love this city. But the words are too narrow. I love this city, but the roads are too narrow. Although/Even Though While however doesn’t connect two clauses, we use although and even though at the beginning of a complete sentence with subject and verb. You can also use them in the middle of a sentence. Here are some examples. Although the book was long, I enjoyed reading it the entire weekend. I enjoyed reading the book the entire weekend, although it was long. Though Though is a looking word found between two phrases. It means despite this or but in the English language. For example: I feel down today. I’m still attending the event, though. She will be my professor this semester, though I don’t know which subject. Yet Use the linker yet when expressing an idea that is surprising because it contradicts what has been mentioned earlier. For example: I helped her, yet she betrayed me. He’s already 94 years old, yet somehow, he is still fit and healthy. Despite/In Spite Of These linking words of contrast are used to contrast ideas. We can also place a noun phrase or an -ing form of the verb after them. For example: Despite her suggestions, Mike still played by his own rules. In spite of being the top artist in the country, she still feels lonely. Despite the Fact That/In Spite of the Fact That Statements like despite the fact that and in spite of the fact that are similar to despite and in spite. But these phrases have a subject and verb after them. For example: I enjoyed the party despite the fact that we arrived late. He was grateful in spite of the fact that he lost the competition. While/Whereas/Unlike While, whereas, and unlike show two things that differ. Use whereas and while with two complete clauses and unlike with a noun. Consider these sentences. Joan is happy, while John is excited. Whereas Kat looks excited in this photo, Sheila seems upset. Unlike Michael, Miguel excels in performing arts. On the Other Hand The contrast linking word on the other hand compares two different ways, facts, or ideas of the same situation. Here are some examples. On one hand, I want to be an engineer. But on the other hand, I also like my current education degree. I enjoy jazz music, but on the other hand, my friend loves rock music. Practice Using Contrast Linking Words Contrast linking words join contradicting ideas, whether they’re clauses, sentences, or phrases. These words may have similar uses but different meanings and placements in sentences. Some contrast words like although and however can be at the beginning of a sentence. Meanwhile, on the other hand, although, and other words can be in the middle of a sentence. 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1849	https://www.nice.org.uk/guidance/ng196	Cookies on the NICE website and services Cookies are files saved on your phone, tablet or computer when you visit a website. We use cookies to store information about how you use the NICE website and services, such as the pages you visit. For more information, view our cookie statement.(Opens in a new window) Essential cookies These cookies enable basic functions like page navigation and access to secure areas of the website. Our website cannot function properly without these cookies and they can only be deactivated by changing your browser preferences. Website usage cookies We use tools such as Google Analytics, Hotjar, Google Optimize and Loop11 to help us anonymously measure how you use our websites. This allows us to make improvements based on our users' needs. These tools set cookies that store anonymised information about how you got to the site, and how you interact with the site. Marketing and advertising cookies We use Google Ads, LinkedIn, Facebook, Twitter, Stackadapt, Amazon DSP, Xandr and Microsoft Ads to show adverts on external sites to promote NICE services, events and content. These services may use cookies to help make advertising more effective. These cookies are used for things like showing relevant adverts based on website visits, preventing the same ad from continuously re-appearing, or by measuring how many times people click on these adverts. Skip to content Accessibility help Sign in Atrial fibrillation: diagnosis and management NICE guideline Reference number: NG196 Published: Last updated: Download guidance (PDF) Overview This guideline covers diagnosing and managing atrial fibrillation in adults. It includes guidance on providing the best care and treatment for people with atrial fibrillation, including assessing and managing risks of stroke and bleeding. See the MHRA advice on warfarin and other anticoagulants – monitoring of patients during the COVID-19 pandemic, which includes reports of supratherapeutic anticoagulation with warfarin. Last reviewed:30 June 2021 We amended our recommendation on using the ORBIT score to assess bleeding risk to reinstate the previous link to an appropriate calculation tool, which was removed in error on 10 June 2021. Next review:This guideline will be reviewed if there is new evidence that is likely to change the recommendations. How we prioritise updating our guidance Decisions about updating our guidance are made by NICE’s prioritisation board. For more information on the principles and process, see NICE-wide topic prioritisation: the manual. For information about individual topics, including any decisions affecting this guideline, see the summary table of prioritisation board decisions. This guideline updates and replaces NICE guideline CG180 (June 2014). Recommendations This guideline includes new and updated recommendations on: detection and diagnosis assessment of stroke and bleeding risks stroke prevention rate control left atrial ablation preventing recurrence after ablation preventing and managing postoperative atrial fibrillation It also includes recommendations on: assessment of cardiac function personalised package of care and information referral rhythm control management of acute presentations initial management of stroke and atrial fibrillation stopping anticoagulation Who is it for? Healthcare professionals Commissioners and providers People with atrial fibrillation, their families and carers Guideline development process How we develop NICE guidelines This guideline was commissioned by NICE and developed in partnership with the Royal College of Physicians (RCP). Your responsibility The recommendations in this guideline represent the view of NICE, arrived at after careful consideration of the evidence available. When exercising their judgement, professionals and practitioners are expected to take this guideline fully into account, alongside the individual needs, preferences and values of their patients or the people using their service. It is not mandatory to apply the recommendations, and the guideline does not override the responsibility to make decisions appropriate to the circumstances of the individual, in consultation with them and their families and carers or guardian. All problems (adverse events) related to a medicine or medical device used for treatment or in a procedure should be reported to the Medicines and Healthcare products Regulatory Agency using the Yellow Card Scheme. Local commissioners and providers of healthcare have a responsibility to enable the guideline to be applied when individual professionals and people using services wish to use it. They should do so in the context of local and national priorities for funding and developing services, and in light of their duties to have due regard to the need to eliminate unlawful discrimination, to advance equality of opportunity and to reduce health inequalities. Nothing in this guideline should be interpreted in a way that would be inconsistent with complying with those duties. Commissioners and providers have a responsibility to promote an environmentally sustainable health and care system and should assess and reduce the environmental impact of implementing NICE recommendations wherever possible. Related quality standards Atrial fibrillation
1850	https://www.quora.com/How-many-times-can-a-permutation-matrix-mutiply-itself-can-be-itself-again-In-other-words-suppose-P-is-a-n-n-matrix-P-k-P-what-is-the-limitation-of-k	Something went wrong. Wait a moment and try again. Permutation Table Matrix Multiplication Linear Algebra Permutation Groups Linear Permutation Efficient Permutation Matrix Computations Permutation Cycle 5 How many times can a permutation matrix mutiply itself can be itself again? In other words, suppose P is a n n matrix, P^k = P what is the limitation of k? Alon Amit 30 years of Linear Algebra. · Upvoted by Nathan Hannon , Ph. D. Mathematics, University of California, Davis (2021) and Vance Faber , Ph. D. Mathematics · Author has 8.8K answers and 173.8M answer views · 2y It’s hard to guess what is meant by “what is the limitation of k”, but let me try. For starters, just to be clear: there’s no upper limit on the value of k, if that’s what you mean. For any permutation P you will find infinitely many k’s for which Pk=P. But we can say things about the possible values of k for a given permutation or for any permutation on n elements. An n×n permutation matrix represents a permutation of n elements. We can use the same letter P to denote either the matrix or the permutation itself, since statements like Pm=1 or Pk=P are equivalent between these interpret It’s hard to guess what is meant by “what is the limitation of k”, but let me try. For starters, just to be clear: there’s no upper limit on the value of k, if that’s what you mean. For any permutation P you will find infinitely many k’s for which Pk=P. But we can say things about the possible values of k for a given permutation or for any permutation on n elements. An n×n permutation matrix represents a permutation of n elements. We can use the same letter P to denote either the matrix or the permutation itself, since statements like Pm=1 or Pk=P are equivalent between these interpretations. Every permutation has an order. This is the smallest positive integer m such that Pm=1, the identity permutation. Once you know the order, you can determine those numbers k for which Pk=P: they are 1, m+1, 2m+1, 3m+1 and so on. This is because the multiples am of m are precisely those powers for which Pam=1. For example, if P has order 2 then Pk=P precisely when k is odd, and if P has order 10 then Pk=P whenever k=1,11,21,31,41 and so on. A permutation can be represented as a product of disjoint cycles, and its order is the least common multiple of the lengths of those cycles. For example, the permutation on 9 elements which swaps A and B, cycles CDE around, and cycles FGHI around has order 12, which is the LCM of 2, 3 and 4. The smallest possible order of a permutation P is 1: it is the order of the identity permutation itself. And of course, for the identity permutation, 1k=1 for any k. The order of a permutation on n elements must divide n!. This provides a simple upper bound, but it is generally very crude. When n=12, for example, the largest possible order of a permutation is 60, much less than 12!. The precise value of this maximum given n is called Landau’s function. So if you get asked about the possible values of k for which Pk=P and P is an n×n permutation matrix, if n is small you can enumerate the possible cycle lengths, find the LCM, and determine the possible orders of permutations on n elements. The values of k are then 1 more than the multiples of those numbers. Related questions What is a proof of 'If A is a permutation matrix, then A^T=A'? How would you prove permutation matrix A of size n to the nth power is Identity matrix? What is the proof that permutation and elimination do not change the null space of a matrix? What are some examples of linear permutation if the formula is P=n! =(n-1) (n-2)? How does one prove ∑ n k = 0 ( n k ) ( − 1 ) k P ( k ) = 0 ? Alexey Godin Ph.D. in Mathematics & Economics, Moscow State University (Graduated 1998) · Author has 2.7K answers and 3.9M answer views · 2y Note that Pk=P⇒Pk2=Pk=P⇒Pkn=P so there is no upper limit for such k. The lower limit is also obvious, it is 0 for the identity matrix. So what we want to find is minimal k≠1:Pk=P and then the maximal such k for given n. If your matrix is a permutation matrix it corresponds to a unique permutation. Every permutation of n elements can be represented as a product of independent cycles. σk=σ⇔σk−1=e. If you have a cycle θ=(a1…at) then the minimal power non-zero power θt=e is t. It is called the order of the Note that Pk=P⇒Pk2=Pk=P⇒Pkn=P so there is no upper limit for such k. The lower limit is also obvious, it is 0 for the identity matrix. So what we want to find is minimal k≠1:Pk=P and then the maximal such k for given n. If your matrix is a permutation matrix it corresponds to a unique permutation. Every permutation of n elements can be represented as a product of independent cycles. σk=σ⇔σk−1=e. If you have a cycle θ=(a1…at) then the minimal power non-zero power θt=e is t. It is called the order of the element. Thus k−1=LCM(l1,…,lm) where li are lengths of independent cycles forming our cycle. So for example if n=10 we can have a cycle σ=(1 2)(3 4 5)(6 7 8 9 10) such that k−1=2⋅3⋅5=30⇒k=31 The relation of such k (actually k-1) to given n is a well-known function, called Landau's function. So if I get you right your answer is k=g(n)+1 David Vanderschel PhD in Mathematics & Physics, Rice (Houston neighborhood) (Graduated 1970) · Author has 37.6K answers and 50.1M answer views · 2y A2A: The representation of the permutation does not matter. Look at the representation in terms of cycles. (See the Cycle Notation section of the Wikipedia article.) If n is the LCM of the cycle lengths, then n repetitions of the permutation produce the identity permutation. One more, and you are back to the original permutation. David Joyce Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979) · Upvoted by Valerian Mahdi Pratama , M.S. Mathematics, Bandung Institute of Technology (2022) and Peter Ferguson , MSci Theoretical Physics & Mathematics, Lancaster University (2018) · Author has 9.9K answers and 68.4M answer views · 5y Related What is a simple way to show that for any permutation matrix, P, its inverse equals its transpose? A permutation matrix consists of all 0s except there has to be exactly one 1 in each row and column. Here’s an example of a 5×5 permutation matrix. This matrix then operates on a column vector by permuting its entries. Note that the first entry in the column vector changes from a1 to a2. That’s because the entry p12 in the permutation matrix is 1. More generally, if pij=1 in the permutation matrix, then ith entry of the column vector is replaced by the jth entry. Naturally, if you want the replacement to go the other way, just take the transpose of the matrix to replace A permutation matrix consists of all 0s except there has to be exactly one 1 in each row and column. Here’s an example of a 5×5 permutation matrix. This matrix then operates on a column vector by permuting its entries. Note that the first entry in the column vector changes from a1 to a2. That’s because the entry p12 in the permutation matrix is 1. More generally, if pij=1 in the permutation matrix, then ith entry of the column vector is replaced by the jth entry. Naturally, if you want the replacement to go the other way, just take the transpose of the matrix to replace each pij by pji. Sponsored by CDW Corporation How can AI help your teams make faster decisions? CDW’s AI solutions offer retrieval-augmented generation (RAG) to expedite info with stronger insights. Related questions How can one prove this ∏ p − 1 k = 0 n − k p − k = ( n p ) ? What is P (P (P (P({}))))? What is n ∑ k = 0 ( − 1 ) k n n + k ( n k ) ? How can we prove that n ! n n ( n ∑ k = 0 n k k ! − ∞ ∑ k = 4 n n k k ! ) − 4 3 = O ( 1 n ) ? How do you prove that ∑ n k = 0 ( − 1 ) k ( n k ) ( n − k ) n = n ! ? Richard Goldstone PhD in Mathematics, The Graduate Center, CUNY (Graduated 1995) · Author has 1.8K answers and 3.9M answer views · Updated 2y Related How would you prove permutation matrix A of size n to the nth power is Identity matrix? Perhaps there is a typo and you meant n!, but if you really mean n that’s false. The collection of n×n permutation matrices is a faithful representation of the symmetric group Sn. You are suggesting that n is the exponent for Sn, which would mean that every element of Sn has order dividing n. In fact, every element of Sn has order dividing n!, but that's very different from dividing n—for example there can be elements of order dividing (n−1)! whose order is relatively prime to n. Every element of Sn can be written as a product of disjoint cycles, and the order of the element is the Perhaps there is a typo and you meant n!, but if you really mean n that’s false. The collection of n×n permutation matrices is a faithful representation of the symmetric group Sn. You are suggesting that n is the exponent for Sn, which would mean that every element of Sn has order dividing n. In fact, every element of Sn has order dividing n!, but that's very different from dividing n—for example there can be elements of order dividing (n−1)! whose order is relatively prime to n. Every element of Sn can be written as a product of disjoint cycles, and the order of the element is the least common multiple of the orders of the cycles. So for example, in S5, the element σ=(12)(345) has order 6, which is bigger than n=5. Corresponding to any partition n=c1+c2+⋯cn, there is a product of disjoint cycles σ=C1C2⋯Cn with Ci of order ci, and \|σ\|=lcm(c1,c2,…,cn). As noted above for n=5, this least common multiple can be (and usually is) bigger than n. To convert the product of disjoint cycles representing σ to a matrix, we let σ−1 act on the indices of the standard basis {e1,e2,…,en} for Rn and write the matrix for the corresponding transformation. (Yeah, in order to permute the ei as specified by σ, we have to use σ−1 on the subscripts). The result for σ=C1C2⋯Cn is a block diagonal matrix [σ]=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣P1P2⋱Pn⎤⎥ ⎥ ⎥ ⎥ ⎥⎦ Where each Pi is a ci×ci permutation matrix corresponding to Ci acting on the subset of basis elements actually moved by Ci. So in the n=5 case above with σ=(12)(345), corresponding to C1=(12) acting on {e1,e2} we have P1=, and corresponding to C2=(345) acting on {e3,e4,e5}we have P2=⎡⎢⎣010001100⎤⎥⎦. These give a matrix of “size 5” whose order is not 5 but 6 [σ]=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0110010001100⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦, and illustrates how to make up infinitely many counterexamples in which the order of an n×n permutation matrix will be bigger than n. But counterexamples don’t have to have order bigger than n, they just have to have an order that isn’t a divisor of n. For example, any cycle of length k<n, where k is not a divisor of n, will do. The smallest example of this is S3=D3, the dihedral group of symmetries of an equilateral triangle. This has elements of order 2 corresponding to reflections in an altitude, and elements of order 3 corresponding to rotations. If we take σ=(12), then [σ]=⎡⎢⎣01101⎤⎥⎦, is a 3×3 permutation matrix that is of order 2. Since [σ]2=I, we have [σ]3=[σ], not I as proposed in the original question. The previous example generalizes almost without change to S2k+1. The element of order 2 corresponding to (12) is [σ]=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣01101⋱1⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦, and, as in the case n=3, we have[σ]2k+1=[σ], not I. Remark. The discussion raises the question about elements of maximal order in Sn. The function n↦g(n) that associates, with each natural number n, the maximal order g(n) of elements of Sn is called Landau’s function. The values of Landau’s function corresponding in order to the natural numbers are an infinite sequence beginning with 1, 2, 3, 4, 6, 6, 12, 15, 20, 30, 30, 60, 60, 84, 105, 140, 210, 210, 420, 420, 420, 420, 840, 840, 1260, 1260, 1540, 2310, 2520, 4620, 4620, 5460, 5460, 9240, 9240, 13860, 13860, 16380, 16380, 27720, 30030, 32760, 60060, 60060, 60060, 60060, 120120… See A000793 - OEIS , which however gives an additional 1 at the beginning corresponding to S0, the group of permutations of the empty set, which is a group with a single element, necessarily the identity element, whose order is 1. Raziman T.V. IOI silver medallist (2007), ACM ICPC world finalist (2011, 2012) · Author has 2.6K answers and 28.1M answer views · 12y Related Combinatorics: Let P be a random permutation of 1 to n. A pair (i,j) is called an inversion in P if i > j, but i appears before j in P. How many permutation of 1 to n have exactly two inversions? (N−2)(N+1)2 Let F(N,k) be the number of permutations of {1..N} with exactly k inversions. Obviously, F(N,0) = 1 and F(N,1)=N-1 for all N. For finding F(N,2), we can look at which position N is located in the permutation. Every number which comes after N in the permutation will result in an inversion. Hence N has to appear in position N,N-1 or N-2. If N appears in position N, then it cannot be part of any inversion and the total number of sequences with 2 inversions is same as number of sequence with 2 inversions because of first N-1 elements. This is equal to F(N-1,2). If N appears in p (N−2)(N+1)2 Let F(N,k) be the number of permutations of {1..N} with exactly k inversions. Obviously, F(N,0) = 1 and F(N,1)=N-1 for all N. For finding F(N,2), we can look at which position N is located in the permutation. Every number which comes after N in the permutation will result in an inversion. Hence N has to appear in position N,N-1 or N-2. If N appears in position N, then it cannot be part of any inversion and the total number of sequences with 2 inversions is same as number of sequence with 2 inversions because of first N-1 elements. This is equal to F(N-1,2). If N appears in position N-1, it will cause exactly 1 inversion with the element at position N. Hence we need to count the number of permutations of remaining elements which give exactly one inversion. This is F(N-1,1)=N-2. If N appears at position N-2, it causes 2 inversions and hence there should not be any further inversions caused by the remaining elements. The number of such permutations is F(N-1,0)=1. Combining, we obtain F(N,2) = F(N-1,2) + N-1 Using the base cases F(1,2) = F(2,2) = 0, we can calculate the remaining values recursively. In fact, it is possible to find the closed form solution F(N,2) = (N-2)(N+1)/2 for N>1, which can be proved by induction. Sponsored by CDW Corporation Ready for the latest in AI-powered capability? AMD Ryzen™ and Windows 11-powered x86 PCs from CDW drive business productivity with AI intuition. Peter v. Elbing Algebra rocks, kind of · Author has 718 answers and 943.1K answer views · 2y Related How would you prove permutation matrix A of size n to the nth power is Identity matrix? This is in general not true. The permutations of n items constitute a group called symmetric group S(n). It has n! elements. The n×n-permutation matrices are a representation of the symmetric group S(n). Every element g of a group has an order k, which is the minimum number one can take g to the power to get the neutral element, in matrix terms the identity matrix: gk=1 where 1 is shorthand for the neutral element. There is a group theoretical theorem that says k must divide the order of the group, i.e. the size of the group (Lagrange's theorem). As \|S(n)\|=n! the number k must divide n!. This is in general not true. The permutations of n items constitute a group called symmetric group S(n). It has n! elements. The n×n-permutation matrices are a representation of the symmetric group S(n). Every element g of a group has an order k, which is the minimum number one can take g to the power to get the neutral element, in matrix terms the identity matrix: gk=1 where 1 is shorthand for the neutral element. There is a group theoretical theorem that says k must divide the order of the group, i.e. the size of the group (Lagrange's theorem). As \|S(n)\|=n! the number k must divide n!. (Don't confuse order of the group and order of an element of the group. They are connected but not the same!) According to Cauchy's theorem (group theory) for every prime number p dividing the group order there must be a subgroup with order p. This is even true for every power of p that divides the group order (Sylow's theorem). As every prime p less or equal n divides n! the symmetric group S(n) has a subgroup of order 2≤p≤n. S(n) has also elements with order n, the permutations that in a closed chain of n items rotate the chain one position to the left or the right. As the order of any element g must divide n! the n!-th power of g must be the neutral element, but not always the n-th power. Saikat Ghosh Lives in Bengaluru, Karnataka, India · Author has 73 answers and 175.3K answer views · 7y Related How many permutations of [N] are there where the substring ( k , k + 1 ) ∀ k ∈ [ n − 1 ] is avoided? How many permutations of the first n integers exist, such that every substring (k,k+1)∀k∈[n−1] is avoided ? Call a permutation good if the substring (k,k+1) is always avoided. Let f(n) represent the number of good permutations of length n. Now, let us imagine we have a good permutation of length n. What happens when we delete the element n from this good permutation ? Either it remains a good permutation or it becomes a bad permutation. Case 1 : A good permutation remains good after n is deleted. This means that n was inserted into a good permutation of length n−1 in any position n How many permutations of the first n integers exist, such that every substring (k,k+1)∀k∈[n−1] is avoided ? Call a permutation good if the substring (k,k+1) is always avoided. Let f(n) represent the number of good permutations of length n. Now, let us imagine we have a good permutation of length n. What happens when we delete the element n from this good permutation ? Either it remains a good permutation or it becomes a bad permutation. Case 1 : A good permutation remains good after n is deleted. This means that n was inserted into a good permutation of length n−1 in any position not to the right of (n−1). There are f(n−1) good permutations of size (n−1). A permutation of length (n−1) has n free spaces. We can insert n anywhere other than the right of (n−1). So, the total number of ways to get a good permutation of length n from a good permutation of length (n−1)=(n−1)×f(n−1) Case 2: A good permutation becomes bad after n is deleted. If the permutation becomes bad after removing n, it means n was in between two consecutive elements (k,k+1). How many permutations of length (n−1) exist with exactly one substring (k,k+1)? Let us combine (k,k+1) and treat it as a single element x. Consider any good permutation of length (n−2). Then replace one of the (n−2) elements , by our . Increment all the values in the permutation by . 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Assuming is a randomly chosen function from to itself, if we let denote the expected number of elements of that survive, or continue showing up, after any number of iterations of , then we have . I don’t see any simpler closed form for the value of this sum, but it appears that the ratio decreases as increases, as , , , , , and . To find this formula, we co Assuming is a randomly chosen function from to itself, if we let denote the expected number of elements of that survive, or continue showing up, after any number of iterations of , then we have . I don’t see any simpler closed form for the value of this sum, but it appears that the ratio decreases as increases, as , , , , , and . To find this formula, we consider an arbitrary element and find the probability , as ranges over all possible functions , that survives under all iterations of , i.e. that is in the image of the iterate of for all . This happens if and only if lies in a -cycle under for some . (See the edit below for an elaboration on this point.) For any , lies in a -cycle if and only if are all distinct and . The probability of this occurring is . So if is the random variable giving value if survives all iterates of and value otherwise, then the expected value of is . And then is the random variable giving the number of elements of which survive all iterates of , and its expected value is . Edit: It occurs to me that some justification is needed for the assertion that lies in the image of for all if and only if lies in a -cycle under for some . Clearly if lies in such a -cycle under then it lies in the image of each . For the converse, suppose merely that lies in the image of . Then there is some with . Let , , and generally for all . Then can’t all be distinct, so for some . Then is a -cycle under (with ), and each with lies within this -cycle. In particular, does since . Alexander Farrugia Used matrices extensively in his PhD thesis. · Author has 3.2K answers and 27.5M answer views · 7y Related Can you transpose a matrix by multiplying by some type of permutation matrix? Not exactly. But we can still salvage something. Permutation matrices work fine on vectors. So we can first write down our matrix as an matrix, basically listing each column underneath each other. For example, we may write as or as Then we may permute the entries of this vector accordingly so that the resulting vector corresponds to the transpose matrix. Of course, an appropriate Not exactly. But we can still salvage something. Permutation matrices work fine on vectors. So we can first write down our matrix as an matrix, basically listing each column underneath each other. For example, we may write as or as Then we may permute the entries of this vector accordingly so that the resulting vector corresponds to the transpose matrix. Of course, an appropriate permutation matrix is fine for this purpose. Lastly, we want to write down the resulting vector as an matrix to obtain our transpose matrix. We still need to perform the transformations from our matrix to an vector and from an vector to an matrix as matrix operations though. Let us start from the first one. This is called vectorizing a matrix , and may be expressed using the Kronecker product , usually denoted by : Here, is an matrix and are the first, second, , last columns of the identity matrix. For example We now move on to the reverse construction: we want an vector to be converted into an matrix, by writing the first entries as the first column, the second entries as the second column, and so on. This is done in the following way. In our example, the vector would have been permuted into by our appropriate ‘transposing’ permutation matrix. We then take this vector and do the following to it: In our example, this would be Essentially, then, we have ‘cheated’ by first transforming our matrix into an vector, then permuting the entries of the vector accordingly using an permutation matrix, and finally rewriting this resulting vector into an matrix. Footnotes Vectorization (mathematics) - Wikipedia Kronecker product - Wikipedia Enrico Gregorio Associate professor in Algebra · Author has 18.4K answers and 16M answer views · 2y Related How would you prove permutation matrix A of size n to the nth power is Identity matrix? This is false even in matrices. Indeed the permutation matrix satisfies (the identity matrix) and therefore Nick Shales MPhys in Theoretical Physics, The University of York (Graduated 2004) · Author has 419 answers and 1.5M answer views · 7y Related How many permutations of [N] are there where the substring is avoided? Call the set of permutations of with at least the substring for . Then define Note that and that is the set of all permutations on . The polynomial with coefficients that are the sums given by may be written Now call the number of permutations belonging to exactly of the sets we have Call the set of permutations of with at least the substring for . Then define Note that and that is the set of all permutations on . The polynomial with coefficients that are the sums given by may be written Now call the number of permutations belonging to exactly of the sets we have since any permutation of belonging to exactly sets is counted precisely times in the sum . Where, clearly is the polynomial whose coefficients count the number of permutations of belonging to exactly sets. i.e. those with exactly occurrences of substrings . We have or by substituting Now, can be easily calculated, for we can choose adjacent pairs in ways, then, since each pair is forcibly joined there are effectively objects which we can permute in ways, hence Equating coefficients at the extremes of and using gives us a generalisation We want to count those permutations belonging to exactly none of these sets , i.e. : This has a sequence Of course it is now trivial to count those permutations with at least adjacent pairs : where the second line becomes the third due to a well known binomial identity. Rizwan Hudda Doing competitive programming since 2009 · 12y Related Combinatorics: Let P be a random permutation of 1 to n. A pair (i,j) is called an inversion in P if i > j, but i appears before j in P. How many permutation of 1 to n have exactly two inversions? A permutation of 1..n can have exactly two inversions in following ways: (1,.., i+2, i, i + 1,..,n) where 1 <= i <= n - 2 (A1) (1,..,i + 1, i +2, i,..,n) where 1 <= i <= n - 2 (A2) (1,.., i + 1, i, .., j + 1, j,..) where 2 <= i + 1 < j <= n - 1 (B) Number of permutations of type (A1) = Number of permutations of type (A2) = Number of values of i >= 1 and <= n - 2 = n -2 Number of permutations of type (B) = Number of pairs (i, j) such that 2 <= i + 1 < j <= n - 1 = (club i, i+1 together and j, j+1 together so there are ways of choosing i, j) Total number of A permutation of 1..n can have exactly two inversions in following ways: (1,.., i+2, i, i + 1,..,n) where 1 <= i <= n - 2 (A1) (1,..,i + 1, i +2, i,..,n) where 1 <= i <= n - 2 (A2) (1,.., i + 1, i, .., j + 1, j,..) where 2 <= i + 1 < j <= n - 1 (B) Number of permutations of type (A1) = Number of permutations of type (A2) = Number of values of i >= 1 and <= n - 2 = n -2 Number of permutations of type (B) = Number of pairs (i, j) such that 2 <= i + 1 < j <= n - 1 = (club i, i+1 together and j, j+1 together so there are ways of choosing i, j) Total number of permutations = num of type A permutations + num of type B permutations = EDIT: Thanks to Raziman T.V. (റസിമാൻ ടി.വി.) for pointing out a bug in my solution, and suggesting a simplification. I have updated this answer accordingly. Related questions What is a proof of 'If A is a permutation matrix, then A^T=A'? How would you prove permutation matrix A of size n to the nth power is Identity matrix? What is the proof that permutation and elimination do not change the null space of a matrix? What are some examples of linear permutation if the formula is P=n! =(n-1) (n-2)? How does one prove ? How can one prove this ? What is P (P (P (P({}))))? What is ? How can we prove that ? How do you prove that ? How do I find all primes and such that ? How do you prove that ? Can you transpose a matrix by multiplying by some type of permutation matrix? Why does p/q^m = q/p^n? What is a decorated permutation? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1851	https://courses.lumenlearning.com/suny-physics/chapter/8-4-elastic-collisions-in-one-dimension/	Elastic Collisions in One Dimension Learning Objectives By the end of this section, you will be able to: Describe an elastic collision of two objects in one dimension. Define internal kinetic energy. Derive an expression for conservation of internal kinetic energy in a one dimensional collision. Determine the final velocities in an elastic collision given masses and initial velocities. Let us consider various types of two-object collisions. These collisions are the easiest to analyze, and they illustrate many of the physical principles involved in collisions. The conservation of momentum principle is very useful here, and it can be used whenever the net external force on a system is zero. We start with the elastic collision of two objects moving along the same line—a one-dimensional problem. An elastic collision is one that also conserves internal kinetic energy. Internal kinetic energy is the sum of the kinetic energies of the objects in the system. Figure 1 illustrates an elastic collision in which internal kinetic energy and momentum are conserved. Truly elastic collisions can only be achieved with subatomic particles, such as electrons striking nuclei. Macroscopic collisions can be very nearly, but not quite, elastic—some kinetic energy is always converted into other forms of energy such as heat transfer due to friction and sound. One macroscopic collision that is nearly elastic is that of two steel blocks on ice. Another nearly elastic collision is that between two carts with spring bumpers on an air track. Icy surfaces and air tracks are nearly frictionless, more readily allowing nearly elastic collisions on them. Elastic Collision An elastic collision is one that conserves internal kinetic energy. Internal Kinetic Energy Internal kinetic energy is the sum of the kinetic energies of the objects in the system. Figure 1. An elastic one-dimensional two-object collision. Momentum and internal kinetic energy are conserved. Now, to solve problems involving one-dimensional elastic collisions between two objects we can use the equations for conservation of momentum and conservation of internal kinetic energy. First, the equation for conservation of momentum for two objects in a one-dimensional collision is p1 + p2 = p′1 + p′2 (Fnet = 0) or m1v1 + m2v2 = m1v′1 + m2v′2 (Fnet = 0), where the primes (′) indicate values after the collision. By definition, an elastic collision conserves internal kinetic energy, and so the sum of kinetic energies before the collision equals the sum after the collision. Thus, 12m1v12+12m2v22=12m1v'12+12m2v'22(two-object elastic collision) expresses the equation for conservation of internal kinetic energy in a one-dimensional collision. Example 1. Calculating Velocities Following an Elastic Collision Calculate the velocities of two objects following an elastic collision, given that m1 = 0.500 kg, m2 = 3.50 kg, v1 = 4.00 m/s, and v2 = 0. Strategy and Concept First, visualize what the initial conditions mean—a small object strikes a larger object that is initially at rest. This situation is slightly simpler than the situation shown in Figure 1 where both objects are initially moving. We are asked to find two unknowns (the final velocities v′1 and v′2). To find two unknowns, we must use two independent equations. Because this collision is elastic, we can use the above two equations. Both can be simplified by the fact that object 2 is initially at rest, and thus v2=0. Once we simplify these equations, we combine them algebraically to solve for the unknowns. Solution For this problem, note that v2=0 and use conservation of momentum. Thus, p1 = p′1 + p′2 or m1v1=m1v′1+m2v′2. Using conservation of internal kinetic energy and that v2=0, 12m1v12=12m1v'12+12m2v'22 Solving the first equation (momentum equation) for v′2, we obtain v'2=m1m2(v1−v'1). Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable v′2, leaving only v′1 as an unknown (the algebra is left as an exercise for the reader). There are two solutions to any quadratic equation; in this example, they are v′1 = 4 . 00 m/s and v′1=−3.00 m/s. As noted when quadratic equations were encountered in earlier chapters, both solutions may or may not be meaningful. In this case, the first solution is the same as the initial condition. The first solution thus represents the situation before the collision and is discarded. The second solution (v′1=−3.00 m/s) is negative, meaning that the first object bounces backward. When this negative value of v′1 is used to find the velocity of the second object after the collision, we get v'2=m1m2(v1−v'1)=0.500 kg3.50 kg[4.00−(−3.00)]m/s or v′2=1.00 m/s. Discussion The result of this example is intuitively reasonable. A small object strikes a larger one at rest and bounces backward. The larger one is knocked forward, but with a low speed. (This is like a compact car bouncing backward off a full-size SUV that is initially at rest.) As a check, try calculating the internal kinetic energy before and after the collision. You will see that the internal kinetic energy is unchanged at 4.00 J. Also check the total momentum before and after the collision; you will find it, too, is unchanged. The equations for conservation of momentum and internal kinetic energy as written above can be used to describe any one-dimensional elastic collision of two objects. These equations can be extended to more objects if needed. Making Connections: Take-Home Investigation—Ice Cubes and Elastic Collision Find a few ice cubes which are about the same size and a smooth kitchen tabletop or a table with a glass top. Place the ice cubes on the surface several centimeters away from each other. Flick one ice cube toward a stationary ice cube and observe the path and velocities of the ice cubes after the collision. Try to avoid edge-on collisions and collisions with rotating ice cubes. Have you created approximately elastic collisions? Explain the speeds and directions of the ice cubes using momentum. PhET Explorations: Collision Lab Investigate collisions on an air hockey table. Set up your own experiments: vary the number of discs, masses and initial conditions. Is momentum conserved? Is kinetic energy conserved? Vary the elasticity and see what happens. Click to run the simulation. Section Summary An elastic collision is one that conserves internal kinetic energy. Conservation of kinetic energy and momentum together allow the final velocities to be calculated in terms of initial velocities and masses in one dimensional two-body collisions. Conceptual Questions What is an elastic collision? Problems & Exercises Two identical objects (such as billiard balls) have a one-dimensional collision in which one is initially motionless. After the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic energy are conserved. Professional Application.Two manned satellites approach one another at a relative speed of 0.250 m/s, intending to dock. The first has a mass of 4.00 × 103 kg, and the second a mass of 7.50 × 10 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity? A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. What would their final velocities be in this case? Glossary elastic collision: a collision that also conserves internal kinetic energy internal kinetic energy: the sum of the kinetic energies of the objects in a system Selected Solutions to Problems & Exercises 2. 0.250 m/s Candela Citations CC licensed content, Shared previously College Physics. Authored by: OpenStax College. Located at: License: CC BY: Attribution. License Terms: Located at License PhET Interactive Simulations . Provided by: University of Colorado Boulder . Located at: License: CC BY: Attribution Licenses and Attributions CC licensed content, Shared previously College Physics. Authored by: OpenStax College. Located at: License: CC BY: Attribution. License Terms: Located at License PhET Interactive Simulations . Provided by: University of Colorado Boulder . Located at: License: CC BY: Attribution
1852	https://paradigmmath.weebly.com/uploads/8/1/7/9/81799270/math_3__p_p_part_2_sequences_patterns_frequency_tables___venn_diagrams.pdf	1 Math 3 P&P Part 2 January 12, 2018 Name:______ Date:___ Math 3 Proportion & Probability Part 2 Sequences, Patterns, Frequency Tables & Venn Diagrams MATH 2 REVIEW ARITHMETIC SEQUENCES In an Arithmetic Sequence the difference between one term and the next is a constant. In other words, we just add the same value each time ... infinitely. In general, we could write an arithmetic sequence like this: {a, a+d, a+2d, a+3d, ... } where: a is the first term, and d is the difference between the terms (called the "common difference") In the example: a = 1 (the first term) and d = 3 (the "common difference" between terms) And we get: {a, a+d, a+2d, a+3d, ... } {1, 1+3, 1+2×3, 1+3×3, ... } {1, 4, 7, 10, ... } EXPLICIT FORMULA FOR ARITHMETIC SEQUENCES We can write an Arithmetic Sequence as an explicit formula: an = a1 + d(n-1) (We use "n-1" because d is not used in the 1st term). Example: Write the explicit formula for arithmetic sequence, and calculate the 4th term for 3, 8, 13, 18, 23, 28, 33, 38, . . . . Solution: This sequence has a difference of 5 between each number. The values of a and d are: a = 3 (the first term) and d = 5 (the "common difference") The explicit formula can be calculated: an = a1 + d(n-1) = 3 + 5(n-1) = 3 + 5n – 5 = 5n - 2 So, the 4th term is: a4 = 5(4) - 2 = 18 RECURSIVE FORMULA FOR ARITHMETIC SEQUENCES We can also write an Arithmetic Sequence as a recursive formula: an = an-1 + d (We use "n-1" because d is not used in the 1st term). Example: Write the recursive formula for arithmetic sequence, and calculate the 4th term for 3, 8, 13, 18, 23, 28, 33, 38, . . . . Solution: As from above , this sequence has a difference of 5 between each number. The values of a and d are: a = 3 (the first term) and d = 5 (the "common difference") The recursive formula can be calculated: an = an-1 + d a2 = a1 + 5 = 3 + 5 = 8 a3 = a2 + 5 = 8 + 5 = 13 So, the 4th term is: a4 = a3 + 5 = 13 + 5 = 18 Sample Question: 1. Which of the following statements is NOT true about the arithmetic sequence 16, 11, 6, 1. . .? a. The fifth term is -4. b. The sum of the first 5 terms is 30. c. The seventh term is -12 d. The common difference of consecutive integers is -5. e. The sum of the first 7 terms is 7. 2. The greatest integer of a set of consecutive even integers is 12. If the sum of these integers is 40, how many integers are in this set? 3. What is the sum of the first 4 terms of the arithmetic sequence in which the 6th term is 8 and the 10th term is 13? 2 Math 3 P&P Part 2 January 12, 2018 GEOMETRIC SEQUENCE In a Geometric Sequence each term is found by multiplying the previous term by a constant. For illustration, look at the sequence: 2, 4, 8, 16, 32, 64, 128, 256, . . . This sequence has a factor of 2 between each number. Each term (except the first term) is found by multiplying the previous term by 2. In general, we write a Geometric Sequence like this: {a, ar, ar2, ar3, ... } where: a is the first term, and r is the factor between the terms (called the "common ratio") Example: {1,2,4,8,...} Solution: The sequence starts at 1 and doubles each time, so a=1 (the first term) and r=2 (the "common ratio" between terms is a doubling) And we get: {a, ar, ar2, ar3, ... } = {1, 1×2, 1×22, 1×23, ... } = {1, 2, 4, 8, ... } But be careful, r should not be 0: When r = 0, we get the sequence {a, 0, 0,...} which is not geometric EXPLICIT FORMULA FOR GEOMETRIC SEQUENCES We can also calculate any term using the explicit formula: an = a1r(n-1) (We use "n-1" because ar0 is for the 1st term) Example: 10, 30, 90, 270, 810, 2430, . . . . Solution: This sequence has a factor of 3 between each number. The values of a and r are: a = 10 (the first term) and r = 3 (the "common ratio") The Rule for any term is: an = 10 × 3(n-1) So, the 4th term is: a4 = 10 × 3(4-1) = 10 × 33 = 10 × 27 = 270 And the 10th term is: a10 = 10 × 3(10-1) = 10 × 39 = 10 × 19683 = 196830 RECURSIVE FORMULA FOR GEOMETRIC SEQUENCES We can also calculate a geometric sequence using the recursive formula: an = an-1(r) Example: 10, 30, 90, 270, 810, 2430, . . . . Solution: As listed above, this sequence has a factor of 3 between each number. The values of a and r are: a = 10 (the first term) and r = 3 (the "common ratio") a2 = a1(r) = 10(3) = 30 a3 = a2(r) = 30(3) = 90 a4 = a3(r) = 90(3) = 270 Example: 4, 2, 1, 0.5, 0.25, . . . . Solution: This sequence has a factor of 0.5 (a half) between each number. Using the explicit formula: an = 4 × (0.5)n-1 Using the recursive formula: an = an-1(0.5) Example Questions: 4. The first term is 1 in the geometric sequence 1, -3, 9, -27, . . . . What is the SEVENTH term in the geometric sequence? 5. The first and second terms of a geometric sequence are a and ab, in that order. What is the 643rd term of the sequence? 6. Which of the following statements is NOT true about the geometric sequence 26, 18, 9, . . . ? a. The fourth term is 4.5 b. The sum of the first five terms is 59.75 c. Each consecutive term is ½ of the previous term d. Each consecutive term is evenly divisible by 3 e. The common ratio of consecutive terms is 2:1 3 Math 3 P&P Part 2 January 12, 2018 COUNTING CONSECUTIVE INTEGERS To find the number of consecutive integers between two values, subtract the smallest from the largest and add 1. So to find the number of consecutive integers from 13 through 31, subtract: 31 - 13 = 18. Then add 1: 18 + 1 = 19. There are 19 consecutive integers from 13 through 31. USING THE SLOT METHOD TO FIND CONSECUTIVE INTEGERS First write the sequence including the unknown numbers by leaving slots to fill in the numbers. Then find the “common difference” between the first and last terms in the sequence. Divide the difference by the number of “jumps” between the first and last integer. Example: What two numbers should be placed in the blanks below so that the difference between successive entries is the same? 26, _, , 53 a. 36, 43 b. 35, 44 c. 34, 45 d. 33, 46 e. 30, 49 Solution: Common Difference = 53-26 = 27. There are three jumps (26 to 2nd term, 2nd term to 3rd term, 3rd term to 53) so divide 27 by 3 = 9. The second term would be 26 + 9 = 35. The third term would be 35 + 9 = 44 (B). USING ARITHMETIC SEQUENCE TO FIND CONSECUTIVE INTEGERS Example: What two numbers should be placed in the blanks below so that the difference between successive entries is the same? 26, , _, 53 Solution: In the arithmetic sequence, you can think of the terms as 26, 26 + s, 26 + s + s, and 26 + s + s + s. In the example, s represents the difference between successive terms. The final term is 53, so set up an algebraic equation: 26 + s + s + s = 53. Solve the equation for s, by first combining like terms: 26 + 3s = 53. Subtract 26 from both sides to get 3s = 27. Divide both sides by 3 to find that s, the difference between terms is 9. Therefore, the terms are 26, 26 + 9, 26 + 9 + 9, and 53 or 26, 35, 44, and 53 (B) Sample Questions: 7. For the difference between consecutive numbers to be the same, which 3 numbers must be placed in the blanks below? 11, , , _, 47 8. What two numbers should be placed in the blanks below so that the difference between the consecutive numbers is the same? 13, , ___, 34 4 Math 3 P&P Part 2 January 12, 2018 SUMMING AN ARITHMETIC SERIES USING A FORMULA To sum up the terms of this arithmetic sequence: a + (a+d) + (a+2d) + (a+3d) + ... ∑(𝑎+ 𝑘𝑑) = 𝑛 2 (2𝑎+ (𝑛−1)𝑑) 𝑛−1 𝑘=0 a is the first term d is the "common difference" between terms n is the number of terms to add up ∑ (called Sigma)means "sum up" And below and above it are shown the starting and ending values: It says "Sum up n where n goes from 1 to 4. Answer=10 Example: Add up the first 10 terms of the arithmetic sequence: { 1, 4, 7, 10, 13, ... } Solution: The values of a, d and n are: a = 1 (the first term) d = 3 (the "common difference" between terms) n = 10 (how many terms to add up) So: ∑ (𝑎+ 𝑘𝑑) = 𝑛 2 (2𝑎+ (𝑛−1)𝑑) 𝑛−1 𝑘=0 Becomes: = 5(2+9·3) = 5(29) = 145 Check: Add up the terms yourself using a calculator, and see if it comes to 145 Example: Cynthia decorates the ceiling of her bedroom with stars that glow in the dark. She puts 1 star on the ceiling on the first day of decorating, 2 stars on the ceiling on the 2nd day of decorating, 3 stars on the 3rd day, and so on. If she puts stars on the ceiling in this pattern for 30 days (so she puts 30 stars on the ceiling on the 30th day), then what will be the total number of stars on the ceiling at the end of 30 days? Solution: You can use your calculator to add 1 + 2 + 3 + 4 +. . .+ 30 = 465. Or if you identified this as an arithmetic series, you can use the equation: ∑(𝑎+ 𝑘𝑑) = 𝑛 2 (2𝑎+ (𝑛−1)𝑑) 𝑛−1 𝑘=0 Where a = 1, d = 1, n = 30 Becomes: ∑ (1 + 𝑘1) = 30 2 [2(1) + (30 −1)(1)] 𝑛−1 𝑘=0 = 15 x 31 = 465 The simplified equation for finding the sum of arithmetic sequences is: S = (a1 + an) x 𝑛 2 a1 is the first term an is the last term n is the number of terms to add up 5 Math 3 P&P Part 2 January 12, 2018 Example: Cynthia decorates the ceiling of her bedroom with stars that glow in the dark. She puts 1 star on the ceiling on the first day of decorating, 2 stars on the ceiling on the 2nd day of decorating, 3 stars on the 3rd day, and so on. If she puts stars on the ceiling in this pattern for 30 days (so she puts 30 stars on the ceiling on the 30th day), then what will be the total number of stars on the ceiling at the end of 30 days? Solution: This time, use the simplified formula S = (a1 + an) x 𝑛 2, where n is the number of terms (30), 1a is the first term (1), an is the nth term (30). (1 + 30) x 30 2 = 465. Sample Questions: 9. A theater has 50 rows of seats. There are 18 seats in the first row, 20 seats in the second, 22 in the third and so on. How many seats are in the theater? 10. The eleventh term of an arithmetic sequence is 30 and the sum of the first eleven terms is 55. What is the common difference? 11. On the first day of school, Mr. Vilani gave his third-grade students 5 new words to spell. On each day of school after that, he gave the students 3 new words to spell. In the first 20 days of school, how many new words had he given the students to spell? 12. How many terms of the arithmetic sequence 2, 8, 14, 20, ..are required to give a sum of 660? SUMMING AN ARITHMETIC SERIES WITH SLOT FORMULA OR CONSECUTIVE INTEGERS To solve via slot method, draw appropriate number of slots and figure out how the numbers in the slots are related. Figure out a formula or other solution to the problem and solve. Consecutive integers can be written as n, n+1, n+2, etc. Consecutive odd or even integers can be written as n, n+2, n+4, etc. Example: The average of 7 consecutive numbers is 16. What is the sum of the least and greatest of the 7 integers? Solution: n + (n + 1) + (n + 2) + (n + 3) + (n + 4) + (n + 5) + (n + 6) = 16 or 7𝑛+21 7 = 16 Multiply both sides by 7 and subtract 21 from each side. 7n = 91 Divide both sides by 7 and get n = 13. So, the first term is 13. The series is 13, 14, 15, 16, 17, 18, 19. The sum of 13 + 19 = 32 Sample Questions: 13. The 6 consecutive integers below add up to 513. n – 2 n – 1 n n + 1 n + 2 n + 3 What is the value of n? 14. What is the smallest possible value for the product of 2 real numbers that differ by 6? 15. The product of two consecutive integers is between 137 and 159. Which of the following CAN be one of the integers? a. 15 b. 13 c. 11 d. 10 e. 7 6 Math 3 P&P Part 2 January 12, 2018 SUMMING A GEOMETRIC SERIES When we need to sum a Geometric Sequence, there is a handy formula. To sum: a + ar + ar2 + ... + arn-1 Each term is ark, where k starts at 0 and goes up to n-1 Use this formula: a is the first term r is the "common ratio" between terms n is the number of terms Example: Sum the first 4 terms of 10, 30, 90, 270, 810, 2430, . . . This sequence has a factor of 3 between each number. The values of a, r and n are: a = 10 (the first term) r = 3 (the "common ratio") n = 4 (we want to sum the first 4 terms) So: Becomes: And, yes, it is easier to just add them in this example, as there are only 4 terms. But imagine adding 50 terms ... then the formula is much easier. The simplified equation for finding the sum of geometric sequences looks very similar. If Sn represents the sum of the first n terms of a geometric series, then Sn = a + ar + ar2 + ar3 + ar4 + . . . .+ arn-1 Where a = the first term, r = the common ratio, and the nth term is not arn, but arn-1 The sum of the first n terms is given by Sn where Sn = 𝑎 (1−𝑟𝑛) 1−𝑟 The sum to infinity, written as 𝑆∞ is given by 𝑆∞ = 𝑎 1−𝑟 only if -1 < r < 1 Example: The sum of an infinite geometric sequence series with first term x and common ratio y < 1 is given by 𝑥 (1−𝑦). The sum of a given infinite geometric series is 200, and the common ratio is 0.15. What is the second term of this series? Solution: 𝑆∞ = 𝑎 1−𝑟 200 = 𝑥 1−0.15 = 170 170 0.15 = 25.5 Sample Questions: 16. Find the sum of each of the geometric series 17. 18. Find the sum of the first ten terms of the geometric sequence: an=132(−2)n−1. 19. If the sum of the first seven terms in a geometric series is 2158 and r = −12, find the first term. 7 Math 3 P&P Part 2 January 12, 2018 TYPES OF DATA AND TABLES Variables that take on values that are names or labels are considered categorical data. These are data that cannot be averaged or represented by a scatter plot as they have no numerical meaning. Variables that represent a measurable quantity are numerical or quantitative data. Example: Answer the question. If it can’t be answered, explain why. The temperature at the park over a 12-hour period was: 60, 64, 66, 71, 75, 77, 78, 80, 78, 77, 73, and 65. Find the average temperature over the 12-hour period. Solution: The data is quantitative so it is possible to find the average of 72 Example: Once a week, Maria has to fill her car up with gas. She records the day of the week each time she fills her car for three months. Monday, Friday, Tuesday, Thursday, Monday, Wednesday, Monday, Tuesday, Wednesday, Tuesday, and Thursday. Find the average day of the week that Maria had to fill her car with gas. Solution: The data is categorical so the average cannot be calculated. However, it is clear that Monday was the day that she most often filled her car. Sample Questions: 20. A group of students has ages of 12, 15, 11, 14, 12, 11, 15, 13, 12, 12, 11, 14, and 13. Find the average age of the students. 21. Suppose 11-year-olds are in 6th grade, 12-year-olds are in 7th grade, 13-year-olds are in 8th grade, 14-year olds are in 9th grade, and 15-year-olds are in 10th grade. A group of students has ages of 12, 15, 11, 14, 12, 11, 15, 13, 12, 12, 11, 14, and 13. Find the average grade of the students. TWO-WAY FREQUENCE TABLE A two-way frequency table is a useful tool for examining the relationships between categorical variables. The entries in the cells of a two-way table can be frequency counts or relative frequencies. The table summarizes and shows how often a value occurs. Generally, conditional relative frequencies are written as a decimal or percentage. It is the ratio of the observed number of a particular event to the total number of events, often taken as an estimate of probability. You can use the relative frequency to determine how often a value may occur in the future. In a two-way table, entries in the “total” row and “total” columns are called marginal frequencies. Entries in the body of the table are called joint frequencies. The two-way frequency table below shows the favorite leisure activities for 20 men and 30 women using frequency counts. The two-way frequency table below shows the same information represented as conditional relative frequencies. 8 Math 3 P&P Part 2 January 12, 2018 Two-way tables can show relative frequencies for the whole table, for rows, or for columns. Each type of relative frequency table makes a different contribution to understanding the relationship between gender and preferences for leisure activity. For example, the “Relative Frequency of Rows” table most clearly shows the probability that each gender will prefer a particular leisure activity. For instance, it is easy to see that the probability that a man will prefer movies is 40% while it is 27% for women. Example: A public opinion survey explored the relationship between age and support for increasing the minimum wage. The results are found in the following two-way frequency table. In the 41 to 60 age group, what percentage supports increasing the minimum wage? Answer: A total of 75 people in the 41 to 60 age group were surveyed. Of those, 30 were for increasing the minimum wage. Thus, 40% supported increasing the minimum wage. Example: Create two-way frequency tables to organize the information from the situations described. A bank teller splits transactions into two categories: deposits and withdrawals. In one shift, the bank teller has 72 transactions. Of those, 12 males make deposits and 30 make withdrawals. While 20 females make deposits. Solution: First, draw an empty frequency table and add the appropriate labels. Deposits Withdrawals Total Men Women Total Then add the numbers that are listed in the question to the table. Deposits Withdrawals Total Men 12 30 Women 20 Total 72 Finally, calculate the missing numbers and add them to the table. Verify that all columns and rows add up correctly. Total Deposits: 12 + 20 = 32. If the total transactions is 72, there must be 72 – 32 = 40 withdrawals. To get a total of 40 withdrawals, there must be 40 – 30 = 10 by women. That makes a total of 20 + 10 = 30 transactions by women. Total transactions by men = 12 + 30 = 42. If you add the total transactions by men and women, you get 42 + 30 = 72. Which is verified by adding the total deposits and withdrawals 32 + 40 =72 Deposits Withdrawals Total Men 12 30 42 Women 20 10 30 Total 32 40 72 9 Math 3 P&P Part 2 January 12, 2018 Sample Questions: 22. Complete the two-way frequency table and answer the questions about the information in the table Hollywood Junior students were interviewed about their eating habits. Male Female Total Eat Breakfast Regularly 110 300 Don’t Eat Breakfast Regularly 130 165 Total 275 a. Of the total males, what percentage do not eat breakfast regularly? b. Of the total people who eat breakfast regularly, what percentage of them are males? c. How many females eat breakfast regularly? d. How many females were included in the survey? e. How many females eat breakfast out of the total number of females? f. How many people were included in this survey? g. How many males do not eat breakfast regularly? h. How many males and females do not eat breakfast regularly? i. Which group of people eat breakfast more regularly? 23. Complete the two-way frequency table and answer the questions about the information in the table. Jersey High students were surveyed about the type of transportation they use the most often. Male Female Total Walk 46 Car 28 45 Bus 12 27 Cycle 17 69 Total 129 92 a. Of the total males, what percentage cycle? b. Of the total people who use the bus, what percentage are female? c. How many more females than males walk? d. What is the percent of males that walk compared to ride something? e. How many females ride in a car? f. How many total people surveyed? G. How many total males cycle? 10 Math 3 P&P Part 2 January 12, 2018 VENN DIAGRAMS An event is an activity or experiment which is usually represented by a capital letter. A sample space is a set of all possible outcomes for an activity or experiment. A smaller set of outcomes from the sample space is called a subset. The complement of a subset is all outcomes in the sample space that are not part of the subset. A subset and its complement make up the entire sample space. If a subset is represented by A, the complement can be represented by any of the following: not A, ~A, or Ac. 11 Math 3 P&P Part 2 January 12, 2018 Sample Questions: 24. Given all the letters of the English Alphabet a. Draw a Venn Diagram of the following: i. List a subset of the letters in your first name. ii. List a subset of the letters in your last name. b. Find the union of the subsets of your first name and last name. c. Find the intersection of the subsets of your first name and last name. 25. Given the sample space S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} with event A = {3, 4, 5, 6, 7} and event B = {1, 2, 3, 4, 5}. a. Draw a Venn diagram representing the sample space with events A and B. b. List all the outcomes for A∪B. c. List all the outcomes for A∩B. 26. Given a standard deck of 52 cards, event A is defined as a red card and event B is defined as the card is a diamond. a. List all the outcomes for A∪B. b. List all the outcomes for A∩B. c. What is A ∩ Ac? (Ac is the compliment or opposite data of A) 12 Math 3 P&P Part 2 January 12, 2018 Answer Key 1. C 2. 5 numbers 3. 14.5 4. 729 5. ab642 6. D 7. 20, 29, 38 8. 20, 27 9. 3,350 10. Common Difference = 5 11. 62 words 12. 15 terms 13. 85 14. -9 15. 13 16. . 13 Math 3 P&P Part 2 January 12, 2018 17. S∞ = ½ 18. -341 32 19. . 20. The data is quantitative so it is possible to find the average age of 12.69. 21. Even though the grade level is a number the data is categorical so the average grade level cannot be found. However, is clear that 7th grade has the highest number of students. 22. Male Female Total Eat Breakfast Regularly 190 110 300 Don’t Eat Breakfast Regularly 130 165 295 Total 320 275 595 a. 41% b. 63% c. 110 d. 300 e. 110 f. 595 g. 130 h. 320 i. male 23. Male Female Total Walk 46 Car 28 45 Bus 12 27 Cycle 17 69 Total 129 92 a. 40% b. 44% c. 12 d. 36% e. 17 f. 221 g. 52 14 Math 3 P&P Part 2 January 12, 2018 24. Example: Jennifer Wright a. A B C D K L M O P Q S U V X Y Z b. Union: E, F, G, H, I, J, N, R, T, W c. Intersection: R, I 25. d. 0 8 9 e. {1, 2, 3, 4, 5, 6, 7} f. {3, 4, 5} g. {0, 1, 2, 8, 9} J E N F W G H T R I 6 7 1 2 3 4 5 15 Math 3 P&P Part 2 January 12, 2018 26. A = ace K = king Q = queen J = jack H = heart D = diamond S = spade C = club AS, 2S, 3S, 4S, 5S, 6S, 7S, 8S, 9S, JS, QS, KS, AC, 2C, 3C, 4D, 5C, 6C, 7C, 8C, 9C, JC, QC, KC a. AH, 2H, 3H, 4H, 5H, 6H, 7H, 8H, 9H, JH, QH, KH, AD, 2D, 3D, 4D, 5D, 6D, 7D, 8D, 9D, JD, QD, KD b. AD, 2D, 3D, 4D, 5D, 6D, 7D, 8D, 9D, JD, QD, KD c. AD, 2D, 3D, 4D, 5D, 6D, 7D, 8D, 9D, JD, QD, KD, AS, 2S, 3S, 4S, 5S, 6S, 7S, 8S, 9S, JS, QS, KS, AC, 2C, 3C, 4D, 5C, 6C, 7C, 8C, 9C, JC, QC, KC d. NONE AH, 2H, 3H, 4H, 5H, 6H, 7H, 8H, 9H, JH, QH, KH AD, 2D, 3D, 4D, 5D, 6D, 7D, 8D, 9D, JD, QD, KD
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Please enter a valid web address About Blog Projects Help Donate Contact Jobs Volunteer People Sign up for free Log in Search metadata Search text contents Search TV news captions Search radio transcripts Search archived web sites Advanced Search About Blog Projects Help Donate Contact Jobs Volunteer People Full text of "Frank White Fluid Mechanics Mc Graw Hill Education ( 2015)" See other formats Me Graw Hill Education Fluid Mechanics Fluid Mechanics Eighth Edition Frank M. White University of Rhode Island Me Graw Hill Education Me Graw Hill Education FLUID MECHANICS, EIGHTH EDITION Published by McGraw-Hill Education, 2 Penn Plaza, New York, NY 10121. Copyright © 2016 by McGraw-Hill Education. All rights reserved. Printed in the United States of America. Previous editions © 2011, 2008, and 2003. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of McGraw-Hill Education, including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1234567890 DOC/DOC 1 0 9 8 7 6 5 ISBN 978-0-07-339827-3 MHID 0-07-339827-6 Senior Vice President, Products & Markets: Kurt L. Strand Vice President, General Manager, Products & Markets: Marty Lange Vice President, Content Design & Delivery: Kimberly Meriwether David Managing Director: Thomas Timp Brand Manager: Thomas Scaife Director, Product Development: Rose Koos Product Developer: Lorraine Buezek Marketing Manager: Nick McFadden Director, Content Design & Delivery: Litida Avenarius Program Manager: Lora Neyens Content Project Manager: Lisa Bruflodt Content Project Manager: Tammy Juran Buyer: Susan K. Culbertson Content Licensing Specialists: Deanna Dausener Cover Image: Doug Sherman/Geofile Compositor: Aptara®, Inc. Printer: R. R. Donnelley All credits appearing on page or at the end of the book are considered to be an extension of the copyright page. Library of Congress Cataloging-in-Publication Data White, Frank M. Fluid mechanics/Frank M. White, University of Rhode Island. — Eighth edition. pages cm Includes index. ISBN 978-0-07-339827-3 (alk. paper)— ISBN 0-07-339827-6 (alk. paper) Fluid mechanics. 1. Title. TA357.W48 2016 620.r06— dc23 2014034259 The Internet addresses listed in the text were accurate at the time of publication. The inclusion of a website does not indicate an endorsement by the authors or McGraw-Hill Education, and McGraw-Hill Education does not guarantee the accuracy of the information presented at these sites. www.mhhe.com About the Author Frank M. White is Professor Emeritus of Mechanical and Ocean Engineering at the University of Rhode Island. He studied at Georgia Tech and M.I.T. In 1966 he helped found, at URI, the first department of ocean engineering in the country. Known primarily as a teacher and writer, he has received eight teaching awards and has written four textbooks on fluid mechanics and heat transfer. From 1979 to 1990, he was editor-in-chief of the ASME Journal of Fluids Engineering and then served from 1991 to 1997 as chairman of the ASME Board of Editors and of the Publications Committee. He is a Fellow of ASME and in 1991 received the ASME Fluids Engineering Award. He lives with his wife, Jeanne, in Narragansett, Rhode Island. v To Jeanne Contents Preface xi Chapter 1 Introduction 3 1.1 Preliminary Remarks 3 1.2 The Concept of a Fluid 4 1.3 The Fluid as a Continuum 6 1.4 Dimensions and Units 7 1.5 Properties of the Velocity Field 15 1.6 Thermodynamic Properties of a Fluid 15 1.7 Viscosity and Other Secondary Properties 23 1.8 Basic Flow Analysis Techniques 39 1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines 39 1.10 The Fundamentals of Engineering (FE) Examination 43 1.11 The History of Eluid Mechanics 43 Summary 44 Problems 45 Eundamentals of Engineering Exam Problems 52 Comprehensive Problems 53 References 56 Chapter 2 Pressure Distrihution in a Fluid 59 2.1 Pressure and Pressure Gradient 59 2.2 Equilibrium of a Eluid Element 61 2.3 Hydrostatic Pressure Distributions 62 2.4 Application to Manometry 69 2.5 Hydrostatic Eorces on Plane Surfaces 72 2.6 Hydrostatic Eorces on Curved Surfaces 80 2.7 Hydrostatic Eorces in Layered Fluids 83 2.8 Buoyancy and Stability 85 2.9 Pressure Distribution in Rigid-Body Motion 91 2.10 Pressure Measurement 99 Summary 103 Problems 103 Word Problems 126 Fundamentals of Engineering Exam Problems 126 Comprehensive Problems 127 Design Projects 129 References 130 Chapter 3 Integral Relations for a Control Volume 133 3.1 Basic Physical Laws of Eluid Mechanics 133 3.2 The Reynolds Transport Theorem 137 3.3 Conservation of Mass 144 3.4 The Linear Momentum Equation 149 3.5 Erictionless Flow: The Bernoulli Equation 163 3.6 The Angular Momentum Theorem 172 3.7 The Energy Equation 178 Summary 189 Problems 189 W ord Problems 216 Eundamentals of Engineering Exam Problems 217 Comprehensive Problems 218 Design Project 219 References 219 Chapter 4 Differential Relations for Fluid Flow 221 4.1 The Acceleration Field of a Fluid 222 4.2 The Differential Equation of Mass Conservation 224 viii Contents 4.3 The Differential Equation of Linear Momentum 230 4.4 The Differential Equation of Angular Momentum 237 4.5 The Differential Equation of Energy 238 4.6 Boundary Conditions for the Basic Equations 241 4.7 The Stream Function 246 4.8 Vorticity and Irrotationality 253 4.9 Frictionless Irrotational Flows 255 4.10 Some Illustrative Incompressible Viscous Flows 261 Summary 269 Problems 269 Word Problems 280 Fundamentals of Engineering Exam Problems 28 1 Comprehensive Problems 28 1 References 282 Chapter 5 Dimensional Analysis and Similarity 285 5.1 Introduction 285 5.2 The Principle of Dimensional Homogeneity 288 5.3 The Pi Theorem 294 5.4 Nondimensionalization of the Basic Equations 304 5.5 Modeling and Similarity 313 Summary 325 Problems 325 Word Problems 333 Fundamentals of Engineering Exam Problems 334 Comprehensive Problems 334 Design Projects 335 References 336 Chapter 6 Viscous Flow in Ducts 339 6.1 Reynolds Number Regimes 339 6.2 Internal versus External Viscous Flows 344 6.3 Head Loss — The Friction Factor 347 6.4 Laminar Fully Developed Pipe Flow 349 6.5 Turbulence Modeling 35 1 6.6 Turbulent Pipe Flow 358 6.7 Four Types of Pipe Flow Problems 366 6.8 Flow in Noncircular Ducts 371 6.9 Minor or Local Losses in Pipe Systems 380 6.10 Multiple-Pipe Systems 389 6.11 Experimental Duct Flows: Diffuser Performance 395 6.12 Fluid Meters 400 Summary 421 Problems 422 Word Problems 440 Fundamentals of Engineering Exam Problems 441 Comprehensive Problems 442 Design Projects 444 References 444 Chapter 7 Flow Past Immersed Bodies 449 7.1 Reynolds Number and Geometry Effects 449 7.2 Momentum Integral Estimates 453 7.3 The Boundary Layer Equations 456 7.4 The Flat-Plate Boundary Layer 459 7.5 Boundary Layers with Pressure Gradient 468 7.6 Experimental External Flows 474 Summary 501 Problems 502 Word Problems 515 Fundamentals of Engineering Exam Problems 515 Comprehensive Problems 516 Design Project 517 References 517 Chapter 8 Potential Flow and Compntational Fluid Dynamics 521 8.1 Introduction and Review 521 8.2 Elementary Plane Flow Solutions 524 8.3 Superposition of Plane Flow Solutions 531 8.4 Plane Flow Past Closed-Body Shapes 537 8.5 Other Plane Potential Flows 547 8.6 Images 551 8.7 Airfoil Theory 554 8.8 Axisymmetric Potential Flow 562 8.9 Numerical Analysis 568 Summary 577 Problems 577 W ord Problems 588 Contents ix Comprehensive Problems 588 Design Projects 589 References 590 Chapter 9 Compressible Flow 593 9.1 Introduction: Review of Thermodynamics 593 9.2 The Speed of Sound 598 9.3 Adiabatic and Isentropic Steady Flow 600 9.4 Isentropic Flow with Area Changes 606 9.5 The Normal Shock Wave 613 9.6 Operation of Converging and Diverging Nozzles 621 9.7 Compressible Duct Flow with Friction 626 9.8 Frictionless Duct Flow with Heat Transfer 637 9.9 Mach Waves and Oblique Shock Waves 642 9.10 Prandtl-Meyer Expansion Waves 652 Summary 664 Problems 665 Word Problems 678 Fundamentals of Engineering Exam Problems 678 Comprehensive Problems 679 Design Projects 680 References 68 1 Chapter 10 Open-Channel Flow 683 10.1 Introduction 683 10.2 Uniform Flow; The Chezy Formula 689 10.3 Efficient Uniform-Flow Channels 695 10.4 Specific Energy; Critical Depth 697 10.5 The Hydraulic Jump 704 10.6 Gradually Varied Flow 708 10.7 Flow Measurement and Control by Weirs 716 Summary 723 Problems 724 Word Problems 736 Fundamentals of Engineering Exam Problems 736 Comprehensive Problems 736 Design Projects 738 References 738 Chapter 11 Turbomachinery 741 11.1 Introduction and Classification 741 11.2 The Centrifugal Pump 744 11.3 Pump Performance Curves and Similarity Rules 750 11.4 Mixed- and Axial-Flow Pumps: The Specific Speed 760 11.5 Matching Pumps to System Characteristics 767 11.6 Turbines 775 Summary 789 Problems 791 Word Problems 804 Comprehensive Problems 804 Design Project 806 References 806 Appendix A Physical Properties of Fluids 808 Appendix B Compressible Flow Tables 813 Appendix C Conversion Factors 820 Appendix D Equations of Motion in Cylindrical Coordinates 822 Appendix E Estimating Uncertainty in Experimental Data 824 Answers to Selected Problems 826 Index 833 General Approach Learning Tools Content Changes Preface The eighth edition of Fluid Mechanics sees some additions and deletions but no philosophical change. The basic outline of eleven chapters, plus appendices, remains the same. The triad of integral, differential, and experimental approaches is retained. Many problem exercises, and some fully worked examples, have been changed. The informal, student-oriented style is retained. A number of new photographs and figures have been added. Many new references have been added, for a total of 445. The writer is a firm believer in “further reading,” especially in the postgraduate years. The total number of problem exercises continues to increase, from 1089 in the first edition, to 1683 in this eighth edition. There are approximately 20 new problems in each chapter. Most of these are basic end-of-chapter problems, classified according to topic. There are also Word Problems, multiple-choice Fundamentals of Engineering Problems, Comprehensive Problems, and Design Projects. The appendix lists approx¬ imately 700 Answers to Selected Problems. The example problems are structured in the text to follow the sequence of recom¬ mended steps outlined in Section 1.7. Most of the problems in this text can be solved with a hand calculator. Some can even be simply explained in words. A few problems, especially in Chapters 6, 9, and 10, involve solving complicated algebraic expressions, laborious for a hand calculator. Check to see if your institution has a license for equation-solving software. Here the writer solves complicated example problems by using the iterative power of Microsoft Office Excel, as illustrated, for example, in Example 6.5. Eor further use in your work. Excel also contains several hundred special mathematical functions for engineering and statistics. Another benefit: Excel is free. There are some revisions in each chapter. Chapter 1 has been substantially revised. The pre-reviewers felt, correctly, that it was too long, too detailed, and at too high a level for an introduction. Eormer Section 1 .2, History of Eluid Mechanics, has been shortened and moved to the end of the chapter. Former Section 1.3, Problem-Solving Techniques, has been moved to appear just before Example 1.7, where these techniques are first used. Eulerian and Lagrangian descriptions have been moved to Chapter 4. A temperature-entropy chart for steam XI xii Preface has been added, to illustrate when steam can and cannot be approximated as an ideal gas. Former Section 1.11, Flow Patterns, has been cut sharply and mostly moved to Chapter 4. Former Section 1.13, Uncertainty in Experimental Data, has been moved to a new Appendix E. No one teaches “uncertainty” in introductory fluid mechanics, but the writer feels it is extremely important in all engineering fields involving exper¬ imental or numerical data. Chapter 2 adds a brief discussion of the fact that pressure is a thermodynamic property, not a force, has no direction, and is not a vector. The arrow, on a surface force caused by pressure, causes confusion for beginning students. The subsection of Section 2.8 entitled Stability Related to Waterline Area has been shortened to omit the complicated derivations. The final metacenter formula is retained; the writer does not think it is sufficient just to show a sketch of a floating body falling over. This book should have reference value. Chapter 3 was substantially revised in the last edition, especially by moving Bernoulli’s equation to follow the linear momentum section. This time the only changes are improvements in the example problems. Chapter 4 now discusses the Eulerian and Lagrangian systems, moved from Chapter 1. The no-slip and no-temperature-jump boundary conditions are added, with problem assignments. Chapter 5 explains a bit more about drag force before assigning dimensional analysis problems. It retains Ipsen’s method as an interesting alternative which, of course, may be skipped by pi theorem adherents. Chapter 6 downplays the Moody chart a bit, suggesting that students use either iteration or Excel. For rough walls, the chart is awkward to read, although it gives an approximation for use in iteration. The author’s fancy rearrangement of pi groups to solve type 2, flow rate, and type 3, pipe diameter problems is removed from the main text and assigned as problems. For noncircular ducts, the hydraulic radius is omitted and moved to Chapter 10. There is a new Example 6.11, which solves for pipe diameter and determines if Schedule 40 pipe is strong enough. A general discussion of pipe strength is added. There is a new subsection on laminar-flow minor losses, appropriate for micro- and nano-tube flows. Chapter 7 has more treatment of vehicle drag and rolling resistance, and a rolling resistance coefficient is defined. There is additional discussion of the Kline-Fogelman airfoil, extremely popular now for model aircraft. Chapter 8 has backed off from extensive discussion of CFD methods, as proposed by the pre-reviewers. Only a few CFD examples are now given. The inviscid duct-expansion example and the implicit boundary layer method are now omitted, but the explicit method is retained. For airfoil theory, the writer considers thin-airfoil vortex-sheet theory to be obsolete and has deleted it. Chapter 9 now has a better discussion of the normal shock wave. New supersonic wave photographs are added. The “new trend in aeronautics” is the Air Force X-35 Joint Strike Fighter. Chapter 10 improves the definition of normal depth of a channel. There is a new subsection on the water-channel compressible flow analogy, and problems are assigned to And the oblique wave angle for supercritical water flow past a wedge. Chapter 11 greatly expands the discussion of wind turbines, with examples and problems taken from the author’s own experience. Preface xiii Adaptive Online Learning Tools ■ LEARNSMART' B SMARTBDDK® ■oormect [engineering Online Supplements Appendices B and D are unchanged. Appendix A adds a list of liquid kinematic viscosities to Table A. 4. A few more conversion factors are added to Appendix C. There is a new Appendix E, Estimating Uncertainty in Experimental Data, which was moved from its inappropriate position in Chapter 1. The writer believes that “uncer¬ tainty” is vital to reporting measurements and always insisted upon it when he was an engineering journal editor. McGraw-Hill LearnSmart® is available as a standalone product or an integrated feature of McGraw-Hill Connect Engineering. It is an adaptive learning system designed to help students learn faster, study more efficiently, and retain more knowl¬ edge for greater success. LearnSmart assesses a student’s knowledge of course content through a series of adaptive questions. It pinpoints concepts the student does not understand and maps out a personalized study plan for success. This innovative study tool also has features that allow instructors to see exactly what students have accom¬ plished and a built-in assessment tool for graded assignments. Visit the following site for a demonstration: www.LearnSmartAdvantage.com Powered by the intelligent and adaptive LearnSmart engine, SmartBook^'^ is the first and only continuously adaptive reading experience available today. Distin¬ guishing what students know from what they don’t, and honing in on concepts they are most likely to forget, SmartBook personalizes the reading experience for each student. Reading is no longer a passive and linear experience but an engaging and dynamic one, where students are more likely to master and retain important concepts, coming to class better prepared. SmartBook includes powerful reports that identify specific topics and learning objectives students need to study, www. LearnSmartAdvantage.com McGraw-Hill’s Connect Engineering offers a number of powerful tools and features to make managing assignments easier, so you can spend more time teaching. Students engage with their coursework anytime from anywhere in a personalized way, making the learning process more accessible and efficient. Connect Engineering optimizes your time and energy, enabling you to focus on course content and learning outcomes, teaching, and student learning. A number of supplements are available to instructors at McGraw-Hill’s Connect Engi¬ neering®. Instructors may obtain the text images in PowerPoint format and the full Solutions Manual in PDF format. The solutions manual provides complete and detailed solutions, including problem statements and artwork, to the end-of-chapter problems. Instructors can also obtain access to the Complete Online Solutions Manual Organiza¬ tion System (C.O.S.M.O.S.) for Fluid Mechanics, 8th edition. Instructors can use C.O.S.M.O.S. to create exams and assignments, to create custom content, and to edit supplied problems and solutions. Acknowledgments As usual, so many people have helped me that I may fail to list them all. Material help, in the form of photos, articles, and problems, came from Scott Larwood of the Univer¬ sity of the Pacific; Sukanta Dash of the Indian Institute of Technology at Kharagpur; Mark Coffey of the Colorado School of Mines; Mac Stevens of Oregon State University; Stephen Carrington of Malvern Instruments; Carla Cioffi of NASA; Lisa Lee and Robert Pacquette of the Rhode Island Department of Environmental Management; Vanessa Blakeley and Samuel Schweighart of Terrafugia Inc.; Beric Skews of the University of the Witwatersrand, South Africa; Kelly Irene Knorr and John Merrill of the School of Oceanography at the University of Rhode Island; Adam Rein of Altaeros Energies Inc.; Dasari Abhinav of Anna University, India; Kris Allen of Transcanada Corporation; Bruce Findlayson of the University of Washington; Wendy Koch of USA Today; Liz Boardman of the South County Independent; Beth Darchi and Colin McAteer of the American Society of Mechanical Engineers; Catherine Hines of the William Beebe Web Site; Laura Garrison of York College of Pennsylvania. The following pre-reviewers gave many excellent suggestions for improving the manuscript: Steve Baker, Naval Postgraduate School; Suresh Aggarwal, University of Illinois at Chicago; Edgar Caraballo, Miami University; Chang-Hwan Choi, Stevens Institute of Technology; Drazen Fabris, Santa Clara University; James Liburdy, Oregon State University; Daniel Maynes, Brigham Young University; Santosh Sahu, Indian Institute of Technology Indore; Brian Savilonis, Worcester Polytechnic Institute; Eric Savory, University of Western Ontario; Rick Sellens, Queen’s University; Gordon Stubley, University of Waterloo. Many others have supported me, throughout my revision efforts, with comments and suggestions: Gordon Holloway of the University of New Brunswick; David Taggart, Donna Meyer, Arun Shukla, and Richard Lessmann of the University of Rhode Island; Debendra K. Das of the University of Alaska-Fairbanks; Elizabeth Kenyon of Math- works; Deborah V. Pence of Oregon State University; Sheldon Green of the University of British Columbia; Elena Bingham of the DuPont Corporation; Jane Bates of Broad Rock School; Kim Mather of West Kingston School; Nancy Dreier of Curtiss Comer School; Richard Kline, co-inventor of the Kline-Fogelman airfoil. The McGraw-Hill staff was, as usual, very helpful. Thanks are due to Bill Stenquist, Katherine Neubauer, Lorraine Buczek, Samantha Donisi-Hamm, Raghu Srinivasan, Tammy Juran, Thomas Scaife, and Lisa Bruflodt. Finally, I am thankful for the continuing support of my family, especially Jeanne, who remains in my heart, and my sister Sally White GNSH, my dog Jack, and my cats Cole and Kerry. XIV Fluid Mechanics Falls on the Nesowadnehunk Stream in Baxter State Park, Maine, which is the northern terminus of the Appalachian Trail. Such flows, open to the atmosphere, are driven simply by gravity and do not depend much upon fluid properties such as density and viscosity. They are discussed later in Chap. 10. To the writer, one of the joys of fluid mechanics is that visualization of a fluid-flow process is simple and beautiful [Photo Credit: Design Pics/Natural Selection Robert Cable]. 1.1 Preliminary Remarks Chapter 1 Introduction Fluid mechanics is the study of fluids either in motion (fluid dynamics) or at rest (fluid statics). Both gases and liquids are classified as fluids, and the number of fluid engi¬ neering applications is enormous: breathing, blood flow, swimming, pumps, fans, turbines, airplanes, ships, rivers, windmills, pipes, missiles, icebergs, engines, filters, jets, and sprinklers, to name a few. When you think about it, almost everything on this planet either is a fluid or moves within or near a fluid. The essence of the subject of fluid flow is a judicious compromise between theory and experiment. Since fluid flow is a branch of mechanics, it satisfies a set of well- documented basic laws, and thus a great deal of theoretical treatment is available. However, the theory is often frustrating because it applies mainly to idealized situations, which may be invalid in practical problems. The two chief obstacles to a workable theory are geometry and viscosity. The basic equations of fluid motion (Chap. 4) are too difficult to enable the analyst to attack arbitrary geometric configurations. Thus most textbooks concentrate on flat plates, circular pipes, and other easy geometries. It is possible to apply numerical computer techniques to complex geometries, and specialized textbooks are now available to explain the new computational fluid dynamics (CFD) approximations and methods [1-4].^ This book will present many theoretical results while keeping their limitations in mind. The second obstacle to a workable theory is the action of viscosity, which can be neglected only in certain idealized flows (Chap. 8). First, viscosity increases the dif¬ ficulty of the basic equations, although the boundary-layer approximation found by Ludwig Prandtl in 1904 (Chap. 7) has greatly simplified viscous-flow analyses. Second, viscosity has a destabilizing effect on all fluids, giving rise, at frustratingly small velocities, to a disorderly, random phenomenon called turbulence. The theory of tur¬ bulent flow is crude and heavily backed up by experiment (Chap. 6), yet it can be quite serviceable as an engineering estimate. This textbook only introduces the standard experimental correlations for turbulent time-mean flow. Meanwhile, there are advanced texts on both time-mean turbulence and turbulence modeling [5, 6] and on the newer, computer-intensive direct numerical simulation (DNS) of fluctuating turbulence [7, 8]. ’Numbered references appear at the end of each chapter. 3 4 Chapter 1 Introduction 1.2 The Concept of a Fluid Thus there is theory available for fluid flow problems, but in all cases it should be backed up by experiment. Often the experimental data provide the main source of information about specific flows, such as the drag and lift of immersed bodies (Chap. 7). Fortunately, fluid mechanics is a highly visual subject, with good instru¬ mentation [9-11], and the use of dimensional analysis and modeling concepts (Chap. 5) is widespread. Thus experimentation provides a natural and easy complement to the theory. You should keep in mind that theory and experiment should go hand in hand in all studies of fluid mechanics. From the point of view of fluid mechanics, all matter consists of only two states, fluid and solid. The difference between the two is perfectly obvious to the layperson, and it is an interesting exercise to ask a layperson to put this difference into words. The technical distinction lies with the reaction of the two to an applied shear or tangential stress. A solid can resist a shear stress by a static deflection; a fluid cannot. Any shear stress applied to a fluid, no matter how small, will result in motion of that fluid. The fluid moves and deforms continuously as long as the shear stress is applied. As a corollary, we can say that a fluid at rest must be in a state of zero shear stress, a state often called the hydrostatic stress condition in structural analysis. In this condi¬ tion, Mohr’s circle for stress reduces to a point, and there is no shear stress on any plane cut through the element under stress. Given this definition of a fluid, every layperson also knows that there are two classes of fluids, liquids and gases. Again the distinction is a technical one concerning the effect of cohesive forces. A liquid, being composed of relatively close-packed molecules with strong cohesive forces, tends to retain its volume and will form a free surface in a gravitational field if unconfined from above. Free-surface flows are domi¬ nated by gravitational effects and are studied in Chaps. 5 and 10. Since gas molecules are widely spaced with negligible cohesive forces, a gas is free to expand until it encounters confining walls. A gas has no definite volume, and when left to itself without confinement, a gas forms an atmosphere that is essentially hydrostatic. The hydrostatic behavior of liquids and gases is taken up in Chap. 2. Gases cannot form a free surface, and thus gas flows are rarely concerned with gravitational effects other than buoyancy. Figure 1 . 1 illustrates a solid block resting on a rigid plane and stressed by its own weight. The solid sags into a static deflection, shown as a highly exaggerated dashed line, resisting shear without flow. A free-body diagram of element A on the side of the block shows that there is shear in the block along a plane cut at an angle 9 through A. Since the block sides are unsupported, element A has zero stress on the left and right sides and compression stress cr = —p on the top and bottom. Mohr’s circle does not reduce to a point, and there is nonzero shear stress in the block. By contrast, the liquid and gas at rest in Fig. 1.1 require the supporting walls in order to eliminate shear stress. The walls exert a compression stress of —p and reduce Mohr’s circle to a point with zero shear everywhere — that is, the hydrostatic condition. The liquid retains its volume and forms a free surface in the container. If the walls are removed, shear develops in the liquid and a big splash results. If the container is tilted, shear again develops, waves form, and the free surface seeks a horizontal configuration, pouring out over the lip if necessary. Meanwhile, the gas is unrestrained 1.2 The Concept of a Fluid 5 -(j = p Fig. 1.1 A solid at rest can resist shear, (a) Static deflection of the solid; {b) equilibrium and Mohr’s circle for solid element A. A fluid cannot resist shear, (c) Containing walls are needed; {d) equilibrium and Mohr’s circle for fluid element A. T Hydrostatic condition -P a (d) and expands out of the container, filling all available space. Element A in the gas is also hydrostatic and exerts a compression stress —p on the walls. In the previous discussion, clear decisions could be made about solids, liquids, and gases. Most engineering fluid mechanics problems deal with these clear cases — that is, the common liquids, such as water, oil, mercury, gasoline, and alcohol, and the common gases, such as air, helium, hydrogen, and steam, in their common tempera¬ ture and pressure ranges. There are many borderline cases, however, of which you should be aware. Some apparently “solid” substances such as asphalt and lead resist shear stress for short periods but actually deform slowly and exhibit definite fluid behavior over long periods. Other substances, notably colloid and slurry mixtures, resist small shear stresses but “yield” at large stress and begin to flow as fluids do. Specialized textbooks are devoted to this study of more general deformation and flow, a field called rheology . Also, liquids and gases can coexist in two-phase mixtures, such as steam-water mixtures or water with entrapped air bubbles. Specialized text¬ books present the analysis of such multiphase flows . Finally, in some situations the distinction between a liquid and a gas blurs. This is the case at temperatures and 6 Chapter 1 Introduction 1.3 The Fluid as a Continuum Fig. 1.2 The limit definition of continuum fluid density: {a) an elemental volume in a fluid region of variable continuum density; (fc) calculated density versus size of the elemental volume. pressures above the so-called critical point of a substance, where only a single phase exists, primarily resembling a gas. As pressure increases far above the critical point, the gashke substance becomes so dense that there is some resemblance to a liquid, and the usual thermodynamic approximations like the perfect-gas law become inaccurate. The critical temperature and pressure of water are = 647 K and = 219 atm (atmosphere)^ so that typical problems involving water and steam are below the critical point. Air, being a mixture of gases, has no distinct critical point, but its principal component, nitrogen, has = 126 K and p^ = 34 atm. Thus typical problems involving air are in the range of high temperature and low pressure where air is distinctly and definitely a gas. This text will be concerned solely with clearly identifiable liquids and gases, and the borderline cases just discussed will be beyond our scope. We have already used technical terms such as fluid pressure and density without a rigorous discussion of their definition. As far as we know, fluids are aggregations of molecules, widely spaced for a gas, closely spaced for a liquid. The distance between molecules is very large compared with the molecular diameter. The molecules are not fixed in a lattice but move about freely relative to each other. Thus fluid density, or mass per unit volume, has no precise meaning because the number of molecules occupying a given volume continually changes. This effect becomes unimportant if the unit volume is large compared with, say, the cube of the molecular spacing, when the number of molecules within the volume will remain nearly constant in spite of the enormous interchange of particles across the boundaries. If, however, the chosen unit volume is too large, there could be a noticeable variation in the bulk aggregation of the particles. This situation is illustrated in Fig. 1 .2, where the “density” as calculated from molecular mass 5m within a given volume 5Y is plotted versus the size of the unit volume. There is a limiting volume dY below which molecular variations may be important and above which aggregate variations may be important. The density p of a fluid is best defined as P = lim dm (1.1) Elemental («) (b) ^One atmosphere equals 2116 Ibf/ft^ = 101,300 Pa. 1.4 Dimensions and Units 7 1.4 Dimensions and Units Primary Dimensions The limiting volume tJT is about 10~® mm^ for all liquids and for gases at atmo¬ spheric pressure. For example, 10”^ mm^ of air at standard conditions contains approximately 3 X 10^ molecules, which is sufficient to define a nearly constant density according to Eq. (1.1). Most engineering problems are concerned with physical dimensions much larger than this limiting volume, so that density is essentially a point function and fluid properties can be thought of as varying continually in space, as sketched in Fig. 1.2a. Such a fluid is called a continuum, which simply means that its variation in properties is so smooth that differential calculus can be used to analyze the substance. We shall assume that continuum calculus is valid for all the analyses in this book. Again there are borderline cases for gases at such low pressures that molecular spacing and mean free path^ are comparable to, or larger than, the physical size of the system. This requires that the continuum approximation be dropped in favor of a molecular theory of rarefied gas flow . In principle, all fluid mechanics problems can be attacked from the molecular viewpoint, but no such attempt will be made here. Note that the use of continuum calculus does not preclude the possibility of discontinuous jumps in fluid properties across a free surface or fluid interface or across a shock wave in a compressible fluid (Chap. 9). Our calculus in analyzing fluid flow must be flexible enough to handle discontinuous boundary conditions. A dimension is the measure by which a physical variable is expressed quantitatively. A unit is a particular way of attaching a number to the quantitative dimension. Thus length is a dimension associated with such variables as distance, displacement, width, deflection, and height, while centimeters and inches are both numerical units for expressing length. Dimension is a powerful concept about which a splendid tool called dimensional analysis has been developed (Chap. 5), while units are the numerical quantity that the customer wants as the final answer. In 1872 an international meeting in France proposed a treaty called the Metric Convention, which was signed in 1875 by 17 countries including the United States. It was an improvement over British systems because its use of base 10 is the founda¬ tion of our number system, learned from childhood by all. Problems still remained because even the metric countries differed in their use of kiloponds instead of dynes or newtons, kilograms instead of grams, or calories instead of joules. To standardize the metric system, a General Conference of Weights and Measures, attended in 1960 by 40 countries, proposed the International System of Units (SI). We are now under¬ going a painful period of transition to SI, an adjustment that may take many more years to complete. The professional societies have led the way. Since July 1, 1974, SI units have been required by all papers published by the American Society of Mechanical Engineers, and there is a textbook explaining the SI . The present text will use SI units together with British gravitational (BG) units. In fluid mechanics there are only four primary dimensions from which all other dimen¬ sions can be derived: mass, length, time, and temperature.' These dimensions and ^The mean distance traveled by molecules between collisions (see Prob. PI. 5). "'if electromagnetic effects are important, a fifth primary dimension must be included, electric current {/}, whose SI unit is the ampere (A). 8 Chapter 1 Introduction Table 1.1 Primary Dimensions in SI and BG Systems The International System (SI) The British Gravitational (BG) System Other Unit Systems Primary dimension SI unit BG unit Conversion factor Mass [M] Kilogram (kg) Slug 1 slug = 14.5939 kg Length {L} Meter (m) Foot (ft) 1 ft = 0.3048 m Time {T} Second (s) Second (s) 1 s = 1 s Temperature {©} Kelvin (K) Rankine (°R) 1 K = 1.8°R their units in both systems are given in Table 1.1. Note that the Kelvin unit uses no degree symbol. The braces around a symbol like {M] mean “the dimension” of mass. All other variables in fluid mechanics can be expressed in terms of (M), {L}, {T}, and {0}. For example, acceleration has the dimensions {LT~^}. The most crucial of these secondary dimensions is force, which is directly related to mass, length, and time by Newton’s second law. Force equals the time rate of change of momentum or, for constant mass, F = ma (1-2) From this we see that, dimensionally, {F} = {MLT~^}. The use of a constant of proportionality in Newton’s law, Eq. (1.2), is avoided by defining the force unit exactly in terms of the other basic units. In the SI sys¬ tem, the basic units are newtons {F}, kilograms {M}, meters {L}, and seconds {F}. We define 1 newton of force = IN = Ikg-lm/s^ The newton is a relatively small force, about the weight of an apple (0.225 Ibf). In addition, the basic unit of temperature { 0 } in the SI system is the degree Kelvin, K. Use of these SI units (N, kg, m, s, K) will require no conversion factors in our equations. In the BG system also, a constant of proportionality in Eq. (1.2) is avoided by defin¬ ing the force unit exactly in terms of the other basic units. In the BG system, the basic units are pound-force {F}, slugs {M}, feet {L}, and seconds {F}. We define 1 pound of force = 1 Ibf = 1 slug • 1 ft/s^ One Ibf ~ 4.4482 N and approximates the weight of four apples. We will use the abbreviation Ibf for pound-force and Ibm for pound-mass. The slug is a rather hefty mass, equal to 32.174 Ibm. The basic unit of temperature {0} in the BG system is the degree Rankine, °R. Recall that a temperature difference 1 K = 1.8°R. Use of these BG units (Ibf, slug, ft, s, °R) will require no conversion factors in our equations. There are other unit systems still in use. At least one needs no proportionality constant: the CGS system (dyne, gram, cm, s, K). However, CGS units are too small for most applications (1 dyne = 10”^ N) and will not be used here. 1.4 Dimensions and Units 9 Table 1.2 Secondary Dimensions in Fluid Mechanics Secondary dimension SI unit BG unit Conversion factor Area [l}] fF 1 = 10.764 fF Volume {U} fF 1 = 35.315 fF Velocity [LT~f m/s ft/s 1 ft/s = 0.3048 m/s Acceleration {LT”^} m/s^ ft/s^ 1 ft/s^ = 0.3048 m/s^ Pressure or stress {ML~^T~^] Pa = N/m^ Ibf/fF 1 Ibf/ft^ = 47.88 Pa Angular velocity s-‘ s-‘ 1 s“‘ = 1 s”‘ Energy, heat, work [MI}T~^] J = N ■ m ft ■ Ibf 1 ft ■ Ibf = 1.3558 J Power [ML^T~^} W = J/s ft ■ Ibf/s 1 ft ■ Ibf/s = 1.3558 W Density {ML~^] kg/m^ slugs/ft^ 1 slug/ft^ = 515.4 kg/m^ Viscosity kg/(m ■ s) slugs/(ft • s) 1 slug/(ft ■ s) = 47.88 kg/(m ■ s) Specific heat m^/(s^ ■ K) fF/(s^ ■ °R) 1 m^/(s^ ■ K) = 5.980 fF/(s^ ■ °R) In the USA, some still use the English Engineering system (Ibf, Ibm, ft, s, °R), where the basic mass unit is the pound of mass. Newton’s law (1.2) must be rewritten: ma ft • Ibm F = — , where g, = 32.174 - ^ (1.3) 8c Ibf-s^ The constant of proportionality, g„ has both dimensions and a numerical value not equal to 1.0. The present text uses only the SI and BG systems and will not solve problems or examples in the English Engineering system. Because Americans still use them, a few problems in the text will be stated in truly awkward units: acres, gallons, ounces, or miles. Your assignment will be to convert these and solve in the SI or BG systems. The Principle of Dimensional Homogeneity In engineering and science, all equations must be dimensionally homogeneous, that is, each additive term in an equation must have the same dimensions. Eor example, take Bernoulli’s incompressible equation, to be studied and used throughout this text: pgZ = constant Each and every term in this equation must have dimensions of pressure {ML~^T~^}. We will examine the dimensional homogeneity of this equation in detail in Example 1.3. A list of some important secondary variables in fluid mechanics, with dimensions derived as combinations of the four primary dimensions, is given in Table 1.2. A more complete list of conversion factors is given in App. C. EXAMPLE 1.1 A body weighs 1000 Ibf when exposed to a standard earth gravity g = 32.174 ft/s^. (a) What is its mass in kg? (b) What will the weight of this body be in N if it is exposed to the moon’s standard acceleration gmoon = 1-62 m/s^? (c) How fast will the body accelerate if a net force of 400 Ibf is applied to it on the moon or on the earth? 10 Chapter 1 Introduction Part (a) Part (b) Part (c) Part (a) Solution We need to find the (a) mass; (b) weight on the moon; and (c) acceleration of this body. This is a fairly simple example of conversion factors for differing unit systems. No property data is needed. The example is too low-level for a sketch. Newton’s law (1.2) holds with known weight and gravitational acceleration. Solve for m: , 1000 Ibf F = W = 1000 Ibf = mg = (m)(32.174ft/s^), or m = - r = 31.08 slugs 32.174 ft/s^ Convert this to kilograms: m = 31.08 slugs = (31.08 slugs) (14.5939 kg/slug) = 454 kg Ans. (a) The mass of the body remains 454 kg regardless of its location. Equation (1.2) applies with a new gravitational acceleration and hence a new weight: F = W^oon = = (454 kg) (1.62 lu/s") = 735 N = 165 Ibf Ans. (b) This part does not involve weight or gravity or location. It is simply an application of Newton’s law with a known mass and known force: Solve for F = 400 Ibf = ma = (31.08 slugs) a 400 Ibf 31.08 slugs 12.87 0.3048 Ans. (c) Comment (c): This acceleration would be the same on the earth or moon or anywhere. Many data in the literature are reported in inconvenient or arcane units suitable only to some industry or specialty or country. The engineer should convert these data to the SI or BG system before using them. This requires the systematic application of conversion factors, as in the following example. EXAMPLE 1.2 Industries involved in viscosity measurement [27, 29] continue to use the CGS system of units, since centimeters and grams yield convenient numbers for many fluids. The absolute viscosity (fi) unit is the poise, named after J. L. M. Poiseuille, a French physician who in 1840 performed pioneering experiments on water flow in pipes; 1 poise = 1 g/(cm-s). The kinematic viscosity {u) unit is the stokes, named after G. G. Stokes, a British physicist who in 1845 helped develop the basic partial differential equations of fluid momentum; 1 stokes = 1 cmVs. Water at 20°C has fi ~ 0.01 poise and also u ~ 0.01 stokes. Express these results in (a) SI and (b) BG units. Solution • Approach: Systematically change grams to kg or slugs and change centimeters to meters or feet. 1.4 Dimensions and Units 11 Part (b) ■ Property values: Given fi = 0.01 g/(cm-s) and i/ = 0.01 cm^/s. • Solution steps: (a) For conversion to SI units, p = 0.01 u = 0.01 g g(l kg/1000 g) kg = 0.01 = 0.001 cm s cm(0.01 m/cm)s m s cm^ cm^(0.01 m/cm)^ = 0.01 — ^ = 0.000001 — s s s Ans. (a) • For conversion to BG units u = 0.01 = 0.01 cm ■ s g(l kg/1000 g)(l slug/14.5939 kg) (0.01 m/cm)(l ft/0.3048 m)s = 0.0000209 cm^ cm^(0.01m/cm)^(l ft/0.3048 m)" ft" V = 0.01 - = 0.01 - ^ = 0.0000108 — s s s slug ft ■ s Ans. (b) • Comments: This was a laborious conversion that could have been shortened by using the direct viscosity conversion factors in App. C or the inside front cover. For example, Mbg ~ /Xsi/47.88. We repeat our advice: Faced with data in unusual units, convert them immediately to either SI or BG units because (1) it is more professional and (2) theoretical equa¬ tions in fluid mechanics are dimensionally consistent and require no further conversion factors when these two fundamental unit systems are used, as the following example shows. EXAMPLE 1.3 A useful theoretical equation for computing the relation between pressure, velocity, and altitude in a steady flow of a nearly inviscid, nearly incompressible fluid with negligible heat transfer and shaft work^ is the Bernoulli relation, named after Daniel Bernoulli, who published a hydrodynamics textbook in 1738: Po= P + kpV^ + pgZ (1) where po = stagnation pressure p = pressure in moving fluid V = velocity p = density Z = altitude g = gravitational acceleration (a) Show that Eq. (1) satisfies the principle of dimensional homogeneity, which states that all additive terms in a physical equation must have the same dimensions, (b) Show that consistent units result without additional conversion factors in SI units, (c) Repeat (b) for BG units. ^That’s an awful lot of assumptions, which need further study in Chap. 3. 12 Chapter 1 Introduction Consistent Units Part (a) Part (b) Part (c) Solution We can express Eq. (1) dimensionally, using braces, by entering the dimensions of each term from Table 1.2: {ML~^}{L^T~'^} + {ML~^}{LT~^}{L} = for all terms Ans. (a) Enter the SI units for each quantity from Table 1.2: {N/m^} = {N/m^} + {kg/m^}{mVs^} + {k;g/m^}{m/s^}{m} = {N/m^} + {kg/(m ■ s^)} The right-hand side looks bad until we remember from Eq. (1.3) that 1 kg = 1 N ■ s^/m. {N ■ s^/m} {kg/(m ■ s^)} = — - ^ = {N/m^} {m-s } Ans. (b) Thus all terms in Bernoulli’s equation will have units of pascals, or newtons per square meter, when SI units are used. No conversion factors are needed, which is true of all theo¬ retical equations in fluid mechanics. Introducing BG units for each term, we have {Ibf/ft^} = {Ibf/ft^} -f {slugs/ft^}{ft^/s^} -f {slugs/ft^}{ft/s^}{ft} = {Ibf/ft^} -f {slugs/(ft ■ s^)} But, from Eq. (1.3), 1 slug = 1 Ibf ■ sVft, so that {slugs/(ft ■ s^)} {Ibf-sVft} = {Ibf/ft-} Ans. (c) All terms have the unit of pounds-force per square foot. No conversion factors are needed in the BG system either. There is still a tendency in English-speaking countries to use pound-force per square inch as a pressure unit because the numbers are more manageable. For exam¬ ple, standard atmospheric pressure is 14.7 Ibf/in^ = 2116 Ibf/ft^ = 101,300 Pa. The pascal is a small unit because the newton is less than \| Ibf and a square meter is a very large area. Note that not only must all (fluid) mechanics equations be dimensionally homoge¬ neous, one must also use consistent units; that is, each additive term must have the same units. There is no trouble doing this with the SI and BG systems, as in Example 1.3, but woe unto those who try to mix colloquial English units. For example, in Chap. 9, we often use the assumption of steady adiabatic compressible gas flow: h + = constant 1.4 Dimensions and Units 13 where h is the fluid enthalpy and V'^/2 is its kinetic energy per unit mass. Colloquial thermodynamic tables might list h in units of British thermal units per pound mass (Btu/lb), whereas V is likely used in ft/s. It is completely erroneous to add Btu/lb to ft^/s^. The proper unit for h in this case is ft • Ibf/slug, which is identical to ft^/s^. The conversion factor is 1 Btu/lb ~ 25,040 ft^/s^ = 25,040 ft • Ibf/slug. Homogeneous versus Dimensionally Inconsistent Equations All theoretical equations in mechanics (and in other physical sciences) are dimensionally homogeneous; that is, each additive term in the equation has the same dimensions. However, the reader should be warned that many empirical formulas in the engineering literature, arising primarily from correlations of data, are dimensionally inconsistent. Their units cannot be reconciled simply, and some terms may contain hidden vari¬ ables. An example is the formula that pipe valve manufacturers cite for liquid volume flow rate Q (m^/s) through a partially open valve: Q Table 1.3 Convenient Prefixes for Engineering Units Multiplicative factor Prefix Symbol 10‘^ tera T 10^' giga G 10’ mega M 10’ kilo k 10’ hecto h 10 deka da 10"‘ deci d 10"’ centi c 10"’ mini m lO'^ micro p 10"’ nano n 10"” pico p 10-’’ femto f 10"’ atto a where Ap is the pressure drop across the valve and SG is the specific gravity of the liquid (the ratio of its density to that of water). The quantity Cy is the valve flow coefficient, which manufacturers tabulate in their valve brochures. Since SG is dimensionless {!}, we see that this formula is totally inconsistent, with one side being a flow rate {L?/T} and the other being the square root of a pressure drop It follows that Cy must have dimensions, and rather odd ones at that: {L™/M'^^}. Nor is the resolution of this discrepancy clear, although one hint is that the values of Cy in the literature increase nearly as the square of the size of the valve. The presentation of experimental data in homogeneous form is the subject of dimensional analysis (Chap. 5). There we shall learn that a homogeneous form for the valve flow relation is Q — /ApV^ \p) where p is the liquid density and A the area of the valve opening. The discharge coefficient is dimensionless and changes only moderately with valve size. Please believe — until we establish the fact in Chap. 5 — that this latter is a much better for¬ mulation of the data. Meanwhile, we conclude that dimensionally inconsistent equations, though they occur in engineering practice, are misleading and vague and even dangerous, in the sense that they are often misused outside their range of applicability. Convenient Prefixes in Engineering results often are too small or too large for the common units, with too Powers of 10 many zeros one way or the other. For example, to write p = 114,000,000 Pa is long and awkward. Using the prefix “M” to mean 10®, we convert this to a concise p = 114 MPa (megapascals). Similarly, t = 0.000000003 s is a proofreader’s nightmare compared to the equivalent f = 3 ns (nanoseconds). Such prefixes are common and convenient, in both the SI and BG systems. A complete list is given in Table 1.3. 14 Chapter 1 Introduction EXAMPLE 1.4 In 1890 Robert Manning, an Irish engineer, proposed the following empirical formula for the average velocity W in uniform flow due to gravity down an open channel (BG units): V = — (1) n where R = hydraulic radius of channel (Chaps. 6 and 10) S = channel slope (tangent of angle that bottom makes with horizontal) n = Manning’s roughness factor (Chap. 10) and M is a constant for a given surface condition for the walls and bottom of the channel. (a) Is Manning’s formula dimensionally consistent? (b) Equation (1) is commonly taken to be valid in BG units with n taken as dimensionless. Rewrite it in SI form. Solution • Assumption: The channel slope S is the tangent of an angle and is thus a dimensionless ratio with the dimensional notation ( 1 } — that is, not containing M, L, or T. • Approach (a): Rewrite the dimensions of each term in Manning’s equation, using brackets { } : ^ {y} = {^}{L^'^}{1} This formula is incompatible unless {1.49/n) = {L'^^/T}. If n is dimensionless (and it is never listed with units in textbooks), the number 1.49 must carry the dimensions of (L'^VT). Ans. (a) ■ Comment (a): Formulas whose numerical coefficients have units can be disastrous for engineers working in a different system or another fluid. Manning’s formula, though popular, is inconsistent both dimensionally and physically and is valid only for water flow with certain wall roughnesses. The effects of water viscosity and density are hidden in the numerical value 1.49. • Approach (b): Part (a) showed that 1.49 has dimensions. If the formula is valid in BG units, then it must equal 1.49 ft'^^/s. By using the SI conversion for length, we obtain (1.49 ft‘'^/s) (0.3048 m/ft) = 1.00m‘'Vs Therefore, Manning’s inconsistent formula changes form when converted to the SI system: SI units: V= — Ans. (b) n with R in meters and V in meters per second. • Comment (b): Actually, we misled you: This is the way Manning, a metric user, first proposed the formula. It was later converted to BG units. Such dimensionally inconsistent formulas are dangerous and should either be reanalyzed or treated as having very limited application. 1.6 Thermodynamic Properties of a Fluid 15 1.5 Properties of the Velocity Field The Velocity Field The Acceleration Field 1.6 Thermodynamic Properties of a Fluid In a given flow situation, the determination, by experiment or theory, of the properties of the fluid as a function of position and time is considered to be the solution to the problem. In almost all cases, the emphasis is on the space-time distribution of the fluid properties. One rarely keeps track of the actual fate of the specific fluid particles. This treatment of properties as continuum-field functions distinguishes fluid mechan¬ ics from solid mechanics, where we are more likely to be interested in the trajectories of individual particles or systems. Foremost among the properties of a flow is the velocity field V(x, y, z, t). In fact, determining the velocity is often tantamount to solving a flow problem, since other properties follow directly from the velocity field. Chapter 2 is devoted to the calcula¬ tion of the pressure field once the velocity field is known. Books on heat transfer (for example. Ref. 20) are largely devoted to finding the temperature field from known velocity fields. In general, velocity is a vector function of position and time and thus has three components u, v, and w, each a scalar field in itself: V(x, y, z, t) = iu(.x, y, z, t) + jv(x, y, z, t) + kwCv, y, z, t) (1.4) The use of u, v, and w instead of the more logical component notation V„ Vy, and I4 is the result of an almost unbreakable custom in fluid mechanics. Much of this text¬ book, especially Chaps. 4, 7, 8, and 9, is concerned with finding the distribution of the velocity vector V for a variety of practical flows. The acceleration vector, a = dVIdt, occurs in Newton’s law for a fluid and thus is very important. In order to follow a particle in the Eulerian frame of reference, the final result for acceleration is nonlinear and quite complicated. Here we only give the formula: dV dY dY dY dY a = — = h U - V — -1- w - dt dt dx dy dz (1.5) where (m, v, w) are the velocity components from Eq. (1.4). We shall study this for¬ mula in detail in Chap. 4. The last three terms in Eq. (1.5) are nonlinear products and greatly complicate the analysis of general fluid motions, especially viscous flows. While the velocity field V is the most important fluid property, it interacts closely with the thermodynamic properties of the fluid. We have already introduced into the discussion the three most common such properties: Pressure p Density p Temperature T 16 Chapter 1 Introduction These three are constant companions of the velocity vector in flow analyses. Four other intensive thermodynamic properties become important when work, heat, and energy balances are treated (Chaps. 3 and 4): Internal energy u Enthalpy h = u + p/p Entropy s Specific heats Cp and In addition, friction and heat conduction effects are governed by the two so-called transport properties: Coefficient of viscosity p Thermal conductivity k All nine of these quantities are true thermodynamic properties that are determined by the thermodynamic condition or state of the fluid. For example, for a single-phase substance such as water or oxygen, two basic properties such as pressure and tem¬ perature are sufficient to fix the value of all the others: P = Pip, T) h = hip, T) p = pip, T) and so on for every quantity in the list. Note that the specific volume, so imporfanf in thermodynamic analyses, is omitted here in favor of its inverse, the density p. Recall that thermodynamic properties describe the state of a system — that is, a collection of matter of fixed identify that interacts with its surroundings. In most cases here the system will be a small fluid element, and all properties will be assumed to be continuum properties of the flow field: p = p(x, y, z, t), and so on. Recall also fhat thermodynamics is normally concerned with static systems, whereas fluids are usually in variable motion with constantly changing properties. Do the properties retain their meaning in a fluid flow that is technically not in equilibrium? The answer is yes, from a statistical argument. In gases at normal pressure (and even more so for liquids), an enormous number of molecular collisions occur over a very short distance of the order of 1 pm, so that a fluid subjected to sudden changes rapidly adjusts itself toward equilibrium. We therefore assume that all the thermodynamic properties just listed exist as point functions in a flowing fluid and follow all the laws and state relations of ordinary equilibrium thermodynamics. There are, of course, important nonequilibrium effects such as chemical and nuclear reactions in flowing fluids, which are not treated in this text. Pressure Pressure is the (compression) stress at a point in a static fluid (Fig. 1.3). Next to velocity, the pressure p is the most dynamic variable in fluid mechanics. Differences or gradients in pressure often drive a fluid flow, especially in ducts. In low-speed flows, the actual magnitude of the pressure is often not important, unless it drops so low as to cause vapor bubbles to form in a liquid. For convenience, we set many such problem assignments at the level of 1 atm = 2116 Ibf/ft^ = 101,300 Pa. High-speed (compressible) gas flows (Chap. 9), however, are indeed sensitive to the magnitude of pressure. 1.6 Thermodynamic Properties of a Fluid 17 Temperature Density Specific Weight Specific Gravity Temperature T is related to the internal energy level of a fluid. It may vary consider¬ ably during high-speed flow of a gas (Chap. 9). Although engineers often use Celsius or Fahrenheit scales for convenience, many applications in this text require absolute (Kelvin or Rankine) temperature scales: °R = °F -F 459.69 K = °C + 273.16 If temperature differences are strong, heat transfer may be important , but our concern here is mainly with dynamic effects. The density of a fluid, denoted by p (lowercase Greek rho), is its mass per unit volume. Density is highly variable in gases and increases nearly proportionally to the pressure level. Density in liquids is nearly constant; the density of water (about 1000 kg/m^) increases only 1 percent if the pressure is increased by a factor of 220. Thus most liquid flows are treated analytically as nearly “incompressible.” In general, liquids are about three orders of magnitude more dense than gases at atmospheric pressure. The heaviest common liquid is mercury, and the lightest gas is hydrogen. Compare their densities at 20°C and 1 atm: Mercury: p = 13,580 kg/m^ Hydrogen: p = 0.0838 kg/m^ They differ by a factor of 162,000! Thus, the physical parameters in various liquid and gas flows might vary considerably. The differences are often resolved by the use of dimensional analysis (Chap. 5). Other fluid densities are listed in Tables A. 3 and A. 4 (in App. A), and in Ref. 21. The specific weight of a fluid, denoted by 7 (lowercase Greek gamma), is its weight per unit volume. Just as a mass has a weight W = mg, density and specific weight are simply related by gravity: l = Pg (1-6) The units of 7 are weight per unit volume, in Ibf/ft^ or N/m^. In standard earth grav- ity, g = 32.174 ft/s^ = 9.807 m/s^. Thus, for example, the specific weights of air and water at 20°C and 1 atm are approximately 7air = (1.205 kg/m^) (9.807 m/s^) = 11.8 N/m^ = 0.0752 Ibf/ft^ Twater = (998 kg/m^) (9.807 m/s^) = 9790 N/m^ = 62.41bf/ft^ Specific weight is very useful in the hydrostatic pressure applications of Chap. 2. Specific weights of other fluids are given in Tables A. 3 and A.4. Specific gravity, denoted by SG, is the ratio of a fluid density to a standard reference fluid, usually water at 4°C (for liquids) and air (for gases): SGgas - SGiiquiij Pgas Pgas Pair 1.205 kg/m^ Pliquid Pliquid Pwater 1000 kg/m (1.7) 18 Chapter 1 Introduction Potential and Kinetic Energies State Relations for Gases For example, the specific gravity of mercury (Hg) is SGng = 13,580/1000 ~ 13.6. Engineers find these dimensionless ratios easier to remember than the actual numerical values of density of a variety of fluids. In thermostatics the only energy in a substance is that stored in a system by molecular activity and molecular bonding forces. This is commonly denoted as internal energy u. A commonly accepted adjustment to this static situation for fluid flow is to add two more energy terms that arise from newtonian mechanics: potential energy and kinetic energy. The potential energy equals the work required to move the system of mass ni from the origin to a position vector r = ix + jy + kz against a gravity field g. Ifs value is — /ng ■ r, or — g ■ r per unif mass. The kinetic energy equals the work required to change the speed of the mass from zero to velocity V. Its value is or per unit mass. Then by common convention the total stored energy e per unit mass in fluid mechanics is the sum of three terms: e = u + + (-g • r) (1-8) Also, throughout this book we shall define z as upward, so fhaf g = — and g ■ r = —gz- Then Eq. (1.8) becomes e = M + iy2 + (1.9) The molecular infernal energy m is a function of T and p for fhe single-phase pure substance, whereas the potential and kinetic energies are kinematic quantities. Thermodynamic properties are found both theoretically and experimentally to be related to each other by state relations that differ for each substance. As mentioned, we shall confine ourselves here to single-phase pure substances, such as water in its liquid phase. The second most common fluid, air, is a mixture of gases, but since the mixture ratios remain nearly constant between 160 and 2200 K, in this temperature range air can be considered to be a pure substance. All gases at high temperatures and low pressures (relative to their critical point) are in good agreement with the perfect-gas law p = pRT R = Cp — = gas constant (1.10) where the specific heafs Cp and are defined in Eqs. (1.14) and (1.15). Since Eq. (1.10) is dimensionally consistent, R has the same dimensions as specific heat, or velocity squared per temperature unit (kelvin or degree Rankine). Each gas has its own constant R, equal to a universal constant A divided by the molecular weight Rpafi (1.11) 1.6 Thermodynamic Properties of a Fluid 19 where A = 49,700 ft-lbf/(slugmol • °R) = 8314 J/(kmol • K). Most applications in this book are for air, whose molecular weight is M = 28.97/mol: 49,700 ft ■ lbf/(slugmol ■ °R) ft ■ Ibf R„r = - rtrrrz— 7 - ^ = 1716 - — = 1716 28.97/mol slug ■ °R fk s^“R = 287 m s^- K (1.12) Standard atmospheric pressure is 2116 Ibf/ft^ = 2116 slug/(ft • s^), and standard tem¬ perature is 60°F = 520°R. Thus standard air density is 2116slug/(ft-s^) 1716ft^/(s^-°R) 0.00237 slug/ft^ 1.22 kg/m^ (1.13) This is a nominal value suitable for problems. For other gases, see Table A.4. Most of the common gases — oxygen, nitrogen, hydrogen, helium, argon — are nearly ideal. This is not so true for steam, whose simplified temperature-entropy chart is shown in Fig. 1.3. Unless you are sure that the steam temperature is “high” and the pressure “low,” it is best to use the Steam Tables to make accurate calculations. One proves in thermodynamics that Eq. (1.10) requires that the internal molecular energy m of a perfect gas vary only with temperature: u = u(T). Therefore, the specific heat c„ also varies only with temperature: Cy du dT c.(T) or du = c,{T)dT (1.14) In like manner h and c„ of a perfect gas also vary only with temperature: h = u -\ — = u + RT = h(T) P f dh\ dh Cp = — 1 = — = CpiT) dT (1.15) dh = Cp(T)dT Fig. 1.3 Temperature-entropy chart for steam. The critical point is = 22,060 kPa, T, = 374°C, S, = 4.41 kj/(kg ■ K). Except near the critical point, the smooth isobars tempt one to assume, often incorrectly, that the ideal-gas law is valid for steam. It is not, except at low pressure and high temperature: the upper right of the graph. 20 Chapter 1 Introduction The ratio of specific heats of a perfect gas is an important dimensionless parameter in compressible flow analysis (Chap. 9) k=— = k{T)>l (1.16) Cv As a first approximation in airflow analysis we commonly take Cp, c„ and k to be constant: R 4293 ft^/(s^ • °R) 6010 ft^/(s^ • °R) 718mV(s^-K) 1005 m^/(s^ • K) (1.17) Actually, for all gases, Cp and c„ increase gradually with temperature, and k decreases gradually. Experimental values of the specific-heat ratio for eight common gases are shown in Fig. 1.4. Nominal values are in Table A.4. Fig. 1.4 Specific-heat ratio of eight common gases as a function of temperature. (Data from Ref. 22.) 1.6 Thermodynamic Properties of a Fluid 21 Many flow problems involve steam. Typical steam operating conditions are often close to the critical point, so that the perfect-gas approximation is inaccurate. Then we must turn to the steam tables, either in tabular or CD-ROM form or as online software . Most online steam tables require a license fee, but the writer, in Example 1.5 that follows, suggests a free online source. Sometimes the error of using the perfect-gas law for steam can moderate, as the following example shows. EXAMPLE 1.5 Estimate p and Cp of steam at 100 Ibf/in^ and 400°F, in English units, (a) by the perfect-gas approximation and {b) by the ASME Steam Tables . Solution • Approach (a) — the perfect-gas law: Although steam is not an ideal gas, we can estimate these properties with moderate accuracy from Eqs. (1.10) and (1.17). First convert pressure from 100 Ibf/in^ to 14,400 Ibf/ft^, and use absolute temperature, (400°F -I- 460) = 860°R. Then we need the gas constant for steam, in English units. From Table A.4, the molecular weight of Fl20 is 18.02, whence A, English M, H2O 49,700 ft • lbf/(slugmol “R) ft ■ Ibf — — 2 - - = 2758 - 18.02/mol slug “R Then the density estimate follows from the perfect-gas law, Eq. (1.10): _ P _ 14,400 Ibf/ft^ _ slug 2758 ft ■ lbf/(slug ■ °R) ~ Ans. (a) At 860°R, from Fig. 1.5, /:steam ~ Cp/c^, ~ 1.30. Then, from Eq. (1.17) kR k - 1 (1.3)(2758 ft ■ lbf/(slug °R)) ft ■ Ibf « 12,000 ■ (1.3 - 1) slug °R Ani’. {a) • Approach (b) — tables or software: One can either read the ASME Steam Tables or use online software. Online software, such as , calculates the properties of steam without reading a table. Most of these require a license fee, which your institution may or may not possess. Eor work at home, the writer has found success with this free com¬ mercial online site: www.spiraxsarco.com/esc/SH_Properties.aspx The software calculates superheated steam properties, as required in this example. The Spirax Sarco Company makes many types of steam equipment: boilers, condensers, valves, pumps, regulators. This site provides many steam properties — density, specific heat, enthalpy, speed of sound — in many different unit systems. Here we need the density and specific heat of steam at 100 Ibf/in^ and 400°F. You enter these two inputs and it will calculate not only p and Cp but also many other properties of interest, in English or metric units. The software results are: p(1001bf/in^400“E) = 0.2027 Ibm/tf = 3.247 kg/m^ Cp(1001bf/in^400°F) = 0.5289 Btu/(lbm-F) = 2215 J/(kg-K) Ans. (b) Ans. (b) 22 Chapter 1 Introduction State Relations for Liquids Comments: These are quite accurate and compare well to other steam tables. The perfect gas estimate of p is 4 percent low, and the estimate of Cp is 9 percent low. The chief reason for the discrepancy is that this temperature and pressure are rather close to the critical point and saturation line of steam. At higher temperatures and lower pressures, say, 800°F and 50 Ibf/in^, the perfect-gas law has an accuracy of about ±1 percent. See Fig. 1.3. Again let us warn that English units (psia, Ibm, Btu) are awkward and must be converted to SI or BG units in almost all fluid mechanics formulas. The writer knows of no “perfect-liquid law” comparable to that for gases. Liquids are nearly incompressible and have a single, reasonably constant specific heat. Thus an idealized state relation for a liquid is p ~ const Cp ~ Cy ~ const dh ~ Cp dT (1-18) Most of the flow problems in this book can be attacked with these simple assumptions. Water is normally taken to have a density of 998 kg/m^ and a specific heat Cp = 4210 m^/(s^ • K). The steam tables may be used if more accuracy is required. The density of a liquid usually decreases slightly with temperature and increases moderately with pressure. If we neglect the temperature effect, an empirical pressure- density relation for a liquid is — = (5 + l)f— ) - B (1.19) Pa \PaJ where B and n are dimensionless parameters that vary slightly with temperature and Pa and Pa are standard atmospheric values. Water can be fitted approximately to the values B ~ 3000 and n ~ 1. Seawater is a variable mixture of water and salt and thus requires three thermody¬ namic properties to define its state. These are normally taken as pressure, temperature, and the salinity S, defined as the weight of the dissolved salt divided by the weight of the mixture. The average salinity of seawater is 0.035, usually written as 35 parts per 1000, or 35 %c. The average density of seawater is 2.00 slugs/ft^ ~ 1030 kg/m^. Strictly speaking, seawater has three specific heats, all approximately equal to the value for pure water of 25,200 ftV(s^ • °R) = 4210 m^/(s^ • K). EXAMPLE 1.6 The pressure at the deepest part of the ocean is approximately 1 100 atm. Estimate the density of seawater in slug/fL at this pressure. Solution Equation (1.19) holds for either water or seawater. The ratio pipa is given as 1100; 1100 « (3001)1 (0 - ™ 1.7 Viscosity and Other Secondary Properties 23 1.7 Viscosity and Other Secondary Properties Viscosity Fig. 1.5 Shear stress causes continuous shear deformation in a fluid: (a) a fluid element straining at a rate 60l5t\ (b) newtonian shear distribution in a shear layer near a wall. or Pa AiooV" Vsooiy 1.046 Assuming an average surface seawater density = 2.00 slugs/ft^, we compute p « 1.046(2.00) = 2.09 slugs/ft^ Ans. Even at these immense pressures, the density increase is less than 5 percent, which justifies the treatment of a liquid flow as essentially incompressible. The quantities such as pressure, temperature, and density discussed in the previous section are primary thermodynamic variables characteristic of any system. Certain secondary variables also characterize specific fluid mechanical behavior. The most important of these is viscosity, which relates the local stresses in a moving fluid to the strain rate of the fluid element. Viscosity is a quantitative measure of a fluid’s resistance to flow. More specifically, it determines the fluid strain rate that is generated by a given applied shear stress. We can easily move through air, which has very low viscosity. Movement is more difficult in water, which has 50 times higher viscosity. Still more resistance is found in SAE 30 oil, which is 300 times more viscous than water. Try to slide your hand through glycerin, which is five times more viscous than SAE 30 oil, or blackstrap molasses, another factor of five higher than glycerin. Eluids may have a vast range of viscosities. Consider a fluid element sheared in one plane by a single shear stress r, as in Eig. 1.5a. The shear strain angle 69 will continuously grow with time as long as the stress T is maintained, the upper surface moving at speed 5u larger than the lower. Such common fluids as water, oil, and air show a linear relation between applied shear and resulting strain rate: T OC 5t (1.20) T u = 8u V (a) (b) 24 Chapter 1 Introduction From the geometry of Fig. 1 .5a, we see that j. Su 5t tan (50 = — — (1-21) 6y In the limit of infinitesimal changes, this becomes a relation between shear strain rate and velocity gradient; dO du dt dy (1.22) From Eq. (1.20), then, the applied shear is also proportional to the velocity gradi¬ ent for the common linear fluids. The constant of proportionality is the viscosity coefficient /i: dd du (1.23) Equation (1.23) is dimensionally consistent; therefore /i has dimensions of stress-time: {FTIl}} or {MULT)}. The BG unit is slugs per foot-second, and the SI unit is kilo¬ grams per meter-second. The linear fluids that follow Eq. (1.23) are called newtonian fluids, after Sir Isaac Newton, who first postulated this resistance law in 1687. We do not really care about the strain angle G(t) in fluid mechanics, concentrating instead on the velocity distribution u{y), as in Fig. 1.5b. We shall use Eq. (1.23) in Chap. 4 to derive a differential equation for finding the velocity distribution u(y) — and, more generally, V(x, y, z, f) — in a viscous fluid. Figure 1.5b illustrates a shear layer, or boundary layer, near a solid wall. The shear stress is proportional to the slope of the velocity profile and is greatest at the wall. Further, at the wall, the velocity u is zero relative to the wall: This is called the no-slip condition and is characteristic of all viscous fluid flows. The viscosity of newtonian fluids is a true thermodynamic property and varies with temperature and pressure. At a given state (p, T) there is a vast range of values among the common fluids. Table 1.4 lists the viscosity of eight fluids at standard pressure and temperature. There is a variation of six orders of magnitude from hydrogen up to glycerin. Thus there will be wide differences between fluids subjected to the same applied stresses. Table 1.4 Viscosity and Kinematic Viscosity of Eight Fluids at 1 atm and 20°C Fluid kg/(in • s)^ Ratio pjpm 3 kg/m^ V mVs' Ratio t/MHg) Hydrogen 9.0 E-6 1.0 0.084 1.05 E-A 910 Air 1.8 E-5 2.1 1.20 1.50 E-5 130 Gasoline 2.9 E^ 33 680 4.22 E-7 3.7 Water 1.0 E-3 114 998 1.01 E-6 8.7 Ethyl alcohol 1.2 E-3 135 789 1.52 E-6 13 Mercury 1.5 E-3 170 13,550 1.16 E-7 1.0 SAE 30 oil 0.29 33,000 891 3.25 E^ 2,850 Glycerin 1.5 170,000 1,260 1.18 E-3 10,300 T kg/(m ■ s) = 0.0209 slug/(ft ■ s); 1 m^/s = 10.76 ft7s. 1.7 Viscosity and Other Secondary Properties 25 Generally speaking, the viscosity of a fluid increases only weakly with pressure. For example, increasing p from 1 to 50 atm will increase /i of air only 10 percent. Tem¬ perature, however, has a strong effect, with /i increasing with T for gases and decreasing for liquids. Figure A.l (in App. A) shows this temperature variation for various common fluids. It is customary in most engineering work to neglect the pressure variation. The variation fi(p, T) for a typical fluid is nicely shown by Fig. 1.6, from Ref. 25, which normalizes the data with the critical-point state (/i^, Pc, T^. This behavior, called the principle of corresponding states, is characteristic of all fluids, but the actual numerical values are uncertain to ±20 percent for any given fluid. For example, values of li(T) for air at 1 atm, from Table A. 2, fall about 8 percent low compared to the “low-density limit” in Fig. 1.6. Note in Fig. 1 .6 that changes with temperature occur very rapidly near the critical point. In general, critical-point measurements are extremely difficult and uncertain. The Reynolds Number The primary parameter correlating the viscous behavior of all newtonian fluids is the dimensionless Reynolds number: pVL VL Re = - = — /i u (1.24) Fig. 1.6 Fluid viscosity nondimensionalized by critical-point properties. This generalized chart is characteristic of all fluids but is accurate only to ±20 percent. (From Ref. 25.) 10 9 8 7 6 5 4 3 2 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.4 0.6 0.8 1 2 3 4 5 6 7 8 9 10 -1 Lie uid 1 1 1 De nse g as T iVO- reg Tha ion se 25 ho c riti Toil :;al It ^ —2 5 .5 1 Pr = p/p 12 1 1 1 0 i. ow-densit y limit T = ^ r X Tc 26 Chapter 1 Introduction where V and L are characteristic velocity and length scales of the flow. The second form of Re illustrates that the ratio of /i to p has its own name, the kinematic viscosity: M i/=- (1.25) P It is called kinematic because the mass units cancel, leaving only the dimensions [L^IT], Generally, the first thing a fluids engineer should do is estimate the Reynolds number range of the flow under study. Very low Re indicates viscous creeping motion, where inertia effects are negligible. Moderate Re implies a smoothly varying laminar flow. High Re probably spells turbulent flow, which is slowly varying in the time- mean but has superimposed strong random high-frequency fluctuations. Explicit numerical values for low, moderate, and high Reynolds numbers cannot be stated here. They depend on flow geometry and will be discussed in Chaps. 5 through 7. Table 1.4 also lists values of i/for the same eight fluids. The pecking order changes considerably, and mercury, the heaviest, has the smallest viscosity relative to its own weight. All gases have high u relative to thin liquids such as gasoline, water, and alcohol. Oil and glycerin still have the highest v, but the ratio is smaller. For given values of V and L in a flow, these fluids exhibit a spread of four orders of magnitude in the Reynolds number. Flow between Plates A classic problem is the flow induced between a fixed lower plate and an upper plate moving steadily at velocity V, as shown in Fig. 1.7. The clearance between plates is h, and the fluid is newtonian and does not slip at either plate. If the plates are large, this steady shearing motion will set up a velocity distribution u(y), as shown, with V = w = 0. The fluid acceleration is zero everywhere. With zero acceleration and assuming no pressure variation in the flow direction, you should show that a force balance on a small fluid element leads to the result that the shear stress is constant throughout the fluid. Then Eq. (1.23) becomes du T — = — = const dy /i which we can integrate to obtain u = a + by Fig. 1.7 Viscous flow induced by relative motion between two parallel plates. 1.7 Viscosity and Other Secondary Properties 27 Problem-Solving Techniques The velocity distribution is linear, as shown in Fig. 1.7, and the constants a and b can be evaluated from the no-slip condition at the upper and lower walls: r 0 = a + (0) at y = 0 \v = a + b (h) aty = h Hence a = 0 and b = V/h. Then the velocity profile between the plates is given by u=V- (1.26) h as indicated in Fig. 1.7. Turbulent flow (Chap. 6) does not have this shape. Although viscosity has a profound effect on fluid motion, the actual viscous stresses are quite small in magnitude even for oils, as shown in the next example. Fluid flow analysis is packed with problems to be solved. The present text has more than 1700 problem assignments. Solving a large number of these is a key to learning the subject. One must deal with equations, data, tables, assumptions, unit systems, and solution schemes. The degree of difficulty will vary, and we urge you to sample the whole spectrum of assignments, with or without the answers in the Appendix. Here are the recommended steps for problem solution: Read the problem and restate it with your summary of the results desired. From tables or charts, gather the needed property data: density, viscosity, etc. Make sure you understand what is asked. Students are apt to answer the wrong question — for example, pressure instead of pressure gradient, lift force instead of drag force, or mass flow instead of volume flow. Read the problem carefully. Make a detailed, labeled sketch of the system or control volume needed. Think carefully and list your assumptions. You must decide if the flow is steady or unsteady, compressible or incompressible, viscous or inviscid, and whether a control volume or partial differential equations are needed. Find an algebraic solution if possible. Then, if a numerical value is needed, use either the SI or BG unit systems reviewed in Sec. 1.4. Report your solution, labeled, with the proper units and the proper number of significant figures (usually two or three) that the data uncertainty allows. We shall follow these steps, where appropriate, in our example problems. EXAMPLE 1.7 Suppose that the fluid being sheared in Fig. 1.7 is SAE 30 oil at 20°C. Compute the shear stress in the oil if V = 3 m/s and h = 2 cm. Solution • System sketch: This is shown earlier in Fig. 1.7. ■ Assumptions: Linear velocity profile, laminar newtonian fluid, no slip at either plate surface. 28 Chapter 1 Introduction ■ Approach: The analysis of Fig. 1.7 leads to Eq. (1.26) for laminar flow. • Property values: From Table 1.4 for SAE 30 oil, the oil viscosity /x = 0.29 kg/(m-s). • Solution steps: In Eq. (1.26), the only unknown is the fluid shear stress: t = pI = [ 0.29 kg \ (3 m/s) m ■ s/(0.02 m) = 43.5 kg ■ m/s^ = 43.5 N ^ 44Pa Ans. ■ Comments: Note the unit identities, 1 kg-m/s^ = 1 N and 1 N/m^ = 1 Pa. Although oil is very viscous, this shear stress is modest, about 2400 times less than atmospheric pres¬ sure. Viscous stresses in gases and thin (watery) liquids are even smaller. Variation of Viscosity with Temperature Nonnewtonian Fluids Temperature has a strong effect and pressure a moderate effect on viscosity. The viscosity of gases and most liquids increases slowly with pressure. Water is anomalous in showing a very slight decrease below 30°C. Since the change in viscosity is only a few percent up to 100 atm, we shall neglect pressure effects in this book. Gas viscosity increases with temperature. Two common approximations are the power law and the Sutherland law: Mo {T/Tof\To + S) T + S power law Sutherland law (1.27) where /Xq is a known viscosity at a known absolute temperature Tq (usually 273 K). The constants n and S are fit to the data, and both formulas are adequate over a wide range of temperatures. For air, n ~ 0.7 and S ~ 110 K = 199°R. Other values are given in Ref. 26. Liquid viscosity decreases with temperature and is roughly exponential, /x ~ but a better fit is the empirical result that In /x is quadratic in l/T) where T is absolute temperature: For water, with Tq = 273.16 K, = 0.001792 kg/(m • s), suggested values are a = —1.94, b = —4.80, and c = 6.74, with accuracy about ±1 percent. The viscos¬ ity of water is tabulated in Table A.l. For further viscosity data, see Refs. 21, 28, and 29. Fluids that do not follow the linear law of Eq. (1.23) are called nonnewtonian and are treated in books on rheology . Figure l.So compares some examples to a newtonian fluid. For the nonlinear curves, the slope at any point is called the apparent viscosity. 1.7 Viscosity and Other Secondary Properties 29 Fig. 1.8 Rheological behavior of various viscous materials: (a) stress versus strain rate; (b) effect of time on applied stress. Fig. 1.9 A rotating parallel-disk rheometer (Image of Kinexus rheometer, used with kind permission of Malvern Instruments). Rheometers dt (a) Strain rate 0 Time — ► (b) Dilatant. This fluid is shear-thickening, increasing its resistance with increasing strain rate. Examples are suspensions of corn starch or sand in water. The classic case is quicksand, which stiffens up if one thrashes about. Pseudoplastic. A shear-thinning fluid is less resistant at higher strain rates. A very strong thinning is called plastic. Some of the many examples are polymer solutions, colloidal suspensions, paper pulp in water, latex paint, blood plasma, syrup, and molasses. The classic case is paint, which is thick when poured but thin when brushed at a high strain rate. Bingham plastic. The limiting case of a plastic substance is one that requires a finite yield stress before it begins to flow. Figure 1.8a shows yielding followed by linear behavior, but nonlinear flow can also occur. Some examples are clay suspensions, drilling mud, toothpaste, mayonnaise, chocolate, and mustard. The classic case is catsup, which will not come out of the bottle until you stress it by shaking. A further complication of nonnewtonian behavior is the transient effect shown in Fig. 1.8h. Some fluids require a gradually increasing shear stress to maintain a con¬ stant strain rate and are called rheopectic. The opposite case of a fluid that thins out with time and requires decreasing stress is termed thixotropic. We neglect nonnew¬ tonian effects in this book; see Ref. 16 for further study. There are many commercial devices for measuring the shear stress versus strain rate behavior of both newtonian and nonnewtonian fluids. They are generically called rheometers and have various designs: parallel disks, cone-plate, rotating coaxial cyl¬ inders, torsion, extensional, and capillary tubes. Reference 29 gives a good discussion. A popular device is the parallel-disk rheometer, shown in Fig. 1.9. A thin layer of fluid is placed between the disks, one of which rotates. The resisting torque on the 30 Chapter 1 Introduction rotating disk is proportional to the viscosity of the fluid. A simplified theory for this device is given in Example 1.10. Surface Tension A liquid, being unable to expand freely, will form an interface with a second liquid or gas. The physical chemistry of such interfacial surfaces is quite complex, and whole textbooks are devoted to this specialty . Molecules deep within the liquid repel each other because of their close packing. Molecules at the surface are less dense and attract each other. Since half of their neighbors are missing, the mechanical effect is that the surface is in tension. We can account adequately for surface effects in fluid mechanics with the concept of surface tension. If a cut of length dL is made in an interfacial surface, equal and opposite forces of magnitude Y dL are exposed normal to the cut and parallel to the surface, where Y is called the coefficient of surface tension. The dimensions of Y are {F/L}, with SI units of newtons per meter and BG units of pounds-force per foot. An alternate concept is to open up the cut to an area t/A; this requires work to be done of amount Y dA. Thus the coefficient Y can also be regarded as the surface energy per unit area of the interface, in N • m/m^ or ft • lbf/ft“. The two most common interfaces are water-air and mercury-air. For a clean sur¬ face at 20°C = 68°F, the measured surface tension is 0.0050 Ibf/ft = 0.073 N/m air-water 0.033 Ibf/ft = 0.48 N/m air-mercury (1.29) These are design values and can change considerably if the surface contains con¬ taminants like detergents or slicks. Generally Y decreases with liquid temperature and is zero at the critical point. Values of Y for water are given in Fig. 1.10 and Table A. 5. If the interface is curved, a mechanical balance shows that there is a pressure dif¬ ference across the interface, the pressure being higher on the concave side, as illus¬ trated in Fig. 1.11. In Fig. 1.1 la, the pressure increase in the interior of a liquid cylinder is balanced by two surface-tension forces: 2RL \p = 2YL or (1.30) 0.080 0.070 s z 0.060 Fig. 1.10 Surface tension of a clean air-water interface. Data from Table A.5. 0.050 0 10 20 30 40 50 60 70 80 90 100 T, °C 1.7 Viscosity and Other Secondary Properties 31 IRLAp ■kR^ Ap ApdA I-kRT Fig. 1.11 Pressure change across a curved interface due to surface tension: (a) interior of a liquid cylinder; (b) interior of a spherical droplet; (c) general curved interface. We are not considering the weight of the liquid in this calculation. In Fig. 1.1 Ih, the pressure increase in the interior of a spherical droplet balances a ring of surface- tension force: ttR^ Ap = IttRY or = T We can use this result to predict the pressure increase inside a soap buhhle, which has two interfaces with air, an inner and outer surface of nearly the same radius R\ _ 4Y ^Pbubble ^ ^Pdroplet ^ (1-32) Figure 1.11c shows the general case of an arbitrarily curved interface whose principal radii of curvature are Ri and /?2. A force balance normal to the surface will show that the pressure increase on the concave side is Ap = + ^2“‘) (1.33) Equations (1.30) to (1.32) can all be derived from this general relation; for example, in Eq. (1.30), Ri = R and R2 = °°. A second important surface effect is the contact angle 9, which appears when a liquid interface intersects with a solid surface, as in Eig. 1.12. The force balance would then involve both Y and 9. If the contact angle is less than 90°, the liquid is said to wet the solid; if 9 > 90°, the liquid is termed nonwetting. For example, water wets soap but does not wet wax. Water is extremely wetting to a clean glass surface, with 9 ~ 0°. Like Y, the contact angle 9 is sensitive to the actual physi¬ cochemical conditions of the solid-liquid interface. For a clean mercury-air-glass interface, 9 = 130°. 32 Chapter 1 Introduction Gas Liquid Fig. 1.12 Contact-angle effects at liquid-gas-solid interface. If 0 < 90°, the liquid “wets” the solid; if 6» > 90°, the liquid is nonwetting. Solid Example 1.8 illustrates how surface tension causes a fluid interface to rise or fall in a capillary tube. EXAMPLE 1.8 Derive an expression for the change in height h in a circular tube of a liquid with surface tension Y and contact angle 9, as in Fig. El. 8. Solution The vertical component of the ring surface-tension force at the interface in the tube must balance the weight of the column of fluid of height h: IttRY cos 9 = 'jTTR^h Solving for h, we have the desired result: h = 2Y cos 9 Ans. Thus the capillary height increases inversely with tube radius R and is positive if 0 < 90° (wetting liquid) and negative (capillary depression) if 9> 90°. Suppose that 1? = 1 mm. Then the capillary rise for a water-air-glass interface, 9 ~ 0°, Y = 0.073 N/m, and p = 1000 kg/m^ is h = 2(0.073 N/m) (cos 0°) ( 1000 kg/m^) (9.81 m/s^)(0.001 m) = 0.015 (N ■ s^)/kg = 0.015 m = 1.5 cm For a mercury-air-glass interface, with 9 = 130°, Y = 0.48 N/m, and p = 13,600 kg/m^, the capillary rise is h = 2(0.48)(cos 130°) 13,600(9.81)(0.001) = —0.0046 m = —0.46 cm When a small-diameter tube is used to make pressure measurements (Chap. 2), these capillary effects must be corrected for. 1.7 Viscosity and Other Secondary Properties 33 Vapor Pressure Vapor pressure is the pressure at which a liquid boils and is in equilibrium with its own vapor. For example, the vapor pressure of water at 68°F is 49 lbf/ft“, while that of mercury is only 0.0035 Ibf/ft^. If the liquid pressure is greater than the vapor pres¬ sure, the only exchange between liquid and vapor is evaporation at the interface. If, however, the liquid pressure falls below the vapor pressure, vapor bubbles begin to appear in the liquid. If water is heated to 212°F, its vapor pressure rises to 2116 lbf/ft“, and thus water at normal atmospheric pressure will boil. When the liquid pressure is dropped below the vapor pressure due to a flow phenomenon, we call the process cavitation. If water is accelerated from rest to about 50 ft/s, its pressure drops by about 15 Ibf/in^, or 1 atm. This can cause cavitation . The dimensionless parameter describing flow-induced boiling is the cavitation number Ca (1.34) where = ambient pressure Py = vapor pressure V = characteristic flow velocity p = fluid density Depending on the geometry, a given flow has a critical value of Ca below which the flow will begin to cavitate. Values of surface tension and vapor pressure of water are given in Table A.5. The vapor pressure of water is plotted in Fig. 1.13. Figure 1.14a shows cavitation bubbles being formed on the low-pressure surfaces of a marine propeller. When these bubbles move into a higher-pressure region, they collapse implosively. Cavitation collapse can rapidly spall and erode metallic surfaces and eventually destroy them, as shown in Fig. .\Ab. Fig. 1.13 Vapor pressure of water. Data from Table A. 5. too 80- 60- 40 ■ 20- 20 40 60 80 T, °C too 34 Chapter 1 Introduction Fig. 1.14 Two aspects of cavitation bubble formation in liquid flows: (a) Beauty: spiral bubble sheets form from the surface of a marine propeller (courtesy of the Garfield Thomas Water Tunnel, Pennsylvania State University); (b) ugliness: collapsing bubbles erode a propeller surface ( courtesy of Thomas T. Huang, David Taylor Research Center). 1.7 Viscosity and Other Secondary Properties 35 No-Slip and No-Temperature- Jump Conditions EXAMPLE 1.9 A certain torpedo, moving in fresh water at 10°C, has a minimum-pressure point given hy the formula Pmin = Po - 0.35 (1) where po = 115 kPa, p is the water density, and V is the torpedo velocity. Estimate the velocity at which cavitation bubbles will form on the torpedo. The constant 0.35 is dimensionless. Solution • Assumption: Cavitation bubbles form when the minimum pressure equals the vapor pressure p„. • Approach: Solve Eq. (1) above, which is related to the Bernoulli equation from Example 1.3, for the velocity when p-^^ = Use SI units (m, N, kg, s). • Property values: At 10°C, read Table A.l for p = 1000 kg/m^ and Table A. 5 for Pv = 1.227 kPa. • Solution steps: Insert the known data into Eq. (1) and solve for the velocity, using SI units: Pmin = Pv = 1227 Pa = 115,000 Pa - 0.35^1000 with V in m/s , (115,000 - 1227) m^ , _ Solve ^ = 325 -p or P = V325 « IS.Om/s Ans. 0.35(1000) • Comments: Note that the use of SI units requires no conversion factors, as discussed in Example 1.3P. Pressures must be entered in pascals, not kilopascals. When a fluid flow is bounded by a solid surface, molecular interactions cause the fluid in contact with the surface to seek momentum and energy equilibrium with that surface. All liquids essentially are in equilibrium with the surfaces they contact. All gases are, too, except under the most rarefied conditions . Excluding rarefied gases, then, all fluids at a point of contact with a solid take on the velocity and tem¬ perature of that surface: '^uid = ^wall (1.35) These are called the no-slip and no-temperature-jump conditions, respectively. They serve as boundary conditions for analysis of fluid flow past a solid surface. Figure 1.15 illustrates the no-slip condition for water flow past the top and bottom surfaces of a fixed thin plate. The flow past the upper surface is disorderly, or turbulent, while the lower surface flow is smooth, or laminar.® In both cases there is clearly no slip at the wall, where the water takes on the zero velocity of the fixed plate. The velocity pro¬ file is made visible by the discharge of a line of hydrogen bubbles from the wire shown stretched across the flow. ^Laminar and turbulent flows are studied in Chaps. 6 and 7. 36 Chapter 1 Introduction Fig. 1.15 The no-slip condition in water flow past a thin fixed plate. The upper flow is turbulent; the lower flow is laminar. The velocity profile is made visible by a line of hydrogen bubbles discharged from the wire across the flow. (National Committee for Fluid Mechanics Films, Education Development Center, Inc., © 1972.) To decrease the mathematical difficulty, the no-slip condition is partially relaxed in the analysis of inviscid flow (Chap. 8). The flow is allowed to “slip” past the surface but not to permeate through the surface ^normal ^normal (1.36) while the tangential velocity V, is allowed to be independent of the wall. The analysis is much simpler, but the flow patterns are highly idealized. For high-viscosity newtonian fluids, the linear velocity assumption and the no-slip conditions can yield some sophisticated approximate analyses for two- and three- dimensional viscous flows. The following example, for a type of rotating-disk vis¬ cometer, will illustrate. EXAMPLE 1.10 An oil film of viscosity fi and thickness h ^ R lies between a solid wall and a circular disk, as in Fig. ELIO. The disk is rotated steadily at angular velocity fl. Noting that both velocity and shear stress vary with radius r, derive a formula for the torque M required to rotate the disk. Neglect air drag. Solution System sketch: Figure ELIO shows a side view (a) and a top view (b) of the system. 1.7 Viscosity and Other Secondary Properties 37 Slip Flow in Gases El.lO • Assumptions: Linear velocity profile, laminar flow, no-slip, local shear stress given by Eq. (1.23). • Approach: Estimate the shear stress on a circular strip of width dr and area dA = litr dr in Eig. El.lOfe, then find the moment dM about the origin caused by this shear stress. Integrate over the entire disk to find the total moment M. • Property values: Constant oil viscosity fi. In this steady flow, oil density is not relevant. • Solution steps: At radius r, the velocity in the oil is tangential, varying from zero at the fixed wall (no-slip) to m = fir at the disk surface (also no-slip). The shear stress at this position is thus du fir This shear stress is everywhere perpendicular to the radius from the origin (see Eig. E1.10f>). Then the total moment about the disk origin, caused by shearing this circular strip, can be found and integrated: / iiflr\ dM = {T){dA)r = ( ^ j(27rrdr)r, M = 0 ■KptflPf' Ih Ans. ■ Comments: This is a simplified engineering analysis, which neglects possible edge effects, air drag on the top of the disk, and the turbulence that might ensue if the disk rotates too fast. The “free slip” boundary condition, Eq. (1.36), is an unrealistic mathematical artifice to enable inviscid-flow solutions. However, actual, realistic wall slip occurs in rarefied gases, where there are too few molecules to establish momentum equilibrium with the wall. In 1879, the physicist James Clerk Maxwell used the kinetic theory of gases to predict a slip velocity at the wall: ^“wall ~ T Iwall (1-37) dy where L is the mean free path of the gas, and u and x are along the wall. If L is very small compared to the lateral scale L of the flow, the Knudsen number, Kn = L/L, is 38 Chapter 1 Introduction small, and the slip velocity is near zero. We will assign a few slip problems, but the details of rarebed gas flow are left for further reading in Refs. 18 and 52. Speed of Sound In gas flow, one must be aware of compressibility effects (significant density changes caused by the flow). We shall see in Sec. 4.2 and in Chap. 9 that compressibility becomes important when the flow velocity reaches a significant fraction of the speed of sound of the fluid. The speed of sound a of a fluid is the rate of propagation of small-disturbance pressure pulses (“sound waves”) through the fluid. In Chap. 9 we shall show, from momentum and thermodynamic arguments, that the speed of sound is defined by a pressure-density derivative proportional to the isentropic bulk modulus'. 2 a = /3 where /3 = isentropic bulk modulus = p( = k\ /dp [dp k = — T This is true for either a liquid or a gas, but it is for gases that the problem of com¬ pressibility occurs. For an ideal gas, Eq. (1.10), we obtain the simple formula n1/2 ^ideal gas {kRT) (1.38) where R is the gas constant, Eq. (1.11), and T the absolute temperature. For example, for air at 20°C, a = {(1.40)287 m^/(s^ • K)}'^ « 343 m/s (1126 ft/s = 768 mi/h). If, in this case, the air velocity reaches a significant fraction of a, say, 100 m/s, then we must account for compressibility effects (Chap. 9). Another way to state this is to account for compressibility when the Mach number Ma = Via of the flow reaches about 0.3. The speed of sound of water is tabulated in Table A.5. For near perfect gases, like air, the speed of sound is simply calculated by Eq. (1.38). Many liquids have their bulk modulus listed in Table A.3. Note, however, as discussed in Ref. 51, even a very small amount of dissolved gas in a liquid can reduce the mixture speed of sound by up to 80 percent. EXAMPLE 1.11 A commercial airplane flies at 540 mi/h at a standard altitude of 30,000 ft. What is its Mach number? Solution • Approach: Find the “standard” speed of sound; divide it into the velocity, using proper units. • Property values: From Table A.6, at 30,000 ft (9144 m), a ~ 303 m/s. Check this against the standard temperature, estimated from the table to be 229 K. From Eq. (1.38) for air, a = [kR^,Tf'^ = [1.4(287) (229)]'^ « 303 m/s. • Solution steps: Convert the airplane velocity to m/s: V = (540mi/h)[0.44704m/s/(mi/h)] « 241 m/s. 1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines 39 1.8 Basic Flow Analysis Techniques 1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines Then the Mach number is given by Ma = Via = (241 m/s)/(303 m/s) = 0.80 Ans. • Comments: This value, Ma = 0.80, is typical of present-day commercial airliners. There are three basic ways to attack a fluid flow problem. They are equally important for a student learning the subject, and this book tries to give adequate coverage to each method: Control-volume, or integral analysis (Chap. 3). Infinitesimal system, or differential analysis (Chap. 4). Experimental study, or dimensional analysis (Chap. 5). In all cases, the flow must satisfy the three basic laws of mechanics plus a thermo¬ dynamic state relation and associated boundary conditions: Conservation of mass (continuity). Linear momentum (Newton’s second law). First law of thermodynamics (conservation of energy). A state relation like p = p(p, T). Appropriate boundary conditions at solid surfaces, interfaces, inlets, and exits. In integral and differential analyses, these five relations are modeled mathemati¬ cally and solved by computational methods. In an experimental study, the fluid itself performs this task without the use of any mathematics. In other words, these laws are believed to be fundamental to physics, and no fluid flow is known to violate them. Fluid mechanics is a highly visual subject. The patterns of flow can be visualized in a dozen different ways, and you can view these sketches or photographs and learn a great deal qualitatively and often quantitatively about the flow. Four basic types of line patterns are used to visualize flows: 1 . A streamline is a line everywhere tangent to the velocity vector at a given instant. A pathline is the actual path traversed by a given fluid particle. A streakline is the locus of particles that have earlier passed through a prescribed point. A timeline is a set of fluid particles that form a line at a given instant. The streamline is convenient to calculate mathematically, while the other three are easier to generate experimentally. Note that a streamline and a timeline are instanta¬ neous lines, while the pathline and the streakline are generated by the passage of time. The velocity profile shown in Fig. 1.15 is really a timeline generated earlier by a single discharge of bubbles from the wire. A pathline can be found by a time exposure of a single marked particle moving through the flow. Streamlines are difficult to 40 Chapter 1 Introduction Fig. 1.16 The most common method of flow-pattern presentation: {a) Streamlines are everywhere tangent to the local velocity vector; (&) a streamtube is formed by a closed collection of streamlines. generate experimentally in unsteady flow unless one marks a great many particles and notes their direction of motion during a very short time interval . In steady flow, where velocity varies only with position, the situation simplifies greatly: Streamlines, pathlines, and streaklines are identical in steady flow. In fluid mechanics the most common mathematical result for visualization purposes is the streamline pattern. Figure 1.16a shows a typical set of streamlines, and Fig. 1.16b shows a closed pattern called a streamtube. By definition the fluid within a streamtube is confined there because it cannot cross the streamlines; thus the streamtube walls need not be solid but may be fluid surfaces. Figure 1.17 shows an arbitrary velocity vector. If the elemental arc length dr of a streamline is to be parallel to V, their respective components must be in proportion: Streamline: dx dy dz dr u V w V (1.39) Fig. 1.17 Geometric relations for defining a streamline. z 1.9 Flow Patterns: Streamlines, Streaklines, and Pathlines 41 Flow Visualization If the velocities (m, v, w) are known functions of position and time, Eq. (1.39) can be integrated to find the streamline passing through the initial point (xq, yo> Zo, ffl)- The method is straightforward for steady flows but may be laborious for unsteady flow. Clever experimentation can produce revealing images of a fluid flow pattern, as shown earlier in Figs. 1.14a and 1.15. For example, streaklines are produced by the continu¬ ous release of marked particles (dye, smoke, or bubbles) from a given point. If the flow is steady, the streaklines will be identical to the streamlines and pathlines of the flow. Some mefhods of flow visualizafion include the following [34—36]: Dye, smoke, or bubble discharges. Surface powder or flakes on liquid flows. Floating or neutral-density particles. Optical techniques that detect density changes in gas flows: shadowgraph, schlieren, and inferferomefer. Tufts of yarn attached to boundary surfaces. Evaporative coatings on boundary surfaces. Luminescent fluids, additives, or bioluminescence. Particle image velocimetry (PIV). Figures 1.14a and 1.15 were both visualized by bubble releases. Another example is the use of particles in Fig. 1.18 to visualize a flow negotiating a 180° turn in a ser¬ pentine channel . Figure 1.18a is at a low, laminar Reynolds number of 1000. The flow is steady, and the particles form streaklines showing that the flow cannof make the sharp turn without separating away from the bottom wall. Figure 1.1 87> is at a higher, turbulent Reynolds number of 30,000. The flow is unsteady, and the streaklines would be chaotic and smeared, unsuitable for visualiza¬ tion. The image is thus produced by the new technique of particle image velocime¬ try . In PIV, hundreds of particles are tagged and photographed at two closely spaced times. Particle movements thus indicate local velocity vectors. These hundreds of vectors are then smoothed by repeated computer operations until the time-mean flow pattern in Fig. 1.1 8fo is achieved. Modern flow experiments and numerical mod¬ els use computers extensively to create their visualizations, as described in the text by Yang . Mathematical details of streamline/streakline/pathline analysis are given in Ref. 33. References 39^1 are beautiful albums of flow photographs. References 34-36 are monographs on flow visualization techniques. Fluid mechanics is a marvelous subject for visualization, not just for still (steady) patterns, but also for moving (unsteady) motion studies. An outstanding list of avail¬ able flow movies and videotapes is given by Carr and Young . 42 Chapter 1 Introduction Fig. 1.18 Two visualizations of flow making a 1 80° turn in a serpentine channel: (a) particle streaklines at a Reynolds number of 1000; (b) time- mean particle image velocimetry (PIV) at a turbulent Reynolds number of 30,000. (From Ref. 42, by permission of the American Society of Mechanical Engineers.) 1.11 The Histoiy of Fluid Mechanics 43 1.10 The Fundamentals of Engineering (FE) Examination 1.11 The History of Fluid Mechanics The road toward a professional engineer’s license has a first stop, the Fundamentals of Engineering Examination, known as the EE exam. It was formerly known as the Engineer-in-Training (E-I-T) Examination. This eight-hour national test will prob¬ ably soon be required of all engineering graduates, not just for licensure, but as a student assessment tool. The 120-problem, four-hour morning session covers many general studies: Mathematics — 1 5 % Engineering probability and statistics — 7% Chemistry — 9% Computers — 7 % Ethics and business practices — 7% Engineering economics — 8% Engineering mechanics — 10% Strength of materials — 7% Material properties — 7% Fluid mechanics — 7% Electricity and magnetism — 9% Thermodynamics — 7 % For the 60-problem, four-hour afternoon session you may choose one of seven mod¬ ules: chemical, civil, electrical, environmental, industrial, mechanical, and other/ general engineering. Note that fluid mechanics is an integral topic of the examination. Therefore, for practice, this text includes a number of end-of-chapter FE problems where appropriate. The format for the FE exam questions is multiple-choice, usually with five selec¬ tions, chosen carefully to tempt you with plausible answers if you used incorrect units, forgot to double or halve something, are missing a factor of tt, or the like. In some cases, the selections are unintentionally ambiguous, such as the following example from a previous exam: Transition from laminar to turbulent flow occurs at a Reynolds number of (A) 900 (B) 1200 (C) 1500 (D) 2100 (E) 3000 The “correct” answer was graded as (D), Re = 2100. Clearly the examiner was think¬ ing, but forgot to specify, Re^ for flow in a smooth circular pipe, since (see Chaps. 6 and 7) transition is highly dependent on geometry, surface roughness, and the length scale used in the definition of Re. The moral is not to get peevish about the exam but simply to go with the flow (pun intended) and decide which answer best fits an undergraduate training situation. Every effort has been made to keep the FE exam questions in this text unambiguous. Many distinguished workers have contributed to the development of fluid mechanics. If you are a sfudent, however, fhis is probably not the time to be studying history. Later, during your career, you will enjoy reading about the history of, not just fluid mechanics, but all of science. Here are some names that will be mentioned as we encounter their contributions in the rest of this book. 44 Chapter 1 Introduction Name Important Contribution Archimedes (285-212 BC) Leonardo da Vinci (1452-1519) Isaac Newton (1642-1727) Leonhard Euler (1707-1783) L. M. H. Navier (1785-1836) Jean Louis Poiseuille (1799-1869) Osborne Reynolds (1842-1912) Ludwig Prandtl (1875-1953) Theodore von Karman (1881-1963) Established laws of buoyancy and floating bodies. Formulated the first equation of continuity. Postulated the law of linear viscous stresses. Developed Bernoulli’s equation by solving the basic equations. Formulated the basic differential equations of viscous flow. Performed first experiments on laminar flow in tubes. Explained the phenomenon of transition to turbulence. Formulated boundary layer theory, predicting flow separation. Major advances in aerodynamics and turbulence theory. References through provide a comprehensive treatment of the history of fluid mechanics. Summary This chapter has discussed the behavior of a fluid — which, unlike a solid, must move if subjected to a shear stress — and the important fluid properties. The writer believes the most important property to be the velocity vector held Y{x, y, z, t). Following closely are the pressure p, density p, and temperature T. Many secondary properties enter into various flow problems: viscosity p, thermal conductivity k, speciflc weight 7, surface tension Y, speed of sound a, and vapor pressure p^. You must learn to locate and use all these properties to become proflcient in fluid mechanics. There was a brief discussion of the five different kinds of mathematical relations we will use to solve flow problems — mass conservation, linear momentum, first law of thermodynamics, equations of state, and appropriate boundary conditions at walls and other boundaries. Flow patterns are also discussed briefly. The most popular, and useful, scheme is to plot the field of streamlines, that is, lines everywhere parallel to the local velocity vector. Since the earth is 75 percent covered with water and 100 percent covered with air, the scope of fluid mechanics is vast and touches nearly every human endeavor. The sciences of meteorology, physical oceanography, and hydrology are concerned with naturally occurring fluid flows, as are medical studies of breathing and blood circulation. All transportation problems involve fluid motion, with well-developed specialties in aerodynamics of aircraft and rockets and in naval hydrodynamics of ships and submarines. Almost all our electric energy is developed either from water flow, air flow through wind turbines, or from steam flow through turbine generators. All combustion problems involve fluid motion as do the more classic problems of irrigation, flood control, water supply, sewage disposal, projectile motion, and oil and gas pipelines. The aim of this book is to present enough fundamental concepts and practical applications in fluid mechanics to prepare you to move smoothly into any of these specialized fields of the science of flow — and then be prepared to move out again as new technologies develop. Problems 45 Problems Dimensions and units Most of the problems herein are fairly straightforward. More diffi¬ cult or open-ended assignments are labeled with an asterisk as in Prob. 1.18. Problems labeled with a computer icon may require the use of a computer. The standard end-of-chapter problems 1 . 1 to 1.86 (categorized in the problem list below) are followed by funda¬ mentals of engineering (FE) exam problems FEl.l to EE 1.10 and comprehensive problems Cl.l to Cl. 12. Problem Distribution Section Topic Problems 1.1, 1.4, 1.5 Fluid continuum concept 1.1-1 .4 1.6 Dimensions and units 1.5-1.23 1.8 Thermodynamic properties 1.24-1.37 1.9 Viscosity, no-slip condition 1.38-1.61 1.9 Surface tension 1.62-1.71 1.9 Vapor pressure; cavitation 1.72-1.74 1.9 Speed of sound, Mach number 1.75-1.80 1.11 Streamlines 1.81-1.83 1.2 History of fluid mechanics 1.84-1.85a-n 1.13 Experimental uncertainty 1.86-1.90 The concept of a fluid Pl.l A gas at 20°C may be considered rarefied, deviating from the continuum concept, when it contains less than lO’^ molecules per cubic millimeter. If Avogadro’s number is 6.023 E23 molecules per mole, what absolute pressure (in Pa) for air does this represent? P1.2 Table A. 6 lists the density of the standard atmosphere as a function of altitude. Use these values to estimate, crudely — say, within a factor of 2 — the number of molecules of air in the entire atmosphere of the earth. P1.3 For the triangular element in Fig. PI. 3, show that a tilted free liquid surface, in contact with an atmosphere at pres¬ sure must undergo shear stress and hence begin to flow. Hint: Account for the weight of the fluid and show that a no-shear condition will cause horizontal forces to be out of balance. P1.3 P1.4 Sand, and other granular materials, appear io flow; that is, you can pour them from a container or a hopper. There are whole textbooks on the “transport” of granular materials . Therefore, is sand a fluids Explain. Fluid density p P1.5 P1.6 The mean free path of a gas, I, is defined as the average distance traveled by molecules between collisions. A pro¬ posed formula for estimating I of an ideal gas is / = 1.26 fJ- pVRT What are the dimensions of the constant 1.26? Use the for¬ mula to estimate the mean free path of air at 20°C and 7 kPa. Would you consider air rarefied at this condition? Henri Darcy, a French engineer, proposed that the pressure drop Ap for flow at velocity V through a tube of length L could be correlated in the form Ap = aLW If Darcy’s formulation is consistent, what are the dimen¬ sions of the coefficient a? P1.7 Convert the following inappropriate quantities into SI units: (a) 2.283 E7 U.S. gallons per day; fb) 4.5 furlongs per minute (racehorse speed); and (c) 72,800 avoirdupois ounces per acre. P1.8 Suppose we know little about the strength of materials but are told that the bending stress cr in a beam is proportional to the beam half-thickness y and also depends on the bending moment M and the beam area moment of inertia I. We also learn that, for the particular case M = 2900 in ■ Ibf, y = 1.5 in, and I = 0.4 in', the predicted stress is 75 MPa. Using this information and dimensional rea¬ soning only, find, to three significant figures, the only possible dimensionally homogeneous formula cr = yf{M,I). P1.9 A hemispherical container, 26 inches in diameter, is filled with a liquid at 20°C and weighed. The liquid weight is found to be 1617 ounces, (a) What is the density of the fluid, in kg/m^? {b) What fluid might this be? Assume stan¬ dard gravity, g = 9.807 m/s^. PI. 10 The Stokes-Oseen formula for drag force F on a sphere of diameter D in a fluid stream of low velocity V, density p, and viscosity p, is Q'jY F = SirpDV + —pV^D^ 16 Is this formula dimensionally homogeneous? PI. 11 In English Engineering units, the specific heat Cp of air at room temperature is approximately 0.24 Btu/(lbm-°F). When working with kinetic energy relations, it is more ap¬ propriate to express Cp as a velocity-squared per absolute degree. Give the numerical value, in this form, of Cp for air in (a) SI units, and (b) BG units. 46 Chapter 1 Introduction PI. 12 For low-speed (laminar) steady flow through a circular pipe, as shown in Fig. PI. 12, the velocity u varies with radius and takes the form Ap , , u = B - (ro - r) M where p, is the fluid viscosity and Ap is the pressure drop from entrance to exit. What are the dimensions of the constant 5? Pipe wall r PI. 15 The height H that fluid rises in a liquid barometer tube depends upon the liquid density p, the barometric pressure p, and the acceleration of gravity g. (a) Arrange these four variables into a single dimensionless group, (b) Can you deduce (or guess) the numerical value of your group? PI. 16 Algebraic equations such as Bernoulli’s relation, Eq. (1) of Example 1.3, are dimensionally consistent, but what about differential equations? Consider, for example, the boundary-layer x-momentum equation, first derived by Ludwig Prandtl in 1904: du du + pv— dy dx dp Pgx + c).X dT dy where t is the boundary-layer shear stress and g^ is the com¬ ponent of gravity in the x direction. Is this equation dimen¬ sionally consistent? Can you draw a general conclusion? PI. 17 The Hazen-Williams hydraulics formula for volume rate of flow Q through a pipe of diameter D and length L is given by P1.12 PI. 13 The efficiency p of a pump is defined as the (dimension¬ less) ratio of the power developed by the flow to the power required to drive the pump: P1.14 eAp V = : - ^ - input power where Q is the volume rate of flow and Ap is the pressure rise produced by the pump. Suppose that a certain pump develops a pressure rise of 35 Ibf/in^ when its flow rate is 40 L/s. If the input power is 16 hp, what is the efficiency? Eigure PI. 14 shows the flow of water over a dam. The volume flow Q is known to depend only on crest width B, acceleration of gravity g, and upstream water height H above the dam crest. It is further known that Q is propor¬ tional to B. What is the form of the only possible dimen¬ sionally homogeneous relation for this flow rate? P1.18 P1.19 P1.20 P1.21 where Ap is the pressure drop required to drive the flow. What are the dimensions of the constant 61.9? Can this formula be used with confidence for various liquids and gases? Eor small particles at low velocities, the first term in the Stokes-Oseen drag law, Prob. 1.10, is dominant; hence, F ~ KV, where A" is a constant. Suppose a particle of mass m is constrained to move horizontally from the initial posi¬ tion X = 0 with initial velocity Vq- Show (a) that its velocity will decrease exponentially with time and (b) that it will stop after traveling a distance x = mV^IK. In his study of the circular hydraulic jump formed by a faucet flowing into a sink, Watson proposed a param¬ eter combining volume flow rate Q, density p, and viscos¬ ity fi of the fluid, and depth h of the water in the sink. He claims that his grouping is dimensionless, with Q in the numerator. Can you verify this? Books on porous media and atomization claim that the vis¬ cosity fi and surface tension Y of a fluid can be combined with a characteristic velocity U to form an important di¬ mensionless parameter, (a) Verify that this is so. (b) Evalu¬ ate this parameter for water at 20°C and a velocity of 3.5 cm/s. Note: You get extra credit if you know the name of this parameter. Aeronautical engineers measure the pitching moment Mq of a wing and then write it in the following form for use in other cases: P1.14 Mo = pV^ACp Problems 47 where V is the wing velocity, A the wing area, C the wing chord length, and p the air density. What are the dimen¬ sions of the coefficient fil P1.22 The Ekman number, Ek, arises in geophysical fluid dynamics. It is a dimensionless parameter combining sea¬ water density p, a characteristic length L, seawater viscosity p, and the Coriolis frequency fl sim/J, where fi is the rotation rate of the earth and ip is the latitude angle. Determine the correct form of Ek if the viscosity is in the numerator. P1.23 During World War II, Sir Geoffrey Taylor, a British fluid dynamicist, used dimensional analysis to estimate the energy released by an atomic bomb explosion. He assumed that the energy released E, was a function of blast wave radius R, air density p, and time t. Arrange these variables into a single dimensionless group, which we may term the blast wave number. Thermodynamic properties P1.24 Air, assumed to be an ideal gas with k = 1.40, flows isen- tropically through a nozzle. At section 1, conditions are sea level standard (see Table A.6). At section 2, the tempera¬ ture is — 50°C. Estimate (a) the pressure, and (b) the den¬ sity of the air at section 2. P1.25 On a summer day in Narragansett, Rhode Island, the air tem¬ perature is 74°E and the barometric pressure is 14.5 Ibf/in^. Estimate the air density in kg/m^. P1.26 When we in the United States say a car’s tire is filled “to 32 lb,” we mean that its internal pressure is 32 Ibf/in^ above the ambient atmosphere. If the tire is at sea level, has a volume of 3.0 ft^, and is at 75°E, estimate the total weight of air, in Ibf, inside the tire. P1.27 Eor steam at a pressure of 45 atm, some values of tempera¬ ture and specific volume are as follows, from Ref. 23: T, “F 500 600 700 800 900 V, ft^/lbm 0.7014 0.8464 0.9653 1.074 1.177 Eind an average value of the predicted gas constant R in mV(s^ • K). Does this data reasonably approximate an ideal gas? If not, explain. P1.28 Wet atmospheric air at 100 percent relative humidity con¬ tains saturated water vapor and, by Dalton’s law of partial pressures, Patm Pdry air Pwaler vapor Suppose this wet atmosphere is at 40°C and 1 atm. Calcu¬ late the density of this 100 percent humid air, and compare it with the density of dry air at the same conditions. P1.29 A compressed-air tank holds 5 ft^ of air at 120 Ibf/in^ “gage,” that is, above atmospheric pressure. Estimate the energy, in ft-lbf, required to compress this air from the atmosphere, assuming an ideal isothermal process. P1.30 Repeat Prob. 1.29 if the tank is filled with compressed water instead of air. Why is the result thousands of times less than the result of 215,000 ft ■ Ibf in Prob. 1.29? P1.31 One cubic foot of argon gas at 10°C and 1 atm is com¬ pressed isentropically to a pressure of 600 kPa. (a) What will be its new pressure and temperature? (b) If it is allowed to cool at this new volume back to 10°C, what will be the final pressure? P1.32 A blimp is approximated by a prolate spheroid 90 m long and 30 m in diameter. Estimate the weight of 20°C gas within the blimp for (a) helium at 1.1 atm and (b) air at 1.0 atm. What might the dijference between these two values represent (see Chap. 2)? P1.33 A tank contains 9 kg of CO2 at 20°C and 2.0 MPa. Estimate the volume of the tank, in m^. P1.34 Consider steam at the following state near the saturation line: (pi, Tj) = (1.31 MPa, 290°C). Calculate and com¬ pare, for an ideal gas (Table A.4) and the steam tables (a) the density pi and {b) the density P2 if the steam ex¬ pands isentropically to a new pressure of 414 kPa. Discuss your results. P1.35 In Table A.4, most common gases (air, nitrogen, oxygen, hydrogen) have a specific heat ratio k ~ 1.40. Why do argon and helium have such high values? Why does NH3 have such a low value? What is the lowest k for any gas that you know of? P1.36 Experimental data for the density of n-pentane liquid for high pressures, at 50°C, are listed as follows: Pressure, kPa 100 10,230 20,700 34,310 Density, kg/m^ 586.3 604.1 617.8 632.8 (a) Fit this data to reasonably accurate values of B and n from Eq. (1.19). (b) Evaluate p at 30 MPa. P1.37 A near-ideal gas has a molecular weight of 44 and a spe¬ cific heat c„ = 610 J/(kg ■ K). What are (a) its specific heat ratio, k, and {b) its speed of sound at 100°C? Viscosity, no-slip condition P1.38 In Fig. 1.7, if the fluid is glycerin at 20°C and the width between plates is 6 mm, what shear stress (in Pa) is required to move the upper plate at 5.5 m/s? What is the Reynolds number if L is taken to be the distance between plates? P1.39 Knowing p for air at 20°C from Table 1.4, estimate its viscosity at 500°C by (a) the power law and (jb) the Suther¬ land law. Also make an estimate from (c) Fig. 1.6. Com¬ pare with the accepted value of p ~ 3.58 E-5 kg/m ■ s. 48 Chapter 1 Introduction P1.40 Glycerin at 20°C fills the space between a hollow sleeve of diameter 12 cm and a fixed coaxial solid rod of diameter 11.8 cm. The outer sleeve is rotated at 120 rev/min. Assuming no temperature change, estimate the torque re¬ quired, in N ■ m per meter of rod length, to hold the inner rod fixed. P1.41 An aluminum cylinder weighing 30 N, 6 cm in diameter and 40 cm long, is falling concentrically through a long vertical sleeve of diameter 6.04 cm. The clearance is filled with SAE 50 oil at 20°C. Estimate the terminal (zero acceleration) fall velocity. Neglect air drag and assume a linear velocity distribution in the oil. Hint: You are given diameters, not radii. P1.42 Helium at 20°C has a viscosity of 1.97 E-5 k;g/(m ■ s). Use the data of Table A. 4 to estimate the temperature, in °C, at which helium’s viscosity will double. P1.43 Eor the flow of gas between two parallel plates of Eig. 1.7, reanalyze for the case of slip flow at both walls. Use the simple slip condition, Su^^n = i {du/dy)^.^\, where ^ is the mean free path of the fluid. Sketch the expected veloc¬ ity profile and find an expression for the shear stress at each wall. P1.44 One type of viscometer is simply a long capillary tube. A commercial device is shown in Prob. Cl. 10. One measures the volume flow rate Q and the pressure drop Ap and, of course, the radius and length of the tube. The theoreti¬ cal formula, which will be discussed in Chap. 6, is Ap ~ SpQL/iirR). For a capillary of diameter 4 mm and length 10 inches, the test fluid flows at 0.9 m^/h when the pressure drop is 58 VoU'vc?. Find the predicted viscosity in kg/m ■ s. P1.45 A block of weight VT slides down an inclined plane while lubricated by a thin film of oil, as in Fig. P1.45. The film contact area is A and its thickness is h. Assuming a linear velocity distribution in the film, derive an expression for the “terminal” (zero-acceleration) velocity V of the block. Find the terminal velocity of the block if the block mass is 6 kg, A = 35 cm^, 6 = 15°, and the film is 1-mm-thick SAE 30 oil at 20°C. Liquid film of P1.46 A simple and popular model for two nonnewtonian fluids in Fig. 1.8a is the power-law. where C and n are constants fit to the fluid . From Fig. 1.8a, deduce the values of the exponent n for which the fluid is (a) newtonian, {b) dilatant, and (c) pseudoplas¬ tic. Consider the specific model constant C = 0.4 N ■ sYm^, with the fluid being sheared between two parallel plates as in Fig. 1.7. If the shear stress in the fluid is 1200 Pa, find the velocity V of the upper plate for the cases (d) n = 1.0, (e) n = 1.2, and (f ) n = 0.8. P1.47 Data for the apparent viscosity of average human blood, at normal body temperature of 37°C, varies with shear strain rate, as shown in the following table. Strain rate, s 1 10 100 1000 Apparent viscosity, 0.011 0.009 0.006 0.004 kg/(m ■ s) (a) Is blood a nonnewtonian fluid? (b) If so, what type of fluid is it? (c) How do these viscosities compare with plain water at 37°C? P1.48 A thin plate is separated from two fixed plates by very viscous liquids pi and (12, respectively, as in Fig. PI. 48. The plate spacings hi and h2 are unequal, as shown. The contact area is A between the center plate and each fluid, (a) Assuming a linear velocity distribution in each fluid, de¬ rive the force F required to pull the plate at velocity V. (b) Is there a necessary relation between the two viscosities, pi and /i2? hi Pi t h p2 P1.48 P1.49 An amazing number of commercial and laboratory devices have been developed to measure fluid viscosity, as described in Refs. 29 and 49. Consider a concentric shaft, fixed axially and rotated inside the sleeve. Let the inner and outer cylinders have radii r, and r„, respectively, with total sleeve length L. Let the rotational rate be Fl (rad/s) and the P1.45 Problems 49 applied torque be M. Using these parameters, derive a theo¬ retical relation for the viscosity /i of the fluid between the cylinders. P1.50 A simple viscometer measures the time t for a solid sphere to fall a distance L through a test fluid of density p. The fluid viscosity p, is then given by where D is the sphere diameter and lV„et is the sphere net weight in the fluid, {a) Prove that both of these formulas are dimensionally homogeneous, {b) Suppose that a 2.5 mm diameter aluminum sphere (density 2700 kg/m^) falls in an oil of density 875 kg/m^. If the time to fall 50 cm is 32 s, estimate the oil viscosity and verify that the inequality is valid. P1.51 An approximation for the boundary-layer shape in Figs. 1.51? and PI. 51 is the formula u(y) ~ U 0 < y < where U is the stream velocity far from the wall and <5 is the boundary layer thickness, as in Fig. P1.51. If the fluid is helium at 20°C and 1 atm, and if ?7 = 10.8 m/s and <5=3 cm, use the formula to (a) estimate the wall shear stress in Pa, and (b) find the position in the boundary layer where r is one-half of T„. y P1.51 P1.52 The belt in Fig. PI. 52 moves at a steady velocity V and skims the top of a tank of oil of viscosity p, as shown. Assuming a linear velocity profile in the oil, develop a simple formula for the required belt-drive power P as a function of (h, L, V, b, p). What belt-drive power P, in watts, is required if the belt moves at 2.5 m/s over SAE 30W oil at 20°C, with L = 2 m, = 60 cm, and h = 3 cm? 1 Q Moving belt, width b D t Oil, depth h P1.52 A solid cone of angle 29, base rg, and density is rotating with initial angular velocity lOq inside a conical seat, as shown in Fig. PI. 53. The clearance h is filled with oil of viscosity p. Neglecting air drag, derive an analytical expression for the cone’s angular velocity a;(r) if there is no applied torque. Base Cj_^ ^ (0 P1.53 P1.54 A disk of radius R rotates at an angular velocity U inside a disk-shaped container filled with oil of viscosity p, as shown in Fig. PI .54. Assuming a linear velocity profile and neglecting shear stress on the outer disk edges, derive a formula for the viscous torque on the disk. P1.54 P1.55 A block of weight W is being pulled over a table by another weight Wa, as shown in Fig. P1.55. Find an algebraic for¬ mula for the steady velocity U of the block if it slides on an 37rDL if t> 2pDL P1.53 50 Chapter 1 Introduction oil film of thickness h and viscosity fi. The block bottom area A is in contact with the oil. Neglect the cord weight and the pulley friction. Assume a linear velocity profile in the oil film. P1.56 The device in Fig. PI. 56 is called a cone-plate viscometer . The angle of the cone is very small, so that sin 0—0, and the gap is filled with the test liquid. The torque M to rotate the cone at a rate fl is measured. Assuming a linear velocity profile in the fluid film, derive an expression for fluid viscosity /i as a function of (M, R, ft, 9). P1.56 P1.57 Extend the steady flow between a fixed lower plate and a moving upper plate, from Fig. 1.7, to the case of two immiscible liquids between the plates, as in Fig. PI. 57. P1.58 The laminar pipe flow example of Prob. 1.12 can be used to design a capillary viscometer . If Q is the volume flow rate, L is the pipe length, and Ap is the pressure drop from entrance to exit, the theory of Chap. 6 yields a formula for viscosity: TTroAp Pipe end effects are neglected . Suppose our capillary has ro = 2 mm and L = 25 cm. The following flow rate and pressure drop data are obtained for a certain fluid: Q, m^/h 0.36 0.72 1.08 1.44 1.80 Ap, kPa 159 318 477 1274 1851 What is the viscosity of the fluid? Note: Only the first three points give the proper viscosity. What is peculiar about the last two points, which were measured accurately? P1.59 A solid cylinder of diameter D, length L, and density falls due to gravity inside a tube of diameter Dq. The clear¬ ance, Df) — D « D, is filled with fluid of density p and viscosity p. Neglect the air above and below the cylinder. Derive a formula for the terminal fall velocity of the cylin¬ der. Apply your formula to the case of a steel cylinder, D = 2 cm, Dq = 2.04 cm, L = 15 cm, with a film of SAE 30 oil at 20°C. P1.60 Pipelines are cleaned by pushing through them a close- fitting cylinder called a pig. The name comes from the squealing noise it makes sliding along. Reference 50 describes a new nontoxic pig, driven by compressed air, for cleaning cosmetic and beverage pipes. Suppose the pig diameter is 5-15/16 in and its length 26 in. It cleans a 6-in-diameter pipe at a speed of 1.2 m/s. If the clearance is filled with glycerin at 20°C, what pressure difference, in pascals, is needed to drive the pig? Assume a linear veloc¬ ity profile in the oil and neglect air drag. P1.61 An air-hockey puck has a mass of 50 g and is 9 cm in diam¬ eter. When placed on the air table, a 20°C air film, of 0.12-mm thickness, forms under the puck. The puck is struck with an initial velocity of 10 m/s. Assuming a linear velocity distribution in the air film, how long will it take the puck to (a) slow down to 1 m/s and (b) stop completely? Also, (c) how far along this extremely long table will the puck have traveled for condition (a)? P1.57 (a) Sketch the expected no-slip velocity distribution ii(y) between the plates, (b) Find an analytic expression for the velocity U at the interface between the two liquid layers, (c) What is the result of (b) if the viscosities and layer thicknesses are equal? Surface tension P1.62 The hydrogen bubbles that produced the velocity profiles in Fig. 1.15 are quite small, D — 0.01 mm. If the hydrogen- water interface is comparable to air-water and the water temperature is 30°C, estimate the excess pressure within the bubble. Problems 51 P1.63 P1.64 P1.65 P1.66 P1.67 P1.68 Derive Eq. (1.33) by making a force balance on the fluid interface in Fig. 1.11c. Pressure in a water container can be measured by an open ver¬ tical tube — see Fig. P2.ll for a sketch. If the expected water rise is about 20 cm, what tube diameter is needed to ensure that the error due to capillarity will be less than 3 percent? The system in Fig. PI. 65 is used to calculate the pressure pi in the tank by measuring the 15-cm height of liquid in the 1-mm-diameter tube. The fluid is at 60°C. Calculate the true fluid height in the tube and the percentage error due to capillarity if the fluid is (a) water or (b) mercuiy. P1.65 A thin wire ring, 3 cm in diameter, is lifted from a water sur¬ face at 20°C. Neglecting the wire weight, what is the force required to lift the ring? Is this a good way to measure surface tension? Should the wire be made of any particular material? A vertical concentric annulus, with outer radius rg and in¬ ner radius p, is lowered into a fluid of surface tension Y and contact angle 0 < 90°. Derive an expression for the capil¬ lary rise h in the annular gap if the gap is very narrow. Make an analysis of the shape Tqfr) of the water-air inter¬ face near a plane wall, as in Fig. PI. 68, assuming that the slope is small, R~^ ~ cFr\ldx^. Also assume that the pres¬ sure difference across the interface is balanced by the specific weight and the interface height, Ap « pgiq. The boundary conditions are a wetting contact angle P at jc = 0 and a horizontal surface ry = 0 as x— >€». What is the maximum height h at the wall? y P1.69 A solid cylindrical needle of diameter d, length L, and den¬ sity p„ may float in liquid of surface tension Y. Neglect buoyancy and assume a contact angle of 0°. Derive a for¬ mula for the maximum diameter d^^ able to float in the liquid. Calculate d,^^^ for a steel needle (SG = 7.84) in water at 20°C. P1.70 Derive an expression for the capillary height change h for a fluid of surface tension Y and contact angle 0 between two vertical parallel plates a distance W apart, as in Fig. PI. 70. What will h be for water at 20°C if IT = 0.5 mm? P1.71 A soap bubble of diameter Di coalesces with another bub¬ ble of diameter D2 to form a single bubble D3 with the same amount of air. Assuming an isothermal process, derive an expression for finding D3 as a function of Di, D2, Patm> and Y. Vapor pressure P1.72 Early mountaineers boiled water to estimate their altitude. If they reach the top and find that water boils at 84°C, approximately how high is the mountain? P1.73 A small submersible moves at velocity V, in fresh water at 20°C, at a 2-m depth, where ambient pressure is 131 kPa. Its critical cavitation number is known to be C„ = 0.25. At what velocity will cavitation bubbles begin to form on the body? Will the body cavitate if P = 30 m/s and the water is cold (5°C)? P1.74 Oil, with a vapor pressure of 20 kPa, is delivered through a pipeline by equally spaced pumps, each of which increases the oil pressure by 1.3 MPa. Friction losses in the pipe are 150 Pa per meter of pipe. What is the maximum possible pump spacing to avoid cavitation of the oil? Speed of sound, Mach number P1.75 An airplane flies at 555 mi/h. At what altitude in the stan¬ dard atmosphere will the airplane’s Mach number be exactly 0.8? 52 Chapter 1 Introduction P1.76 Derive a formula for the bulk modulus of an ideal gas, with constant specific heats, and calculate it for steam at 300°C and 200 kPa. Compare your result to the steam tables. P1.77 Assume that the n-pentane data of Prob. PI. 36 represents isentropic conditions. Estimate the value of the speed of sound at a pressure of 30 MPa. [Hint: The data approxi¬ mately fit Eq. (1.19) with B = 260 and n = 11.] P1.78 Sir Isaac Newton measured the speed of sound by timing the difference between seeing a cannon’s puff of smoke and hearing its boom. If the cannon is on a mountain 5.2 mi away, estimate the air temperature in degrees Celsius if the time difference is (a) 24.2 s and (b) 25.1 s. P1.79 From Table A.3, the density of glycerin at standard conditions is about 1260 kg/m^. At a very high pressure of 8000 Ib/in^, its density increases to approximately 1275 kg/m^. Use this data to estimate the speed of sound of glycerin, in ft/s. P1.80 In Problem P1.24, for the given data, the air velocity at sec¬ tion 2 is 1180 ft/s. What is the Mach number at that section? Streamlines P1.81 Use Eq. (1.39) to find and sketch the streamlines of the fol¬ lowing flow field: u = Kx\ V = —Ky; w = 0, where A" is a constant. P1.82 A velocity field is given hy u = V cos0, v = V sin0, and w = 0, where V and 0 are constants. Derive a formula for the streamlines of this flow. P1.83 Use Eq. (1.39) to find and sketch the streamlines of the following flow field: u = K{x^ — y^); v = —2Kxy; w = 0, where is a constant. Hint: This is a first-order exact differential equation. Fundamentals of Engineering Exam Problems FEl.l The absolute viscosity /i of a fluid is primarily a function of (a) Density, (b) Temperature, (c) Pressure, (d) Velocity, (e) Surface tension FE1.2 Carbon dioxide, at 20°C and 1 atm, is compressed isen- tropically to 4 atm. Assume CO2 is an ideal gas. The final temperature would be (a) 130°C, (b) 162°C, (c) 171°C, (d) 237°C, (e) 313°C EE1.3 Helium has a molecular weight of 4.003. What is the weight of 2 m^ of helium at 1 atm and 20°C? (a) 3.3 N, (b) 6.5 N, (c) 11.8 N, (d) 23.5 N, (e) 94.2 N EE1.4 An oil has a kinematic viscosity of 1.25 E-4 mVs and a specific gravity of 0.80. What is its dynamic (absolute) viscosity in kg/(m ■ s)? (a) 0.08, (b) 0.10, (c) 0.125, (d) 1.0, (e) 1.25 History of fluid mechanics P1.84 In the early 1900s, the British chemist Sir Cyril Hinshelwood quipped that fluid dynamics study was di¬ vided into “workers who observed things they could not explain and workers who explained things they could not observe.” To what historic situation was he referring? P1.85 Do some reading and report to the class on the life and achievements, especially vis-a-vis fluid mechanics, of (a) Evangelista Torricelli (1608-1647) (fe) Henri de Pitot (1695-1771) (c) Antoine Chezy (1718-1798) {d) Gotthilf Heinrich Ludwig Hagen (1797-1884) (e) Julius Weisbach ( 1 806-187 1 ) (f) George Gabriel Stokes (1819-1903) (g) Moritz Weber (1871-1951) (h) Theodor von Karman (1881-1963) (i) Paul Richard Heinrich Blasius (1883-1970) (/■) Ludwig Prandtl (1875-1953) (k) Osborne Reynolds ( 1 842- 1912) (/) John William Strutt, Lord Rayleigh ( 1 842- 1919) (m) Daniel Bernoulli (1700-1782) (m) Leonhard Euler (1707-1783) Experimental uncertainty P1.86 A right circular cylinder volume v is to be calculated from the measured base radius R and height H. If the uncertainty in R is 2 percent and the uncertainty in // is 3 percent, esti¬ mate the overall uncertainty in the calculated volume. Hint: Read Appendix E. FE1.5 Consider a soap bubble of diameter 3 mm. If the surface tension coefficient is 0.072 N/m and external pressure is 0 Pa gage, what is the bubble’s internal gage pressure? (a) -24 Pa, (b) -f48 Pa, (c) +96 Pa, (d) +192 Pa, (e) -192 Pa FE1.6 The only possible dimensionless group that combines ve¬ locity V, body size L, fluid density p, and surface tension coefficient ct is (a) Lpcj/V, (b) pVL^Ia, (c) paV^/L, (d) aLV^/p, (e) pLV^/a FE1.7 Two parallel plates, one moving at 4 m/s and the other fixed, are separated by a 5-mm-thick layer of oil of spe¬ cific gravity 0.80 and kinematic viscosity 1.25 E-4 mVs. What is the average shear stress in the oil? (a) 80 Pa, (b) 100 Pa, (c) 125 Pa, (d) 160 Pa, (e) 200 Pa Comprehensive Problems 53 FE1.8 Carbon dioxide has a specific heat ratio of 1.30 and a gas constant of 189 J/(kg ■ °C). If its temperature rises from 20 to 45°C, what is its internal energy rise? (a) 12.6 kJ/kg, (b) 15.8 kj/kg, (c) 17.6 kj/kg, (d) 20.5 kj/kg, (e) 25.1 kJ/kg FE1.9 A certain water flow at 20°C has a critical cavitation number, where bubbles form, Ca ~ 0.25, where Ca = '^(Pa~ PvapVpk'^- If Pa = 1 the vapor pressure is 0.34 pounds per square inch absolute (psia), for what water velocity will bubbles form? (a) 12 mi/h, (b) 28 mi/h, (c) 36 mi/h, (d) 55 mi/h, (e) 63 mi/h Comprehensive Problems Cl.l Sometimes we can develop equations and solve practical problems by knowing nothing more than the dimensions of the key parameters in the problem. For example, consider the heat loss through a window in a building. Window effi¬ ciency is rated in terms of “P value,” which has units of (ft^ ■ h ■ °F)/Btu. A certain manufacturer advertises a double¬ pane window with an R value of 2.5. The same company produces a triple-pane window with an R value of 3.4. In either case the window dimensions are 3 ft by 5 ft. On a given winter day, the temperature difference between the inside and outside of the building is 45°F. (a) Develop an equation for the amount of heat lost in a given time period Af, through a window of area A, with a given R value, and temperature difference AT. How much heat (in Btu) is lost through the double-pane window in one 24-h period? {b) How much heat (in Btu) is lost through the triple-pane window in one 24-h period? (c) Suppose the building is heated with propane gas, which costs $3.25 per gallon. The propane burner is 80 percent efficient. Propane has approximately 90,000 Btu of available energy per gallon. In that same 24-h period, how much money would a homeowner save per window by installing triple-pane rather than double¬ pane windows? {d) Finally, suppose the homeowner buys 20 such triple¬ pane windows for the house. A typical winter has the equivalent of about 120 heating days at a temperature difference of 45°F. Each triple-pane window costs $85 more than the double-pane window. Ignoring interest and inflation, how many years will it take the home- owner to make up the additional cost of the triple-pane windows from heating bill savings? EEl.lO Example 1.10 gave an analysis that predicted that the viscous moment on a rotating disk M = 7r(jIlR/{2h). If the uncertainty of each of the four variables (/i, H, R, h) is 1.0 percent, what is the estimated overall uncertainty of the moment M2 (a) 4.0 percent (b) 4.4 percent (c) 5.0 percent (d) 6.0 percent (e) 7.0 percent C1.2 When a person ice skates, the surface of the ice actually melts beneath the blades, so that he or she skates on a thin sheet of water between the blade and the ice. (a) Find an expression for total friction force on the bottom of the blade as a function of skater velocity V, blade length L, water thickness (between the blade and the ice) h, water viscosity /i, and blade width W. (b) Suppose an ice skater of total mass m is skating along at a constant speed of Vq when she suddenly stands stiff with her skates pointed directly forward, allowing herself to coast to a stop. Neglecting fric¬ tion due to air resistance, how far will she travel before she comes to a stop? (Remember, she is coasting on two skate blades.) Give your answer for the total distance traveled, x, as a function of Vo^ L, h, (1, and W. (c) Find for the case where Vo = 4.0 m/s, m = 100 kg, L = 30 cm, W = 5.0 mm, and h = 0.10 mm. Do you think our assumption of negligible air resistance is a good one? C1.3 Two thin flat plates, tilted at an angle a, are placed in a tank of liquid of known surface tension Y and contact angle 0, as shown in Fig. Cl. 3. At the free surface of the liquid in the tank, the two plates are a distance L apart and have width b into the page. The liquid rises a distance h between the plates, as shown. (a) What is the total upward (z-directed) force, due to sur¬ face tension, acting on the liquid column between the plates? (b) If the liquid density is p, find an expression for surface tension Y in terms of the other variables. 54 Chapter 1 Introduction C1.3 V C1.5 Viscosity can be measured by flow through a thin-bore or capillary tube if the flow rate is low. For length L, (small) diameter D L, pressure drop Ap, and (low) volume flow rate Q, the formula for viscosity is p = D\p/{CLQ), where C is a constant. (a) Verify that C is dimensionless. The following data are for water flowing through a 2-mm-diameter tube which is 1 meter long. The pressure drop is held constant at Ap = 5 kPa. T,°C 10.0 40.0 70.0 s Q, L/min 0.091 0.179 0.292 C1.6 C1.4 Oil of viscosity fi and density p drains steadily down the side of a tall, wide vertical plate, as shown in Fig. Cl. 4. In the region shown, /n/Zy developed conditions exist; that is, the velocity profile shape and the film thickness 8 are inde¬ pendent of distance z along the plate. The vertical velocity w becomes a function only of x, and the shear resistance from the atmosphere is negligible. Plate C1.4 (a) Sketch the approximate shape of the velocity profile w(x), considering the boundary conditions at the wall and at the film surface. (b) Suppose film thickness A and the slope of the velocity profile at the wall, (dw/t&)„aii, are measured by a laser Doppler anemometer (to be discussed in Chap. 6). Find an expression for the viscosity of the oil as a function of p, d, (dw/dx)„^ii, and the gravitational acceleration g. Note that, for the coordinate system given, both w and (dw/dx)^^i are negative. {b) Using proper SI units, determine an average value of C by accounting for the variation with temperature of the viscosity of water. The rotating-cylinder viscometer in Fig. C1.6 shears the fluid in a narrow clearance Ar, as shown. Assuming a linear velocity distribution in the gaps, if the driving torque M is measured, find an expression for fi by (a) neglecting and (b) including the bottom friction. C1.6 Viscous fluid p Ar«R C1.7 C1.8 Make an analytical study of the transient behavior of the sliding block in Prob. 1.45. (fl) Solve for V(t) if the block starts from rest, V = 0 at t = 0. (b) Calculate the time when the block has reached 98 percent of its terminal velocity. A mechanical device that uses the rotating cylinder of Fig. Cl. 6 is the S former viscometer . Instead of being driven at constant fl, a cord is wrapped around the shaft and attached to a falling weight W. The time t to turn the shaft a given number of revolutions (usually five) is mea¬ sured and correlated with viscosity. The formula is Apt W - B Comprehensive Problems 55 where A and B are constants that are determined by cali¬ brating the device with a known fluid. Here are calibration data for a Stormer viscometer tested in glycerol, using a weight of 50 N: where C is a calibration constant. For u in the range of 100-500 mm^/s, the recommended constant is C = 0.50 mmVs^, with an accuracy less than 0.5 percent. p, kg/(m-s) 0.23 0.34 0.57 0.84 1.15 t, sec 15 23 38 56 77 (a) Find reasonable values of A and B to fit this calibration data. Hint: The data are not very sensitive to the value ofB. (b) A more viscous fluid is tested with a 100 N weight and the measured time is 44 s. Estimate the viscosity of this fluid. C1.9 The lever in Fig. C 1 .9 has a weight W at one end and is tied to a cylinder at the left end. The cylinder has negligible weight and buoyancy and slides upward through a film of heavy oil of viscosity fi. (a) If there is no acceleration (uni¬ form lever rotation), derive a formula for the rate of fall V2 of the weight. Neglect the lever weight. Assume a linear velocity profile in the oil film, (b) Estimate the fall velocity of the weight if VT = 20 N, Lj = 75 cm, L2 = 50 cm, D = 10 cm, L = 22 cm, AR = 1 mm, and the oil is glycerin at 20°C. Vi pivot Cylinder, diameter D, length L, in an oil film of thickness AR. IT ^2? C1.9 Cl.lO A popular gravity-driven instrument is the Cannon- Ubbelohde viscometer, shown in Fig. Cl.lO. The test liquid is drawn up above the bulb on the right side and allowed to drain by gravity through the capillary tube below the bulb. The time t for the meniscus to pass from upper to lower timing marks is recorded. The kinematic viscosity is computed by the simple formula: V = Ct Upper timing mark Lower timing mark Cl.lO The Cannon- Ubbelohde viscometer. Source: Courtesy of Cannon Instrument Company. Capillary tube (a) What liquids from Table A. 3 are in this viscosity range? (b) Is the calibration formula dimensionally consistent? (c) What system properties might the constant C depend upon? (d) What problem in this chapter hints at a formula for estimating the viscosity? Cl.ll Mott [Ref. 49, p. 38] discusses a simple falling-ball vis¬ cometer, which we can analyze later in Chap. 7. A small ball of diameter D and density p/, falls through a tube of test liquid (p, p). The fall velocity V is calculated by the time to fall a measured distance. The formula for calculating the viscosity of the fluid is iPb - P)gD^ This result is limited by tbe requirement that the Reynolds number (pVD/p) be less than 1.0. Suppose a steel ball (SG = 7.87) of diameter 2.2 mm falls in SAE 25W oil (SG = 0.88) at 20°C. The measured fall velocity is 8.4 cm/s. (a) What is the viscosity of the oil, in kg/m-s? (b) Is the Reynolds num¬ ber small enough for a valid estimate? C1.12 A solid aluminum disk (SG = 2.7) is 2 in in diameter and 3/16 in thick. It slides steadily down a 14° incline that is coated with a castor oil (SG = 0.96) film one hundredth of an inch thick. The steady slide velocity is 2 cm/s. Using Figure A.l and a linear oil velocity profile assumption, estimate the temperature of the castor oil. 56 Chapter 1 Introduction References T. J. Chung, Computational Fluid Dynamics, 2d ed., Cambridge University Press, New York, 2010. J. D. Anderson, Computational Fluid Dynamics: An Intro¬ duction, 3d ed.. Springer, New York, 2010. H. Lomax, T. H. Pulliam, and D. W. Zingg, Fundamentals of Computational Fluid Dynamics, Springer, New York, 2011. B. Andersson, L. Hakansson, and M. Mortensen, Computa¬ tional Fluid Dynamics for Engineers, Cambridge University Press, New York, 2012. D. C. Wilcox, Turbulence Modeling for CFD, 3d ed., DCW Industries, La Canada, California, 2006. P. S. Bernard and J. M. Wallace, Turbulent Flow: Analysis, Measurement and Prediction, Wiley, New York, 2002. X. Jiang and C. H. Lai, Numerical Techniques for Direct and Large-Eddy Simulations, CRC Press, Boca Raton, Florida, 2009. B. Geurts, Elements of Direct and Large Eddy Simulation, R. T. Edwards Inc., Flourtown, PA, 2003. S. Tavoularis, Measurement in Eluid Mechanics, Cambridge University Press, New York, 2005. R. C. Baker, Introductory Guide to Elow Measurement, Wiley, New York, 2002. R. W. Miller, Flow Measurement Engineering Handbook, 3d ed., McGraw-Hill, New York, 1996. H. Rouse and S. Ince, History of Hydraulics, Iowa Institute of Hydraulic Research, Univ. of Iowa, Iowa City, lA, 1957; reprinted by Dover, New York, 1963. H. Rouse, Hydraulics in the United States 1776-1976, Iowa Institute of Hydraulic Research, Univ. of Iowa, Iowa City, lA, 1976. G. A. Tokaty, A History and Philosophy of Eluid Mechanics, Dover Publications, New York, 1994. Cambridge University Press, “Ludwig Prandtl — Father of Modem Fluid Mechanics,” URL . R. 1. Tanner, Engineering Rheology, 2d ed., Oxford Univer¬ sity Press, New York, 2000. C. E. Brennen, Eundamentals of Multiphase Flow, Cambridge University Press, New York, 2009. C. Shen, Rarefied Gas Dynamics, Springer, New York, 2010. F. Carderelli and M. J. Shields, Scientific Unit Conversion: A Practical Guide to Metrification, 2d ed., Springer-Verlag, New York, 1999. J. P. Holman, Heat Transfer, 10th ed., McGraw-Hill, New York, 2009. B. E. Poling, J. M. Prausnitz, and J. P. O’Connell, The Prop¬ erties of Gases and Liquids, 5th ed., McGraw-Hill, New York, 2000. J. Hilsenrath et ah, “Tables of Thermodynamic and Transport Properties,” U.S. Nat. Bur. Standards Circular 564, 1955; reprinted by Pergamon, New York, 1960. W. T. Parry, ASME International Steam Tables for Industrial Use, 2d ed., ASME Press, New York, 2009. Steam Tables URL: O. A. Hougen and K. M. Watson, Chemical Process Princi¬ ples Charts, Wiley, New York, 1960. E. M. White, Viscous Fluid Flow, 3d ed., McGraw-Hill, New York, 2005. M. Bourne, Food Texture and Viscosity: Concept and Mea¬ surement, 2d ed.. Academic Press, Salt Lake City, Utah, 2002. SAE Euels and Lubricants Standards Manual, Society of Automotive Engineers, Warrendale, PA, 2001. C. L. Yaws, Handbook of Viscosity, 3 vols., Elsevier Science and Technology, New York, 1994. A. W. Adamson and A. P. Gast, Physical Chemistry of Sur¬ faces, Wiley, New York, 1999. C. E. Brennen, Eundamentals of Multiphase Flow, Cambridge University Press, New York, 2009. National Committee for Fluid Mechanics Films, Illustrated Experiments in Fluid Mechanics, M.I.T. Press, Cambridge, MA, 1972. G. Currie, Fundamental Mechanics of Fluids, 3d ed., Marcel Dekker, New York, 2003. W.-J. Yang (ed.). Handbook of Flow Visualization, 2d ed., Taylor and Francis, New York, 2001. F.T. Nieuwstadt (ed.). Flow Visualization and Image Analy¬ sis, Springer, New York, 2007. A. J. Smits and T. T. Lim, Flow Visualization: Techniques and Examples, 2d ed.. Imperial College Press, London, R. J. Adrian and J. Westerweel, Particle Image Velocimetry, Cambridge University Press, New York, 2010. Wen-Jai Yang, Computer-Assisted Flow Visualization, Begell House, New York, 1994. M. van Dyke, An Album of Fluid Motion, Parabolic Press, Stanford, CA, 1982. Y. Nakayama and Y. Tanida (eds.), Visualized Flow, vol. 1, Elsevier, New York, 1993; vols. 2 and 3, CRC Press, Boca Raton, FL, 1996. M. Samimy, K. S. Breuer, L. G. Leal, and P. H. Steen, A Gal¬ lery of Fluid Motion, Cambridge University Press, New York, 2003. S. Y. Son et ah, “Coolant Flow Field Measurements in a Two-Pass Channel Using Particle Image Velocimetry,” 1999 Heat Transfer Gallery, Journal of Heat Transfer, vol. 121, August, 1999. References 57 B. Carr and V. E. Young, “Videotapes and Movies on Fluid Dynamics and Fluid Machines,” in Handbook of Fluid Dynamics and Fluid Machinery, vol. II, J. A. Schetz and A. E. Fuhs (eds.), Wiley, New York, 1996, pp. 1171-1189. Online Steam Tables URL: SH_Properties.aspx. H. W. Coleman and W. G. Steele, Experimentation and Uncertainty Analysis for Engineers, 3d ed., Wiley, New York, 2009. I. Hughes and T. Hase, Measurements and Their Uncertain¬ ties, Oxford University Press, New York, 2010. A. Thom, “The Flow Past Circular Cylinders at Low Speeds,” Proc. Royal Society, A141, London, 1933, pp. 651-666. S. J. Kline and F. A. McClintock, “Describing Uncertainties in Single-Sample Experiments,” Mechanical Engineering, January, 1953, pp. 3-9. R. L. Mott, Applied Eluid Mechanics, Pearson Prentice-Hall, Upper Saddle River, NJ, 2006. “Putting Porky to Work,” Technology Focus, Mechanical Engineering, August 2002, p. 24. R. M. Olson and S. J. Wright, Essentials of Engineering Eluid Mechanics, 5th ed., HarperCollins, New York, 1990. B . Kirby, Micro- and Nanoscale Fluid Mechanics, Cambridge University Press, New York, 2010. E. J. Watson, “The Spread of a Liquid Jet over a Horizontal Plane,” J. Fluid Mechanics, vol. 20, 1964, pp. 481-499. J. Dean and A. Reisinger, Sands, Powders, and Grains: An Introduction to the Physics of Granular Materials, Springer- Verlag, New York, 1999. E. Kiran and Y. L. Sen, “High Pressure Density and Viscosity of n-alkanes,” Int. J. Thermophysics, vol. 13, no. 3, 1992, pp. 411-442. 58 _ uJ^ The bathyscaphe Trieste was built by explorers August and Jacques Picard in 1953. They made many deep-sea dives over the next five years. In 1958 the U.S. Navy purchased it and, in 1960, descended to the deepest part of the ocean, the Marianas Trench, near Guam. The recorded depth was 10,900 m, or nearly seven miles. The photo shows the vessel being lifted from the water. The pressure sphere is on the bottom, and the boat-shaped structure above is filled with gasoline, which provides buoyancy even at those depths. [Image courtesy of the U.S. Navy.] 2.1 Pressure and Pressure Gradient Chapter 2 Pressure Distribution in a Fluid Motivation. Many fluid problems do not involve motion. They concern the pressure distribution in a static fluid and its effect on solid surfaces and on floating and sub¬ merged bodies. When the fluid velocity is zero, denoted as the hydrostatic condition, the pressure variation is due only to the weight of the fluid. Assuming a known fluid in a given gravity field, the pressure may easily be calculated by integration. Important applications in this chapter are (1) pressure distribution in the atmosphere and the oceans, (2) the design of manometer, mechanical, and electronic pressure instruments, (3) forces on submerged flat and curved surfaces, (4) buoyancy on a submerged body, and (5) the behavior of floating bodies. The last two result in Archimedes’ principles. If the fluid is moving in rigid-body motion, such as a tank of liquid that has been spinning for a long time, the pressure also can be easily calculated because the fluid is free of shear stress. We apply this idea here to simple rigid-body accelerations in Sec. 2.9. Pressure measurement instruments are discussed in Sec. 2.10. As a matter of fact, pressure also can be analyzed in arbitrary (nonrigid-body) motions Y{x, y, z, f), but we defer that subject to Chap. 4. In Fig. 1.1 we saw that a fluid at rest cannot support shear stress and thus Mohr’s circle reduces to a point. In other words, the normal stress on any plane through a fluid element at rest is a point property called the fluid pressure p, taken positive for compression by common convention. This is such an important concept that we shall review it with another approach. First let us emphasize that pressure is a thermodynamic property of the fluid, like temperature or density. It is not a. force. Pressure has no direction and is not a vector. The concept of force only arises when considering a surface immersed in fluid pressure. The pressure creates a force, due to fluid molecules bombarding the surface, and it is normal to that surface. 59 60 Chapter 2 Pressure Distribution in a Fluid Fig. 2.1 Equilibrium of a small wedge of fluid at rest. Pressure Force on a Fluid Element z(up) Figure 2.1 shows a small wedge of fluid at rest of size Ax by Az by As and depth b into the paper. There is no shear by definition, but we postulate that the pressures Pj;, Pj, and p„ may be different on each face. The weight of the element also may be important. The element is assumed to be small, so the pressure is constant on each face. Summation of forces must equal zero (no acceleration) in both the x and Z directions. SfZj = 0 = p^b Az — Pnb As sin 6 , (2.1) = 0 = pj? Ax — p„b As cos 6 — 2Pgb Ax Az But the geometry of the wedge is such that As sin 0 = Az As cos 9 = Ax (2.2) Substitution into Eq. (2.1) and rearrangement give Px = Pn Pz= P„ + zPg (2-3) These relations illustrate two important principles of the hydrostatic, or shear-free, condition: (1) There is no pressure change in the horizontal direction, and (2) there is a vertical change in pressure proportional to the density, gravity, and depth change. We shall exploit these results to the fullest, starting in Sec. 2.3. In the limit as the fluid wedge shrinks to a “point,” Az— >0 and Eq. (2.3) become Px = Pz = Pn = P (2-4) Since 9 is arbitrary, we conclude that the pressure /? in a static fluid is a point property, independent of orientation. Pressure (or any other stress, for that matter) causes a net force on a fluid element. To see this, consider the pressure acting on the two x faces in Eig. 2.2. Let the pressure vary arbitrarily P = Pix, y, z, t) 2.2 Equilibrium of a Fluid Element 61 Fig. 2.2 Net X force on an element due to pressure variation. 2.2 Equilibrium of a Fluid Element dp dx dx) dy dz The net force in the x direction on the element in Fig. 2.2 is given by , dp \ dp dF^ = p dy dz — [p - dx ]dy dz = - dx dy dz dx J dx In like manner the net force dFy involves —dpidy, and the net force dF^ concerns —dp/dz- The total net-force vector on the element due to pressure is r/F = ^ press .dp .dp i - j - k — dx dy dz dx ^ dy dz) :^i, (2.5) We recognize the term in parentheses as the negative vector gradient of p. Denoting f as the net force per unit element volume, we rewrite Eq. (2.5) as fpress = -Vp (2.6) where V = gradient operator = i - hj - h k — dx dy dz Thus it is not the pressure but the pressure gradient causing a net force that must be balanced by gravity or acceleration or some other effect in the fluid. The pressure gradient is a surface force that acts on the sides of the element. There may also be a body force, due to electromagnetic or gravitational potentials, acting on the entire mass of the element. Here we consider only the gravity force, or weight of the element: ‘^Ferav = Pg dx dy dz (2.7) or fgrav = Pg In addition to gravity, a fluid in motion will have surface forces due to viscous stresses. By Newton’s law, Eq. (1.2), the sum of these per-unit- volume forces equals the mass per unit volume (density) times the acceleration a of the fluid element: 2^ ^ fpress + fgrav + fvisc = -V p + fg + fyisc = PU (2.8) 62 Chapter 2 Pressure Distribution in a Fluid Fig. 2.3 Illustration of absolute, gage, and vacuum pressure readings. p (Pascals) 120,000 90,000 t 30,000 I High pressure: p = 120,000 Pa abs = 30,000 Pa gage Local atmosphere: p = 90,000 Pa abs = 0 Pa gage = 0 Pa vacuum 50,000 0 40,000 I Vacuum pressure: I p = 50,000 Pa abs = 40,000 Pa vacuum 50,000 "J" (Tension) Absolute zero reference: p = 0 Pa abs = 90,000 Pa vacuum Gage Pressure and Vacuum Pressure: Relative Terms 2.3 Hydrostatic Pressure Distributions This general equation will be studied in detail in Chap. 4. Note that Eq. (2.8) is a vector relation, and the acceleration may not be in the same vector direction as the velocity. For our present topic, hydrostatics, the viscous stresses and the acceleration are zero. Before embarking on examples, we should note that engineers are apt to specify pres¬ sures as (1) the absolute or total magnitude or (2) the value relative to the local ambient atmosphere. The second case occurs because many pressure instruments are of dijferential type and record, not an absolute magnitude, but the difference between the fluid pressure and the atmosphere. The measured pressure may be either higher or lower than the local atmosphere, and each case is given a name; p > Pa Gage pressure: f>(gage) = p - Pa P < Pa Vacuum pressure: p(vacuum) = Pa ~ P This is a convenient shorthand, and one later adds (or subtracts) atmospheric pressure to determine the absolute fluid pressure. A typical situation is shown in Fig. 2.3. The local atmosphere is at, say, 90,000 Pa, which might reflect a storm condition in a sea-level location or normal conditions at an altitude of 1000 m. Thus, on this day, pa = 90,000 Pa absolute = 0 Pa gage = 0 Pa vacuum. Suppose gage 1 in a laboratory reads pi = 120,000 Pa absolute. This value may be reported as a gage pressure, pi = 120,000 — 90,000 = 30,000 Pa gage. (One must also record the atmospheric pressure in the laboratory, since pa changes gradually.) Suppose gage 2 reads p2 = 50,000 Pa absolute. Locally, this is a vacuum pressure and might be reported as p2 = 90,000 — 50,000 = 40,000 Pa vacuum. Occasionally, in the problems section, we will specify gage or vacuum pressure to keep you alert to this common engineering practice. If a pressure is listed without the modifier gage or vacuum, we assume it is absolute pressure. If the fluid is at rest or at constant velocity, a = 0 and = 0. Equation (2.8) for the pressure distribution reduces to Vp = pg (2.9) 2.3 Hydrostatic Pressure Distributions 63 Effect of Variable Gravity This is a hydrostatic distribution and is correct for all fluids at rest, regardless of their viscosity, because the viscous term vanishes identically. Recall from vector analysis that the vector Vp expresses the magnitude and direc¬ tion of the maximum spatial rate of increase of the scalar property p. As a result, Vp is perpendicular everywhere to surfaces of constant p. Thus Eq. (2.9) states that a fluid in hydrostatic equilibrium will align its constant-pressure surfaces everywhere normal to the local-gravity vector. The maximum pressure increase will be in the direction of gravity — that is, “down.” If the fluid is a liquid, its free surface, being at atmospheric pressure, will be normal to local gravity, or “horizontal.” You probably knew all this before, but Eq. (2.9) is the proof of it. In our customary coordinate system z is “up.” Thus the local-gravity vector for small-scale problems is g = -^k (2.10) where g is the magnitude of local gravity, for example, 9.807 m/s^. For these coor¬ dinates Eq. (2.9) has the components dp dx = 0 dp -P8= -7 dz (2.11) the first two of which tell us that p is independent of x and y. Hence dp/dz can be replaced by the total derivative dp/dz, and the hydrostatic condition reduces to dz = -7 or P2- Pi = - 'jdz (2.12) •'i Equation (2.12) is the solution to the hydrostatic problem. The integration requires an assumption about the density and gravity distribution. Gases and liquids are usually treated differently. We state the following conclusions about a hydrostatic condition; Pressure in a continuously distributed uniform static fluid varies only with vertical distance and is independent of the shape of the container. The pressure is the same at all points on a given horizontal plane in the fluid. The pressure increases with depth in the fluid. An illustration of this is shown in Fig. 2.4. The free surface of the container is atmospheric and forms a horizontal plane. Points a, b, c, and d are at equal depth in a horizontal plane and are interconnected by the same fluid, water; therefore, all points have the same pressure. The same is tme of points A, B, and C on the bottom, which all have the same higher pressure than at a, b, c, and d. However, point D, although at the same depth as A, B, and C, has a different pressure because it lies beneath a different fluid, mercury. For a spherical planet of uniform density, the acceleration of gravity varies inversely as the square of the radius from its center g = go r (2.13) 64 Chapter 2 Pressure Distribution in a Fluid Fig. 2.4 Hydrostatic-pressure distribution. Points a, b, c, and d are at equal depths in water and therefore have identical pressures. Points A, B, and C are also at equal depths in water and have identical pressures higher than a, b, c, and d. Point D has a different pressure from A, B, and C because it is not connected to them by a water path. Hydrostatic Pressure in Liquids Table 2.1 Specific Weight of Some Common Fluids Atmospheric pressure Mercury where Tq is the planet radius and gQ is the surface value of g. For earth, Tq ~ 3960 statute mi ~ 6400 km. In typical engineering problems the deviation from extends from the deepest ocean, about 1 1 km, to the atmospheric height of supersonic trans¬ port operation, about 20 km. This gives a maximum variation in g of (6400/6420)^, or 0.6 percent. We therefore neglect the variation of g in most problems. Liquids are so nearly incompressible that we can neglect their density variation in hydrostatics. In Example 1.6 we saw that water density increases only 4.6 percent at the deepest part of the ocean. Its effect on hydrostatics would be about half of this, or 2.3 percent. Thus we assume constant density in liquid hydrostatic calculations, for which Eq. (2.12) integrates to Liquids: Pl- P\ = -7(22 - Z\|) Pi P\ or Z\ - z-) - - 7 7 (2.14) We use the first form in most problems. The quantity 7 is called the specific weight of the fluid, with dimensions of weight per unit volume; some values are tabulated in Table 2.1. The quantity p!') is a length called the pressure head of the fluid. Specilic weight 7 at 68“F = 20°C Fluid Ibf/fF nW Air (at 1 atm) 0.0752 11.8 Ethyl alcohol 49.2 7,733 SAE 30 oil 55.5 8,720 Water 62.4 9,790 Seawater 64.0 10,050 Glycerin 78.7 12,360 Carbon tetrachloride 99.1 15,570 Mercury 846 133,100 2.3 Hydrostatic Pressure Distributions 65 Fig. 2.5 Hydrostatic-pressure distribution in oceans and atmospheres. The Mercury Barometer z +b p~p- Air V 0 Free surface: Z = 0, p = I Water P ^ Pa"^ ^3^water For lakes and oceans, the coordinate system is usually chosen as in Fig. 2.5, with Z = 0 at the free surface, where p equals the surface atmospheric pressure p^. When we introduce the reference value (pi, zi) = {p^, 0), Eq. (2.14) becomes, for p at any (negative) depth z, Lakes and oceans: P = Pa ~ Jz (2.15) where 7 is the average specific weight of the lake or ocean. As we shall see, Eq. (2.15) holds in the atmosphere also with an accuracy of 2 percent for heights z up to 1000 m. EXAMPLE 2.1 Newfound Lake, a freshwater lake near Bristol, New Hampshire, has a maximum depth of 60 m, and the mean atmospheric pressure is 91 kPa. Estimate the absolute pressure in kPa at this maximum depth. Solution • System sketch: Imagine that Fig. 2.5 is Newfound Lake, with /? = 60 m and z = 0 at the surface. • Property values: From Table 2.1, Tjvater = 9790 N/m^. We are given that Patmos = 91 kPa. • Solution steps: Apply Eq. (2.15) to the deepest point. Use SI units, pascals, not kilopascals: Pmax = Pa - lz = 91,000 Pa - (9790 ^)(-60 m) = 678,400 Pa « 678 kPa Ans. m • Comments: Kilopascals are awkward. Use pascals in the formula, then convert the answer. The simplest practical application of the hydrostatic formula (2.14) is the barometer (Fig. 2.6), which measures atmospheric pressure. A tube is filled with mercury and inverted while submerged in a reservoir. This causes a near vacuum in the closed upper end because mercury has an extremely small vapor pressure at room temperatures 66 Chapter 2 Pressure Distribution in a Fluid Fig. 2.6 A barometer measures local absolute atmospheric pressure: (a) the height of a mercury column is proportional to (^) ^ modern portable barometer, with digital readout, uses the resonating silicon element of Fig. 2.28c. (Courtesy of Paul Lupke, Druck, Inc.) (0.16 Pa at 20°C). Since atmospheric pressure forces a mercury column to rise a distance h into the tube, the upper mercury surface is at zero pressure. From Fig. 2.6, Eq. (2.14) applies with pi = Q a.i zi= h and p2 = Pa at Z2 = 0: Pa - 0 = -7m (0 - h) or h = — (2.16) 7m At sea-level standard, with = 101,350 Pa and 7m = 133,100 N/m^ from Table 2.1, the barometric height is h = 101,350/133,100 = 0.761 m or 761 mm. In the United States the weather service reports this as an atmospheric “pressure” of 29.96 inHg (inches of mercury). Mercury is used because it is the heaviest common liquid. A water barometer would be 34 ft high. Hydrostatic Pressure in Gases Gases are compressible, with density nearly proportional to pressure. Thus density must be considered as a variable in Eq. (2.12) if the integration carries over large pressure changes. It is sufficiently accurate to introduce the perfect-gas law p = pRT in Eq. (2.12): dif dz = -Pg = - _P^ RT 8 2.3 Hydrostatic Pressure Distributions 67 The Standard Atmosphere Separate the variables and integrate between points 1 and 2: = In — = -- — P Pi R T (2.17) The integral over z requires an assumption about the temperature variation T{z). One common approximation is the isothermal atmosphere, where T = Tq. Pi = Pi exp gjzi - Zl) RTa (2.18) The quantity in brackets is dimensionless. (Think that over; it must be dimensionless, right?) Equation (2.18) is a fair approximation for earth, but actually the earth’s mean atmospheric temperature drops off nearly linearly with z up to an altitude of about 36,000 ft (11,000 m): T^Tq-Bz (2.19) Here Tq is sea-level temperature (absolute) and B is the lapse rate, both of which vary somewhat from day to day. By international agreement the following standard values are assumed to apply from 0 to 36,000 ft: Tq = 518.69°R = 288.16 K = 15°C B = 0.003566°R/ft = 0.00650 K/m This lower portion of the atmosphere is called the troposphere. Introducing Eq. (2.19) into Eq. (2.17) and integrating, we obtain the more accurate relation in the troposphere, with z = 0 at sea level. The exponent g/{RB) is dimensionless (again it must be) and has the standard value of 5.26 for air, with R = 287 m^/(s^ • K). The U.S. standard atmosphere is sketched in Eig. 2.7. The pressure is seen to be nearly zero at z = 30 km. For tabulated properties see Table A.6. EXAMPLE 2.2 If sea-level pressure is 101,350 Pa, compute the standard pressure at an altitude of 5000 m, using (a) the exact formula and (b) an isothermal assumption at a standard sea-level tem¬ perature of 15°C. Is the isothermal approximation adequate? 68 Chapter 2 Pressure Distribution in a Fluid Part (a) Part (b) Is the Linear Formula Adequate for Gases? Fig. 2.7 Temperature and pressure distribution in the U.S. standard atmosphere. Source: U.S. Standard Atmosphere, 1976, Government Printing Office, Washington DC, 1976. Solution Use absolute temperature in the exact formula, Eq. (2.20): P (0.00650 K/m) (5000 m) ' L 288.16 K 101,350(0.5328) = 54,000 Pa 5.26 (101,350 Pa)(0.8872)“® Ans. (a) This is the standard-pressure result given at z = 5000 m in Table A.6. If the atmosphere were isothermal at 288.16 K, Eq. (2.18) would apply: ‘ Pa exp RT = (101,350 Pa) exp \ - (9.807 m/s^) (5000 m) 287m^/(s^ ■ K) = (101,350 Pa) exp(-0.5929) « 56,000 Pa Ans. (b) This is 4 percent higher than the exact result. The isothermal formula is inaccurate in the troposphere. The linear approximation from Eq. (2.14), Sp ~ —pg 5z, is satisfactory for liquids, which are nearly incompressible. For gases, it is inaccurate unless 5z is rather small. Problem P2.26 asks you to show, by binomial expansion of Eq. (2.20), that the error in using constant gas density to estimate dp from Eq. (2.14) is small if 5z < 2To (n - 1)B (2.21) Temperature, °C 60 50 - 2.4 Application to Manometry 69 2.4 Application to Manometry Pressure Increases Downward Fig. 2.8 Evaluating pressure changes through a column of multiple fluids. where Tg is the local absolute temperature, B is the lapse rate from Eq. (2.19), and n = g/{RB) is the exponent in Eq. (2.20). The error is less than 1 percent if 6z < 200 m. Erom the hydrostatic formula (2.14), a change in elevation Z2 — Zi of a liquid is equivalent to a change in pressure {p2 — P)l^- Thus a static column of one or more liquids or gases can be used to measure pressure differences between two points. Such a device is called a manometer. If multiple fluids are used, we must change the density in the formula as we move from one fluid to another. Figure 2.8 illustrates the use of the formula with a column of multiple fluids. The pressure change through each fluid is calculated separately. If we wish to know the total change — pi, we add the successive changes p2 — P\, Ps — P2, Pa ~ Pi^ ^nd p^ — p^. The intermediate values of p cancel, and we have, for the example of Fig. 2.8, P5 - Pi = -7o(z2 - Zi) - IwiZi - Z2) - 7c(z4 - Z3) - 7m(z5 - Z4) (2.22) No additional simplification is possible on the right-hand side because of the different densities. Notice that we have placed the fluids in order from the lightest on top to the heaviest at bottom. This is the only stable configuration. If we attempt to layer them in any other manner, the fluids will overturn and seek the stable arrangement. The basic hydrostatic relation, Eq. (2. 14), is mathematically correct but vexing to engi¬ neers because it combines two negative signs to have the pressure increase downward. When calculating hydrostatic pressure changes, engineers work instinctively by simply having the pressure increase downward and decrease upward. If point 2 is a distance h below point 1 in a uniform liquid, then P2 = P\ + Pgh. In the meantime, Eq. (2.14) remains accurate and safe if used properly. For example, Eq. (2.22) is correct as shown, or it could be rewritten in the following “multiple downward increase” mode: P5= P+ 7o \|zi - 22! + Iw \Z2 - ZsI + 7g ks - Z4I + 7m 1^4 - ZsI That is, keep adding on pressure increments as you move down through the layered fluid. A different application is a manometer, which involves both “up” and “down” calculations. Known pressure p. z = Zi ^ Oil,p„ P2-Pl=-PoS(z^-zP Water, p ^3 Glycerin, Mercury, Z5 P2-P2 = -P«S{Z-i-zP Pa-P2 = - PaS^^A-zP /’5-P4=- V(Z5-Z4) Sum = Pj -P[ 70 Chapter 2 Pressure Distribution in a Fluid Open, Fig. 2.9 Simple open manometer for measuring relative to atmo¬ spheric pressure. Application: A Simple Manometer Za^Pa — [ a z,,p, - Jump across — h-P2~Pa p = Pj at z = Zj in fluid 2 Figure 2.9 shows a simple U-tube open manometer that measures the gage pressure Pa relative to the atmosphere, p^- The chamber fluid pi is separated from the atmo¬ sphere by a second, heavier fluid po, perhaps because fluid A is corrosive, or more likely because a heavier fluid p2 will keep Z2 small and the open tube can be shorter. We first apply the hydrostatic formula (2.14) from A down to Zi- Note that we can then go down to the bottom of the U-tube and back up on the right side to Zi, and the pressure will be the same, p = P. Thus we can “jump across” and then up to level Zz- Pa + 7i - Zil - 72 ki - 22! = P2^ Fatm (2-23) Another physical reason that we can “jump across” at section 1 is that a continuous length of the same fluid connects these two equal elevations. The hydrostatic relation (2.14) requires this equality as a form of Pascal’s law: Any two points at the same elevation in a continuous mass of the same static fluid will be at the same pressure. This idea of jumping across to equal pressures facilitates multiple-fluid problems. It will be inaccurate, however, if there are bubbles in the fluid. EXAMPLE 2.3 The classic use of a manometer is when two U-tube legs are of equal length, as in Fig. E2.3, and the measurement involves a pressure difference across two horizontal points. The typi¬ cal application is to measure pressure change across a flow device, as shown. Derive a formula for the pressure difference — pi, in terms of the system parameters in Fig. E2.3. Flow device 2.4 Application to Manometry 71 Solution Using Eq. (2.14), start at (a), evaluate pressure changes around the U-tube, and end up at (h): Pa + PigL + Pigh - P2gh - pigL = Pi, or Pa- Pb= iPi - Pi)gh Ans. The measurement only includes /i, the manometer reading. Terms involving L drop out. Note the appearance of the dijference in densities between manometer fluid and working fluid. It is a common student error to fail to subtract out the working fluid density pi — a serious error if both fluids are liquids and less disastrous numerically if fluid 1 is a gas. Academically, of course, such an error is always considered serious by fluid mechanics instructors. Although Example 2.3, because of its popularity in engineering experiments, is sometimes considered to be the “manometer formula,” it is best not to memorize it but rather to adapt Eq. (2.14) to each new multiple-fluid hydrostatics problem. For example, Fig. 2.10 illustrates a multiple-fluid manometer problem for flnding the dif¬ ference in pressure between two chambers A and B. We repeatedly apply Eq. (2.14), jumping across at equal pressures when we come to a continuous mass of the same fluid. Thus, in Fig. 2.10, we compute four pressure differences while making three jumps: Pa- Pb= (Pa - Pi) + (Pi - Pi) + iPi - Pi) + iPi - Pb) (2.24) = -hiZA - Zi) - 72(^1 “ Zi) - Jiizi - Zi) - 74(23 - Zb) The intermediate pressures pixi cancel. It looks complicated, but really it is merely sequential. One starts at A, goes down to 1, jumps across, goes up to 2, jumps across, goes down to 3, jumps across, and Anally goes up to B. Fig. 2.10 A complicated multiple- fluid manometer to relate to pg. This system is not especially practical but makes a good homework or examination problem. Jump across ^2’ Pi “bh'" - ■" - [“[“ ^2-^2 Za-Pa[ a 'Pi Zl.Pl--- Jump across --f f- 7'Pi ZyPi -- 'Pi y Pi Jump across ^3, P3 'P4 B ]Zb.Pb 72 Chapter 2 Pressure Distribution in a Fluid 2.5 Hydrostatic Forces on Plane Surfaces EXAMPLE 2.4 Pressure gage B is to measure the pressure at point A in a water flow. If the pressure at B is 87 kPa, estimate the pressure at A in kPa. Assume all fluids are at 20°C. See Fig. E2.4. Solution ■ System sketch: The system is shown in Fig. E2.4. ■ Assumptions: Hydrostatic fluids, no mixing, vertical “up” in Eig. E2.4. ■ Approach: Sequential use of Eq. (2.14) to go from A to B. ■ Property values: From Table 2.1 or Table A.3: Twater = 9790 N/m^; 'Jra.ctcmy = 133,100 N/m^; 7oq = 8720 N/m^ • Solution steps: Proceed from A to B, “down” then “up,” jumping across at the left mercury meniscus: Pa + Pw I Az\|„ - 7,„ I ^Zm\ - I Az U = Pb or Pa + (9790 N/m^) (0.05m) - (133,100 N/m^)(0.07 m) - (8720 N/m^) (0.06m) = 87,000 or Pa + 490 - 9317 - 523 = 87,000 Solve for Pa = 96,350 N/m^ « 96.4 kPa Ans. ■ Comments: Note that we abbreviated the units N/m^ to pascals, or Pa. The intermediate five-figure result, Pa = 96,350 Pa, is unrealistic, since the data are known to only about three significant figures. In making these manometer calculations we have neglected the capillary height changes due to surface tension, which were discussed in Example 1.8. These effects cancel if there is a fluid interface, or meniscus, between similar fluids on both sides of the U-tube. Otherwise, as in the right-hand U-tube of Fig. 2.10, a capillary correction can be made or the effect can be made negligible by using large-bore (> 1 cm) tubes. The design of containment structures requires computation of the hydrostatic forces on various solid surfaces adjacent to the fluid. These forces relate to the weight of fluid bearing on the surface. For example, a container with a flat, horizontal bottom 2.5 Hydrostatic Forces on Plane Surfaces 73 Fig. 2.11 Hydrostatic force and center of pressure on an arbitrary plane surface of area A inclined at an angle 0 below the free surface. of area A/, and water depth H will experience a downward bottom force Fi, = jHAi,. If the surface is not horizontal, additional computations are needed to find the hori¬ zontal components of the hydrostatic force. If we neglect density changes in the fluid, Eq. (2.14) applies and the pressure on any submerged surface varies linearly with depth. For a plane surface, the linear stress distribution is exactly analogous to combined bending and compression of a beam in strength-of-materials theory. The hydrostatic problem thus reduces to simple formulas involving the centroid and moments of inertia of the plate cross- sectional area. Figure 2.11 shows a plane panel of arbitrary shape completely submerged in a liquid. The panel plane makes an arbitrary angle 9 with the horizontal free surface, so that the depth varies over the panel surface. If h is the depth to any element area dA of the plate, from Eq. (2.14) the pressure there is p = p„ + 'yh. To derive formulas involving the plate shape, establish an xy coordinate system in the plane of the plate with the origin at its centroid, plus a dummy coordinate ^ down from the surface in the plane of the plate. Then the total hydrostatic force on one side of the plate is given by F = p dA iPa + Jh) dA PaA + 7 hdA (2.25) The remaining integral is evaluated by noticing from Fig. 2.11 that h = ^ sin 9 and, by definition, the centroidal slant distance from the surface to the plate is ^CG ~ A ^dA 74 Chapter 2 Pressure Distribution in a Fluid Therefore, since 0 is constant along the plate, Eq. (2.25) becomes F = PaA + 7 sin 6 ^ d4 = ftA + 7 sin 6» Finally, unravel this by noticing that i^cg sin 6 = Hqq, the depth straight down from the surface to the plate centroid. Thus F = pA + jhccA = {pa + 7hcc)A = pccA (2.26) The force on one side of any plane submerged surface in a uniform fluid equals the pressure at the plate centroid times the plate area, independent of the shape of the plate or the angle 9 at which it is slanted. Equation (2.26) can be visualized physically in Fig. 2.12 as the resultant of a linear stress distribution over the plate area. This simulates combined compression and bend¬ ing of a beam of the same cross section. It follows that the “bending” portion of the stress causes no force if its “neutral axis” passes through the plate centroid of area. Thus the remaining “compression” part must equal the centroid stress times the plate area. This is the result of Eq. (2.26). However, to balance the bending-moment portion of the stress, the resultant force F acts not through the centroid but below it toward the high-pressure side. Its line of action passes through the center of pressure CP of the plate, as sketched in Fig. 2.11. To find the coordinates (xcp, ycp); we sum moments of the elemental force p dA about the centroid and equate to the moment of the resultant F. To compute Vcp, we equate Fy CP yp dA yiPa + 7^ sin 9) dA 7 sin 9 yCdA Fig. 2.12 The hydrostatic pressure force on a plane surface is equal, regardless of its shape, to the resultant of the three-dimensional linear pressure distrihution on that surface F = PcA- 2.5 Hydrostatic Forces on Plane Surfaces 75 The term f p^y dA vanishes by definition of centroidal axes. Introducing ^ = ^ca ~ y, we obtain Fycp 7 sin 0 y dA — -7 sin dl^ where again f y dA = 0 and is the area moment of inertia of the plate area about its centroidal x axis, computed in the plane of the plate. Substituting for F gives the result ycp = -7sin 0— - PCG^ (2.27) The negative sign in Eq. (2.27) shows that y^p is below the centroid at a deeper level and, unlike F, depends on angle 9. If we move the plate deeper, y^p approaches the centroid because every term in Eq. (2.27) remains constant except Pqq, which increases. The determination of x^p is exactly similar: FXqp xp dA x[Pa + 7(^cg - }') sin 9] dA = —7 sin 9 xy dA -7 sin 91^ where is the product of inertia of the plate, again computed in the plane of the plate. Substituting for F gives (2.28) For positive xcp is negative because the dominant pressure force acts in the third, or lower left, quadrant of the panel. If 7^, = 0, usually implying symmetry, xcp = 0 and the center of pressure lies directly below the centroid on the y axis. Gage Pressure Formulas In most cases the ambient pressure p^ is neglected because it acts on both sides of the plate; for example, the other side of the plate is inside a ship or on the dry side of a gate or dam. In this case Pqq = 7/icg> the center of pressure becomes inde¬ pendent of specific weight: F = jhccA 7cp /„ sin 9 hcG^ Xcp 7,^, sin 9 hcG^ (2.29) Figure 2.13 gives the area and moments of inertia of several common cross sections for use with these formulas. Note that 9 is the angle between the plate and the horizon. 76 Chapter 2 Pressure Distribution in a Fluid Fig. 2.13 Centroidal moments of inertia for various cross sections: {a) rectangle, {b) circle, (c) triangle, and (d) semicircle. EXAMPLE 2.5 The gate in Fig. E2.5a is 5 ft wide, is hinged at point B, and rests against a smooth wall at point A. Compute (a) the force on the gate due to seawater pressure, (b) the horizontal force P exerted by the wall at point A, and (c) the reactions at the hinge B. Wall 2.5 Hydrostatic Forces on Plane Surfaces 77 Part (a) Part (b) Part (c) Solution By geometry the gate is 10 ft long from A to S, and its centroid is halfway between, or at elevation 3 ft above point B. The depth Hqq is thus 15 — 3 = 12 ft. The gate area is 5(10) = 50 . Neglect as acting on both sides of the gate. From Eq. (2.26) the hydrostatic force on the gate is F = PcqA = jhcaA = (64 lbf/tf)(12 ft)(50 ft^) = 38,400 Ibf Ans. (a) First we must find the center of pressure of F. A free-body diagram of the gate is shown in Fig. E2.5b. The gate is a rectangle, hence 1^ = 0 and 1„ bl? 12 (5ft)(10ft)^ 12 = 417 ft'' The distance I from the CG to the CP is given by Eqs. (2.29) since is neglected. I ~ Tcp ~ + sin 9 hcoA (417ft‘')(^) (12ft)(50ft^) 0.417 ft E2.5b A The distance from point B to force F is thus 10 — / — 5 = 4.583 ft. Summing the moments counterclockwise about B gives PL sm9 - F(5 - I) = P(6 ft) - (38,400 Ibf) (4.583 ft) = 0 or P = 29,300 Ibf Ans. (b) With F and P known, the reactions Bj. and B^ are found by summing forces on the gate: F^ = 0 = B^ + FsinO - P = B, + 38,400 Ibf (0.6) - 29,300 Ibf or B^ = 6300 Ibf 2 = 0 = B, - F cos 6» = S, - 38,400 Ibf (0.8) or B, = 30,700 Ibf Ans. (c) This example should have reviewed your knowledge of statics. 78 Chapter 2 Pressure Distribution in a Fluid The solution of Example 2.5 was achieved with the moment of inertia formulas, Eqs. (2.29). They simplify the calculations, but one loses a physical feeling for the forces. Let us repeat Parts (a) and (b) of Example 2.5 using a more visual approach. EXAMPLE 2.6 Repeat Example 2.5 to sketch the pressure distribution on plate AB, and break this distribu¬ tion into rectangular and triangular parts to solve for (a) the force on the plate and (b) the center of pressure. Solution Part (a) Point A is 9 ft deep, hence = 76^ = (64 lbf/ft^)(9 ft) = 576 Ibf/ft^. Similarly, Point B is 15 ft deep, hence pg = 'yhg = (64 lbf/fT)(15 ft) = 960 Ibf/fL. This defines the linear pressure distribution in Fig. E2.6. The rectangle is 576 by 10 ft by 5 ft into the paper. The triangle is (960 — 576) = 384 Ibf/ft^ X 10 ft by 5 ft. The centroid of the rectangle is 5 ft down the plate from A. The centroid of the triangle is 6.67 ft down from A. The total force is the rectangle force plus the triangle force: E2.6 F = (^576^j(10ft)(5ft) = 28,800 Ibf -f 9600 Ibf /384 lbf\ = 38,400 Ibf Ans. (a) 576 lbf/ft2 Part (b) The moments of these forces about point A are SAL = (28,800 Ibf) (5 ft) -f (9600 Ibf) (6.67 ft) = 144,000 -f 64,000 = 208,000 ft ■ Ibf AL 208,000 ft ■ Ibf Then 5 ft -f / = — = - - = 5.417 ft hence / = 0.417 ft Ans. (b) F 38,400 Ibf Comment: We obtain the same force and center of pressure as in Example 2.5 but with more understanding. However, this approach is awkward and laborious if the plate is not a rectangle. It would be difficult to solve Example 2.7 with the pressure distribu¬ tion alone because the plate is a triangle. Thus moments of inertia can be a useful simplification. 2.5 Hydrostatic Forces on Plane Surfaces 79 Part (a) Part (b) EXAMPLE 2.7 A tank of oil has a right-triangular panel near the bottom, as in Fig. E2.7. Omitting p^, find the (a) hydrostatic force and (b) CP on the panel. Solution The triangle has properties given in Fig. 2.13c. The centroid is one-third up (4 m) and one-third over (2 m) from the lower left comer, as shown. The area is \|(6m)(12m) = 36 m^ The moments of inertia are and bL? (6m)(12m)^ . = ^ = ^ = 288 m^ 36 36 b(b - 2s)L^ (6 m)6 m - 2(6 m)^ = - = - = —72 m 72 72 The depth to the centroid is hco = 5 -I- 4 = 9 m; thus the hydrostatic force from Eq. (2.26) is F = pghcoA = (800kg/m^)(9.807m/s^)(9m)(36m^) = 2.54 X 10® (kg • m)/s^ = 2.54 X 10® N = 2.54 MN Ans. (a) The CP position is given by Eqs. (2.29): Fcp -fcp — sin 0 her, A sin 9 hcnA (288 m') (sin 30°) (9 m)(36 m^) (-72 m') (sin 30°) (9 m)(36m2) = -0.444 m = -fO.lll m Ans. (b) The resultant force F = 2.54 MN acts through this point, which is down and to the right of the centroid, as shown in Pig. E2.7. 80 Chapter 2 Pressure Distribution in a Fluid 2.6 Hydrostatic Forces on Curved Surfaces Fig. 2.14 Computation of hydrostatic force on a curved surface: {a) submerged curved surface; {b) free-body diagram of fluid above the curved surface. The resultant pressure force on a curved surface is most easily computed by separating it into horizontal and vertical components. Consider the arbitrary curved surface sketched in Fig. 2.14a. The incremental pressure forces, being normal to the local area element, vary in direction along the surface and thus cannot be added numerically. We could sum the separate three components of these elemental pressure forces, but it turns out that we need not perform a laborious three-way integration. Figure 2.\Ab shows a free-body diagram of the column of fluid contained in the vertical projection above the curved surface. The desired forces Fh and Fy are exerted by the surface on the fluid column. Other forces are shown due to fluid weight and horizontal pressure on the vertical sides of this column. The column of fluid must be in static equilibrium. On the upper part of the column bcde, the horizontal components Fi exactly balance and are not relevant to the discussion. On the lower, irregular portion of fluid abc adjoining the surface, summation of horizontal forces shows that the desired force Fh due to the curved surface is exactly equal to the force Fh on the vertical left side of the fluid column. This left-side force can be computed by the plane surface formula, Eq. (2.26), based on a vertical projection of the area of the curved surface. This is a general rule and simplifies the analysis: The horizontal component of force on a curved surface equals the force on the plane area formed by the projection of the curved surface onto a vertical plane normal to the component. If there are two horizontal components, both can be computed by this scheme. Sum¬ mation of vertical forces on the fluid free body then shows that Fy=W, + W2 + Wai, (2.30) We can state this in words as our second general rule: The vertical component of pressure force on a curved surface equals in magnitude and direction the weight of the entire column of fluid, both liquid and atmosphere, above the curved surface. Curved surface projection onto ia) (b) 2.6 Hydrostatic Forces on Curved Surfaces 81 Thus the calculation of Fy involves little more than finding centers of mass of a col¬ umn of fluid — perhaps a little integration if the lower portion abc in Fig. 2.14& has a particularly vexing shape. EXAMPLE 2.8 A dam has a parabolic shape zIzq = (jc/jcq)^ as shown in Fig. E2.8a, with Xq = 10 ft and Zq = 24 ft. The fluid is water, 7 = 62.4 lbf/ft'\ and atmospheric pressure may be omitted. Compute the forces Fh and Fy on the dam and their line of action. The width of the dam is 50 ft. p^ = 0 Ibf/fF gage Solution • System sketch: Figure E2.8h shows the various dimensions. The dam width is h = 50 ft. ■ Approach: Calculate Fh and its line of action from Eqs. (2.26) and (2.29). Calculate Fy and its line of action by finding the weight of fluid above the parabola and the centroid of this weight. • Solution steps for the horizontal component: The vertical projection of the parabola lies along the z axis in Eig. E2.8h and is a rectangle 24 ft high and 50 ft wide. Its centroid is halfway down, or h^Q = 24/2 = 12 ft. Its area is Ap„j = (24 ft)(50 ft) = 1200 ft^. Then, from Eq. (2.26), Fh = T/tcoAproj = (^62.4^j(12ft)(1200tf) = 898,560 Ibf « 899 X lOMbf The line of action of Fh is below the centroid of Ap^j, as given by Eq. (2.29): /,„sin6» (1/I2)(50ft)(24ft)^sin90“ Fcp. proj - ~ ( 12 ft) ( 1200 ft^) ~ Thus is 12 -f 4 = 16 ft, or two-thirds of the way down from the surface (8 ft up from the bottom). • Comments: Note that you calculate Fh and its line of action from the vertical projection of the parabola, not from the parabola itself. Since this projection is vertical, its angle 9 = 90°. • Solution steps for the vertical component: The vertical force Fy equals the weight of water above the parabola. Alas, a parabolic section is not in Fig. 2.13, so we had to look 82 Chapter 2 Pressure Distribution in a Fluid it up in another book. The area and centroid are shown in Fig. E2.8fc. The weight of this parabolic amount of water is ^ ^section^ ( 62.4 Ibf ft^ (24 ft) (10 ft) (50 ft) = 499,200 Ibf « 499 X 10^ Ibf This force acts downward, through the centroid of the parabolic section, or at a distance 3xo/8 = 3.75 ft over from the origin, as shown in Figs. E2.8fc,c. The resultant hydrostatic force on the dam is F = (FI + Fyf^ = [(899E3 Ibf)^ + (499E3 lbf)^]‘® = 1028 X 10^ Ibf at^^9^ Ans. This resultant is shown in Eig. E2.8c and passes through a point 8 ft up and 3.75 ft over from the origin. It strikes the dam at a point 5.43 ft over and 7.07 ft up, as shown. • Comments: Note that entirely different formulas are used to calculate Fu and Fy- The concept of center of pressure CP is, in the writer’s opinion, stretched too far when applied to curved surfaces. z 2.7 Hydrostatic Forces in Layered Fluids 83 E2.9 2.7 Hydrostatic Forces in Layered Fluids Fig. 2.15 Hydrostatic forces on a surface immersed in a layered fluid must be summed in separate pieces. EXAMPLE 2.9 Find an algebraic formula for the net vertical force F on the submerged semicircular project¬ ing structure CDE in Fig. E2.9. The structure has uniform width b into the paper. The liquid has specific weight 7. Solution The net force is the difference between the upward force F/, on the lower surface DE and the downward force Fy on the upper surface CD, as shown in Fig. E2.9. The force Fy equals 7 times the volume ABDC above surface CD. The force F^ equals 7 times the volume ABDEC above surface DE. The latter is clearly larger. The difference is 7 times the volume of the structure itself. Thus the net upward fluid force on the semicylinder is TT , F = Tnuid (volume CDE) = y F b Ans. This is the principle upon which the laws of buoyancy. Sec. 2.8, are founded. Note that the result is independent of the depth of the structure and depends upon the specific weight of ihe fluid, not the material within the structure. The formulas for plane and curved surfaces in Secs. 2.5 and 2.6 are valid only for a fluid of uniform density. If the fluid is layered with different densities, as in Fig. 2.15, a single formula cannot solve the problem because the slope of the linear pressure z 84 Chapter 2 Pressure Distribution in a Fluid distribution changes between layers. However, the formulas apply separately to each layer, and thus the appropriate remedy is to compute and sum the separate layer forces and moments. Consider the slanted plane surface immersed in a two-layer fluid in Fig. 2.15. The slope of the pressure distribution becomes steeper as we move down into the denser second layer. The total force on the plate does not equal the pressure at the centroid times the plate area, but the plate portion in each layer does satisfy the formula, so that we can sum forces to And the total: ^=2^, = 2pcgA (2.31) Similarly, the centroid of the plate portion in each layer can be used to locate the center of pressure on that portion: Fcp, Pig sin PcaAi .CPi Pig sin 61, 4, PcoAi (2.32) These formulas locate the center of pressure of that particular F", with respect to the centroid of that particular portion of plate in the layer, not with respect to the centroid of the entire plate. The center of pressure of the total force F" = S F, can then be found by summing moments about some convenient point such as the surface. The following example will illustrate this. EXAMPLE 2.10 A tank 20 ft deep and 7 ft wide is layered with 8 ft of oil, 6 ft of water, and 4 ft of mercury. Compute (a) the total hydrostatic force and (b) the resultant center of pressure of the fluid on the right-hand side of the tank. Solution Part (a) Divide the end panel into three parts as sketched in Fig. E2. 10, and find the hydrostatic pressure at the centroid of each part, using the relation (2.26) in steps as in Fig. E2.10: Pea, = (55.01bf/fF)(4ft) = 220 Ibf/fF Pea, = (55.0)(8) -f 62.4(3) = 627 Ibf/ft^ Pea, = (55.0)(8) -f 62.4(6) -f 846(2) = 2506 Ibf/ft- These pressures are then multiplied by the respective panel areas to find the force on each portion: Fi = PeaAi = (2201bf/ft^)(8ft)(7ft) = 12,300 Ibf Fi = PcaAi = 627(6)(7) = 26,300 Ibf Fs = Pca,A-i = 2506(4) (7) = 70,200 Ibf F = 'ZFi^ 108,800 Ibf Ans. (a) 2.8 Buoyancy and Stability 85 Part (b) 2.8 Buoyancy and Stability Equations (2.32) can be used to locate the CP of each force F„ noting that 6 = 90° and sin 6 = 1 for all parts. The moments of inertia are = (7 ft)(8 ft)^/12 = 298.7 ft", = 7(6)^/12 = 126.0 ft', and = 7(4)^/12 = 37.3 ft"^. The centers of pressure are thus at Tcp, “ Pigl^ (55.0 Ibf/ft^) (298.7 ft'^) TcPj Fi 62.4(126.0) 26,300 12,300 Ibf = —0.30 ft ycp, = = -1.33 ft 846(37.3) 70,200 = -0.45 ft This locates zcp, = —4 — 1.33 = —5.33 ft, zcp^ =—11 — 0.30 = —11.30 ft, and zcp, = 0.45 = — 16.45 ft. Summing moments about the surface then gives -16- 2F,Zcp, = Fzcp 12,300(-5.33) + 26,300(- 11.30) + 70,200(- 16.45) = 108,800zcp 1,518,000 Zcp — 108,800 = -13.95 ft Ans. (b) The center of pressure of the total resultant force on the right side of the tank lies 13.95 ft below the surface. The same principles used to compute hydrostatic forces on surfaces can be applied to the net pressure force on a completely submerged or floating body. The results are the two laws of buoyancy discovered by Archimedes in the third century b.c.: A body immersed in a fluid experiences a vertical buoyant force equal to the weight of the fluid it displaces. A floating body displaces its own weight in the fluid in which it floats. 86 Chapter 2 Pressure Distribution in a Fluid Fig. 2.16 Two different approaches to the buoyant force on an arbitrary immersed body: (a) forces on upper and lower curved surfaces; (b) summation of elemental vertical-pressure forces. (a) Horizontal (b) Archimedes (287-212 b.c.) was born and lived in the Greek city-state of Syracuse, on what is now the island of Sicily. He was a brilliant mathematician and engineer, two millennia ahead of his time. He calculated an accurate value for pi and approxi¬ mated areas and volumes of various bodies by summing elemental shapes. In other words, he invented the integral calculus. He developed levers, pulleys, catapults, and a screw pump. Archimedes was the first to write large numbers as powers of 10, avoiding Roman numerals. And he deduced the principles of buoyancy, which we study here, when he realized how light he was when sitting in a bathtub. Archimedes’ two laws are easily derived by referring to Fig. 2.16. In Fig. 2.16a, the body lies between an upper curved surface 1 and a lower curved surface 2. From Eq. (2.30) for vertical force, the body experiences a net upward force Fb = Fy(2) -Fy(l) = (fluid weight above 2) — (fluid weight above 1) = weight of fluid equivalent to body volume (2.33) Alternatively, from Fig. 2.16ft, we can sum the vertical forces on elemental vertical slices through the immersed body: Fb iPi - Pi) dAu = ^body (Z2 ^l) dAjq (7) (body volume) (2.34) These are identical results and equivalent to Archimedes’ law 1. Equation (2.34) assumes that the fluid has uniform specific weight. The line of action of the buoyant force passes through the center of volume of the displaced body; that is, its center of mass computed as if it had uniform density. This point through which Fb acts is called the center of buoyancy, commonly labeled B or CB on a drawing. Of course, the point B may or may not correspond to the actual center of mass of the body’s own material, which may have variable density. 2.8 Buoyancy and Stability 87 Fig. 2.17 Static equilibrium of a floating body. Equation (2.34) can be generalized to a layered fluid (LF) by summing the weights of each layer of density p, displaced by the immersed body: (^b)lf = 2 p,g(displaced volume),- (2.35) Each displaced layer would have its own center of volume, and one would have to sum moments of the incremental buoyant forces to And the center of buoyancy of the immersed body. Since liquids are relatively heavy, we are conscious of their buoyant forces, but gases also exert buoyancy on any body immersed in them. For example, human beings have an average specific weight of about 60 Ibf/ft^. We may record the weight of a person as 180 Ibf and thus estimate the person’s total volume as 3.0 ft^. However, in so doing we are neglecting the buoyant force of the air surrounding the person. At standard conditions, the specihc weight of air is 0.0763 Ibf/ft^; hence the buoyant force is approximately 0.23 Ibf. If measured in a vacuum, the person would weigh about 0.23 Ibf more. For balloons and blimps the buoyant force of air, instead of being negligible, is the controlling factor in the design. Also, many flow phenomena, such as natural convection of heat and vertical mixing in the ocean, are strongly dependent on seemingly small buoyant forces. Floating bodies are a special case; only a portion of the body is submerged, with the remainder poking up out of the free surface. This is illustrated in Fig. 2.17, where the shaded portion is the displaced volume. Equation (2.34) is modified to apply to this smaller volume: Fg = (7) (displaced volume) = floating-body weight (2.36) Not only does the buoyant force equal the body weight, but also they are colUnear since there can be no net moments for static equilibrium. Equation (2.36) is the math¬ ematical equivalent of Archimedes’ law 2, previously stated. EXAMPLE 2.11 A block of concrete weighs 100 Ibf in air and “weighs” only 60 Ibf when immersed in fresh water (62.4 Ibf/ft^). What is the average specific weight of the block? 88 Chapter 2 Pressure Distribution in a Fluid eOlbf IF= lOOlbf E2.11 Stability Solution A free-body diagram of the submerged block (see Fig. E2.11) shows a balance between the apparent weight, the buoyant force, and the actual weight: 2 F, = 0 = 60 + Fg - 100 or Fg = 40 Ibf = (62.4 Ibf/ft^) (block volume, ft^) Solving gives the volume of the block as 40/62.4 = 0.641 ft^. Therefore, the specific weight of the block is Tblock 100 Ibf 0.641 fr'' = 1561bf/tf Ans. Occasionally, a body will have exactly the right weight and volume for its ratio to equal the specific weight of the fluid. If so, the body will be neutrally buoyant and will remain at rest at any point where it is immersed in the fluid. Small, neutrally buoyant particles are sometimes used in flow visualization, and a neutrally buoyant body called a Swallow float is used to track oceanographic currents. A submarine can achieve positive, neutral, or negative buoyancy by pumping water into or out of its ballast tanks. A floating body as in Fig. 2.17 may not approve of the position in which it is floating. If so, it will overturn at the first opportunity and is said to be statically unstable, like a pencil balanced on its point. The least disturbance will cause it to seek another equilib¬ rium position that is stable. Engineers must design to avoid floating instability. The only way to tell for sure whether a floating position is stable is to “disturb” the body a slight amount mathematically and see whether it develops a restoring moment that will return it to its original position. If so, it is stable; if not, unstable. Such calculations for arbitrary floating bodies have been honed to a fine art by naval architects , but we can at least outline the basic principle of the static stability calculation. Figure 2.18 illustrates the computation for the usual case of a symmetric floating body. The steps are as follows: The basic floating position is calculated from Eq. (2.36). The body’s center of mass G and center of buoyancy B are computed. The body is tilted a small angle A9, and a new waterline is established for the body to float at this angle. The new position B' of the center of buoyancy is calculated. A vertical line drawn upward from B' intersects the line of symmetry at a point M, called the metacenter, which is independent of A6 for small angles. If point M is above G (that is, if the metacentric height MG is positive), a restoring moment is present and the original position is stable. If M is below G (negative MG), the body is unstable and will overturn if disturbed. Stability increases with increasing MG. 2.8 Buoyancy and Stability 89 Fig. 2.18 Calculation of the metacenter M of the floating body shown in (a). Tilt the body a small angle Aft Either (b) B' moves far out (point M above G denotes stability); or (c) B’ moves slightly (point M below G denotes instability). Small Small Either Restoring moment or Overturning moment (a) (b) ic) Thus the metacentric height is a property of the cross section for the given weight, and its value gives an indication of the stability of the body. For a body of varying cross section and draft, such as a ship, the computation of the metacenter can be very involved. Stability Related to Waterline Area^ Naval architects have developed the general stability concepts from Fig. 2.18 into a simple computation involving the area moment of inertia of the water¬ line area (as seen from above) about the axis of tilt. The derivation — see for details — assumes that the body has a smooth shape variation (no discontinuities) near the waterline. Recall that M is the metacenter, B is the center of buoyancy, and G is the center of gravity. The final elegant formula relates the distances between these points: /o — MG = -- GB Vmh (2.37) Where Iq is the area moment of inertia of the waterline area about the tilt axis O and Vsub is the volume of the submerged portion of the floating body. It is desirable, of course, that MG be positive for the body to be stable. The engineer locates G and B from the basic shape and design of the floating body and then calculates Iq and Vjub to determine if MG is positive. Engineering design counts upon effective operation of the results. A stability analy¬ sis is useless if the floating body runs aground on rocks, as in Fig. 2.19. ^This section may be omitted without loss of continuity. 90 Chapter 2 Pressure Distribution in a Fluid Fig. 2.19 The Italian liner Costa Concordia aground on January 14, 2012. Stability analysis may fail when operator mistakes occur {Associated Press photo/Gregorio Borgia). EXAMPLE 2.12 A barge has a uniform rectangular cross section of width 2L and vertical draft of height H, as in Fig. E2.12. Determine (a) the metacentric height for a small tilt angle and (b) the range of ratio UH for which the barge is statically stable if G is exactly at the waterline as shown. Solution If the barge has length b into the paper, the waterline area, relative to tilt axis O, has a base b and a height 2L; therefore, Iq = b(2L)^/12. Meanwhile, = 2LbH. Equation (2.37) predicts MG 4 - 8Z>LV12 -- GB = - ■Wsub '2-LbH 2~ m 2 Arts, (a) The barge can thus be stable only if 3//V2 or 2L > 2.45// Ans. (b) The wider the barge relative to its draft, the more stable it is. Lowering G would help also. 2.9 Pressure Distribution in Rigid-Body Motion 91 Fig. 2.20 A North Atlantic iceberg formed by calving from a Greenland glacier. These, and their even larger Antarctic sisters, are the largest floating bodies in the world. Note the evidence of further calving fractures on the front surface. (© Corbis.) 2.9 Pressure Distribution in Rigid-Body Motion Even an expert will have difficulty determining the floating stability of a buoyant body of irregular shape. Such bodies may have two or more stable positions. For example, a ship may float the way we like it, so that we can sit on the deck, or it may float upside down (capsized). An interesting mathematical approach to floating stability is given in Ref. 1 1 . The author of this reference points out that even simple shapes, such as a cube of uniform density, may have a great many stable floating orientations, not necessarily symmetric. Homogeneous circular cylinders can float with the axis of symmetry tilted from the vertical. Floating instability occurs in nature. Fish generally swim with their planes of sym¬ metry vertical. After death, this position is unstable and they float with their flat sides up. Giant icebergs may overturn after becoming unstable when their shapes change due to underwater melting. Iceberg overturning is a dramatic, rarely seen event. Figure 2.20 shows a typical North Atlantic iceberg formed by calving from a Greenland glacier that protruded into the ocean. The exposed surface is rough, indicat¬ ing that it has undergone further calving. Icebergs are frozen fresh, bubbly, glacial water of average density 900 kg/m^. Thus, when an iceberg is floating in seawater, whose average density is 1025 kg/m^, approximately 900/1025, or seven-eighths, of its volume lies below the water. In rigid-body motion, all particles are in combined translation and rotation, and there is no relative motion between particles. With no relative motion, there are no strains or strain rates, so that the viscous term in Eq. (2.8) vanishes, leaving a balance between pressure, gravity, and particle acceleration: Vp = p(g - a) (2.38) The pressure gradient acts in the direction g — a, and lines of constant pressure (including the free surface, if any) are perpendicular to this direction. The general case of combined translation and rotation of a rigid body is discussed in Chap. 3, Fig. 3.11. 92 Chapter 2 Pressure Distribution in a Fluid Fig. 2.21 Tilting of constant-pressure surfaces in a tank of liquid in rigid- body acceleration. z Fluids can rarely move in rigid-body motion unless restrained by confining walls for a long time. For example, suppose a tank of water is in a car that starts a constant acceleration. The water in the tank would begin to slosh about, and that sloshing would damp out very slowly until finally the particles of water would be in approximately rigid-body acceleration. This would take so long that the car would have reached hypersonic speeds. Nevertheless, we can at least discuss the pressure distribution in a tank of rigidly accelerating water. Uniform Linear Acceleration In the case of uniform rigid-body acceleration, Eq. (2.38) applies, a having the same magnitude and direction for all particles. With reference to Fig. 2.21, the parallelo¬ gram sum of g and —a gives the direction of the pressure gradient or greatest rate of increase of p. The surfaces of constant pressure must be perpendicular to this and are thus tilted at a downward angle 0 such that e = tan”'— ^ (2.39) g + a. One of these tilted lines is the free surface, which is found by the requirement that the fluid retain its volume unless it spills out. The rate of increase of pressure in the direction g — a is greater than in ordinary hydrostatics and is given by — = pG where G = [al + {g + (2.40) as These results are independent of the size or shape of the container as long as the fluid is continuously connected throughout the container. EXAMPLE 2.13 A drag racer rests her coffee mug on a horizontal tray while she accelerates at 7 m/s^. The mug is 10 cm deep and 6 cm in diameter and contains coffee 7 cm deep at rest, {a) Assum¬ ing rigid-body acceleration of the coffee, determine whether it will spill out of the mug. (b) Calculate the gage pressure in the comer at point A if the density of coffee is 1010 kg/m^. 2.9 Pressure Distribution in Rigid-Body Motion 93 Solution • System sketch: Figure E2.13 shows the coffee tilted during the acceleration. • Assumptions: Rigid-body horizontal acceleration, = 7 m/s^. Symmetric coffee cup. • Property values: Density of coffee given as 1010 kg/m^. • Approach (a): Determine the angle of tilt from the known acceleration, then find the height rise. • Solution steps: From Eq. (2.39), the angle of tilt is given by 0 = tan ‘ Clx Y 7.0 m/s^ tan - r 9.81 m/s^ 35.5“ If the mug is symmetric, the tilted surface will pass through the center point of the rest position, as shown in Fig. E2.13. Then the rear side of the coffee free surface will rise an amount Az given by Az = (3 cm)(tan 35.5°) = 2.14 cm < 3 cm therefore no spilling Ans. (a) • Comment (a): This solution neglects sloshing, which might occur if the start-up is uneven. ■ Approach (b): The pressure at A can be computed from Eq. (2.40), using the perpen¬ dicular distance As from the surface to A. When at rest, p^ = pgh^cst = (1010 kg/m^) (9.81 m/s^)(0.07 m) = 694 Pa. Pa = pG As = 1010 kg' V(9.81)^ -f (7.0)^ [(0.07 -f 0.0214) cos35.5°] »= 906 Pa Ans. (b) • Comment (b): The acceleration has increased the pressure at A by 31 percent. Think about this alternative: why does it work? Since a^ = 0, we may proceed vertically down the left side to compute Pa = P^(Zsurf - Za) = (1010kg/m^)(9.81 m/s2)(0.0214 m -f 0.07 m) = 906 Pa 94 Chapter 2 Pressure Distribution in a Fluid Fig. 2.22 Development of paraboloid constant-pressure surfaces in a fluid in rigid-body rotation. The dashed line along the direction of maximum pressure increase is an exponential curve. z, k Rigid-Body Rotation As a second special case, consider rotation of the fluid about the z axis without any translation, as sketched in Fig. 2.22. We assume that the container has been rotating long enough at constant (1 for the fluid to have attained rigid-body rotation. The fluid acceleration will then be a centripetal term. In the coordinates of Fig. 2.22, the angular- velocity and position vectors are given by fl = kd ro = ii-r (2.41) Then the acceleration is given by d X (il X ro) = -rD,\ (2.42) as marked in the figure, and Eq. (2.38) for the force balance becomes dp dp , Vp = i,— + k— = pig - a) = pi-gk + rflX) dr dz Equating like components, we find the pressure field by solving two first-order partial differential equations: dr prU^ (2.43) The right-hand sides of (2.43) are known functions of r and z- One can proceed as follows: Integrate the first equation “partially,” holding z constant, with respect to r. The result is p = \pr^n^ + fiz) (2.44) where the “constant” of integration is actually a function /(z).^ Now differentiate this with respect to z and compare with the second relation of (2.43): dp ^ = 0+/'(z) = -7 dz ^This is because /(z) vanishes when differentiated with respect to r. If you don’t see this, you should review your calculus. 2.9 Pressure Distribution in Rigid-Body Motion 95 Fig. 2.23 Determining the free- surface position for rotation of a cylinder of fluid about its central axis. Still¬ water level Volume = Y R^h T h = 1 2g or /(z) = -7z + C where C is a constant. Thus Eq. (2.44) now becomes p = const — 7z + (2.45) This is the pressure distribution in the fluid. The value of C is found by specifying the pressure at one point. \i p = Pq at (r, z) = (0, 0), then C = p^. The final desired distribution is P = Pq- 1Z + kpr^il- (2.46) The pressure is linear in z and parabolic in r. If we wish to plot a constant-pressure surface, say, p = pi, Eq. (2.45) becomes Po - Pi z = - 7 -b 2g = a + br^ (2.47) Thus the surfaces are paraboloids of revolution, concave upward, with their minimum points on the axis of rotation. Some examples are sketched in Fig. 2.22. As in the previous example of linear acceleration, the position of the free surface is found by conserving the volume of fluid. For a noncircular container with the axis of rotation off-center, as in Fig. 2.22, a lot of laborious mensuration is required, and a single problem will take you all weekend. However, the calculation is easy for a cylinder rotating about its central axis, as in Fig. 2.23. Since the volume of a parabo¬ loid is one-half the base area times its height, the still-water level is exactly halfway between the high and low points of the free surface. The center of the fluid drops an amount hl2 = n^/?^/(4g), and the edges rise an equal amount. EXAMPLE 2.14 The coffee cup in Example 2.13 is removed from the drag racer, placed on a turntable, and rotated about its central axis until a rigid-body mode occurs. Find (a) the angular velocity that will cause the coffee to just reach the lip of the cup and (b) the gage pressure at point A for this condition. 96 Chapter 2 Pressure Distribution in a Fluid Solution Part (a) The cup contains 7 cm of coffee. The remaining distance of 3 cm up to the lip must equal the distance hH in Fig. 2.23. Thus Solving, we obtain h 2 = 0.03 m = 4.? n^(0.03 m)^ 4(9.81 m/s^) = 1308 or D = 36.2 rad/s = 345 r/min Ans. (a) To compute the pressure, it is convenient to put the origin of coordinates r and z at the bottom of the free-surface depression, as shown in Fig. E2.14. The gage pressure here is = 0, and point A is at (r, z) = (3 cm, —4 cm). Equation (2.46) can then be evaluated: Pa = 0- (1010 kg/m^) (9.81 m/s^)(-0.04 m) i(1010 kg/m^)(0.03 m)^(1308 radV) = 396 N/m^ + 594 N/m^ = 990 Pa Ans. (b) This is about 43 percent greater than the still-water pressure p^ = 694 Pa. Here, as in the linear acceleration case, it should be emphasized that the paraboloid pressure distribution (2.46) sets up in any fluid under rigid-body rotation, regardless of the shape or size of the container. The container may even be closed and filled with fluid. It is only necessary that the fluid be continuously interconnected throughout the container. The following example will illustrate a peculiar case in which one can visualize an imaginary free surface extending outside the walls of the container. E2.15 EXAMPLE 2.15 A U-tube with a radius of 10 in and containing mercury to a height of 30 in is rotated about its center at 180 r/min until a rigid-body mode is achieved. The diameter of the tubing is negligible. Atmospheric pressure is 2116 Ibf/ft^. Find the pressure at point A in the rotating condition. See Fig. E2.15. Solution Convert the angular velocity to radians per second: „ 27rrad/r H = (180 r/min) - = 18.85 rad/s 60 s/mm From Table 2.1 we find for mercury that 7 = 846 Ibf/ft^ and hence p = 846/32.2 = 26.3 slugs/ft^. At this high rotation rate, the free surface will slant upward at a fierce angle [about 84°; check this from Eq. (2.47)], but the tubing is so thin that the free surface will remain at approximately the same 30-in height, point B. Placing our origin of coordinates 2.9 Pressure Distribution in Rigid-Body Motion 97 at this height, we can calculate the constant C in Eq. (2.45) from the condition = 2116 Ibf/ft^ at (r, z) = (10 in, 0): Pb = 21161bf/ft^ = C - 0 -f j(26.3 slugs/ft^)(n ft)2(18.85 rad/s)^ or C = 2116 - 3245 = -1129 Ibf/ft^ We then obtain p/^ by evaluating Eq. (2.46) at (r, z) = (0, —30 in): Pa= -1129 - (8461bf/ft^)(-ff ft) = -1129 -f 2115 = 986 Ibf/ft^ Ans. This is less than atmospheric pressure, and we can see why if we follow the free-surface paraboloid down from point B along the dashed line in the figure. It will cross the horizon¬ tal portion of the U-tuhe (where p will be atmospheric) and fall below point A. From Fig. 2.23 the actual drop from point B will be h = (18.85)"(1§)" 2g 2(32.2) = 3.83 ft = 46 in Thus is about 16 inHg below atmospheric pressure, or about 11(846) = 1128 Ibf/ft^ helow Pa = 2116 Ibf/ft^, which checks with the answer above. When the tube is at rest. Pa = 2116 - 846(-ff) = 4231 Ibf/tf Hence rotation has reduced the pressure at point A by 77 percent. Further rotation can reduce Pa to near-zero pressure, and cavitation can occur. An interesting by-product of this analysis for rigid-body rotation is that the lines everywhere parallel to the pressure gradient form a family of curved surfaces, as sketched in Fig. 2.22. They are everywhere orthogonal to the constant-pressure surfaces, and hence their slope is the negative inverse of the slope computed from Eq. (2.47): dz dr GL 1 idzldr)p=^„^,t where GL stands for gradient line or tfe ^ _ ^ dr rfi^ 1 rfl^/g (2.48) Separating the variables and integrating, we find the equation of the pressure-gradient surfaces: r = Cl exp (2.49) Notice that this result and Eq. (2.47) are independent of the density of the fluid. In the absence of friction and Coriolis effects, Eq. (2.49) defines the lines along which the apparent net gravitational held would act on a particle. Depending on its density, a small particle or bubble would tend to rise or fall in the huid along these Chapter 2 Pressure Distribution in a Fluid Fig. 2.24 Experimental demonstration with buoyant streamers of the fluid force field in rigid-body rotation: (top) fluid at rest (streamers hang vertically upward); (bottom) rigid-body rotation (streamers are aligned with the direction of maximum pressure gradient). (© The American Association of Physics Teachers. Reprinted with permission from ‘‘The Apparent Field of Gravity in a Rotating Fluid System” by R. Ian Fletcher. American Journal of Physics vol. 40, pp. 959-965, July 1972.) 2.10 Pressure Measurement 99 2.10 Pressure Measurement exponential lines, as demonstrated experimentally in Ref. 5. Also, buoyant streamers would align themselves with these exponential lines, thus avoiding any stress other than pure tension. Figure 2.24 shows the configuration of such streamers before and during rotation. Pressure is a derived property. It is the force per unit area as related to fluid molecular bombardment of a surface. Thus most pressure instruments only infer the pressure by calibration with a primary device such as a deadweight piston tester. There are many such instruments, for both a static fluid and a moving stream. The instrumentation texts in Refs. 7 to 10, 12, 13, and 16-17 list over 20 designs for pressure measurement instruments. These instruments may be grouped into four categories: Gravity-based: barometer, manometer, deadweight piston. Elastic deformation: bourdon tube (metal and quartz), diaphragm, bellows, strain-gage, optical beam displacement. Gas behavior: gas compression (McLeod gage), thermal conductance (Pirani gage), molecular impact (Knudsen gage), ionization, thermal conductivity, air piston. Electric output: resistance (Bridgman wire gage), diffused strain gage, capacitative, piezoelectric, potentiometric, magnetic inductance, magnetic reluctance, linear variable differential transformer (LVDT), resonant frequency. Luminescent coatings for surface pressures . The gas-behavior gages are mostly special-purpose instruments used for certain scientific experiments. The deadweight tester is the instrument used most often for calibrations; for example, it is used by the U.S. National Institute for Standards and Technology (NIST). The barometer is described in Fig. 2.6. The manometer, analyzed in Sec. 2.4, is a simple and inexpensive hydrostatic- principle device with no moving parts except the liquid column itself. Manometer measurements must not disturb the flow. The best way to do this is to take the mea¬ surement through a static hole in the wall of the flow, as illustrated in Fig. 2.25a. The hole should be normal to the wall, and burrs should be avoided. If the hole is small enough (typically 1-mm diameter), there will be no flow into the measuring tube once the pressure has adjusted to a steady value. Thus the flow is almost undis¬ turbed. An oscillating flow pressure, however, can cause a large error due to possible dynamic response of the tubing. Other devices of smaller dimensions are used for dynamic -pressure measurements. The manometer in Fig. 2.25fl measures the gage pressure p^. The instrument in Fig. 2.25b is a digital differential manometer, which can measure the difference between two different points in the flow, with stated accuracy of 0. 1 percent of full scale. The world of instrumentation is moving quickly toward digital readings. 100 Chapter 2 Pressure Distribution in a Fluid Flow Fig. 2.25 Two types of accurate manometers for precise measure¬ ments: (a) tilted tube with eyepiece; {b) a capacitive-type digital manometer of rated accuracy ±0.1 percent. (Courtesy of Dwyer Instruments, Inc.) Pi In category 2, elastic-deformation instruments, a popular, inexpensive, and reliable device is the bourdon tube, sketched in Fig. 2.26. When pressurized internally, a curved tube with flattened cross section will deflect outward. The deflection can be measured by a linkage attached to a calibrated dial pointer, as shown. Or the deflec¬ tion can be used to drive electric-output sensors, such as a variable transformer. Similarly, a membrane or diaphragm will deflect under pressure and can either be sensed directly or used to drive another sensor. An interesting variation of Fig. 2.26 is the fused-quartz, force-balanced bourdon tube, shown in Fig. 2.27, whose spiral-tube deflection is sensed optically and returned to a zero reference state by a magnetic element whose output is proportional to the Fig. 2.26 Schematic of a bourdon- tube device for mechanical measurement of high pressures. I High pressure 2.10 Pressure Measurement 101 Fig. 2.27 The fused-quartz, force- balanced bourdon tube is the most accurate pressure sensor used in commercial applications today. ( Courtesy ofRuska Instrument Corporation, Houston, TX.) fluid pressure. The fused-quartz, force-balanced bourdon tube is reported to be one of the most accurate pressure sensors ever devised, with uncertainty on the order of ±0.003 percent. The quartz gages, both the bourdon type and the resonant type, are expensive but extremely accurate, stable, and reliable . They are often used for deep- ocean pressure measurements, which detect long waves and tsunami activity over extensive time periods. The last category, electric-output sensors, is extremely important in engineering because the data can be stored on computers and freely manipulated, plotted, and analyzed. Three examples are shown in Fig. 2.28, the first being the capacitive sensor in Fig. 2.28fl. The differential pressure deflects the silicon diaphragm and changes the capacitance of the liquid in the cavity. Note that the cavity has spherical end caps to prevent overpressure damage. In the second type. Fig. 2.28i>, strain gages and other sensors are chemically diffused or etched onto a chip, which is stressed by the applied pressure. Finally, in Fig. 2.28c, a micromachined silicon sensor is arranged to deform under pressure such that its natural vibration frequency is proportional to the pressure. An oscillator excites the element’s resonant frequency and converts it into appropriate pressure units. Another kind of dynamic electric-output sensor is the piezoelectric transducer, shown in Fig. 2.29. The sensing elements are thin layers of quartz, which generate an electric charge when subjected to stress. The design in Fig. 2.29 is flush-mounted on a solid surface and can sense rapidly varying pressures, such as blast waves. Other designs are of the cavity type. This type of sensor primarily detects transient pressures, not steady stress, but if highly insulated can also be used for short-term static events. Note also that it measures gage pressure — that is, it detects only a change from ambi¬ ent conditions. Cover flange Seal diaphragm (a) Fig. 2.28 Pressure sensors with electric output: (a) a silicon diaphragm whose deflection changes the cavity capacitance (b) a silicon strain gage that is stressed by applied pressure; (c) a micromachined silicon element that resonates at a frequency proportional to applied pressure. Source: (a) Courtesy of Yokogawa Corporation of America, (b) and (c) are courtesy of Druck, Inc., Fairfield, CT. Temperature sensor On-chip diode for optimum temperature performance Etched cavity Micromachined silicon sensor Strain gages Diffused into integrated silicon chip Wire bonding Stitch bonded connections from chip to body plug 102 Problems 103 Fig. 2.29 A piezoelectric transducer measures rapidly changing pressures. Source: Courtesy ofPCB Piezorronics, Inc, Depew, New York. Summary This chapter has been devoted entirely to the computation of pressure distributions and the resulting forces and moments in a static fluid or a fluid with a known veloc¬ ity held. All hydrostatic (Secs. 2.3 to 2.8) and rigid-body (Sec. 2.9) problems are solved in this manner and are classic cases that every student should understand. In arbitrary viscous flows, both pressure and velocity are unknowns and are solved together as a system of equations in the chapters that follow. Problems Most of the problems herein are fairly straightforward. More diffi¬ cult or open-ended assignments are indicated with an asterisk, as in Prob. 2.9. Problems labeled with a computer icon may require the use of a computer. The standard end-of-chapter problems 2.1 to 2.161 (categorized in the problem distribution) are followed by word problems W2. 1 to W2.9, fundamentals of engineering exam problems FE2.1 to FE2.10, comprehensive problems C2.1 to C2.9, and design projects D2.1 to D2.3. Problem Distribution Section Topic Problems 2.1, 2.2 Stresses; pressure gradient; gage pressure 1-2.6 2.3 Hydrostatic pressure; barometers 2.7-2.23 2.3 The atmosphere 2.24-2.29 2.4 Manometers; multiple fluids 2.30-2.47 2.5 Forces on plane surfaces 2.48-2.80 2.6 Forces on curved surfaces 2.81-2.100 2.7 Forces in layered fluids 2.101-2.102 2.8 Buoyancy; Archimedes' principles 2.103-2.126 2.8 Stability of floating bodies 2.127-2.136 2.9 Uniform acceleration 2.137-2.151 2.9 Rigid-body rotation 2.152-2.159 2.10 Pressure measurements 2.160-2.161 Stresses; pressure gradient; gage pressure P2.1 Eor the two-dimensional stress field shown in Fig. P2. 1 it is found that = 3000 Ibf/ft^ Uyy = 2000 Ibf/ft^ = 500 Ibf/ft^ Find the shear and normal stresses (in Ibf/ft^) acting on plane AA cutting through the element at a 30° angle as shown. 104 Chapter 2 Pressure Distribution in a Fluid jy P2.2 For the two-dimensional stress field shown in Fig. P2.1 suppose that (T„ = 2000 Ibf/ft^ = 3000 Ibf/ft^ a„(AA) = 2500 Ibf/ft^ Compute (a) the shear stress errand (b) the shear stress on plane AA. P2.3 A vertical, clean, glass piezometer tube has an inside diameter of 1 mm. When pressure is applied, water at 20°C rises into the tube to a height of 25 cm. After cor¬ recting for surface tension, estimate the applied pressure in Pa. P2.4 Pressure gages, such as the bourdon gage in Fig. P2.4, are calibrated with a deadweight piston. If the bourdon gage is designed to rotate the pointer 10 degrees for every 2 psig of internal pressure, how many degrees does the pointer rotate if the piston and weight together total 44 newtons? \ \ P2.5 Quito, Ecuador, has an average altitude of 9350 ft. On a stan¬ dard day, pressure gage A in a laboratory experiment reads 63 kPa and gage B reads 105 kPa. Express these readings in gage pressure or vacuum pressure, whichever is appropriate. P2.6 Any pressure reading can be expressed as a length or head, h = pipg. What is standard sea-level pressure expressed in (a) ft of glycerin, (b) inHg, (c) m of water, and (d) mm of ethanol? Assume all fluids are at 20°C. Hydrostatic pressure; barometers P2.7 La Paz, Bolivia, is at an altitude of approximately 12,000 ft. Assume a standard atmosphere. How high would the liquid rise in a methanol barometer, assumed at 20°C? Hint: Don’t forget the vapor pressure. P2.8 Suppose, which is possible, that there is a half-mile deep lake of pure ethanol on the surface of Mars. Esti¬ mate the absolute pressure, in Pa, at the bottom of this speculative lake. P2.9 A storage tank, 26 ft in diameter and 36 ft high, is hlled with SAE 30W oil at 20°C. (a) What is the gage pressure, in Ibf/in^, at the bottom of the tank? {b) How does your re¬ sult in {a) change if the tank diameter is reduced to 15 ft? (c) Repeat {a) if leakage has caused a layer of 5 ft of water to rest at the bottom of the (full) tank. P2.10 A large open tank is open to sea-level atmosphere and hlled with liquid, at 20°C, to a depth of 50 ft. The absolute pres¬ sure at the bottom of the tank is approximately 221.5 kPa. From Table A.3, what might this liquid be? P2.ll In Fig. P2.ll, pressure gage A reads 1.5 kPa (gage). The fluids are at 20°C. Determine the elevations z, in meters, of the liquid levels in the open piezometer tubes B and C. P2.12 In Fig. P2.12 the tank contains water and immiscible oil at 20°C. What is h in cm if the density of the oil is 898 kg/m^? Problems 105 12 cm 8 cm 6 cm Oil Water P2.12 P2.13 In Fig. P2. 1 3 the 20°C water and gasoline surfaces are open to the atmosphere and at the same elevation. What is the height h of the third liquid in the right leg? V V Gasoline 5 m Water m Liquid, SG = 1.60 P2.13 P2.14 For the three-liquid system shown, compute hi and h2. Neglect the air density. Mercury 8 cm hi 5 cm P2.14 Oil, SG = 0.78 P2.15 The air-oil-water system in Fig. P2.15 is at 20°C. Know¬ ing that gage A reads 15 Ihf/in^ absolute and gage B reads 1.25 Ibf/in^ less than gage C, compute (a) the specific weight of the oil in Ihf/ft^ and (b) the actual reading of gage C in Ihf/in^ absolute. 15 Ibf/in^ abs P2.16 If the absolute pressure at the interface between water and mercuiy in Fig. P2.16 is 93 kPa, what, in Ibf/ft^, is (a) the pressure at the surface and (b) the pressure at the bottom of the container? P2.16 I P2.17 The system in Fig. P2.17 is at 20°C. Determine the height h of the water in the left side. P2.17 0 Pa (gage) Air, 200 Pa (gage) Oil, SG = 0.8 Water 25 cm 20 cm 106 Chapter 2 Pressure Distribution in a Fluid P2.18 The system in Fig. P2.18 is at 20°C. If atmospheric pressure is 101.33 kPa and the pressure at the bottom of the tank is 242 kPa, what is the specific gravity of fluid X? SAE 30 oil 1 1 m Water Fluid X 3 m P2.18 Mercury 0.5 m t P2.19 The U-tube in Fig. P2.19 has a 1-cm ID and contains mer¬ cury as shown. If 20 cm^ of water is poured into the right- hand leg, what will the free-surface height in each leg be after the sloshing has died down? P2.21 P2.19 10 cm / Mercury 10 cm \|-« — 10 cm - »J P2.20 The hydraulic jack in Fig. P2.20 is filled with oil at 56 Ibf/ft^. Neglecting the weight of the two pistons, what force F on the handle is required to support the 2000-lbf weight for this design? P2.21 At 20°C gage A reads 350 kPa absolute. What is the height h of the water in cm? What should gage B read in kPa absolute? See Fig. P2.21. P2.22 The fuel gage for a gasoline tank in a car reads propor¬ tional to the bottom gage pressure as in Fig. P2.22. If the tank is 30 cm deep and accidentally contains 2 cm of water plus gasoline, how many centimeters of air remain at the top when the gage erroneously reads “full”? P2.23 In Fig. P2.23 both fluids are at 20'’C. If surface ten¬ sion effects are negligible, what is the density of the oil, in kgW? Problems 107 The atmosphere P2.24 In Prob. 1.2 we made a crude integration of the density distribution p(z) in Table A. 6 and estimated the mass of the earth’s atmosphere to be m ~ 6 El 8 kg. Can this result be used to estimate sea-level pressure on the earth? Conversely, can the actual sea-level pressure of 101.35 kPa be used to make a more accurate estimate of the atmospheric mass? P2.25 As measured by NASA’s Viking landers, the atmosphere of Mars, where g =3.71 m/s^, is almost entirely carbon dioxide, and the surface pressure averages 700 Pa. The tem¬ perature is cold and drops off exponentially: T ~ where C = 1.3E-5 m“' and T^, = 250 K. For example, at 20,000 m altitude, T ~ 193 K. (a) Find an analytic formula for the variation of pressure with altitude. {b) Find the altitude where pressure on Mars has dropped to 1 pascal. P2.26 For gases that undergo large changes in height, the linear approximation, Eq. (2.14), is inaccurate. Expand the tropo¬ sphere power-law, Eq. (2.20), into a power series, and show that the linear approximation p ~ Pa ~ Pa gz is ade¬ quate when 6z < 2To (n - l)B where n = — RB P2.27 Conduct an experiment to illustrate atmospheric pressure. Note: Do this over a sink or you may get wet! Find a drinking glass with a very smooth, uniform rim at the top. Fill the glass nearly full with water. Place a smooth, light, flat plate on top of the glass such that the entire rim of the glass is covered. A glossy postcard works best. A small index card or one flap of a greeting card will also work. See Fig. P2.27a. (a) Hold the card against the rim of the glass and turn the glass upside down. Slowly release pressure on the card. Does the water fall out of the glass? Record your experimental observations, {b) Find an expression for the pressure at points 1 and 2 in Fig. P2.27&. Note that the glass is now inverted, so the original top rim of the glass is at the bottom of the picmre, and the original bottom of the glass is at the top of the picture. The weight of the card can be neglected, (c) Estimate the theo¬ retical maximum glass height at which this experiment could still work, such that the water would not fall out of the glass. P2.28 A correlation of computational fluid dynamics results indi¬ cates that, all other things being equal, the distance traveled by a well-hit baseball varies inversely as the 0.36 power of the air density. If a home-run ball hit in Citi Field in New York travels 400 ft, estimate the distance it would travel in (a) Quito, Ecuador, and (b) Colorado Springs, CO. P2.29 Follow up on Prob. P2.8 by estimating the altitude on Mars where the pressure has dropped to 20 percent of its surface value. Assume an isothermal atmosphere, not the exponen¬ tial variation of P2.25. Manometers; multiple fluids P2.30 For the traditional equal-level manometer measurement in Fig. E2.3, water at 20°C flows through the plug device from a to b. The manometer fluid is mercury. If L = 12 cm and h = 24 cm, (a) what is the pressure drop through the device? (b) If the water flows through the pipe at a velocity V = 18 ft/s, what is the dimensionless loss coefficient of the device, defined by A" = Ap/(pV^)2 We will study loss coefficients in Chap. 6. 108 Chapter 2 Pressure Distribution in a Fluid P2.31 In Fig. P2.31 all fluids are at 20°C. Determine the pressure P2.34 difference (Pa) between points A and B. P2.32 For the inverted manometer of Fig. P2.32, all fluids are at 20°C. If Pb~ Pa ~ 97 kPa, what must the height H be in cm? P2.33 In Fig. P2.33 the pressure at point A is 25 Ibf/in^. All fluids are at 20°C. What is the air pressure in the closed chamber B, in Pa? Sometimes manometer dimensions have a significant effect. In Fig. P2.34 containers (a) and (b) are cylindrical and conditions are such that = p^. Derive a formula for the pressure difference — pi, when the oil-water interface on the right rises a distance A/t < h, for {a)d<D and (b) d = 0.1 5D. What is the percentage change in the value of Ap? Water flows upward in a pipe slanted at 30°, as in Fig. P2.35. The mercury manometer reads h = 12 cm. Both fluids are at 20°C. What is the pressure difference pi — p2 in the pipe? P2.35 ^ - 2 m - - In Fig. P2.36 both the tank and the tube are open to the atmosphere. If L = 2.13 m, what is the angle of tilt 8 of the tube? The inclined manometer in Fig. P2.37 contains Meriam red manometer oil, SG = 0.827. Assume that the reservoir is very large. If the inclined arm is fitted with graduations 1 in apart, what should the angle 6 be if each graduation corre¬ sponds to 1 Ibf/ft^ gage pressure forp^? P2.33 Problems 109 P2.37 P2.38 If the pressure in container A in Fig. P2.38 is 200 kPa, compute the pressure in container B. P2.40 P2.41 The system in Fig. P2.41 is at 20°C. Compute the pressure at point A in Ihf/ft^ absolute. Water P2.41 P2.39 In Fig. P2.39 the right leg of the manometer is open to the atmosphere. Find the gage pressure, in Pa, in the air gap in the tank. P2.40 In Fig. P2.40, if pressure gage A reads 20 Ibf/in^ absolute, find the pressure in the closed air space B. The manometer fluid is Meriam red oil, SG = 0.827. P2.42 Very small pressure differences pj^ — pg can be mea¬ sured accurately by the two-fluid differential manometer in Fig. P2.42. Density p2 is only slightly larger than that of the upper fluid pi. Derive an expression for the pro¬ portionality between h and — pg if the reservoirs are very large. 110 Chapter 2 Pressure Distribution in a Fluid P2.43 The traditional method of measuring blood pressure uses a sphygmomanometer, first recording the highest (systolic) and then the lowest (diastolic) pressure from which flow¬ ing “Korotkoff” sounds can be heard. Patients with danger¬ ous hypertension can exhibit systolic pressures as high as 5 Ibf/in^. Normal levels, however, are 2.7 and 1.7 Ibf/in^, respectively, for systolic and diastolic pressures. The manometer uses mercury and air as fluids. (a) How high in cm should the manometer tube be? (b) Express normal systolic and diastolic blood pressure in millimeters of mercury. P2.44 Water flows downward in a pipe at 45°, as shown in Fig. P2.44. The pressure drop pi — p2 is partly due to gravity and partly due to friction. The mercury manometer reads a 6-in height difference. What is the total pressure drop pi — p2 in Ibf/in^? What is the pressure drop due to friction only between 1 and 2 in Ibf/in^? Does the manometer reading correspond only to friction drop? Why? P2.45 In Fig. P2.45, determine the gage pressure at point A in Pa. Is it higher or lower than atmospheric? P2.46 In Fig. P2.46 both ends of the manometer are open to the atmosphere. Estimate the specific gravity of fluid X. P2.47 The cyhndrical tank in Fig. P2.47 is being filled with water at 20°C by a pump developing an exit pressure of 175 kPa. At the instant shown, the air pressure is 1 10 kPa and H = 35 cm. The pump stops when it can no longer raise the water pressure. For isothermal air compression, estimate H at that time. Problems 111 50 cm Air 20° C 75 cm — X H I Water - P2.47 J _ _ - PuniP P2.48 The system in Fig. P2.48 is open to 1 atm on the right side, (a) \f L = 120 cm, what is the air pressure in con¬ tainer A? {b) Conversely, if Pa = 135 kPa, what is the length L? Forces on plane surfaces P2.49 Conduct the following experiment to illustrate air pres¬ sure. Find a thin wooden ruler (approximately 1 ft in length) or a thin wooden paint stirrer. Place it on the edge of a desk or table with a little less than half of it hanging over the edge lengthwise. Get two full-size sheets of newspaper; open them up and place them on top of the ruler, covering only the portion of the ruler resting on the desk as illustrated in Fig. P2.49. (a) Estimate the total force on top of the newspaper due to air pressure in the room, (fc) Careful! To avoid potential injury, make sure nobody is standing directly in front of the desk. Perform a karate chop on the portion of the ruler sticking out over the edge of the desk. Record your results, (c) Explain your results. P2.50 A small submarine, with a hatch door 30 in in diameter, is submerged in seawater, (a) If the water hydrostatic force on the hatch is 69,000 Ibf, how deep is the sub? (b) If the sub is 350 ft deep, what is the hydrostatic force on the hatch? P2.51 Gate AB in Eig. P2.5I is 1.2 m long and 0.8 m into the paper. Neglecting atmospheric pressure, compute the force F on the gate and its center-of-pressure position X. P2.52 Example 2.5 calculated the force on plate AB and its line of action, using the moment-of-inertia approach. Some teachers say it is more instructive to calculate these by direct integration of the pressure forces. Using Eigs. P2.52 and E2.5fl, (a) find an expression for the pressure variation p(^) along the plate; (b) integrate this expression to find the total force F; (c) integrate the moments about point A to find the position of the center of pressure. 112 Chapter 2 Pressure Distribution in a Fluid \ P2.57 P2.53 The Hoover Dam, in Arizona, encloses Lake Mead, which contains 10 trillion gallons of water. The dam is 1200 ft wide and the lake is 500 ft deep, (a) Estimate the hydrostatic force on the dam, in MN. (b) Explain how you might analyze the stress in the dam due to this hydro¬ static force. P2.54 In Eig. P2.54, the hydrostatic force F is the same on the bottom of all three containers, even though the weights of liquid above are quite different. The three bottom shapes and the fluids are the same. This is called the hydrostatic paradox. Explain why it is true and sketch a free body of each of the liquid columns. P2.58 P2.54 (c) P2.59 P2.55 Gate AB in Eig. P2.55 is 5 ft wide into the paper, hinged at A, and restrained by a stop at B. The water is at 20°C. Compute (a) the force on stop B and (b) the reactions at A if the water depth h = 9.5 ft. P2.55 V Water 4 ft P2.60 P2.56 In Fig. P2.55, gate AB is 5 ft wide into the paper, and stop B will break if the water force on it equals 9200 Ibf. For what water depth h is this condition reached? The square vertical panel ASCD in Fig. P2.57 is submerged in water at 20°C. Side AB is at least 1.7 m below the surface. Determine the difference between the hydrostatic forces on subpanels ABD and BCD. P2.57 In Fig. P2.58, the cover gate AB closes a circular opening 80 cm in diameter. The gate is held closed by a 200-kg mass as shown. Assume standard gravity at 20°C. At what water level h will the gate be dislodged? Neglect the weight of the gate. Gate AB has length L and width b into the paper, is hinged at B, and has negligible weight. The liquid level h remains at the top of the gate for any angle 6. Find an analytic expression for the force P, perpendicular to AB, required to keep the gate in equilibrium in Fig. P2.59. P2.59 B In Fig. P2.60, vertical, unsymmetrical trapezoidal panel ABCD is submerged in fresh water with side AB 12 ft below the surface. Since trapezoid formulas are complicated, (a) estimate, reasonably, the water force on the panel, in Ibf, neglecting atmospheric pressure. For extra credit, {b) look up the formula and compute the exact force on the panel. Problems 113 P2.61 Gate AB in Fig. P2.61 is a homogeneous mass of 180 kg, 1.2 m wide into the paper, hinged at A, and resting on a smooth bottom at B. All fluids are at 20°C. For what water depth h will the force at point B he zero? P2.62 P2.61 Gate AB in Fig. P2.62 is 15 ft long and 8 ft wide into the paper and is hinged at B with a stop at A. The water is at 20°C. The gate is 1-in-thick steel, SG = 7.85. Compute the water level h for which the gate will start to fall. P2.63 P2.64 Gate ABC in Fig. P2.64 has a fixed hinge line at B and is 2 m wide into the paper. The gate will open at A to release water if the water depth is high enough. Compute the depth h for which the gate will begin to open. 1 m V 20 cm Water at 20°C P2.64 P2.62 P2.63 The tank in Fig. P2.63 has a 4-cm-diameter plug at the bottom on the right. All fluids are at 20°C. The plug will pop out if the hydrostatic force on it is 25 N. For this condition, what will be the reading h on the mercury manometer on the left side? P2.65 Gate AB in Fig. P2.65 is semicircular, hinged at B, and held by a horizontal force P at A. What force P is required for equilibrium? P2.65 114 Chapter 2 Pressure Distribution in a Fluid P2.66 Dam ABC in Fig. P2.66 is 30 m wide into the paper and made of concrete (SG = 2.4). Find the hydrostatic force on surface AB and its moment about C. Assuming no seepage of water under the dam, could this force tip the dam over? How does your argument change if there is seepage under the dam? P2.67 Generalize Prob. P2.66 as follows. Denote length AB as H, length BC as L, and angle ABC as 0. Let the dam mate¬ rial have specific gravity SG. The width of the dam is b. Assume no seepage of water under the dam. Find an ana- p2 ~j\ lytic relation between SG and the critical angle 0^ for which the dam will just tip over to the right. Use your relation to compute 0^ for the special case SG = 2.4 (concrete). P2.68 Isosceles triangle gate AB in Fig. P2.68 is hinged at A and weighs 1500 N. What horizontal force P is required at point B for equilibrium? P2.68 P2.69 Consider the slanted plate AB of length L in Fig. P2.69. (a) Is the hydrostatic force F on the plate equal to the weight of the missing water above the plate? If not, correct this hypothesis. Neglect the atmosphere, (b) Can a “missing water” theory be generalized to curved surfaces of this type? The swing-check valve in Fig. P2.70 covers a 22.86-cm diameter opening in the slanted wall. The hinge is 15 cm from the centerline, as shown. The valve will open when the hinge moment is 50 N ■ m. Find the value of h for the water to cause this condition. P2.70 In Fig. P2.71 gate AB is 3 m wide into the paper and is connected by a rod and pulley to a concrete sphere (SG = 2.40). What diameter of the sphere is just sufficient to keep the gate closed? In Fig. P2.72, gate AB is circular. Find the moment of the hydrostatic force on this gate about axis A. P2.72 Water A B T' 3 m i t ■ 2m Problems 115 P2.73 Gate AB is 5 ft wide into the paper and opens to let fresh water out when the ocean tide is dropping. The hinge at A is 2 ft above the freshwater level. At what ocean level h will the gate first open? Neglect the gate weight. P2.73 P2.74 Find the height H in Fig. P2.74 for which the hydrostatic force on the rectangular panel is the same as the force on the semicircular panel helow. P2.74 V P2.75 The cap at point B on the 5-cm-diameter tube in Fig. P2.75 will be dislodged when the hydrostatic force on its base reaches 22 Ibf. For what water depth h does this occur? P2.77 Pa V i ^ Water A Y BO C r Pa t 1 m Oil, SG = 0.8 P2.76 Panel BC in Fig. P2.76 is circular. Compute {a) the hydro¬ static force of the water on the panel, (b) its center of pres¬ sure, and (c) the moment of this force about point B. P2.77 The circular gate ABC in Fig. P2.77 has a 1 -m radius and is hinged at B. Compute the force P just sufficient to keep the gate from opening when /t = 8 m. Neglect atmospheric pressure. P2.78 Panels AB and CD in Fig. P2.78 are each 120 cm wide into the paper, (a) Can you deduce, by inspection, which panel has the larger water force? (b) Even if your deduction is brilliant, calculate the panel forces anyway. P2.78 P2.79 Gate ABC in Fig. P2.79 is 1 m square and is hinged at B. It will open automatically when the water level h becomes high enough. Determine the lowest height for which the gate will open. Neglect atmospheric pressure. Is this result independent of the liquid density? 116 Chapter 2 Pressure Distribution in a Fluid V Water 60 cm 40 cm P2.80 P2.79 A concrete dam (SG = 2.5) is made in the shape of an isosceles triangle, as in Fig. P2.80. Analyze this geometry to find the range of angles 6 for which the hydrostatic force will tend to tip the dam over at point B. The width into the paper is b. P2.80 Forces on curved surfaces P2.81 P2.82 For the semicircular cylinder CDE in Example 2.9, find the vertical hydrostatic force by integrating the vertical com¬ ponent of pressure around the surface from 9 = 0 to 9 = n. The dam in Fig. P2.82 is a quarter circle 50 m wide into the paper. Determine the horizontal and vertical components of the hydrostatic force against the dam and the point CP where the resultant strikes the dam. P2.82 P2.83 Gate AB in Fig. P2.83 is a quarter circle 10 ft wide into the paper and hinged at B. Find the force F just sufficient to keep the gate from opening. The gate is uniform and weighs 3000 Ibf P2.83 P2.84 Panel AB in Fig. P2.84 is a parabola with its maximum at point A. It is 150 cm wide into the paper. Neglect atmo¬ spheric pressure. Find (a) the vertical and (b) the horizontal water forces on the panel. P2.84 P2.85 Compute the horizontal and vertical components of the hydrostatic force on the quarter-circle panel at the bottom of the water tank in Fig. P2.85. P2.86 The quarter circle gate BC in Fig. P2.86 is hinged at C. Find the horizontal force P required to hold the gate stationary. Neglect the weight of the gate. P2.86 Problems 117 P2.87 The bottle of champagne (SG = 0.96) in Fig. P2.87 is under pressure, as shown by the mercury-manometer reading. Compute the net force on the 2-in-radius hemispherical end cap at the bottom of the bottle. P2.88 Gate ABC is a circular arc, sometimes called a Tainter gate, which can be raised and lowered by pivoting about point O. See Fig. P2.88. For the position shown, determine (a) the hydrostatic force of the water on the gate and {b) its line of action. Does the force pass through point O? P2.89 The tank in Fig. P2.89 contains benzene and is pressurized to 200 kPa (gage) in the air gap. Determine the vertical hydrostatic force on circular-arc section AB and its line of action. P2.90 The tank in Fig. P2.90 is 120 cm long into the paper. Deter¬ mine the horizontal and vertical hydrostatic forces on the quarter-circle panel AB. The fluid is water at 20°C. Neglect atmospheric pressure. P2.89 A P2.91 The hemispherical dome in Fig. P2.91 weighs 30 kN and is hlled with water and attached to the floor by six equally spaced bolts. What is the force in each bolt required to hold down the dome? P2.91 P2.92 A 4-m-diameter water tank consists of two half cylinders, each weighing 4.5 kN/m, bolted together as shown in Fig. P2.92. If the support of the end caps is neglected, determine the force induced in each bolt. 118 Chapter 2 Pressure Distribution in a Fluid Bolt spacing 25 cm P2.92 P2.93 In Fig. P2.93, a one-quadrant spherical shell of radius R is submerged in liquid of specific weight 7 and depth h > R. Find an analytic expression for the resultant hydrostatic force, and its line of action, on the shell surface. z P2.94 Find an analytic formula for the vertical and horizontal forces on each of the semicircular panels AB in Fig. P2.94. The width into the paper is b. Which force is larger? Why? P2.94 i P2.96 In Fig. P2.96, curved section AS is 5 m wide into the paper and is a 60° circular arc of radius 2 m. Neglecting atmo¬ spheric pressure, calculate the vertical and horizontal hydrostatic forces on arc AS. P2.97 The contractor ran out of gunite mixture and finished the deep comer of a 5-m-wide swimming pool with a quarter- circle piece of PVC pipe, labeled AS in Fig. P2.97. Compute the horizontal and vertical water forces on the curved panel AS. P2.97 2m 1 m Water ._J _ P2.95 The uniform body A in Fig. P2.95 has width b into the paper and is in static equilibrium when pivoted about hinge O. What is the specific gravity of this body if (a) h = 0 and (b) h = S? P2.98 The curved surface in Fig. P2.98 consists of two quarter- spheres and a half cylinder. A side view and front view are shown. Calculate the horizontal and vertical forces on the surface. Problems 119 1.5 m \|l m “t” 2 m I 1 m P2.99 The mega-magnum cylinder in Fig. P2.99 has a hemispher¬ ical bottom and is pressurized with air to 75 kPa (gage) at the top. Determine {a) the horizontal and (b) the vertical hydrostatic forces on the hemisphere, in Ibf. P2.99 P2.100 Pressurized water fills the tank in Fig. P2.100. Compute the net hydrostatic force on the conical surface ABC. Forces on layered surfaces P2.101 The closed layered box in Fig. P2.101 has square horizon¬ tal cross sections everywhere. All fluids are at 20°C. Estimate the gage pressure of the air if (a) the hydrostatic force on panel AB is 48 kN or (b) the hydrostatic force on the bottom panel BC is 97 kN. P2.102 A cubical tank is3mX3mX3m and is layered with 1 meter of fluid of specific gravity 1.0, 1 meter of fluid with SG = 0.9, and 1 meter of fluid with SG = 0.8. Neglect atmospheric pressure. Find (a) the hydrostatic force on the bottom and (b) the force on a side panel. Buoyancy; Archimedes’ principles P2.103 A solid block, of specific gravity 0.9, floats such that 75 percent of its volume is in water and 25 percent of its volume is in fluid X, which is layered above the water. What is the specific gravity of fluid X? P2.104 The can in Fig. P2.104 floats in the position shown. What is its weight in N? 3 cm 8 cm P2.104 -D = 9 cm- P2.105 It is said that Archimedes discovered the buoyancy laws when asked by King Hiero of Syracuse to determine whether his new crown was pure gold (SG = 19.3). Archimedes measured the weight of the crown in air to be 1 1 .8 N and its weight in water to be 10.9 N. Was it pure gold? P2.106 A spherical helium balloon has a total mass of 3 kg. It set¬ tles in a calm standard atmosphere at an altitude of 5500 m. Estimate the diameter of the balloon. P2.107 Repeat Prob. 2.62, assuming that the 10,000-lbf weight is alu¬ minum (SG = 2.71) and is hanging submerged in the water. 120 Chapter 2 Pressure Distribution in a Fluid P2.108 A 7 -cm-diameter solid aluminum ball (SG = 2.7) and a solid brass ball (SG = 8.5) balance nicely when submerged in a liquid, as in Fig. P2.108. (a) If the fluid is water at 20°C, what is the diameter of the brass ball? (b) If the brass ball has a diameter of 3.8 cm, what is the density of the fluid? P2.108 2 pulleys P2.109 A hydrometer floats at a level that is a measure of the specific gravity of the liquid. The stem is of constant diam¬ eter D, and a weight in the bottom stabilizes the body to float vertically, as shown in Fig. P2.109. If the position = 0 is pure water (SG = I.O), derive a formula for as a function of total weight W, D, SG, and the specific weight 7o of water. P2.109 P2.110 A solid sphere, of diameter 18 cm, floats in 20°C water with 1527 cubic centimeters exposed above the surface. (a) What are the weight and specific gravity of this sphere? (b) Will it float in 20°C gasoline? If so, how many cubic centimeters will be exposed? P2.111 A solid wooden cone (SG = 0.729) floats in water. The cone is 30 cm high, its vertex angle is 90°, and it floats with vertex down. How much of the cone protrudes above the water? P2.112 The uniform 5-m-long round wooden rod in Fig. P2. 1 12 is tied to the bottom by a string. Determine (a) the tension in the string and (b) the specific gravity of the wood. Is it pos¬ sible for the given information to determine the inclination angle 01 Explain. P2.112 P2.113 A spar buoy is a buoyant rod weighted to float and protrude vertically, as in Fig. P2.113. It can be used for measure¬ ments or markers. Suppose that the buoy is maple wood (SG = 0.6), 2 in by 2 in by 12 ft, floating in seawater (SG = 1.025). How many pounds of steel (SG = 7.85) should be added to the bottom end so that /t = 1 8 in? P2.114 The uniform rod in Fig. P2.1 14 is hinged at point B on the waterline and is in static equilibrium as shown when 2 kg of lead (SG = 11.4) are attached to its end. What is the specific gravity of the rod material? What is peculiar about the rest angle 9 = 30°? Problems 121 P2.115 The 2-in by 2-in by 12-ft spar buoy from Fig. P2.113 has 5 Ibm of steel attached and has gone aground on a rock, as in Fig. P2.1 15. Compute the angle 6 at which the buoy will lean, assuming that the rock exerts no moments on the spar. P2.116 The bathysphere of the chapter-opener photo is steel, SG ~ 7.85, with inside diameter 54 inches and wall thick¬ ness 1.5 inches. Will the empty sphere float in seawater? P2.117 The solid sphere in Fig. P2.117 is iron (SG « 7.9). The tension in the cable is 600 Ibf. Estimate the diameter of the sphere, in cm. P2.117 P2.118 An intrepid treasure-salvage group has discovered a steel box, containing gold doubloons and other valuables, rest¬ ing in 80 ft of seawater. They estimate the weight of the box and treasure (in air) at 7000 Ibf. Their plan is to attach the box to a sturdy balloon, inflated with air to 3 atm pres¬ sure. The empty balloon weighs 250 Ibf. The box is 2 ft wide, 5 ft long, and 18 in high. What is the proper diameter of the balloon to ensure an upward lift force on the box that is 20 percent more than required? P2.119 When a 5-lbf weight is placed on the end of the uniform floating wooden beam in Fig. P2.119, the beam tilts at an angle 0 with its upper right corner at the surface, as shown. Determine (a) the angle 6 and (b) the specific gravity of the wood. Hint: Both the vertical forces and the moments about the beam centroid must be balanced. P2.119 P2.120 A uniform wooden beam (SG = 0.65) is 10 cm by 10 cm by 3 m and is hinged at A, as in Fig. P2.120. At what angle 9 will the beam float in the 20°C water? P2.120 P2.121 The uniform beam in Fig. P2.121, of size Lhy hhy b and with specific weight -ji,, floats exactly on its diagonal when a heavy uniform sphere is tied to the left comer, as shown. Width b «L P2.121 Show that this can happen only (a) when 7^, = 7/3 and (fe) when the sphere has size 7r(SG - 1) P2.122 A uniform block of steel (SG = 7.85) will “float” at a mercury-water interface as in Fig. P2.122. What is the ratio of the distances a and b for this condition? V Water V Steel a block h Mercury: SG = 13.56 122 Chapter 2 Pressure Distribution in a Fluid P2.123 A barge has the trapezoidal shape shown in Fig. P2.123 and is 22 m long into the paper. If the total weight of barge and cargo is 350 tons, what is the draft H of the barge when floating in seawater? P2.123 \ V V ■ t 2.5 m {601 _ L h - 8 m P2.128 Specific gravity V Water S= 1.0 P2.124 A balloon weighing 3.5 Ibf is 6 ft in diameter. It is hlled with hydrogen at 18 Ibf/in^ absolute and 60°F and is released. At what altitude in the U.S. standard atmosphere will this balloon be neutrally buoyant? P2.125 A uniform cylindrical white oak log, p = 710 kg/m^, floats lengthwise in fresh water at 20°C. Its diameter is 24 inches. What height of the log is visible above the surface? P2.126 A block of wood (SG = 0.6) floats in fluid X in Fig. P2. 1 26 such that 75 percent of its volume is submerged in fluid X. Estimate the vacuum pressure of the air in the tank. P2.126 Air = 0 kPa gage 40 cm 70 cm Air pressure? Wood Fluid X Stability of floating bodies P2.127 Consider a cylinder of specific gravity S < 1 floating verti¬ cally in water (S = 1), as in Fig. P2.127. Derive a formula for the stable values of DIL as a function of S and apply it to the case DIL = 1.2. P2.128 An iceberg can be idealized as a cube of side length L, as in Fig. P2.128. If seawater is denoted by S = 1.0, then glacier ice (which forms icebergs) has 5 = 0.88. Determine if this “cubic” iceberg is stable for the position shown in Fig. P2.128. P2.129 The iceberg idealization in Prob. P2.128 may become unstable if its sides melt and its height exceeds its width. In Fig. P2.128 suppose that the height is L and the depth into the paper is L, but the width in the plane of the paper is H < L. Assuming S = 0.88 for the iceberg, find the ratio HIL for which it becomes neutrally stable (about to overturn). P2.130 Consider a wooden cylinder (SG = 0.6) 1 m in diameter and 0.8 m long. Would this cylinder be stable if placed to float with its axis vertical in oil (SG = 0.8)? P2.131 A barge is 15 ft wide and 40 ft long and floats with a draft of 4 ft. It is piled so high with gravel that its center of grav¬ ity is 3 ft above the waterline. Is it stable? P2.132 A solid right circular cone has SG = 0.99 and floats vertically as in Fig. P2.132. Is this a stable position for the cone? P2.132 P2.133 Consider a uniform right circular cone of specific gravity S < 1, floating with its vertex down in water (S = 1). The base radius is R and the cone height is H. Calculate and plot the stability MG of this cone, in dimensionless form, versus HIR for a range of S < 1 . P2.134 When floating in water (SG = 1.0), an equilateral triangu¬ lar body (SG = 0.9) might take one of the two positions shown in Fig. P2.134. Which is the more stable position? Assume large width into the paper. (a) (b) P2.134 Problems 123 P2.135 Consider a homogeneous right circular cylinder of length L, radius R, and specific gravity SG, floating in water (SG = 1). Show that the body will be stable with its axis vertical if I > [2SG(1 - SG)]''^ P2.136 Consider a homogeneous right circular cylinder of length L, radius R, and specific gravity SG = 0.5, floating in water (SG = 1). Show that the body will be stable with its axis horizontal if L/R > 2.0. Uniform acceleration P2.137 A tank of water 4 m deep receives a constant upward acceleration a^. Determine (a) the gage pressure at the tank bottom if = 5 m^/s and (b) the value of that causes the gage pressure at the tank bottom to be 1 atm. P2.138 A 12-fl-oz glass, of 3-in diameter, partly full of water, is attached to the edge of an 8-ft-diameter merry-go-round, which is rotated at 12 r/min. How full can the glass be before water spills? Hint: Assume that the glass is much smaller than the radius of the merry-go-round. P2.139 The tank of liquid in Fig. P2.139 accelerates to the right with the fluid in rigid-body motion, (a) Compute Oj, in m/s^. (b) Why doesn’t the solution to part {a) depend on the den¬ sity of the fluid? (c) Determine the gage pressure at point A if the fluid is glycerin at 20°C. P2.139 P2.140 The U-tube in Fig. P2. 140 is moving to the right with vari¬ able velocity. The water level in the left tube is 6 cm, and the level in the right tube is 16 cm. Determine the accelera¬ tion and its direction. P2.141 The same tank from Prob. P2.139 is now moving with con¬ stant acceleration up a 30° inclined plane, as in Fig. P2.141. Assuming rigid-body motion, compute (a) the value of the acceleration a, (b) whether the acceleration is up or down, and (c) the gage pressure at point A if the fluid is mercury at 20°C. P2.141 P2.142 The tank of water in Fig. P2. 142 is 12 cm wide into the paper. If the tank is accelerated to the right in rigid-body motion at 6.0 m/s^, compute (a) the water depth on side AB and (b) the water-pressure force on panel AB. Assume no spilling. B V Water at 20°C A P2.142 24 cm 9 cm P2.143 The tank of water in Fig. P2. 143 is full and open to the atmo¬ sphere at point A. For what acceleration in ft/s^ will the pressure at point B be (a) atmospheric and (b) zero absolute? 2 ft 1 ft Water p =15 Ibf/in^ abs 4 a -1 ft- -2 ft- P2.140 P2.143 124 Chapter 2 Pressure Distribution in a Fluid P2.144 Consider a hollow cube of side length 22 cm, Filed com¬ pletely with water at 20°C. The top surface of the cube is horizontal. One top comer, point A, is open through a small hole to a pressure of 1 atm. Diagonally opposite to point A is top comer B. Determine and discuss the various rigid- body accelerations for which the water at point B begins to cavitate, for {a) horizontal motion and {b) vertical motion. P2.145 A hsh tank 14 in deep by 16 by 27 in is to be earned in a car that may experience accelerations as high as 6 m/s^. What is the maximum water depth that will avoid spilling in rigid-body motion? What is the proper alignment of the tank with respect to the car motion? P2.146 The tank in Fig. P2.146 is filled with water and has a vent hole at point A. The tank is 1 m wide into the paper. Inside the tank, a 10-cm balloon. Filed with helium at 130 kPa, is tethered centrally by a string. If the tank accelerates to the right at 5 m/s^ in rigid-body motion, at what angle will the balloon lean? Will it lean to the right or to the left? P2.146 P2.147 The tank of water in Fig. P2.147 accelerates uniformly by freely rolling down a 30° incline. If the wheels are friction¬ less, what is the angle 92 Can you explain this interesting result? P2.147 P2.148 A child is holding a string onto which is attached a helium- Flled balloon, (a) The child is standing still and suddenly accelerates forward. In a frame of reference moving with the child, which way will the balloon tilt, forward or back¬ ward? Explain, (b) The child is now sitting in a car that is stopped at a red light. The helium-Flled balloon is not in contact with any part of the car (seats, ceiling, etc.) but is held in place by the string, which is in turn held by the child. All the windows in the car are closed. When the traf- Fc light turns green, the car accelerates forward. In a frame of reference moving with the car and child, which way will the balloon tilt, forward or backward? Explain, (c) Purchase or borrow a helium-Flled balloon. Conduct a scientiFc experiment to see if your predictions in parts (a) and (b) above are correct. If not, explain. P2.149 The 6-ft-radius waterwheel in Pig. P2.149 is being used to lift water with its 1-ft-diameter half-cylinder blades. If the wheel rotates at 10 r/min and rigid-body motion is assumed, what is the water surface angle 9 at position A? P2.150 A cheap accelerometer, probably worth the price, can be made from a U-tube as in Fig. P2.150. If L = 18 cm and D = 5 mm, what will h he if = 6 m/s^? Can the scale markings on the tube be linear multiples of a/! P2.150 r - ^ M P2.151 The U-tube in Fig. P2.151 is open at A and closed at D. If accelerated to the right at uniform a„ what acceleration Problems 125 will cause the pressure at point C to be atmospheric? The fluid is water (SG = 1.0). P2.151 Rigid-body rotation P2.152 A 16-cm-diameter open cylinder 27 cm high is full of water. Compute the rigid-body rotation rate about its central axis, in r/min, (a) for which one-third of the water will spill out and (b) for which the bottom will be barely exposed. P2.153 A tall cylindrical container, 14 in in diameter, is used to make a mold for forming 14-in salad bowls. The bowls are to be 8 in deep. The cylinder is half-filled with molten plas¬ tic, fi = 1.6 kg/(m-s), rotated steadily about the central axis, then cooled while rotating. What is the appropriate rotation rate, in r/min? P2.154 A very tall 10-cm-diameter vase contains 1 178 cm^ of water. When spun steadily to achieve rigid-body rotation, a 4-cm- diameter dry spot appears at the bottom of the vase. What is the rotation rate, in r/min, for this condition? P2.155 For what uniform rotation rate in r/min about axis C will the U-tube in Fig. P2. 155 take the configuration shown? The fluid is mercury at 20°C. 12 cm P2.155 10 cm c5 cm -I P2.156 Suppose that the U-tube of Fig. P2.151 is rotated about axis DC. If the fluid is water at 122°F and atmospheric pressure is 2116 Ibf/ft^ absolute, at what rotation rate will the fluid within the tube begin to vaporize? At what point will this occur? P2.157 The 45° V-tube in Fig. P2. 157 contains water and is open at A and closed at C. What uniform rotation rate in r/min about axis AB will cause the pressure to be equal at points B and C? For this condition, at what point in leg BC will the pressure be a minimum? P2.157 P2.158 It is desired to make a 3-m-diameter parabolic telescope mirror by rotating molten glass in rigid-body motion until the desired shape is achieved and then cooling the glass to a solid. The focus of the mirror is to be 4 m from the mirror, measured along the centerline. What is the proper mirror rotation rate, in r/min, for this task? P2.159 The three-legged manometer in Fig. P2.159 is filled with water to a depth of 20 cm. All tubes are long and have equal small diameters. If the system spins at angular veloc¬ ity n about the central tube, (a) derive a formula to find the change of height in the tubes; (b) find the height in cm in each tube if fl = 120 r/min. Hint: The central tube must supply water to both the outer legs. P2.159 Pressure measurements P2.160 Figure P2.160 shows a gage for very low pressures, in¬ vented in 1874 by Herbert McLeod, (a) Can you deduce, from the figure, how it works? (b) If not, read about it and explain it to the class. 126 Chapter 2 Pressure Distribution in a Fluid P2.161 Figure P2.161 shows a sketch of a commercial pressure gage, (a) Can you deduce, from the figure, how it works? P2.161 Word Problems W2.1 Consider a hollow cone with a vent hole in the vertex at the top, along with a hollow cylinder, open at the top, with the same base area as the cone. Fill both with water to the top. The hydrostatic paradox is that both containers have the same force on the bottom due to the water pressure, although the cone contains 67 percent less water. Can you explain the paradox? W2.2 Can the temperature ever rise with altitude in the real atmo¬ sphere? Wouldn’t this cause the air pressure to increase upward? Explain the physics of this situation. W2.3 Consider a submerged curved surface that consists of a two-dimensional circular arc of arbitrary angle, arbitrary depth, and arbitrary orientation. Show that the resultant hydro¬ static pressure force on this surface must pass through the center of curvature of the arc. W2.4 Fill a glass approximately 80 percent with water, and add a large ice cube. Mark the water level. The ice cube, having SG ~ 0.9, sticks up out of the water. Let the ice cube melt with negligible evaporation from the water surface. Will the water level be higher than, lower than, or the same as before? W2.5 A ship, carrying a load of steel, is trapped while floating in a small closed lock. Members of the crew want to get out. Fundamentals of Engineering Exam Problems FE2.1 A gage attached to a pressurized nitrogen tank reads a gage pressure of 28 in of mercury. If atmospheric pressure is 14.4 psia, what is the absolute pressure in the tank? (a) 95 kPa, {b) 99 kPa, (c) 101 kPa, {d) 194 kPa, (e) 203 kPa but they can’t quite reach the top wall of the lock. A crew member suggests throwing the steel overboard in the lock, claiming the ship will then rise and they can climb out. Will this plan work? W2.6 Consider a balloon of mass m floating neutrally in the atmosphere, carrying a person/basket of mass M > m. Discuss the stability of this system to disturbances. W2.7 Consider a helium balloon on a string tied to the seat of your stationary car. The windows are closed, so there is no air motion within the car. The car begins to accelerate for¬ ward. Which way will the balloon lean, forward or back¬ ward? Hint: The acceleration sets up a horizontal pressure gradient in the air within the car. W2.8 Repeat your analysis of Prob. W2.7 to let the car move at constant velocity and go around a curve. Will the balloon lean in, toward the center of curvature, or out? W2.9 The deep submersible vehicle ALVIN weighs approxi¬ mately 36,000 Ibf in air. It carries 800 Ibm of steel weights on the sides. After a deep mission and return, two 400-lbm piles of steel are left on the ocean floor. Can you explain, in terms relevant to this chapter, how these steel weights are used? FE2.2 On a sea-level standard day, a pressure gage, moored below the surface of the ocean (SG = 1.025), reads an absolute pressure of 1.4 MPa. How deep is the instrument? (a) 4 m, (b) 129 m, (c) 133 m, (d) 140 m, (e) 2080 m Comprehensive Problems 127 FE2.3 In Fig. FE2.3, if the oil in region B has SG = 0.8 and the absolute pressure at point A is 1 atm, what is the absolute pressure at point fi? (fl) 5.6 kPa, {b) 10.9 kPa, (c) 107 kPa, (rf) 1 12 kPa, (e) 157 kPa FE2.4 In Fig. FE2.3, if the oil in region B has SG = 0.8 and the absolute pressure at point S is 14 psia, what is the absolute pressure at point A? (a) 11 kPa, (b) 41 kPa, (c) 86 kPa, {d) 91 kPa, (e) 101 kPa FE2.5 A tank of water (SG = 1.0) has a gate in its vertical wall 5 m high and 3 m wide. The top edge of the gate is 2 m below the surface. What is the hydrostatic force on the gate? (fl) 147 kN, {b) 367 kN, (c) 490 kN, (rf) 661 kN, (e) 1028 kN Comprehensive Problems C2.1 Some manometers are constructed as in Fig. C2.1, where one side is a large reservoir (diameter D) and the other side is a small tube of diameter d, open to the atmosphere. In such a case, the height of manometer liquid on the reservoir side does not change appreciably. This has the advantage that only one height needs to be measured rather than two. The manometer liquid has density while the air has den¬ sity p^. Ignore the effects of surface tension. When there is no pressure difference across the manometer, the eleva¬ tions on both sides are the same, as indicated by the dashed line. Fleight h is measured from the zero pressure level as shown, (a) When a high pressure is applied to the left side, the manometer liquid in the large reservoir goes down, while that in the tube at the right goes up to conserve mass. Write an exact expression for Pigage^ taking into account the movement of the surface of the reservoir. Your equation should give Pigage as a function of h, p„, and the physical parameters in the problem, h, d, D, and gravity constant g. (b) Write an approximate expression for Pigage, neglecting the change in elevation of the surface of the reservoir liquid. FE2.6 In Prob. FE2.5, how far below the surface is the center of pressure of the hydrostatic force? (a) 4.50 m, (b) 5.46 m, (c) 6.35 m, (d) 5.33 m, (e) 4.96 m FE2.7 A solid 1-m-diameter sphere floats at the interface between water (SG = 1.0) and mercury (SG = 13.56) such that 40 percent is in the water. What is the specific gravity of the sphere? (a) 6.02, (b) 7.28, (c) 7.78, (d) 8.54, (e) 12.56 FE2.8 A 5-m-diameter balloon contains helium at 125 kPa absolute and 15°C, moored in sea-level standard air. If the gas constant of helium is 2077 mV(s^ ■ K) and balloon material weight is neglected, what is the net lifting force of the balloon? (fl) 67 N, (b) 134 N, (c) 522 N, (d) 653 N, (e) 787 N FE2.9 A square wooden (SG = 0.6) rod, 5 cm by 5 cm by 10 m long, floats vertically in water at 20°C when 6 kg of steel (SG = 7.84) are attached to one end. How high above the water surface does the wooden end of the rod protrude? (a) 0.6 m, (b) 1.6 m, (c) 1.9 m, (d) 2.4 m, (e) 4.0 m FE2.10 A floating body will be stable when its (a) center of gravity is above its center of buoyancy, (b) center of buoyancy is below the waterline, (c) center of buoyancy is above its metacenter, (d) metacenter is above its center of buoyancy, (e) metacenter is above its center of gravity. (c) Suppose h = 0.26 m in a certain application. If = 101,000 Pa and the manometer liquid has a density of 820 kg/m^, estimate the ratio Did required to keep the error of the approximation of part {b) within 1 percent of the exact measurement of paid {a). Repeat for an error within 0.1 percent. To pressure measurement location 128 Chapter 2 Pressure Distribution in a Fluid C2.2 A prankster has added oil, of specific gravity SGq, to the left leg of the manometer in Fig. C2.2. Nevertheless, the U-tube is still useful as a pressure-measuring device. It is attached to a pressurized tank as shown in the figure. (a) Find an expression for h as a function of H and other parameters in the problem. (&) Find the special case of your result in (a) when = p^. (c) Suppose H = 5.0 cm, Pa is 101.2 kPa,p,a„i; is 1.82 kPa higher than and SGq = 0.85. Calculate h in cm, ignoring surface tension effects and neglecting air density effects. C2.3 Professor F. Dynamics, riding the merry-go-round with his son, has brought along his U-tube manometer. (You never know when a manometer might come in handy.) As shown in Fig. C2.3, the merry-go-round spins at constant angu¬ lar velocity and the manometer legs are 7 cm apart. The manometer center is 5.8 m from the axis of rotation. Determine the height difference h in two ways: (a) approx¬ imately, by assuming rigid-body translation with a equal to the average manometer acceleration; and {b) exactly, using rigid-body rotation theory. How good is the approximation? C2.4 A student sneaks a glass of cola onto a roller coaster ride. The glass is cylindrical, twice as tall as it is wide, and filled to the brim. He wants to know what percent of the cola he should drink before the ride begins, so that none of it spills during the big drop, in which the roller coaster achieves 0.55-g acceleration at a 45° angle below the horizontal. Make the calculation for him, neglecting sloshing and assuming that the glass is vertical at all times. C2.5 Dry adiabatic lapse rate (DALR) is defined as the negative value of atmospheric temperature gradient, dTIdz, when temperature and pressure vary in an isentropic fashion. Assuming air is an ideal gas, DALR = —dT/dz when T = Tdip/poY, where exponent a = (k — l)/k, k = CjJc, is the ratio of specific heats, and Tq and po are the temperature and pressure at sea level, respectively, (a) Assuming that hydrostatic conditions exist in the atmosphere, show that the dry adiabatic lapse rate is constant and is given by DALR = g{k — l)/(kR), where R is the ideal gas constant for air. {b) Calculate the numerical value of DALR for air in units of °C/km. C2.6 In “soft” liquids (low bulk modulus j3), it may be necessary to account for liquid compressibility in hydrostatic calcula¬ tions. An approximate density relation would be dp ~ ^ dp = a^dp or p ~ Po + a^(p — Po) where a is the speed of sound and (po, Po) the conditions at the liquid surface z = 0. Use this approximation to show that the density variation with depth in a soft liquid is p = poe ^"" where g is the acceleration of gravity and z is positive upward. Then consider a vertical wall of width b, extending from the surface (z = 0) down to depth z = —h. Find an analytic expression for the hydrostatic force F on this wall, and compare it with the incompressible result F = poigl^bll. Would the center of pressure be below the incompressible position z = —2/2/3? C2.7 Venice, Italy, is slowly sinking, so now, especially in winter, plazas and walkways are flooded during storms. The proposed solution is the floating levee of Fig. C2.7. When filled with air, it rises to block off the sea. The levee is 30 m high, 5 m wide, and 20 m deep. Assume a uniform density of 300 kg/m^ when floating. For the 1-m sea- lagoon difference shown, estimate the angle at which the levee floats. H = 6.00 rpm R = 5.80 m (to center of manometer) 7.00 cm C2.3 Center of rotation Design Projects 129 C2.7 C2.8 In the U.S. Standard Atmosphere, the lapse rate B may vary from day to day. It is not a fundamental quantity like, say, Planck’s constant. Suppose that, on a certain day in Rhode Island, with = 288 K, the following pressures are mea¬ sured by weather balloons: Altitude z, km 0 2 5 8 Pressure p, kPa 100 78 53 34 Estimate the best-fit value of B for this data. Explain any difficulties. [Hint: EES is recommended.] C2.9 The ALVIN submersible vehicle has a passenger compart¬ ment which is a titanium sphere of inside diameter 78.08 in and thickness 1 .93 in. If the vehicle is submerged to a depth of 3850 m in the ocean, estimate (a) the water pressure outside the sphere, (b) the maximum elastic stress in the sphere, in Ibf/in^, and (c) the factor of safety of the titanium alloy (6% aluminum, 4% vanadium). Design Projects D2.1 It is desired to have a bottom-moored, floating system that creates a nonlinear force in the mooring line as the water level rises. The design force F need only be accurate in the range of seawater depths h between 6 and 8 m, as shown in the accompanying table. Design a buoyant system that will provide this force distribution. The system should be prac¬ tical (of inexpensive materials and simple construction). h, m f,N h, m f,N 6.00 400 7.25 554 6.25 437 7.50 573 6.50 471 7.75 589 6.75 502 8.00 600 7.00 530 D2.2 A laboratory apparatus used in some universities is shown in Fig. D2.2. The purpose is to measure the hydrostatic force on the flat face of the circular-arc block and compare it with the theoretical value for given depth h. The counter¬ weight is arranged so that the pivot arm is horizontal when the block is not submerged, whence the weight W can be correlated with the hydrostatic force when the submerged arm is again brought to horizontal. First show that the apparatus concept is valid in principle; then derive a for¬ mula for VT as a function of h in terms of the system param¬ eters. Finally, suggest some appropriate values of Y, L, and so on for a suitable apparatus and plot theoretical W versus h for these values. Counterweight Side view of block face D2.2 130 Chapter 2 Pressure Distribution in a Fluid D2.3 The Leary Engineering Company (see Popular Science, November 2000, p. 14) has proposed a ship hull with hinges that allow it to open into a flatter shape when enter¬ ing shallow water. A simplified version is shown in Fig. D2.3. In deep water, the hull cross section would be triangular, with large draft. In shallow water, the hinges would open to an angle as high as 8 = 45°. The dashed line indicates that the bow and stern would be closed. Make a parametric study of this configuration for various 9, assum¬ ing a reasonable weight and center of gravity location. Show how the draft, the metacentric height, and the ship’s stability vary as the hinges are opened. Comment on the effectiveness of this concept. References U.S. Standard Atmosphere, 1976, Government Printing Office, Washington, DC, 1976. J. A. Knauss, Introduction to Physical Oceanography, 2d ed., Waveland Press, Long Grove, IL, 2005. E. C. Tupper, Introduction to Naval Architecture, 4th ed., Elsevier, New York, 2004. D. T. Greenwood, Advanced Dynamics, Cambridge University Press, New York, 2006. R. I. Fletcher, “The Apparent Field of Gravity in a Rotating Fluid System,” Am. J. Phys., vol. 40, July 1972, pp. 959-965. National Committee for Fluid Mechanics Films, Illustrated Experiments in Fluid Mechanics, M.I.T. Press, Cambridge, MA, 1972. J. P. Holman, Experimental Methods for Engineers, 8th ed., McGraw-Hill, New York, 2011. R. C. Baker, Flow Measurement Handbook, Cambridge University Press, New York, 2005. T. G. Beckwith, R. G. Marangoni, and J. H. Lienhard V, Mechanical Measurements, 6th ed., Prentice-Hall, Upper Saddle River, NJ, 2006. J. W. Dally, W. F. Riley, and K. G. McConnell, Instrumenta¬ tion for Engineering Measurements, 2d ed., Wiley, New York, E. N. Gilbert, “How Things Float,” Am. Math. Monthly, vol. 98, no. 3, 1991, pp. 201-216. R. J. Figliola and D. E. Beasley, Theory and Design for Mechanical Measurements, 4th ed., Wiley, New York, 2005. R. W. Miller, Flow Measurement Engineering Handbook, 3d ed., McGraw-Hill, New York, 1996. L. D. Clayton, E. P. EerNisse, R. W. Ward, and R. B. Wiggins, “Miniature Crystalline Quartz Electromechanical Structures,” Sensors and Actuators, vol. 20, Nov. 15, 1989, pp. 171-177. A. Kitai (ed.). Luminescent Materials and Applications, John Wiley, New York, 2008. B. G. Liptak (ed.). Instrument Engineer’s Handbook: Process Measurement and Analysis, 4th ed., vol. 1, CRC Press, Boca Raton, FL, 2003. A. von Beckerath, WIKA Handbook — Pressure and Temper¬ ature Measurement, WIKA Instrument Corp., Lawrenceville, GA, 2008. On July 16, 1969, a massive Saturn V rocket lifted Apollo 11 from the NASA Kennedy Space Center, carrying astronauts Neil Armstrong, Michael Collins, and Edwin Aldrin to the hrst landing on the moon, four days later. The photo is filled with fluid momentum. In this chapter we learn how to analyze both the thrust of the rocket and the force of the exit jet on the solid surface. [Photo credit: NASA] 132 3.1 Basic Physical Laws of Fluid Mechanics Chapter 3 Integral Relations for a Control Volume Motivation. In analyzing fluid motion, we might take one of two paths: (1) seeking to describe the detailed flow pattern at every point {x, y, z) in the held or (2) working with a finite region, making a balance of flow in versus flow out, and determining gross flow effects such as the force or torque on a body or the total energy exchange. The second is the “control volume” method and is the subject of this chapter. The first is the “differential” approach and is developed in Chap. 4. We first develop the concept of the control volume, in nearly the same manner as one does in a thermodynamics course, and we find the rate of change of an arbitrary gross fluid property, a result called the Reynolds transport theorem. We then apply this theorem, in sequence, to mass, linear momentum, angular momentum, and energy, thus deriving the four basic control volume relations of fluid mechanics. There are many applications, of course. The chapter includes a special case of frictionless, shaft-work-free momentum and energy: the Bernoulli equation. The Bernoulli equation is a wonderful, historic relation, but it is extremely restrictive and should always be viewed with skepticism and care in applying it to a real (viscous) fluid motion. It is time now to really get serious about flow problems. The fluid statics applications of Chap. 2 were more like fun than work, at least in this writer’s opinion. Statics problems basically require only the density of the fluid and knowledge of the position of the free surface, but most flow problems require the analysis of an arbitrary state of variable fluid motion defined by the geometry, the boundary conditions, and the laws of mechanics. This chapter and the next two outline the three basic approaches to the analysis of arbitrary flow problems: Control volume, or large-scale, analysis (Chap. 3). Differential, or small-scale, analysis (Chap. 4). Experimental, or dimensional, analysis (Chap. 5). 133 134 Chapter 3 Integral Relations for a Control Volume The three approaches are roughly equal in importance. Control volume analysis, the present topic, is accurate for any flow distribution but is often based on average or “one-dimensional” property values at the boundaries. It always gives useful “engineering” estimates. In principle, the differential equation approach of Chap. 4 can be applied to any problem. Only a few problems, such as straight pipe flow, yield to exact analytical solutions. But the differential equations can be modeled numerically, and the flourishing field of computational fluid dynamics (CFD) can now be used to give good estimates for almost any geometry. Finally, the dimensional analysis of Chap. 5 applies to any problem, whether analytical, numerical, or experimental. It is particularly useful to reduce the cost of experimentation. Differential analysis of hydrodynamics began with Euler and d’Alembert in the late eighteenth century. Lord Rayleigh and E. Buckingham pioneered dimensional analysis at the end of the nineteenth century. The control volume was described in words, on an ad hoc one-case basis, by Daniel Bernoulli in 1753. Ludwig Prandtl, the celebrated founder of modem fluid mechanics, developed the control volume as a systematic tool in the early 1900s. The writer’s teachers at M.I.T. introduced control volume analysis into American textbooks, for thermodynamics by Keenan in 1941 , and for fluids by Hunsaker and Rightmire in 1947 . Eor a complete history of the control volume, see Vincent! . Systems versus Control Volumes All the laws of mechanics are written for a system, which is defined as an arbitrary quantity of mass of fixed identity. Everything external to this system is denoted by the term surroundings, and the system is separated from its surroundings by its bound¬ aries. The laws of mechanics then state what happens when there is an interaction between the system and its surroundings. First, the system is a fixed quantity of mass, denoted by m. Thus the mass of the system is conserved and does not change.' This is a law of mechanics and has a very simple mathematical form, called conservation of mass: or OTsyst = const (3.1) This is so obvious in solid mechanics problems that we often forget about it. In fluid mechanics, we must pay a lot of attention to mass conservation, and it takes a little analysis to make it hold. Second, if the surroundings exert a net force F on the system, Newton’s second law states that the mass in the system will begin to accelerate:^ dY d F = ma = m - = — (mV) (3.2) dt dt In Eq. (2.8) we saw this relation applied to a differential element of viscous incom¬ pressible fluid. In fluid mechanics Newton’s second law is called the linear momentum relation. Note that it is a vector law that implies the three scalar equations = ma„ Fy = mOy, and = ma^. ’We are neglecting nuclear reactions, where mass can be changed to energy. ^We are neglecting relativistic effects, where Newton’s law must be modified. 3.1 Basic Physical Laws of Fluid Mechanics 135 Third, if the surroundings exert a net moment M about the center of mass of the system, there will be a rotation effect (3.3) where H = S(r X Y)5m is the angular momentum of the system about its center of mass. Here we call Eq. (3.3) the angular momentum relation. Note that it is also a vector equation implying three scalar equations such as = dHJdt. For an arbitrary mass and arbitrary moment, H is quite complicated and contains nine terms (see, for example. Ref. 1). In elementary dynamics we commonly treat only a rigid body rotating about a fixed x axis, for which Eq. (3.3) reduces to M, = 4^(u;J (3.4) dt where is the angular velocity of the body and 4 is its mass moment of inertia about the x axis. Unfortunately, fluid systems are not rigid and rarely reduce to such a simple relation, as we shall see in Sec. 3.6. Fourth, if heat 5Q is added to the system or work dW is done by the system, the system energy dE must change according to the energy relation, or first law of thermodynamic s : 6Q - SW = dE an, or Q-W=— (3.5) dt Like mass conservation, Eq. (3.1), this is a scalar relation having only a single component. Finally, the second law of thermodynamics relates entropy change dS to heat added dQ and absolute temperature T: dS^-^ (3.6) This is valid for a system and can be written in control volume form, but there are almost no practical applications in fluid mechanics except to analyze flow loss details (see Sec. 9.5). All these laws involve thermodynamic properties, and thus we must supplement them with state relations p = p{p, T) and e = e{p, T) for the particular fluid being studied, as in Sec. 1.8. Although thermodynamics is not the main topic of this book, it is very important to the general study of fluid mechanics. Thermodynamics is crucial to compressible flow. Chap. 9. The student should review the first law and the state relations, as discussed in Refs. 6 and 7. The purpose of this chapter is to put our four basic laws into the control volume form suitable for arbitrary regions in a flow: Conservation of mass (Sec. 3.3). The linear momentum relation (Sec. 3.4). The angular momentum relation (Sec. 3.6). The energy equation (Sec. 3.7). 136 Chapter 3 Integral Relations for a Control Volume Wherever necessary to complete the analysis we also introduce a state relation such as the perfect-gas law. Equations (3.1) to (3.6) apply to either fluid or solid systems. They are ideal for solid mechanics, where we follow the same system forever because it represents the product we are designing and building. For example, we follow a beam as it deflects under load. We follow a piston as it oscillates. We follow a rocket system all the way to Mars. But fluid systems do not demand this concentrated attention. It is rare that we wish to follow the ultimate path of a specific particle of fluid. Instead it is likely that the fluid forms the environment whose effect on our product we wish to know. For the three examples just cited, we wish to know the wind loads on the beam, the fluid pressures on the piston, and the drag and lift loads on the rocket. This requires that the basic laws be rewritten to apply to a specific region in the neighborhood of our product. In other words, where the fluid particles in the wind go after they leave the beam is of little interest to a beam designer. The user’s point of view underlies the need for the control volume analysis of this chapter. In analyzing a control volume, we convert the system laws to apply to a specific region, which the system may occupy for only an instant. The system passes on, and other systems come along, but no matter. The basic laws are reformulated to apply to this local region called a control volume. All we need to know is the flow field in this region, and often simple assumptions will be accurate enough (such as uniform inlet and/or outlet flows). The flow conditions away from the control volume are then irrel¬ evant. The technique for making such localized analyses is the subject of this chapter. Volume and Mass Rate of Flow All the analyses in this chapter involve evaluation of the volume flow Q or mass flow rh passing through a surface (imaginary) defined in the flow. Suppose that the surface S in Fig. 3.1a is a sort of (imaginary) wire mesh through which the fluid passes without resistance. How much volume of fluid passes through S in unit time? If, typically, V varies with position, we must integrate over the elemental surface dA in Fig. 3.1a. Also, typically V may pass through dA at an angle 9 off the normal. Fet n be defined as the unit vector normal to dA. Then the amount of fluid swept through dA in time dt is the volume of the slanted parallelepiped in Fig. 3.lb\ dT = VdtdA cos 9= (\ ■ n) dA dt Fig. 3.1 Volume rate of flow through an arbitrary surface: (a) an elemental area dA on the surface; (b) the incremental volume swept through dA equals V dt dA cos 9. Unit normal n n 3.2 The Reynolds Transport Theorem 137 3.2 The Reynolds Transport Theorem Fig. 3.2 Fixed, moving, and deformable control volumes: {a) fixed control volume for nozzle stress analysis; {b) control volume moving at ship speed for drag force analysis; (c) control volume deforming within cylinder for transient pressure variation analysis. The integral of d¥/dt is the total volume rate of flow Q through the surface S'. Q (V • n) dA V„dA (3.7) We could replace V • n by its equivalent, V„, the component of V normal to dA, but the use of the dot product allows Q to have a sign to distinguish between inflow and outflow. By convention throughout this book we consider n to be the outward normal unit vector. Therefore V • n denotes outflow if it is positive and inflow if negative. This will be an extremely useful housekeeping device when we are computing volume and mass flow in the basic control volume relations. Volume flow can be multiplied by density to obtain the mass flow rh. If density varies over the surface, it must be part of the surface integral: m p(V • n) dA pV„ dA If density and velocity are constant over the surface S, a simple expression results: One-dimensional approximation: rk = pQ = pAV Typical units for Q are m^/s and for rh kg/s. To convert a system analysis to a control volume analysis, we must convert our mathematics to apply to a specific region rather than to individual masses. This con¬ version, called the Reynolds transport theorem, can be applied to all the basic laws. Examining the basic laws (3.1) to (3.3) and (3.5), we see that they are all concerned with the time derivative of fluid properties m, V, H, and E. Therefore what we need is to relate the time derivative of a system property to the rate of change of that property within a certain region. The desired conversion formula differs slightly according to whether the con¬ trol volume is fixed, moving, or deformable. Figure 3.2 illustrates these three cases. The fixed control volume in Fig. 3.2fl encloses a stationary region of inter¬ est to a nozzle designer. The control surface is an abstract concept and does not hinder the flow in any way. It slices through the jet leaving the nozzle, encloses the surrounding atmosphere, and slices through the flange bolts and the fluid within the nozzle. This particular control volume exposes the stresses in the flange bolts, Control surface Control surface Control surface {b) (c) 138 Chapter 3 Integral Relations for a Control Volume Arbitrary Fixed Control Volume Fig. 3.3 An arbitrary control volume with an arbitrary flow pattern. which contribute to applied forces in the momentum analysis. In this sense the control volume resembles the free-body concept, which is applied to systems in solid mechanics analyses. Figure 3.2b illustrates a moving control volume. Here the ship is of interest, not the ocean, so that the control surface chases the ship at ship speed V. The control volume is of fixed volume, but the relative motion between water and ship must be considered. If V is constant, this relative motion is a steady flow pattern, which sim¬ plifies the analysis.^ If V is variable, the relative motion is unsteady, so that the computed results are time-variable and certain terms enter the momentum analysis to reflect the noninertial (accelerating) frame of reference. Figure 3.2c shows a deforming control volume. Varying relative motion at the boundaries becomes a factor, and the rate of change of shape of the control volume enters the analysis. We begin by deriving the fixed control volume case, and we con¬ sider the other cases as advanced topics. An interesting history of control volume analysis is given by Vincenti . Figure 3.3 shows a fixed control volume with an arbitrary flow pattern passing through. There are variable slivers of inflow and outflow of fluid all about the control surface. In general, each differential area dA of surface will have a different velocity V making a different angle 9 with the local normal to dA. Some elemental areas will have ’a wind tunnel uses a fixed model to simulate flow over a body moving through a fluid. A tow tank uses a moving model to simulate the same situation. 3.2 The Reynolds Transport Theorem 139 inflow volume (VA cos dt, and others will have outflow volume (VA cos dt, as seen in Fig. 3.3. Some surfaces might correspond to streamlines (9 = 90°) or solid walls (V = 0) with neither inflow nor outflow. Let B be any property of the fluid (energy, momentum, enthalpy, etc.) and let /3 = dB/dm be the intensive value, or the amount of B per unit mass in any small element of the fluid. The total amount of B in the control volume (the solid curve in Fig. 3.3) is thus Bcv f3 dm •'cv I3pdr (3 ■’ey dB dm (3.8) Examining Fig. 3.3, we see three sources of changes in B relating to the control volume: A change within the control volume Outflow of /3 from the control volume (3pV cos 9 rMout -’cs (3.9) Inflow of (5 to the control volume fdpV cos 9 dAi^ •'cs The notations CV and CS refer to the control volume and control surface, respec¬ tively. Note, in Fig. 3.3, that the system has moved a bit. In the limit as dt^O, the instantaneous change of B in the system is the sum of the change within, plus the outflow, minus the inflow: d d / I3pdr] + ■>cv (3pY cos 9 dAou, — -'cs /3pVcos 6»dAi„ (3.10) •"cs This is the Reynolds transport theorem for an arbitrary fixed control volume. By let¬ ting the property B be mass, momentum, angular momentum, or energy, we can rewrite all the basic laws in control volume form. Note that all three of the integrals are concerned with the intensive property /3. Since the control volume is fixed in space, the elemental volumes dY do not vary with time, so that the time derivative of the volume integral vanishes unless either /3 or p varies with time (unsteady flow). Equation (3.10) expresses the basic formula that a system derivative equals the rate of change of B within the control volume plus the flow of B out of the control surface minus the flow of B into the control surface. The quantity B (or (3) may be any vector or scalar property of the fluid. Two alternate forms are possible for the flow terms. First we may notice that V cos 9 is the component of V normal to the area element of the control surface. Thus we can write Flow terms (3pVyi fL4out •'cs PpV„ dAi„ •'cs (3 dnia^i •'cs (3 dmia (3.10fl) •'cs where dm = pV^dA is the differential mass flow through the surface. Form (3.10fl) helps us visualize what is being calculated. 140 Chapter 3 Integral Relations for a Control Volume Control Volume Moving at Constant Velocity Control Volume of Constant Shape but Variable Velocity' A second, alternative form offers elegance and compactness as advantages. If n is defined as the outward normal unit vector everywhere on the control surface, then V • n = y„ for outflow and V • n = —V,, for inflow. Therefore the flow terms can be represented by a single integral involving V • n that accounts for both positive outflow and negative inflow: Flow terms (3p{\ • n) dA •'cs The compact form of the Reynolds transport theorem is thus dt (5pdY] + •'ey /3p(V • n) dA •'cs (3.11) (3.12) This is beautiful but only occasionally useful, when the coordinate system is ide¬ ally suited to the control volume selected. Otherwise the computations are easier when the flow of B out is added and the flow of B in is subtracted, according to Eqs. (3.10) or (3.11). The time derivative term can be written in the equivalent form dt f3p dr •’ey -’ey W dr ot (3.13) for the fixed control volume since the volume elements do not vary. If the control volume is moving uniformly at velocity V^, as in Fig. 3.2b, an observer fixed to the control volume will see a relative velocity V^ of fluid crossing the control surface, defined by V, = V - V, (3.14) where V is the fluid velocity relative to the same coordinate system in which the control volume motion V^ is observed. Note that Eq. (3.14) is a vector subtraction. The flow terms will be proportional to V^, but the volume integral of Eq. (3.12) is unchanged because the control volume moves as a fixed shape without deforming. The Reynolds transport theorem for this case of a uniformly moving control volume is dt (f^syst) d dt (3pdr] + PpCVr • n) dA ■’cs (3.15) which reduces to Eq. (3.12) if = 0. If the control volume moves with a velocity V/t) that retains its shape, then the volume elements do not change with time, but the boundary relative velocity = V(r, t) — Vj(f) becomes a somewhat more complicated function. Equation (3.15) is unchanged in form, but the area integral may be more laborious to evaluate. This section may be omitted without loss of continuity. 3.2 The Reynolds Transport Theorem 141 Arbitrarily Moving and Deformable Control Volume The most general situation is when the control volume is both moving and deforming arbitrarily, as illustrated in Fig. 3.4. The flow of volume across the control surface is again proportional to the relative normal velocity component • n, as in Eq. (3.15). However, since the control surface has a deformation, its velocity Vj = V/r, f), so that the relative velocity = V(r, f) — V/r, t) is or can be a complicated function, even though the flow integral is the same as in Eq. (3.15). Meanwhile, the volume integral in Eq. (3.15) must allow the volume elements to distort with time. Thus the time derivative must be applied after integration. For the deforming control volume, then, the transport theorem takes the form dt (3p dr ] + •'ey (5p{\r ■ n) dA •'cs (3.16) This is the most general case, which we can compare with the equivalent form for a fixed control volume: dt (^syst) im dr + (5p{r • n) dA ■’cs (3.17) The moving and deforming control volume, Eq. (3.16), contains only two complica¬ tions: (1) The time derivative of the first integral on the right must be taken outside, and (2) the second integral involves the relative velocity between the fluid system and the control surface. These differences and mathematical subtleties are best shown by examples. Fig. 3.4 Relative velocity effects between a system and a control volume when both move and deform. The system boundaries move at velocity V, and the control surface moves at velocity Vj. System at CV at time t + dt time t + dt ^This section may be omitted without loss of continuity. 142 Chapter 3 Integral Relations for a Control Volume Fig. 3.5 A control volume with simplihed one-dimensional inlets and exits. One-Dimensional Flux Term Approximations I t i@ J L I I Section 2: uniform ^2 , A2 , /O2 ’ /^2 ’ All sections i: y, approximately normal to area Aj In many situations, the flow crosses the boundaries of the control surface only at simplified inlets and exits that are approximately one-dimensional; that is, flow prop¬ erties are nearly uniform over the cross section. For a fixed control volume, the surface integral in Eq. (3.12) reduces to a sum of positive (outlet) and negative (inlet) product terms for each cross section: 4 (^syst) = 4( (3dmj 2 I out “ 2 I m where m,- = (3.18) To the writer, this is an attractive way to set up a control volume analysis without using the dot product notation. An example of multiple one-dimensional flows is shown in Fig. 3.5. There are inlet flows at sections 1 and 4 and outflows at sections 2, 3, and 5. Equation (3.18) becomes dt Mpmi + /33(pAy)3 + /35(pAy)5 (3i(pAV)i /34 (pA VO4 (3.19) with no contribution from any other portion of the control surface because there is no flow across the boundary. EXAMPLE 3.1 A fixed control volume has three one-dimensional boundary sections, as shown in Fig. E3.1. The flow within the control volume is steady. The flow properties at each section are tabu¬ lated below. Find the rate of change of energy of the system that occupies the control volume at this instant. E3.1 3.2 The Reynolds Transport Theorem 143 Section Type p, kg/m^ V, m/s A, e, J/kg 1 Inlet 800 5.0 2.0 300 2 Inlet 800 8.0 3.0 100 3 Outlet 800 17.0 2.0 150 Solution • System sketch: Figure E3.1 shows two inlet flows, 1 and 2, and a single outlet flow, 3. ■ Assumptions: Steady flow, fixed control volume, one-dimensional inlet and exit flows. • Approach: Apply Eq. (3.17) with energy as the property, where B = E and = dE/dm = e. Use the one-dimensional flow approximation and then insert the data from the table. • Solution steps: Outlet 3 contributes a positive term, and inlets 1 and 2 are negative. The appropriate form of Eq. (3.12) is (f 1,, ^ + e,m,-e,m,-e,m, cv Since the flow is steady, the time-derivative volume integral term is zero. Introducing (pAV)i as the mass flow grouping, we obtain Introducing the numerical values from the table, we have = -(300J/kg)(800k;g/m^)(2m^)(5m/s) - 100(800)(3)(8) -f 150(800)(2)(17) = (-2,400,000 - 1,920,000 -f 4,080,000) J/s = -240,000 J/s = -0.24 MJ/s Ans. Thus the system is losing energy at the rate of 0.24 MJ/s = 0.24 MW. Since we have accounted for all fluid energy crossing the boundary, we conclude from the first law that there must be heat loss through the control surface, or the system must be doing work on the environment through some device not shown. Notice that the use of SI units leads to a consistent result in joules per second without any conversion factors. We promised in Chap. 1 that this would be the case. • Comments: This problem involves energy, but suppose we check the balance of mass also. Then B = mass m, and /3 = dm/dm = unity. Again the volume integral vanishes for steady flow, and Eq. (3.17) reduces to j p(\ • n) dA = —piAiVi — P2A2V2 + P2A2V2 'cs = -(800kg/m^)(2m^)(5m/s) - 800(3)(8) -f 800(17)(2) = (-8000 - 19,200 -f 27,200) kg/s = 0 kg/s Thus the system mass does not change, which correctly expresses the law of conservation of system mass, Eq. (3.1). 144 Chapter 3 Integral Relations for a Control Volume CV EXAMPLE 3.2 Compressed air in a rigid tank of volume T exhausts through a small nozzle as in Fig. E3.2. Air properties change through the nozzle, and the flow exits at A^. Find an expression for the rate of change of tank density. Solution • System sketch: Fig. E3.2 shows one exit, no inlets. The constant exit area is A^. • Control volume: As shown, we choose a CV that encircles the entire tank and nozzle. • Assumptions: Unsteady flow (the tank mass decreases), one-dimensional exit flow. • Approach: Apply Eq. (3.16) for mass, B = m and fi = dm/dm = unity. • Solution steps: Write out the Reynolds transport relation (3.16) for this problem: / dm\ ^ d V tit J sysj dt pdT + p{\ dp n)dA =r^ + p,VAo dt Solve for the rate of change of tank density: tip Po — = - Ans. dt T • Comments: This is a first-order ordinary differential equation for the tank density. If we account for changes in p^ and Vo from the compressible-flow theories of Chap. 9, we can readily solve this equation for the tank density p(t). For advanced study, many more details of the analysis of deformable control vol¬ umes can be found in Hansen and Potter et al. . 3.3 Conservation of Mass The Reynolds transport theorem, Eq. (3.16) or (3.17), establishes a relation between system rates of change and control volume surface and volume integrals. But system derivatives are related to the basic laws of mechanics, Eqs. (3.1) to (3.5). Eliminating system derivatives between the two gives the control volume, or integral, forms of the laws of mechanics of fluids. The dummy variable B becomes, respectively, mass, linear momentum, angular momentum, and energy. Eor conservation of mass, as discussed in Examples 3.1 and 3.2, B = m and f3 = dm! dm = 1. Equation (3.1) becomes / dm\ ^ d V tit J dt -f p{Nr • n) dA ■’cs (3.20) This is the integral mass conservation law for a deformable control volume. Eor a fixed control volume, we have do — dY + dt p(V •n)dA = 0 . 'cv 'cs (3.21) 3.3 Conservation of Mass 145 If the control volume has only a number of one-dimensional inlets and outlets, we can write (3.22) Other special cases occur. Suppose that the flow within the control volume is steady; then dp! St = 0, and Eq. (3.21) reduces to p(V -11)^14 = 0 •'cs (3.23) This states that in steady flow the mass flows entering and leaving the control volume must balance exactly.® If, further, the inlets and outlets are one-dimensional, we have for steady flow 2 (P; = 2 (PiAiVdout (3.24) i i This simple approximation is widely used in engineering analyses. For example, refer¬ ring to Fig. 3.5, we see that if the flow in that control volume is steady, the three outlet mass flows balance the two inlets: Outflow = Inflow P2A2V2 + P2A3V2, + PsAsVs = PxAyVi + P4A4V4 (3.25) The quantity pAV is called the mass flow m passing through the one-dimensional cross section and has consistent units of kilograms per second (or slugs per second) for SI (or BG) units. Equation (3.25) can be rewritten in the short form m2 + /W3 + m5 = mj 4- nin (3.26) and, in general, the steady-flow-mass-conservation relation (3.23) can be written as 2 = 2 (3-27) If the inlets and outlets are not one-dimensional, one has to compute m by integration over the section = p(V • n) dA (3.28) where “cs” stands for cross section. An illustration of this is given in Example 3.4. ^Throughout this section we are neglecting sources or sinks of mass that might he embedded in the control volume. Equations (3.20) and (3.21) can readily be modified to add source and sink terms, but this is rarely necessary. 146 Chapter 3 Integral Relations for a Control Volume Incompressible Flow Still further simplification is possible if the fluid is incompressible, which we may deflne as having density variations that are negligible in the mass conservation require¬ ment.^ As we saw in Chap. 1, all liquids are nearly incompressible, and gas flows can behave as if they were incompressible, particularly if the gas velocity is less than about 30 percent of the speed of sound of the gas. Again consider the fixed control volume. For nearly incompressible flow, the term dpidt is small, so the time-derivative volume integral in Eq. (3.21) can be neglected. The constant density can then be removed from the surface integral for a nice simplification: or p(V • n) dA = 0 = p(V • n) dA cs (V • n) t/A = 0 . CS p . cs (V • n) dA (3.29) If the inlets and outlets are one-dimensional, we have 2 (V;A,.)out = 2 i i or 22out = 22in (3.30) where Qj = VjAi is called the volume flow passing through the given cross section. Again, if consistent units are used, Q = VA will have units of cubic meters per second (SI) or cubic feet per second (BG). If the cross section is not one-dimensional, we have to integrate Qcs (V • n) dA ■'cs (3.31) Equation (3.31) allows us to define an average velocity that, when multiplied by the section area, gives the correct volume flow: Q A A. (V • n) dA (3.32) This could be called the volume-average velocity. If the density varies across the sec¬ tion, we can define an average density in the same manner: P&v (3.33) But the mass flow would contain the product of density and velocity, and the average product (/9V)av would in general have a different value from the product of the averages: (pV)av A p(V • n) dA ~ PavKv (3.34) ’Be warned that there is subjectivity in specifying incompressibility. Oceanographers consider a 0. 1 percent density variation very significant, while aerodynamicists may neglect density vaiiations in highly compressible, even hypersonic, gas hows. Your task is to jushfy the incompressible approximahon when you make it. 3.3 Conservation of Mass 147 We illustrate average velocity in Example 3.4. We can often neglect the difference or, if necessary, use a correction factor between mass average and volume average. V • n = 0 EXAMPLE 3.3 Write the conservation-of-mass relation for steady flow through a streamtube (flow every¬ where parallel to the walls) with a single one-dimensional inlet 1 and exit 2 (Fig. E3.3). Solution For steady flow Eq. (3.24) applies with the single inlet and exit: rii = piAiVi = P2A2V2 = const Thus, in a streamtube in steady flow, the mass flow is constant across every section of the tube. If the density is constant, then Ai Q = AiVi = A2V2 = const or V2 = — Vi Ai The volume flow is constant in the tube in steady incompressible flow, and the velocity increases as the section area decreases. This relation was derived by Leonardo da Vinci in 1500. r = R E3.4 EXAMPLE 3.4 For steady viscous flow through a circular tube (Fig. E3.4), the axial velocity profile is given approximately by so that u varies from zero at the wall (r = R), or no slip, up to a maximum u = Uq at the centerline r = 0. For highly viscous (laminar) flow m ~ while for less viscous (turbulent) flow m ~ 7. Compute the average velocity if the density is constant. Solution The average velocity is defined by Eq. (3.32). Here V = in and n = i, and thus V • n = u. Since the flow is symmetric, the differential area can be taken as a circular strip dA = litr dr. Equation (3.32) becomes 1 2 1 m ) Inr dr (1 + m) {2 + m) or Eav = Uo Ans. 148 Chapter 3 Integral Relations for a Control Volume Tank area A, E3.5 For the laminar flow approximation, m ~ \ and ~ 0.53 I/q. (The exact laminar theory in Chap. 6 gives = 0.50Uq.) For turbulent flow, m and ~ O.SIUq. (There is no exact turbulent theory, and so we accept this approximation.) The turbulent velocity profile is more uniform across the section, and thus the average velocity is only slightly less than maximum. EXAMPLE 3.5 The tank in Fig. E3.5 is being filled with water by two one-dimensional inlets. Air is trapped at the top of the tank. The water height is h. (a) Find an expression for the change in water height dhidt. (b) Compute dhidt if Di = 1 in, D2 = 3 in, Vi = 3 ft/s, V2 = 2 ft/s, and A, = 2 ft^, assuming water at 20°C. Solution Part (a) A suggested control volume encircles the tank and cuts through the two inlets. The flow within is unsteady, and Eq. (3.22) applies with no outlets and two inlets: dt P PiAiVi P2A2F2 — H (1) Now if A, is the tank cross-sectional area, the unsteady term can be evaluated as follows: d_ dt P d d ^ ^ dh — (p^.A,h) + — [PaA,(H - h)]= p„Afj- at at at (2) The term vanishes because it is the rate of change of air mass and is zero because the air is trapped at the top. Substituting (2) into (1), we find the change of water height dh piAiT] + P2A2V2 dt p„A, Ans. (a) Eor water, P\ = P2 = Pw, and this result reduces to dh __ A^Vi + A2V2 _ Qi 3- Q2 dt A, A, Part (b) The two inlet volume flows are Qi = AiVi = J7r(n ft)^(3 ft/s) = 0.016 ftVs (3) Q2 = A2V2 = ?7r(n ft)^(2 ft/s) = 0.098 ftVs Then, from Eq. (3), dh (0.016 -f 0.098) ftVs = 0.057 ft/s dt 2 ft^ Suggestion: Repeat this problem with the top of the tank open. Ans. (b) 3.4 The Linear Momentum Equation 149 3.4 The Linear Momentum Equation The control volume mass relations, Eq. (3.20) or (3.21), are fundamental to all fluid flow analyses. They involve only velocity and density. Vector directions are of no consequence except to determine the normal velocity at the surface and hence whether the flow is in or out. Although your specific analysis may concern forces or moments or energy, you must always make sure that mass is balanced as part of the analysis; otherwise the results will be unrealistic and probably incorrect. We shall see in the examples that follow how mass conservation is constantly checked in perform¬ ing an analysis of other fluid properties. In Newton’s second law, Eq. (3.2), the property being differentiated is the linear momentum mV. Therefore our dummy variable is B = mV and (3 = dR/dm = V, and application of the Reynolds transport theorem gives the linear momentum relation for a deformable control volume: \pdr] + Jcv Vp(V, • n) dA (3.35) Jcs The following points concerning this relation should be strongly emphasized: The term V is the fluid velocity relative to an inertial (nonaccelerating) coordinate system; otherwise Newton’s second law must be modified to include noninertial relative acceleration terms (see the end of this section). The term X F is the vector sum of all forces acting on the system material considered as a free body; that is, it includes surface forces on all fluids and solids cut by the control surface plus all body forces (gravity and electromagnetic) acting on the masses within the control volume. The entire equation is a vector relation; both the integrals are vectors due to the term V in the integrands. The equation thus has three components. If we want only, say, the x component, the equation reduces to 2^. d dt upiYr ' n) dA ■’cs (3.36) and similarly, X Fy and X F^ would involve v and w, respectively. Failure to account for the vector nature of the linear momentum relation (3.35) is probably the greatest source of student error in control volume analyses. For a fixed control volume, the relative velocity V,. = V, and Eq. (3.35) becomes 2F dt VpdY + Jcv Vp(V • n) dA -CS (3.37) Again we stress that this is a vector relation and that V must be an inertial-frame velocity. Most of the momentum analyses in this text are concerned with Eq. (3.37). 150 Chapter 3 Integral Relations for a Control Volume One-Dimensional Momentum Flux Net Pressure Force on a Closed Control Surface Fig. 3.6 Pressure force computation by subtracting a uniform distribution: (a) uniform pressure, F = —p^ n (iA = 0; By analogy with the term mass flow used in Eq. (3.28), the surface integral in Eq. (3.37) is called the momentum flow term. If we denote momentum by M, then Mcs V\|0(V • n) dA ^sec (3.38) Because of the dot product, the result will be negative for inlet momentum flow and positive for outlet flow. If the cross section is one-dimensional, V and p are uniform over the area and the integrated result is Msec; = (3.39) for outlet flow and — m,V, for inlet flow. Thus if the control volume has only one¬ dimensional inlets and outlets, Eq. (3.37) reduces to (3.40) This is a commonly used approximation in engineering analyses. It is crucial to realize that we are dealing with vector sums. Equation (3.40) states that the net vector force on a fixed control volume equals the rate of change of vector momentum within the control volume plus the vector sum of outlet momentum flows minus the vector sum of inlet flows. Generally speaking, the surface forces on a control volume are due to (1) forces exposed by cutting through solid bodies that protrude through the surface and (2) forces due to pressure and viscous stresses of the surrounding fluid. The computation of pressure force is relatively simple, as shown in Fig. 3.6. Recall from Chap. 2 that (b) (b) nonuniform pressure, F = - (p - Pa)ndA. (a) 3.4 The Linear Momentum Equation 151 the external pressure force on a surface is normal to the surface and inward. Since the unit vector n is defined as outward, one way to write the pressure force is F press p{—n) dA ■’cs (3.41) Now if the pressure has a uniform value all around the surface, as in Fig. 3.7a, the net pressure force is zero: Fitp — p^(-n)dA = ndA = Q (3.42) where the subscript UP stands for uniform pressure. This result is independent of the shape of the surface^ as long as the surface is closed and all our control volumes are closed. Thus a seemingly complicated pressure force problem can be simplified by subtracting any convenient uniform pressure p^ and working only with the pieces of gage pressure that remain, as illustrated in Fig. 3.6b. So Eq. (3.41) is entirely equivalent to F = ^ press (P - Pa)i-ti)dA = JCS Pgage( tl) dA ■’CS This trick can mean quite a savings in computation. EXAMPLE 3.6 A control volume of a nozzle section has surface pressures of 40 Ibf/in^ absolute at section 1 and atmospheric pressure of 15 Ibf/in^ absolute at section 2 and on the external rounded part of the nozzle, as in Fig. E3.6a. Compute the net pressure force if Dj = 3 in and D2 = I in. Solution • System sketch: The control volume is the outside of the nozzle, plus the cut sections (1) and (2). There would also be stresses in the cut nozzle wall at section 1, which we are neglecting here. The pressures acting on the control volume are shown in Fig. E3.6a. Figure E3.6fc shows the pressures after 15 Ibf/in^ has been subtracted from all sides. Here we compute the net pressure force only. Jet exit pressure is atmospheric E3.6 («) (^) Can you prove this? It is a consequence of Gauss’s theorem from vector analysis. 152 Chapter 3 Integral Relations for a Control Volume Pressure Condition at a Jet Exit Fig. 3.7 Net force on a one¬ dimensional streamtube in steady flow: (a) streamtube in steady flow; (b) vector diagram for computing net force. ■ Assumptions: Known pressures, as shown, on all surfaces of the control volume. • Approach: Since three surfaces have p = 15 Ibf/in^, subtract this amount everywhere so that these three sides reduce to zero “gage pressure” for convenience. This is allowable because of Eq. (3.42). • Solution steps: For the modified pressure distribution. Fig. E3.6h, only section 1 is needed: IbA ^press Pgage,l ( -^1 ^ J -(-i) [-(3m)^J = 177ilbf Ans. • Comments: This “uniform subtraction” artifice, which is entirely legal, has greatly sim¬ plified the calculation of pressure force. Note: We were a bit too informal when multiply¬ ing pressure in Ibf/in^ times area in square inches. We achieved Ibf correctly, but it would be better practice to convert all data to standard BG units. Further note: In addition to Fpress. there are other forces involved in this flow, due to tension stresses in the cut nozzle wall and the fluid weight inside the control volume. Figure E3.6 illustrates a pressure boundary condition commonly used for jet exit flow problems. When a fluid flow leaves a confined internal duct and exits into an ambient “atmosphere,” its free surface is exposed to that atmosphere. Therefore the jet itself will essentially be at atmospheric pressure also. This condition was used at section 2 in Fig. E3.6. Only two effects could maintain a pressure difference between the atmosphere and a free exit jet. The first is surface tension, Eq. (1.31), which is usually negligible. The second effect is a supersonic jet, which can separate itself from an atmosphere with expansion or compression waves (Chap. 9). For the majority of applications, therefore, we shall set the pressure in an exit jet as atmospheric. EXAMPLE 3.7 A fixed control volume of a streamtube in steady flow has a uniform inlet flow (pi, Ai, Vi) and a uniform exit flow (p2, A2, V2)> shown in Fig. 3.7. Find an expression for the net force on the control volume. 3.4 The Linear Momentum Equation 153 Solution Equation (3.40) applies with one inlet and exit: 2f = - miVi = (P2A2L2)V2 - (MiViiVi The volume integral term vanishes for steady flow, but from conservation of mass in Example 3.3 we saw that rill = m2 = rh = const Therefore a simple form for the desired result is 2r = m(V2-Vi) Ans. This is a vector relation and is sketched in Fig. 3.1b. The term 2 F represents the net force acting on the control volume due to all causes; it is needed to balance the change in momen¬ tum of the fluid as it turns and decelerates while passing through the control volume. EXAMPLE 3.8 As shown in Fig. 3.8a, a fixed vane turns a water jet of area A through an angle 0 without changing its velocity magnitude. The flow is steady, pressure is everywhere, and friction on the vane is negligible, (a) Find the components and Fy of the applied vane force. (b) Find expressions for the force magnitude F and the angle (p between F and the horizontal; plot them versus 9. Fig. 3.8 Net applied force on a fixed jet-tuming vane: (a) geometry of the vane turning the water jet; (b) vector diagram for the net force. Solution Part (a) The control volume selected in Fig. 3.8a cuts through the inlet and exit of the jet and through the vane support, exposing the vane force F. Since there is no cut along the vane-jet interface, vane friction is internally self-canceling. The pressure force is zero in the uniform atmosphere. We neglect the weight of fluid and the vane weight within the control volume. Then Eq. (3.40) reduces to Fvane 1112^2 - miVi 154 Chapter 3 Integral Relations for a Control Volume But the magnitude Vi = V2 = V as given, and conservation of mass for the streamtuhe requires ihi = m2 = lit = pAV. The vector diagram for force and momentum change becomes an isos¬ celes triangle with legs mV and base F, as in Fig. 3.8h. We can readily find the force components from this diagram: Fj, = mV(cos 0 — 1) Fy = mV sin 0 Ans. (a) where tiiV = pAV^ for this case. This is the desired result. Part (b) The force magnitude is obtained from part (a): 0 F = (FI + Fyf^ = mV[sin^6» -f (cos 6 - 1)^]‘® = 2mVsin- Ans. (b) E3.8 From the geometry of Fig. 3.8/? we obtain 1 ^ d) = 180° - tan” — = 90° + - Ans. (b) F, 2 These can be plotted versus 6 as shown in Fig. E3.8. Two special cases are of interest. First, the maximum force occurs al 6 = 180° — that is, when the jet is turned around and thrown back in the opposite direction with its momentum completely reversed. This force is 27nV' and acts to the left\ that is, (p = 180°. Second, at very small turning angles {6 < 10°) we obtain approximately F « mV9 (p « 90° The force is linearly proportional to the turning angle and acts nearly normal to the Jet. This is the principle of a lifting vane, or airfoil, which causes a slight change in the oncoming flow direction and thereby creates a lift force normal to the basic flow. EXAMPLE 3.9 A water jet of velocity V, impinges normal to a flat plate that moves to the right at velocity 14, as shown in Fig. 3.9a. Find the force required to keep the plate moving at constant veloc¬ ity if the jet density is 1000 kg/m^, the jet area is 3 cm^, and V, and I4 are 20 and 15 m/s, respectively. Neglect the weight of the jet and plate, and assume steady flow with respect to the moving plate with the jet splitting into an equal upward and downward half-jet. 3.4 The Linear Momentum Equation 155 Solution The suggested control volume in Fig. 3.9a cuts through the plate support to expose the desired forces and Ry. This control volume moves at speed 14 and thus is fixed relative to the plate, as in Fig. 3.9b. We must satisfy both mass and momentum conservation for the assumed steady flow pattern in Fig. 3.9b. There are two outlets and one inlet, and Eq. (3.30) applies for mass conservation: Ulout ^in or PiAiVi + P2A2V2 = PjAjiVj - K) (1) We assume that the water is incompressible Pi = P2 = Pp and we are given that Ai = A2 = pAj. Therefore Eq. (1) reduces to Li + L2 = 2(Vj - VJ (2) Strictly speaking, this is all that mass conservation tells us. However, from the symmetry of the jet deflection and the neglect of gravity on the fluid trajectory, we conclude that the two velocities Vi and 1^2 must be equal, and hence Eq. (2) becomes Vi = V2 = Vj - V, (3) This equality can also be predicted by Bernoulli’s equation in Sec. 3.5. For the given numeri¬ cal values, we have Li = V2 = 20 - 15 = 5 m/s Now we can compute R,. and Ry from the two components of momentum conservation. Equation (3.40) applies with the unsteady term zero: = R^ = riiiUi + m2U2 — rrijUj (4) where from the mass analysis, = m2 ~ = kPjAj{Vj — 14). Now check the flow direc¬ tions at each section: uj = M2 = 0, and Uj = Vj — = 5 m/s. Thus Eq. (4) becomes R, = -mjUy = -[PjAjiVy - V,)]iVj - K) (5) Fig. 3.9 Force on a plate moving at constant velocity: (a) jet striking a moving plate normally; (b) control volume fixed relative to the plate. Nozzle (a) CS @A2=\|a, (b) 156 Chapter 3 Integral Relations for a Control Volume For the given numerical values we have = -(1000 kg/m^) (0.0003 m^)(5 m/s)^ = -7.5 (kg ■ m)/s^ = -7.5 N Ans. This acts to the left\ that is, it requires a restraining force to keep the plate from accelerating to the right due to the continuous impact of the jet. The vertical force is Fy = Ry = riiiVi + m2V2 — thjVj Check directions again: rti = Vi, V2 = —V2, Vj = 0. Thus Ry = mi(Vi) + m2(-V2) = im/Vi - V2) (6) But since we found earlier that Vi = V2, this means that Ry = 0, as we could expect from the symmetry of the jet deflection.^ Two other results are of interest. First, the relative velocity at section 1 was found to he 5 m/s up, from Eq. (3). If we convert this to absolute motion hy adding on the control- volume speed F,- = 15 m/s to the right, we find that the absolute velocity Vi = 151 + 5j m/s, or 15.8 m/s at an angle of 18.4° upward, as indicated in Fig. 3.9a. Thus the absolute jet speed changes after hitting the plate. Second, the computed force Rj. does not change if we assume the jet deflects in all radial directions along the plate surface rather than just up and down. Since the plate is normal to the x axis, there would still be zero outlet x-momentum flow when Eq. (4) was rewritten for a radial deflection condition. EXAMPLE 3.10 The sluice gate in Fig. E3.10a controls flow in open channels. At sections 1 and 2, the flow is uniform and the pressure is hydrostatic. Neglecting bottom friction and atmospheric pres¬ sure, derive a formula for the horizontal force F required to hold the gate. Express your final formula in terms of the inlet velocity Vi, eliminating ^2- E3.10fl A I Sluice gate, width b F 2 ^2 Solution Choose a control volume, Eig. E3.10b, that cuts through known regions (section 1 and sec¬ tion 2 just above the bottom, and the atmosphere) and that cuts along regions where unknown information is desired (the gate, with its force F). ^Symmetry can be a powerful tool if used properly. Try to learn more about the uses and misuses of symmetry conditions. 3.4 The Linear Momentum Equation 157 Assume steady incompressible flow with no variation across the width b. The inlet and outlet mass flows balance: m = pVihib = pV2h2b or V2 = Vi(hifh2) We may use gage pressures for convenience because a uniform atmospheric pressure causes no force, as shown earlier in Fig. 3.6. With x positive to the right, equate the net horizontal force to the x-directed momentum change: = -F„te + -ghi(hib) - -glhiMb) = m (Vj “ Vi) m = phibVi Solve for Fg3,e, and eliminate V'2 using the mass flow relation. The desired result is: ^gate ™ g^^l 1 - phibV\ alh /12 1 Ans. This is a powerful result from a relatively simple analysis. Later, in Sec. 10.4, we will be able to calculate the actual flow rate from the water depths and the gate opening height. EXAMPLE 3.11 Example 3.9 treated a plate at normal incidence to an oncoming flow. In Fig. 3.10 the plate is parallel to the flow. The stream is not a jet but a broad river, ov free stream, of uniform veloc¬ ity V = C/fli. The pressure is assumed uniform, and so it has no net force on the plate. The plate does not block the flow as in Fig. 3.9, so the only effect is due to boundary shear, which was neglected in the previous example. The no-slip condition at the wall brings the fluid there to a halt, and these slowly moving particles retard their neighbors above, so that at the end of the plate there is a significant retarded shear layer, or boundary layer, of thickness y = d. The viscous stresses along the wall can sum to a finite drag force on the plate. These effects are illustrated in Fig. 3.10. The problem is to make an integral analysis and find the drag force D in terms of the flow properties p, Uq, and d and the plate dimensions L and Solution Like most practical cases, this problem requires a combined mass and momentum balance. A proper selection of control volume is essential, and we select the four-sided region from °The general analysis of such wall shear problems, called boundary-layer theory, is treated in Sec. 7.3. 158 Chapter 3 Integral Relations for a Control Volume Fig. 3.10 Control volume analysis of drag force on a flat plate due to boundary shear. The control volume is bounded by sections 1, 2, 3, and 4. Oncoming stream parallel to plate Streamline just outside the shear-layer region P=Pa V = 6 > = h Boundary layer where shear stress is significant i<y) Plate of width b 0 to h to ^ to L and back to the origin 0, as shown in Fig. 3.10. Had we chosen to cut across horizontally from left to right along the height y = h, we would have cut through the shear layer and exposed unknown shear stresses. Instead we follow the streamline passing through {x, y) = (0, h), which is outside the shear layer and also has no mass flow across it. The four control volume sides are thus From (0, 0) to (0, h): a one-dimensional inlet, V • n = -Uo. From (0, h) to (L, d)'. a streamline, no shear, V • n = 0. From (L, d) to (L, 0): a two-dimensional outlet, V • n = +ii(y). From (L, 0) to (0, 0): a streamline just above the plate surface, V • n = 0, shear forces summing to the drag force —Di acting from the plate onto the retarded fluid. The pressure is uniform, and so there is no net pressure force. Since the flow is assumed incompressible and steady, Eq. (3.37) applies with no unsteady term and flows only across sections 1 and 3: 2 f'x = -D = p \ u{0, y)(V • n) rfA -f p\ u(L, y) (V • n) dA •^1 •'3 = p\ Ua{-Uti)b dy + p\ u{L,y)[ + u{L,y)]b dy Evaluating the first integral and rearranging give D = pUobh — pb u dy 1^=^ (1) This could be considered the answer to the problem, but it is not useful because the height h is not known with respect to the shear layer thickness 6. This is found by applying mass conservation, since the control volume forms a streamtube: p\ (V • n) r/A = 0 = p I (-(/o)hrfy + p \| ubdy\^=L •'cs Ugh = \ udy (2) 3.4 The Linear Momentum Equation 159 after canceling b and p and evaluating the first integral. Introduce this value of h into Eq. (1) for a much cleaner result: D = pb\ u{Uo - u) dy Ans. (3) This result was first derived by Theodore von Karman in 1921." It relates the friction drag on one side of a flat plate to the integral of the momentum deficit pu(Uo — u) across the trailing cross section of the flow past the plate. Since Uq — u vanishes as y increases, the integral has a finite value. Equation (3) is an example of momentum integral theory for boundary layers, which is treated in Chap. 7. Momentum Flux Correction Factor For flow in a duct, the axial velocity is usually nonuniform, as in Example 3.4. For this case the simple momentum flow calculation JupiY • n) dA = rhV = pAV^ is somewhat in error and should be corrected to ^pAV^, where Q is the dimensionless momentum flow correction factor, C — 1- The factor Q accounts for the variation of across the duct section. That is, we compute the exact flow and set it equal to a flow based on average velocity in the duct: or ^dA = (pit Vav = CpAVl dA (3.43fl) Values of can be computed based on typical duct velocity profiles similar to those in Example 3.4. The results are as follows: 2N Laminar flow: Turbulent flow: M = C/n 1 R‘ c = t/n 1 - — - ^ m < - R 9 c = (1 + mf(2 + mf 2(1 + 2m) (2 + 2m) (3A3b) (3.43c) The turbulent correction factors have the following range of values: Turbulent flow: m 1 1 1 1 1 5 6 7 8 9 c 1.037 1.027 1.020 1.016 1.013 These are so close to unity that they are normally neglected. The laminar correction is often important. "The autobiography of this great twentieth-century engineer and teacher is recommended for its historical and scientific insight. 160 Chapter 3 Integral Relations for a Control Volume Linear Momentum Tips Noninertial Reference Frame^^ To illustrate a typical use of these correction factors, the solution to Example 3.8 for nonuniform velocities at sections 1 and 2 would be modified as = m(C2V2 - CiVi) (3.43t;) Note that the basic parameters and vector character of the result are not changed at all by this correction. The previous examples make it clear that the vector momentum equation is more difficult to handle than the scalar mass and energy equations. Here are some momen¬ tum tips to remember; • The momentum relation is a vector equation. The forces and the momentum terms are directional and can have three components. A sketch of these vectors will be indispensable for the analysis. • The momentum flow terms, such as fV(pV ■ n)dA, link two different sign conventions, so special care is needed. First, the vector coefficient V will have a sign depending on its direction. Second, the mass flow term (pV • n) will have a sign (+ , — ) depending on whether it is (out, in). For example, in Fig. 3.8, the jc- components of V2 and Vj, U2 and Uj, are both positive; that is, they both act to the right. Meanwhile, the mass flow at (2) is positive (out) and at (1) is negative (in). • The one-dimensional approximation, Eq. (3.40), is glorious, because non- uniform velocity distributions require laborious integration, as in Eq. (3.11). Thus the momentum flow correction factors are very useful in avoiding this integration, especially for pipe flow. • The applied forces XF act on all the material in the control volume — that is, the surfaces (pressure and shear stresses), the solid supports that are cut through, and the weight of the interior masses. Stresses on non-control- surface parts of the interior are self-canceling and should be ignored. • If the fluid exits subsonically to an atmosphere, the fluid pressure there is atmospheric. • Where possible, choose inlet and outlet surfaces normal to the flow, so that pressure is the dominant force and the normal velocity equals the actual velocity. Clearly, with that many helpful tips, substantial practice is needed to achieve momen¬ tum skills. All previous derivations and examples in this section have assumed that the coordinate system is inertial — that is, at rest or moving at constant velocity. In this case the rate of change of velocity equals the absolute acceleration of the system, and Newton’s law applies directly in the form of Eqs. (3.2) and (3.35). In many cases it is convenient to use a noninertial, or accelerating, coordinate system. An example would be coordinates fixed to a rocket during takeoff. A second example is any flow on the earth’s surface, which is accelerating relative to the fixed ’^This section may be omitted without loss of continuity. 3.4 The Linear Momentum Equation 161 Fig. 3.11 Geometry of fixed versus accelerating coordinates. stars because of the rotation of the earth. Atmospheric and oceanographic flows expe¬ rience the so-called Coriolis acceleration, outlined next. It is typically less than 10~^g, where g is the acceleration of gravity, but its accumulated effect over distances of many kilometers can be dominant in geophysical flows. By contrast, the Coriolis acceleration is negligible in small-scale problems like pipe or airfoil flows. Suppose that the fluid flow has velocity V relative to a noninertial xyz coordinate system, as shown in Fig. 3.11. Then dVldt will represent a noninertial acceleration that must be added vectorially to a relative acceleration Hrei to give the absolute accel¬ eration a, relative to some inertial coordinate system XYZ, as in Fig. 3.11. Thus dY a; = — + a^ei (3.44) at Since Newton’s law applies to the absolute acceleration, 2r ^ dY or 2 ^ -45) Thus Newton’s law in noninertial coordinates xyz is analogous to adding more “force” terms — marei to account for noninertial effects. In the most general case, sketched in Fig. 3.11, the term arei contains four parts, three of which account for the angular velocity Ylit) of the inertial coordinates. By inspection of Fig. 3.11, the absolute dis¬ placement of a particle is Si = r -f R (3.46) Differentiation gives the absolute velocity Vi = V + — + ft X r dt (3.47) 162 Chapter 3 Integral Relations for a Control Volume A second differentiation gives the absolute acceleration: a, = dN ctR dil ^ ^ ^ h — ^ H - xr + 2ilxV + ilx(flxr) dt df- dt (3.48) By comparison with Eq. (3.44), we see that the last four terms on the right represent the additional relative acceleration: 1 . d^RJdt^ is the acceleration of the noninertial origin of coordinates xyz- (dCl/dt) X r is the angular acceleration effect. 2Q. X V is the Coriolis acceleration. ft X (ft X r) is the centripetal acceleration, directed from the particle normal to the axis of rotation with magnitude ft^L, where L is the normal distance to the axis.^ Equation (3.45) differs from Eq. (3.2) only in the added inertial forces on the left-hand side. Thus the control volume formulation of linear momentum in noninertial coordi¬ nates merely adds inertial terms by integrating the added relative acceleration over each differential mass in the control volume: (3.49) where This is the noninertial analog of the inertial form given in Eq. (3.35). To analyze such problems, one must know the displacement R and angular velocity ft of the nonin¬ ertial coordinates. If the control volume is fixed in a moving frame, Eq. (3.49) reduces to 2F a^i dm •'ey dt \pdr] + •’ey Vp(V • n) dA •'cs (3.50) In other words, the right-hand side reduces to that of Eq. (3.37). EXAMPLE 3.12 A classic example of an accelerating control volume is a rocket moving straight up, as in Fig. E3.12. Let the initial mass be Mg, and assume a steady exhaust mass flow m and exhaust velocity 1/, relative to the rocket, as shown. If the flow pattern within the rocket motor is steady and air drag is neglected, derive the differential equation of vertical rocket motion V(0 and integrate using the initial condition F = 0 at t = 0. complete discussion of these noninertial coordinate terms is given, for example, in Ref. 4, pp. 49-51. 3.5 Frictionless Flow: The Bernoulli Equation 163 E3.12 3.5 Frictionless Flow: The Bernoulli Equation Fig. 3.12 The Bernoulli equation for frictionless flow along a streamline: (a) forces and flows; (b) net pressure force after uniform subtraction of p. Solution The appropriate control volume in Fig. E3. 12 encloses the rocket, cuts through the exit jet, and accelerates upward at rocket speed V{t). The z-momentum Eq. (3.49) becomes \a,Adm = — w dm] + (mw)g dV -mg — m — = 0 + m( — Fj) dt with m = m(t) = Mg — tilt The term = dV/dt of the rocket. The control volume integral vanishes because of the steady rocket flow conditions. Separate the variables and integrate, assuming F = 0 at f = 0: ■ V dt C ( mt\ dV = m Vc\ - ^ - g \ dt or V(t) = — F^ln 1 - — gt Ans. Jo Jo Mo - mt Jo V MgJ This is a classic approximate formula in rocket dynamics. The first term is positive and, if the fuel mass burned is a large fraction of initial mass, the final rocket velocity can exceed Fg. A classic linear momentum analysis is a relation between pressure, velocity, and eleva¬ tion in a frictionless flow, now called the Bernoulli equation. It was stated (vaguely) in words in 1738 in a textbook by Daniel Bernoulli. A complete derivation of the equation was given in 1755 by Leonhard Euler. The Bernoulli equation is very famous and very widely used, but one should be wary of its restrictions — all fluids are viscous and thus all flows have friction to some extent. To use the Bernoulli equation correctly, one must confine it to regions of the flow that are nearly frictionless. This section (and, in more detail, Chap. 8) will address the proper use of the Bernoulli relation. Consider Fig. 3.12, which is an elemental fixed streamtube control volume of variable area A(s) and length ds, where s is the streamline direction. The properties (p, V, p) may vary with s and time but are assumed to be uniform over the cross section A. The streamtube orientation 9 is arbitrary, with an elevation change dz = ds sin 9. Friction on the streamtube walls is shown and then neglected — a very 164 Chapter 3 Integral Relations for a Control Volume restrictive assumption. Note that the limit of a vanishingly small area means that the streamtube is equivalent to a streamline of the flow. Bernoulli’s equation is valid for both and is usually stated as holding “along a streamline” in frictionless flow. Conservation of mass [Eq. (3.20)] for this elemental control volume yields d f [ \ . . dn — p d°V ) + tMout ~ ntin = 0 ~ — dY + dm where m = pAV and dV ~ A ds. Then our desired form of mass conservation is dm = d{pAV) = — -Ads (3.51) This relation does not require an assumption of frictionless flow. Now write the linear momentum relation [Eq. (3.37)] in the streamwise direction; d dt •'ey Vpdr]-^ (mV)^ (fnV)in — ipV) Ads + d{mV) dt where = V itself because 5 is the streamline direction. If we neglect the shear force on the walls (frictionless flow), the forces are due to pressure and gravity. The streamwise gravity force is due to the weight component of the fluid within the control volume: '^^i.grav = —dW sin 9 = —'jA ds sin 6 = —jA dz The pressure force is more easily visualized, in Eig. 3.12^?, by first subtracting a uniform value p from all surfaces, remembering from Eig. 3.6 that the net force is not changed. The pressure along the slanted side of the streamtube has a streamwise component that acts not on A itself but on the outer ring of area increase dA. The net pressure force is thus dF s,press = \ dp dA — dp(A + dA) ~ —A dp to first order. Substitute these two force terms into the linear momentum relation: 2 dF^ = — jA dz — Adp = — (pV) Ads + d{mV) dp dV . = — VA ds H - pA ds + m dV + V dm dt dt ^ The first and last terms on the right cancel by virtue of the continuity relation [Eq. (3.51)]. Divide what remains by pA and rearrange into the final desired relation: dV dp — ds H - h V dV + g dz = 0 (3.52) dt p ^ This is Bernoulli’s equation for unsteady frictionless flow along a streamline. It is in dif¬ ferential form and can be integrated between any two points 1 and 2 on the streamline: ds + ~dp 1 , , ^ + -(v?- y?) + g{z2-z^) = 0 J, P ^ (3.53) 3.5 Frictionless Flow: The Bernoulli Equation 165 Steady Incompressible Flow Bernoulli Interpreted as an Energy Relation Restrictions on the Bernoulli Equation To evaluate the two remaining integrals, one must estimate the unsteady effect dVIdt and the variation of density with pressure. At this time we consider only steady (dVIdt = 0) incompressible (constant-density) flow, for which Eq. (3.53) becomes -(vl- y?) + g(z2 - zi) = 0 P ^ Pi 1 9 P2 1 9 or — -F -V\ + ^^1 = ^ + 2^^ + SZi = const (3.54) This is the Bernoulli equation for steady frictionless incompressible flow along a streamline. The Bernoulli relation, Eq. (3.54), is a classic momentum result, Newton’s law for a frictionless, incompressible fluid. It may also be interpreted, however, as an idealized energy relation. The changes from 1 to 2 in Eq. (3.54) represent reversible pressure work, kinetic energy change, and potential energy change. The fact that the total remains the same means that there is no energy exchange due to viscous dissipation, heat transfer, or shaft work. Section 3.7 will add these effects by making a control volume analysis of the first law of thermodynamics. The Bernoulli equation is a momentum-based force relation and was derived using the following restrictive assumptions: 1 . Steady flow: a common situation, application to most flows in this text. Incompressible flow: appropriate if the flow Mach number is less than 0.3. This restriction is removed in Chap. 9 by allowing for compressibility. Frictionless flow: restrictive — solid walls and mixing introduce friction effects. Flow along a single streamline: different streamlines may have different “Bernoulli constants” Wq = p/p + V^I2 -I- gz, but this is rare. In most cases, as we shall prove in Chap. 4, a frictionless flow region is irrotationab, that is, curl(V) = 0. For irrotational flow, the Bernoulli constant is the same everywhere. The Bernoulli derivation does not account for possible energy exchange due to heat or work. These thermodynamic effects are accounted for in the steady flow energy equation. We are thus warned that the Bernoulli equation may be modified by such an energy exchange. Figure 3.13 illustrates some practical limitations on the use of Bernoulli’s equation (3.54). For the wind tunnel model test of Fig. 3.13a, the Bernoulli equation is valid in the core flow of the tunnel but not in the tunnel wall boundary layers, the model surface boundary layers, or the wake of the model, all of which are regions with high friction. In the propeller flow of Fig. 3.13fo, Bernoulli’s equation is valid both upstream and downstream, but with a different constant Wq = p/p + V^/2 + gz, caused by the work addition of the propeller. The Bernoulli relation (3.54) is not valid near the propeller blades or in the helical vortices (not shown, see Fig. 1.14) shed downstream of the blade edges. Also, the Bernoulli constants are higher in the flowing “slipstream” than in the ambient atmosphere because of the slipstream kinetic energy. 166 Chapter 3 Integral Relations for a Control Volume Fig. 3.13 Illustration of regions of validity and invalidity of the Bernoulli equation: (a) tunnel model, (b) propeller, (c) chimney. Jet Exit Pressure Equals Atmospheric Pressure Stagnation, Static, and Dynamic Pressures Valid, new constant Valid, new constant For the chimney flow of Fig. 3.13c, Eq. (3.54) is valid before and after the fire, but with a change in Bernoulli constant that is caused by heat addition. The Bernoulli equation is not valid within the fire itself or in the chimney wall boundary layers. When a subsonic jet of liquid or gas exits from a duct into the free atmosphere, it immediately takes on the pressure of that atmosphere. This is a very important bound¬ ary condition in solving Bernoulli problems, since the pressure at that point is known. The interior of the free jet will also be atmospheric, except for small effects due to surface tension and streamline curvature. In many incompressible-flow Bernoulli analyses, elevation changes are negligible. Thus Eq. (3.54) reduces to a balance between pressure and kinetic energy. We can write this as 1 2 1 2 Pi + l^pyi = Pi + -pVi = Po = constant The quantity is the pressure at any point in the frictionless flow where the velocity is zero. It is called the stagnation pressure and is the highest pressure possible in the flow, if elevation changes are neglected. The place where zero-velocity occurs is called a stagnation point. For example, on a moving aircraft, the front nose and the wing leading edges are points of highest pressure. The pressures pi and p2 are called static pressures, in the moving fluid. The grouping {l/2)pV^ has dimensions of pres¬ sure and is called the dynamic pressure. A popular device called a Pitot-static tube (Fig. 6.30) measures (po ~ p) and then calculates V from the dynamic pressure. Fig. 3.14 Hydraulic and energy grade lines for frictionless flow in a duct. Hydraulic and Energy Grade Lines 3.5 Frictionless Flow: The Bernoulli Equation 167 Note, however, that one particular zero-velocity condition, no-slip flow along a fixed wall, does not result in stagnation pressure. The no-slip condition is 3. frictional effect, and the Bernoulli equation does not apply. A useful visual interpretation of Bernoulli’s equation is to sketch two grade lines of a flow. The energy grade line (EGL) shows the height of the total Bernoulli constant h(i = z + pi'y + V^Kflg). In frictionless flow with no work or heat transfer [Eq. (3.54)] the EGL has constant height. The hydraulic grade line (HGL) shows the height cor¬ responding to elevation and pressure head z -I- p!^ — that is, the EGL minus the velocity head V^l{2g). The HGL is the height to which liquid would rise in a piezom¬ eter tube (see Proh. 2.11) attached to the flow. In an open-channel flow the HGL is identical to the free surface of the water. Figure 3.14 illustrates the EGL and HGL for frictionless flow at sections 1 and 2 of a duct. The piezometer tubes measure the static pressure head z + pl'^ and thus outline the HGL. The pitot stagnation-velocity tubes measure the total head z + p!') -I- V^Hflg), which corresponds to the EGL. In this particular case the EGL is constant, and the HGL rises due to a drop in velocity. In more general flow conditions, the EGL will drop slowly due to friction losses and will drop sharply due to a substantial loss (a valve or obstruction) or due to work extrac¬ tion (to a turbine). The EGL can rise only if there is work addition (as from a pump or propeller). The HGL generally follows the behavior of the EGL with respect to losses or work transfer, and it rises and/or falls if the velocity decreases and/or increases. As mentioned before, no conversion factors are needed in computations with the Bernoulli equation if consistent SI or BG units are used, as the following examples will show. In all Bernoulli-type problems in this text, we consistently take point 1 upstream and point 2 downstream. 168 Chapter 3 Integral Relations for a Control Volume EXAMPLE 3.13 Find a relation between nozzle discharge velocity V2 tank free surface height h as in Fig. E3.13. Assume steady frictionless flow. 2g Solution As mentioned, we always choose point 1 upstream and point 2 downstream. Try to choose points 1 and 2 where maximum information is known or desired. Here we select point 1 as the tank free surface, where elevation and pressure are known, and point 2 as the nozzle exit, where again pressure and elevation are known. The two unknowns are Vi and V2- Mass conservation is usually a vital part of Bernoulli analyses. If Ai is the tank cross section and A2 the nozzle area, this is approximately a one-dimensional flow with constant density, Eq. (3.30): AiVi = A2V2 Bernoulli’s equation (3.54) gives (1) ^ i 1^1 + - ^ + 5 E2 + gZ2 P P But since sections 1 and 2 are both exposed to atmospheric pressure pi = P2 — Pm the pressure terms cancel, leaving Vi - Vj = 2g(z: - Z2) = 2gh (2) Eliminating Vi between Eqs. (1) and (2), we obtain the desired result: 2gh 1 - AilAi Ans. (3) Generally the nozzle area A2 is very much smaller than the tank area Ai, so that the ratio A^/Af is doubly negligible, and an accurate approximation for the outlet velocity is 1/2 F2 « i2gh) Ans. (4) 3.5 Frictionless Flow: The Bernoulli Equation 169 Surface Velocity Condition for a Large Tank This formula, discovered hy Evangelista Torricelli in 1644, states that the discharge veloc¬ ity equals the speed that a frictionless particle would attain if it fell freely from point 1 to point 2. In other words, the potential energy of the surface fluid is entirely converted to kinetic energy of efflux, which is consistent with the neglect of friction and the fact that no net pressure work is done. Note that Eq. (4) is independent of the fluid density, a char¬ acteristic of gravity-driven flows. Except for the wall boundary layers, the streamlines from 1 to 2 all hehave in the same way, and we can assume that the Bernoulli constant /tg is the same for all the core flow. However, the outlet flow is likely to he nonuniform, not one-dimensional, so that the aver¬ age velocity is only approximately equal to Torricelli’s result. The engineer will then adjust the formula to include a dimensionless discharge coefficient Cj: (V'2)av = = Q(2g/t)‘“ (5) A2 As discussed in Sec. 6.12, the discharge coefficient of a nozzle varies from about 0.6 to 1.0 as a function of (dimensionless) flow conditions and nozzle shape. Many Bernoulli, and also steady flow energy, problems involve liquid flow from a large tank or reservoir, as in Example 3.13. If the outflow is small compared to the volume of the tank, the surface of the tank hardly moves. Therefore these problems are analyzed assuming zero velocity at the tank surface. The pressure at the top of the tank or reservoir is assumed to be atmospheric. Before proceeding with more examples, we should note carefully that a solution by Bernoulli’s equation (3.54) does not require a second control volume analysis, only a selection of two points 1 and 2 along a given streamline. The control volume was used to derive the differential relation (3.52), but the integrated form (3.54) is valid all along the streamline for frictionless flow with no heat transfer or shaft work, and a control volume is not necessary. A classic Bernoulli application is the familiar process of siphoning a fluid from one container to another. No pump is involved; a hydrostatic pressure difference provides the motive force. We analyze this in the following example. EXAMPLE 3.14 Consider the water siphon shown in Fig. E3.14. Assuming that Bernoulli’s equation is valid, {a) find an expression for the velocity V2 exiting the siphon tube, {b) If the tube is 1 cm in diam¬ eter and zi = 60 cm, Z2 = “25 cm, Z3, = 90 cm, and 24 = 35 cm, estimate the flow rate in cmVs. z = 0 - \ V2 Z2 E3.14 170 Chapter 3 Integral Relations for a Control Volume Solution ■ Assumptions: Frictionless, steady, incompressible flow. Write Bernoulli’s equation start¬ ing from where information is known (the surface, zi) and proceeding to where informa¬ tion is desired (the tube exit, Z2)- Note that the velocity is approximately zero at zi, and a streamline goes from zi to Z2- Note further that pi and p2 are both atmospheric, p = p^^, and therefore cancel, (a) Solve for the exit velocity from the tube: V2 = V2g(zi - Z2) Ans. (a) The velocity exiting the siphon increases as the tube exit is lowered below the tank surface. There is no siphon effect if the exit is at or above the tank surface. Note that Z3 and Z4 do not directly enter the analysis. However, Z3 should not be too high because the pressure there will be lower than atmospheric, and the liquid might vaporize, {b) For the given numerical information, we need only zi and Z2 and calculate, in SI units. V2 = V'2(9.81 m/s2)[0.6 m - (-0.25) m] = 4.08 m/s Q = V2A2 = (4.08 m/s) (7r/4) (0.01 m)^ = 321 E-6 mVs = 321 cmVs Ans. (b) ■ Comments: Note that this result is independent of the density of the fluid. As an exercise, you may check that, for water (998 kg/m^), p^ is 11,300 Pa below atmospheric pressure. In Chap. 6 we will modify this example to include friction effects. EXAMPLE 3.15 A constriction in a pipe will cause the velocity to rise and the pressure to fall at section 2 in the throat. The pressure difference is a measure of the flow rate through the pipe. The smoothly necked-down system shown in Fig. E3.15 is called a venturi tube. Find an expres¬ sion for the mass flow in the tube as a function of the pressure change. E3.15 Solution Bernoulli’s equation is assumed to hold along the center streamline: 3.5 Frictionless Flow: The Bernoulli Equation 171 If the tube is horizontal, zi = Z2 and we can solve for ¥2- , , 2 Ap Vj- Vj = - ^ Ap = Pi - p2 We relate the velocities from the incompressible continuity relation: (1) AiVi = A2V2 or Combining (1) 2 ^2 Fi = /3V2 13 = -^ and (2), we obtain a formula for the velocity in the throat: 2Ap F, = Lpd -/3")J The mass flow is given by til — PA2F2 — A2 2p Ap 1 - /3V (2) (3) (4) This is the ideal frictionless mass flow. In practice, we measure fflactuai = QtWyjai and correlate the dimensionless discharge coefficient Cj. EXAMPLE 3.16 A 10-cm fire hose with a 3-cm nozzle discharges 1.5 m^/min to the atmosphere. Assuming frictionless flow, find the force fg exerted by the flange bolts to hold the nozzle on the hose. Solution We use Bernoulli’s equation and continuity to find the pressure pi upstream of the noz¬ zle, and then we use a control volume momentum analysis to compute the bolt force, as in Fig. E3.16. E3.16 Control volume ib) 172 Chapter 3 Integral Relations for a Control Volume The flow from 1 to 2 is a constriction exactly similar in effect to the venturi in Example 3.15, for which Eq. (1) gave Pi=P2 + \p{Vl-V) (1) The velocities are found from the known flow rate 2 = 1.5 mVmin or 0.025 mVs: ^2 = — = 0.025 m7s ^ ^2 (7r/4)(0.03m)^ = 35.4 m/s 0.025 m7s = 3.2 m/s Ai (7r/4)(0.1 m)^ We are given P2 = Pa ~ ^ gage pressure. Then Eq. (1) becomes Pi = 1(1000 kg/m^)[(35.42 - 3.22)mV] = 620,000 kg/(m ■ s^) = 620,000 Pa gage The control volume force balance is shown in Eig. E3.I6/2: -Fb + PiM and the zero gage pressure on all other surfaces contributes no force. The x-momentum flow is + mV2 at the outlet and —mVi at the inlet. The steady flow momentum relation (3.40) thus gives -Fb + PiAi = rhiVi - Vi) or Fb = PiAi - m{V2 - Vi) (2) Substituting the given numerical values, we find til = pQ = (1000 kg/m^) (0.025 mVs) = 25 kg/s Ai = ^D\ = ^(0.1 m)^ = 0.00785 m^ Fb = (620,000 N/m^) (0.00785 m^) - (25 kg/s)[(35.4 - 3.2)m/s] = 4872 N - 805 (kg ■ m)/s^ = 4067 N (915 Ibf) Ans. Notice from these examples that the solution of a typical problem involving Bernoulli’s equation almost always leads to a consideration of the continuity equation as an equal partner in the analysis. The only exception is when the complete velocity distribution is already known from a previous or given analysis, but that means the continuity relation has already been used to obtain the given information. The point is that the continuity relation is always an important element in a flow analysis. 3.6 The Angular Momentum A control volume analysis can be applied to the angular momentum relation, Eq. (3.3), Theorem^" by letting our dummy variable B be the angular-momentum vector H. However, since the system considered here is typically a group of nonrigid fluid particles of variable velocity, the concept of mass moment of inertia is of no help, and we have to calculate '"'This section may be omitted without loss of continuity. 3.6 The Angular Momentum Theorem 173 the instantaneous angular momentum by integration over the elemental masses dm. If O is the point about which moments are desired, the angular momentum about O is given by H O (r X V) dm ^syst (3.55) where r is the position vector from 0 to the elemental mass dm and V is the velocity of that element. The amount of angular momentum per unit mass is thus seen to be /3 dm = r X V The Reynolds transport theorem (3.16) then tells us that dt syst d_ dt (r X V)p dY ■’ey (r X V)p(V, • n) dA (3.56) •’cs for the most general case of a deformable control volume. But from the angular momentum theorem (3.3), this must equal the sum of all the moments about point O applied to the control volume dt 2m, 2 (r X F). Note that the total moment equals the summation of moments of all applied forces about point O. Recall, however, that this law, like Newton’s law (3.2), assumes that the particle velocity V is relative to an inertial coordinate system. If not, the moments about point O of the relative acceleration terms in Eq. (3.49) must also be included: 2 (r X F), (r X a^i) dm ■’ey (3.57) where the four terms constituting ajei are given in Eq. (3.49). Thus the most general case of the angular momentum theorem is for a deformable control volume associated with a noninertial coordinate system. We combine Eqs. (3.56) and (3.57) to obtain X F), d ' (r X a, el) dm = — Jcv dt (r X )p d9 . . cv ^ . (r X V)p(Vr • n) dA (3.58) For a nondeformable inertial control volume, this reduces to (r X )pdr ■'ey ■+■ . (r X V)p(V • n) dA cs (3.59) Further, if there are only one-dimensional inlets and exits, the angular momentum flow terms evaluated on the control surface become (r X V)p(V • n)dA = '^ (r x V)out'«out - 2 x V)i„iWin (3.60) •'cs Although at this stage the angular momentum theorem can be considered a supple¬ mentary topic, it has direct application to many important fluid flow problems 174 Chapter 3 Integral Relations for a Control Volume involving torques or moments. A particularly important case is the analysis of rotating fluid flow devices, usually called turbomachines (Chap. 11). EXAMPLE 3.17 As shown in Fig. E3.17a, a pipe hend is supported at point A and connected to a flow system by flexible couplings at sections 1 and 2. The fluid is incompressible, and ambient pressure is zero, (a) Find an expression for the torque T that must be resisted by the support at A, in terms of the flow properties at sections 1 and 2 and the distances hi and h2. (b) Compute this torque if Di = £>2 = 3 in, pi = 100 Ibf/in^ gage, p2 = 80 Ibf/in^ gage, Vi = 40 ft/s, hi = 2 in, h2 = 10 in, and p = 1.94 slugs/ft^. Pi, Vi, Ai E3.17a Solution Part (a) The control volume chosen in Fig. E3.17& cuts through sections 1 and 2 and through the sup¬ port at A, where the torque Tji is desired. The flexible couplings description specifies that there is no torque at either section 1 or 2, and so the cuts there expose no moments. For the angular momentum terms r X V, r should be taken from point A to sections 1 and 2. Note that the gage pressure forces pAi ™d P2A2 both have moments about A. Equation (3.59) with one¬ dimensional flow terms becomes 2 M,4 = T^ -f Ti X (— pjAiIIj) -I- r2 X (—^2^2112) = (r2 X VjK+mout) + (ri x Vi)(-niin) (1) Figure E3.17c shows that all the cross products are associated with either ri sin 0i = hi or r2 sin O2 = h2, the perpendicular distances from point A to the pipe axes at 1 and 2. Remem¬ ber that = hia^i from the steady flow continuity relation. In terms of counterclockwise moments, Eq. (1) then becomes Ta + PiA-ihi — P2A2^2 ~ ;w(/j2V2 ~ hiVi) (2) Rewriting this, we find the desired torque to be Ta = ^(^2^2 + MV2) ~ hiipiAi + tiiVi) Ans. (a) (3) counterclockwise. The quantities pi and p2 are gage pressures. Note that this result is inde¬ pendent of the shape of the pipe bend and varies only with the properties at sections 1 and 2 and the distances hi and /i2.^ '^Indirectly, the pipe bend shape probably affects the pressure change from pi to p2- 3.6 The Angular Momentum Theorem 175 Part (b) For the numerical example, convert all data to BG units: Ihf Ihf Ihf Di = Z)2 = 3 in = 0.25 ft pi = 100^^ = 14,400 ^ P2 = 80 — 11,520 Ibf 2 10 /;. = 2 in = — ft ho = 10 in = — ft p ‘ 12 ^ 12 1.94 slug ft^ The inlet and exit areas are the same, Ai = A2 = (7r/4)(0.25 ft)^ = 0.0491 ft^. Since the density is constant, we conclude from mass conservation, pAiVi = PA2V2, that Vi = V'2 = 40 ft/s. The mass flow is m = pAiVi slug (0.0491 fF) 40 3.81 slug s Evaluation of the torque: The data can now he substituted into Eq. (3): T, = 10 — 1 12 Ibf ft" 11,520^ (0.0491 ft") + 3.81 — ^ 40 slug\ s / ft 12 lbf\ 2 ( slugV ft 14,400 ^1(0.0491 fF) + ( 3.81 ^ )( 40 = 598 ft ■ Ibf — 143 ft ■ Ibf = 455 ft ■ Ibf counterclockwise Ans. (b) ■ Comments: The use of standard BG units is crucial when combining dissimilar terms, such as pressure times area and mass flow times velocity, into proper additive units for a numerical solution. EXAMPLE 3.18 Figure 3.15 shows a schematic of a centrifugal pump. The fluid enters axially and passes through the pump blades, which rotate at angular velocity UJ\ the velocity of the fluid is changed from Vi to V2 and its pressure from pi to p2. (a) Find an expression for the torque To that must be applied to these blades to maintain this flow, (b) The power supplied to the pump would be P = To illustrate numerically, suppose ri = 0.2 m, r2 = 0.5 m, and 176 Chapter 3 Integral Relations for a Control Volume b = 0.15 m. Let the pump rotate at 600 r/min and deliver water at 2.5 mVs with a density of 1000 kg/m^. Compute the torque and power supplied. Fig. 3.15 Schematic of a simplihed centrifugal pump. Solution Part (a) The control volume is chosen to he the annular region between sections 1 and 2 where the flow passes through the pump blades (see Fig. 3.15). The flow is steady and assumed incompress¬ ible. The contribution of pressure to the torque about axis O is zero since the pressure forces at 1 and 2 act radially through O. Equation (3.59) becomes 2^0 = T„ = (r2 X V2)mout “ (ri x Vi)m;^ (1) where steady flow continuity tells us that 'tiin = pVni^Tvrib = = pV„227rr2b = pQ The cross product r X V is found to be clockwise about O at both sections: fi ^ V2 = r2Vi2 sin 90° k = r2V',2k clockwise ri X Vi = riV,ik clockwise Equation (1) thus becomes the desired formula for torque: To ~ PGi^i^ti ~ clockwise Ans. (a) (2a) This relation is called Euler’s turbine formula. In an idealized pump, the inlet and outlet tangential velocities would match the blade rotational speeds V,i = tori and 1^,2 = LOr2. Then the formula for torque supplied becomes To — pQ^^iti ~ fi) clockwise (2b) 3.6 The Angular Momentum Theorem 177 Part (b) Convert uo to 600(27r/60) = 62.8 rad/s. The normal velocities are not needed here but follow from the flow rate = = Inrib Q 2.5 mVs 27r(0.2m)(0.15 m) 2.5 = 13.3 m/s ■ = 5.3 m/s iTTrib 27r(0.5)(0.15) For the idealized inlet and outlet, tangential velocity equals tip speed: V,i = LJri = (62.8 rad/s) (0.2 m) = 12.6 m/s V,2 = ii’r2 = 62.8(0.5) = 31.4 m/s Equation (2a) predicts the required torque to be = (1000kg/m^)(2.5mVs)[(0.5m)(31.4m/s) - (0.2 m)( 12.6 m/s)] = 33,000 (kg ■ m^)/s^ = 33,000 N • m The power required is P = ljT„= (62.8 rad/s) (33,000 N ■ m) = 2,070,000 (N ■ m)/s = 2.07 MW (2780 hp) Ans. Ans. In actual practice the tangential velocities are considerably less than the impeller-tip speeds, and the design power requirements for this pump may be only 1 MW or less. Absolute outlet Fig. 3.16 View from above of a single arm of a rotating lawn sprinkler. EXAMPLE 3.19 Figure 3.16 shows a lawn sprinkler arm viewed from above. The arm rotates about O at constant angular velocity UJ. The volume flow entering the arm at O is Q, and the fluid is incompressible. There is a retarding torque at O, due to bearing friction, of amount —T^k. Find an expression for the rotation ui in terms of the arm and flow properties. Solution The entering velocity is Vok, where Vq = Equation (3.59) applies to the control vol¬ ume sketched in Fig. 3.16 only if V is the absolute velocity relative to an inertial frame. Thus the exit velocity at section 2 is V2 = Foi - RllA Equation (3.59) then predicts that, for steady flow, = -r„k = (rj X Vziniou, - (rj x \ (1) where, from continuity, ntom = = pQ. The cross products with reference to point O are r2 X V2 = /y X (Fq — Ruj)i = (R^u! — /?Fo)k Ti X Vi = Oj X Fflk = 0 178 Chapter 3 Integral Relations for a Control Volume Equation (1) thus becomes -r„k = pQiR^LO - RVo)k V T '' o ^ o u = - r R pQR- Ans. The result may surprise you: Even if the retarding torque is negligible, the arm rotational speed is limited to the value Vg/R imposed by the outlet speed and the arm length. 3.7 The Energy Equation^ As our fourth and final basic law, we apply the Reynolds transport theorem (3.12) to the first law of thermodynamics, Eq. (3.5). The dummy variable B becomes energy E, and the energy per unit mass is /3 = dE/dm = e. Equation (3.5) can then be written for a fixed control volume as follows: dQ dt dW _ dE dt dt d_ dt epdr ]+ ep{Y ■ n) dA (3.61) -'ey ' •'cs Recall that positive Q denotes heat added to the system and positive W denotes work done by the system. The system energy per unit mass e may be of several types: ^ ^internal T t^kinetic T ^^potential 5“ ^^other where gother could encompass chemical reactions, nuclear reactions, and electrostatic or magnetic field effects. We neglect Cotker here and consider only the first three terms as discussed in Eq. (1.9), with z defined as “up”: e = u + + gz (3.62) The heat and work terms could be examined in detail. If this were a heat transfer book, dQldt would be broken down into conduction, convection, and radiation effects and whole chapters written on each (see, for example. Ref. 3). Here we leave the term untouched and consider it only occasionally. Using for convenience the overdot to denote the time derivative, we divide the work term into three parts: ^ ^shaft T Wpress "h fTyiscous stresses T T The work of gravitational forces has already been included as potential energy in Eq. (3.62). Other types of work, such as those due to electromagnetic forces, are excluded here. The shaft work isolates the portion of the work that is deliberately done by a machine (pump impeller, fan blade, piston, or the like) protruding through the control surface into the control volume. No further specification other than VE, is desired at this point, but calculations of the work done by turbomachines will be performed in Chap. 11. '®This section should be read for information and enrichment even if you lack formal background in thermodynamics. ’The energy equation for a deformable control volume is rather complicated and is not discussed here. See Refs. 4 and 5 for further details. 3.7 The Energy Equation 179 The rate of work Wp done by pressure forces occurs at the surface only; all work on internal portions of the material in the control volume is by equal and opposite forces and is self-canceling. The pressure work equals the pressure force on a small surface element dA times the normal velocity component into the control volume: dWp = —{p £/A)y„ in = —p{—\ • n) dA The total pressure work is the integral over the control surface: p(V ■ n) dA •’cs (3.63) A cautionary remark: If part of the control surface is the surface of a machine part, we prefer to delegate that portion of the pressure to the shaft work term not to Wp, which is primarily meant to isolate the fluid flow pressure work terms. Finally, the shear work due to viscous stresses occurs at the control surface and consists of the product of each viscous stress (one normal and two tangential) and the respective velocity component: or dW^ = -T • V dA T-\dA ■'cs (3.64) where r is the stress vector on the elemental surface dA. This term may vanish or be negligible according to the particular type of surface at that part of the control volume: Solid surface. For all parts of the control surface that are solid confining walls, V = 0 from the viscous no-slip condition; hence = zero identically. Surface of a machine. Here the viscous work is contributed by the machine, and so we absorb this work in the term VTj. An inlet or outlet. At an inlet or outlet, the flow is approximately normal to the element dA\ hence the only viscous work term comes from the normal stress dA. Since viscous normal stresses are extremely small in all but rare cases, such as the interior of a shock wave, it is customary to neglect viscous work at inlets and outlets of the control volume. Streamline surface. If the control surface is a streamline, such as the upper curve in the boundary layer analysis of Fig. 3.11, the viscous work term must be evaluated and retained if shear stresses are significant along this line. In the particular case of Fig. 3.11, the streamline is outside the boundary layer, and viscous work is negligible. The net result of this discussion is that the rate-of-work term in Eq. (3.61) consists essentially of W =W,-- p(V •n)dA- {T-Y),,dA (3.65) ■'cs ■'cs where the subscript SS stands for stream surface. When we introduce (3.65) and (3.62) into (3.61), we find that the pressure work term can be combined with the energy flow 180 Chapter 3 Integral Relations for a Control Volume term since both involve surface integrals of V • n. The control volume energy equa¬ tion thus becomes Q-W,-W^ d dt ep dY 1-1- •'cv )p(V • n) dA (3.66) Using e from (3.62), we see that the enthalpy h = u + pip occurs in the control surface integral. The final general form for the energy equation for a fixed control volume becomes (3.67) As mentioned, the shear work term is rarely important. One-Dimensional Energy-Flux If the control volume has a series of one-dimensional inlets and outlets, as in Fig. 3.5, Terms the surface integral in (3.67) reduces to a summation of outlet flows minus inlet flows: (h + + gz)p(V • n) dA -'cs = + + gz)out»^out - + + gz)inmin (3.68) where the values of h, \V^, and gz are taken to be averages over each cross section. EXAMPLE 3.20 A steady flow machine (Fig. E3.20) takes in air at section 1 and discharges it at sections 2 and 3. The properties at each section are as follows: Section A, fF Q, fF/s T op p, Ibf/in^ abs z, ft 1 0.4 100 70 20 1.0 2 1.0 40 100 30 4.0 3 0.25 50 200 ? 1.5 Work is provided to the machine at the rate of 150 hp. Find the pressure p-^ in VoUir? abso¬ lute and the heat transfer Q in Btu/s. Assume that air is a perfect gas with R = 1716 and Cp = 6003 ft-lbf/(slug • °R). Solution • System sketch: Figure E3.20 shows inlet 1 (negative flow) and outlets 2 and 3 (positive flows). ■ Assumptions: Steady flow, one-dimensional inlets and outlets, ideal gas, negligible shear work. The flow is not incompressible. Note that Qi Q2 + 63 because the densities are different. E3.20 3.7 The Energy Equation 181 • Approach: Evaluate the velocities and densities and enthalpies and substitute into Eq. (3.67). Use BG units for all properties, including the pressures. With given, we evaluate Vi = QtlAj. a _ 100 ftVs Ai 0.4 ft^ 1^2 40 ftVs 1.0 ft^ 50 ftVs 0.25 ft^ 200 ft s The densities at sections 1 and 2 follow from the ideal-gas law: _ Pi _ (20 X 144) Ibf/ft^ _ slug [1716ft-lbf/(slug°R)][(70 + 460)°R] ~ ^.003 17— (30 X 144) (1716)(100 + 460) 0.00450 slug ft^ However, p^ is unknown, so how do we find Use the steady flow continuity relation: tn^ = m2 + hii or piQi = P2Q2 + psgs (1) / slugV ft^\ slug (0.00317 100 —1 = 0.00450(40) + P3(50) solve for p^ = 0.00274 Knowing p^ enables us to find p^ from the ideal-gas law: / slugV ft-lbf \ Ibf Ibf Pi = P2RT3 = 0.00274 1716 - — — (200 + 460“R) = 3100 -r = 21.5^ Ans. \ ft J\ slug °R/ ft in ■ Final solution steps: For an ideal gas, simply approximate enthalpies as hj = CpTi. The shaft work is negative (into the control volume) and viscous work is neglected for this solid-wall machine: ( ft-lbA ft-lbf Wi, « 0 W,= (-150hp) 550 - = -82,500 - (workonthe system) V s-hp / s For steady flow, the volume integral in Eq. (3.67) vanishes, and the energy equation becomes Q - W,= -rhiiCpTi 3- jV? + gzi) + m2(CpT2 + + gZ2) + mtCpTi + jUf + SZ3) (2) From our continuity calculations in Eq. (1) above, the mass flows are slug slug mi = PiQi = (0.003 17) (100) = 0.317 — ^ m2 = P1Q2 = 0.180 — ^ s “ s slug ™3 = PsGs = 0.137 - s It is instructive to separate the flow terms in the energy equation (2) for examination: Enthalpy flow = Cp(—rh{ri + m2T2 + m^T^) = (6003)[(-0.317)(530) -f (0.180)(560) -f (0.137)(660)] = -1,009,000 -f 605,000 -f 543,000 « -f 139,000 ft-lbf/s Kinetic energy flow = + hi2V2 + ih^Vl) = i[-0.317(250)^ -f (0.180)(40)^ -f (0.137)(200)^] = -9900 -f 140 -f 2740 « -7000 ft-lbf/s Potential energy flow = g(—miZi + hi2Z2 + rii^zs) = (32.2)[-0.317(1.0) -f 0.180(4.0) -f 0.137(1.5)] = -10 -f 23 -f 7 « -f 20 ft-lbf/s 182 Chapter 3 Integral Relations for a Control Volume The Steady Flow Energy Equation Friction and Shaft Work in Low-Speed Flow Equation (2) may now be evaluated for the heat transfer: or (-82,500) = 139,000 - 7,000 + 20 49,520 - ft-lbf 1 Btu 778.2 ft-lbf = -f 64- Btu Ans. • Comments: The heat transfer is positive, which means into the control volume. It is typical of gas flows that potential energy flow is negligible, enthalpy flow is dominant, and kinetic energy flow is small unless the velocities are very high (that is, high subsonic or supersonic). For steady flow with one inlet and one outlet, both assumed one-dimensional, Eq. (3.67) reduces to a celebrated relation used in many engineering analyses. Let section 1 be the inlet and section 2 the outlet. Then Q - W, - W^, = -liiiihi + + gzi) + m2(h2 + ^Vl + gzj) (3.69) But, from continuity, rhi = m2 = m, we can rearrange (3.69) as follows: h + + gzi = (h2 + kvl + gZ2) - q + w, + (3.70) where q = Qhn = dQldm, the heat transferred to the fluid per unit mass. Similarly, Wj = ITj/m = dWJdm and = W^lm = dWJdm. Equation (3.70) is a general form of the steady flow energy equation, which states that the upstream stagnation enthalpy Hi = {h + -f gz)i differs from the downstream value H2 only if there is heat trans¬ fer, shaft work, or viscous work as the fluid passes between sections 1 and 2. Recall that q is positive if heat is added to the control volume and that Wj and are positive if work is done by the fluid on the surroundings. Each term in Eq. (3.70) has the dimensions of energy per unit mass, or velocity squared, which is a form commonly used by mechanical engineers. If we divide through by g, each term becomes a length, or head, which is a form preferred by civil engineers. The traditional symbol for head is h, which we do not wish to confuse with enthalpy. Therefore we use internal energy in rewriting the head form of the energy relation: Ih 7 -(- Ml 8 V\ Pi ^2 ^2 h - 1" - - Z2 ~ hg hs 7 8 2g ^ ^ ^ (3.71) where = q/g, = wjg, and h^, = wflg are the head forms of the heat added, shaft work done, and viscous work done, respectively. The term p/7 is called pressure head, and the term V^/2g is denoted as velocity head. A common application of the steady flow energy equation is for low-speed (incom¬ pressible) flow through a pipe or duct. A pump or turbine may be included in the pipe system. The pipe and machine walls are solid, so the viscous work is zero. Equation (3.71) may be written as -f U2 — Ui — q 8 (3.72) 3.7 The Energy Equation 183 Every term in this equation is a length, or head. The terms in parentheses are the upstream (1) and downstream (2) values of the useful or available head or total head of the flow, denoted by h^. The last term on the right is the difference (/toi — h^^), which can include pump head input, turbine head extraction, and the friction head loss hf, always positive. Thus, in incompressible flow with one inlet and one outlet, we may write p r — f - — f z 7 2g (- + f + V 7 2g /out h friction — h pump h turbine (3.73) Most of our internal flow problems will be solved with the aid of Eq. (3.73). The h terms are all positive; that is, friction loss is always positive in real (viscous) flows, a pump adds energy (increases the left-hand side), and a turbine extracts energy from the flow. If hp and/or /i, are included, the pump and/or turbine must lie between points 1 and 2. In Chaps. 5 and 6 we shall develop methods of correlating hf losses with flow parameters in pipes, valves, fittings, and other internal flow devices. EXAMPLE 3.21 Gasoline at 20°C is pumped through a smooth 12-cm-diameter pipe 10 km long, at a flow rate of 75 mVh (330 gal/min). The inlet is fed by a pump at an absolute pressure of 24 atm. The exit is at standard atmospheric pressure and is 150 m higher. Estimate the frictional head loss hf and compare it to the velocity head of the flow V^l(2g). (These numbers are quite realistic for liquid flow through long pipelines.) Solution • Property values: From Table A. 3 for gasoline at 20°C, p = 680 kg/m^, or 7 = (680)(9.81) = 6670 N/ml • Assumptions: Steady flow. No shaft work, thus hp = h, = 0. If zi = 0, then Z2 — 150 m. • Approach: Find the velocity and the velocity head. These are needed for comparison. Then evaluate the friction loss from Eq. (3.73). • Solution steps: Since the pipe diameter is constant, the average velocity is the same everywhere: Q Q (75 m7h)/(3600 s/h) m Tin = Kn, = ^ = - ^ = , « 1.84 - A (7r/4)D^ (7r/4)(0.12m)^ s (1.84 m/s)^ Velocity head = — = - ^ ~ 0.173 m 2g 2(9.81 m/s^) Substitute into Eq. (3.73) and solve for the friction head loss. Use pascals for the pressures and note that the velocity heads cancel because of the constant-area pipe. 7 -I- Pout Zin - - f 7 Zout hf (24)(101.350N/m^) 6670 N/m^ -I- 0.173 m -I- 0 m = 101,350 N/m^ 6670 N/m^ -I- 0.173 m -I- 150 m -I- hf or hf = 364.7 - 15.2 - 150 « 199 m Ans. 184 Chapter 3 Integral Relations for a Control Volume The friction head is larger than the elevation change Az, and the pump must drive the flow against both changes, hence the high inlet pressure. The ratio of friction to velocity head is VV(2g) 199 m 0.173 m 1150 Ans. ■ Comments: This high ratio is typical of long pipelines. (Note that we did not make direct use of the 10,000-m pipe length, whose effect is hidden within hf) In Chap. 6 we can state this problem in a more direct fashion: Given the flow rate, fluid, and pipe size, what inlet pres¬ sure is needed? Our correlations for hf will lead to the estimate ~ 24 atm, as stated here. EXAMPLE 3.22 Air [R = 1716, Cp = 6003 ft ■ lbf/(slug ■ °R)] flows steadily, as shown in Fig. E3.22, through a turbine that produces 700 hp. For the inlet and exit conditions shown, estimate {a) the exit velocity V2 and {b) the heat transferred Q in Btu/h. E3.22 Solution Part (a) The inlet and exit densities can be computed from the perfect-gas law: Pi Pi 150(144) RTi 1716(460 -f 300) Pi Pi 40(144) RT2 1716(460 -f 35) = 0.0166 slug/fr’ 0.00679 slug/ft^ The mass flow is determined by the inlet conditions m PiAiVi = (0.0166) TT 4 I (100) = 0.325 slug/s Knowing mass flow, we compute the exit velocity rh = 0.325 = P2A2V2 = (0.00679) ^ or V2 = 244 ft/s Ans. (a) 3.7 The Energy Equation 185 Part (b) Kinetic Energy Correction Factor The steady flow energy equation (3.69) applies with IVj, = 0, zi = Z2, and h = CpT\ Q-W, = m{CpT2 + Ivi - CpTi - iy?) Convert the turbine work to foot-pounds-force per second with the conversion factor 1 hp = 550 ft • Ihf/s. The turbine work IVj is positive Q - 700(550) = 0.325[6003(495) + ^(244 f - 6003(760) - ^(100)^] = -510,000 ft- Ibf/s or 2 = -125,000 ft- Ibf/s Convert this to British thermal units as follows: 3600 s/h Q = (- 125,000 ft -Ibf/s) 778.2 ft - Ibf/Btu = -578,000 Btu/h Ans. (b) The negative sign indicates that this heat transfer is a loss from the control volume. Often the flow entering or leaving a port is not strictly one-dimensional. In particular, the velocity may vary over the cross section, as in Fig. E3.4. In this case the kinetic energy term in Eq. (3.68) for a given port should be modified by a dimensionless correction factor a so that the integral can be proportional to the square of the average velocity through the port: where (^V^)p(V ■ n) dA ^ a(^vl)m ^port u dA for incompressible flow If the density is also variable, the integration is very cumbersome; we shall not treat this complication. By letting u be the velocity normal to the port, the first equation above becomes, for incompressible flow. or 2P u^dA = I paVlA (3.74) The term a is the kinetic energy correction factor, having a value of about 2.0 for fully developed laminar pipe flow and from 1 .04 to 1.11 for turbulent pipe flow. The complete incompressible steady flow energy equation (3.73), including pumps, tur¬ bines, and losses, would generalize to P a — -h — P8 2g f P a 2 ^ 1 3"_F-l-z) "1" ^hurbine ^^pump 3" /Zfnction \pg 2g /out (3.75) 186 Chapter 3 Integral Relations for a Control Volume where the head terms on the right (h„ h^, h^) are all numerically positive. All additive terms in Eq. (3.75) have dimensions of length [L], In problems involving turbulent pipe flow, it is common to assume that a ~ 1.0. To compute numerical values, we can use these approximations to be discussed in Chap. 6: Laminar flow: from which = 0.5 C/q and a = 2.0 Turbulent flow: from which, in Example 3.4, 2Uo V = ■ (1 + m){2 + m) Substituting into Eq. (3.74) gives (1 + m)^(2 + m)^ “ ~ 4(1 + 3m) (2 + 3m) and numerical values are as follows: (3.76) (3.77) Turbulent flow: m 1 1 I 1 1 5 6 7 8 9 a 1.106 1.077 1.058 1.046 1.037 These values are only slightly different from unity and are often neglected in ele¬ mentary turbulent flow analyses. However, a should never be neglected in laminar flow. EXAMPLE 3.23 A hydroelectric power plant (Fig. E3.23) takes in 30 mVs of water through its turbine and discharges it to the atmosphere at ¥2 = 2 m/s. The head loss in the turbine and penstock system is hf = 20 m. Assuming turbulent flow, a ~ 1.06, estimate the power in MW extracted by the turbine. Solution We neglect viscous work and heat transfer and take section 1 at the reservoir surface (Fig. E3.23), where Vi ~ Q, pi = Patm. and zi = 100 m. Section 2 is at the turbine outlet. 3.7 The Energy Equation 187 The steady flow energy equation (3.75) hecomes, in head form, Pi OliVi P2 a2V2 r r 7 2g 7 2g ^ ^ ^ 1.06(0)^ 7 2(9.81) 100 m = — + ■ 7 2(9.81 m/s^) ' + 0 m + A, + 20 m The pressure terms cancel, and we may solve for the turbine head (which is positive): h, = 100 - 20 - 0.2 « 79.8 m The turbine extracts about 79.8 percent of the 100-m head available from the dam. The total power extracted may be evaluated from the water mass flow: P = mw,= (pQ)(gh,) = (998kg/m^)(30mVs)(9.81 m/s^)(79.8m) = 23.4 E6 kg • mVs^ = 23.4 E6 N • m/s = 23.4 MW Ans. The turbine drives an electric generator that probably has losses of about 15 percent, so the net power generated by this hydroelectric plant is about 20 MW. EXAMPLE 3.24 The pump in Fig. E3.24 delivers water (62.4 Ibf/fr’) at 1.5 ftVs to a machine at section 2, which is 20 ft higher than the reservoir surface. The losses between 1 and 2 are given by 188 Chapter 3 Integral Relations for a Control Volume hf = KV\l{2g), where K ~ 7.5 is a dimensionless loss coefficient (see Sec. 6.7). Take a ~ 1.07. Find the horsepower required for the pump if it is 80 percent efficient. Solution • System sketch: Figure E3.24 shows the proper selection for sections 1 and 2. • Assumptions: Steady flow, negligible viscous work, large reservoir (Vi ~ 0). • Approach: First find the velocity V2 at the exit, then apply the steady flow energy equation. • Solution steps: Use BG units, pi = 14.7(144) = 2117 Ibf/tf and p, = 10(144) = 1440 Ibf/ftl Find V2 from the known flow rate and the pipe diameter: Q l.SftVs A2 ~ (7r/4)(3/12ft)^ 30.6 ft/s The steady flow energy equation (3.75), with a pump (no turbine) plus ~ 0 and Vi ~ 0, becomes or Pi 7 0(2^2 Z2- hp + hf, hf - K vl Z2 + {0:2 + K) — 2^ V\| 2g ■ Comment: The pump must balance four different effects: the pressure change, the eleva¬ tion change, the exit jet kinetic energy, and the friction losses. ■ Final solution: For the given data, we can evaluate the required pump head: (1440 - 2117)lbf/ft 62.4 Ibf/tf 2 -f 20 + (1.07 -f 7.5) (30.6 ft/s)^ 2(32.2 ft/s^) = -11 -f 20 + 124 = 133 ft With the pump head known, the delivered pump power is computed similar to the turbine in Example 3.23: = mw, = jQlip = ( 62.4 ^ )( 1.5 — 1(133 ft) Ibf = 12450- ft - Ibf fC /V s 12,450 ft-lbf/s 550ft-lbf/(s-hp) = 22.6 hp If the pump is 80 percent efficient, then we divide by the efficiency to find the input power required: Ppump 22.6 hp Rinput = - ^ = 28.3 hp ^ efficiency 0.80 Ans. • Comment: The inclusion of the kinetic energy correction factor a in this case made a difference of about 1 percent in the result. The friction loss, not the exit jet, was the dominant parameter. Problems 189 Summary This chapter has analyzed the four basic equations of fluid mechanics: conservation of (1) mass, (2) linear momentum, (3) angular momentum, and (4) energy. The equa¬ tions were attacked “in the large” — that is, applied to whole regions of a flow. As such, the typical analysis will involve an approximation of the flow field within the region, giving somewhat crude but always instructive quantitative results. However, the basic control volume relations are rigorous and correct and will give exact results if applied to the exact flow field. There are two main points to a control volume analysis. The first is the selection of a proper, clever, workable control volume. There is no substitute for experience, but the following guidelines apply. The control volume should cut through the place where the information or solution is desired. It should cut through places where maximum information is already known. If the momentum equation is to be used, it should not cut through solid walls unless absolutely necessary, since this will expose possible unknown stresses and forces and moments that make the solution for the desired force difficult or impossible. Finally, every attempt should be made to place the control volume in a frame of reference where the flow is steady or quasi-steady, since the steady formulation is much simpler to evaluate. The second main point to a control volume analysis is the reduction of the analy¬ sis to a case that applies to the problem at hand. The 24 examples in this chapter give only an introduction to the search for appropriate simplifying assumptions. You will need to solve 24 or 124 more examples to become truly experienced in simplifying the problem just enough and no more. In the meantime, it would be wise for the beginner to adopt a very general form of the control volume conservation laws and then make a series of simplifications to achieve the final analysis. Starting with the general form, one can ask a series of questions: Is the control volume nondeforming or nonaccelerating? Is the flow field steady? Can we change to a steady flow frame? Can friction be neglected? Is the fluid incompressible? If not, is the perfect-gas law applicable? Are gravity or other body forces negligible? Is there heat transfer, shaft work, or viscous work? Are the inlet and outlet flows approximately one-dimensional? Is atmospheric pressure important to the analysis? Is the pressure hydrostatic on any portions of the control surface? Are there reservoir conditions that change so slowly that the velocity and time rates of change can be neglected? In this way, by approving or rejecting a list of basic simplifications like these, one can avoid pulling Bernoulli’s equation off the shelf when it does not apply. Problems Most of the problems herein are fairly straightforward. More diffi¬ cult or open-ended assignments are labeled with an asterisk. Prob¬ lems labeled with a computer icon may require the use of a computer. The standard end-of-chapter problems P3.1 to P3.185 (categorized in the problem list here) are followed by word prob¬ lems W3.1 to W3.7, fundamentals of engineering (FE) exam prob¬ lems FE3.1 to EE3.10, comprehensive problems C3.1 to C3.5, and design project D3. 1 . 190 Chapter 3 Integral Relations for a Control Volume Problem Distribution Section Topic Problems 3.1 Basic physical laws; volume flow P3.UP3.5 3.2 The Reynolds transport theorem P3.6-P3.9 3.3 Conservation of mass P3.10-P3.38 3.4 The linear momentum equation P3.39-P3.109 3.5 The Bernoulli equation P3.110-P3.148 3.6 The angular momentum theorem P3.149-P3.164 3.7 The energy equation P3.165-P3.185 Basic physical laws; volume flow P3.1 Discuss Newton’s second law (the linear momentum rela¬ tion) in these three forms: = ma dt I \pdr) •^system '' Are they all equally valid? Are they equivalent? Are some forms better for fluid mechanics as opposed to solid mechanics? P3.2 Consider the angular momentum relation in the form dt (r X )pdr system What does r mean in this relation? Is this relation valid in both solid and fluid mechanics? Is it related to the linear momentum equation (Prob. 3.1)? In what manner? P3.3 For steady low-Reynolds-number (laminar) flow through a long tube (see Prob. 1.12), the axial velocity distribution is given by M = C{R^ — r^), where R is the tube radius and r^R. Integrate u(r) to hnd the total volume flow Q through the tube. P3.4 Water at 20°C flows through a long elliptical duct 30 cm wide and 22 cm high. What average velocity, in m/s, would cause the weight flow to be 500 Ibf/s? P3.5 Water at 20°C flows through a 5-in-diameter smooth pipe at a high Reynolds number, for which the velocity profile is approximated by u ~ Ugiy/K)^'^, where Uo is the centerline velocity, R is the pipe radius, and y is the distance mea¬ sured from the wall toward the centerline. If the centerline velocity is 25 ft/s, estimate the volume flow rate in gallons per minute. The Reynolds transport theorem P3.6 Water fills a cylindrical tank to depth h. The tank has diameter D. The water flows out at average velocity V,, from a hole in the bottom of area Ag. Use the Reynolds transport theorem to find an expression for the instanta¬ neous depth change dh/dt. P3.7 A spherical tank, of diameter 35 cm, is leaking air through a 5-mm-diameter hole in its side. The air exits the hole at 360 m/s and a density of 2.5 kg/m^. Assuming uniform mixing, (a) find a formula for the rate of change of average density in the tank and (b) calculate a numerical value for (dpidt) in the tank for the given data. P3.8 Three pipes steadily deliver water at 20°C to a large exit pipe in Fig. P3.8. The velocity 1^2 5 m/s, and the exit flow rate <24 = 120 m^/h. Find (a) Vi, (b) V^, and (c) V4 if it is known that increasing by 20 percent would increase Q4 by 10 percent. P3.9 A laboratory test tank contains seawater of salinity S and density p. Water enters the tank at conditions (5i, pi, Ai, Vi) and is assumed to mix immediately in the tank. Tank water leaves through an outlet A2 at velocity V'2- If salt is a “conservative” property (neither created nor destroyed), use the Reynolds transport theorem to find an expression for the rate of change of salt mass within the tank. Conservation of mass P3.10 Water flowing through an 8-cm-diameter pipe enters a porous section, as in Fig. P3. 10, which allows a uniform radial veloc¬ ity through the wall surfaces for a distance of 1.2 m. If the entrance average velocity Vi is 12 m/s, find the exit velocity V2 if (a) v„, = 15 cm/s out of the pipe walls or (b) v„, = 10 cm/s into the pipe, (c) What value of v„, will make V2 = 9 m/s? _ J Vw ^k2 £) = 8 cm P3.10 P3.ll Water flows from a faucet into a sink at 3 U.S. gallons per minute. The stopper is closed, and the sink has two Problems 191 rectangular overflow drains, each Vg in by 114 in. If the sink water level remains constant, estimate the average over¬ flow velocity, in ft/s. P3.12 The pipe flow in Fig. P3. 12 fills a cylindrical surge tank as shown. At time f = 0, the water depth in the tank is 30 cm. Estimate the time required to fill the remainder of the tank. -• — D = 75 cm — ► V 1 Vj = 2.5 m/s d=12cm V'2 = T9 m/s P3.12 P3.13 The cylindrical container in Fig. P3 . 1 3 is 20 cm in diameter and has a conical contraction at the bottom with an exit hole 3 cm in diameter. The tank contains fresh water at standard sea-level conditions. If the water surface is falling at the nearly steady rate dh/dt ~ —0.072 m/s, estimate the average velocity V out of the bottom exit. (1> 23 = 0.01 mVs 9 V Z)i = 5 cm P3.14 1 Water d - ► 9 D-j = l cm X = 0 x = L u(r) P3.15 D h{t) P3.13 P3.14 The open tank in Fig. P3.14 contains water at 20°C and is being filled through section 1. Assume incompressible flow. First derive an analytic expression for the water-level change dh/dt in terms of arbitrary volume flows (Qi, Q2, Q2) and tank diameter d. Then, if the water level h is con¬ stant, determine the exit velocity V2 for the given data Vi — 3 m/s and = 0.01 mVs. P3.15 Water, assumed incompressible, flows steadily through the round pipe in Fig. P3.15. The entrance velocity is con¬ stant, u = Uq, and the exit velocity approximates turbulent flow, u = Umax(l “ r/R)^''' . Determine the ratio for this flow. P3.16 An incompressible fluid flows past an impermeable flat plate, as in Fig. P3.16, with a uniform inlet profile u = Uq and a cubic polynomial exit profile f3r] - rf\ y u~Uq I - ^ - I where R = Compute the volume flow Q across the top surface of the control volume. P3.16 P3.17 Incompressible steady flow in the inlet between parallel plates in Fig. P3.17 is uniform, u = Uq = i cm/s, while downstream the flow develops into the parabolic laminar pro¬ file u = az/zo — z), where a is a constant. If zq = 4 cm and the fluid is SAE 30 oil at 20°C, what is the value of in cm/s? 192 Chapter 3 Integral Relations for a Control Volume P3.18 Gasoline enters section 1 in Fig. P3. 1 8 at 0.5 mVs. It leaves section 2 at an average velocity of 12 m/s. What is the aver¬ age velocity at section 3? Is it in or out? P3.19 P3.20 P3.21 P3.22 (2) Water from a storm drain flows over an outfall onto a porous hed that absorbs the water at a uniform vertical velocity of 8 mm/s, as shown in Fig. P3.19. The system is 5 m deep into the paper. Find the length L of the bed that will completely absorb the storm water. ^ Initial depth = 20 cm ► 2 m/s Oil (SG = 0.89) enters at section 1 in Fig. P3.20 at a weight flow of 250 N/h to lubricate a thrust bearing. The steady oil flow exits radially through the narrow clearance between thrust plates. Compute (a) the outlet volume flow in mL/s and (b) the average outlet velocity in cm/s. For the two-port tank of Fig. E3.5, assume Di = 4 cm, Fi = 18 m/s, D2 = 7 cm, and V2 = 8 m/s. If the tank surface is rising at 17 mm/s, estimate the tank diameter. The converging-diverging nozzle shown in Fig. P3.22 expands and accelerates dry air to supersonic speeds at the exit, where P2 = 8 kPa and T2 = 240 K. At the throat, pi = 284 kPa, Ti = 665 K, and Vi = 517 m/s. For steady compressible flow of an ideal gas, estimate (a) the mass flow in kg/h, (b) the velocity V2, and (c) the Mach number Ma2. D2 = 2.5 cm P3.22 P3.23 The hypodermic needle in Fig. P3.23 contains a liquid serum (SG = 1.05). If the serum is to be injected steadily at 6 cm Vs, how fast in in/s should the plunger be advanced (a) if leakage in the plunger clearance is neglected and (b) if leakage is 10 percent of the needle flow? Di = 0.75 in P3.24 Water enters the bottom of the cone in Fig. P3.24 at a uniformly increasing average velocity V = Kt. If d is very small, derive an analytic formula for the water surface rise h{t) for the condition /; = 0 at f = 0. Assume incompressible flow. P3.24 Problems 193 P3.25 As will be discussed in Chaps. 7 and 8, the flow of a stream Uq past a blunt flat plate creates a broad low-velocity wake behind the plate. A simple model is given in Fig. P3.25, with only half of the flow shown due to symmetry. The velocity profile behind the plate is idealized as “dead air” (near-zero velocity) behind the plate, plus a higher velocity, decaying vertically above the wake according to the variation u ~ Uo + AU e2^^, where L is the plate height and z = 0 is the top of the wake. Find At/ as a function of stream speed Uq. Uo z Width b into paper Exponential curve 2 Dead air (negligible velocity) -<L- P3.25 P3.26 A thin layer of liquid, draining from an inclined plane, as in Fig. P3.26, will have a laminar velocity profile u ~ U(,{2ylh — y^lh^), where Uo is the surface velocity. If the plane has width b into the paper, determine the volume rate of flow in the film. Suppose that h = 0.5 in and the flow rate per foot of channel width is 1 .25 gal/min. Estimate Uo in ft/s. P3.26 P3.27 Consider a highly pressurized air tank at conditions (po, po. To) and volume Vo- In Chap. 9 we will learn that, if the tank is allowed to exhaust to the atmosphere through a well-designed converging nozzle of exit area A, the outgo¬ ing mass flow rate will be apoA m = — ; - where a ~ 0.685 for air vm This rate persists as long as po is at least twice as large as the atmospheric pressure. Assuming constant To and an ideal gas, (a) derive a formula for the change of density Po(t) within the tank, (b) Analyze the time At required for the density to decrease by 25 percent. P3.28 Air, assumed to be a perfect gas from Table A.4, flows through a long, 2-cm-diameter insulated tube. At section 1, the pressure is 1.1 MPa and the temperature is 345 K. At section 2, 67 meters further downstream, the density is 1.34 kg/m^, the temperature 298 K, and the Mach number is 0.90. For one-dimensional flow, calculate (a) the mass flow; (b) p2, (c) V2\ and id) the change in entropy between 1 and 2. (e) How do you explain the entropy change? P3.29 In elementary compressible flow theory (Chap. 9), com¬ pressed air will exhaust from a small hole in a tank at the mass flow rate m ~ Cp, where p is the air density in the tank and C is a constant. If po is the initial density in a tank of volume T, derive a formula for the density change p{t) after the hole is opened. Apply your formula to the follow¬ ing case: a spherical tank of diameter 50 cm, with initial pressure 300 kPa and temperature 100°C, and a hole whose initial exhaust rate is 0.01 kg/s. Find the time required for the tank density to drop by 50 percent. P3.30 For the nozzle of Fig. P3.22, consider the following data forair,k= 1.4. At the throat, pi = 1000 kPa, Vi = 491 m/s, and Ti = 600 K. At the exit, p2 = 28.14 kPa. Assuming isentropic steady flow, compute (a) the Mach number Maj; (b) T2, (c) the mass flow; and (d) V2- P3.31 A bellows may be modeled as a deforming wedge-shaped volume as in Fig. P3.31. The check valve on the left (pleated) end is closed during the stroke. If b is the bellows width into the paper, derive an expression for outlet mass flow iho as a function of stroke 0{t). P3.31 194 Chapter 3 Integral Relations for a Control Volume P3.32 Water at 20°C flows steadily through the piping junction in Fig. P3.32, entering section 1 at 20 gal/min. The average velocity at section 2 is 2.5 m/s. A portion of the flow is diverted through the showerhead, which contains 100 holes of 1-mm diameter. Assuming uniform shower flow, esti¬ mate the exit velocity from the showerhead jets. P3.35 In contrast to the hquid rocket in Fig. P3.34, the solid-propellant rocket in Fig. P3.35 is self-contained and has no entrance ducts. Using a control volume analysis for the conditions shown in Fig. P3.35, compute the rate of mass loss of the propellant, assuming that the exit gas has a molecular weight of 28. Propellant Trrrrr Combustion: 1500 K, 950 kPa Propellant Exit section = 1 8 cm Pc = 90 kPa Vc = 1150 m/s Tc = 750 K P3.35 P3.33 In some wind tunnels the test section is perforated to suck out fluid and provide a thin viscous boundary layer. The test section wall in Fig. P3.33 contains 1200 holes of 5-mm diameter each per square meter of wall area. The suction velocity through each hole is = 8 m/s, and the test- section entrance velocity is Vi = 35 m/s. Assuming incom¬ pressible steady flow of air at 20°C, compute (a) Vq, (b) V2, and (c) Vf, in m/s. Test section Dj = 0.8 m P3.34 A rocket motor is operating steadily, as shown in Fig. P3.34. The products of combustion flowing out the exhaust nozzle approximate a perfect gas with a molecular weight of 28. For the given conditions calculate V2 in ft/s. P3.36 ThejetpumpinFig.P3.36injectswaterat(/i = 40m/sthrough a 3-in pipe and entrains a secondary flow of water 1/2 = 3 m/s in the annular region around the small pipe. The two flows become fully mixed downstream, where U3 is approximately constant. For steady incompressible flow, compute t/3 in m/s. Mixing Fully P3.37 If the rectangular tank full of water in Fig. P3.37 has its right-hand wall lowered by an amount <5, as shown, water will flow out as it would over a weir or dam. In Prob. PI . 14 we deduced that the outflow Q would be given by Q = where b is the tank width into the paper, g is the accelera¬ tion of gravity, and C is a dimensionless constant. Assume that the water surface is horizontal, not slightly curved as in the figure. Let the initial excess water level be 6o- Derive a formula for the time required to reduce the excess water level to (a) (^o/lO and {b) zero. Problems 195 P3.38 An incompressible fluid in Fig. P3.38 is being squeezed outward between two large circular disks by the uniform downward motion Vq of the upper disk. Assuming one¬ dimensional radial outflow, use the control volume shown to derive an expression for V(r). ■ 1 ^ 1— y(r)? 1 ^ Fixed circular disk P3.38 The linear momentum equation P3.39 A wedge splits a sheet of 20°C water, as shown in Fig. P3.39. Both wedge and sheet are very long into the paper. If the force required to hold the wedge stationary is F = 124 N per meter of depth into the paper, what is the angle 9 of the wedge? P3.40 The water jet in Fig. P3.40 strikes normal to a fixed plate. Neglect gravity and friction, and compute the force F in newtons required to hold the plate fixed. _ Plate P3.41 In Fig. P3.4I the vane turns the water jet completely around. Find an expression for the maximum jet velocity Vq if the maximum possible support force is Fq. P3.41 P3.42 A liquid of density p flows through the sudden contraction in Fig. P3.42 and exits to the atmosphere. Assume uniform conditions (pi, Vj, Dj) at section 1 and (p2, V2, £>2) section 2. Find an expression for the force F exerted by the fluid on the contraction. Atmosphere I © P3.42 © P3.43 Water at 20°C flows through a 5-cm-diameter pipe that has a 180° vertical hend, as in Fig. P3.43. The total length of pipe between flanges 1 and 2 is 75 cm. When the weight flow rate is 230 N/s, pi = 165 kPa and p2 = 134 kPa. Neglecting pipe weight, determine the total force that the flanges must withstand for this flow. © P3.40 P3.44 When a uniform stream flows past an immersed thick cyl¬ inder, a hroad low-velocity wake is created downstream, idealized as a V shape in Fig. P3.44. Pressures pi and p2 are 196 Chapter 3 Integral Relations for a Control Volume approximately equal. If the flow is two-dimensional and incompressible, with width b into the paper, derive a for¬ mula for the drag force F on the cylinder. Rewrite your result in the form of a dimensionless drag coefficient based on body length = F/{p\J^bL). o P3.44 P3.45 Water enters and leaves the 6-cm-diameter pipe bend in Fig. P3.45 at an average velocity of 8.5 m/s. The horizontal force to support the bend against momentum change is 300 N. Find (a) the angle 0; and (b) the vertical force on the bend. P3.46 aQ, V © P3.45 P3.46 When a jet strikes an inclined hxed plate, as in Fig. P3.46, it breaks into two jets at 2 and 3 of equal velocity V = Vj\|.t but unequal flows aQ at 2 and (1 — a)Q at section 3, a being a fraction. The reason is that for frictionless flow the fluid can exert no tangential force F, on the plate. The con¬ dition F, = 0 enables us to solve for a. Perform this analy¬ sis, and find a as a function of the plate angle 9. Why doesn’t the answer depend on the properties of the jet? P3.47 A liquid jet of velocity V, and diameter Dj strikes a fixed hollow cone, as in Fig. P3.47, and deflects back as a conical sheet at the same velocity. Find the cone angle 0 for which the restraining force F = jpAjVf. P3.48 The small boat in Fig. P3.48 is driven at a steady speed Vq by a jet of compressed air issuing from a 3-cm-diameter hole at Vg = 343 m/s. Jet exit conditions arep^ = 1 atm and Tg = 30°C. Air drag is negligible, and the hull drag is kVl, where k~ 19 N ■ sVm^. Estimate the boat speed Vq in m/s. Hull drag kV^ P3.48 P3.49 The horizontal nozzle in Fig. P3.49 has Di = 12 in and D2 = 6 in, with inlet pressure pi = 38 Ibf/in^ absolute and V2 = 56 ft/s. For water at 20°C, compute the horizontal force provided by the flange bolts to hold the nozzle fixed. /},= 15 Ibf/in^abs Problems 197 P3.50 The jet engine on a test stand in Fig. P3.50 admits air at 20°C and 1 atm at section 1, where Ai = 0.5 m^ and Vi = 250 m/s. The fuel-to-air ratio is 1 :30. The air leaves section 2 at atmospheric pressure and higher temperature, where V2 = 900 m/s and A2 = 0.4 m^. Compute the horizontal test stand reaction needed to hold this engine fixed. '-F (b) Turbulent: U2 ~ 1 - R P3.54 For the pipe-flow-reducing section of Fig. P3.54, Di = 8 cm, D2 = 5 cm, and P2= I atm. All fluids are at 20°C. If Fi = 5 m/s and the manometer reading is /i = 58 cm, esti¬ mate the total force resisted hy the flange bolts. P3.50 P3.51 A liquid jet of velocity V, and area Aj strikes a single 180° bucket on a turbine wheel rotating at angular velocity H, as in Fig. P3.51. Derive an expression for the power P deliv¬ ered to this wheel at this instant as a function of the system parameters. At what angular velocity is the maximum power delivered? How would your analysis differ if there were many, many buckets on the wheel, so that the jet was continually striking at least one bucket? ,= 101 kPa P3.55 In Fig. P3.55 the jet strikes a vane that moves to the right at constant velocity I4 on n frictionless cart. Compute (a) the force fj. required to restrain the cart and (b) the power P delivered to the cart. Also find the cart velocity for which (c) the force is a maximum and (d) the power P is a maximum. p,Vj,Aj Vt. = constant P3.52 A large commercial power washer delivers 21 gal/min of water through a nozzle of exit diameter one-third of an inch. Estimate the force of the water jet on a wall normal to the jet. P3.53 Consider incompressible flow in the entrance of a circular tube, as in Fig. P3.53. The inlet flow is uniform, = Uq. The flow at section 2 is developed pipe flow. Find the wall drag force F as a function of (pi, p2, p, Uq, R) if the flow at section 2 is / ..2' (a) Laminar: U2 = P3.55 P3.56 Water at 20°C flows steadily through the box in Fig. P3.56, entering station (1) at 2 m/s. Calculate the (a) horizontal and (b) vertical forces required to hold the box stationary against the flow momentum. P3.57 Water flows through the duct in Fig. P3.57, which is 50 cm wide and 1 m deep into the paper. Gate BC completely closes the duct when /3 = 90°. Assuming one-dimensional flow, for what angle (3 will the force of the exit jet on the plate be 3 kN? 198 Chapter 3 Integral Relations for a Control Volume F=3kN P3.57 P3.58 The water tank in Fig. P3.58 stands on a frictionless cart and feeds a jet of diameter 4 cm and velocity 8 m/s, which is deflected 60° by a vane. Compute the tension in the supporting cable. P3.58 P3.60 Water at 20°C flows through the elbow in Fig. P3.60 and exits to the atmosphere. The pipe diameter is Di = 10 cm, while D2 = 3 cm. At a weight flow rate of 150 N/s, the pressure pi = 2.3 atm (gage). Neglecting the weight of water and elbow, estimate the force on the flange bolts at section 1. © P3.61 A 20°C water jet strikes a vane mounted on a tank with frictionless wheels, as in Fig. P3.61. The jet turns and falls into the tank without spilling out. If 6 = 30°, evaluate the horizontal force F required to hold the tank stationary. P3.59 When a pipe flow suddenly expands from Ai to A2, as in Fig. P3.59, low-speed, low-friction eddies appear in the comers and the flow gradually expands to A2 downstream. Using the suggested control volume for incompressible steady flow and assuming that p ~ pi on the comer annu¬ lar ring as shown, show that the downstream pressure is given by Neglect wall friction. P3.61 Problems 199 P3.62 Water at 20°C exits to the standard sea-level atmosphere through the split nozzle in Fig. P3.62. Duct areas are Ai = 0.02 m^ and A2 = A3 = 0.008 m^. If pi = 135 kPa (abso¬ lute) and the flow rate is ^2 = Gs = 275 mVh, compute the force on the flange bolts at section 1 . P3.63 Water flows steadily through the box in Fig. P3 .63 . Average velocity at all ports is 7 m/s. The vertical momentum force on the box is 36 N. What is the inlet mass flow? P3.64 The 6-cm-diameter 20°C water jet in Fig. P3.64 strikes a plate containing a hole of 4-cm diameter. Part of the jet passes through the hole, and part is deflected. Determine the horizontal force required to hold the plate. P3.65 The box in Fig. P3.65 has three 0.5-in holes on the right side. The volume flows of 20°C water shown are steady, but the details of the interior are not known. Compute the force, if any, that this water flow causes on the box. P3.66 The tank in Fig. P3.66 weighs 500 N empty and contains 600 L of water at 20°C. Pipes 1 and 2 have equal diameters of 6 cm and equal steady volume flows of 300 m^/h. What should the scale reading VP be in N? P3.66 P3.67 For the boundary layer of Fig. 3.10, for air, p = 1.2 kg/m^, let h = 1 cm, Uo = 12 m/s, b = 2m, and L = 1 m. Let the velocity at the exit, x = L, approximate a turbulent flow: ulUo ~ . Calculate (a) S; and (b) the friction drag D. P3.68 The rocket in Fig. P3.68 has a supersonic exhaust, and the exit pressure is not necessarily equal to p^. Show that the force F required to hold this rocket on the test stand is F = + Ag(pg — Pa). Is this force F what we term the thrust of the rocket? Fuel P3.64 P3.68 200 Chapter 3 Integral Relations for a Control Volume P3.69 A uniform rectangular plate, 40 cm long and 30 cm deep into the paper, hangs in air from a hinge at its top (the 30-cm side). It is struck in its center hy a horizontal 3-cm-diameter jet of water moving at 8 m/s. If the gate has a mass of 16 kg, estimate the angle at which the plate will hang from the vertical. P3.70 The dredger in Fig. P3.70 is loading sand (SG = 2.6) onto a harge. The sand leaves the dredger pipe at 4 ft/s with a weight flow of 850 Ihf/s. Estimate the tension on the moor¬ ing line caused by this loading process. P3.71 Suppose that a deflector is deployed at the exit of the jet engine of Prob. P3.50, as shown in Fig. P3.71. What will the reaction on the test stand be now? Is this reac¬ tion sufficient to serve as a braking force during airplane landing? P3.72 P3.73 P3.72 When immersed in a uniform stream, a thick elliptical cylinder creates a broad downstream wake, as idealized in Fig. P3.72. The pressure at the upstream and downstream sections are approximately equal, and the fluid is water at 20°C. If l/o “ 4 m/s and L = 80 cm, estimate the drag force on the cylinder per unit width into the paper. Also compute the dimensionless drag coefficient Co = IFKpU^abL). P3.73 A pump in a tank of water at 20°C directs a jet at 45 ft/s and 200 gal/min against a vane, as shown in Fig. P3.73. Com¬ pute the force F to hold the cart stationary if the jet follows (a) path A or {b) path B. The tank holds 550 gal of water at this instant. P3.74 Water at 20°C flows down through a vertical, 6-cm- diameter tube at 300 gal/min, as in Fig. P3.74. The flow then turns horizontally and exits through a 90° radial duct segment 1 cm thick, as shown. If the radial outflow is uni¬ form and steady, estimate the forces (F„ Fy, F,) required to support this system against fluid momentum changes. P3.75 A jet of liquid of density p and area A strikes a block and splits into two jets, as in Fig. P3.75. Assume the same velocity V for all three jets. The upper jet exits at an angle 0 and area oA. The lower jet is turned 90° downward. Neglecting fluid weight, {a) derive a formula for the forces (F„ Fy) required to support the block against fluid Problems 201 momentum changes, {b) Show that = 0 only if a £ 0.5. (c) Find the values of a and 9 for which both Fj. and Fy are zero. P3.75 P3.76 A two-dimensional sheet of water, 10 cm thick and moving at 7 m/s, strikes a hxed wall inclined at 20° with respect to the jet direction. Assuming frictionless flow, find (a) the normal force on the wall per meter of depth, and find the widths of the sheet deflected (b) upstream and (c) down¬ stream along the wall. P3.77 Water at 20°C flows steadily through a reducing pipe bend, as in Fig. P3.77. Known conditions are pi = 350 kPa, Di = 25 cm, Vi = 2.2 m/s, pi = 120 kPa, and D2 = S cm. Neglecting bend and water weight, estimate the total force that must he resisted by the flange bolts. © P3.78 P3.80 A river of width b and depth hi passes over a submerged obstacle, or “drowned weir,” in Fig. P3.80, emerging at a new flow condition (V2, h2). Neglect atmospheric pressure, and assume that the water pressure is hydrostatic at both sections 1 and 2. Derive an expression for the force exerted by the river on the obstacle in terms of Fi, fii, h2, b, p, and g. Neglect water friction on the river bottom. Width b into paper V2, ^2 P3.80 P3.81 Torricelli’s idealization of efflux from a hole in the side of a tank is F = V2 gh, as shown in Fig. P3.81. The cylindri¬ cal tank weighs 150 N when empty and contains water at 20°C. The tank bottom is on very smooth ice (static friction coefficient ^ ~ 0.01). The hole diameter is 9 cm. For what water depth h will the tank just begin to move to the right? P3.77 P3.78 A fluid jet of diameter Di enters a cascade of moving blades at absolute velocity Fj and angle /3i, and it leaves at absolute velocity F2 and angle /32, as in Fig. P3.78. The blades move at velocity u. Derive a formula for the power P delivered to the blades as a function of these parameters. P3.79 The Saturn V rocket in the chapter opener photo was pow¬ ered by five F-1 engines, each of which burned 3945 Ibm/s of liquid oxygen and 1738 Ibm of kerosene per second. The exit velocity of burned gases was approximately 8500 ft/s. In the spirit of Prob. P3.34, neglecting external pressure P3.82 forces, estimate the total thrust of the rocket, in Ibf. P3.81 Static friction The model car in Fig. P3.82 weighs 17 N and is to be accelerated from rest by a 1-cm-diameter water jet moving 202 Chapter 3 Integral Relations for a Control Volume at 75 m/s. Neglecting air drag and wheel friction, estimate the velocity of the car after it has moved forward 1 m. P3.82 P3.83 Gasoline at 20°C is flowing at Vi = 12 m/s in a 5-cm- diameter pipe when it encounters a 1-m length of uniform radial wall suction. At the end of this suction region, the average fluid velocity has dropped to V2 = 10 m/s. If pi = 120 kPa, estimate p2 if the wall friction losses are neglected. P3.84 Air at 20°C and 1 atm flows in a 25-cm-diameter duct at 15 m/s, as in Fig. P3.84. The exit is choked by a 90° cone, as shown. Estimate the force of the airflow on the cone. 1 cm \ \ - f 40 cm P3.84 P3.85 The thin-plate orifice in Fig. P3.85 causes a large pressure drop. For 20°C water flow at 500 gal/min, with pipe D = 10 cm and orifice d = 6 cm, pi — P2~ 145 kPa. If the wall friction is negligible, estimate the force of the water on the orifice plate. P3.86 For the water jet pump of Prob. P3.36, add the following data: Pi = P2 = 25 Ibf/in^, and the distance between sec¬ tions 1 and 3 is 80 in. If the average wall shear stress between sections 1 and 3 is 7 Ibf/ft^, estimate the pressure P3. Why is it higher than pi? P3.87 A vane turns a water jet through an angle a, as shown in Fig. P3.87. Neglect friction on the vane walls, (a) What is the angle a for the support force to be in pure compression? (b) Calculate this compression force if the water velocity is 22 ft/s and the jet cross section is 4 in^. P3.88 The boat in Fig. P3.88 is jet-propelled by a pump that develops a volume flow rate Q and ejects water out the stem at velocity V,. If the boat drag force is F = kV^, where k is a constant, develop a formula for the steady forward speed V of the boat. P3.88 P3.89 Consider Fig. P3.36 as a general problem for analysis of a mixing ejector pump. If all conditions (p, p, V) are known at sections 1 and 2 and if the wall friction is negligible, derive formulas for estimating (a) V3 and (jb) p^. P3.90 As shown in Fig. P3.90, a liquid column of height h is confined in a vertical tube of cross-sectional area A by a stopper. At f = 0 the stopper is suddenly removed, expos¬ ing the bottom of the liquid to atmospheric pressure. Using a control volume analysis of mass and vertical momentum, derive the differential equation for the downward motion V(t) of the liquid. Assume one-dimensional, incompress¬ ible, frictionless flow. P3.91 Extend Prob. P3.90 to include a linear (laminar) average wall shear stress resistance of the form r ~ cV, where c is a constant. Eind the differential equation for dV/dt and then solve for V(t), assuming for simplicity that the wall area remains constant. Problems 203 P3.90 Stopper P3.92 A more involved version of Prob. P3.90 is the elbow- shaped tube in Fig. P3.92, with constant cross-sectional area A and diameter D <S h, L. Assume incompressible flow, neglect friction, and derive a differential equation for dV/dt when the stopper is opened. Hint: Combine two control volumes, one for each leg of the tube. P3.92 P3.93 According to Torricelli’s theorem, the velocity of a fluid draining from a hole in a tank is F « (Tg/t)'®, where h is the depth of water above the hole, as in Fig. P3.93. Let the hole have area A„ and the cylindrical tank have cross-section area Aj, 3> A„. Derive a formula for the time to drain the tank completely from an initial depth h„. P3.94 A water jet 3 in in diameter strikes a concrete (SG = 2.3) slab which rests freely on a level floor. If the slab is 1 ft wide into the paper, calculate the jet velocity which will just begin to tip the slab over. P3.95 A tall water tank discharges through a well-rounded orifice, as in Fig. P3.95. Use the Torricelli formula of Prob. P3.81 to estimate the exit velocity, (a) If, at this instant, the force F required to hold the plate is 40 N, what is the depth h? (b) If the tank surface is dropping at the rate of 2.5 cm/s, what is the tank diameter D? P3.95 P3.96 Extend Prob. P3.90 to the case of the liquid motion in a frictionless U-tube whose liquid column is displaced a dis¬ tance Z upward and then released, as in Fig. P3.96. — h 1 F t J V — h2~0 - z h — Equilibrium position Liquid -column length Z, = + h2 + P3.96 204 Chapter 3 Integral Relations for a Control Volume Neglect the short horizontal leg, and combine control volume analyses for the left and right legs to derive a single differential equation for V{t) of the liquid column. P3.97 Extend Proh. P3.96 to include a linear (laminar) average wall shear stress resistance of the form T = SfiV/D, where fi is the fluid viscosity. Find the differential equation for dV/dt and then solve for V(l), assuming an initial displace¬ ment z — Za, V = Q t = 0. The result should be a damped oscillation tending toward z = 0. P3.98 As an extension of Example 3.9, let the plate and its cart (see Fig. 3.9fl) be unrestrained horizontally, with friction¬ less wheels. Derive (a) the equation of motion for cart velocity Vc(t) and (h) a formula for the time required for the cart to accelerate from rest to 90 percent of the jet velocity (assuming the jet continues to strike the plate horizontally), (c) Compute numerical values for part {b) using the condi¬ tions of Example 3.9 and a cart mass of 2 kg. P3.99 Let the rocket of Fig. E3.12 start at z = 0, with constant exit velocity and exit mass flow, and rise vertically with zero drag, (a) Show that, as long as fuel burning continues, the vertical height S(t) reached is given by S = — ^ — fOnC ~ C + 1]^ where C = 1 - m (b) Apply this to the case V,, = 1500 m/s andM„ = 1000 kg to find the height reached after a bum of 30 seconds, when the final rocket mass is 400 kg. P3.100 Suppose that the solid-propellant rocket of Proh. P3.35 is built into a missile of diameter 70 cm and length 4 m. The system weighs 1800 N, which includes 700 N of propel¬ lant. Neglect air drag. If the missile is fired vertically from rest at sea level, estimate (a) its velocity and height at fuel burnout and (b) the maximum height it will attain. P3.101 Water at 20°C flows steadily through the tank in Fig. P3.101. Known conditions are Di = 8 cm, Pi = 6 m/s, and D2 = 4 cm. A rightward force F = 70 N is required to keep the tank fixed, (a) What is the velocity leaving section 2? (b) If the tank cross section is 1.2 m^, how fast is the water surface h{t) rising or falling? 9 9 h 't) — — F - ► P3.102 As can often be seen in a kitchen sink when the faucet is running, a high-speed channel flow (Pi, hi) may “jump” to a low-speed, low-energy condition (P2, ^2) in Fig. P3. 102. The pressure at sections 1 and 2 is approximately hydro¬ static, and wall friction is negligible. Use the continuity and momentum relations to find /t2 and P2 in terms of (hi. Pi). Hydraulic jump ' P3.102 P3.103 Suppose that the solid-propellant rocket of Proh. P3.35 is mounted on a 1000-kg car to propel it up a long slope of 15°. The rocket motor weighs 900 N, which includes 500 N of propellant. If the car starts from rest when the rocket is fired, and if air drag and wheel friction are neglected, estimate the maximum distance that the car will travel up the hill. P3.104 A rocket is attached to a rigid horizontal rod hinged at the origin as in Fig. P3.104. Its initial mass is Mq, and its exit properties are m and P^ relative to the rocket. Set up the differential equation for rocket motion, and solve for the angular velocity iij(t) of the rod. Neglect gravity, air drag, and the rod mass. X P3.104 P3.105 Extend Proh. P3.104 to the case where the rocket has a linear air drag force F = cV, where c is a constant. Assum¬ ing no burnout, solve for Ld(f) and find the terminal angular velocity — that is, the final motion when the angular accel¬ eration is zero. Apply to the case Mq = 6 kg, /? = 3 m, m = 0.05 kg/s, Pe = 1 100 m/s, and c = 0.075 N • s/m to find the angular velocity after 12 s of burning. P3.106 Actual airflow past a parachute creates a variable distribu¬ tion of velocities and directions. Let us model this as a circular air jet, of diameter half the parachute diameter, which is turned completely around by the parachute, as in Fig. P3.106. (a) Find the force F required to support Problems 205 the chute, (b) Express this force as a dimensionless drag coefficient, Q, = F/[(!/2)pV'^(7r/4)D‘] and compare with Table 7.3. L P^V — r D/2 P3.106 P3.107 The cart in Fig. P3.107 moves at constant velocity Vq = 12 m/s and takes on water with a scoop 80 cm wide that dips h = 2.5 cm into a pond. Neglect air drag and wheel friction. Estimate the force required to keep the cart moving. P3.107 The Bernoulli Equation P3.110 Repeat Prob. P3.49 by assuming that pi is unknown and using Bernoulli’s equation with no losses. Compute the new bolt force for this assumption. What is the head loss between 1 and 2 for the data of Prob. P3.49? P3.111 As a simpler approach to Prob. P3.96, apply the unsteady Bernoulli equation between 1 and 2 to derive a differential equation for the motion z(t). Neglect friction and com¬ pressibility. P3.112 A jet of alcohol strikes the vertical plate in Fig. P3.112. A force F ~ 425 N is required to hold the plate stationary. Assuming there are no losses in the nozzle, estimate {a) the mass flow rate of alcohol and fb) the absolute pressure at section 1. P3.108 A rocket sled of mass M is to be decelerated by a scoop, as in Fig. P3.108, which has width b into the paper and dips into the water a depth h, creating an upward jet at 60°. The rocket thrust is T to the left. Let the initial velocity be Vq, and neglect air drag and wheel friction. Find an expression for V{t) of the sled for (a) T = 0 and (b) finite T ^ 0. 60° P3.109 For the boundary layer flow in Fig. 3.10, let the exit velocity profile, 3tx = L, simulate turbulent flow, u ~ Uo{y/S)^''’. (a) Find a relation between h and 6. (b) Find an expression for the drag force F on the plate between 0 and L. P3.113 An airplane is flying at 300 mi/h at 4000 m standard altitude. As is typical, the air velocity relative to the upper surface of the wing, near its maximum thickness, is 26 per¬ cent higher than the plane’s velocity. Using Bernoulli’s equation, calculate the absolute pressure at this point on the wing. Neglect elevation changes and compressibility. P3.114 Water flows through a circular nozzle, exits into the air as a jet, and strikes a plate, as shown in Fig. P3. 1 14. The force required to hold the plate steady is 70 N. Assuming steady, frictionless, one-dimensional flow, estimate {a) the velocities at sections (1) and (2) and fb) the mercury manometer reading h. P3.114 V. 206 Chapter 3 Integral Relations for a Control Volume P3.115 A free liquid jet, as in Fig. P3.115, has constant ambient pressure and small losses; hence from Bernoulli’s equation z + V^l{2g) is constant along the jet. For the fire nozzle in the figure, what are (a) the minimum and (b) the maximum values of 0 for which the water jet will clear the comer of the building? For which case will the jet velocity be higher when it strikes the roof of the building? P3.116 For the container of Fig. P3. 116 use Bernoulli’s equation to derive a formula for the distance X where the free jet leav¬ ing horizontally will strike the floor, as a function of h and H. For what ratio hlH will X be maximum? Sketch the three trajectories for hlH = 0.25, 0.5, and 0.75. P3.116 P3.117 Water at 20°C, in the pressurized tank of Fig. P3. 1 17, flows out and creates a vertical jet as shown. Assuming steady frictionless flow, determine the height H to which the jet rises. m ,t^ Air 75 kPa (gage) Water 85 cm P3.118 Bernoulli’s 1738 treatise Hydrodynamica contains many excellent sketches of flow patterns related to his friction¬ less relation. One, however, redrawn here as Fig. P3.118, seems physically misleading. Can you explain what might be wrong with the figure? P3.118 Jet Jet P3.119 A long fixed tube with a rounded nose, aligned with an oncoming flow, can be used to measure velocity. Measure¬ ments are made of the pressure at (1) the front nose and (2) a hole in the side of the tube further along, where the pressure nearly equals stream pressure. (a) Make a sketch of this device and show how the velocity is calculated, (b) For a particular sea-level airflow, the dif¬ ference between nose pressure and side pressure is 1 .5 Ibf/in^. What is the air velocity, in mi/h? P3.120 The manometer fluid in Fig. P3.120 is mercury. Estimate the volume flow in the tube if the flowing fluid is (a) gaso¬ line and (b) nitrogen, at 20°C and 1 atm. P3.121 In Fig. P3. 121 the flowing fluid is CO2 at 20°C. Neglect losses. If Pi = 170 kPa and the manometer fluid is Meriam red oil (SG = 0.827), estimate (a) p2 and (b) the gas flow rate in m^/h. P3.122 The cylindrical water tank in Fig. P3.122 is being filled at a volume flow 2i = 1-0 gal/min, while the water also drains from a bottom hole of diameter d = 6 mm. At time f = 0, /t = 0. Find and plot the variation h{t) and the even¬ tual maximum water depth Assume that Bernoulli’s steady-flow equation is valid. P3.117 Problems 207 P3.122 \v2 Diameter D = 20 cm h P3.123 The air-cushion vehicle in Fig. P3.123 brings in sea-level standard air through a fan and discharges it at high velocity through an annular skirt of 3-cm clearance. If the vehicle weighs 50 kN, estimate (a) the required airflow rate and (b) the fan power in kW. P3.124 A necked-down section in a pipe flow, called a venturi, develops a low throat pressure that can aspirate fluid upward from a reservoir, as in Fig. P3.124. Using Bernoulli’s equation with no losses, derive an expression for the velocity Vi that is just sufficient to bring reservoir fluid into the throat. P3.124 P3.125 Suppose you are designing an air hockey table. The table is 3.0 X 6.0 ft in area, with ^-in-diameter holes spaced every inch in a rectangular grid pattern (2592 holes total). The required jet speed from each hole is estimated to be 50 ft/s. Your job is to select an appropriate blower that will meet the requirements. Estimate the volumetric flow rate (in ft^/min) and pressure rise (in Ih/in^) required of the hlower. Hint: Assume that the air is stagnant in the large volume of the manifold under the table surface, and neglect any frictional losses. P3.126 The liquid in Fig. P3.126 is kerosene at 20°C. Estimate the flow rate from the tank for (a) no losses and (b) pipe losses /i^«4.5FV(2g). P3.126 Air: p = 20 Ibf/in^ abs Y 5 ft p = 14.7 Ibf/in^ abs ‘ a D = 1 in y P3.127 In Fig. P3.I27 the open jet of water at 20°C exits a nozzle into sea-level air and strikes a stagnation tube as shown. 208 Chapter 3 Integral Relations for a Control Volume If the pressure at the centerline at section 1 is 1 10 kPa, and losses are neglected, estimate (a) the mass flow in kg/s and {b) the height H of the fluid in the stagnation tube. P3.128 A venturi meter, shown in Fig. P3.128, is a carefully designed constriction whose pressure difference is a mea¬ sure of the flow rate in a pipe. Using Bernoulli’s equation for steady incompressible flow with no losses, show that the flow rate Q is related to the manometer reading h by ^ ^ Ai l^ghjpM - p) Vl - (D2/Di)W P where Puf is the density of the manometer fluid. P3.129 A water stream flows past a small circular cylinder at 23 ft/s, approaching the cylinder at 3000 Ibf/ft^. Measure¬ ments at low (laminar flow) Reynolds numbers indicate a maximum surface velocity 60 percent higher than the stream velocity at point B on the cylinder. Estimate the pressure at B. P3.130 In Fig. P3.130 the fluid is gasoline at 20°C at a weight flow of 120 N/s. Assuming no losses, estimate the gage pressure at section 1 . 5 cm P3.131 In Fig. P3.131 both fluids are at 20^. If Vi = 1.7 ft/s and losses are neglected, what should the manometer reading h ft be? P3.132 Extend the siphon analysis of Example 3. 14 to account for friction in the tube, as follows. Let the friction head loss in the tube be correlated as 5.4(V,,,\|,e)^/(2g), which approxi¬ mates turbulent flow in a 2-m-long tube. Calculate the exit velocity in m/s and the volume flow rate in cmVs, and com¬ pare to Example 3.14. P3.133 If losses are neglected in Fig. P3.133, for what water level h will the flow begin to form vapor cavities at the throat of the nozzle? P3.133 Open jet P3.134 For the 40°C water flow in Fig. P3. 134, estimate the volume flow through the pipe, assuming no losses; then explain what is wrong with this seemingly innocent question. If the actual flow rate is 2 = 40 m^/h, compute (a) the head loss in ft and (b) the constriction diameter D that causes cavitation, assuming that the throat divides the head loss equally and that changing the constriction causes no addi¬ tional losses. Problems 209 P3.135 The 35°C water flow of Fig. P3.135 discharges to sea-level standard atmosphere. Neglecting losses, for what nozzle diameter D will cavitation begin to occur? To avoid cavitation, should you increase or decrease D from this critical value? V P3.135 P3.136 Air, assumed frictionless, flows through a tube, exiting to sea-level atmosphere. Diameters at 1 and 3 are 5 cm, while D2 = 3 cm. What mass flow of air is required to suck water up 10 cm into section 2 of Fig. P3.136? P3.137 In Fig. P3.137 the piston drives water at 20°C. Neglecting losses, estimate the exit velocity V2 ft/s. If D2 is further constricted, what is the limiting possible value of F2? P3.138 For the sluice gate flow of Example 3.10, use Bernoulli’s equation, along the surface, to estimate the flow rate 2 as a function of the two water depths. Assume constant width b. P3.139 In the spillway flow of Fig. P3.139, the flow is assumed uniform and hydrostatic at sections 1 and 2. If losses are neglected, compute (a) V2 (b) the force per unit width of the water on the spillway. V P3.140 For the water channel flow of Fig. P3.140, hi = 1.5 m, ^5. // = 4 m, and Vi = 3 m/s. Neglecting losses and assuming uniform flow at sections 1 and 2, find the downstream depth /t2, and show that two realistic solutions are possible. Jv ^2 J^V2 P3.140 P3.141 For the water channel flow of Fig. P3.141, hi = 0.45 ft, ^p. H = 2.2 ft, and Fi = 16 ft/s. Neglecting losses and assum¬ ing uniform flow at sections 1 and 2, find the downstream depth h2; show that two realistic solutions are possible. 210 Chapter 3 Integral Relations for a Control Volume hi 1 ^^2 P3.142 A cylindrical tank of diameter D contains liquid to an ini¬ tial height Hq. At time t = Oa small stopper of diameter d is removed from the bottom. Using Bernoulli’s equation with no losses, derive (a) a differential equation for the free- surface height h(t) during draining and (b) an expression for the time 1q to drain the entire tank. P3.143 The large tank of incompressible liquid in Fig. P3.143 is at rest when, at f = 0, the valve is opened to the atmosphere. Assuming h ~ constant (negligible velocities and accelera¬ tions in the tank), use the unsteady frictionless Bernoulli equation to derive and solve a differential equation for V(t) in the pipe. r - L P3.143 P3.144 A fire hose, with a 2-in-diameter nozzle, delivers a water jet straight up against a ceiling 8 ft higher. The force on the ceiling, due to momentum change, is 25 Ibf. Use Bernoulli’s equation to estimate the hose flow rate, in gal/min. [Hint: The water jet area expands upward.] P3.145 The incompressible flow form of Bernoulli’s relation, Eq. (3.54), is accurate only for Mach numbers less than about 0.3. At higher speeds, variable density must be accounted for. The most common assumption for compressible fluids is isentropic flow of an ideal gas, or p = Cp'‘, where k = Cp/Cjj. Substitute this relation into Eq. (3.52), integrate, and eliminate the constant C. Compare your compressible result with Eq. (3.54) and comment. P3.146 The pump in Fig. P3.146 draws gasoline at 20°C from a reservoir. Pumps are in big trouble if the liquid vaporizes (cavitates) before it enters the pump, (a) Neglecting losses and assuming a flow rate of 65 gal/min, find the limitations on (x, y, z) for avoiding cavitation, (b) If pipe friction losses are included, what additional limitations might be important? P3.147 The very large water tank in Fig. P3.147 is discharging through a 4-in-diameter pipe. The pump is running, with a performance curve hp ~ 40 — 4 Q^, with hp in feet and Q in ft^/s. Estimate the discharge flow rate in ftVs if the pipe friction loss is 1.5(V^/2g). P3.148 By neglecting friction, (a) use the Bernoulli equation between surfaces 1 and 2 to estimate the volume flow through the orihce, whose diameter is 3 cm. (b) Why is the result to part (a) absurd? (c) Suggest a way to resolve this paradox and find the true flow rate. Problems 211 The angular momentum theorem P3.149 The horizontal lawn sprinkler in Fig. P3.149 has a water flow rate of 4.0 gal/min introduced vertically through the center. Estimate (a) the retarding torque required to keep the arms from rotating and (b) the rotation rate (r/min) if there is no retarding torque. P3.150 In Proh. P3.60 find the torque caused around flange 1 if the center point of exit 2 is 1.2 m directly helow the flange center. P3.151 The wye joint in Fig. P3.151 splits the pipe flow into equal amounts QI2, which exit, as shown, a distance from the axis. Neglect gravity and friction. Find an expression for the torque T about the x axis required to keep the system rotating at angular velocity fl. Q P3.151 P3.152 Modify Example 3.19 so that the arm starts from rest and spins up to its final rotation speed. The moment of inertia of the arm about O is Iq. Neglecting air drag, find dtoldt and integrate to determine the angular velocity LJ{f), assuming a; = 0 at t = 0. P3.153 The three-arm lawn sprinkler of Fig. P3.153 receives 20°C water through the center at 2.7 m^/h. If collar friction is negligible, what is the steady rotation rate in r/min for (fl) 0 = 0° and (b) 0 = 40°? P3.154 Water at 20°C flows at 30 gal/min through the 0.75-in¬ diameter double pipe bend of Fig. P3.154. The pressures are pi = 30 Ibf/in^ and p2 = 24 Ibf/in^. Compute the torque T at point B necessary to keep the pipe from rotating. B exits the impeller at an angle 02 relative to the blades, as shown. The fluid enters axially at section 1. Assuming incompressible flow at shaft angular velocity iO, derive a formula for the power P required to drive the impeller. 212 Chapter 3 Integral Relations for a Control Volume P3.156 A simple turbomachine is constructed from a disk with two internal ducts that exit tangentially through square holes, as in Fig. P3.156. Water at 20°C enters normal to the disk at the center, as shown. The disk must drive, at 250 r/min, a small device whose retarding torque is 1.5 N ■ m. What is the proper mass flow of water, in kg/s? P3.161 Extend Prob. P3.46 to the problem of computing the center of pressure L of the normal face F„, as in Fig. P3.161. (At the center of pressure, no moments are required to hold the plate at rest.) Neglect friction. Express your result in terms of the sheet thickness hi and the angle 9 between the plate and the oncoming jet 1. 2 cm i 2 cm t P3.156 P3.157 Reverse the flow in Fig. P3. 155, so that the system operates as a radial-inflow turbine. Assuming that the outflow into section 1 has no tangential velocity, derive an expression for the power P extracted by the turbine. P3.158 Revisit the turbine cascade system of Prob. P3.78, and derive a formula for the power P delivered, using the angular momentum theorem of Eq. (3.59). P3.159 A centrifugal pump impeller delivers 4000 gal/min of water at 20°C with a shaft rotation rate of 1750 r/min. Neglect losses. If ri = 6 in, r2 = 14 in, bi = b2 = 1.75 in, V,i = 10 ft/s, and Vi2 =110 ft/s, compute the absolute velocities (a) Vi and (b) V2 and (c) the horsepower required. (d) Compare with the ideal horsepower required. P3.160 The pipe bend of Fig. P3.160 has Di = 27 cm and D2 = 13 cm. When water at 20°C flows through the pipe at 4000 gal/min. Pi = 194 kPa (gage). Compute the torque required at point B to hold the bend stationary. P3.161 P3.162 The waterwheel in Fig. P3.162 is being driven at 200 r/min by a 150-ft/s jet of water at 20°C. The jet diameter is 2.5 in. Assuming no losses, what is the horsepower developed by the wheel? For what speed Cl r/min will the horsepower developed be a maximum? Assume that there are many buckets on the waterwheel. P3.163 A rotating dishwasher arm delivers at 60°C to six nozzles, as in Fig. P3.163. The total flow rate is 3.0 gal/min. Each nozzle has a diameter of ^ in. If the nozzle flows are equal and friction is neglected, estimate the steady rotation rate of the arm, in r/min. Problems 213 P3.163 P3.164 A liquid of density p flows through a 90° bend as shown in Fig. P3.164 and issues vertically from a uniformly porous section of length L. Neglecting pipe and liquid weight, derive an expression for the torque M at point 0 required to hold the pipe stationary. y." - « - ► A 0 7 ^ ^ Closed valve ■« - d«R, L upstream are 2, = 2.5 mVs and T, = 18°C. The river is 45 m wide and 2.7 m deep. If heat losses to the atmosphere and ground are negligible, estimate the downstream river conditions {Qq, Tq). Q P3.164 The energy equation P3.165 There is a steady isothermal flow of water at 20°C through the device in Fig. P3.165. Heat-transfer, gravity, and tem¬ perature effects are negligible. Known data are Di = 9 cm, 2i = 220 m^/h,pi = 150 kPa, 0^ = 1 cm, gj = 100 m^/h, P2 = 225 kPa, D3 = 4 cm, and p3 = 265 kPa. Compute the rate of shaft work done for this device and its direction. P3.166 A power plant on a river, as in Fig. P3. 166, must eliminate 55 MW of waste heat to the river. The river conditions P3.167 For the conditions of Prob. P3.166, if the power plant is to heat the nearby river water by no more than 12°C, what should be the minimum flow rate Q, in m^/s, through the plant heat exchanger? How will the value of Q affect the downstream conditions (Qq, Tq)7 P3.168 Multnomah Falls in the Columbia River Gorge has a sheer drop of 543 ft. Using the steady flow energy equation, estimate the water temperature change in °F caused by this drop. P3.169 When the pump in Fig. P3.169 draws 220 m^/h of water at 20°C from the reservoir, the total friction head loss is 5 m. The flow discharges through a nozzle to the atmosphere. Estimate the pump power in kW delivered to the water. P3.169 214 Chapter 3 Integral Relations for a Control Volume P3.170 A steam turbine operates steadily under the following conditions. At the inlet, p = 2.5 MPa, T = 450°C, and V = 40 m/s. At the outlet, p = 22 kPa, T = 70°C, and V = 225 m/s. (a) If we neglect elevation changes and heat trans¬ fer, how much work is delivered to the turbine blades, in kj/kg? {b) If the mass flow is 10 kg/s, how much total power is delivered? (c) Is the steam wet as it leaves the exit? P3.171 Consider a turbine extracting energy from a penstock in a dam, as in Fig. P3.171. For turbulent pipe flow (Chap. 6), the friction head loss is approximately hf = CQ^, where the constant C depends on penstock dimensions and the prop¬ erties of water. Show that, for a given penstock geometry and variable river flow Q, the maximum turbine power pos¬ sible in this case is = 2pgHQ/3 and occurs when the flow rate is 2 = P3.172 The long pipe in Fig. P3.172 is filled with water at 20°C. When valve A is closed, P\ — P2 = 25 kPa. When the valve is open and water flows at 500 m^/h, Pi — po = 160 kPa. What is the friction head loss between 1 and 2, in m, for the flowing condition? P3.173 A 36-in-diameter pipeline carries oil (SG = 0.89) at 1 million barrels per day (bbl/day) (1 bbl = 42 U.S. gal). The friction head loss is 13 ft/1000 ft of pipe. It is planned to place pumping stations every 10 mi along the pipe. Estimate the horsepower that must be delivered to the oil by each pump. P3.174 The pump-turbine system in Fig. P3.174 draws water from the upper reservoir in the daytime to produce power for a city. At night, it pumps water from lower to upper reser¬ voirs to restore the situation. For a design flow rate of 15,000 gal/min in either direction, the friction head loss is 17 ft. Estimate the power in kW (a) extracted by the turbine and (b) delivered by the pump. P3.175 Water at 20°C is delivered from one reservoir to another through a long 8-cm-diameter pipe. The lower reservoir has a surface elevation Z2 = 80 m. The friction loss in the pipe is correlated by the formula ~ 17.5(V'V2g), where V is the average velocity in the pipe. If the steady flow rate through the pipe is 500 gallons per minute, estimate the surface elevation of the higher reservoir. P3.176 A fireboat draws seawater (SG = 1.025) from a submerged pipe and discharges it through a nozzle, as in Fig. P3.176. The total head loss is 6.5 ft. If the pump efficiency is 75 percent, what horsepower motor is required to drive it? Pump P3.177 A device for measuring liquid viscosity is shown in Fig. P3.177. With the parameters (p, L, H, d) known, the flow rate Q is measured and the viscosity calculated, assuming a laminar-flow pipe loss from Chap. 6, hf = (32p,LV)/(pgd^). Heat transfer and all other losses are negligible, (a) Derive a formula for the viscosity p of the fluid, (b) Calculate p for the case d = 2 mm, p = 800 kg/m^, L = 95 cm, H = 30 cm. Problems 215 and Q = 760 cm^/h. (c) What is your guess of the fluid in part (b)! (d) Verify that the Reynolds number Re^ is less than 2000 (laminar pipe flow). Water level H V P3.181 A typical pump has a head that, for a given shaft rotation rate, varies with the flow rate, resulting in a pump perfor¬ mance curve as in Fig. P3.181. Suppose that this pump is 75 percent efficient and is used for the system in Prob. 3.180. Estimate (a) the flow rate, in gal/min, and (b) the horsepower needed to drive the pump. L P3.177 e P3.178 The horizontal pump in Fig. P3.178 discharges 20°C water at 57 mVh. Neglecting losses, what power in kW is deliv¬ ered to the water by the pump? 300 200 100 0 0 12 3 4 Flow rate, ft^/s Pump performance P3.181 P3.179 Steam enters a horizontal turbine at 350 Ibf/in^ absolute, 580°C, and 12 ft/s and is discharged at 110 ft/s and 25°C saturated conditions. The mass flow is 2.5 Ibm/s, and the heat losses are 7 Btu/lb of steam. If head losses are negli¬ gible, how much horsepower does the turbine develop? P3.180 Water at 20°C is pumped at 1500 gal/min from the lower to the upper reservoir, as in Fig. P3.180. Pipe friction losses are approximated by hf ~ 27V^/(2g), where V is the average velocity in the pipe. If the pump is 75 percent efficient, what horsepower is needed to drive it? P3.182 The insulated tank in Fig. P3. 1 82 is to be filled from a high- pressure air supply. Initial conditions in the tank are T = 20°C and p = 200 kPa. When the valve is opened, the ini¬ tial mass flow rate into the tank is 0.013 kg/s. Assuming an ideal gas, estimate the initial rate of temperature rise of the air in the tank. P3.183 The pump in Fig. P3.183 creates a 20°C water jet oriented to travel a maximum horizontal distance. System friction head losses are 6.5 m. The jet may be approximated by the trajectory of frictionless particles. What power must be delivered by the pump? 216 Chapter 3 Integral Relations for a Control Volume P3.184 The large turbine in Fig. P3 . 1 84 diverts the river flow under a dam as shown. System friction losses are hf = 3.5V^/{2g), where V is the average velocity in the supply pipe. For what river flow rate in mVs will the power extracted he 25 MW? Which of the two possible solutions has a better “conversion efficiency”? P3.185 Kerosine at 20°C flows through the pump in Fig. P3.185 at 2.3 ft^/s. Head losses between 1 and 2 are 8 ft, and the pump delivers 8 hp to the flow. What should the mercury manometer reading h ft be? £>2 = 6 in Word Problems W3.1 Derive a control volume form of the second law of thermo¬ dynamics. Suggest some practical uses for your relation in analyzing real fluid flows. W3.2 Suppose that it is desired to estimate volume flow 2 in a pipe by measuring the axial velocity u{r) at specific points. For cost reasons only three measuring points are to be used. What are the best radii selections for these three points? W3.3 Consider water flowing by gravity through a short pipe connecting two reservoirs whose surface levels differ by an amount Az. Why does the incompressible frictionless Ber¬ noulli equation lead to an absurdity when the flow rate through the pipe is computed? Does the paradox have something to do with the length of the short pipe? Does the paradox disappear if we round the entrance and exit edges of the pipe? W3.4 Use the steady flow energy equation to analyze flow through a water faucet whose supply pressure is po. What physical mechanism causes the flow to vary continuously from zero to maximum as we open the faucet valve? W3.5 Consider a long sewer pipe, half full of water, sloping downward at angle 6. Antoine Chezy in 1768 determined that the average velocity of such an open channel flow should be F « CVR tan 0, where R is the pipe radius and C is a constant. How does this famous formula relate to the steady flow energy equation applied to a length L of the channel? W3.6 Put a table tennis ball in a funnel, and attach the small end of the funnel to an air supply. You probably won’t be able to blow the ball either up or down out of the funnel. Explain why. W3.7 How does a siphon work? Are there any limitations (such as how high or how low can you siphon water away from a tank)? Also, how far — could you use a flexible tube to siphon water from a tank to a point 100 ft away? Fundamentals of Engineering Exam Problems 217 Fundamentals of Engineering Exam Problems FE3.1 In Eig. EE3.1 water exits from a nozzle into atmospheric pressure of 101 kPa. If the flow rate is 160 gal/min, what is the average velocity at section 1 ? (a) 2.6 m/s, (b) 0.81 m/s, (c) 93 m/s, (d) 23 m/s, (e) 1.62 m/s FE3.2 In Fig. FE3.1 water exits from a nozzle into atmospheric pressure of 101 kPa. If the flow rate is 160 gal/min and fric¬ tion is neglected, what is the gage pressure at section 1? (a) 1.4 kPa, (b) 32 kPa, (c) 43 kPa, (d) 29 kPa, (e) 123 kPa EE3.3 In Fig. FE3.1 water exits from a nozzle into atmospheric pressure of 101 kPa. If the exit velocity is 1^2 = 8 m/s and friction is neglected, what is the axial flange force required to keep the nozzle attached to pipe 1 ? (a) 1 1 N, (b) 56 N, (c) 83 N, (d) 123 N, (e) 1 10 N EE3.4 In Fig. FE3.1 water exits from a nozzle into atmospheric pressure of 101 kPa. If the manometer fluid has a specific gravity of 1.6 and h = 66 cm, with friction neglected, what is the average velocity at section 2? (a) 4.55 m/s, (b) 2.4 m/s, (c) 2.95 m/s, (d) 5.55 m/s, (e) 3.4 m/s EE3.5 A jet of water 3 cm in diameter strikes normal to a plate as in Fig. FE3.5. If the force required to hold the plate is 23 N, what is the Jet velocity? (a) 2.85 m/s, (b) 5.7 m/s, (c) 8.1 m/s, (d) 4.0 m/s, (e) 23 m/s I 3 cm FE3.6 A fireboat pump delivers water to a vertical nozzle with a 3:1 diameter ratio, as in Fig. FE3.6. If friction is neglected and the flow rate is 500 gal/min, how high will the outlet water jet rise? (a) 2.0 m, (b) 9.8 m, (c) 32 m, (d) 64 m, (e) 98 m FE3.7 A fireboat pump delivers water to a vertical nozzle with a 3:1 diameter ratio, as in Fig. FE3.6. If friction is neglected and the pump increases the pressure at section 1 to 51 kPa (gage), what will be the resulting flow rate? (a) 187 gal/min, (b) 199 gaPmin, (c) 214 gaPmin, (d) 359 gal/min, (e) 141 gal/min FE3.8 A fireboat pump delivers water to a vertical nozzle with a 3:1 diameter ratio, as in Fig. FE3.6. If duct and nozzle fric¬ tion are neglected and the pump provides 12.3 ft of head to the flow, what will be the outlet flow rate? (a) 85 gal/min, (b) 120 gal/min, (c) 154 gal/min, (d) 217 gal/min, (e) 285 gal/min FE3.9 Water flowing in a smooth 6-cm-diameter pipe enters a venturi contraction with a throat diameter of 3 cm. Upstream pressure is 120 kPa. If cavitation occurs in the throat at a flow rate of 155 gal/min, what is the estimated fluid vapor pressure, assuming ideal frictionless flow? (fl) 6 kPa, (b) 12 kPa, (c) 24 kPa, (d) 31 kPa, (e) 52 kPa FE3.10 Water flowing in a smooth 6-cm-diameter pipe enters a venturi contraction with a throat diameter of 4 cm. Upstream pressure is 120 kPa. If the pressure in the throat is 50 kPa, what is the flow rate, assuming ideal frictionless flow? (a) 7.5 gal/min, (b) 236 gal/min, (c) 263 gal/min, (d) 745 gal/min, (e) 1053 gal/min FE3.5 218 Chapter 3 Integral Relations for a Control Volume Comprehensive Problems C3.1 In a certain industrial process, oil of density p flows through the inclined pipe in Fig. C3.1. A 17-tube manometer, with fluid density /9„, measures the pressure difference between points 1 and 2, as shown. The pipe flow is steady, so that the fluids in the manometer are stationary, (a) Find an analytic expression for pi — p2 in terms of the system parameters, (b) Discuss the conditions on h necessary for there to be no flow in the pipe, (c) What about flow up, from 1 to 2? (d) What about flow down, from 2 to 1? C3.3 C3.4 The airflow underneath an air hockey puck is very com¬ plex, especially since the air jets from the air hockey table impinge on the underside of the puck at various points non- symmetrically. A reasonable approximation is that at any given time, the gage pressure on the bottom of the puck is halfway between zero (atmospheric pressure) and the stag¬ nation pressure of the impinging jets. (Stagnation pressure is dehned as Pq = jpVjf) Find the jet velocity Vjet required to support an air hockey puck of weight W and diameter d. Give your answer in terms of W, d, and the density p of the air. (b) For W = 0.05 Ibf and d = 2.5 in, estimate the required jet velocity in ft/s. C3.2 A rigid tank of volume Y = 1 .0 m^ is initially filled with air at 20°C and po = 100 kPa. At time t = 0, a vacuum pump is turned on and evacuates air at a constant volume flow rate 2 = 80 L/min (regardless of the pressure). Assume an ideal gas and an isothermal process, {a) Set up a differen¬ tial equation for this flow, {b) Solve this equation for f as a function of (T, Q, p, po). (c) Compute the time in minutes to pump the tank down to p = 20 kPa. Hint: Your answer should lie between 15 and 25 min. C3.3 Suppose the same steady water jet as in Prob. P3.40 (jet velocity 8 m/s and jet diameter 10 cm) impinges instead on a cup cavity as shown in Fig. C3.3. The water is turned 180° and exits, due to friction, at lower velocity, = 4 m/s. (Looking from the left, the exit jet is a circular annulus of outer radius R and thickness h, flowing toward the viewer.) The cup has a radius of curvature of 25 cm. Find (a) the thickness h of the exit jet and (b) the force F required to hold the cupped object in place, (c) Compare part (b) to Prob. 3.40, where F ~ 500 N, and give a physi¬ cal explanation as to why F has changed. C3.5 Atmosphere © t ■ Zl References 219 C3.5 Neglecting friction sometimes leads to odd results. You are asked to analyze and discuss the following example in Fig. C3.5. A fan blows air through a duct from section 1 to section 2, as shown. Assume constant air density p. Neglecting frictional losses, find a relation between the required fan head hp and the flow rate and the eleva¬ tion change. Then explain what may be an unexpected result. Design Project D3.1 Let us generalize Probs. P3.180 and P3.181, in which a pump performance curve was used to determine the flow rate between reservoirs. The particular pump in Fig. P3. 1 8 1 is one of a family of pumps of similar shape, whose dimen¬ sionless performance is as follows: where hp is the pump head (ft), n is the shaft rotation rate (r/s), and Dp is the impeller diameter (ft). The range of validity is 0 < C < 0.027. The pump of Fig. P3.181 had Dp = 2 ft in diameter and rotated at n = 20 r/s (1200 r/min). The solution to Prob. P3.181, namely, Q ~ 2.57 ftVs and hp ~ 172 ft, corresponds to (p~ 3.46, ^ ~ 0.016, 7] ~ 0.75 (or 75 percent), and power to the water = pgQhp ~ 27,500 ft ■ Ibf/s (50 hp). Please check these numerical values before beginning this project. Now revisit Prob. P3.181 and select a low-cost pump that rotates at a rate no slower than 600 r/min and delivers no less than 1.0 ftVs of water. Assume that the cost of the pump is linearly proportional to the power input required. Comment on any limitations to your results. Head: 4) « 6.04 - 161C 0 = and C = ^ n Dp nDp Efficiency: , power to water References D. T. GreomNood, Advanced Dynamics, Cambridge University Press, New York, 2006. T. von Karman, The Wind and Beyond, Little, Brown, Boston, J. P. Holman, Heat Transfer, 10th ed., McGraw-Hill, New York, 2009. A. G. Hansen, Fluid Mechanics, Wiley, New York, 1967. M. C. Potter, D. C. Wiggert, and M. Hondzo, Mechanics of Fluids, Brooks/Cole, Chicago, 2001. S. Klein and G. Nellis, Thermodynamics, Cambridge University Press, New York, 2011. Y. A. Cengel and M. A. Boles, Thermodynamics: An Engineering Approach, 7th ed., McGraw-Hill, New York, J. F. Wendt, Computational Fluid Dynamics: An Introduction, Springer, 3d ed.. New York, 2009. W. G. Vincenti, “Control Volume Analysis: A Difference in Thinking between Engineering and Physics,” Technology and Culture, vol. 23, no. 2, 1982, pp. 145-174. J. Keenan, Thermodynamics, Wiley, New York, 1941. J. Hunsaker and B. Rightmire, Engineering Applications of Fluid Mechanics, McGraw-Hill, New York, 1947. The differential equations to be studied in this chapter can be modeled numerically by compu¬ tational fluid dynamics (CFD). This study, from Ref. 21, models turbulent flow near a rotating cylinder at a Reynolds number Re^ ~ 8960. The grid (a) contains 3.1 million nodes, very finely spaced near the cylinder. The results (b) show turbulent velocity fluctuations, obtained by direct numerical simulation (DNS), at = 10 away from the cylinder surface. Source: From ASME J. Fluids Engineering, J-Y. Flwang, K-S. Yang, and K. Bremhorst, “Direct Numerical Simulation of Turbulent Flow Around a Rotating Circular Cylinder, " Vol. 129, Jan. 2007, pp 40-47, by permission of the American Society of Mechanical Engineers. 220 Chapter 4 Differential Relations for Fluid Flow Motivation. In analyzing fluid motion, we might take one of two paths; (1) seeking an estimate of gross effects (mass flow, induced force, energy change) over a finite region or control volume or (2) seeking the point-by-point details of a flow pattern by analyzing an infinitesimal region of the flow. The former or gross-average view¬ point was the subject of Chap. 3. This chapter treats the second in our trio of techniques for analyzing fluid motion: small-scale, or differential, analysis. That is, we apply our four basic conservation laws to an infinitesimally small control volume or, alternately, to an infinitesimal fluid system. In either case the results yield the basic differential equations of fluid motion. Appropriate boundary conditions are also developed. In their most basic form, these differential equations of motion are quite difficult to solve, and very little is known about their general mathematical properties. How¬ ever, certain things can be done that have great educational value. First, as shown in Chap. 5, the equations (even if unsolved) reveal the basic dimensionless parameters that govern fluid motion. Second, as shown in Chap. 6, a great number of useful solutions can be found if one makes two simplifying assumptions: (1) steady flow and (2) incompressible flow. A third and rather drastic simplification, frictionless flow, makes our old friend the Bernoulli equation valid and yields a wide variety of ideal¬ ized, or perfect-fluid, possible solutions. These idealized flows are treated in Chap. 8, and we must be careful to ascertain whether such solutions are in fact realistic when compared with actual fluid motion. Finally, even the difficult general differential equa¬ tions now yield to the approximating technique known as computational fluid dynam¬ ics (CFD) whereby the derivatives are simulated by algebraic relations between a finite number of grid points in the flow field, which are then solved on a computer. Reference 1 is an example of a textbook devoted entirely to numerical analysis of fluid motion. 221 222 Chapter 4 Differential Relations for Fluid Flow 4.1 The Acceleration Field of a Fluid In Sec. 1.7 we established the cartesian vector form of a velocity field that varies in space and time: V(r, t) = m{x, y, z, t) + Sv{x, y, z, t) + ktv(x, y, z, t) (1.4) This is the most important variable in fluid mechanics: Knowledge of the velocity vector held is nearly equivalent to solving a fluid flow problem. Our coordinates are fixed in space, and we observe the fluid as it passes by — as if we had scribed a set of coordinate fines on a glass window in a wind tunnel. This is the Eulerian frame of reference, as opposed to the Lagrangian frame, which follows the moving position of individual particles. The Eulerian system can be visualized as a window through which we watch a flow. The coordinates (x, y, z) are fixed, and the flow passes by. A fixed instrument placed in the flow takes an Eulerian measurement. In contrast, Lagrangian coordinates follow the moving particles and are common in solid mechanics. Almost all articles and books about fluid mechanics use the Eulerian system. Writers often use traffic as an example. A traffic engineer will remain fixed and will measure the flow of cars going by — an Eulerian viewpoint. Conversely, the police will follow specific cars as a function of time — a Lagrangian viewpoint. To write Newton’s second law for an infinitesimal fluid system, we need to cal¬ culate the acceleration vector field a of the flow. Thus, we compute the total time derivative of the velocity vector: dW , du , dv dw a = — = 1 - hi - h k — dt dt dt dt Since each scalar component (m, v, w) is a function of the four variables (x, y, z, t), we use the chain rule to obtain each scalar time derivative. Eor example, du{x, y, z, t) _ du ^ du dx ^ du dy ^ du dz dt dt dx dt dy dt dz dt But, by definition, dx/dt is the local velocity component u, and dy/dt = v, and dz/dt = w. The total time derivative of u may thus be written as follows, with exactly similar expressions for the time derivatives of v and w: dy du du du du du = = - u - hi; h W — = dt dt dx dy dz dv dv dv dv dv = = - u - hi; - hw - = dt dt dx dy dz dw dw dw dw dw = = - U - -h V - h W - dt dt dx dy dz du — 4 dt dV dt (V • V)D dw dt (V • V)w (4.1) Summing these into a vector, we obtain the total acceleration: cN d\ { dy dy dy\ 3V a = = — + « ^ V - h w — = — + (V • V)V dt dt Local V dx dy Convective dz J dt (4.2) The term dNIdt is called the local acceleration, which vanishes if the flow is steady — that is, independent of time. The three terms in parentheses are called the convective 1 The Acceleration Field of a Fluid 223 acceleration, which arises when the particle moves through regions of spatially vary¬ ing velocity, as in a nozzle or diffuser. Flows that are nominally “steady” may have large accelerations due to the convective terms. Note our use of the compact dot product involving V and the gradient operator V: d d d „ „ d d d u - h V - h w — = V • V where v = i - h i - h k — dx dy dz dx dy dz The total time derivative — sometimes called the substantial or material derivative — concept may be applied to any variable, such as the pressure: dp dt dp dp dp dp dp hM - V - h W - = - h (V • V)p dt dx dy dz dt (4.3) Wherever convective effects occur in the basic laws involving mass, momentum, or energy, the basic differential equations become nonlinear and are usually more com¬ plicated than flows that do not involve convective changes. We emphasize that this total time derivative follows a particle of fixed identity, making it convenient for expressing laws of particle mechanics in the eulerian fluid field description. The operator didt is sometimes assigned a special symbol such as DIDt as a further reminder that it contains four terms and follows a fixed particle. As another reminder of the special nature of cUdt, some writers give it the name substantial or material derivative. EXAMPLE 4.1 Given the Eulerian velocity vector field V = 3d -f jcjj -f ty\ find the total acceleration of a particle. Solution • Assumptions: Given three known unsteady velocity components, u = 3t,v = xz, and w = ty^. • Approach: Carry out all the required derivatives with respect to (x, y, z, t), substitute into the total acceleration vector, Eq. (4.2), and collect terms. • Solution step 1: First work out the local acceleration flV/cff: d\ du dV dw = 1 - hi - h k - dt dt ■’ dt dt 3i -h Oj -h k • Solution step 2: In a similar manner, the convective acceleration terms, from Eq. (4.2), are d\ d 9 u — =(3t) — (3/1 + XZ] + ry^k) = (3?)(0i -h zj -h Ok) = 3tzj dx dx d\ d 2 V — =ixz) — (3u + xzj + ty) = (xz)(0i -h Oj -h 2/yk) = Itxyz k dy dy 224 Chapter 4 Differential Relations for Fluid Flow 4.2 The Differential Equation of Mass Conservation Fig. 4.1 Elemental cartesian fixed control volume showing the inlet and outlet mass flows on the x faces. ■ Solution step 3: Combine all four terms above into the single “total” or “substantial” derivative: rl\ rlV rl\ — = — + u— +t; — + W— =(3i + y) + 3tq + Itxyzls. + txy^j at at ax ay az = 3i + (3fz + txy^)j + (y^ + 2txyz)^ Ans. • Comments: Assuming that V is valid everywhere as given, this total acceleration vector dV/dt applies to all positions and times within the flow field. Conservation of mass, often called the continuity relation, states that the fluid mass cannot change. We apply this concept to a very small region. All the basic differential equations can be derived by considering either an elemental control volume or an elemental system. We choose an infinitesimal fixed control volume {dx, dy, dz), as in Fig. 4.1, and use our basic control volume relations from Chap. 3. The flow through each side of the element is approximately one-dimensional, and so the appropriate mass conservation relation to use here is ^ r/y + 2 (pi Ai Vi )ou, - 2 (Pi Ay,)in = 0 (3.22) Jcv The element is so small that the volume integral simply reduces to a differential term: dp dp — dT ~ — dxdy dz Jcv The mass flow terms occur on all six faces, three inlets and three outlets. We make use of the field or continuum concept from Chap. 1, where all fluid properties are considered to be uniformly varying functions of time and position, such as p = p(x, y, z, t). Thus, if T is the temperature on the left face of the element in Fig. 4.1, the right face will have a slightly different temperature T + (dTIdx) dx. For mass conservation, if pu is known on the left face, the value of this product on the right face is pu -I- (dpu/dx) dx. ^{pu)dx] dydz dx 4.2 The Differential Equation of Mass Conservation 225 Figure 4.1 shows only the mass flows on the x or left and right faces. The flows on the y (bottom and top) and the z (back and front) faces have been omitted to avoid cluttering up the drawing. We can list all these six flows as follows: Face Inlet mass flow Outlet mass flow X pu dy dz a pw + — (pu) dx dy dz y pv dx dz a pv + -^(pv) dy dx dz Z pw dx dy a pw H - ipw) dz dz dx dy Introduce these terms into Eq. (3.22) and we have — dx dy dz H - (pu) dx dy dz - ipv) dx dy dz - (pvv) dx dy dz = 0 dt dx dy dz The element volume cancels out of all terms, leaving a partial differential equation involving the derivatives of density and velocity: dp d — + ^ (P«) + dt dx d dy ipv) + ^ ipw) = 0 dz (4.4) This is the desired result: conservation of mass for an infinitesimal control volume. It is often called the equation of continuity because it requires no assumptions except that the density and velocity are continuum functions. That is, the flow may be either steady or unsteady, viscous or frictionless, compressible or incompressible.' However, the equation does not allow for any source or sink singularities within the element. The vector gradient operator ^ . 3 . d d V — 1 - h j - h k — dx dy dz enables us to rewrite the equation of continuity in a compact form, not that it helps much in finding a solution. The last three terms of Eq. (4.4) are equivalent to the divergence of the vector pV ^ ipu) + ^ ipv) + ^ ipw) = V • (pV) dx dy dz so the compact form of the continuity relation dp IS V • (pV) = 0 (4.5) (4.6) In this vector form the equation is still quite general and can readily be converted to other coordinate systems. 'One case where Eq. (4.4) might need special care is two-phase flow, where the density is discontinuous between the phases. For further details on this case, see Ref. 2, for example. 226 Chapter 4 Differential Relations for Fluid Flow Fig. 4.2 Definition sketch for the cylindrical coordinate system. Cylindrical Polar Coordinates The most common alternative to the cartesian system is the cylindrical polar coordi¬ nate system, sketched in Fig. 4.2. An arbitrary point P is defined by a distance z along the axis, a radial distance r from the axis, and a rotation angle 9 about the axis. The three independent orthogonal velocity components are an axial velocity v,, a radial velocity v^, and a circumferential velocity Vg, which is positive counterclockwise — that is, in the direction of increasing 9. In general, all components, as well as pressure and density and other fluid properties, are continuous functions of r, 9, z, and t. The divergence of any vector function A(r, 9, z, t) is found by making the trans¬ formation of coordinates r= {x^ + 9 = tan“' - z = z (4.7) and the result is given here without proof^ V • A = - ^ (M,) + - ^ (Ag) + ^ (A,) (4.8) r dr r d9 dz The general continuity equation (4.6) in cylindrical polar coordinates is thus dp I d la d 7 ^ ^ (P'^z) = 0 (4.9) dt r dr r d9 dz There are other orthogonal curvilinear coordinate systems, notably spherical polar coordinates, which occasionally merit use in a fluid mechanics problem. We shall not treat these systems here except in Prob. P4.12. There are also other ways to derive the basic continuity equation (4.6) that are interesting and instructive. One example is the use of the divergence theorem. Ask your instructor about these alternative approaches. ^See, for example. Ref. 3. 4.2 The Differential Equation of Mass Conservation 227 Steady Compressible Flow Incompressible Flow If the flow is steady, didt = 0 and all properties are functions of position only. Equa¬ tion (4.6) reduces to d d d Cartesian: — {pu) + — {pv) -f — (pw) = 0 dx ay az Cylindrical: -^{rpv,) + -^(pVe) + ^ (pv,) = 0 (4.10) r dr r da dz Since density and velocity are both variables, these are still nonlinear and rather formidable, but a number of special-case solutions have been found. A special case that affords great simplification is incompressible flow, where the density changes are negligible. Then dp/dt ~ 0 regardless of whether the flow is steady or unsteady, and the density can be slipped out of the divergence in Eq. (4.6) and divided out. The result V • V = 0 (4.11) is valid for steady or unsteady incompressible flow. The two coordinate forms are Cartesian: Cylindrical: du dV dw h - 1 - — 0 dx dy dz 13 13 3 — (rv,) + - — iVg) + — (D,) = 0 r dr r dd dz (4.12fl) (4.12(7) These are linear differential equations, and a wide variety of solutions are known, as discussed in Chaps. 6 to 8. Since no author or instructor can resist a wide variety of solutions, it follows that a great deal of time is spent studying incompressible flows. Fortunately, this is exactly what should be done, because most practical engineering flows are approximately incompressible, the chief exception being the high-speed gas flows treated in Chap. 9. When is a given flow approximately incompressible? We can derive a nice criterion by using some density approximations. In essence, we wish to slip the density out of the divergence in Eq. (4.6) and approximate a typical term such as This is equivalent to the strong 3 — (pw) OX inequality or dp u — dx p V (4.13) (4.14) As shown in Eq. (1.38), the pressure change is approximately proportional to the density change and the square of the speed of sound a of the fluid: Sp ~ 5p (4.15) 228 Chapter 4 Differential Relations for Fluid Flow Meanwhile, if elevation changes are negligible, the pressure is related to the velocity change by Bernoulli’s equation (3.52): Sp^-pVSV (4.16) Combining Eqs. (4.14) to (4.16), we obtain an explicit criterion for incompressible flow: 2 = 1 (4.17) a where Ma = Via is the dimensionless Mach number of the flow. How small is small? The commonly accepted limit is Ma < 0.3 (4.18) For air at standard conditions, a flow can thus be considered incompressible if the velocity is less than about 100 m/s (330 ft/s). This encompasses a wide variety of airflows: automobile and train motions, light aircraft, landing and takeoff of high¬ speed aircraft, most pipe flows, and turbomachinery at moderate rotational speeds. Further, it is clear that almost all liquid flows are incompressible, since flow velocities are small and the speed of sound is very large. ^ Before attempting to analyze the continuity equation, we shall proceed with the derivation of the momentum and energy equations, so that we can analyze them as a group. A very clever device called the stream function can often make short work of the continuity equation, but we shall save it until Sec. 4.7. One further remark is appropriate: The continuity equation is always important and must always be satisfied for a rational analysis of a flow pattern. Any newly discov¬ ered momentum or energy “solution” will ultimately fail when subjected to critical analysis if it does not also satisfy the continuity equation. EXAMPLE 4.2 Under what conditions does the velocity field V = (aix + biy + Ciz)i -I- (a2X + b^y + C2z)j + (ape + b^y + C3z)k where fli, bi, etc. = const, represent an incompressible flow that conserves mass? Solution Recalling that V = id -I- wj -I- wk, we see that u = {aix + by + Ciz), etc. Substituting into Eq. (4.12a) for incompressible continuity, we obtain d d d — (ay + by + cy) H - (a2X + P2.V + C2Z) H - (ayx + by + cy) = 0 dx dy dz or Oi + b2 + C} = 0 Ans. At least two of constants Oi, b2, and C3 must have opposite signs. Continuity imposes no restrictions whatever on constants bi, Cj, 03, C2, 03, and b^, which do not contribute to a volume increase or decrease of a differential element. ^An exception occurs in geophysical flows, where a density change is imposed thermally or mechan¬ ically rather than by the flow conditions themselves. An example is fresh water layered upon saltwater or warm air layered upon cold air in the atmosphere. We say that the fluid is stratified, and we must account for vertical density changes in Eq. (4.6) even if the velocities are small. 4.2 The Differential Equation of Mass Conservation 229 EXAMPLE 4.3 An incompressible velocity field is given by u = a(x^ — y^) t; unknown w = b where a and b are constants. What must the form of the velocity component v be? Solution Again Eq. (4.12a) applies: or c) T T — (ax^ - ay^) + dx dV db dy dz = 0 dV dy = —2ax (1) This is easily integrated partially with respect to y: V (x, y, z, t) = —laxy + f (x, z, t) Ans. This is the only possible form for v that satisfies the incompressible continuity equation. The function of integration / is entirely arbitrary since it vanishes when v is differentiated with respect to y.' EXAMPLE 4.4 A centrifugal impeller of 40-cm diameter is used to pump hydrogen at 15°C and 1-atm pressure. Estimate the maximum allowable impeller rotational speed to avoid compress¬ ibility effects at the blade tips. Solution • Assumptions: The maximum fluid velocity is approximately equal to the impeller tip speed: V'max = flfmax where r„ax = DU = 0.20 m • Approach: Find the speed of sound of hydrogen and make sure that is much less. • Property values: From Table A.4 for hydrogen, R = 4124 m^/(s^ — K) and k = 1.41. From Eq. (1.39) at 15°C = 288K, compute the speed of sound: Oh, = = Vl.414124mV(s^ - K) « 1294 m/s • Final solution step: Use our rule of thumb, Eq. (4.18), to estimate the maximum impel¬ ler speed: V = ^^'■max ^ 0.3a or £1(0.2 m) < 0.3(1294 m/s) „ rad rev Solve for O < 1940 - « 18,500 — Ans. s mm • Comments: This is a high rate because the speed of sound of hydrogen, a light gas, is nearly four times greater than that of air. An impeller moving at this speed in air would create tip shock waves. “'This is a very realistic flow that simulates the turning of an inviscid fluid through a 60° angle; see Examples 4.7 and 4.9. 230 Chapter 4 Differential Relations for Fluid Flow 4.3 The Differential Equation of Linear Momentum This section uses an elemental volume to derive Newton’s law for a moving fluid. An alternate approach, which the reader might pursue, would be a force balance on an elemental moving particle. Having done it once in Sec. 4.2 for mass conservation, we can move along a little faster this time. We use the same elemental control volume as in Fig. 4.1, for which the appropriate form of the linear momentum relation is 2f = dt •'cv Vp ufT + 2 - 2 (3.40) Again the element is so small that the volume integral simply reduces to a derivative term: — (VptfT) = — (pV) dx dy dz (4.19) dt dt The momentum fluxes occur on all six faces, three inlets and three outlets. Refer¬ ring again to Fig. 4.1, we can form a table of momentum fluxes by exact analogy with the discussion that led up to the equation for net mass flux: Faces Inlet momentum flux Outlet momentum flux X puV dy dz 5 poV -1- —(pu) dx dy dz y pvS dx dz pv\ + —{pv) dy Sy dx dz Z pw\ dx dy pw\ H - {pw) dz dz dx dy Introduce these terms and Eq. (4.19) into Eq. (3.40), and get this intermediate result: 2 F = dx dy dz \|:(pV) + ^ (p«V) + ^ (pdV) + ^ (pwV) dt dx dy dz (4.20) Note that this is a vector relation. A simplification occurs if we split up the term in brackets as follows: f (pV) + ^ (pwV) + ^ (pvY) + f (pwV) dt dx dy dz = V (dY dY dY dY\ — -f V (pV) P[ — U - h V — w — 1 dt ^\dt dx dy dzj (4.21) The term in brackets on the right-hand side is seen to be the equation of continuity, Eq. (4.6), which vanishes identically. The long term in parentheses on the right-hand side is seen from Eq. (4.2) to be the total acceleration of a particle that instantaneously occupies the control volume: dY dY av dY dV h U - h V - h W - = - dt dx dy dz dt (4.2) Thus, we have now reduced Eq. (4.20) to 4.3 The Differential Equation of Linear Momentum 231 It might be good for you to stop and rest now and think about what we have just done. What is the relation between Eqs. (4.22) and (3.40) for an infinitesimal control volume? Could we have begun the analysis at Eq. (4.22)7 Equation (4.22) points out that the net force on the control volume must be of differential size and proportional to the element volume. These forces are of two types, body forces and surface forces. Body forces are due to external helds (gravity, mag¬ netism, electric potential) that act on the entire mass within the element. The only body force we shall consider in this book is gravity. The gravity force on the dif¬ ferential mass p dx dy dz within the control volume is “^grav = pg dx dy dz (4.23) where g may in general have an arbitrary orientation with respect to the coordinate sys¬ tem. In many applications, such as Bernoulli’s equation, we take z “up,” and g = — gk. The surface forces are due to the stresses on the sides of the control surface. These stresses are the sum of hydrostatic pressure plus viscous stresses that arise from motion with velocity gradients: -P + Tx) ' yx -p + T,, -P + Tu (4.24) The subscript notation for stresses is given in Eig. 4.3. Unlike velocity V, which is a three-component vector, stresses CTij and and strain rates e,-,- are nine-component tensors and require two subscripts to define each component. Eor further study of tensor analysis, see Refs. 6, 11, or 13. It is not these stresses but their gradients, or differences, that cause a net force on the differential control surface. This is seen by referring to Fig. 4.4, which shows only y Fig. 4.3 Notation for stresses. 232 Chapter 4 Differential Relations for Fluid Flow Fig. 4.4 Elemental cartesian fixed control volume showing the surface forces in the x direction only. — — dy) dx dz y dy dx) dy dz the x-directed stresses to avoid cluttering up the drawing. For example, the leftward force CTjj dy dz on the left face is balanced by the rightward force cr^ dy dz on the right face, leaving only the net rightward force (dcTj^/dx) dx dy dz on the right face. The same thing happens on the other four faces, so the net surface force in the x direction is given by ^f^x.surf d d d — iCTxx) + — (CTyJ + — iCFzx) ax ay az dx dy dz (4.25) We see that this force is proportional to the element volume. Notice that the stress terms are taken from the top row of the array in Eq. (4.24). Splitting this row into pressure plus viscous stresses, we can rewrite Eq. (4.25) as dl\ dr dp d d d ■— + — (T«) + — (Tyx) + — (T^) OX ox oy oz (4.26) where dY = dx dy dz. In an exactly similar manner, we can derive the y and z forces per unit volume on the control surface: dFy dF, dr dp d d d to dp d d d 1 - (r,.) H - (t_) H - (t„) OZ ox oy oz (4.27) Now we multiply Eqs. (4.26) and (4.27) by i, j, and k, respectively, and add to obtain an expression for the net vector surface force: VISCOUS (4.28) 4.3 The Differential Equation of Linear Momentum 233 where the viscous force has a total of nine terms: W'V/ viscous ^\dx dy dz J j dT^ dy dz J V dx dy dz J (4.29) Since each term in parentheses in Eq. (4.29) represents the divergence of a stress component vector acting on the x, y, and z faces, respectively, Eq. (4.29) is sometimes expressed in divergence form: 1 dF \dr /viscous = V • '^ij (4.30) T ' XX T,, where '^ii = Tyy 'Ey (4.31) -Txz -Tyz '^ZZ- is the viscous stress tensor acting on the element. The surface force is thus the sum of the pressure gradient vector and the divergence of the viscous stress tensor. Substituting into Eq. (4.22) and utilizing Eq. (4.23), we have the basic differential momentum equation for an infinitesimal element: pg-Vp + V.r,.p- dV d\ d\ d\ d\ where - = - h u - h v - h w — dt dt dx dy dz We can also express Eq. (4.32) in words: Gravity force per unit volume + pressure force per unit volume + viscous force per unit volume = density X acceleration (4.32) (4.33) (4.34) Equation (4.32) is so brief and compact that its inherent complexity is almost invisible. It is a vector equation, each of whose component equations contains nine terms. Let us therefore write out the component equations in full to illustrate the mathematical difficulties inherent in the momentum equation: (4.35) 234 Chapter 4 Differential Relations for Fluid Flow Inviscid Flow: Euler’s Equation Newtonian Fluid: Navier-Stokes Equations This is the differential momentum equation in its full glory, and it is valid for any fluid in any general motion, particular fluids being characterized by particular viscous stress terms. Note that the last three “convective” terms on the right-hand side of each component equation in (4.35) are nonlinear, which complicates the general mathematical analysis. Equation (4.35) is not ready to use until we write the viscous stresses in terms of velocity components. The simplest assumption is frictionless flow = 0, for which Eq. (4.32) reduces to dy pg - Vp = p— (4.36) This is Euler’s equation for inviscid flow. We show in Sec. 4.9 that Euler’s equation can be integrated along a streamline to yield the frictionless Bernoulli equation, (3.52) or (3.54). The complete analysis of inviscid flow fields, using continuity and the Bernoulli relation, is given in Chap. 8. Eor a newtonian fluid, as discussed in Sec. 1.9, the viscous stresses are proportional to the element strain rates and the coefficient of viscosity. Eor incompressible flow, the generalization of Eq. (1.23) to three-dimensional viscous flow is^ du dV dw = 2p — Tyy = 2n— T,, = 2p — dx dy dz ( du dV T, = T,, - + dx ( dw du (4.37) / dV dw where p, is the viscosity coefficient. Substitution into Eq. (4.35) gives the differential momentum equation for a newtonian fluid with constant density and viscosity: These are the incompressible flow Navier-Stokes equations, named after C. L. M. H. Navier (1785-1836) and Sir George G. Stokes (1819-1903), who are credited with ^When compressibility is significant, additional small terms arise containing the element volume expansion rate and a second coefficient of viscosity; see Refs. 4 and 5 for details. 4.3 The Differential Equation of Linear Momentum 235 their derivation. They are second-order nonlinear partial differential equations and are quite formidable, but solutions have been found to a variety of interesting viscous flow problems, some of which are discussed in Sec. 4.11 and in Chap. 6 (see also Refs. 4 and 5). For compressible flow, see Eq. (2.29) of Ref. 5. Equations (4.38) have four unknowns: p, u, v, and w. They should be combined with the incompressible continuity relation [Eqs. (4.12)] to form four equations in these four unknowns. We shall discuss this again in Sec. 4.6, which presents the appropriate boundary conditions for these equations. Even though the Navier-Stokes equations have only a limited number of known analytical solutions, they are amenable to fine-gridded computer modeling . The held of CED is maturing fast, with many commercial software tools available. It is possible now to achieve approximate, but realistic, CED results for a wide variety of complex two- and three-dimensional viscous flows. EXAMPLE 4.5 Take the velocity field of Example 4.3, with b = 0 for algebraic convenience u = a(j^ — y^) V = —2axy w = 0 and determine under what conditions it is a solution to the Navier-Stokes momentum equa¬ tions (4.38). Assuming that these conditions are met, determine the resulting pressure distribution when z is “up” (g^. = 0, g^, = 0, g^ = — g). Solution • Assumptions: Constant density and viscosity, steady flow (u and v independent of time). • Approach: Substitute the known {u, v, w) into Eqs. (4.38) and solve for the pressure gradients. If a unique pressure function p{x, y, z) can then be found, the given solution is exact. • Solution step 1: Substitute (u, v, w) into Eqs. (4.38) in sequence: p(0) - f u{2a — 2a + 0) = p[ u dx \ P(0) - ' du dx V V Rearrange and solve for the three pressure gradients: (1) Comment 1: The vertical pressure gradient is hydrostatic. [Could you have predicted this by noting in Eqs. (4.38) that w = 0?] However, the pressure is velocity-dependent in the xy plane. 236 Chapter 4 Differential Relations for Fluid Flow • Solution step 2: To determine if the x and y gradients of pressure in Eq. (1) are compatible, evaluate the mixed derivative (d^p/dx dy); that is, cross-differentiate these two equations: ^ [-2aV(^}' + /)] = -4a^pxy dx\dy J dx • Comment 2: Since these are equal, the given velocity distribution is indeed an exact solution of the Navier-Stokes equations. • Solution step 3: To hnd the pressure, integrate Eqs. (1), collect, and compare. Start with dp/dx. The procedure requires care! Integrate partially with respect to x, holding y and z constant: P = "" I + V) dx\y,z = -2.a^p + fiiy, z) (2) Note that the “constant” of integration/i is a function of the variables that were not integrated. Now differentiate Eq. (2) with respect to y and compare with dp/dy from Eq. (1): ^ 1(2) = -la^px^y + ^ = 1(1) = -2a^p(x^y + y^) dy dydy 5/1 9 c, f 3/1 I 9 y‘^ Compare: — = -2a^p/ or /i = —dy\^ = -2a^p— +/2(z) dy j dy 4 Collect terms: So far p = —2a^p{ — H — ^ — h — j +/2(z) (3) This time the “constant” of integration /2 is a function of z only (the variable not integrated). Now differentiate Eq. (3) with respect to z and compare with dp/dz from Eq. (1): dp df2 dp — 1(3) = ^ = — 1(1) = -pg or f2=-pgz + C (4) dz az dz where C is a constant. This completes our three integrations. Combine Eqs. (3) and (4) to obtain the full expression for the pressure distribution in this flow: p{x, y, z) = -pgz - ^ a^pix + y^ + 2x^y^) + C Ans. (5) This is the desired solution. Do you recognize it? Not unless you go back to the beginning and square the velocity components: u^ + i/ + = a^(x + y^ + 2x^y^) (6) Comparing with Eq. (5), we can rewrite the pressure distribution as p + \pV^ + pgz = C (7) ■ Comment: This is Bernoulli’s equation (3.54). That is no accident, because the velocity distribution given in this problem is one of a family of flows that are solutions to the Navier-Stokes equations and that satisfy Bernoulli’s incompressible equation everywhere in the flow field. They are called irrotational flows, for which curl V = V X V = 0. This subject is discussed again in Sec. 4.9. 4.4 The Differential Equation of Angular Momentum 237 4.4 The Differential Equation of Angular Momentum Fig. 4.5 Elemental cartesian fixed control volume showing shear stresses that may cause a net angular acceleration about axis O. Having now been through the same approach for both mass and linear momentum, we can go rapidly through a derivation of the differential angular momentum relation. The appropriate form of the integral angular momentum equation for a fixed control volume is 2m, d dt (r X V)p dr •’cv (r X V)p(V • n) dA ■’cs (3.59) We shall confine ourselves to an axis through O that is parallel to the z axis and passes through the centroid of the elemental control volume. This is shown in Fig. 4.5. Let 8 be the angle of rotation about O of the fluid within the control volume. The only stresses that have moments about O are the shear stresses and r^,^. We can evaluate the moments about O and the angular momentum terms about O. A lot of algebra is involved, and we give here only the result: 15 15 Tw + zT — (t„) dx — — — ^ 2dx ^ 2 dy {Tyx) dy dx dy dz 1 . . 2 1 d^d = —p{dx dy dz){dx + dy ) -jj (4.39) Assuming that the angular acceleration d^dldf' is not infinite, we can neglect all higher-order differential terms, which leaves a finite and interesting result: Txy = Tyx (4.40) Had we summed moments about axes parallel to y or x, we would have obtained exactly analogous results: '^xz Ty^ (4.41) There is no differential angular momentum equation. Application of the integral theo¬ rem to a differential element gives the result, well known to students of stress analysis or strength of materials, that the shear stresses are symmetric: Ty = Tp. This is the '^yx+ ^iryx)‘iy nv 'Tyx 238 Chapter 4 Differential Relations for Fluid Flow only result of this section.® There is no differential equation to rememher, which leaves room in your brain for the next topic, the differential energy equation. 4.5 The Differential Equation of Energy’ We are now so used to this type of derivation that we can race through the energy equation at a bewildering pace. The appropriate integral relation for the hxed control volume of Fig. 4.1 is Q-W,-W^ d dt p{\ ■ n) dA (3.66) where VF, = 0 because there can be no inhnitesimal shaft protruding into the con¬ trol volume. By analogy with Eq. (4.20), the right-hand side becomes, for this tiny element. Q-w^ = ^ (pe) + ^ (puO + ^{pvO + ^ ipwO dt dx dy dz dx dy dz where (^ = e + pip. When we use the continuity equation by analogy with Eq. (4.21), this becomes e - V • V/7 -f pV • Y jdxdy dz (4.42) Thermal Conductivity; Fourier’s Law To evaluate Q, we neglect radiation and consider only heat conduction through the sides of the element. Experiments for both fluids and solids show that the vector heat transfer per unit area, q, is proportional to the vector gradient of temperature, YT. This proportionality is called Fourier’s law of conduction, which is analogous to Newton’s viscosity law: or: q = -kYT (4.43) where k is called the thermal conductivity, a fluid property that varies with tempera¬ ture and pressure in much the same way as viscosity. The minus sign satisfies the convention that heat flux is positive in the direction of decreasing temperature. Fourier’s law is dimensionally consistent, and k has SI units of joules per (sec-meter- kelvin) and can be correlated with T in much the same way as Eqs. (1.27) and (1.28) for gases and liquids, respectively. Figure 4.6 shows the heat flow passing through the x faces, the y and z heat flows being omitted for clarity. We can list these six heat flux terms: ^We are neglecting the possibility of a finite couple being applied to the element by some powerful external force field. See, for example, Ref. 6. ^This section may be omitted without loss of continuity. 4.5 The Differential Equation of Energy 239 Fig. 4.6 Elemental cartesian control volume showing heat flow and viscous work rate terms in the X direction. 9x + d_ dx (q^)dx dx {w^ dx Faces Inlet heat flux Outlet heat flux X qx dy dz d qx + — (qx) dx dx dy dz y qy dx dz d 4, + — (qy) dy dx dz Z dx dy d 4z + — (qz) dz Sz J dx dy By adding the inlet terms and subtracting the outlet terms, we obtain the net heat added to the element: Q = d d d — iqx) + — (.qy) + — iqz) dx dy dz dxdy dz = — V • q dx dy dz (4.44) As expected, the heat flux is proportional to the element volume. Introducing Fourier’s law from Eq. (4.43), we have Q = V • ik\^T) dxdydz (4.45) The rate of work done by viscous stresses equals the product of the stress compo¬ nent, its corresponding velocity component, and the area of the element face. Figure 4.6 shows the work rate on the left x face is ^ w^dy dz where = —(uT^ + VT^ + wt^^) (4.46) (where the subscript LF stands for left face) and a slightly different work on the right face due to the gradient in w^. These work fluxes could be tabulated in exactly the same manner as the heat fluxes in the previous table, with w^. replacing q„ and so on. After outlet terms are subtracted from inlet terms, the net viscous work rate becomes IF„ = d d — (UT^^ + VT^ + -1- — {uTy^ + VTyy + WTy^) d — (mT^ + VT^ 4- WT,j) dz = —V • (V • Ty) dx dy dz dx dy dz (4.47) 240 Chapter 4 Differential Relations for Fluid Flow We now substitute Eqs. (4.45) and (4.47) into Eq. (4.43) to obtain one form of the differential energy equation: de p — + V • Vp + pV • V = V • (A: VT) + V • (V • Ty) where e = u + + gz A more useful form is obtained if we split up the viscous work term: V • (V • r^) - V • (V • T,^) + D where is short for the viscous-dissipation function.^ Eor a newtonian incompressible viscous fluid, this function has the form (4.48) (4.49) <1> = /i dx dw dV dy dz \dy V dz dx dy du dw — 4 — dz dx 21 (4.50) Since all terms are quadratic, viscous dissipation is always positive, so a viscous flow always tends to lose its available energy due to dissipation, in accordance with the second law of thermodynamics. Now substitute Eq. (4.49) into Eq. (4.48), using the linear momentum equation (4.32) to eliminate V • Ty. This will cause the kinetic and potential energies to cancel, leaving a more customary form of the general differential energy equation: P^ + .(V-V) V • (kVT) + If) (4.5 1) This equation is valid for a newtonian fluid under very general conditions of unsteady, compressible, viscous, heat-conducting flow, except that it neglects radiation heat transfer and internal sources of heat that might occur during a chemical or nuclear reaction. Equation (4.51) is too difficult to analyze except on a digital computer . It is customary to make the following approximations: du ~ Cj,dT Cj,, p, k, p ~ const (4.52) Equation (4.51) then takes the simpler form, for V • V = 0, dT T \|0c„— = kVV-f d) (4.53) dt which involves temperature T as the sole primary variable plus velocity as a second¬ ary variable through the total time-derivative operator: dT dT dT dT dT = - U - V - hw - dt dt dx dy dz (4.54) Tor further details, see, Ref. 5, p. 72. 4.6 Boundary Conditions for the Basic Equations 241 A great many interesting solutions to Eq. (4.53) are known for various flow condi¬ tions, and extended treatments are given in advanced books on viscous flow [4, 5] and books on heat transfer [7, 8]. One well-known special case of Eq. (4.53) occurs when the fluid is at rest or has negligible velocity, where the dissipation and convective terms become negligible: = (4.55) The change from c^, to Cp is correct and justified by the fact that, when pressure terms are neglected from a gas flow energy equation [4, 5], what remains is approximately an enthalpy change, not an internal energy change. This is called the heat conduction equation in applied mathematics and is valid for solids and fluids at rest. The solution to Eq. (4.55) for various conditions is a large part of courses and books on heat transfer. This completes the derivation of the basic differential equations of fluid motion. 4.6 Boundary Conditions for the Basic Equations There are three basic differential equations of fluid motion, just derived. Let us sum¬ marize those here: Continuity: Momentum: Energy: ^ + V • ipY) = 0 p — = pg-yp + v-Tij du p — + p(V • V) = V • (k VT) -f <1) dt (4.56) (4.57) (4.58) where $ is given by Eq. (4.50). In general, the density is variable, so these three equations contain five unknowns, p, V, p, u, and T. Therefore, we need two additional relations to complete the system of equations. These are provided by data or algebraic expressions for the state relations of the thermodynamic properties: p = p(/7, T) a = Hp, T) (4.59) Eor example, for a perfect gas with constant specific heats, we complete the system with P RT u Cp, dT ~ Cp,T + const (4.60) It is shown in advanced books [4, 5] that this system of equations (4.56) to (4.59) is well posed and can be solved analytically or numerically, subject to the proper bound¬ ary conditions. What are the proper boundary conditions? Eirst, if the flow is unsteady, there must be an initial condition or initial spatial distribution known for each variable: At f = 0: p, V, p,u,T = known /(x, y, z) (4.61) Thereafter, for all times t to be analyzed, we must know something about the variables at each boundary enclosing the flow. 242 Chapter 4 Differential Relations for Fluid Flow Figure 4.7 illustrates the three most common types of boundaries encountered in fluid flow analysis: a solid wall, an inlet or outlet, and a liquid-gas interface. First, for a solid, impermeable wall, there can be slip and temperature jump in a viscous heat-conducting fluid: No-slip: ^fluid ^wall (4.62) where, for the rarefied gas, n is normal to the wall, u is parallel to the wall, I is the mean free path of the gas [see Eq. (1.37)], and C, denotes, just this one time, the specific heat ratio. The above so-called temperature-jump relation for gases is given here only for completeness and will not be studied (see page 48 of Ref. 5). A few velocity-jump assignments will be given. Second, at any inlet or outlet section of the flow, the complete distribution of velocity, pressure, and temperature must be known for all times: Inlet or outlet: Known V, p, T (4.63) These inlet and outlet sections can be and often are at ± simulating a body immersed in an infinite expanse of fluid. Finally, the most complex conditions occur at a liquid-gas interface, or free surface, as sketched in Fig. 4.7. Let us denote the interface by Interface: z = ri(x, y, t) (4.64) Liquid-gas interface z = T\|(^, y, t)\ Pliq=Pgas-Tf(R^' +R^') z I drj 'Tliq = Wgas = Equality of q and T across interface Gas Liquid Inlet: known V, p, T Outlet: known V, p. T Solid contact: Fig. 4.7 Typical boundary conditions in a viscous heat- conducting fluid flow analysis. (V , Dfluid - (V , Dwall Solid impermeable wall 4.6 Boundary Conditions for the Basic Equations 243 Simplified Free Surface Conditions Then there must be equality of vertical velocity across the interface, so that no holes appear between liquid and gas: Wliq = Wgas dri dr] dt] drj — = — + M— + V — dt dt dx dy (4.65) This is called the kinematic boundary condition. There must be mechanical equilibrium across the interface. The viscous shear stresses must balance: i't~zy')liq (^zyigas (^zxOliq (4.66) Neglecting the viscous normal stresses, the pressures must balance at the interface except for surface tension effects: Puq = Pgas - + R~^) (4.67) which is equivalent to Eq. (1.33). The radii of curvature can be written in terms of the free surface position rj: drjldx Vl + (drjldxf + {dr]ldyf- d drj/dy ^3^1 Vl + (drjldxf + (drjldyf- r:' + rJ = dx (4.68) Finally, the heat transfer must be the same on both sides of the interface, since no heat can be stored in the infinitesimally thin interface: (?z)liq = (?z)gas (4.69) Neglecting radiation, this is equivalent to (4.70) This is as much detail as we wish to give at this level of exposition. Further and even more complicated details on fluid flow boundary conditions are given in Refs. 5 and 9. In the introductory analyses given in this book, such as open-channel flows in Chap. 10, we shall back away from the exact conditions (4.65) to (4.69) and assume that the upper fluid is an “atmosphere” that merely exerts pressure on the lower fluid, with shear and heat conduction negligible. We also neglect nonlinear terms involving the slopes of the free surface. We then have a much simpler and linear set of condi¬ tions at the surface: Fliq Fgas Yi dx^ dy^J (4.71) 244 Chapter 4 Differential Relations for Fluid Flow Incompressible Flow with Constant Properties Inviscid Flow Approximations In many cases, such as open-channel flow, we can also neglect surface tension, so PViq Pzim (4.72) These are the types of approximations that will be used in Chap. 10. sional forms of these conditions will also be useful in Chap. 5. The nondimen- Flow with constant p, p, and A: is a basic simplification that will be used, for example, throughout Chap. 6. The basic equations of motion (4.56) to (4.58) reduce to Continuity: V • V = 0 (4.73) Momentum: ^/V , p— = pg - Vp -f pV-V (4.74) Energy: dT pcp — = kV^T + (4.75) Since p is constant, there are only three unknowns: p, V, and T. The system is closed.® Not only that, the system splits apart: Continuity and momentum are independent of T. Thus we can solve Eqs. (4.73) and (4.74) entirely separately for the pressure and velocity, using such boundary conditions as Solid surface: V = V,,11 (4.76) Inlet or outlet: Known V, p (4.77) Free surface: dr] P^ Pa ^ at (4.78) Later, usually in another course, we can solve for the temperature distribution from Eq. (4.75), which depends on velocity V through the dissipation $ and the total time- derivative operator cUdt. Chapter 8 assumes inviscid flow throughout, for which the viscosity /i = 0. The momentum equation (4.74) reduces to P d\ dt = pg-Vp (4.79) This is Euler’s equation; it can be integrated along a streamline to obtain Bernoulli’s equation (see Sec. 4.9). By neglecting viscosity we have lost the second-order deriv¬ ative of V in Eq. (4.74); therefore, we must relax one boundary condition on velocity. The only mathematically sensible condition to drop is the no-slip condition at the wall. We let the flow slip parallel to the wall but do not allow it to flow into the wall. The proper inviscid condition is that the normal velocities must match at any solid surface: Inviscid flow: (VJfluid = (V„)waii (4.80) ^For this system, what are the thermodynamic equivalents to Eq. (4.59)? ^^Since temperature is entirely uncoupled by this assumption, we may never get around to solving for it here and may ask you to wait until you take a course on heat transfer. 4.6 Boundary Conditions for the Basic Equations 245 In most cases the wall is fixed; therefore, the proper inviscid flow condition is V„ = 0 (4.81) There is no condition whatever on the tangential velocity component at the wall in inviscid flow. The tangential velocity will he part of the solution to an inviscid flow analysis (see Chap. 8). EXAMPLE 4.6 For steady incompressible laminar flow through a long tube, the velocity distribution is given by V, = [7^1 - Vr = Vg = 0 where U is the maximum, or centerline, velocity and R is the tube radius. If the wall tem¬ perature is constant at T„ and the temperature T = T(r) only, find T{r) for this flow. Solution With T = T(r), Eq. (4.75) reduces for steady flow to PCpVr dT k d / dT dr r dr \ dr m( f dr (1) But since “ 0 for this flow, the convective term on the left vanishes. Introduce into Eq. (1) to obtain dT r dr \ dr -m( (dv^^ dr. (2) Multiply through by rik and integrate once: ’’ dr kR Divide through by r and integrate once again: C, pU^r^ T = - Y + Clin r + C2 4kR‘' (3) (4) Now we are in position to apply our boundary conditions to evaluate Ci and C2. First, since the logarithm of zero is —00, the temperature at r = 0 will be infinite unless Cl = 0 (5) Thus, we eliminate the possibility of a logarithmic singularity. The same thing will happen if we apply the symmetry condition dTIdr = 0 at r = 0 to Eq. (3). The constant C2 is then found by the wall-temperature condition at r = jf: 246 Chapter 4 Differential Relations for Fluid Flow or C2 = (6) 4k The correct solution is thus which is a fourth-order parabolic distrihution with a maximum value Tg = + fj,U^/{4k) at the centerline. 4.7 The Stream Function We have seen in Sec. 4.6 that even if the temperature is uncoupled from our system of equations of motion, we must solve the continuity and momentum equations simultaneously for pressure and velocity. The stream function %[) is a clever device that allows us to satisfy the continuity equation and then solve the momentum equation directly for the single variable ip. Lines of constant xp are streamlines of the flow. The stream function idea works only if the continuity equation (4.56) can be reduced to two terms. In general, we have four terms: Cartesian: Cylindrical: dp d d d — + — (pu) + — ipv) + — ipw) = 0 dt dx dy dz dp I d Id d ^ + 7 ^ ('•pn) + 7 77 (P^s) + 7“ = 0 dt r dr r d& dz (4.82fl) (4.82P) First, let us eliminate unsteady flow, which is a peculiar and unrealistic application of the stream function idea. Reduce either of Eqs. (4.82) to any two terms. The most common application is incompressible flow in the xy plane: du dx (4.83) This equation is satisfied identically if a function tpix, y) is defined such that Eq. (4.83) becomes dx \ dy dy dx = 0 (4.84) Comparison of (4.83) and (4.84) shows that this new function tp must be defined such that dip u = - dy (4.85) or 4.7 The Stream Function 247 Geometric Interpretation of il> Is this legitimate? Yes, it is just a mathematical trick of replacing two variables (m and v) by a single higher-order function The vorticity" or curl V is an interesting function: curlV = — where vV d^lh d^lb dx^ dy^ (4.86) Thus, if we take the curl of the momentum equation (4.74) and utilize Eq. (4.86), we obtain a single equation for ip for incompressible flow: dip d . dip d . . . (V>) (VV) = I^V2(VV) (4.87) dy dx dx dy where v = flip is the kinematic viscosity. This is partly a victory and partly a defeat: Eq. (4.87) is scalar and has only one variable, ip, but it now contains fourth-order derivatives and probably will require computer analysis. There will be four boundary conditions required on ip. Eor example, for the flow of a uniform stream in the X direction past a solid body, the four conditions would be dip dip At infinity: — = U^o — = 0 dy dx dip dip At the body: — = — = 0 (4.88) dy dx Many examples of numerical solution of Eqs. (4.87) and (4.88) are given in Ref. 1. One important application is inviscid, incompressible, irrotational flow^ in the xy plane, where curl V = 0. Equations (4.86) and (4.87) reduce to W^ip d^lp d^lp — — H - — = 0 T 2 3 2 dx dy (4.89) This is the second-order Laplace equation (Chap. 8), for which many solutions and analytical techniques are known. Also, boundary conditions like Eq. (4.88) reduce to At infinity: t/’ = + const (4.90) At the body: Ip = const It is well within our capability to find some useful solutions to Eqs. (4.89) and (4.90), which we shall do in Chap. 8. The fancy mathematics above would serve alone to make the stream function immor¬ tal and always useful to engineers. Even better, though, ip has a beautiful geometric interpretation: Lines of constant ip are streamlines of the flow. This can be shown as follows: From Eq. (1.41) the definition of a streamline in two-dimensional flow is dx _ dy u V or udy — vdx = 0 streamline (4.91) "See Section 4.8. "See Section 4.8. 248 Chapter 4 Differential Relations for Fluid Flow Introducing the stream function from Eq. (4.85), we have dlb dip — dx + — dy = 0 = dip (4.92) dx dy Thus the change in ip is zero along a streamline, or ip = const along a streamline (4.93) Having found a given solution ip(x, y), we can plot lines of constant ip to give the streamlines of the flow. There is also a physical interpretation that relates ip to volume flow. From Fig. 4.8, we can compute the volume flow dQ through an element ds of control surface of unit depth: dQ = • n) dA / dip . dlp\ V dy ^ dx ) V ds ds(l) dip dip = — dx H - dy = dip (4.94) dx dy Thus the change in ip across the element is numerically equal to the volume flow through the element. The volume flow between any two streamlines in the flow field is equal to the change in stream function between those streamlines: Q\^i r2 (V • n) dA h r2 dip = ip2- ipi P (4.95) Further, the direction of the flow can be ascertained by noting whether ip increases or decreases. As sketched in Fig. 4.9, the flow is to the right if lp^ is greater than ip^, where the subscripts stand for upper and lower, as before; otherwise the flow is to the left. Both the stream function and the velocity potential were invented by the French mathematician Joseph Louis Lagrange and published in his treatise on fluid mechanics in 1781. Fig. 4.8 Geometric interpretation of stream function: volume flow through a differential portion of a control surface. 4.7 The Stream Function 249 Fig. 4.9 Sign convention for flow in terms of change in stream function: (a) flow to the right if xpy is greater; (b) flow to the left if ipi is greater. EXAMPLE 4.7 If a stream function exists for the velocity field of Example 4.5 u = a(x^ ~ y^) V = —2axy w = 0 find it, plot it, and interpret it. Solution • Assumptions: Incompressible, two-dimensional flow. • Approach: Use the definition of stream function derivatives, Eqs. (4.85), to find lp{x, y). • Solution step 1: Note that this velocity distribution was also examined in Example 4.3. It satisfies continuity, Eq. (4.83), but let’s check that; otherwise ip will not exist: du dV d j ^ d 1 - = — [a(x “ y )] H - (—2axy) = 2ax -I- { — 2ax) = 0 checks dx dy dx dy Thus we are certain that a stream function exists. • Solution step 2: To find ip, write out Eqs. (4.85) and integrate: dip , u = — = ax" — ay (1) dy dip V = = -2axy (2) dx and work from either one toward the other. Integrate (1) partially Ip = ax^y - ^ -f fix) (3) Differentiate (3) with respect to x and compare with (2) dip — = 2axy + fix) = 2axy (4) dx Therefore /'(x) = 0, or/ = constant. The complete stream function is thus found: Ip = a(^x^y - ^ + C Ans. (5) 250 Chapter 4 Differential Relations for Fluid Flow To plot this, set C = 0 for convenience and plot the function y 3 31/; a (6) for constant values of ip. The result is shown in Fig. E4.7fl to he six 60° wedges of circulat¬ ing motion, each with identical flow patterns except for the arrows. Once the streamlines are labeled, the flow directions follow from the sign convention of Fig. 4.9. Flow can the flow be interpreted? Since there is slip along all streamlines, no streamline can truly represent a solid surface in a viscous flow. Flowever, the flow could represent the impingement of three incoming streams at 60, 180, and 300°. This would be a rather unrealistic yet exact solution to the Navier-Stokes equations, as we showed in Example 4.5. E4.7a Flow around a rounded 60° comer Incoming stream impinging against a 120° corner By allowing the flow to slip as a frictionless approximation, we could let any given streamline be a body shape. Some examples are shown in Fig. E4.1b. 4.7 The Stream Function 251 A stream function also exists in a variety of other physical situations where only two coordinates are needed to define the flow. Three examples are illustrated here. Steady Plane Compressible Flow Suppose now that the density is variable but that w = 0, so that the flow is in the xy plane. Then the equation of continuity becomes ^ (pu) + ^ ipv) = 0 (4.96) ox ay We see that this is in exactly the same form as Eq. (4.84). Therefore a compressible flow stream function can be defined such that pu dip dip — pv = - dy dx (4.97) Again lines of constant ip are streamlines of the flow, but the change in ip is now equal to the mass flow, not the volume flow: or dm = p(V • n) dA r2 p(y ■ n) dA dip 'ipi - 'ipi (4.98) The sign convention on flow direction is the same as in Fig. 4.9. This particular stream function combines density with velocity and must be substituted into not only momen¬ tum but also the energy and state relations (4.58) and (4.59) with pressure and tem¬ perature as companion variables. Thus the compressible stream function is not a great victory, and further assumptions must be made to effect an analytical solution to a typical problem (see, for instance. Ref. 5, Chap. 7). Incompressible Plane Flow in Polar Coordinates Suppose that the important coordinates are r and 9, with = 0, and that the density is constant. Then Eq. (4.82b) reduces to la la --(m,) + - — (t;^) = 0 (4.99) r dr r dO After multiplying through by r, we see that this is the analogous form of Eq. (4.84): ±(dlp\ ^ arVa6i/ dO (4.100) By comparison of (4.99) and (4.100) we deduce the form of the incompressible polar coordinate stream function: V, 1 dip r de (4.101) Once again lines of constant ip are streamlines, and the change in ip is the volume flow Qi^2 ~ '4^2 ~ 'tpi- The sign convention is the same as in Fig. 4.9. This type of stream function is very useful in analyzing flows with cylinders, vortices, sources, and sinks (Chap. 8). 252 Chapter 4 Differential Relations for Fluid Flow Incompressible Axisymmetric Flow As a final example, suppose that the flow is three-dimensional {v,., v^) but with no circumferential variations, Vq = dld9 = 0 (see Fig. 4.2 for definition of coordinates). Such a flow is termed axisymmetric, and the flow pattern is the same when viewed on any meridional plane through the axis of revolution z- For incompressible flow, Eq. (4.82fi) becomes (fD,) + f (D,) = 0 (4.102) r dr dz This doesn’t seem to work: Can’t we get rid of the one r outside? But when we realize that r and z are independent coordinates, Eq. (4.102) can be rewritten as ^ (rv,) + ^ (rv,) = 0 or dz By analogy with Eq. (4.84), this has the form dr\ dz J dz\dr J (4.103) (4.104) By comparing (4.103) and (4.104), we deduce the form of an incompressible axisym¬ metric stream function ipir, z) I dip I dip r dz r dr (4.105) Here again lines of constant ip are streamlines, but there is a factor (27r) in the volume flow: Qi-^2 = 27r('02 — ipi). The sign convention on flow is the same as in Eig. 4.9. EXAMPLE 4.8 Investigate the stream function in polar coordinates Ip = (/ sin 6 ( r — R- (1) where U and R are constants, a velocity and a length, respectively. Plot the streamlines. What does the flow represent? Is it a realistic solution to the basic equations? Solution The streamlines are lines of constant ip, which has units of square meters per second. Note that ipt(UR) is dimensionless. Rewrite Eq. (1) in dimensionless form r = sm 0 u - 11 = ^ (2) UR \' r]J ' R Of particular interest is the special line ip = 0. From Eq. (1) or (2) this occurs when (a) 9 = 0 or 180° and (b) r = R. Case (a) is the x axis, and case (b) is a circle of radius R, both of which are plotted in Fig. E4.8. 4.8 Vorticity and Irrotationality 253 4.8 Vorticity and Irrotationality For any other nonzero value of ip it is easiest to pick a value of r and solve for 0: sin 9 ip/iUR) r/R - R/r (3) In general, there will be two solutions for 6 because of the symmetry about the y axis. For example, take ip/iUR) = +1.0: E4.8 Streamlines converge, at origin UR = +l 1 2 0 2 -1 Guess HR 3.0 2.5 2.0 1.8 1.7 1.618 Compute 6 22° 28° 42° 53° 64° SO O 0 158° 152° 138° 127° 116° This line is plotted in Fig. E4.8 and passes over the circle r = R.Be careful, though, because there is a second curve for lp/(UR) = + 1.0 for small r < R below the x axis: Guess HR 0.618 0.6 0.5 0.4 0.3 0.2 0.1 Compute 6 -90° -70° -110° 1 1 -28° -152° -19° -161° -12° -168° -6° -174° This second curve plots as a closed curve inside the circle r = R. There is a singularity of infi¬ nite velocity and indeterminate flow direction at the origin. Figure E4.8 shows the full pattern. The given stream function, Eq. (1), is an exact and classic solution to the momentum equation (4.38) for frictionless flow. Outside the circle r = R it represents two-dimensional inviscid flow of a uniform stream past a circular cylinder (Sec. 8.4). Inside the circle it represents a rather unrealistic trapped circulating motion of what is called a line doublet. The assumption of zero fluid angular velocity, or irrotationality, is a very useful simplification. Here we show that angular velocity is associated with the curl of the local velocity vector. The differential relations for deformation of a fluid element can be derived by examining Fig. 4.10. Two fluid lines AB and BC, initially perpendicular at time f, 254 Chapter 4 Differential Relations for Fluid Flow dy ' Fig. 4.10 Angular velocity and strain rate of two fluid lines deforming in the xy plane. 0 X move and deform so that at t + dt they have slightly different lengths A'B' and B'C and are slightly off the perpendicular by angles da and dfd. Such deformation occurs kinematically because A, B, and C have slightly different velocities when the velocity field V has spatial gradients. All these differential changes in the motion of A, B, and C are noted in Fig. 4.10. We define the angular velocity ui^ about the z axis as the average rate of counter¬ clockwise turning of the two lines: 1 fda uj=- \ — " 2\dt (4.106) But from Fig. 4.10, da and d(3 are each directly related to velocity derivatives in the limit of small dt: da = d(3 = lim lim dt^O tan tan {dv/dx) dx dt dx + (duldx) dx dt (du/dy) dy dt dy + (dv/dy) dy dt Combining Eqs. (4.106) and (4.107) gives the desired result: (4.107) 4.9 Frictionless Irrotational Flows 255 In exactly similar manner we determine the other two rates: I f dw OOr = 2\dy dV dz ^y=2 \ f du dz dw dx (4.109) The vector u = + joJy + is thus one-half the curl of the velocity vector i j k 1 1 a a d cu = - (curl V) = - 2 2 dx dy dz u V w (4.110) Since the factor of \| is annoying, many workers prefer to use a vector twice as large, called the vorticity. C=2aJ = curlV (4.111) Many flows have negligible or zero vorticity and are called irrotational: curlV^O (4.112) The next section expands on this idea. Such flows can be incompressible or compress¬ ible, steady or unsteady. We may also note that Fig. 4.10 demonstrates the shear strain rate of the element, which is defined as the rate of closure of the initially perpendicular lines: da dd dv du — -f — = — -f — dt dt dx dy (4.113) When multiplied by viscosity /i, this equals the shear stress in a newtonian fluid, as discussed earlier in Eqs. (4.37). Appendix D lists strain rate and vorticity compo¬ nents in cylindrical coordinates. 4.9 Frictionless Irrotational Flows When a flow is both frictionless and irrotational, pleasant things happen. First, the momentum equation (4.38) reduces to Euler’s equation: p — = pg-^p (4.114) dt Second, there is a great simplification in the acceleration term. Recall from Sec. 4.1 that acceleration has two terms: dV dY — = — + (V • V)V (4.2) dt dt A beautiful vector identity exists for the second term : (V • V)V = V(\|y‘) -f C X V (4.115) where ^ = curl V from Eq. (4.111) is the fluid vorticity. 256 Chapter 4 Differential Relations for Fluid Flow Now combine (4.1 14) and (4.1 15), divide by p, and rearrange on the left-hand side. Dot-product the entire equation into an arbitrary vector displacement dr: d_ dt V 1 V ^ xV-l-^Vp — g dr = 0 Nothing works right unless we can get rid of the third term. We want (C X V) • (dr) - 0 This will be true under various conditions: (4.116) (4.117) V is zero; trivial, no flow (hydrostatics). ^ is zero; irrotational flow. (fr is perpendicular to ^ X V; this is rather specialized and rare. dr is parallel to V; we integrate along a streamline (see Sec. 3.5). Condition 4 is the common assumption. If we integrate along a streamline in friction¬ less compressible flow and take, for convenience, g = — gk, Eq. (4.116) reduces to av / 1 tip dr + d(-v^] + — + gdz = 0 (4.118) dt \2jp Except for the first term, these are exact differentials. Integrate between any two points 1 and 2 along the streamline: dV dp I , , — ds+ — + -(Vl- y?) + g(z2 - Zi) = 0 (4.119) dt Jj P 2 where ds is the arc length along the streamline. Equation (4.119) is Bernoulli’s equa¬ tion for frictionless unsteady flow along a streamline and is identical to Eq. (3.53). For incompressible steady flow, it reduces to P P 1 2 y^ + gz constant along streamline (4.120) The constant may vary from streamline to streamline unless the flow is also irrota¬ tional (assumption 2). For irrotational flow (^ = 0, the offending term Eq. (4.117) vanishes regardless of the direction of dr, and Eq. (4.120) then holds all over the flow field with the same constant. Velocity Potential Irrotationality gives rise to a scalar function (p similar and complementary to the stream function tp. From a theorem in vector analysis , a vector with zero curl must be the gradient of a scalar function If V X V ^ 0 then Y = Vcp (4.121) where (p = (p(x, y, z, t) is called the velocity potential function. Knowledge of (p thus immediately gives the velocity components dtp dtp dtp U = - V = - w = — d.x dy dz (4.122) Lines of constant tp are called the potential lines of the flow. 4.9 Frictionless Irrotational Flows 257 Orthogonality of Streamlines and Potential Lines Note that (j), unlike the stream function, is fully three-dimensional and is not limited to two coordinates. It reduces a velocity problem with three unknowns u, v, and w to a single unknown potential 0; many examples are given in Chap. 8. The velocity potential also simplifies the unsteady Bernoulli equation (4.118) because if (p exists, we obtain av a „ /a0\ dr = -{V(P)-dr = d[^) (4.123) at at \ at / along any arbitrary direction. Equation (4.118) then becomes a relation between <p and p\ dd) (dp 1 , . h - 1 — + gz = const (4.124) at } p 2 This is the unsteady irrotational Bernoulli equation. It is very important in the analy¬ sis of accelerating flow fields (see Refs. 10 and 15), but the only application in this text will be in Sec. 9.3 for steady flow. If a flow is both irrotational and described by only two coordinates, xp and (p both exist, and the streamlines and potential lines are everywhere mutually perpendicular except at a stagnation point. For example, for incompressible flow in the xy plane, we would have dtp d(p dy dx dxp d(p dx dy (4.125) (4.126) Can you tell by inspection not only that these relations imply orthogonality but also that (p and xp satisfy Laplace’s equation?'^ A line of constant (p would be such that the change in <p is zero: d(p d(p d(p = — dx H - dy = 0 = u dx + xj dy dx dy Solving, we have / dy\ u 1 const ^ const (4.127) (4.128) Equation (4.128) is the mathematical condition that lines of constant (p and xp be mutually orthogonal. It may not be true at a stagnation point, where both u and v are zero, so their ratio in Eq. (4.128) is indeterminate. '^Equations (4.125) and (4.126) are called the Cauchy-Riemcmn equations and are studied in complex variable theory. 258 Chapter 4 Differential Relations for Fluid Flow Generation of Rotationality' This is the second time we have discussed Bernoulli’s equation under different circumstances (the first was in Sec. 3.5). Such reinforcement is useful, since this is probably the most widely used equation in fluid mechanics. It requires frictionless flow with no shaft work or heat transfer between sections 1 and 2. The flow may or may not be irrotational, the former being an easier condition, allowing a universal Bernoulli constant. The only remaining question is this: When is a flow irrotational? In other words, when does a flow have negligible angular velocity? The exact analysis of fluid rota- tionality under arbitrary conditions is a topic for advanced study (for example. Ref. 10, Sec. 8.5; Ref. 9, Sec. 5.2; and Ref. 5, Sec. 2.10). We shall simply state those results here without proof. A fluid flow that is initially irrotational may become rotational if 1 . There are significant viscous forces induced by jets, wakes, or solid boundaries. In this case Bernoulli’s equation will not be valid in such viscous regions. There are entropy gradients caused by curved shock waves (see Fig. 4.11b). There are density gradients caused by stratification (uneven heating) rather than by pressure gradients. There are significant noninertial effects such as the earth’s rotation (the Coriolis acceleration). In cases 2 to 4, Bernoulli’s equation still holds along a streamline if friction is negligible. We shall not study cases 3 and 4 in this book. Case 2 will be treated briefly in Chap. 9 on gas dynamics. Primarily we are concerned with case 1, where rotation is induced by viscous stresses. This occurs near solid surfaces, where the no-slip condition creates a boundary layer through which the stream velocity drops to zero, and in jets and wakes, where streams of different velocities meet in a region of high shear. Internal flows, such as pipes and ducts, are mostly viscous, and the wall layers grow to meet in the core of the duct. Bernoulli’s equation does not hold in such flows unless it is modified for viscous losses. External flows, such as a body immersed in a stream, are partly viscous and partly inviscid, the two regions being patched together at the edge of the shear layer or boundary layer. Two examples are shown in Fig. 4.11. Figure 4.11a shows a low- speed subsonic flow past a body. The approach stream is irrotational; that is, the curl of a constant is zero, but viscous stresses create a rotational shear layer beside and downstream of the body. Generally speaking (see Chap. 7), the shear layer is laminar, or smooth, near the front of the body and turbulent, or disorderly, toward the rear. A separated, or deadwater, region usually occurs near the trailing edge, followed by an unsteady turbulent wake extending far downstream. Some sort of laminar or turbulent viscous theory must be applied to these viscous regions; they are then patched onto the outer flow, which is frictionless and irrotational. If the stream Mach number is less than about 0.3, we can combine Eq. (4.122) with the incompressible continuity equation (4.73): V • V = V • (V0) = 0 ’"'This section may be omitted without loss of continuity. 4.9 Frictionless Irrotational Flows 259 Fig. 4.11 Typical flow patterns illustrating viscous regions patched onto nearly frictionless regions: (a) low subsonic flow past a body ({/ ^ a); frictionless, irrotational potential flow outside the boundary layer (Bernoulli and Laplace equations valid); (b) supersonic flow past a body (U > a)\ frictionless, rotational flow outside the boundary layer (Bernoulli equation valid, potential flow invalid). Viscous regions where Bernoulli's equation fails: Uniform approach flow (irrotational) Laminar Turbulent boundary boundary .Senaratp.d (a) Curved shock wave introduces rotationality Viscous regions where Bernoulli is invalid: Laminar Turbulent boundary boundary Slight layer layer separated Wake (b) or (4.129) This is Laplace’s equation in three dimensions, there being no restraint on the number of coordinates in potential flow. A great deal of Chap. 8 will be concerned with solv¬ ing Eq. (4.129) for practical engineering problems; it holds in the entire region of Fig. 4.11fl outside the shear layer. Figure 4.11b shows a supersonic flow past a round-nosed body. A curved shock wave generally forms in front, and the flow downstream is rotational due to entropy gradients (case 2). We can use Euler’s equation (4.114) in this frictionless region but not potential theory. The shear layers have the same general character as in Fig. 4.1 la except that the separation zone is slight or often absent and the wake is usually thin¬ ner. Theory of separated flow is presently qualitative, but we can make quantitative estimates of laminar and turbulent boundary layers and wakes. 260 Chapter 4 Differential Relations for Fluid Flow EXAMPLE 4.9 If a velocity potential exists for the velocity field of Example 4.5 u = — y^) V = —2axy w = 0 hnd it, plot it, and compare with Example 4.7. Solution Since w = 0, the curl of V has only one z component, and we must show that it is zero: dV du d 5 9 (V X V). = 2lj, = - = — (— 2axy) - {ax~ — ay^) dy dx dy dx = —2ay + 2ay = 0 checks The flow is indeed irrotational. A velocity potential exists. To hnd (p{x, y), set 30 2 2 u = — = ax — ay dx 30 V = — = —2axy 3y Integrate (1) 0 (a) Under what conditions, if any, on (A, B, C, D) can this function be a steady plane-flow velocity potential? (b) If you find a (p(x, y) to satisfy part (a), also find the associated stream function tp{x, y), if any, for this flow. P4.72 Water flows through a two-dimensional narrowing wedge at 9.96 gal/min per meter of width into the paper (Fig. P4.72). If this inward flow is purely radial, find an expression, in SI units, for (a) the stream function and (b) the velocity potential of the flow. Assume one¬ dimensional flow. The included angle of the wedge is 45°. P4.73 A CFD model of steady two-dimensional incompressible flow has printed out the values of velocity potential (p(x, y). Problems 277 in mVs, at each of the four comers of a small 10-cm-by- 10-cm cell, as shown in Fig. P4.73. Use these numbers to estimate the resultant velocity in the center of the cell and its angle a. with respect to the x axis. (j) = 4.8338 m^/s 5.0610 P4.74 Consider the two-dimensional incompressible polar- coordinate velocity potential (p = Br cos 9 + B L 6 where S is a constant and L is a constant length scale. (a) What are the dimensions of fi? (b) Locate the only stagnation point in this flow field, (c) Prove that a stream function exists and then find the function 'lp(r, 9). P4.75 Given the following steady axisymmetric stream function: where B and R are constants in the upper half plane, (b) Prove that a stream function exists, and then find ipix, y), using the hint that f dxl(c^ -I- x^) = (l/a)tan~(x4!). P4.77 Outside an inner, intense-activity circle of radius R, a trop¬ ical storm can be simulated by a polar-coordinate velocity potential (p{r, 9) = UJi 9, where 11^ is the wind velocity at radius R. (a) Determine the velocity components outside r = R. (b) If, at = 25 mi, the velocity is 100 mi/h and the pressure 99 kPa, calculate the velocity and pressure at r = 100 mi. P4.78 An incompressible, irrotational, two-dimensional flow has the following stream function in polar coordinates: xp = A r" sin (nff) where A and n are constants. Find an expression for the velocity potential of this flow. Incompressible viscous flows P4.79 Study the combined effect of the two viscous flows in Fig. 4.12. That is, find u(y) when the upper plate moves at speed V and there is also a constant pressure gradient (dpldx). Is superposition possible? If so, explain why. Plot representative velocity profiles for (a) zero, (b) positive, and (c) negative pressure gradients for the same upper-wall speed V. P4.80 Oil, of density p and viscosity p, drains steadily down the side of a vertical plate, as in Fig. P4.80. After a develop¬ ment region near the top of the plate, the oil film will become independent of z and of constant thickness 6. Assume that w = w{x) only and that the atmosphere offers no shear resistance to the surface of the film, (a) Solve the Navier-Stokes equation for w(x), and sketch its approxi¬ mate shape, (b) Suppose that film thickness 6 and the slope of the velocity profile at the wall [dwldx]^^^ are measured with a laser-Doppler anemometer (Chap. 6). Find an expression for oil viscosity /i as a function of (p, 6, g, valid in the region 0 ^ r ?? and 0 ^ z ^ L. (a) What are the dimensions of the constant B7 (b) Show whether this flow possesses a velocity potential, and, if so, find it. (c) What might this flow represent? Hint: Examine the axial velocity v^. P4.76 A two-dimensional incompressible flow has the velocity potential (p = K(x^ — •^) + C ln(x^ + y^) where K and C are constants. In this discussion, avoid the origin, which is a singularity (infinite velocity), (ti) Find the sole stagnation point of this flow, which is somewhere P4.80 Plate Oil film Air X 278 Chapter 4 Differential Relations for Fluid Flow P4.81 Modify the analysis of Fig. 4.13 to hnd the velocity Ug P4.84 when the inner cylinder is hxed and the outer cylinder rotates at angular velocity fio. May this solution he added to Eq. (4.140) to represent the flow caused when both inner and outer cylinders rotate? Explain your conclusion. P4.82 A solid circular cylinder of radius R rotates at angular velocity fl in a viscous incompressihle fluid that is at rest far from the cylinder, as in Eig. P4.82. Make simplifying assumptions and derive the governing differential equation and boundary conditions for the velocity field Vg in the fluid. Do not solve unless you are obsessed with this prob¬ lem. What is the steady-state flow field for this problem? Vg(r,0,t) P4.82 P4.83 The flow pattern in bearing lubrication can be illustrated by Eig. P4.83, where a viscous oil (p, fi) is forced into the gap h(x) between a fixed slipper block and a wall moving at velocity U. If the gap is thin, h 12. 3. D. Zwillinger, CRC Standard Mathematical Tables and Formulae, 32d ed., CRC Press Inc., Cleveland, Ohio, 2011. 4. Ft. Schlichting and K. Gersten, Boundary Layer Theory, 8th 13. ed.. Springer, New York, 2000. 5. F. M. White, Viscous Fluid Flow, 3d ed., McGraw-Hill, New York, 2005. 14. 6. E. B. Tadmor, R. E. Miller, and R. S. Elliott, Continuum Mechanics and Thermodynamics, Cambridge University 15. Press, New York, 2012. 16. 7. J. P. Holman, Heat Transfer, 10th ed., McGraw-Hill, New York, 2009. 17. 8. W. M. Kays and M. E. Crawford, Convective Heat and Mass Transfer, 4th ed., McGraw-Hill, New York, 2004. 9. G. K. Batchelor, An Introduction to Fluid Dynamics, 18. Cambridge University Press, Cambridge, England, 1967. L. Prandtl and O. G. Tietjens, Fundamentals of Hydro- and Aeromechanics, Dover, New York, 1957. D. Eleisch, A Student’s Guide to Vectors and Tensors, Cambridge University Press, New York, 2011. O. Gonzalez and A. M. Stuart, A First Course in Contin¬ uum Mechanics, Cambridge University Press, New York, 2008. D. A. Danielson, Vectors and Tensors in Engineering and Physics, 2d ed., Westview (Perseus) Press, Boulder, CO, 2003. R. I. Tanner, Engineering Rheology, 2d ed., Oxford University Press, New York, 2000. H. Lamb, Hydrodynamics, 6th ed., Dover, New York, 1945. J. P. Davin, Tribology for Engineers: A Practical Guide, Woodhead Publishing, Philadelphia, 2011. G. 1. Taylor, “Stability of a Viscous Liquid Contained between Two Rotating Cylinders,” Philos. Trans. Roy. Soc. London Ser. A, vol. 223, 1923, pp. 289-343. E. L. Koschmieder, “Turbulent Taylor Vortex Elow,” J. Fluid Meek, vol. 93, pt. 3, 1979, pp. 515-527. References 283 19. M. T. Nair, T. K. Sengupta, and U. S. Chauhan, “Flow Past Rotating Cylinders at High Reynolds Numbers Using Higher Order Upwind Scheme,” Computers and Fluids, vol. 27, no. 1, 1998, pp. 47-70. 20. M. Constanceau and C. Menard, “Influence of Rotation on the Near-Wake Development behind an Impulsively Started Circular Cylinder,” J. Fluid Mechanics, vol. 1258, 1985, pp. 399-446. 21. J-Y. Hwang, K-S. Yang, and K. Bremhorst, “Direct Numerical Simulation of Turbulent Flow Around a Rotating Circular Cylinder.” Fluids Engineering, vol. 129, Jan. 2007, pp. 40-47. A full-scale NASA parachute, which helped lower the vehicle Curiosity to the Mars surface in 2012, was tested in the world’s largest wind tunnel, at NASA Ames Research Center, Moffett Field, California. It is the largest disc-gap-band parachute ever built, with a diameter of 51 feet. In the Mars atmosphere it will generate up to 65,000 Ibf of drag, which leads to a problem assignment in Chapter 7. [Image from NASA/JPL-Caltech.] 284 Chapter 5 Dimensional Analysis and Similarity 5.1 Introduction Motivation. In this chapter we discuss the planning, presentation, and interpretation of experimental data. We shall try to convince you that such data are best presented in dimensionless form. Experiments that might result in tables of output, or even multiple volumes of tables, might be reduced to a single set of curves — or even a single curve — when suitably nondimensionalized. The technique for doing this is dimensional analysis. It is also effective in theoretical studies. Chapter 3 presented large-scale control volume balances of mass, momentum, and energy, which led to global results: mass flow, force, torque, total work done, or heat transfer. Chapter 4 presented infinitesimal balances that led to the basic partial dif¬ ferential equations of fluid flow and some particular solutions for both inviscid and viscous (laminar) flow. These straight analytical techniques are limited to simple geometries and uniform boundary conditions. Only a fraction of engineering flow problems can be solved by direct analytical formulas. Most practical fluid flow problems are too complex, both geometrically and physi¬ cally, to be solved analytically. They must be tested by experiment or approximated by computational fluid dynamics (CFD) . These results are typically reported as experimental or numerical data points and smoothed curves. Such data have much more generality if they are expressed in compact, economic form. This is the motiva¬ tion for dimensional analysis. The technique is a mainstay of fluid mechanics and is also widely used in all engineering fields plus the physical, biological, medical, and social sciences. The present chapter shows how dimensional analysis improves the presentation of both data and theory. Basically, dimensional analysis is a method for reducing the number and complexity of experimental variables that affect a given physical phenomenon, by using a sort of com¬ pacting technique. If a phenomenon depends on n dimensional variables, dimensional analysis will reduce the problem to only k dimensionless variables, where the reduction 285 286 Chapter 5 Dimensional Analysis and Similarity n — k = 1, 2, 3, or 4, depending on the problem complexity. Generally n — k equals the number of different dimensions (sometimes called basic or primary or fundamental dimensions) that govern the problem. In fluid mechanics, the four basic dimensions are usually taken to be mass M, length L, time T, and temperature 0, or an MLT& system for short. Alternatively, one uses an FLT& system, with force F replacing mass. Although its purpose is to reduce variables and group them in dimensionless form, dimensional analysis has several side benefits. The first is enormous savings in time and money. Suppose one knew that the force F on a particular body shape immersed in a stream of fluid depended only on the body length L, stream velocity V, fluid density p, and fluid viscosity /i; that is. F = f{L, V, p, p) (5.1) Suppose further that the geometry and flow conditions are so complicated that our inte¬ gral theories (Chap. 3) and differential equations (Chap. 4) fail to yield the solution for the force. Then we must find the function f{L, V, p, p) experimentally or numerically. Generally speaking, it takes about 10 points to define a curve. To find the effect of body length in Eq. (5.1), we have to mn the experiment for 10 lengths L. For each L we need 10 values of V, 10 values of p, and 10 values of p, making a grand total of 10“, or 10,000, experiments. At $100 per experiment — well, you see what we are getting into. However, with dimensional analysis, we can immediately reduce Eq. (5.1) to the equivalent form (5.2) or That is, the dimensionless force coefficient FI{pV^l}) is a function only of the dimen¬ sionless Reynolds number pVUp. We shall learn exactly how to make this reduction in Secs. 5.2 and 5.3. Equation (5.2) will be useful in Chap. 7. Note that Eq. (5.2) is just an example, not the full story, of forces caused by fluid flows. Some fluid forces have a very weak or negligible Reynolds number dependence in wide regions (Fig. 5.3fl). Other groups may also be important. The force coefficient may depend, in high-speed gas flow, on the Mach number, Ma = V/a, where a is the speed of sound. In free-surface flows, such as ship drag, Cf may depend upon Froude number, Fr = V^/{gL), where g is the acceleration of gravity. In turbulent flow, force may depend upon the roughness ratio, dL, where e is the roughness height of the surface. The function g is different mathematically from the original function f but it con¬ tains all the same information. Nothing is lost in a dimensional analysis. And think of the savings: We can establish g by running the experiment for only 10 values of the single variable called the Reynolds number. We do not have to vary L, V, p, or p separately but only the grouping pVL/p. This we do merely by varying velocity V in, say, a wind tunnel or drop test or water channel, and there is no need to build 10 different bodies or find 100 different fluids with 10 densities and 10 viscosities. The cost is now about $1000, maybe less. A second side benefit of dimensional analysis is that it helps our thinking and plan¬ ning for an experiment or theory. It suggests dimensionless ways of writing equations before we spend money on computer analysis to find solutions. It suggests variables that can be discarded; sometimes dimensional analysis will immediately reject 5.1 Introduction 287 variables, and at other times it groups them off to the side, where a few simple tests will show them to be unimportant. Finally, dimensional analysis will often give a great deal of insight into the form of the physical relationship we are trying to study. A third benefit is that dimensional analysis provides scaling laws that can convert data from a cheap, small model to design information for an expensive, large proto¬ type. We do not build a million-dollar airplane and see whether it has enough lift force. We measure the lift on a small model and use a scaling law to predict the lift on the full-scale prototype airplane. There are rules we shall explain for finding scal¬ ing laws. When the scaling law is valid, we say that a condition of similarity exists between the model and the prototype. In the simple case of Eq. (5.1), similarity is achieved if the Reynolds number is the same for the model and prototype because the function g then requires the force coefficient to be the same also: IfRe„, = Re^ then = Cpp (5.3) where subscripts m and p mean model and prototype, respectively. From the definition of force coefficient, this means that for data taken where PpVpLp/pp = Equation (5.4) is a scaling law: If you measure the model force at the model Reynolds number, the prototype force at the same Reynolds number equals the model force times the density ratio times the veloc¬ ity ratio squared times the length ratio squared. We shall give more examples later. Do you understand these introductory explanations? Be careful; learning dimensional analysis is like learning to play tennis: There are levels of the game. We can establish some ground rules and do some fairly good work in this brief chapter, but dimensional analysis in the broad view has many subtleties and nuances that only time, practice, and maturity enable you to master. Although dimensional analysis has a firm physical and mathematical foundation, considerable art and skill are needed to use it effectively. EXAMPLE 5.1 A copepod is a water cmstacean approximately 1 mm in diameter. We want to know the drag force on the copepod when it moves slowly in fresh water. A scale model 100 times larger is made and tested in glycerin at F = 30 cm/s. The measured drag on the model is 1.3 N. For similar conditions, what are the velocity and drag of the actual copepod in water? Assume that Eq. (5.2) applies and the temperature is 20°C. Solution • Property values: From Table A. 3, the densities and viscosities at 20°C are Water (prototype): Pp = 0.001 kg/(m-s) Pp - 998 kg/m^ Glycerin (model): ffm = 1-5 kg/(m-s) = 1263 kg/m^ • Assumptions: Equation (5.2) is appropriate and similarity is achieved; that is, the model and prototype have the same Reynolds number and, therefore, the same force coefficient. 288 Chapter 5 Dimensional Analysis and Similarity 5.2 The Principle of Dimensional Homogeneity • Approach: The length scales are = 100 mm and Lp = 1 mm. Calculate the Reynolds number and force coefficient of the model and set them equal to prototype values: _ Pny,nL,n _ ( 1263 kg/m^) (0.3 ui/s) (0. 1 m) _ (998 kg/m") Tp(0.001 m) 1.5kg/(m-s) ^ 0.001 kg/(m-s) Solve for Vj, = 0.0253 m/s = 2.53 cm/s Ans. In like manner, using the prototype velocity just found, equate the force coefficients: (1263 kg/m")(0.3 m/s)"(0.1 m)" (998 kg/m")(0.0253 m/s)"(0.001 m)" Solve for Fp = 7.3E-7 N Ans. ■ Comments: Assuming we modeled the Reynolds number correctly, the model test is a very good idea, as it would obviously be difficult to measure such a tiny copepod drag force. Historically, the first person to write extensively about units and dimensional rea¬ soning in physical relations was Euler in 1765. Euler’s ideas were far ahead of his time, as were those of Joseph Eourier, whose 1822 book Analytical Theory of Heat outlined what is now called the principle of dimensional homogeneity and even devel¬ oped some similarity rules for heat flow. There were no further significant advances until Lord Rayleigh’s book in 1877, Theory of Sound, which proposed a “method of dimensions” and gave several examples of dimensional analysis. The final break¬ through that established the method as we know it today is generally credited to E. Buckingham in 1914 , whose paper outlined what is now called the Buckingham Pi Theorem for describing dimensionless parameters (see Sec. 5.3). However, it is now known that a Frenchman, A. Vaschy, in 1892 and a Russian, D. Riabouchinsky, in 1911 had independently published papers reporting results equivalent to the pi theorem. Following Buckingham’s paper, P. W. Bridgman published a classic book in 1922 , outlining the general theory of dimensional analysis. Dimensional analysis is so valuable and subtle, with both skill and art involved, that it has spawned a wide variety of textbooks and treatises. The writer is aware of more than 30 books on the subject, of which his engineering favorites are listed here [3-10]. Dimensional analysis is not confined to fluid mechanics, or even to engineering. Specialized books have been published on the application of dimensional analysis to metrology , astrophysics , economics , chemistry , hydrology , medi¬ cations , clinical medicine , chemical processing pilot plants , social sciences , biomedical sciences , pharmacy , fractal geometry , and even the growth of plants . Clearly this is a subject well worth learning for many career paths. In making the remarkable jump from the five-variable Eq. (5.1) to the two-variable Eq. (5.2), we were exploiting a rule that is almost a self-evident axiom in physics. This rule, the principle of dimensional homogeneity (PDH), can be stated as follows: 5.2 The Principle of Dimensional Homogeneity 289 Variables and Constants If an equation truly expresses a proper relationship between variables in a physical process, it will be dimensionally homogeneous; that is, each of its additive terms will have the same dimensions. All the equations that are derived from the theory of mechanics are of this form. For example, consider the relation that expresses the displacement of a falling body: S = So+Vot + \gt^ (5.5) Each term in this equation is a displacement, or length, and has dimensions {L}. The equation is dimensionally homogeneous. Note also that any consistent set of units can be used to calculate a result. Consider Bernoulli’s equation for incompressible flow: F P + + gz = const (5.6) Each term, including the constant, has dimensions of velocity squared, or {L^T~^}. The equation is dimensionally homogeneous and gives proper results for any consis¬ tent set of units. Students count on dimensional homogeneity and use it to check themselves when they cannot quite remember an equation during an exam. Eor example, which is it: 5 = or S = i/f? (5.7) By checking the dimensions, we reject the second form and back up our faulty mem¬ ory. We are exploiting the principle of dimensional homogeneity, and this chapter simply exploits it further. Equations (5.5) and (5.6) also illustrate some other factors that often enter into a dimensional analysis: Dimensional variables are the quantities that actually vary during a given case and would be plotted against each other to show the data. In Eq. (5.5), they are S and t; in Eq. (5.6) they are p, V, and z- All have dimensions, and all can be nondimensionalized as a dimensional analysis technique. Dimensional constants may vary from case to case but are held constant during a given run. In Eq. (5.5) they are ^o, Vq, and g, and in Eq. (5.6) they are p, g, and C. They all have dimensions and conceivably could be nondimensional¬ ized, but they are normally used to help nondimensionalize the variables in the problem. Pure constants have no dimensions and never did. They arise from mathematical manipulations. In both Eqs. (5.5) and (5.6) they are j and the exponent 2, both of which came from an integration: ft dt = jP, fV dV = Other common dimensionless constants are tt and e. Also, the argument of any mathematical function, such as In, exp, cos, or Jg, is dimensionless. Angles and revolutions are dimensionless. The preferred unit for an angle is the radian, which makes it clear that an angle is a ratio. In like manner, a revolu¬ tion is Itt radians. 290 Chapter 5 Dimensional Analysis and Similarity Ambiguity: The Choice of Variables and Scaling Parameters Counting numbers are dimensionless. For example, if we triple the energy E to 3E, the coefficient 3 is dimensionless. Note that integration and differentiation of an equation may change the dimen¬ sions but not the homogeneity of the equation. For example, integrate or differenti¬ ate Eq. (5.5): S dt = Sot -t- ^Vot^ + ^gt^ (5.8fl) . dS ^ = Vo + g? {5M) dt In the integrated form (5.8fl) every term has dimensions of {LT}, while in the deriva¬ tive form (5.8fo) every term is a velocity {L7^}. Finally, some physical variables are naturally dimensionless by virtue of their defi¬ nition as ratios of dimensional quantities. Some examples are strain (change in length per unit length), Poisson’s ratio (ratio of transverse strain to longitudinal strain), and specific gravity (ratio of density to standard water density). The motive behind dimensional analysis is that any dimensionally homogeneous equation can be written in an entirely equivalent nondimensional form that is more compact. Usually there are multiple methods of presenting one’s dimensionless data or theory. Let us illustrate these concepts more thoroughly by using the falling-body relation (5.5) as an example. Equation (5.5) is familiar and simple, yet it illustrates most of the concepts of dimen¬ sional analysis. It contains hve terms {S, Sq, Vq, t, g), which we may divide, in our thinking, into variables and parameters. The variables are the things we wish to plot, the basic output of the experiment or theory: in this case, S versus t. The parameters are those quantities whose effect on the variables we wish to know: in this case Sq, Vq, and g. Almost any engineering study can be subdivided in this manner. To nondimensionalize our results, we need to know how many dimensions are contained among our variables and parameters: in this case, only two, length {L} and time {T}. Check each term to verify this: {A} = {Ao} = {L} {f} = {T} {Vo} = {LT-^} (gj = {LT-^} Among our parameters, we therefore select two to be scaling parameters (also called repeating variables), used to define dimensionless variables. What remains will be the “basic” parameter(s) whose effect we wish to show in our plot. These choices will not affect the content of our data, only the form of their presentation. Clearly there is ambiguity in these choices, something that often vexes the beginning experimenter. But the ambiguity is deliberate. Its purpose is to show a particular effect, and the choice is yours to make. Eor the falling-body problem, we select any two of the three parameters to be scal¬ ing parameters. Thus, we have three options. Let us discuss and display them in turn. I am indebted to Prof. Jacques Lewalle of Syracuse University for suggesting, outlining, and clari¬ fying this entire discussion. 5.2 The Principle of Dimensional Homogeneity 291 Option 1: Scaling parameters Sq a„d Vq: the effect of gravity g. First use the scaling parameters (Sq, Vq) to define dimensionless () displacement and time. There is only one suitable definition for each:^ (5.9) Substitute these variables into Eq. (5.5) and clean everything up until each term is dimensionless. The result is our first option: S = I + t + -at^ 2 a = g^Q (5.10) This result is shown plotted in Fig. 5.1a. There is a single dimensionless parameter a, which shows here the effect of gravity. It cannot show the direct effects of Sq and Vo, since these two are hidden in the ordinate and abscissa. We see that gravity increases the parabolic rate of fall for t > 0, but not the initial slope at t = 0. We would learn the same from falling-body data, and the plot, within experimental accu¬ racy, would look like Fig. 5.1a. Option 2: Scaling parameters Vq and 8- the effect of initial displacement Sq. Now use the new scaling parameters (Vq, g) to define dimensionless () displace¬ ment and time. Again there is only one suitable definition: ^g vl Vo (5.11) Substitute these variables into Eq. (5.5) and clean everything up again. The result is our second option: S = a + t + -t^ 2 a = gSp vl (5.12) This result is plotted in Fig. 5.1f>. The same single parameter a again appears and here shows the effect of initial displacement, which merely moves the curves upward without changing their shape. Option 3: Scaling parameters Sq and 8- the effect of initial speed Vq. Finally use the scaling parameters (^o, g) to define dimensionless () displace¬ ment and time. Again there is only one suitable definition: S = f Sq Substitute these variables into Eq. (5.5) and clean everything up as usual. The result is our third and final option: 1 9 1 Vo S = 1 + p = _ = (5.14) ,\l/2 (5.13) ^Make them proportional to S and t. Do not define dimensionless terms upside down: SJS or 5o/(VoO- The plots will look funny, users of your data will be confused, and your supervisor will be angry. It is not a good idea. 292 Chapter 5 Dimensional Analysis and Similarity So Vo (a) (b) Fig. 5.1 Three entirely equivalent dimensionless presentations of the falling-body problem, Eq. (5.5): the effect of (a) gravity, (b) initial displacement, and (c) initial velocity. All plots contain the same information. 0 12 3 f = ty/g/So ic) This final presentation is shown in Fig. 5.1c. Once again the parameter a appears, but we have redefined it upside down, /3 = 1/Vtt, so that our display parameter Vq is in the numerator and is linear. This is our free choice and simply improves the display. Figure 5.1c shows that initial velocity increases the falling displacement. Note that, in all three options, the same parameter a appears but has a different mean¬ ing: dimensionless gravity, initial displacement, and initial velocity. The graphs, which contain exactly the same information, change their appearance to reflect these differences. Whereas the original problem, Eq. (5.5), involved five quantities, the dimensionless presentations involve only three, having the form S' = fcn(f', a) a gSp vl (5.15) 5.2 The Principle of Dimensional Homogeneity 293 Selection of Scaling (Repeating) Variables Some Peculiar Engineering Equations The reduction 5 — 3 = 2 should equal the numher of fundamental dimensions involved in the problem {L, T). This idea led to the pi theorem (Sec. 5.3). The selection of scaling variables is left to the user, but there are some guidelines. In Eq. (5.2), it is now clear that the scaling variables were p, V, and L, since they appear in both force coefficient and Reynolds number. We could then interpret data from Eq. (5.2) as the variation of dimensionless force versus dimensionless viscosity, since each appears in only one dimensionless group. Similarly, in Eq. (5.5) the scaling variables were selected from (So, Vo> s)> not (S, t), because we wished to plot S versus t in the final result. The following are some guidelines for selecting scaling variables: 1. They must not form a dimensionless group among themselves, but adding one more variable will form a dimensionless quantity. For example, test powers of p, V, and L\ poyb^c ^ {^ML~^y(LIT)\Ly = M°L°T° only if a = 0, h = 0, c = 0 In this case, we can see why this is so: Only p contains the dimension (M), and only V contains the dimension {T}, so no cancellation is possible. If, now, we add p to the scaling group, we will obtain the Reynolds number. If we add F to the group, we form the force coefficient. 2. Do not select output variables for your scaling parameters. In Eq. (5.1), certainly do not select F, which you wish to isolate for your plot. Nor was p, selected, for we wished to plot force versus viscosity. 3. If convenient, select popular, not obscure, scaling variables because they will appear in all of your dimensionless groups. Select density, not surface tension. Select body length, not surface roughness. Select stream velocity, not speed of sound. The examples that follow will make this clear. Problem assignments might give hints. Suppose we wish to study drag force versus velocity. Then we would not use V as a scaling parameter in Eq. (5.1). We would use (p, p, L) instead, and the hnal dimen¬ sionless function would become pF pVL C; = ^=/(Re) = (5.16) In plotting these data, we would not be able to discern the effect of p or p, since they appear in both dimensionless groups. The grouping C'f again would mean dimension¬ less force, and Re is now interpreted as either dimensionless velocity or size.^ The plot would be quite different compared to Eq. (5.2), although it contains exactly the same information. The development of parameters such as Cf and Re from the initial variables is the subject of the pi theorem (Sec. 5.3). The foundation of the dimensional analysis method rests on two assumptions: (1) The proposed physical relation is dimensionally homogeneous, and (2) all the relevant variables have been included in the proposed relation. If a relevant variable is missing, dimensional analysis will fail, giving either alge¬ braic difficulties or, worse, yielding a dimensionless formulation that does not resolve ^We were lucky to achieve a size effect because in this case L, a scaling parameter, did not appear in the drag coefficient. 294 Chapter 5 Dimensional Analysis and Similarity 5.3 The Pi Theorem the process. A typical case is Manning’s open-channel formula, discussed in Example 1.4 and Chap. 10. 1.49 V = - (1) n Since V is velocity, R is a radius, and n and S are dimensionless, the formula is not dimensionally homogeneous. This should he a warning that (1) the formula changes if the units of V and R change and (2) if valid, it represents a very special case. Equa¬ tion (1) in Example 1.4 predates the dimensional analysis technique and is valid only for water in rough channels at moderate velocities and large radii in BG units. Such dimensionally inhomogeneous formulas abound in the hydraulics literature. Another example is the Hazen-Williams formula for volume flow of water through a straight smooth pipe: Q 61. (5.17) where D is diameter and dpidx is the pressure gradient. Some of these formulas arise because numbers have been inserted for fluid properties and other physical data into perfectly legitimate homogeneous formulas. We shall not give the units of Eq. (5.17) to avoid encouraging its use. On the other hand, some formulas are “constructs” that cannot be made dimension- ally homogeneous. The “variables” they relate cannot be analyzed by the dimensional analysis technique. Most of these formulas are raw empiricisms convenient to a small group of specialists. Here are three examples: 25,000 B = 100 - R - 140 ~ 130 + API 3.74 172 0.0147D£ - = 0.26t,i - (5.18) (5.19) (5.20) Equation (5.18) relates the Brinell hardness B of a metal to its Rockwell hardness R. Equation (5.19) relates the specific gravity S of an oil to its density in degrees API. Equation (5.20) relates the viscosity of a liquid in D^, or degrees Engler, to its viscosity in Saybolt seconds. Such formulas have a certain usefulness when communicated between fellow specialists, but we cannot handle them here. Vari¬ ables like Brinell hardness and Saybolt viscosity are not suited to an MLT& dimen¬ sional system. There are several methods of reducing a number of dimensional variables into a smaller number of dimensionless groups. The first scheme given here was proposed in 1914 by Buckingham and is now called the Buckingham Pi Theorem. The name pi comes from the mathematical notation H, meaning a product of variables. The dimensionless groups found from the theorem are power products denoted by Hi, 112, 133, ^tc. The method allows the pi groups to be found in sequential order without resorting to free exponents. 5.3 The Pi Theorem 295 The first part of the pi theorem explains what reduction in variables to expect: If a physical process satisfies the PDH and involves n dimensional variables, it can be reduced to a relation between k dimensionless variables or Ils. The reduction j = n — k equals the maximum number of variables that do not form a pi among themselves and is always less than or equal to the number of dimensions describing the variables. Take the specific case of force on an immersed body: Eq. (5.1) contains five variables F, L, U, p, and p described by three dimensions {MLT}. Thus n = 5 and j < 3. Therefore it is a good guess that we can reduce the problem to k pi groups, with k = n — 7 > 5 — 3 = 2. And this is exactly what we obtained: two dimensionless variables Hi = Cp and 112 = Rc- On rnre occasions it may take more pi groups than this mini¬ mum (see Example 5.5). The second part of the theorem shows how to find the pi groups one at a time: Find the reduction /, then select j scaling variables that do not form a pi among themselves.^ Each desired pi group will be a power product of these j variables plus one additional vari¬ able, which is assigned any convenient nonzero exponent. Each pi group thus found is independent. To be specific, suppose the process involves five variables: til = fiV2, Vs, V^, Vs) Suppose there are three dimensions {MLT} and we search around and find that indeed 7 = 3. Then k = 5 — 3 = 2 and we expect, from the theorem, two and only two pi groups. Pick out three convenient variables that do not form a pi, and suppose these turn out to be V2, v^, and V4. Then the two pi groups are formed by power products of these three plus one additional variable, either Vi v^: Hi = (T>2)“(T’3)(lt4)"T>i = n2 = (V2)‘‘ (Vs)’’ {V4)‘^Vs = Here we have arbitrarily chosen v^ and v^, the added variables, to have unit expo¬ nents. Equating exponents of the various dimensions is guaranteed by the theorem to give unique values of a, b, and c for each pi. And they are independent because only Hi contains Vi and only 112 contains v^. It is a very neat system once you get used to the procedure. We shall illustrate it with several examples. Typically, six steps are involved: 1 . List and count the n variables involved in the problem. If any important variables are missing, dimensional analysis will fail. 2. List the dimensions of each variable according to {MLT©} or {FLT = {T~^)‘‘{FT^L~y(LY{FTL~^) = This time, a=— 1,1>=— 1, and c = — 2; or 113 = a sort of Reynolds number. Step 5 The original relation between six variables is now reduced to three dimensionless groups: — ^ pCYO^ VflZ)^ pflD^ Comment: These three are the classical coefficients used to correlate pump power in Chap. 11. EXAMPLE 5.4 At low velocities (laminar flow), the volume flow Q through a small-bore tube is a function only of the tube radius R, the fluid viscosity p, and the pressure drop per unit tube length dpidx. Using the pi theorem, find an appropriate dimensionless relationship. Solution Write the given relation and count variables: dp Q = fyR, p,— \ four variables (n = 4) Make a list of the dimensions of these variables from Table 5.1 using the [MLT] system: Q R dpidx IT) {ML'‘r'‘) There are three primary dimensions (A/, L, T), hence y £ 3. By trial and error we determine that R, p, and dpidx cannot be combined into a pi group. Then j = 3, and n — y' = 4 — 3 = 1. There is only one pi group, which we find by combining 2 in a power product with the other three: n, -'■'ll M°L°T° 300 Chapter 5 Dimensional Analysis and Similarity Equate exponents: Mass: b + c = 0 Length: a — fc — 2c + 3 = 0 Time: -fc - 2c - 1 = 0 Solving simultaneously. we obtain a = —4, b = 1, and c = n. = «-v(0'e or rii = . = const lf{dp/dx) Ans. Since there is only one pi group, it must equal a dimensionless constant. This is as far as dimensional analysis can take us. The laminar flow theory of Sec. 4.10 shows that the value of the constant is — f. This result is also useful in Chap. 6. EXAMPLE 5.5 Assume that the tip deflection A of a cantilever beam is a function of the tip load P, beam length L, area moment of inertia I, and material modulus of elasticity E\ that is, 6 = j{P, L, /, E). Rewrite this function in dimensionless form, and comment on its complexity and the peculiar value of j. Solution List the variables and their dimensions: 6 P L I E IL] [MLT~^] {L] {E] There are hve variables (n = 5) and three primary dimensions (M, L, T), hence j ^ 3. But try as we may, we cannot hnd any combination of three variables that does not form a pi group. This is because [M] and {T} occur only in P and E and only in the same form. Thus we have encountered a special case of j = 2, which is less than the number of dimensions (M, L, T). To gain more insight into this peculiarity, you should rework the problem, using the (f, L, T) system of dimensions. You will find that only {F} and {L} occur in these variables, hence j = 2. With j = 2, we select L and E as two variables that cannot form a pi group and then add other variables to form the three desired pis: Oi = lV/‘ = from which, after equating exponents, we find that a = —4, fc = 0, or Hi = HL^. Then Dz = lVp = from which we find a = —2,b= —1, or 112 = P!(.El}), and Os = 5.3 The Pi Theorem 301 from which a = — 1, fc = 0, or 113 = The proper dimensionless function is 113 = /(Ilj, III), or 6 / P /A An.(l) This is a complex three-variable function, but dimensional analysis alone can take us no further. Comments: We can “improve” Eq. (1) by taking advantage of some physical reasoning, as Langhaar points out [4, p. 91]. For small elastic deflections, S is proportional to load P and inversely proportional to moment of inertia I. Since P and I occur separately in Eq. (1), this means that 113 must bs proportional to 112 inversely proportional to Hi. Thus, for these conditions. or — = (const) S = (const) EL^ I PL^ El (2) This could not be predicted by a pure dimensional analysis. Strength-of-materials theory predicts that the value of the constant is j. An Alternate Step-by-Step Method The pi theorem method, just explained and illustrated, is often called the repeating by Ipsen (I960) variable method of dimensional analysis. Select the repeating variables, add one more, and you get a pi group. The writer likes it. This method is straightforward and sys¬ tematically reveals all the desired pi groups. However, there are drawbacks: (1) All pi groups contain the same repeating variables and might lack variety or effectiveness, and (2) one must (sometimes laboriously) check that the selected repeating variables do not form a pi group among themselves (see Prob. P5.21). Ipsen suggests an entirely different procedure, a step-by-step method that obtains all of the pi groups at once, without any counting or checking. One simply successively eliminates each dimension in the desired function by division or multi¬ plication. Let us illustrate with the same classical drag function proposed in Eq. (5.1). Underneath the variables, write out the dimensions of each quantity. F = fcn(L, V, p, p) (5.1) {MLT~^} {L} {LT”'} {ML~^} {ML~^T~^} There are three dimensions, {MET}. Eliminate them successively by division or multiplication by a variable. Start with mass {M}. Pick a variable that contains mass and divide it into all the other variables with mass dimensions. We select p, divide, and rewrite the function (5.1): ^ =fcn(L, y, / (5.1a) {l} ^This method may be omitted without loss of continuity. 302 Chapter 5 Dimensional Analysis and Similarity We did not divide into L or V, which do not contain {M}. Equation (5.1a) at first looks strange, but it contains five distinct variables and the same information as Eq. (5.1). We see that p is no longer important. Thus discard p, and now there are only four variables. Next, eliminate time [T] by dividing the time-containing variables by suit¬ able powers of, say, V. The result is = fcn^L, pV^ \ pvj {L^} {L} {L} (5. lb) Now we see that V is no longer relevant. Finally, eliminate {L) through division by, say, appropriate powers of L itself: — ^ = fcn(^ — ) (5.1c) pV^L^ V pVLj {1} {1} Now L by itself is no longer relevant, and so discard it also. The result is equivalent to Eq. (5.2): — ^ = fcnf— ) (5.2) pVV \pVLj In Ipsen’s step-by-step method, we find the force coefficient is a function solely of the Reynolds number. We did no counting and did not hnd j. We just successively eliminated each primary dimension by division with the appropriate variables. Recall Example 5.5, where we discovered, awkwardly, that the number of repeating variables was less than the number of primary dimensions. Ipsen’s method avoids this preliminary check. Recall the beam-deflection problem proposed in Example 5.5 and the various dimensions: S=f(P, L, I, E) {L} {ML'T^} {L} {L'} For the first step, let us ehminate {M} by dividing by E. We only have to divide into P: i./g. L. {L} {L^} {L} {L^ We see that we may discard E as no longer relevant, and the dimension {T} has vanished along with {M}. We need only eliminate {L} by dividing by, say, powers of L itself: {1} {1} {1} Discard L itself as now irrelevant, and we obtain Answer (1) to Example 5.5: 5.3 The Pi Theorem 303 Ipsen’s approach is again successful. The fact that {M] and {T} vanished in the same division is proof that there are only two repeating variables this time, not the three that would be inferred by the presence of {M}, {L], and {T}. EXAMPLE 5.6 The leading-edge aerodynamic moment M^e on a supersonic airfoil is a function of its chord length C, angle of attack a, and several air parameters: approach velocity V, density p, speed of sound a, and specific heat ratio k (Fig. E5.6). There is a very weak effect of air viscosity, which is neglected here. E5.6 Use Ipsen’s method to rewrite this function in dimensionless form. Solution Write out the given function and list the variables’ dimensions {MLT} underneath: Mee = fcn(C, a, V, p, a, k) {MLVT^} {L} {1) {UT} {M/L^} {UT} {Ij Two of them, a and k, are already dimensionless. Leave them alone; they will be pi groups in the final function. You can eliminate any dimension. We choose mass {M} and divide by p: Mle = fcn(C, a, V, ^ a, k) {L^lt} {L] {1} {UT} {UT} {1} Recall Ipsen’s rules: Only divide into variables containing mass, in this case only Mee, and then discard the divisor, p. Now eliminate time { T} by dividing by appropriate powers of a: ^LE „ y A — ^ = fcnl C, a, — , ^ k I pa \ a ) {L^} (L) {Ij {1) {1} 304 Chapter 5 Dimensional Analysis and Similarity 5.4 Nondimensionalization of the Basic Equations Finally, eliminate {L} on the left side hy dividing hy C^: Mle pa^C^ {1) fcn^jST a, 11} {1} {1} We end up with four pi groups and recognize V/a as the Mach number, Ma. In aerodynam¬ ics, the dimensionless moment is often called the moment coefficient. Cm- Thus our final result could he written in the compact form Cm = fcn(a, Ma, k) Ans. Comments: Our analysis is fine, but experiment and theory and physical reasoning all indicate that M^e varies more strongly with V than with a. Thus aerodynamicists commonly define the moment coefficient as Cm = MeeKpV^C^) or something similar. We will study the analysis of supersonic forces and moments in Chap. 9. We could use the pi theorem method of the previous section to analyze problem after problem after problem, finding the dimensionless parameters that govern in each case. Textbooks on dimensional analysis [for example, 5] do this. An alternative and very powerful technique is to attack the basic equations of flow from Chap. 4. Even though these equations cannot be solved in general, they will reveal basic dimensionless parameters, such as the Reynolds number, in their proper form and proper position, giving clues to when they are negligible. The boundary conditions must also be nondimensionalized. Let us briefly apply this technique to the incompressible flow continuity and momentum equations with constant viscosity: Continuity: V • V = 0 (5.21fl) U T 9 Navier-Stokes: p — = pg — Vp + pN V dt Typical boundary conditions for these two equations are (Sect. 4.6) Fixed solid surface: V = 0 i5.2\b) Inlet or outlet: Known V,p (5.22) dr) Free surface, z = rj: ^ ^ ^ P ^ Pa ~ ^ (Rx + Ry ) We omit the energy equation (4.75) and assign its dimensionless form in the problems (Prob. P5.43). Equations (5.21) and (5.22) contain the three basic dimensions M, L, and T. All variables p, V, x, y, z, and f can be nondimensionalized by using density and two reference constants that might be characteristic of the particular fluid flow: Reference velocity = U Reference length = L For example, U may be the inlet or upstream velocity and L the diameter of a body immersed in the stream. 5.4 Nondimensionalization of the Basic Equations 305 Dimensionless Parameters Now define all relevant dimensionless variables, denoting them by an asterisk: y = V U V = LV X X L t P + Pgz pU^ (5.23) All these are fairly obvious except for p, where we have introduced the piezometric pressure, assuming that z is up. This is a hindsight idea suggested by Bernoulli’s equation (3.54). Since p, U, and L are all constants, the derivatives in Eqs. (5.21) can all be handled in dimensionless form with dimensional coefficients. For example, du d{Uu) U du dx d{Lx) L dx Substitute the variables from Eqs. (5.23) into Eqs. (5.21) and (5.22) and divide through by the leading dimensional coefficient, in the same way as we handled Eq. (5.12). Here are the resulting dimensionless equations of motion: Continuity: V • E = 0 (5.24fl) Momentum: dy „ M „ 2 -^P + T7^V^(V) dt put The dimensionless boundary conditions are: Fixed solid surface: Inlet or outlet: V = 0 Known \,p Free surface, z = T]- w dp dt Pa pu^ Y ptJ-L {Rr' + Rr ) (5.24b) (5.25) These equations reveal a total of four dimensionless parameters, one in the Navier- Stokes equation and three in the free-surface-pressure boundary condition. In the continuity equation there are no parameters. The Navier-Stokes equation con¬ tains one, generally accepted as the most important parameter in fluid mechanics: pUL Reynolds number Re = - 306 Chapter 5 Dimensional Analysis and Similarity It is named after Osborne Reynolds (1842-1912), a British engineer who first pro¬ posed it in 1883 (Ref. 4 of Chap. 6). The Reynolds number is always important, with or without a free surface, and can be neglected only in flow regions away from high- velocity gradients — for example, away from solid surfaces, jets, or wakes. The no-slip and inlet-exit boundary conditions contain no parameters. The free- surface-pressure condition contains three: Euler number (pressure coefficient) Eu Pa pu^ This is named after Leonhard Euler (1707-1783) and is rarely important unless the pressure drops low enough to cause vapor formation (cavitation) in a liquid. The Euler number is often written in terms of pressure differences: Eu = Ap/ipiP). If Ap involves vapor pressure p^, it is called the cavitation number Ca = (p^ ~ Pv)KpU^)- Cavitation problems are surprisingly common in many water flows. The second free-surface parameter is much more important: Eroude number Er = — gL It is named after William Eroude (1810-1879), a British naval architect who, with his son Robert, developed the ship-model towing-tank concept and proposed similarity rules for free-surface flows (ship resistance, surface waves, open channels). The Eroude number is the dominant effect in free-surface flows. It can also be important in stratified fiows, where a strong density difference exists without a free surface. For example, see Ref. . Chapter 10 investigates Eroude number effects in detail. The final free-surface parameter is pU^L Weber number We = — — It is named after Moritz Weber (1871-1951) of the Polytechnic Institute of Berlin, who developed the laws of similitude in their modern form. It was Weber who named Re and Fr after Reynolds and Eroude. The Weber number is important only if it is of order unity or less, which typically occurs when the surface curvature is comparable in size to the liquid depth, such as in droplets, capillary flows, ripple waves, and very small hydraulic models. If We is large, its effect may be neglected. If there is no free surface, Fr, Eu, and We drop out entirely, except for the pos¬ sibility of cavitation of a liquid at very small Eu. Thus, in low-speed viscous flows with no free surface, the Reynolds number is the only important dimensionless parameter. Compressibility Parameters In high-speed flow of a gas there are significant changes in pressure, density, and temperature that must be related by an equation of state such as the perfect-gas law, Eq. (1.10). These thermodynamic changes introduce two additional dimensionless parameters mentioned briefly in earlier chapters: Mach number Ma U Cp a Specific-heat ratio k 5.4 Nondimensionalization of the Basic Equations 307 Oscillating Flows Other Dimensionless Parameters The Mach number is named after Ernst Mach (1838-1916), an Austrian physicist. The effect of k is only slight to moderate, but Ma exerts a strong effect on com¬ pressible flow properties if it is greater than about 0.3. These effects are studied in Chap. 9. If the flow pattern is oscillating, a seventh parameter enters through the inlet boundary condition. For example, suppose that the inlet stream is of the form u = U cos ujt Nondimensionalization of this relation results in u ( uL ^ U \U J The argument of the cosine contains the new parameter ujL Strouhal number St = - U The dimensionless forces and moments, friction, and heat transfer, and so on of such an oscillating flow would be a function of both Reynolds and Strouhal numbers. This parameter is named after V. Strouhal, a German physicist who experimented in 1878 with wires singing in the wind. Some flows that you might guess to be perfectly steady actually have an oscillatory pattern that is dependent on the Reynolds number. An example is the periodic vortex shedding behind a blunt body immersed in a steady stream of velocity U. Figure 5.2a shows an array of alternating vortices shed from a circular cylinder immersed in a steady crossflow. This regular, periodic shedding is called a Kdrmdn vortex street, after T. von Karman, who explained it theoreti¬ cally in 1912. The shedding occurs in the range 10^ < Re < 10^, with an average Strouhal number udl{2TTU) ~ 0.21. Figure 5.2b shows measured shedding frequencies. Resonance can occur if a vortex shedding frequency is near a body’s structural vibration frequency. Electric transmission wires sing in the wind, undersea mooring lines gallop at certain current speeds, and slender structures flutter at critical wind or vehicle speeds. A striking example is the disastrous failure of the Tacoma Narrows suspension bridge in 1940, when wind-excited vortex shedding caused resonance with the natural torsional oscillations of the bridge. The problem was magnified by the bridge deck nonlinear stiffness, which occurred when the hangers went slack during the oscillation. We have discussed seven important parameters in fluid mechanics, and there are oth¬ ers. Four additional parameters arise from nondimensionalization of the energy equa¬ tion (4.75) and its boundary conditions. These four (Prandtl number, Eckert number, Grashof number, and wall temperature ratio) are listed in Table 5.2 just in case you fail to solve Prob. P5.43. Another important and perhaps surprising parameter is the 308 Chapter 5 Dimensional Analysis and Similarity Fig. 5.2 Vortex shedding from a circular cylinder: (a) vortex street behind a circular cylinder ( Courtesy ofU.S. Navy): (b) experimental shedding frequencies (data from Refs. 25 and 26). wall roughness ratio elL (in Table 5.2).® Slight changes in surface roughness have a striking effect in the turbulent flow or high-Reynolds-number range, as we shall see in Chap. 6 and in Fig. 5.3. This book is primarily concerned with Reynolds-, Mach-, and Froude-number effects, which dominate most flows. Note that we discovered these parameters (except elV) simply by nondimensionalizing the basic equations without actually solving them. ^Roughness is easy to overlook because it is a slight geometric effect that does not appear in the equations of motion. It is a boundary condition that one might forget. 5.4 Nondimensionalization of the Basic Equations 309 Table 5.2 Dimensionless Groups in Fluid Mechanics Qualitative ratio Parameter Definition of effects Importance Inertia Viscosity Almost always Reynolds number pUL Re = ^ - M Mach number U Ma = — a Froude number Fr = — gL Weber number pU^L We= ^ Rossby number Ro= ^ ^ ^ earth ^ Cavitation number (Euler number) Ca = '^j— ^ Prandtl number k Eckert number Ec = - CpTo Specific-heat ratio Strouhal number ujL Roughness ratio e L Grashof number ^ l3\TgL^p^ Gr = Rayleigh number PMgLYcp Ra = p k Temperature ratio To Pressure coefficient 8 1 II Lift coefficient L Drag coefficient D ~ ^jpU^A Friction factor hf ^ Yl2g)(Lld) Skin friction coefficient "^wall ^ pv^n Flow speed Sound speed Compressible flow Inertia Gravity Free-surface flow Inertia Surface tension Free-surface flow Flow velocity Coriolis effect Geophysical flows Pressure Inertia Cavitation Dissipation Conduction Heat convection Kinetic energy Enthalpy Dissipation Enthalpy Internal energy Compressible flow Oscillation Mean speed Oscillating flow Wall roughness Body length Turbulent, rough walls Buoyancy Viscosity Natural convection Buoyancy Viscosity Natural convection Wall temperature Heat transfer Stream temperature Static pressure Dynamic pressure Aerodynamics, hydrodynamics Lift force Dynamic force Aerodynamics, hydrodynamics Drag force Dynamic force Aerodynamics, hydrodynamics Friction head loss Velocity head Pipe flow Wall shear stress Dynamic pressure Boundary layer flow 310 Chapter 5 Dimensional Analysis and Similarity Fig. 5.3 The proof of practical dimensional analysis: drag coefficients of a cylinder and sphere: (a) drag coefficient of a smooth cylinder and sphere (data from many sources); (b) increased roughness causes earlier transition to a turbulent boundary layer. A Successful Application Cylinder length effect (lO'^ < Re < 10^) Ltd Cd oo 1.20 40 0.98 20 0.91 10 0.82 5 0.74 3 0.72 2 0.68 1 0.64 ib) If the reader is not satiated with the 19 parameters given in Table 5.2, Ref. 29 con¬ tains a list of over 1200 dimensionless parameters in use in engineering and science. Dimensional analysis is fun, but does it work? Yes, if all important variables are included in the proposed function, the dimensionless function found by dimensional analysis will collapse all the data onto a single curve or set of curves. An example of the success of dimensional analysis is given in Fig. 5.3 for the measured drag on smooth cylinders and spheres. The flow is normal to the axis of the cylinder, which is extremely long. Lid — > “. The data are from many sources, for both liquids and gases, and include bodies from several meters in diameter down to fine wires and balls less than 1 mm in size. Both curves in Fig. 5.3fl are entirely experimental; the analysis of immersed body drag is one of the weakest areas of modern fluid mechanics theory. Except for digital computer calculations, there is little theory for cylinder and sphere drag except creeping flow, Re < 1 . 5.4 Nondimensionalization of the Basic Equations 311 The concept of a fluid-caused drag force on bodies is covered extensively in Chap. 7. Drag is the fluid force parallel to the oncoming stream — see Fig. 7.10 for details. The Reynolds number of both bodies is based on diameter, hence the notation Re^. But the drag coefficients are defined differently; Cn drag \pU^U drag 1 _ T r2 1 t2 \ 2 pU ^TTCl cylinder sphere (5.26) They both have a factor ^ because the term occurs in Bernoulli’s equation, and both are based on the projected area — that is, the area one sees when looking toward the body from upstream. The usual definition of Co is thus Co drag ^ pl7^(projected area) (5.27) However, one should carefully check the definitions of Co, Re, and the like before using data in the literature. Airfoils, for example, use the planform area. Figure 5.3a is for long, smooth cylinders. If wall roughness and cylinder length are included as variables, we obtain from dimensional analysis a complex three- parameter function; (5.28) To describe this function completely would require 1000 or more experiments or CFD results. Therefore it is customary to explore the length and roughness effects sepa¬ rately to establish trends. The table with Fig. 5.3a shows the length effect with zero wall roughness. As length decreases, the drag decreases by up to 50 percent. Physically, the pressure is “relieved” at the ends as the flow is allowed to skirt around the tips instead of deflect¬ ing over and under the body. Figure 53b shows the effect of wall roughness for an infinitely long cylinder. The sharp drop in drag occurs at lower Re^ as roughness causes an earlier transition to a turbulent boundary layer on the surface of the body. Roughness has the same effect on sphere drag, a fact that is exploited in sports by deliberate dimpling of golf balls to give them less drag at their flight Re^ ~ 10^. See Fig. D5.2. Figure 5.3 is a typical experimental study of a fluid mechanics problem, aided by dimensional analysis. As time and money and demand allow, the complete three- parameter relation (5.28) could be filled out by further experiments. EXAMPLE 5.7 A smooth cylinder, 1 cm in diameter and 20 cm long, is tested in a wind tunnel for a crossflow of 45 m/s of air at 20°C and 1 atm. The measured drag is 2.2 ± 0. 1 N. (a) Does this data point agree with the data in Fig. 5.3? (b) Can this data point be used to predict the drag of a chimney 1 m in diameter and 20 m high in winds at 20°C and 1 atm? If so, what 312 Chapter 5 Dimensional Analysis and Similarity is the recommended range of wind velocities and drag forces for this data point? (c) Why are the answers to part {b) always the same, regardless of the chimney height, as long as L = 20c/? Solution (a) For air at 20°C and 1 atm, take p = 1.2 kg/m^ and /x = 1.8 E— 5 kg/(m-s). Since the test cylinder is short, L/d = 20, it should be compared with the tabulated value Cd ~ 0.91 in the table to the right of Fig. 5.3c!. First calculate the Reynolds number of the test cylinder: Rerf = pUd (1.2kg/m^)(45 m/s)(0.01m) 1.8E-5 kg/(m - s) = 30,000 Yes, this is in the range 10^ < Re < 10^ listed in the table. Now calculate the test drag coefficient: f^D.test 2.2 N {l/2)pU^U (l/2)(1.2kg/m^)(45m/s)^(0.2m)(0.01 m) = 0.905 Yes, this is close, and certainly within the range of ±5 percent stated by the test results. Ans. (a) (b) Since the chimney has L/d = 20, we can use the data if the Reynolds number range is correct: (1.2 kg/m^) (/chimney! 1 m) 1.8E-5kg/(m • s) ni m if 0.15 f^chimney S S These are negligible winds, so the test data point is not very useful Ans. (b) The drag forces in this range are also negligibly small: P 7 /1.2kg/m^\ F^n = (0.91) ( - 2 - j(0.15in/s)"(20m)(lm) = 0.25 N P ^ /^l-2kg/m^\ /^max = Cn^Ui^U = (0.91) ( - ^ - j (1.5 m/s)"(20 m)(l m) = 25 N (c) Try this yourself. Choose any 20:1 size for the chimney, even something silly like 20 mm:l mm. You will get the same results for U and F as in part (b) above. This is because the product Ud occurs in Re^ and, if L = 20d, the same product occurs in the drag force. For example, for Re = 10", Ud = 10'^- then F = Co- U^Ld = Co- U^(20d)d = 20Co - (Udf = 20Co p 2 2 2 The answer is always Fmm = 0.25 N. This is an algebraic quirk that seldom occurs. EXAMPLE 5.8 Telephone wires are said to “sing” in the wind. Consider a wire of diameter 8 mm. At what sea-level wind velocity, if any, will the wire sing a middle C note? 5.5 Modeling and Similarity 313 5.5 Modeling and Similarity Solution For sea-level air take i/ = 1.5 E— 5 m^/s. For nonmusical readers, middle C is 262 Hz. Measured shedding rates are plotted in Fig. 5.2b. Over a wide range, the Strouhal number is approximately 0.2, which we can take as a hrst guess. Note that (lo/Itt) = /, the shedding frequency. Thus fd (262 s“')(0.008m) St = — = — ii— — — i « 0.2 u u m U « 10.5 — s Now check the Reynolds number to see if we fall into the appropriate range: Rsrf = Ud V (10.5 m/s) (0.008 m) 1.5 E-5 m^/s « 5600 In Fig. 5.2b, at Re = 5600, maybe St is a little higher, at about 0.21. Thus a slightly improved estimate is (/wind = (262)(0.008)/(0.21) « 10.0 m/s Ans. So far we have learned about dimensional homogeneity and the pi theorem method, using power products, for converting a homogeneous physical relation to dimension¬ less form. This is straightforward mathematically, but certain engineering difficulties need to be discussed. First, we have more or less taken for granted that the variables that affect the process can be listed and analyzed. Actually, selection of the important variables requires considerable judgment and experience. The engineer must decide, for exam¬ ple, whether viscosity can be neglected. Are there significant temperature effects? Is surface tension important? What about wall roughness? Each pi group that is retained increases the expense and effort required. Judgment in selecting variables will come through practice and maturity; this book should provide some of the necessary experience. Once the variables are selected and the dimensional analysis is performed, the experimenter seeks to achieve similarity between the model tested and the prototype to be designed. With sufficient testing, the model data will reveal the desired dimen¬ sionless function between variables: U^=f{U2,U„...U,) (5.29) With Eq. (5.29) available in chart, graphical, or analytical form, we are in a position to ensure complete similarity between model and prototype. A formal statement would be as follows: Flow conditions for a model test are completely similar if all relevant dimensionless parameters have the same corresponding values for the model and the prototype. This follows mathematically from Eq. (5.29). If 112^ = — n3p, and so forth, Eq. (5.29) guarantees that the desired output ni,„ will equal Flip. But this is 314 Chapter 5 Dimensional Analysis and Similarity Geometric Similarity Fig. 5.4 Geometric similarity in model testing: (a) prototype; (b) one-tenth-scale model. easier said than done, as we now discuss. There are specialized texts on model test¬ ing [30-32]. Instead of complete similarity, the engineering literature speaks of particular types of similarity, the most common being geometric, kinematic, dynamic, and thermal. Let us consider each separately. Geometric similarity concerns the length dimension {Lj and must be ensured before any sensible model testing can proceed. A formal definition is as follows: A model and prototype are geometrically similar if and only if all body dimensions in all three coordinates have the same linear scale ratio. Note that all length scales must be the same. It is as if you took a photograph of the prototype and reduced it or enlarged it until it fitted the size of the model. If the model is to be made one-tenth the prototype size, its length, width, and height must each be one-tenth as large. Not only that, but also its entire shape must be one-tenth as large, and technically we speak of homologous points, which are points that have the same relative location. For example, the nose of the prototype is homologous to the nose of the model. The left wingtip of the prototype is homologous to the left wingtip of the model. Then geometric similarity requires that all homologous points be related by the same linear scale ratio. This applies to the fluid geometry as well as the model geometry. All angles are preserved in geometric similarity. All flow directions are preserved. The orientations of model and prototype with respect to the surroundings must be identical. Figure 5.4 illustrates a prototype wing and a one-tenth-scale model. The model lengths are all one-tenth as large, but its angle of attack with respect to the free stream is the same for both model and prototype: 10° not 1°. All physical details on the model must be scaled, and some are rather subtle and sometimes overlooked: 1. The model nose radius must be one-tenth as large. 2. The model surface roughness must be one-tenth as large. Homologous points {b) 5.5 Modeling and Similarity 315 Fig. 5.5 Geometric similarity and dissimilarity of flows: (a) similar; (&) dissimilar. Kinematic Similarity -c ^2 czz::> - Large 4:1 ellipsoid (b) Medium 3.5:1 ellipsoid Small 3:1 ellipsoid 3. If the prototype has a 5-mm boundary layer trip wire 1.5 m from the leading edge, the model should have a 0.5-mm trip wire 0.15 m from its leading edge. 4. If the prototype is constructed with protruding fasteners, the model should have homologous protruding fasteners one-tenth as large. And so on. Any departure from these details is a violation of geometric similarity and must be justihed by experimental comparison to show that the prototype behavior was not significantly affected by the discrepancy. Models that appear similar in shape but that clearly violate geometric similarity should not be compared except at your own risk. Figure 5.5 illustrates this point. The spheres in Fig. 5.5a are all geometrically similar and can be tested with a high expec¬ tation of success if the Reynolds number, Froude number, or the like is matched. But the ellipsoids in Fig. 5.5b merely look similar. They actually have different linear scale ratios and therefore cannot be compared in a rational manner, even though they may have identical Reynolds and Froude numbers and so on. The data will not be the same for these ellipsoids, and any attempt to “compare” them is a matter of rough engineering judgment. Kinematic similarity requires that the model and prototype have the same length scale ratio and the same time scale ratio. The result is that the velocity scale ratio will be the same for both. As Langhaar states it: The motions of two systems are kinematically similar if homologous particles lie at homologous points at homologous times. Length scale equivalence simply implies geometric similarity, but time scale equivalence may require additional dynamic considerations such as equivalence of the Reynolds and Mach numbers. One special case is incompressible frictionless flow with no free surface, as sketched in Fig. 5.6a. These perfect- fluid flows are kinematically similar with inde¬ pendent length and time scales, and no additional parameters are necessary (see Chap. 8 for further details). 316 Chapter 5 Dimensional Analysis and Similarity Vun = PVi^ Fig. 5.6 Frictionless low-speed flows are kinematically similar: (a) Flows with no free surface are kinematically similar with independent length and time scale ratios; (b) free-surface flows are kinematically similar with length and time scales related hy the Froude numher. (b) Froude Scaling Frictionless flows with a free surface, as in Fig. 5.6b, are kinematically similar if their Froude numbers are equal: Fr m gL,n gLp (5.30) Note that the Froude number contains only length and time dimensions and hence is a purely kinematic parameter that Axes the relation between length and time. From Eq. (5.30), if the length scale is L„, o:Lp where ct is a dimensionless ratio, the velocity scale is Yu Vp = Va (5.31) 1/2 (5.32) 5.5 Modeling and Similarity 317 Dynamic Similarity Fig. 5.7 Dynamic similarity in sluice gate flow. Model and prototype yield identical homologous force polygons if the Reynolds and Froude numbers are the same corresponding values: (a) prototype; (b) model. and the time scale is T m T L,n/Vm Va (5.33) These Froude-scaling kinematic relations are illustrated in Fig. 5.6b for wave motion modeling. If the waves are related by the length scale a, then the wave period, propa¬ gation speed, and particle velocities are related by Vo:. If viscosity, surface tension, or compressibility is important, kinematic similarity depends on the achievement of dynamic similarity. Dynamic similarity exists when the model and the prototype have the same length scale ratio, time scale ratio, and force scale (or mass scale) ratio. Again geometric similarity is a first requirement; without it, proceed no further. Then dynamic similar¬ ity exists, simultaneous with kinematic similarity, if the model and prototype force and pressure coefficients are identical. This is ensured if 1. For compressible flow, the model and prototype Reynolds number and Mach number and specific-heat ratio are correspondingly equal. 2. For incompressible flow a. With no free surface: model and prototype Reynolds numbers are equal. b. With a free surface; model and prototype Reynolds number, Froude number, and (if necessary) Weber number and cavitation number are correspondingly equal. Mathematically, Newton’s law for any fluid particle requires that the sum of the pres¬ sure force, gravity force, and friction force equal the acceleration term, or inertia force, Fp + Fg + Ff= Fi The dynamic similarity laws listed above ensure that each of these forces will be in the same ratio and have equivalent directions between model and prototype. Figure 5.7 (a) ih) 318 Chapter 5 Dimensional Analysis and Similarity shows an example for flow through a sluice gate. The force polygons at homologous points have exactly the same shape if the Reynolds and Froude numbers are equal (neglecting surface tension and cavitation, of course). Kinematic similarity is also ensured by these model laws. Discrepancies in Water and Air Testing The perfect dynamic similarity shown in Fig. 5.7 is more of a dream than a reality because true equivalence of Reynolds and Froude numbers can be achieved only by dramatic changes in fluid properties, whereas in fact most model testing is simply done with water or air, the cheapest fluids available. First consider hydraulic model testing with a free surface. Dynamic similarity requires equivalent Froude numbers, Eq. (5.30), and equivalent Reynolds numbers: (5.34) But both velocity and length are constrained by the Froude number, Eqs. (5.31) and (5.32). Therefore, for a given length scale ratio a, Eq. (5.34) is true only if ^ = = = (5.35) i^P Lp Vp Eor example, for a one-tenth-scale model, a = 0.1 and 0?^'^ = 0.032. Since Up is undoubtedly water, we need a fluid with only 0.032 times the kinematic viscosity of water to achieve dynamic similarity. Referring to Table 1.4, we see that this is impos¬ sible: Even mercury has only one-ninth the kinematic viscosity of water, and a mercury hydraulic model would be expensive and bad for your health. In practice, water is used for both the model and the prototype, and the Reynolds number similarity (5.34) is unavoidably violated. The Eroude number is held constant since it is the dominant param¬ eter in free-surface flows. Typically the Reynolds number of the model flow is too small by a factor of 10 to 1000. As shown in Fig. 5.8, the low-Reynolds-number model data are used to estimate by extrapolation the desired high-Reynolds-number prototype data. As the figure indicates, there is obviously considerable uncertainty in using such an extrapolation, but there is no other practical alternative in hydraulic model testing. Fig. 5.8 Reynolds-number extrapolation, or scaling, of hydraulic data with equal Froude numbers. log Co Range Range log Re 5.5 Modeling and Similarity 319 Second, consider aerodynamic model testing in air with no free surface. The impor¬ tant parameters are the Reynolds number and the Mach number. Equation (5.34) should be satisfied, plus the compressibility criterion (5.36) Elimination of V„JVp between (5.34) and (5.36) gives Lp Up (5.37) Since the prototype is no doubt an air operation, we need a wind-tunnel fluid of low viscosity and high speed of sound. Hydrogen is the only practical example, but clearly it is too expensive and dangerous. Therefore, wind tunnels normally operate with air as the working fluid. Cooling and pressurizing the air will bring Eq. (5.37) into better agreement but not enough to satisfy a length scale reduction of, say, one-tenth. There¬ fore Reynolds number scaling is also commonly violated in aerodynamic testing, and an extrapolation like that in Fig. 5.8 is required here also. There are specialized monographs devoted entirely to wind tunnel testing: low speed , high speed , and a detailed general discussion . The following example illustrates modeling discrepancies in aeronautical testing. EXAMPLE 5.9 A prototype airplane, with a chord length of 1.6 m, is to fly at Ma = 2 at 10 km standard altitude. A one-eighth scale model is to be tested in a helium wind tunnel at 100°C and 1 atm. Find the helium test section velocity that will match (a) the Mach number or (b) the Reynolds number of the prototype. In each case criticize the lack of dynamic similarity. (c) What high pressure in the helium tunnel will match both the Mach and Reynolds numbers? (d) Why does part (c) still not achieve dynamic similarity? Solution For helium, from Table A. 4, R = 2077 mV(s^-K), k = 1.66, and estimate ^He ~ 2.32 E— 5 kg/(m ■ s) from the power-law, n = 0.67, in the table, (a) Calculate the helium speed of sound and velocity: One = V(kRfhk = V(1.66)(2077mVK) X (373 K) = 1134 m/s Tne Vne Ma^ir = Mane = 2.0 = Vne = 2268 tine m 1134 m/s Ans. (a) For dynamic similarity, the Reynolds numbers should also be equal. From Table A.6 at an altitude of 10,000 m, read = 0.4125 kg/m^, = 299.5 m/s, and estimate ~ 1.48 E— 5 kg/m ■ s from the power-law, n = 0.7, in Table A. 4. The air velocity is F^ir = (Ma)(flaij) = 2(299.5) = 599 m/s. The model chord length is (1.6 m)/8 = 0.2 m. The helium 320 Chapter 5 Dimensional Analysis and Similarity density is pne = (/^/^r)He = (101,350 Pa)/[(2077 mV K)(373 K)] = 0.131 kg/ml Now calculate the two Reynolds numbers: RCrair ~ pVC (0.4125 kg/m^)(599 m/s) (1.6 m) p air 1.48 E-5 kg/(m ■ s) pVC (0.131 kg/m^)(2268 m/s)(0.2 m) p He 2.32 E-5 kg/(m-s) = 26.6 E6 = 2.56 E6 The model Reynolds number is 10 times less than the prototype. This is typical when using small-scale models. The test results must be extrapolated for Reynolds number effects. (b) Now ignore Mach number and let the model Reynolds number match the prototype: Reue = Rcair = 26.6 E6 = (0.131 kg V)THe(0-2m) 2.32E-5kg/(m-s) Vhc = 23,600 — s Ans. (b) This is ridiculous: a hypersonic Mach number of 21, suitable for escaping from the earth’s gravity. One should match the Mach numbers and correct for a lower Reynolds number, (c) Match both Reynolds and Mach numbers by increasing the helium density: Ma matches if Then Solve for m Vne = 2268 - s Rene = 26.6 E6 PHe(2268 m/s) (0.2 m) 2.32 E-5 kg/(m ■ s) kg pHe = 1.36^ pHe = PRT\kc = ( 1 .36) (2077) (373) = 1.05 E6 Pa Ans. (c) m A match is possible if we increase the tunnel pressure by a factor of ten, a daunting task. {d) Even with Ma and Re matched, we are still not dynamically similar because the two gases have different specific heat ratios: = 1.66 and = 1.40. This discrepancy will cause substantial differences in pressure, density, and temperature throughout supersonic flow. Figure 5.9 shows a hydraulic model of the Bluestone Lake Dam in West Virginia. The model itself is located at the U.S. Army Waterways Experiment Station in Vicksburg, MS. The horizontal scale is 1:65, which is sufficient that the vertical scale can also be 1:65 without incurring significant surface tension (Weber number) effects. Velocities are scaled by the Froude number. However, the prototype Reynolds number, which is of order 1 E7, cannot be matched here. The engineers set the Reynolds num¬ ber at about 2 E4, high enough for a reasonable approximation of prototype turbulent flow viscous effects. Note the intense turbulence below the dam. The downstream bed, or apron, of a dam must be strengthened structurally to avoid bed erosion. 5.5 Modeling and Similarity 321 Fig. 5.9 Hydraulic model of the Bluestone Lake Dam on the New River near Hinton, West Virginia. The model scale is 1:65 both vertically and horizontally, and the Reynolds number, though far below the prototype value, is set high enough for the flow to be turbulent. (Courtesy of the U.S. Army Corps of Engineers Waterways Experiment Station.) For hydraulic models of larger scale, such as harbors, estuaries, and embayments, geometric similarity may be violated of necessity. The vertical scale will be distorted to avoid Weber number effects. For example, the horizontal scale may be 1:1000, while the vertical scale is only 1:100. Thus the model channel may be deeper relative to its horizontal dimensions. Since deeper passages flow more efficiently, the model channel bottom may be deliberately roughened to create the friction level expected in the prototype. EXAMPLE 5.10 The pressure drop due to friction for flow in a long, smooth pipe is a function of average flow velocity, density, viscosity, and pipe length and diameter: Ap = fcn(V, p, p, L, D). We wish to know how Ap varies with V. (a) Use the pi theorem to rewrite this function in 322 Chapter 5 Dimensional Analysis and Similarity dimensionless form, {b) Then plot this function, using the following data for three pipes and three fluids: D, cm L, m e, m^/h Ap, Pa p, kg/m’ p, kg/(m ■ s) V, m/s 1.0 5.0 0.3 4,680 680t 2.92 E-4t 1.06 1.0 7.0 0.6 22,300 680t 2.92 E-4t 2.12 1.0 9.0 1.0 70,800 680t 2.92 E-4t 3.54 2.0 4.0 1.0 2,080 9981: o.ooiot 0.88 2.0 6.0 2.0 10,500 9981: 0.00101: 1.77 2.0 8.0 3.1 30,400 9981: o.ooiot 2.74 3.0 3.0 0.5 540 13,550§ 1.56 E-3§ 0.20 3.0 4.0 1.0 2,480 13,550§ 1.56 E-3§ 0.39 3.0 5.0 1.7 9,600 13,550§ 1.56 E-3§ 0.67 y = GM, A = fGasoline. t Water. §Mercury. (c) Suppose it is further known that Ap is proportional to L (which is quite true for long pipes with well-rounded entrances). Use this information to simplify and improve the pi theorem formulation. Plot the dimensionless data in this improved manner and comment on the results. Solution There are six variables with three primary dimensions involved {MLT}. Therefore, we expect thaty = 6 — 3 = 3 pi groups. We are correct, for we can find three variables that do not form a pi product (e.g., p, V, L). Carefully select three (y) repeating variables, but not including Ap or V, which we plan to plot versus each other. We select (p, p, D), and the pi theorem guarantees that three independent power-product groups will occur: or Hi = p‘‘p‘’D‘' A.p n, = pD^\p 02 02 p'^p^I^V pVD Dj = f^p!'D‘L 03 = L D We have omitted the algebra of finding (a, b, c, d, e, /, g, h, i) by setting all exponents to zero L°, f. Therefore, we wish to plot the dimensionless relation pD^ Ap fcn\| pVD L\ At ’ d) Ans. (a) We plot Ill versus 112 with 113 us a parameter. There will be nine data points. For example, the first row in the data here yields pVD At pD-Ap (680) (0.01)^(4680) At (680)(1.06)(0.01) 2.92 E-4 (2.92 E-4)^ = 3.73 E9 = 24,700 L - = 500 D The nine data points are plotted as the open circles in Fig. 5.10. The values of LID are listed for each point, and we see a significant length effect. In fact, if we connect the only two points that have the same LID (= 200), we could see (and cross-plot to verify) that Ap increases linearly with L, as stated in the last part of the problem. Since L occurs only in 5.5 Modeling and Similarity 323 Fig. 5.10 Two different correlations of the data in Example 5.10: Open circles when plotting pD^ versus Re^,, UD is a parameter; once it is known that Ap is proportional to L, a replot (solid circles) of pD^ Apl{Lp^) versus Re^ collapses into a single power-law curve. Hi n. 108 107 900 700 400 2 00 300 133 1 500 ^ = 200 100 \ 0. 155 R er,‘-75 lOS 1011 1010 Hi lOO lO'i Ren 105 113 = UD, the function Hi = fcn(n2, 113) must reduce to Hi = {UD) fcn(n2), or simply a function involving only two parameters: pD^ Ap fpVD\ - 3 — = fen I - 1 flow in a long pipe Ans. (c) \ M / We now modify each data point in Fig. 5.10 hy dividing it hy its UD value. For example, for the first row of data, pD^ Ap/{Lp}) = (3.73 E9)/500 = 7.46 E6. We replot these new data points as solid circles in Fig. 5.10. They correlate almost perfectly into a straight-line power-law function: pD^ Ap Lp} 0.155 pVDV'‘ M ) Ans. (c) All newtonian smooth pipe flows should correlate in this manner. This example is a varia¬ tion of the first completely successful dimensional analysis, pipe-flow friction, performed by PrandtTs student Paul Blasius, who published a related plot in 1911. For this range of (turbulent flow) Reynolds numbers, the pressure drop increases approximately as EXAMPLE 5.11 The smooth sphere data plotted in Fig. 5.3fl represent dimensionless drag versus dimension¬ less viscosity, since {p, V, d) were selected as scaling or repeating variables, {a) Replot these data to display the effect of dimensionless velocity on the drag, {b) Use your new figure to predict the terminal (zero-acceleration) velocity of a 1-cm-diameter steel ball (SG = 7.86) falling through water at 20°C. Solution • Assumptions: Fig 5.3a is valid for any smooth sphere in that Reynolds number range. • Approach (a): Form pi groups from the function F = fcn(d, V, p, p) in such a way that F is plotted versus V. The answer was already given as Eq. (5.16), but let us review the 324 Chapter 5 Dimensional Analysis and Similarity steps. The proper scaling variables are (p, p, d), which do not form a pi. Therefore j = 3, and we expect n — / = 5 — 3 = 2 pi groups. Skipping the algebra, they arise as follows: . pF . pyd rfi = p‘‘p!’d'' F = ^ FI2 = p‘‘u^d‘' V = - Ans. (a) p P We may replot the data of Fig. 5.3a in this new form, noting that IIi = (7r/8)(CD)(Re)^. This replot is shown as Fig. 5.1 1. The drag increases rapidly with velocity up to transition, where there is a slight drop, after which it increases more than ever. If force is known, we may predict velocity from the hgure, and vice versa. • Property values for part (b): p^^aa ^ 998 kg/m^ Mwater = 0.001 kg/(m-s) Fsteei = 7.86p„ater = 7844 kg/ml ■ Solution to part (b): For terminal velocity, the drag force equals the net weight of the sphere in water: F = = (ft - pjg'^d^ = (7840 - 998)(9.81)(f )(0.01)' = 0.0351 N Fig. 5.11 Cross-plot of sphere drag data from Fig. 5.3a to show dimensionless force versus dimensionless velocity. o’ t= 00 k. ' Re = pVd ~P~ Problems 325 Therefore, the ordinate of Fig. 5.11 is known: Falling steel sphere: pF _ (998 kg/m^) (0.0351 N) p? [0.001 kg/(m • s)]^ 3.5 E7 From Fig. 5.11, at pF/p^ ~ 3.5 E7, a magnifying glass reveals that Re^ = 2 E4. Then a crude estimate of the terminal fall velocity is pVd - — « 20,000 or 20,000[0.001 kg/(m • s)] m V « - - - -I- « 2.0 — Ans. (b) (998 kg/m^) (0.01 m) s • Comments: Better accuracy could be obtained by expanding the scale of Fig. 5.11 in the region of the given force coefficient. Flowever, there is considerable uncertainty in pub¬ lished drag data for spheres, so the predicted fall velocity is probably uncertain by at least ±10 percent. Note that we found the answer directly from Fig. 5.11. We could use Fig. 5.3fl also but would have to iterate between the ordinate and abscissa to obtain the final result, since V is contained in both plotted variables. Summary Chapters 3 and 4 presented integral and differential methods of mathematical analysis of fluid flow. This chapter introduces the third and final method: experimentation, as supplemented hy the technique of dimensional analysis. Tests and experiments are used both to strengthen existing theories and to provide useful engineering results when theory is inadequate. The chapter begins with a discussion of some familiar physical relations and how they can be recast in dimensionless form because they satisfy the principle of dimensional homogeneity. A general technique, the pi theorem, is then presented for systematically finding a set of dimensionless parameters by grouping a list of variables that govern any particular physical process. A second technique, Ipsen’s method, is also described. Alternately, direct application of dimensional analysis to the basic equations of fluid mechanics yields the fundamental parameters gov¬ erning flow patterns: Reynolds number, Froude number, Prandtl number, Mach number, and others. It is shown that model testing in air and water often leads to scaling difficulties for which compromises must be made. Many model tests do not achieve true dynamic similarity. The chapter ends by pointing out that classic dimensionless charts and data can be manipulated and recast to provide direct solutions to problems that would otherwise be quite cumbersome and laboriously iterative. Problems Most of the problems herein are fairly straightforward. More diffi¬ cult or open-ended assignments are labeled with an asterisk. Prob¬ lems labeled with a computer icon may require the use of a computer. The standard end-of-chapter problems P5.1 to P5.91 (categorized in the problem list here) are followed by word prob¬ lems W5.1 to W5.10, fundamentals of engineering exam problems FE5.1 to FE5.12, comprehensive applied problems C5.1 to C5.5, and design projects D5.1 and D5.2. 326 Chapter 5 Dimensional Analysis and Similarity Problem Distribution Section Topic Problems 5.1 Introduction P5.1-P5.9 5.2 The principle of dimensional homogeneity P5.10-P5.13 5.3 The pi theorem; Ipsen’s method P5.14-P5.42 5.4 Nondimensionalizing the basic equations P5.43-P5.47 5.4 Data for spheres, cylinders, other bodies P5.48-P5.59 5.5 Scaling of model data P5.60-P5.74 5.5 Froude and Mach number scaling P5.75-P5.84 5.5 Inventive rescaling of the data P5.85-P5.91 Introduction; dynamic similarity P5.1 For axial flow through a circular tube, the Reynolds num¬ ber for transition to turbulence is approximately 2300 [see Eq. (6.2)], based on the diameter and average velocity. If d = 5 cm and the fluid is kerosene at 20°C, find the volume flow rate in m^/h that causes transition. P5.2 A prototype automobile is designed for cold weather in Denver, CO (— 10°C, 83 kPa). Its drag force is to be tested on a one-seventh-scale model in a wind tunnel at 150 mi/h, 20°C, and 1 atm. If the model and proto¬ type are to satisfy dynamic similarity, what prototype velocity, in mi/h, needs to be matched? Comment on your result. P5.3 The transfer of energy by viscous dissipation is dependent upon viscosity fi, thermal conductivity k, stream velocity U, and stream temperature Tq. Group these quantities, if possible, into the dimensionless Brinkman number, which is proportional to /i. P5.4 When tested in water at 20°C flowing at 2 m/s, an 8-cm- diameter sphere has a measured drag of 5 N. What will be the velocity and drag force on a 1 .5-m-diameter weather balloon moored in sea-level standard air under dynami¬ cally similar conditions? P5.5 An automobile has a characteristic length and area of 8 ft and 60 ft^, respectively. When tested in sea-level standard air, it has the following measured drag force versus speed: V, mi/h 20 40 60 Drag, Ibf 31 115 249 The same car travels in Colorado at 65 mi/h at an altitude of 3500 m. Using dimensional analysis, estimate (a) its drag force and (b) the horsepower required to overcome air drag. P5.6 The disk-gap-band parachute in the chapter-opener photo had a drag of 1600 Ibf when tested at 15 mi/h in air at 20°C and 1 atm. (a) What was its drag coefficient? (b) If, as stated, the drag on Mars is 65,000 Ibf and the velocity is 375 mi/h in the thin Mars atmosphere, p ~ 0.020 kg/m^, what is the drag coefficient on Mars? (c) Can you explain the difference between (a) and (/;)? P5.7 A body is dropped on the moon (g = 1.62 m/s^) with an initial velocity of 12 m/s. By using option 2 variables, Eq. (5.11), the ground impact occurs at t = 0.34 and 5 = 0.84. Estimate (a) the initial displacement, (b) the final dis¬ placement, and (c) the time of impact. P5.8 The Archimedes number, Ar, used in the flow of stratified fluids, is a dimensionless combination of gravity g, density difference Ap, fluid width L, and viscosity p. Find the form of this number if it is proportional to g. P5.9 The Richardson number, Ri, which correlates the production of turbulence by buoyancy, is a dimensionless combination of the acceleration of gravity g, the fluid temperature Tg, the local temperature gradient dT/dz, and the local velocity gradient du/dz- Determine the form of the Richardson number if it is proportional to g. The principle of dimensional homogeneity P5.10 Determine the dimension {MLT&} of the following quantities: du d^T (a) pu — (b) {p - Pa) dA (c) pCp — — dx Jj dx dy (^ HI p^dxdydz at All quantities have their standard meanings; for example, p is density. P5.ll During World War II, Sir Geoffrey Taylor, a British fluid dynamicist, used dimensional analysis to estimate the wave speed of an atomic bomb explosion. He assumed that the blast wave radius R was a function of energy released E, air density p, and time t. Use dimensional reasoning to show how wave radius must vary with time. P5.12 The Stokes number, St, used in particle dynamics studies, is a dimensionless combination of five variables: accelera¬ tion of gravity g, viscosity p, density p, particle velocity U, and particle diameter D. (a) If St is proportional to p and inversely proportional to g, find its form, (b) Show that St is actually the quotient of two more traditional dimensionless groups. P5.13 The speed of propagation C of a capillary wave in deep water is known to be a function only of density p, wave¬ length A, and surface tension Y. Find the proper func¬ tional relationship, completing it with a dimensionless constant. For a given density and wavelength, how does the propagation speed change if the surface tension is doubled? Problems 327 The pi theorem or Ipsen’s method P5.14 Flow in a pipe is often measured with an orifice plate, as in Fig. P5.14. The volume flow 2 is a function of the pressure drop Ap across the plate, the fluid density p, the pipe diam¬ eter D, and the orifice diameter d. Rewrite this functional relationship in dimensionless form. P5.14 P5.15 The wall shear stress in a boundary layer is assumed to be a function of stream velocity U, boundary layer thick¬ ness A local turbulence velocity u', density p, and local pressure gradient dp/dx. Using (p, U, S) as repeating vari¬ ables, rewrite this relationship as a dimensionless function. P5.16 Convection heat transfer data are often reported as a heat transfer coefficient h, defined by Q = hA\T where Q = heat flow, J/s A = surface area, m^ AT = temperature difference, K The dimensionless form of h, called the Stanton number, is a combination of h, fluid density p, specific heat Cp, and flow velocity V. Derive the Stanton number if it is propor¬ tional to h. What are the units of hi P5.17 If you disturb a tank of length L and water depth h, the surface will oscillate back and forth at frequency fl, assumed here to depend also upon water density p and the acceleration of gravity g. (a) Rewrite this as a dimension¬ less function, fb) If a tank of water sloshes at 2.0 Hz on earth, how fast would it oscillate on Mars (g ~ 3.7 m/s^)? P5.18 Under laminar conditions, the volume flow Q through a small triangular- section pore of side length b and length L is a function of viscosity /i, pressure drop per unit length Ap/L, and b. Using the pi theorem, rewrite this relation in dimensionless form. How does the volume flow change if the pore size b is doubled? P5.19 The period of oscillation T of a water surface wave is assumed to be a function of density p, wavelength /, depth h. gravity g, and surface tension Y. Rewrite this relationship in dimensionless form. What results if Y is negligible? Hint: Take Z, p, and g as repeating variables. P5.20 A fixed cylinder of diameter D and length L, immersed in a stream flowing normal to its axis at velocity {/, will experi¬ ence zero average lift. However, if the cylinder is rotating at angular velocity fl, a lift force F will arise. The fluid density p is important, but viscosity is secondary and can be neglected. Formulate this lift behavior as a dimension¬ less function. P5.21 In Example 5. 1 we used the pi theorem to develop Eq. (5.2) from Eq. (5.1). Instead of merely listing the primary dimensions of each variable, some workers list the powers of each primary dimension for each variable in an array: M L T F ' 1 1 -2 L U p p, 0 0 1 1 1 1-3-1 0-1 0 -1 This array of exponents is called the dimensional matrix for the given function. Show that the rank of this matrix (the size of the largest nonzero determinant) is equal to ] = n — k, the desired reduction between original variables and the pi groups. This is a general property of dimensional matrices, as noted by Buckingham . P5.22 As will be discussed in Chap. 1 1 , the power P developed by a wind turbine is a function of diameter D, air density p, wind speed V, and rotation rate UJ. Viscosity effects are negligible. Rewrite this relationship in dimensionless form. P5.23 The period T of vibration of a beam is a function of its length L, area moment of inertia /, modulus of elasticity E, density p, and Poisson’s ratio cr. Rewrite this relation in dimensionless form. What further reduction can we make if E and I can occur only in the product form Ell Hint: Take L, p, and E as repeating variables. P5.24 The lift force F on a missile is a function of its length L, velocity V, diameter D, angle of attack a, density p, viscos¬ ity /i, and speed of sound a of the air. Write out the dimen¬ sional matrix of this function and determine its rank. (See Prob. P5.21 for an explanation of this concept.) Rewrite the function in terms of pi groups. P5.25 The thrust F of a propeller is generally thought to be a function of its diameter D and angular velocity fi, the for¬ ward speed V, and the density p and viscosity p of the fluid. Rewrite this relationship as a dimensionless function. P5.26 A pendulum has an oscillation period T which is assumed to depend on its length L, bob mass m, angle of swing 0, and the acceleration of gravity. A pendulum 1 m long, with a bob mass of 200 g, is tested on earth and found to have a period of 2.04 s when swinging at 20°. (a) What is its period when it swings at 45°? A similarly constructed pendulum. 328 Chapter 5 Dimensional Analysis and Similarity with L = 30 cm and ?n = 100 g, is to swing on the moon (g = 1.62 m/s^) at 0 = 20°. (b) What will he its period? P5.27 In studying sand transport hy ocean waves, A. Shields in 1936 postulated that the threshold wave-induced bottom shear stress r required to move particles depends on grav¬ ity g, particle size d and density Pp, and water density p and viscosity p. Find suitable dimensionless groups of this problem, which resulted in 1936 in the celebrated Shields sand transport diagram. P5.28 A simply supported beam of diameter D, length L, and mod¬ ulus of elasticity E is subjected to a fluid crossflow of veloc¬ ity V, density p, and viscosity p. Its center deflection <5 is assumed to be a function of all these variables, (a) Rewrite this proposed function in dimensionless form, (b) Suppose it is known that <5 is independent of p, inversely proportional to E, and dependent only on pV^, not p and V separately. Simplify the dimensionless function accordingly. Elint: Take L, p, and V as repeating variables. P5.29 When fluid in a pipe is accelerated linearly from rest, it begins as laminar flow and then undergoes transition to turbulence at a time that depends on the pipe diame¬ ter D, fluid acceleration a, density p, and viscosity p. Arrange this into a dimensionless relation between ftr and D. P5.30 When a large tank of high-pressure gas discharges through a nozzle, the exit mass flow ?h is a function of tank pressure Po and temperature Tq, gas constant R, specific heat Cp, and nozzle diameter D. Rewrite this as a dimensionless func¬ tion. Check to see if you can use (po, Tq, R, D) as repeating variables. P5.31 The pressure drop per unit length in horizontal pipe flow, Ap/L, depends on the fluid density p, viscosity p, diameter D, and volume flow rate Q. Rewrite this function in terms of pi groups. P5.32 A weir is an obstruction in a channel flow that can be cali¬ brated to measure the flow rate, as in Fig. P5.32. The vol¬ ume flow Q varies with gravity g, weir width b into the paper, and upstream water height H above the weir crest. If it is known that Q is proportional to b, use the pi theorem to find a unique functional relationship Q{g, b, ET). P5.32 P5.33 A spar buoy (see Prob. P2.113) has a period T of vertical (heave) oscillation that depends on the waterline cross- sectional area A, buoy mass m, and fluid specific weight 7. How does the period change due to doubling of (a) the mass and (b) the area? Instrument buoys should have long periods to avoid wave resonance. Sketch a possible long- period buoy design. P5.34 To good approximation, the thermal conductivity k of a gas (see Ref. 21 of Chap. 1) depends only on the density p, mean free path /, gas constant R, and absolute temperature T. For air at 20°C and 1 atm, k « 0.026 W/(m ■ K) and I « 6.5 E-8 m. Use this information to determine k for hydro¬ gen at 20°C and 1 atm if / ~ 1.2 E-7 m. P5.35 The torque M required to turn the cone-plate viscometer in Eig. P5.35 depends on the radius R, rotation rate fi, fluid viscosity p, and cone angle 9. Rewrite this relation in dimensionless form. How does the relation simplify it if it is known that M is proportional to 91 P5.35 P5.36 The rate of heat loss through a window or wall is a function of the temperature difference between inside and outside AT, the window surface area A, and the R value of the window, which has units of (ft^ • h ■ °E)/ Btu. (a) Using the Buckingham Pi Theorem, find an expression for rate of heat loss as a function of the other three parameters in the problem, (b) If the temperature difference AT doubles, by what factor does the rate of heat loss increase? P5.37 The volume flow Q through an orifice plate is a function of pipe diameter D, pressure drop Ap across the orifice, fluid density p and viscosity p, and orifice diameter d. Using D, p, and Ap as repeating variables, express this relationship in dimensionless form. P5.38 The size d of droplets produced by a liquid spray nozzle is thought to depend on the nozzle diameter D, Jet velocity U, and the properties of the liquid p, p, and Y. Rewrite this relation in dimensionless form. Elint: Take D, p, and U as repeating variables. Problems 329 P5.39 The volume flow Q over a certain dam is a function of dam width b, gravity g, and the upstream water depth H above the dam crest. It is known that Q is proportional to b.lfb = 120 ft and H = 15 in., the flow rate is 600 ftVs. What will be the flow rate if H = 3 ft? P5.40 The time tj to drain a liquid from a hole in the bottom of a tank is a function of the hole diameter d, the initial fluid volume -uq, the initial liquid depth hg, and the density p and viscosity p, of the fluid. Rewrite this relation as a dimen¬ sionless function, using Ipsen’s method. P5.41 A certain axial flow turbine has an output torque M that is proportional to the volume flow rate Q and also depends on the density p, rotor diameter D, and rotation rate O. How does the torque change due to a doubling of (a) D and (b) H? P5.42 When disturbed, a floating buoy will bob up and down at frequency/. Assume that this frequency varies with buoy mass m, waterline diameter d, and the specific weight 7 of the liquid, (a) Express this as a dimensionless function, (b) If d and 7 are constant and the buoy mass is halved, how will the frequency change? Nondimensionalizing the basic equations P5.43 P5.44 Nondimensionalize the energy equation (4.75) and its boundary conditions (4.62), (4.63), and (4.70) by defining T = T/Tg, where Tg is the inlet temperature, assumed con¬ stant. Use other dimensionless variables as needed from Eqs. (5.23). Isolate all dimensionless parameters you find, and relate them to the list given in Table 5.2. The differential energy equation for incompressible two- dimensional flow through a “Darcy-type” porous medium is approximately adpdT (jdpdT d^T pcp— — — + pcp— — — + k — T - 0 p dx dx p dy dy dy^ P5.45 where cr is the permeability of the porous medium. All other symbols have their usual meanings, {a) What are the appropriate dimensions for cr? {b) Nondimensionalize this equation, using (L, U, p, Tg) as scaling constants, and dis¬ cuss any dimensionless parameters that arise. A model differential equation, for chemical reaction dynamics in a plug reactor, is as follows: ac dx = D ax dx^ - kC - d£ dt where u is the velocity, D is a diffusion coefficient, k is a reaction rate, x is distance along the reactor, and C is the (dimensionless) concentration of a given chemical in the reactor, (a) Determine the appropriate dimensions of D and k. (b) Using a characteristic length scale L and average velocity V as parameters, rewrite this equation in dimensionless form and comment on any pi groups appearing. P5.46 If a vertical wall at temperature is surrounded by a fluid at temperature Tg, a natural convection boundary layer flow will form. For laminar flow, the momentum equation is du dll d^u p{u— + V—) = pl3{T - Tg)g + p — dx dy dy to be solved, along with continuity and energy, for (u, v, T) with appropriate boundary conditions. The quantity (3 is the thermal expansion coefficient of the fluid. Use p, g, L, and (T,f,—Tg) to nondimensionalize this equation. Note that there is no “stream” velocity in this type of flow. P5.47 The differential equation for small-amplitude vibrations y(x, t) of a simple beam is given by d^y d'^y pA^ + EI—^ = 0 dt^ dx^ where p = beam material density A = cross-sectional area I = area moment of inertia E = Young’s modulus Use only the quantities p, E, and A to nondimensionalize y, x, and t, and rewrite the differential equation in dimensionless form. Do any parameters remain? Could they be removed by further manipulation of the variables? Data for spheres, cylinders, other bodies P5.48 A smooth steel (SG = 7.86) sphere is immersed in a stream of ethanol at 20°C moving at 1.5 m/s. Estimate its drag in N from Fig. 5.3a. What stream velocity would quadruple its drag? Take D = 2.5 cm. P5.49 The sphere in Prob. P5.48 is dropped in gasoline at 20°C. Ignoring its acceleration phase, what will its terminal (con¬ stant) fall velocity be, from Fig. 5.3a? P5.50 The parachute in the chapter-opener photo is, of course, meant to decelerate the payload on Mars. The wind tunnel test gave a drag coefficient of about 1.1, based upon the projected area of the parachute. Suppose it was falling on earth and, at an altitude of 1000 m, showed a steady descent rate of about 18 mi/h. Estimate the weight of the payload. P5.51 A ship is towing a sonar array that approximates a sub¬ merged cylinder 1 ft in diameter and 30 ft long with its axis normal to the direction of tow. If the tow speed is 12 kn (1 kn = 1.69 ft/s), estimate the horsepower required to tow this cylinder. What will be the frequency of vortices shed from the cylinder? Use Figs. 5.2 and 5.3. P5.52 When fluid in a long pipe starts up from rest at a uniform acceleration a, the initial flow is laminar. The flow under¬ goes transition to turbulence at a time t which depends, to first approximation, only upon a, p, and p. Experiments by 330 Chapter 5 Dimensional Analysis and Similarity P5.53 P5.54 P5.55 P5.56 P5.57 P5.58 P. J. Lefebvre, on water at 20°C starting from rest with 1-g acceleration in a 3-cm-diameter pipe, showed transition at t = 1.02 s. Use this data to estimate {a) the transition time and (b) the transition Reynolds number Re^ for water flow accelerating at 35 m/s^ in a 5-cm-diameter pipe. Vortex shedding can be used to design a vortex flowmeter (Fig. 6.34). A blunt rod stretched across the pipe sheds vortices whose frequency is read by the sensor down¬ stream. Suppose the pipe diameter is 5 cm and the rod is a cylinder of diameter 8 mm. If the sensor reads 5400 counts per minute, estimate the volume flow rate of water in m^/h. How might the meter react to other liquids? A hshnet is made of 1-mm-diameter strings knotted into 2X2 cm squares. Estimate the horsepower required to tow 300 ft^ of this netting at 3 kn in seawater at 20°C. The net plane is normal to the flow direction. The radio antenna on a car begins to vibrate wildly at 8 Hz when the car is driven at 45 mi/h over a rutted road that approximates a sine wave of amplitude 2 cm and wave¬ length X. = 2.5 m. The antenna diameter is 4 mm. Is the vibration due to the road or to vortex shedding? Flow past a long cylinder of square cross-section results in more drag than the comparable round cylinder. Here are data taken in a water tunnel for a square cylinder of side length b = 2 cm: V, m/s 1.0 2.0 3.0 4.0 Drag, N/(in of depth) 21 85 191 335 (a) Use these data to predict the drag force per unit depth of wind blowing at 6 m/s, in air at 20°C, over a tall square chimney of side length b = 55 cm. (b) Is there any uncer¬ tainty in your estimate? The simply supported 1040 carbon-steel rod of Fig. P5.57 is subjected to a crossflow stream of air at 20°C and 1 atm. For what stream velocity U will the rod center deflection be approximately 1 cm? P5.57 For the steel rod of Prob. P5.57, at what airstream velocity U will the rod begin to vibrate laterally in resonance in its first mode (a half sine wave)? Hint: Consult a vibration text [34,35] under “lateral beam vibration.” P5.59 A long, slender, smooth 3-cm-diameter flagpole bends alarmingly in 20 mi/h sea-level winds, causing patriotic citizens to gasp. An engineer claims that the pole will bend less if its surface is deliberately roughened. Is she correct, at least qualitatively? Scaling of model data P5.60 The thrust F of a free propeller, either aircraft or marine, depends upon density p, the rotation rate n in r/s, the diame¬ ter D, and the forward velocity V. Viscous effects are slight and neglected here. Tests of a 25-cm-diameter model aircraft propeller, in a sea-level wind tunnel, yield the following thrust data at a velocity of 20 m/s: Rotation rate, r/min 4800 6000 8000 Measured thrust, N 6.1 19 47 (a) Use this data to make a crude but effective dimension¬ less plot, (b) Use the dimensionless data to predict the thrust, in newtons, of a similar 1.6-m-diameter prototype propeller when rotating at 3800 r/min and flying at 225 mi/h at 4000-m standard altitude. P5.61 If viscosity is neglected, typical pump flow results flrom Ex¬ ample 5.3 are shown in Fig. P5.61 for a model pump tested in water. The pressure rise decreases and the power required in¬ creases with the dimensionless flow coefficient. Curve-fit ex¬ pressions are given for the data. Suppose a similar pump of 12-cm diameter is built to move gasoline at 20°C and a flow rate of 25 m^/h. If the pump rotation speed is 30 r/s, find (a) the pressure rise and (b) the power required. - ; = flow coefficient flD^ P5.61 P5.62 For the system of Prob. P5.22, assume that a small model wind turbine of diameter 90 cm, rotating at 1200 r/min, delivers 280 watts when subjected to a wind of 12 m/s. The data is to be used for a prototype of diameter 50 m and winds of 8 m/s. For dynamic similarity, estimate (a) the rotation rate, and (b) the power delivered by the prototype. Assume sea-level air density. Problems 331 P5.63 The Keystone Pipeline in the Chapter 6 opener photo has D = 36 in. and an oil flow rate Q = 590,000 barrels per day (1 barrel = 42 U.S. gallons). Its pressure drop per unit length, A/j/L, depends on the fluid density p, viscosity p, diameter D, and flow rate Q. A water-flow model test, at 20°C, uses a 5-cm-diameter pipe and yields Ap/L « 4000 Pa/m. For dynamic similarity, estimate Ap/L of the pipeline. For the oil take p = 860 kg/m^ and p = 0.005 kg/m ■ s. P5.64 The natural frequency UJ of vibration of a mass M attached to a rod, as in Fig. P5.64, depends only on M LO P5.64 and the stiffness El and length L of the rod. Tests with a 2-kg mass attached to a 1040 carbon steel rod of diameter 12 mm and length 40 cm reveal a natural frequency of 0.9 Hz. Use these data to predict the natural frequency of a 1-kg mass attached to a 2024 aluminum alloy rod of the same size. P5.65 In turbulent flow near a flat wall, the local velocity u varies only with distance y from the wall, wall shear stress and fluid properties p and p. The following data were taken in the University of Rhode Island wind tunnel for airflow, p = 0.0023 slug/ft^, p = 3.81 E-7 slug/(ft • s), and r„ = 0.029 Ibf/ft^: P5.67 A student needs to measure the drag on a prototype of char¬ acteristic dimension dp moving at velocity Up in air at stan¬ dard atmospheric conditions. He constructs a model of characteristic dimension d^, such that the ratio dp/d„ is some factor/. He then measures the drag on the model at dynamically similar conditions (also with air at standard atmospheric conditions). The student claims that the drag force on the prototype will be identical to that measured on the model. Is this claim correct? Explain. P5.68 Eor the rotating-cylinder function of Prob. P5.20, if L >> D, the problem can be reduced to only two groups, F/(pU^LD) versus (UlD/U). Here are experimental data for a cylinder 30 cm in diameter and 2 m long, rotating in sea-level air, with U = 25 m/s. fi, rev/min 0 3000 6000 9000 12000 15000 F, N 0 850 2260 2900 3120 3300 (a) Reduce this data to the two dimensionless groups and make a plot, (b) Use this plot to predict the lift of a cylinder with D = 5 cm, L = 80 cm, rotating at 3800 rev/min in water alU = 4 m/s. P5.69 A simple flow measurement device for streams and chan¬ nels is a notch, of angle a, cut into the side of a dam, as shown in Pig. P5.69. The volume flow Q depends only on a, the acceleration of gravity g, and the height d of the upstream water surface above the notch vertex. Tests of a model notch, of angle a = 55°, yield the following flow rate data: S, cm 10 20 30 40 Q, m^/h 8 47 126 263 (a) Pind a dimensionless correlation for the data, (ft) Use the model data to predict the flow rate of a prototype notch, also of angle a = 55°, when the upstream height is 3.2 m. V, in 0.021 0.035 0.055 0.080 0.12 0.16 u, ft/s 50.6 54.2 57.6 59.7 63.5 65.9 (a) Plot these data in the form of dimensionless u versus dimensionless y, and suggest a suitable power-law curve fit. (b) Suppose that the tunnel speed is increased until « = 90 ft/s at y = 0. 1 1 in. Estimate the new wall shear stress, in Ibf/ftl P5.66 A torpedo 8 m below the surface in 20°C seawater cavitates at a speed of 21 m/s when atmospheric pressure is 101 kPa. If Reynolds number and Proude number effects are negli¬ gible, at what speed will it cavitate when running at a depth of 20 m? At what depth should it be to avoid cavitation at 30 m/s? P5.70 A diamond-shaped body, of characteristic length 9 in, has the following measured drag forces when placed in a wind tunnel at sea-level standard conditions: V, ft/s 30 38 48 56 61 F, Ibf 1.25 1.95 3.02 4.05 4.81 332 Chapter 5 Dimensional Analysis and Similarity Use these data to predict the drag force of a similar 15-in diamond placed at similar orientation in 20°C water flow¬ ing at 2.2 m/s. P5.71 The pressure drop in a venturi meter (Fig. P3 . 1 28) varies only with the fluid density, pipe approach velocity, and diameter ratio of the meter. A model venturi meter tested in water at 20°C shows a 5-kPa drop when the approach velocity is 4 m/s. A geometrically similar prototype meter is used to measure gasoline at 20°C and a flow rate of 9 mVmin. If the prototype pressure gage is most accurate at 15 kPa, what should the upstream pipe diameter be? P5.72 A one-twelfth-scale model of a large commercial aircraft is tested in a wind tunnel at 20°C and 1 atm. The model chord length is 27 cm, and its wing area is 0.63 m^. Test results for the drag of the model are as follows: V, mi/h 50 75 100 125 Drag, N 15 32 53 80 In the spirit of Fig. 5.8, use this data to estimate the drag of the full-scale aircraft when flying at 550 mi/h, for the same angle of attack, at 32,800 ft standard altitude. P5.73 The power P generated by a certain windmill design depends on its diameter D, the air density p, the wind velocity V, the rotation rate fl, and the number of blades n. (a) Write this re¬ lationship in dimensionless form. A model windmill, of diam¬ eter 50 cm, develops 2.7 kW at sea level when V = 40 m/s and when rotating at 4800 r/min. (b) What power will be devel¬ oped by a geometrically and dynamically similar prototype, of diameter 5 m, in winds of 12 m/s at 2000 m standard altitude? (c ) What is the appropriate rotation rate of the prototype? P5.74 A one-tenth-scale model of a supersonic wing tested at 700 m/s in air at 20°C and 1 atm shows a pitching moment of 0.25 kN ■ m. If Reynolds number effects are negligible, what will the pitching moment of the prototype wing be if it is flying at the same Mach number at 8-km standard altitude? Froude and Mach number scaling P5.75 According to the web site USGS Daily Water Data for the Nation, the mean flow rate in the New River near Hinton, WV, is 10,100 ftVs. If the hydraulic model in Fig. 5.9 is to match this condition with Froude number scaling, what is the proper model flow rate? P5.76 A 2-ft-long model of a ship is tested in a freshwater tow tank. The measured drag may be split into “friction” drag (Reynolds scaling) and “wave” drag (Froude scaling). The model data are as follows: Tow speed, ft/s 0.8 1.6 2.4 3.2 4.0 4.8 Friction drag, Ibf 0.016 0.057 0.122 0.208 0.315 0.441 Wave drag, Ibf 0.002 0.021 0.083 0.253 0.509 0.697 The prototype ship is 150 ft long. Estimate its total drag when cruising at 15 kn in seawater at 20°C. P5.77 A dam 75 ft wide, with a nominal flow rate of 260 ft^, is to be studied with a scale model 3 ft wide, using Froude scal¬ ing. (a) What is the expected flow rate for the model? (b) What is the danger of only using Froude scaling for this test? (c) Derive a formula for a force on the model as com¬ pared to a force on the prototype. P5.78 A prototype spillway has a characteristic velocity of 3 m/s and a characteristic length of 10 m. A small model is con¬ structed by using Froude scaling. What is the minimum scale ratio of the model that will ensure that its minimum Weber number is 100? Both flows use water at 20°C. P5.79 An East Coast estuary has a tidal period of 12.42 h (the semidiurnal lunar tide) and tidal currents of approximately 80 cm/s. If a one-five-hundredth-scale model is constructed with tides driven by a pump and storage apparatus, what should the period of the model tides be and what model current speeds are expected? P5.80 A prototype ship is 35 m long and designed to cruise at 1 1 m/s (about 21 kn). Its drag is to be simulated by a 1-m- long model pulled in a tow tank. Eor Eroude scaling find (a) the tow speed, (b) the ratio of prototype to model drag, and (c) the ratio of prototype to model power. P5.81 An airplane, of overall length 55 ft, is designed to fly at 680 m/s at 8000-m standard altitude. A one-thirtieth-scale model is to be tested in a pressurized helium wind tunnel at 20°C. What is the appropriate tunnel pressure in atm? Even at this (high) pressure, exact dynamic similarity is not achieved. Why? P5.82 A one-fiftieth-scale model of a military airplane is tested at 1020 m/s in a wind tunnel at sea-level conditions. The model wing area is 180 cm^. The angle of attack is 3°. If the measured model lift is 860 N, what is the prototype lift, using Mach number scaling, when it flies at 10,000 m stan¬ dard altitude under dynamically similar conditions? Note: Be careful with the area scaling. P5.83 A one-fortieth-scale model of a ship’s propeller is tested in a tow tank at 1200 r/min and exhibits a power output of 1.4 ft ■ Ibf/s. According to Froude scaling laws, what should the revolutions per minute and horsepower output of the proto¬ type propeller be under dynamically similar conditions? P5.84 A prototype ocean platform piling is expected to encounter currents of 150 cm/s and waves of 12-s period and 3-m height. If a one-fifteenth-scale model is tested in a wave channel, what current speed, wave period, and wave height should be encountered by the model? Inventive rescaling of the data P5.85 As shown in Example 5.3, pump performance data can be nondimensionalized. Problem P5.61 gave typical Word Problems 333 P5.86 P5.87 P5.88 P5.89 dimensionless data for centrifugal pump “head,” H = Ap/pg, as follows: gH nW « 6.0 - where Q is the volume flow rate, n the rotation rate in r/s, and D the impeller diameter. This type of correlation allows one to compute H when (p, Q, D) are known, (a) Show how to rearrange these pi groups so that one can size the pump, that is, compute D directly when {Q, H, n) are known, {b) Make a crude but effective plot of your new function, (c) Apply part (b) to the following example: Find D when // = 37 m, g = 0. 14 m^/s, and n = 35 r/s. Find the pump diameter for this condition. Solve Prob. P5.49 for glycerin at 20°C, using the modified sphere-drag plot of Fig. 5.11. In Prob. P5.61 it would be difficult to solve for £1 because it appears in all three of the dimensionless pump coefficients. Suppose that, in Prob. 5.61, fi is unknown but D = 12 cm and Q = 25 m^/h. The fluid is gasoline at 20°C. Rescale the coefficients, using the data of Prob. P5.61, to make a plot of dimensionless power versus dimensionless rotation speed. Enter this plot to find the maximum rotation speed fi for which the power will not exceed 300 W. Modify Prob. P5.61 as follows: Let fi = 32 r/s and g = 24 m^/h for a geometrically similar pump. What is the maxi¬ mum diameter if the power is not to exceed 340 W? Solve this problem by rescaling the data of Fig. P5.61 to make a plot of dimensionless power versus dimensionless diame¬ ter. Enter this plot directly to find the desired diameter. Wall friction t„, for turbulent flow at velocity 17 in a pipe of diameter D, was correlated, in 1911, with a dimensionless correlation by Ludwig Prandtl’s student H. Blasius: T„ ^ 0.632 pU^ ~ ipUDIpf"^ Suppose that (p, (/, p, t„) were all known and it was de¬ sired to find the unknown velocity U. Rearrange and re¬ write the formula so that U can be immediately calculated. P5.90 Knowing that Ap is proportional to L, rescale the data of Ex¬ ample 5.10 to plot dimensionless Ap versus dimensionless viscosity. Use this plot to find the viscosity required in the first row of data in Example 5.10 if the pressure di'op is increased to 10 kPa for the same flow rate, length, and density. P5.91 The traditional “Moody-type” pipe friction correlation in Chap. 6 is of the form ^ pV^L fen (T-i) where D is the pipe diameter, L the pipe length, and £ the wall roughness. Note that pipe average velocity V is used on both sides. This form is meant to find Ap when V is known, (a) Suppose that Ap is known, and we wish to find V. Rearrange the above function so that V is isolated on the left-hand side. Use the following data, for e/D = 0.005, to make a plot of your new function, with your velocity parameter as the ordinate of the plot. / 0.0356 0.0316 0.0308 0.0305 0.0304 pVD/p 15,000 75,000 250,000 900,000 3,330,000 (b) Use your plot to determine V, in m/s, for the following pipe flow: D = 5 cm, £ = 0.025 cm, L = 10 m, for water flow at 20°C and 1 atm. The pressure drop Ap is 1 10 kPa. Word Problems W5.1 In 98 percent of data analysis cases, the “reducing fac¬ tor” j, which lowers the number n of dimensional vari¬ ables to n — j dimensionless groups, exactly equals the number of relevant dimensions (M, L, T, @). In one case (Example 5.5) this was not so. Explain in words why this situation happens. W5.2 Consider the following equation: 1 dollar bill = 6 in. Is this relation dimensionally inconsistent? Does it satisfy the PDH? Why? W5.3 In making a dimensional analysis, what rules do you follow for choosing your scaling variables? W5.4 In an earlier edition, the writer asked the following ques¬ tion about Eig. 5.1: “Which of the three graphs is a more effective presentation?” Why was this a dumb question? W5.5 This chapter discusses the difficulty of scaling Mach and Reynolds numbers together (an airplane) and Eroude and Reynolds numbers together (a ship). Give an example of a flow that would combine Mach and Froude numbers. Would there be scaling problems for common fluids? W5.6 What is different about a very small model of a weir or dam (Fig. P5.32) that would make the test results difficult to relate to the prototype? W5.7 What else are you studying this term? Give an example of a popular equation or formula from another course (ther¬ modynamics, strength of materials, or the like) that does not satisfy the principle of dimensional homogeneity. Explain what is wrong and whether it can be modified to be homogeneous. W5.8 Some colleges (such as Colorado State University) have environmental wind tunnels that can be used to study phe¬ nomena like wind flow over city buildings. What details of scaling might be important in such studies? 334 Chapter 5 Dimensional Analysis and Similarity W5.9 If the model scale ratio is ct = L„/Lp, as in Eq. (5.31), and the Weher number is important, how must the model and prototype surface tension he related to a for dynamic similarity? Fundamentals of Engineering Exam Problems FE5.1 Given the parameters ((/, L, g, p, p) that affect a certain liquid flow problem, the ratio V^/(Lg) is usually known as the (a) velocity head, (b) Bernoulli head, (c) Froude number, (d) kinetic energy, (e) impact energy FE5.2 A ship 150 m long, designed to cruise at 18 kn, is to be tested in a tow tank with a model 3 m long. The appropriate tow velocity is (fl)0.19 m/s, (/i) 0.35 m/s, (c) 1.31 m/s, (rf) 2.55 m/s, (e) 8.35 m/s EE5.3 A ship 150 m long, designed to cruise at 18 kn, is to be tested in a tow tank with a model 3 m long. If the model wave drag is 2.2 N, the estimated full-size ship wave drag is (a) 5500 N, (b) 8700 N, (c) 38,900 N, (i/) 6 1,800 N, (e) 275,000 N EE5.4 A tidal estuary is dominated by the semidiurnal lunar tide, with a period of 12.42 h. If a 1:500 model of the estuary is tested, what should be the model tidal period? (a) 4.0 s, (b) 1.5 min, (c) 17 min, (d) 33 min, (e) 64 min EE5.5 A football, meant to be thrown at 60 mi/h in sea-level air (p = 1.22 kg/m^, p = 1.78 E-5 N • s/m^), is to be tested using a one-quarter scale model in a water tunnel (p = 998 kg/m^, p = 0.0010 N ■ s/m^). For dynamic similarity, what is the proper model water velocity? (a) 7.5 mi/h, (b) 15.0 mi/h, (c) 15.6 mi/h, (d) 16.5 mi/h, (e) 30 mi/h EE5.6 A football, meant to be thrown at 60 mi/h in sea-level air (p = 1.22 kg/m^, p = 1.78 E-5 N • m^), is to be tested using a one-quarter scale model in a water tunnel (p = 998 kg/m^, p = 0.0010 N ■ s/m^). For dynamic similarity, what is the ratio of prototype force to model force? (a) 3.86:1, (b) 16:1, (c) 32:1, (d) 56:1, (e) 64:1 Comprehensive Problems C5.1 Estimating pipe wall friction is one of the most common tasks in fluids engineering. For long circular rough pipes in turbulent flow, wall shear is a function of density p, viscosity p, average velocity V, pipe diameter d, and wall roughness height e. Thus, functionally, we can write T„ = fcn(p, p, V, d, e). (a) Using dimensional analysis, rewrite this function in dimensionless form, (b) A certain W5.10 For a typical incompressible velocity potential analysis in Chap. 8 we solve V^(p = 0, subject to known values of d(p/dn on the boundaries. What dimensionless parameters govern this type of motion? EE5.7 Consider liquid flow of density p, viscosity p, and velocity U over a very small model spillway of length scale L, such that the liquid surface tension coefficient Y is impor¬ tant. The quantity pU^L/Y in this case is important and is called the (a) capillary rise, (b) Froude number, (c) Prandtl number, (d) Weber number, (e) Bond number EE5.8 If a stream flowing at velocity U past a body of length L causes a force F on the body that depends only on U, L, and fluid viscosity p, then F must be proportional to (a) pUL/p, ib) pU^L^, (c) pU/L, (d) pUL, (e) ULip EE5.9 In supersonic wind tunnel testing, if different gases are used, dynamic similarity requires that the model and proto¬ type have the same Mach number and the same (a) Euler number, (b) speed of sound, (c) stagnation enthalpy, (d) Froude number, (e) specific-heat ratio EE5.10 The Reynolds number for a 1-ft-diameter sphere moving at 2.3 mi/h through seawater (specific gravity 1.027, viscosity 1.07 E-3 N ■ s/m^) is approximately (fl) 300, (b) 3000, (c) 30,000, (d) 300,000, (e) 3,000,000 EE5.11 The Ekman number, important in physical oceanography, is a dimensionless combination of p, L, p, and the earth’s rotation rate fi. If the Ekman number is proportional to fi, it should take the form (a) pflV/p, (b) pnUp, (c) pflL/p, (d) pilL^/p, (e) pCULp EE5.12 A valid, but probably useless, dimensionless group is given by (pTQg)/(YLa), where everything has its usual mean¬ ing, except a. What are the dimensions of oP. (a) 0L"'T“‘, (b) 0L"'T“^ (c) 0ML“‘, (d) 0“‘LT“‘, (e) 0LT“‘ pipe has d = 5 cm and e = 0.25 mm. For flow of water at 20°C, measurements show the following values of wall shear stress: Q, gal/min 1.5 3.0 6.0 9.0 12.0 14.0 T„, Pa 0.05 0.18 0.37 0.64 0.86 1.25 Design Projects 335 Plot these data using the dimensionless form obtained in part (a) and suggest a curve-fit formula. Does your plot reveal the entire functional relation obtained in part (a)? C5.2 When the fluid exiting a nozzle, as in Fig. P3.49, is a gas, instead of water, compressibility may be important, espe¬ cially if upstream pressure pi is large and exit diameter <^2 is small. In this case, the difference pi — p2 is no longer controlling, and the gas mass flow m reaches a maximum value that depends on pi and ^2 and also on the absolute upstream temperature and the gas constant R. Thus, functionally, m = fcn(pi, d2, Ti, R). (a) Using dimensional analysis, rewrite this function in dimensionless form. (b) A certain pipe has d2 = I cm. For flow of air, measure¬ ments show the following values of mass flow through the nozzle: Ti, K 300 300 300 500 800 Pi, kPa 200 250 300 300 300 m, kg/s 0.037 0.046 0.055 0.043 0.034 Plot these data in the dimensionless form obtained in part (a). Does your plot reveal the entire functional relation obtained in part (a)? C5.3 Reconsider the fully developed draining vertical oil film problem (see Fig. P4.80) as an exercise in dimensional analysis. Let the vertical velocity be a function only of distance from the plate, fluid properties, gravity, and film thickness. That is, w = fcn(x, p, p,, g, <5). (a) Use the pi theorem to rewrite this function in terms of Design Projects D5.1 We are given laboratory data, taken by Prof. Robert Kircbhoff and his students at the University of Massachusetts, for the spin rate of a 2-cup anemometer. The anemometer was made of ping-pong balls (d = 1.5 in) split in half, facing in opposite directions, and glued to thin (j-in) rods pegged to a center axle. (See Fig. P7.91 for a sketch.) There were four rods, of lengths / = 0.212, 0.322, 0.458, and 0.574 ft. The experimental data, for wind tunnel velocity U and rotation rate fi, are as follows: 1 = 0.212 / = 0.322 1 = 0.458 1 = 0.574 l/,ft/s fi, r/min u,m fi, r/min l/,ft/s n, r/min l/,ft/s fi, r/min 18.95 435 18.95 225 20.10 140 23.21 115 22.20 545 23.19 290 26.77 215 27.60 145 25.90 650 29.15 370 31.37 260 32.07 175 29.94 760 32.79 425 36.05 295 36.05 195 38.45 970 38.45 495 39.03 327 39.60 215 dimensionless parameters, (b) Verify that the exact solution from Prob. P4.80 is consistent with your result in part (a). C5.4 The Taco Inc. model 4013 centrifugal pump has an impel¬ ler of diameter D = 12.95 in. When pumping 20°C water at D = 1160 r/min, the measured flow rate Q and pressure rise Ap are given by the manufacturer as follows: Q, gal/min 200 300 400 500 600 700 Ap, Ib/in^ 36 35 34 32 29 23 (a) Assuming that Ap = fcn(p, Q, D, D), use the pi theorem to rewrite this function in terms of dimensionless parame¬ ters and then plot the given data in dimensionless form. (b) It is desired to use the same pump, running at 900 r/min, to pump 20°C gasoline at 400 gal/min. According to your dimensionless correlation, what pressure rise Ap is expected, in Ibf/in^? C5.5 Does an automobile radio antenna vibrate in resonance due to vortex shedding? Consider an antenna of length L and diameter D. According to beam vibration theory [see or [35, p. 401]], the first mode natural frequency of a solid circular cantilever beam is UJ,, = 3.516[£'//(pAL"^)]'^^, where E is the modulus of elasticity, I is the area moment of inertia, p is the beam material density, and A is the beam cross-section area, (a) Show that is proportional to the antenna radius R. (b) If the antenna is steel, with L = 60 cm and D = 4 mm, estimate the natural vibration frequency, in Hz. (c) Compare with the shedding frequency if the car moves at 65 mi/h. Assume that the angular velocity O of the device is a function of wind speed U, air density p and viscosity p, rod length /, and cup diameter d. For all data, assume air is at 1 atm and 20°C. Define appropriate pi groups for this problem, and plot the data in this dimensionless manner. Comment on the possible uncertainty of the results. As a design application, suppose we are to use this anemometer geometry for a large-scale (d = 30 cm) airport wind anemometer. If wind speeds vary up to 25 m/s and we desire an average rotation rate D = 120 r/min, what should be the proper rod length? What are possible limitations of your design? Predict the expected fi (in r/min) of your design as affected by wind speeds from 0 to 25 m/s. D5.2 By analogy with the cylinder drag data in Fig. 5.3b, spheres also show a strong roughness effect on drag, at least in the Reynolds number range 4 E4 < Reo < 3 E5, which accounts for the dimpling of golf balls to increase their distance traveled. Some experimental data for roughened 336 Chapter 5 Dimensional Analysis and Similarity spheres are given in Fig. D5.2. The figure also shows typical golf hall data. We see that some roughened spheres are better than golf halls in some regions. For the present study, let us neglect the hall’s spin, which causes the very important side-force or Magnus ejfect (see Fig. 8.15) and assume that the hall is hit without spin and follows the equations of motion for plane motion (x, z): mx = —F cos 6 mz = — Fsin 9 — W where F = cJ-—D^{x^ + z^) 9 = tan” ^ 2 4 x The ball has a particular C/j(Reo) curve from Fig. D5.2 and is struck with an initial velocity Vo™d angle 9q. Take the ball’s average mass to be 46 g and its diameter to be 4.3 cm. Assuming sea-level air and a modest but finite range of initial conditions, integrate the equations of motion to com¬ pare the trajectory of “roughened spheres” to actual golf ball calculations. Can the rough sphere outdrive a normal References 1. E. Buckingham, “On Physically Similar Systems: Illustra¬ tions of the Use of Dimensional Equations,” Phys. Rev., vol. 4, no. 4, 1914, pp. 345-376. 2. J. D. Anderson, Computational Fluid Dynamics: The Basics with Applications, McGraw-Hill, New York, 1995. 3. P. W. Bridgman, Dimensional Analysis, Yale University Press, New Haven, CT, 1922, rev. ed., 1963. 4. H. L. Langhaar, Dimensional Analysis and the Theory of Models, Wiley, New York, 1951. 5. E. C. Ipsen, Units, Dimensions, and Dimensionless Numbers, McGraw-Hill, New York, 1960. 6. H. G. Homung, Dimensional Analysis: Examples of the Use of Symmetry, Dover, New York, 2006. 7. E. S. Taylor, Dimensioned Ancdysis for Engineers, Clarendon Press, Oxford, England, 1974. 8. G. I. Barenblatt, Dimensional Analysis, Gordon and Breach, New York, 1987. 9. A. C. Palmer, Dimensional Analysis and Intelligent Experi¬ mentation, World Scientific Publishing, Hackensack, NJ, 2008. 10. T. Szirtes, Applied Dimensional Analysis and Modeling, 2d ed., Butterworth-Heinemann, Burlington, MA, 2006. 11. R. Esnault-Pelterie, Dimensioned Analysis and Metrology, E. Rouge, Lausanne, Switzerland, 1950. 12. R. Kurth, Dimensioned Analysis and Group Theory in Astro¬ physics, Pergamon, New York, 1972. 13. R. Kimball and M. Ross, Die Data Warehouse Toolkit: The Complete Guide to Dimensional Modeling, 2d ed., Wiley, New York, 2002. golf ball for any conditions? What roughness-effect differ¬ ences occur between a low-impact duffer and, say. Tiger Woods? Reynolds number, IJDIv D5.2 14. R. Nakon, Chemical Problem Solving Using Dimensional Analysis, Prentice-Hall, Upper Saddle River, NJ, 1990. 15. D. R. Maidment (ed.). Hydrologic and Hydraulic Modeling Support: With Geographic Information Systems, Environ¬ mental Systems Research Institute, Redlands, CA, 2000. 16. A. M. Curren, Dimensional Analysis for Meds, 4th ed., Delmar Cengage Learning, Independence, KY, 2009. 17. G. P. Craig, Clinical Calculations Made Easy: Solving Problems Using Dimensional Analysis, 4th ed., Lippincott Williams and Wilkins, Baltimore, MD, 2008. 18. M. Zlokarnik, Dimensional Analysis and Scale-Up in Chemical Engineering, Springer- Verlag, New York, 1991. 19. W. G. Jacoby, Data Theory and Dimensional Analysis, Sage, Newbury Park, CA, 1991. 20. B. Schepartz, Dimensional Analysis in the Biomedical Sciences, Thomas, Springfield, IL, 1980. 21. T. Horntvedt, Calculating Dosages Safely: A Dimensional Analysis Approach, F. A. Davis Co., Philadelphia, PA, 2012. 22. J. B. Bassingthwaighte et ah. Fractal Physiology, Oxford Univ. Press, New York, 1994. 23. K. J. Niklas, Plant Allometry: The Scaling of Form and Process, Univ. of Chicago Press, Chicago, 1994. 24. “Flow of Fluids through Valves, Fittings, and Pipes,” Crane Valve Group, Long Beach, CA, 1957 (now updated as a CD- ROM; see ). 25. A. Roshko, “On the Development of Turbulent Wakes from Vortex Streets,” AACA Rep. 1191, 1954. References 337 26. G. W. Jones, Jr., “Unsteady Lift Forces Generated by Vortex Shedding about a Large, Stationary, Oscillating Cylinder at High Reynolds Numbers,” ASME Symp. Unsteady Flow, 1968. 27. O. M. Griffin and S. E. Ramberg, “The Vortex Street Wakes of Vibrating Cylinders,” J. Fluid Mech., vol. 66, pt. 3, 1974, pp. 553-576. 28. Encyclopedia of Science and Technology, 1 1th ed., McGraw- Hill, New York, 2012. 29. J. Kunes, Dimensionless Physical Quantities in Science and Engineering, Elsevier, New York, 2012. 30. V. P. Singh et al. (eds.). Hydraulic Modeling, Water Resources Publications LLC, Highlands Ranch, CO, 1999. 31. L. Armstrong, Hydraulic Modeling and GIS, ESRI Press, La Vergne, TN, 2011. 32. R. Ettema, Hydraulic Modeling: Concepts and Practice, American Society of Civil Engineers, Reston, VA, 2000. 33. R. D. Blevins, Applied Fluid Dynamics Handbook, van Nostrand Reinhold, New York, 1984. 34. W. J. Palm III, Mechanical Vibration, Wiley, New York, 2006. 35. S. S. Rao, Mechanical Vibrations, 5th ed., Prentice-Hall, Upper Saddle River, NJ, 2010. 36. G. I. Barenblatt, Scaling, Cambridge University Press, Cambridge, UK, 2003. 37. L. J. Eingersh, “Unsteady Aerodynamics Experiment,” Journal of Solar Energy Engineering, vol. 123, Nov. 2001, p. 267. 38. J. B. Barlow, W. H. Rae, and A. Pope, Low-Speed Wind Tunnel Testing, Wiley, New York, 1999. 39. B. H. Goethert, Transonic Wind Tunnel Testing, Dover, New York, 2007. 40. American Institute of Aeronautics and Astronautics, Recom¬ mended Practice: Wind Tunnel Testing, 2 vols., Reston, VA, 2003. 41. P. N. Desai, J. T. Schofield, and M. E. Lisano, “Flight Recon¬ struction of the Mars Pathfinder Disk-Gap-Band Parachute Drag Coefficients,” J. Spacecraft and Rockets, vol. 42, no. 4, July-August 2005, pp. 672-676. 42. K.-H. Kim, “Recent Advances in Cavitation Research,” 14th International Symposium on Transport Phenomena, Honolulu, HI, March 2012. This chapter is mostly about flow analysis. The photo shows the 36-inch-diameter Keystone Pipeline, which has been operating since July 2010. This pipeline delivers heavy and light blends of oil from Hardisty, Alberta, Canada, to refineries in Texas. The pipeline is completely buried and emerges occasionally at delivery or pump stations; it can currently deliver up to 700,000 barrels of oil per day. [Image courtesy of TransCanada] 338 6.1 Reynolds Number Regimes Chapter 6 Viscous Flow in Ducts Motivation. This chapter is completely devoted to an important practical fluids engi¬ neering problem: flow in ducts with various velocities, various fluids, and various duct shapes. Piping systems are encountered in almost every engineering design and thus have been studied extensively. There is a small amount of theory plus a large amount of experimentation. The basic piping problem is this: Given the pipe geometry and its added compo¬ nents (such as fittings, valves, bends, and diffusers) plus the desired flow rate and fluid properties, what pressure drop is needed to drive the flow? Of course, it may be stated in alternative form: Given the pressure drop available from a pump, what flow rate will ensue? The correlations discussed in this chapter are adequate to solve most such piping problems. This chapter is for incompressible flow; Chap. 9 treats compressible pipe flow. Now that we have derived and studied the basic flow equations in Chap. 4, you would think that we could just whip off myriad beautiful solutions illustrating the full range of fluid behavior, of course expressing all these educational results in dimensionless form, using our new tool from Chap. 5, dimensional analysis. The fact of the matter is that no general analysis of fluid motion yet exists. There are several dozen known particular solutions, there are many approximate digital computer solutions, and there are a great many experimental data. There is a lot of theory available if we neglect such important effects as viscosity and compressibility (Chap. 8), but there is no general theory and there may never be. The reason is that a profound and vexing change in fluid behavior occurs at moderate Reynolds numbers. The flow ceases being smooth and steady {laminar) and becomes fluctuating and agitated {turbulent). The changeover is called transition to turbulence. In Fig. 5.3a we saw that transition on the cylinder and sphere occurred at about Re = 3 X 10^, where the sharp drop in the drag coefficient appeared. Transition depends on many effects, such as wall roughness (Fig. 5.3b) or fluctuations in the inlet stream, but the primary parameter is the Reynolds number. There are a great many data on transition but only a small amount of theory [1 to 3]. 339 340 Chapter 6 Viscous Flow in Ducts u u u t- Small natural disturbances Fig. 6.1 The three regimes of damp quickly viscous flow: (a) laminar flow at low Re; (b) transition at intermediate Re; (c) turbulent _ flow at high Re. (a) t t Intermittent bursts of turbulence Continuous turbulence (b) (C) ' Turbulence can be detected from a measurement by a small, sensitive instrument such as a hot-wire anemometer (Fig. 6.29e) or a piezoelectric pressure transducer. The flow will appear steady on average but will reveal rapid, random fluctuations if turbu¬ lence is present, as sketched in Fig. 6.1. If the flow is laminar, there may be occasional natural disturbances that damp out quickly (Fig. 6.1a). If transition is occurring, there will he sharp bursts of intermittent turbulent fluctuation (Fig. 6.1h) as the increasing Reynolds number causes a breakdown or instability of laminar motion. At sufficiently large Re, the flow will fluctuate continually (Fig. 6.1c) and is termed /nZ/y turbulent. The fluctuations, typically ranging from 1 to 20 percent of the average velocity, are not strictly periodic but are random and encompass a continuous range, or spectrum, of frequencies. In a typical wind tunnel flow at high Re, the turbulent frequency ranges from 1 to 10,000 Hz, and the wavelength ranges from about 0.01 to 400 cm. EXAMPLE 6.1 The accepted transition Reynolds number for flow in a circular pipe is Re^^j, = 2300. For flow through a 5-cm-diameter pipe, at what velocity will this occur at 20°C for (a) airflow and (b) water flow? Solution Almost all pipe flow formulas are based on the average velocity V = Q/A, not centerline or any other point velocity. Thus transition is specified at pVd/fi ~ 2300. With d known, we introduce the appropriate fluid properties at 20°C from Tables A.3 and A.4: (a) Air: (b) Water: pVd pVd (1.205 kg/m^)V(0.05 m) 1.80 E-5kg/(m-s) (998kg/m^)V(0.05 m) 0.001 kg/(m-s) = 2300 m 0.7 — s = 2300 V = 0.046 These are very low velocities, so most engineering air and water pipe flows are turbulent, not laminar. We might expect laminar duct flow with more viscous fluids such as lubricating oils or glycerin. In free-surface flows, turbulence can be observed directly. Figure 6.2 shows liquid flow issuing from the open end of a tube. The low-Reynolds-number jet (Fig. 6.2a) is smooth and laminar, with the fast center motion and slower wall flow forming different trajectories 6.1 Reynolds Number Regimes 341 Fig. 6.2 Flow issuing at constant speed from a pipe: (a) high- viscosity, low-Reynolds-numher, laminar flow; (b) low-viscosity, high-Reynolds-numher, turbulent flow. Note the ragged, disorderly shape of the jet. (National Committee for Fluid Meclumics Films, Education Development Center, Inc., © 1972.) 342 Chapter 6 Viscous Flow in Ducts Fig. 6.3 Formation of a turbulent puff in pipe flow: (a) and (b) near the entrance; (c) somewhat downstream; (d) far downstream. ( Courtesy of Cambridge University Press-P. R. Bandyopadhyay, “Aspects of the Equilibrium Puff in Transitional Pipe Flow,” Journal of Fluid Mechanics, vol. 163, 1986, pp. 439^58.) Flow joined by a liquid sheet. The higher-Reynolds-nutnber turbulent flow (Fig. 6.2b) is unsteady and irregular but, when averaged over time, is steady and predictable. How did turbulence form inside the pipe? The laminar parabolic flow profile, which is similar to Eq. (4.137), became unstable and, at Re^; ~ 2300, began to form “slugs” or “puffs” of intense turbulence. A puff has a fast-moving front and a slow-moving rear and may be visualized by experimenting with glass tube flow. Figure 6.3 shows a puff as photographed by Bandyopadhyay . Near the entrance (Fig. 6.3a and b) there is an irregular laminar-turbulent interface, and vortex roll-up is visible. Further downstream (Fig. 6.3c) the puff becomes fully turbulent and very active, with helical motions visible. Far downstream (Fig. 6.3d) the puff is cone-shaped and less active, with a fuzzy, ill-defined interface, sometimes called the “relaminarization” region. A complete description of the statistical aspects of turbulence is given in Ref. 1, while theory and data on transition effects are given in Refs. 2 and 3. At this introductory level we merely point out that the primary parameter affecting transition is the Reynolds number. If Re = ULIu, where U is the average stream velocity and L is the “width,” or transverse thickness, of the shear layer, the following approximate ranges occur: 0 < Re < 1: highly viscous laminar “creeping” motion 1 < Re < 100: laminar, strong Reynolds number dependence 100 < Re < 10^: laminar, boundary layer theory useful 10^ < Re < 10": transition to turbulence 10" < Re < 10^: turbulent, moderate Reynolds number dependence 10® < Re < 00 : turbulent, slight Reynolds number dependence These representative ranges vary somewhat with flow geometry, surface roughness, and the level of fluctuations in the inlet stream. The great majority of our analyses are concerned with laminar flow or with turbulent flow, and one should not normally design a flow operation in the transition region. 6.1 Reynolds Number Regimes 343 Historical Outline Fig. 6.4 Experimental evidence of transition for water flow in a j-in smooth pipe 10 ft long. Since turbulent flow is more prevalent than laminar flow, experimenters have observed turbulence for centuries without being aware of the details. Before 1930 flow instru¬ ments were too insensitive to record rapid fluctuations, and workers simply reported mean values of velocity, pressure, force, and so on. But turbulence can change the mean values dramatically, as with the sharp drop in drag coefficient in Fig. 5.3. A German engineer named G. H. L. Hagen first reported in 1839 that there might be two regimes of viscous flow. He measured water flow in long brass pipes and deduced a pressure-drop law: ^p LQ (const) — j- -f entrance effect R (6.1) This is exactly our laminar flow scaling law from Example 5.4, but Hagen did not realize that the constant was proportional to the fluid viscosity. The formula broke down as Hagen increased Q beyond a certain limit — that is, past the critical Reynolds number — and he stated in his paper that there must be a second mode of flow characterized by “strong movements of water for which Ap varies as the second power of the discharge. . . .” He admitted that he could not clarify the reasons for the change. A typical example of Hagen’s data is shown in Fig. 6.4. The pressure drop varies linearly with V = Q/A up to about 1.1 ft/s, where there is a sharp change. Above about 120 344 Chapter 6 Viscous Flow in Ducts Needle Dye filament Tank (a) \ _ / (b) V = 2.2 ft/s the pressure drop is nearly quadratic with V. The actual power Ap oc seems impossible on dimensional grounds but is easily explained when the dimensionless pipe flow data (Fig. 5.10) are displayed. In 1883 Osborne Reynolds, a British engineering professor, showed that the change depended on the parameter pVd/p, now named in his honor. By introducing a dye streak into a pipe flow, Reynolds could observe transition and turbulence. His sketches of the flow behavior are shown in Fig. 6.5. If we examine Hagen’s data and compute the Reynolds number at V = 1.1 ft/s, we obtain Re^ = 2100. The flow became fully turbulent, V = 2.2 ft/s, at Re^ = 4200. The accepted design value for pipe flow transition is now taken to be Re^.^i, = 2300 (6.2) V _ , ==— - - _ ^ (c) Fig. 6.5 Reynolds’ sketches of pipe flow transition: (a) low-speed, laminar flow; (b) high-speed, turbulent flow; (c) spark photograph of condition (b). Source: Reynolds, “An Experimental Investigation of the Circumstances which Determine Whether the Motion of Water Shall Be Direct or Sinuous and of the Law of Resistance in Parallel Channels, " Phil. Trans. R. Soc., vol. 174, 1883, pp. 935-982. This is accurate for commercial pipes (Fig. 6.13), although with special care in pro¬ viding a rounded entrance, smooth walls, and a steady inlet stream. Recent can be delayed until much higher values. The study of transition in pipe flow, both experi¬ mentally and theoretically, continues to be a fascinating topic for researchers, as discussed in a recent review article . Note: The value of 2300 is for transition in pipes. Other geometries, such as plates, airfoils, cylinders, and spheres, have com¬ pletely different transition Reynolds numbers. Transition also occurs in external flows around bodies such as the sphere and cylinder in Fig. 5.3. Ludwig Prandtl, a German engineering professor, showed in 1914 that the thin boundary layer surrounding the body was undergoing transition from laminar to turbulent flow. Thereafter the force coefficient of a body was acknowledged to be a function of the Reynolds number [Eq. (5.2)]. There are now extensive theories and experiments of laminar flow instability that explain why a flow changes to turbulence. Reference 5 is an advanced textbook on this subject. Laminar flow theory is now well developed, and many solutions are known [2, 3], but no analyses can simulate the fine-scale random fluctuations of turbulent flow. Therefore most turbulent flow theory is semiempirical, based on dimensional analysis and physical reasoning; it is concerned with the mean flow properties only and the mean of the fluctuations, not their rapid variations. The turbulent flow “theory” pre¬ sented here in Chaps. 6 and 7 is unbelievably crude yet surprisingly effective. We shall attempt a rational approach that places turbulent flow analysis on a firm physical basis. 6.2 Internal versus External Both laminar and turbulent flow may be either internal (that is, “bounded” by walls) Viscous Flows or external and unbounded. This chapter treats internal flows, and Chap. 7 studies external flows. An internal flow is constrained by the bounding walls, and the viscous effects will grow and meet and permeate the entire flow. Figure 6.6 shows an internal flow in a long duct. There is an entrance region where a nearly inviscid upstream flow con¬ verges and enters the tube. Viscous boundary layers grow downstream, retarding the 'However, direct numerical simulation (DNS) of low-Reynolds-number turbulence is now quite common . 6.2 Internal versus External Viscous Flows 345 Fig. 6.6 Developing velocity profiles and pressure changes in the entrance of a duct flow. axial flow u{r, x) at the wall and thereby accelerating the center core flow to maintain the incompressible continuity requirement Q udA = const (6.3) At a flnite distance from the entrance, the boundary layers merge and the inviscid core disappears. The tube flow is then entirely viscous, and the axial velocity adjusts slightly further until at x = it no longer changes with x and is said to be fully developed, u ~ u(r) only. Downstream of x = the velocity profile is constant, the wall shear is constant, and the pressure drops linearly with x, for either laminar or turbulent flow. All these details are shown in Fig. 6.6. Dimensional analysis shows that the Reynolds number is the only parameter affecting entrance length. If then 4 = m V, p, p) g(Rerf) (6.4) For laminar flow [2, 3], the accepted correlation is 4 — = 0.06 Rerf d laminar (6.5) 346 Chapter 6 Viscous Flow in Ducts The maximum laminar entrance length, at Recent = 2300, is = 138t/, which is the longest development length possible. In turbulent flow, the boundary layers grow faster, and L^, is relatively shorter. For decades, the writer has favored a sixth-power-law estimate, LJd ~ 4.4 Re^®, but recent CFD results, communicated by Fabien Anselmet, and separately by Sukanta Dash, indicate that a better turbulent entrance-length correlation is — =1.6Rey'‘ for Re^ < lO’ (6.6) a Some computed turbulent entrance-length estimates are thus Re. The slope of the logarithm law remains the same, 1/k, but the shift outward causes the constant B to be less by an amount AZ? ~ (1/k) In Nikuradse simulated roughness by gluing uniform sand grains onto the inner walls of the pipes. He then measured the pressure drops and flow rates and correlated friction factor versus Reynolds number in Fig. 6.12h. We see that laminar friction is unaffected, but turbulent friction, after an onset point, increases monotonically with the roughness ratio eld. For any given eld, the friction factor becomes constant (fully rough) at high Reynolds numbers. These points of change are certain values of e^ = eulv. eu^ V < 5: eu 5 < - <70: V eu^ v > 70: hydraulically smooth walls, no effect of roughness on friction transitional roughness, moderate Reynolds number effect fully rough flow, sublayer totally broken up and friction independent of Reynolds number 6.6 Turbulent Pipe Flow 361 Fig. 6.12 Effect of wall roughness on turbulent pipe flow, (a) The logarithmic overlap velocity profile shifts down and to the right; (b) experiments with sand-grain roughness by Nikuradse show a systematic increase of the turbulent friction factor with the roughness ratio. For fully rough flow, > 70, the log law downshift AB in Fig. 6.12a is AB^-\ne^ - 3.5 (6.45) K and the logarithm law modified for roughness becomes , 1 , . 1 y m"" = -Iny"" + B - A5 = -In- + 8.5 (6.46) K K £ The viscosity vanishes, and hence fully rough flow is independent of the Reynolds number. If we integrate Eq. (6.46) to obtain the average velocity in the pipe, we obtain y d — = 2.44 In - + 3.2 M £ 1 £ld or ^=-2.0 log— fully rough flow (6.47) There is no Reynolds number effect; hence the head loss varies exactly as the square of the velocity in this case. Some numerical values of friction factor may be listed: eld 0.00001 0.0001 0.001 0.01 0.05 f 0.00806 0.0120 0.0196 0.0379 0.0716 Friction factor / = 362 Chapter 6 Viscous Flow in Ducts The friction factor increases by 9 times as the roughness increases by a factor of 5000. In the transitional roughness region, sand grains behave somewhat differently from commercially rough pipes, so Fig. 6.\2b has now been replaced by the Moody chart. The Moody Chart In 1939 to cover the transitionally rough range, Colebrook combined the smooth wall [Eq. (6.38)] and fully rough [Eq. (6.47)] relations into a clever interpolation formula: f •1/2 -2.0 log U/d V3-7 + 2.51 \ Re,/''V (6.48) This is the accepted design formula for turbulent friction. It was plotted in 1944 by Moody into what is now called the Moody chart for pipe friction (Eig. 6.13). Values of (Vd) for water at 60°F (velocity, fi/s x diameter, in) 0.1 0.2 0.4 0.6 0.8 1 2 4 6 8 10 20 40 60 80 100 200 400 600 800 1000 2000 4000 6000 10,000 Fig. 6.13 The Moody chart for pipe friction with smooth and rough walls. This chart is identical to Eq. (6.48) for turbulent flow. (From Ref. 8, Source: ASME.) 6.6 Turbulent Pipe Flow 363 Table 6.1 Recommended Roughness Values for Commercial Ducts e Material Condition ft mm Uncertainty, % Steel Sheet metal, new 0.00016 0.05 ±60 Stainless, new 0.000007 0.002 ±50 Commercial, new 0.00015 0.046 ±30 Riveted 0.01 3.0 ±70 Rusted 0.007 2.0 ±50 Iron Cast, new 0.00085 0.26 ±50 Wrought, new 0.00015 0.046 ±20 Galvanized, new 0.0005 0.15 ±40 Asphalted cast 0.0004 0.12 ±50 Brass Drawn, new 0.000007 0.002 ±50 Plastic Drawn tubing 0.000005 0.0015 ±60 Glass — Smooth Smooth Concrete Smoothed 0.00013 0.04 ±60 Rough 0.007 2.0 ±50 Rubber Smoothed 0.000033 0.01 ±60 Wood Stave 0.0016 0.5 ±40 The Moody chart is probably the most famous and useful figure in fluid mechanics. It is accurate to ±15 percent for design calculations over the full range shown in Fig. 6.13. It can be used for circular and noncircular (Sec. 6.6) pipe flows and for open-channel flows (Chap. 10). The data can even be adapted as an approximation to boundary layer flows (Chap. 7). The Moody chart gives a good visual summary of laminar and turbulent pipe fric¬ tion, including roughness effects. When the writer was in college, everyone solved problems by carefully reading this chart. Currently, though, Eq. (6.48), though implicit in /, is easily solved by iteration or a direct solver. If only a calculator is available, the clever explicit formula given by Haaland as -1-8 log varies less than 2 percent from Eq. (6.48). The shaded area in the Moody chart indicates the range where transition from laminar to turbulent flow occurs. There are no reliable friction factors in this range, 2000 < Re^ < 4000. Notice that the roughness curves are nearly horizontal in the fully rough regime to the right of the dashed line. Erom tests with commercial pipes, recommended values for average pipe roughness are listed in Table 6.1. 6.9 fe/dV 11 .Re: + K3.lJ (6.49) EXAMPLE 6.6^ Compute the loss of head and pressure drop in 200 ft of horizontal 6-in-diameter asphalted cast iron pipe carrying water with a mean velocity of 6 ft/s. ^This example was given by Moody in his 1944 paper . 364 Chapter 6 Viscous Flow in Ducts Solution • System sketch: See Fig. 6.7 for a horizontal pipe, with Az = 0 and hf proportional to Ap. • Assumptions: Turbulent flow, asphalted horizontal cast iron pipe, d = 0.5 ft, L = 200 ft. • Approach: Find Re^ and eld\ enter the Moody chart. Fig. 6.13; find/, then hf and Ap. • Property values: From Table A. 3 for water, converting to BG units, p = 998/515.38 = 1.94 slug/fr\ p = 0.001/47.88 = 2.09 E-5 slug/(ft-s). • Solution step 1: Calculate Re^ and the roughness ratio. As a crutch. Moody provided water and air values of “Vd” at the top of Fig. 6.13 to find Re^. Instead, let’s calculate it ourselves: Rerf = pVd M (1.94 slug/fr’)(6ft/s)(0.5ft) 2.09 E-5 slug/(ft ■ s) 279,000 (turbulent) From Table 6.1, for asphalted cast iron, e = 0.0004 ft. Then calculate eld = (0.0004 ft)/(0.5 ft) = 0.0008 • Solution step 2: Find the friction factor from the Moody chart or from Eq. (6.48). If you use the Moody chart, Eig. 6.13, you need practice. Find the line on the right side for eld = 0.0008 and follow it back to the left until it hits the vertical line for Re^ ~ 2.79 E5. Read, approximately, / = 0.02 [or compute /= 0.0198 from Eq. (6.48).] ■ Solution step 3: Ap = pghf = (1.94 slug/ft^)(32.2 ft/s^)(4.5 ft) « 280 Ibf/ft^ Ans. Calculate /t^from Eq. (6.10) and Ap from Eq. (6.8) for a horizontal pipe: L r =f-— = m2) d 2p /200ft\ V 0.5 ft/ (6 ft/s)^ 2(32.2 ft/s^) 4.5 ft Ans. • Comments: In giving this example. Moody stated that this estimate, even for clean new pipe, can be considered accurate only to about ±10 percent. EXAMPLE 6.7 Oil, with p = 900 kg/m^ and v = 0.00001 m^/s, flows at 0.2 m^/s through 500 m of 200-mm- diameter cast iron pipe. Determine (a) the head loss and (6) the pressure drop if the pipe slopes down at 10° in the flow direction. Solution First compute the velocity from the known flow rate: Y _ _ 0.2 mVs ttR” 7r(0.1 m)^ 6.4 m/s Then the Reynolds number is Vd (6.4 m/s)(0.2 m) V 0.00001 mVs = 128,000 6.6 Turbulent Pipe Flow 365 From Table 6.1, £" = 0.26 mm for cast iron pipe. Then e 0.26 mm d 200 mm 0.0013 Enter the Moody chart on the right at eld = 0.0013 (you will have to interpolate), and move to the left to intersect with Re = 128,000. Read/— 0.0225 [from Eq. (6.48) for these values we could compute/ = 0.0227]. Then the head loss is hf = f d 2g = (0.0225) 500 m (6.4 m/s)^ 0.2 m 2(9.81 m/s^) ■ = 117m Ans. (a) From Eq. (6.9) for the inclined pipe. An An hf = - ^ Zi — Z2 = - f L sin 10° Pg Pg or Ap — pg[hf — (500 m) sin 10°] = pg{ 1 17 m — 87 m) = (900 kg/m^) (9.81 m/s^)(30m) = 265,000 kg/(m ■ s^) = 265,000 Pa Ans. (b) EXAMPLE 6.8 Repeat Example 6.4 to see whether there is any possible turbulent flow solution for a smooth-walled pipe. Solution In Example 6.4 we estimated a head loss hf — 1.66 ft, assuming laminar exit flow (a — 2.0). For this condition the friction factor is / = d 2g . (0.004 ft) (2) (32.2 ft/s") (1.66 ft) - — - — - 5 — ■ (1.0 ft)(3.32 ft/s)" 0.0388 For laminar flow, Re^ = 64/f = 64/0.0388 — 1650, as we showed in Example 6.4. However, from the Moody chart (Fig. 6.13), we see that / = 0.0388 also corresponds to a turbulent smooth-wall condition, at Re^ = 4500. If the flow actually were turbulent, we should change our kinetic energy factor to cr « 1.06 [Eq. (3.77)], whence the corrected hf ~ 1.82 ft and/ = 0.0425. With/ known, we can estimate the Reynolds number from our formulas: Rerf« 3250 [Eq. (6.38)] or Re^ « 3400 [Eq. (6.396)] So the flow might have been turbulent, in which case the viscosity of the fluid would have been pVd __ 1.80(3.32)(0.004) Re^ ~ 3300 7.2 X 10”® slug/(ft ■ s) Ans. This is about 55 percent less than our laminar estimate in Example 6.4. The moral is to keep the capillary-flow Reynolds number below about 1000 to avoid such duplicate solutions. 366 Chapter 6 Viscous Flow in Ducts 6.7 Four Types of Pipe Flow Problems Type 2 Problem: Find the Flow Rate The Moody chart (Fig. 6.13) can be used to solve almost any problem involving friction losses in long pipe flows. However, many such problems involve considerable iteration and repeated calculations using the chart because the standard Moody chart is essentially a head loss chart. One is supposed to know all other variables, compute Rej, enter the chart, find /, and hence compute hf. This is one of four fundamental problems which are commonly encountered in pipe flow calculations: 1. Given d, L, and V or Q, p, pt, and g, compute the head loss /t^(head loss problem). 2. Given d, L, hf, p, p, and g, compute the velocity V or flow rate Q (flow rate problem). 3. Given Q, L, hf, p, p, and g, compute the diameter d of the pipe (sizing problem). 4. Given Q, d, hf, p, p, and g, compute the pipe length L. Problems 1 and 4 are well suited to the Moody chart. We have to iterate to compute velocity or diameter because both d and V are contained in the ordinate and the abscissa of the chart. There are two alternatives to iteration for problems of type 2 and type 3: (a) prepa¬ ration of a suitable new Moody-type formula (see Probs. P6.68 and P6.73); or (b) the use of solver software, like Excel. Examples 6.9 and 6.11 include the Excel approach to these problems. Even though velocity (or flow rate) appears in both the ordinate and the abscissa on the Moody chart, iteration for turbulent flow is nevertheless quite fast because /varies so slowly with Re^. In earlier editions, the writer rescaled the Colebrook formula (6.48) into a relation where Q could be calculated directly. That idea is now down¬ sized to Problem P6.68. Example 6.9, which follows, is illustrated both by iteration and by an Excel solution. EXAMPLE 6.9 Oil, with p = 950 kg/m^ and v = 2 E-5 m/s, flows through a 30-cm-diameter pipe 100 m long with a head loss of 8 m. The roughness ratio is eld = 0.0002. Find the average veloc¬ ity and flow rate. Iterative Solution By definition, the friction factor is known except for V: or fV^ « 0.471 (SI units) To get started, we only need to guess /, compute V = VO.471//, then get Re^, compute a better/from the Moody chart, and repeat. The process converges fairly rapidly. A good first guess is the “fully rough” value for eld = 0.0002, or/~ 0.014 from Fig. 6.13. The iteration would be as follows: /= hf = (8m)\| 70.3m\ 2(9.81 m/sG UOOm/ Guess / « 0.014, then V = VO.471/0.014 = 5.80 m/s and Re^ = Vdiv « 87,000. At Re^ = 87,000 and eld = 0.0002, compute /„ew ~ 0.0195 [Eq. (6.48)]. 6.7 Four Types of Pipe Flow Problems 367 Type 3 Problem: Find the Pipe Diameter New / « 0.0195, V = VO.471/0.0195 = 4.91 m/s and Re (6.53) This section may be omitted without loss of continuity. 372 Chapter 6 Viscous Flow in Ducts Flow between Parallel Plates Comparing this to Eq. (6.9b), we see that A/S? takes the place of one-fourth of the pipe diameter for a circular cross section. We define the friction factor in terms of average shear stress; /ncd - (6.54) where NCD stands for noncircular duct and V = Q/A as usual, Eq. (6.53) becomes hf=f- .L r (6.55) This is equivalent to Eq. (6.10) for pipe flow except that d is replaced by Df,. Therefore, we customarily define the hydraulic diameter as 4A 4 X area 91 wetted perimeter (6.56) We should stress that the wetted perimeter includes all surfaces acted upon by the shear stress. For example, in a circular annulus, both the outer and the inner perimeter should be added. We would therefore expect by dimensional analysis that this friction factor/, based on hydraulic diameter as in Eq. (6.55), would correlate with the Reynolds number and roughness ratio based on the hydraulic diameter f=F (VDh V ’ dJ (6.57) and this is the way the data are correlated. But we should not necessarily expect the Moody chart (Fig. 6.13) to hold exactly in terms of this new length scale. And it does not, but it is surprisingly accurate: 64 Reo. ./Moody I D, ±40% laminar flow ±15% turbulent flow Now let us look at some particular cases. (6.58) Probably the simplest noncircular duct flow is fully developed flow between parallel plates a distance 2h apart, as in Fig. 6.14. As noted in the figure, the width b ^ h, so the flow is essentially two-dimensional; that is, u = uiy) only. The hydraulic diameter is AA A(2bh) Du = = lim - 9? 2b + Ah Ah (6.59) that is, twice the distance between the plates. The pressure gradient is constant, (—dp/dx) = Ap/L, where L is the length of the channel along the x axis. 6.8 Flow in Noncircular Ducts 373 Fig. 6.14 Fully developed flow between parallel plates. Laminar Flow Solution Turbulent Flow Solution The laminar solution was given in Sec. 4.10, in connection with Fig. 4.16fo. Let us review those results here: u ^max where 2bh^ Ap Q = - - ^ 3p L _Q_ \^Ap__ 2 ~ A~ 3p L ~ 3 Ap 2fi L ^Ap 3pV y = h ^ ^ ^p _ 3fiLV Pg pgh^ r„ = P du (6.60) Now use the head loss to establish the laminar friction factor: 96/i 96 " (LID,){V^I2g) ^ pV{Ah) ^ ^ Thus, if we could not work out the laminar theory and chose to use the approximation / ~ 64/Reo^, we would be 33 percent low. The hydraulic-diameter approximation is relatively crude in laminar flow, as Eq. (6.58) states. Just as in circular-pipe flow, the laminar solution above becomes unstable at about Refl^ ~ 2000; transition occurs and turbulent flow results. For turbulent flow between parallel plates, we can again use the logarithm law, Eq. (6.28), as an approximation across the entire channel, using not y but a wall coordinate Y, as shown in Eig. 6.14: m(T) 1 Tm = -ln - + B 0 < Y < h (6.62) U K V This distribution looks very much like the flat turbulent profile for pipe flow in Fig. 6.11/?, and the mean velocity is y = h. u dY = w 1 hu 1 -In - + B - K V K 0 (6.63) 374 Chapter 6 Viscous Flow in Ducts Recalling that Vlu = (8//)^^, we see that Eq. (6.63) is equivalent to a parallel-plate friction law. Rearranging and cleaning up the constant terms, we obtain ^ « 2.0 log (Re^, / -1.19 (6.64) where we have introduced the hydraulic diameter £),, = Ah. This is remarkably close to the smooth-wall pipe friction law, Eq. (6.38). Therefore we conclude that the use of the hydraulic diameter in this turbulent case is quite successful. That turns out to be true for other noncircular turbulent flows also. Equation (6.64) can be brought into exact agreement with the pipe law by rewriting it in the form = 2.0 log (0.64 - 0.8 (6.65) Thus the turbulent friction is predicted most accurately when we use an effective diameter Deff equal to 0.64 times the hydraulic diameter. The effect on/itself is much less, about 10 percent at most. We can compare with Eq. (6.66) for laminar flow, which predicted Parallel plates: De« — (6.66) This close resemblance (0.64T);, versus 0.661Dij) occurs so often in noncircular duct flow that we take it to be a general rule for computing turbulent friction in ducts: f^eff Deff — Ai 4A — reasonable accuracy 64 = D, (/Re/5j) laminar theory better accuracy (6.67) Jones shows that the effective-laminar-diameter idea collapses all data for rect¬ angular ducts of arbitrary height-to-width ratio onto the Moody chart for pipe flow. We recommend this idea for all noncircular ducts. EXAMPLE 6.13 Fluid flows at an average velocity of 6 ft/s between horizontal parallel plates a distance of 2.4 in apart. Find the head loss and pressure drop for each 100 ft of length for p = 1.9 slugs/fr’ and (a) v = 0.00002 ftVs and (b) v = 0.002 ftVs. Assume smooth walls. Solution Part (a) The viscosity /i = pr' = 3.8X10 ^ slug/(ft ■ s). The spacing is 2h = 2.4 in = 0.2 ft, and A = Ah = 0.4 ft. The Reynolds number is RSfl. = VDh (6.0 ft/s) (0.4 ft) 120,000 V 0.00002 fr/s 6.8 Flow in Noncircular Ducts 375 Part (a) The flow is therefore turbulent. For reasonable accuracy, simply look on the Moody chart (Fig. 6.13) for smooth walls: L 100 (6.0)^ /■« 0.0173 /!,«/ - = 0.0173 - 2.42 ft ^ Da 2g 0.4 2(32.2) Since there is no change in elevation. Ap = pghf= 1.9(32.2)(2.42) = 148 Ibf/fF Ans. (a) Ans. (a) This is the head loss and pressure drop per 100 ft of channel. For more accuracy, take Deff = ^Da from laminar theory; then Rseff = ^(120,000) = 80,000 and from the Moody chart read / « 0.0189 for smooth walls. Thus a better estimate is 100 (6.0)^ hf = 0.0189 - ^ ^ = 2.64 ft ^ 0.4 2(32.2) and Ap = 1.9(32.2)(2.64) = 161 Ibf/ft^ The more accurate formula predicts friction about 9 percent higher. Better ans. (a) Compute p, = pv = 0.0038 slug/(ft • s). The Reynolds number is 6.0(0.4)70.002 = 1200; therefore the flow is laminar, since Re is less than 2300. You could use the laminar flow friction factor, Eq. (6.61) ./lam 96 Ren, 96 1200 = 0.08 from which and hf = 0.08 100 (6.0)" = 11.2 ft 0.4 2(32.2) Ap = 1.9(32.2)(11.2) = 6841bf/ft" Ans. (b) Alternately you can finesse the Reynolds number and go directly to the appropriate laminar flow formula, Eq. (6.60): V = I^Ap^ 3/i L 3(6.0 ft/s)0.0038 slug/(ft ■ s) Ap = - - , = 684 slugs/(ft ■ s") = 684 Ibf/ft" and (0.1 ft)" Ap hf = 684 pg 1.9(32.2) = 11.2 ft Flow through a Concentric Annulus Consider steady axial laminar flow in the annular space between two concentric cylinders, as in Fig. 6.15. There is no slip at the inner (r = b) and outer radius (r = a). For u = u{r) only, the governing relation is Eq. (D.7) in Appendix D: d dr Kr K = —{p + pgz) ax (6.68) 376 Chapter 6 Viscous Flow in Ducts Fig. 6.15 Fully developed flow through a concentric annulus. r = X a Integrate this twice: 1 .K u = — r - h Cl In r + C2 4 M The constants are found from the two no-slip conditions: 1 ,K u(r = a) = 0 = —a - h Ci In a + Ct 4 /i u(r = b) = 0 = -b^ - 4- Cl In -f C2 4 /i The final solution for the velocity profile is J_ 4/i dx ip + Pg^) + 2 1,2 a — b a - In — In (fi/fl) r (6.69) The volume flow is given by Q = u 27rr dr = 4 TT ip + Pgz) dx 4 4 (a^ - by a — b — In {db) (6.70) The velocity profile u{r) resembles a parabola wrapped around in a circle to form a split doughnut, as in Fig. 6.15. It is confusing to base the friction factor on the wall shear because there are two shear stresses, the inner stress being greater than the outer. It is better to define / with respect to the head loss, as in Eq. (6.55), t = hf - r where V = Q 7r{a b^) The hydraulic diameter for an annulus is 47r(a^ — b^) 27r(a + b) A, =■ = 2(a - b) (6.71) (6.72) It is twice the clearance, rather like the parallel-plate result of twice the distance between plates [Eq. (6.59)]. 6.8 Flow in Noncircular Ducts 377 Substituting hf, Df,, and V into Eq. (6.71), we find that the friction factor for laminar flow in a concentric annulus is of the form 64C (a - b)\a^ - b^) R&d, ^ - b - (a^ - (alb) (6.73) Table 6.3 Laminar Friction Factors for a Concentric Annulus bla /Reo. = Vi 0.0 64.0 1.000 0.00001 70.09 0.913 0.0001 71.78 0.892 0.001 74.68 0.857 0.01 80.11 0.799 0.05 86.27 0.742 0.1 89.37 0.716 0.2 92.35 0.693 0.4 94.71 0.676 0.6 95.59 0.670 0.8 95.92 0.667 1.0 96.0 0.667 The dimensionless term ^ is a sort of correction factor for could rewrite Eq. (6.73) as 64 Concentric annulus: / = - RSeff = RCeff 1 -R. A, the hydraulic diameter. We (6.74) Some numerical values of / Re^^ and = 1/^ are given in Table 6.3. Again, laminar annular flow becomes unstable at Re^^ ~ 2000. For turbulent flow through a concentric annulus, the analysis might proceed by patching together two logarithmic law profiles, one going out from the inner wall to meet the other coming in from the outer wall. We omit such a scheme here and proceed directly to the friction factor. According to the general rule proposed in Eq. (6.58), turbulent friction is predicted with excellent accuracy by replacing d in the Moody chart with Df.ff = 2(a — f>)/^, with values listed in Table 6.3.“^ This idea includes roughness also (replace e/d in the chart with elD^ff). For a quick design number with about 10 percent accuracy, one can simply use the hydraulic diameter Df, = 2(a — b). EXAMPLE 6.14 What should the reservoir level h be to maintain a flow of 0.01 mVs through the commer¬ cial steel annulus 30 m long shown in Fig. E6.14? Neglect entrance effects and take p = 1000 kgW and v = 1.02 X 10“® mVs for water. o Solution • Assumptions: Fully developed annulus flow, minor losses neglected. • Approach: Determine the Reynolds number, then find / and hf and thence h. Tones and Leung show that data for annular flow also satisfy the effective-laminar- diameter idea. 378 Chapter 6 Viscous Flow in Ducts Other Noncircular Cross Sections • Property values: Given p = 1000 kg/m^ and v = 1.02 E-6 m^/s. • Solution step 1: Calculate the velocity, hydraulic diameter, and Reynolds number: Q _ 0.01 mVs A 7r[ (0.05 m)^ - (0.03 m)2] 1.99 m s D = 2(a - b) = 2(0.05 m - 0.03 m) = 0.04 m Reo,^ VDk V (1.99 m/s) (0.04 m) 1.02 E-6 mVs 78,000 (turbulent flow) ■ Solution step 2: Apply the steady flow energy equation between sections 1 and 2: Pi , ol{V\ P2 OfiVi H - f Zl - - f - ^ Z2 + hf pg 2g Pg or 2^ 0:21^2 V\( + hf - cti + f Note that zi = h. For turbulent flow, from Eq. (3.43c), we estimate 0:2 ~ 1.03 • Solution step 3: Determine the roughness ratio and the friction factor. From Table 6.1, for (new) commercial steel pipe, e = 0.046 mm. Then e "Dk 0.046 mm 40 mm 0.00115 For a reasonable estimate, use Re^,^ to estimate the friction factor from Eq. (6.48): 1 /0.00115 2.51 \ — — ~ —2.0 logio - 1 - — solve for/ — 0.0232 V7 V 3.7 78,oooy// For slightly better accuracy, we could use Djff = D;,/C From Table 6.3, for bla = 3/5, 1/^ = 0.67. Then Deft = 0.67(40 mm) = 26.8 mm, whence Rco^,, = 52,300, elD^g = 0.00172, and /ff — 0.0257. Using the latter estimate, we find the required reservoir level from Eq. (1): h Vf 2g (^012 + (1.99 m/s)^ 30 m , 1.03 + 0.0257 - 2(9.81 m/s)^ 0.04 m « 4.1 m Ans. • Comments: Note that we do not replace D), with in the head loss term JLlDj,, which comes from a momentum balance and requires hydraulic diameter. If we used the simpler friction estimate,/ — 0.0232, we would obtain h ~ 3.72 m, or about 9 percent lower. In principle, any duct cross section can be solved analytically for the laminar flow velocity distribution, volume flow, and friction factor. This is because any cross sec¬ tion can be mapped onto a circle by the methods of complex variables, and other powerful analytical techniques are also available. Many examples are given by White [3, pp. 112-115], Berber , and Olson . Reference 34 is devoted entirely to laminar duct flow. In general, however, most unusual duct sections have strictly academic and not commercial value. We list here only the rectangular and isosceles-triangular sections, in Table 6.4, leaving other cross sections for you to find in the references. 6.8 Flow in Noncircular Ducts 379 Fig. 6.16 Illustration of secondary turbulent flow in noncircular ducts: {a) axial mean velocity contours; (&) secondary flow in-plane cellular motions. (After J. Nikuradse, dissertation, Gottingen, 1926.) Table 6.4 Laminar Friction Constants /Re for Rectangular and Triangular Ducts Rectangular Isosceles triangle bta /Reo. e, deg 0.0 96.00 0 48.0 0.05 89.91 10 51.6 0.1 84.68 20 52.9 0.125 82.34 30 53.3 0.167 78.81 40 52.9 0.25 72.93 50 52.0 0.4 65.47 60 51.1 0.5 62.19 70 49.5 0.75 57.89 80 48.3 1.0 56.91 90 48.0 For turbulent flow in a duct of unusual cross section, one should replace d with Df, on the Moody chart if no laminar theory is available. If laminar results are known, such as Table 6.4, replace d with = [64/(fRs)]Di, for the particular geometry of the duct. For laminar flow in rectangles and triangles, the wall friction varies greatly, being largest near the midpoints of the sides and zero in the comers. In turbulent flow through the same sections, the shear is nearly constant along the sides, dropping off sharply to zero in the corners. This is because of the phenomenon of turbulent secondary flow, in which there are nonzero mean velocities v and w in the plane of the cross section. Some measurements of axial velocity and secondary flow patterns are shown in Fig. 6.16, as sketched by Nikuradse in his 1926 dissertation. The secondary flow “cells” drive the mean flow toward the corners, so that the axial velocity contours are similar to the cross section and the wall shear is nearly constant. This is why the hydraulic-diameter concept is so successful for turbulent flow. Laminar flow in a straight noncircular duct has no secondary flow. An accurate theoretical prediction of turbulent secondary flow has yet to be achieved, although numerical models are often successful . EXAMPLE 6.15 Air, with p = 0.00237 slug/ft^ and v = 0.000157 ftVs, is forced through a horizontal square 9-by-9-in duct 100 ft long at 25 ft^/s. Find the pressure drop if e = 0.0003 ft. Solution Compute the mean velocity and hydraulic diameter: 25 frVs V = (0.75 ft)2 = 44.4 ft/s 4A 4(81 in^ Dh = T;r= \ . = 9 in = 0.75 ft S' 36 in 380 Chapter 6 Viscous Flow in Ducts 6.9 Minor or Local Losses in Pipe Systems From Table 6.4, for bla = 1.0, the effective diameter is 64 D.ff — A, = 0.843 ft whence 56.91 VDeff 44.4(0.843) Re,ff = — ^ ^ ^ = 239,000 V 0.000157 e 0.0003 A) 0.843 = 0.000356 From the Moody chart, read / = 0.0177. Then the pressure drop is Ap = pghf = 0.00237(32.2) 0.0177 , 100 44.4" 0.75 2(32.2) or Ap = 5.5 Ibf/ft" Pressure drop in air ducts is usually small because of the low density. Ans. For any pipe system, in addition to the Moody-type friction loss computed for the length of pipe, there are additional so-called minor losses or local losses due to 1. Pipe entrance or exit. 2. Sudden expansion or contraction. 3. Bends, elbows, tees, and other fittings. 4. Valves, open or partially closed. 5. Gradual expansions or contractions. The losses may not be so minor; for example, a partially closed valve can cause a greater pressure drop than a long pipe. Since the flow pattern in fittings and valves is quite complex, the theory is very weak. The losses are commonly measured experimentally and correlated with the pipe flow parameters. The data, especially for valves, are somewhat dependent on the particular manufacturer’ s design, so that the values listed here must be taken as aver¬ age design estimates [15, 16, 35, 43, 46]. The measured minor loss is usually given as a ratio of the head loss = Apl(pg) through the device to the velocity head V^/(2g) of the associated piping system: Loss coefficient K = , ^ ^ (6.75) V\2g) Although K is dimensionless, it often is not correlated in the literature with the Reyn¬ olds number and roughness ratio but rather simply with the raw size of the pipe in, say, inches. Almost all data are reported for turbulent flow conditions. A single pipe system may have many minor losses. Since all are correlated with V^l(2g), they can be summed into a single total system loss if the pipe has constant diameter: A/Jtoi = hf+ ^h,„ = (6.76) 6.9 Minor or Local Losses in Pipe Systems 381 Fig. 6.17 Typical commercial valve geometries: (a) gate valve; (b) globe valve; (c) angle valve; (d) swing- check valve; (e) disk-type gate valve. Note, however, that we must sum the losses separately if the pipe size changes so that changes. The length L in Eq. (6.76) is the total length of the pipe axis. There are many different valve designs in commercial use. Figure 6.17 shows hve typical designs; (a) the gate, which slides down across the section; {b) the globe, which closes a hole in a special insert; (c) the angle, similar to a glohe but with a 90° turn; {d) the swing-check valve, which allows only one-way flow; and (e) the disk, which closes the section with a circular gate. The globe, with its tortuous flow path, has the highest losses when fully open. Many excellent details about these and other valves are given in the handbooks by Skousen and Crane Co. . Table 6.5 lists loss coefficients K for four types of valve, three angles of elbow fitting, and two tee connections. Fittings may be connected by either internal screws or flanges, hence the two listings. We see that K generally decreases with pipe size, which is consistent with the higher Reynolds number and decreased roughness ratio of large pipes. We stress that Table 6.5 represents losses averaged among various manufacturers, so there is an uncertainty as high as ±50 percent. In addition, most of the data in Table 6.5 are relatively old [15, 16] and therefore based on fittings manufactured in the 1950s. Modern forged and molded fittings may yield somewhat different loss factors, often less than those listed in Table 6.5. An example, shown in Fig. 6.18a, gives recent data for fairly short (bend-radius/elbow- diameter = 1.2) flanged 90° elbows. The elbow diameter was 1.69 in. Notice first that K is plotted versus Reynolds number, rather than versus the raw (dimensional) pipe 382 Chapter 6 Viscous Flow in Ducts Table 6.5 Resistance Coefficients K = hJ[V^I{2g)\ for Open Valves, Elbows, and Tees Fig. 6.18a Recent measured loss coefficients for 90° elbows. These values are less than those reported in Table 6.5. (From Ref. 48, Source of Data R. D. Coffield.) Nominal diameter, in Screwed Flanged 1 241 248 20 Valves (fully open): Globe 14 Gate 0.30 Swing check 5.1 Angle 9.0 Elbows: 45° regular 0.39 45° long radius 90° regular 2.0 90° long radius 1.0 180° regular 2.0 180° long radius Tees: Line flow 0.90 Branch flow 2.4 8.2 6.9 5.7 13 0.24 0.16 0.11 0.80 2.9 2.1 2.0 2.0 4.7 2.0 1.0 4.5 0.32 0.30 0.29 0.21 1.5 0.95 0.64 0.50 0.72 0.41 0.23 0.40 1.5 0.95 0.64 0.41 0.40 0.90 0.90 0.90 0.24 1.8 1.4 1.1 1.0 8.5 6.0 5.8 5.5 0.35 0.16 0.07 0.03 2.0 2.0 2.0 2.0 2.4 2.0 2.0 2.0 0.20 0.19 0.16 0.14 0.39 0.30 0.26 0.21 0.30 0.19 0.15 0.10 0.35 0.30 0.25 0.20 0.30 0.21 0.15 0.10 0.19 0.14 0.10 0.07 0.80 0.64 0.58 0.41 diameters in Table 6.5, and therefore Fig. 6.18a has more generality. Then notice that the K values of 0.23 ± 0.05 are significantly less than the values for 90° elbows in Table 6.5, indicating smoother walls and/or better design. One may conclude that (1) Table 6.5 data are probably conservative and (2) loss factors are highly dependent on actual design and manufacturing factors, with Table 6.5 serving only as a rough guide. The valve losses in Table 6.5 are for the fully open condition. Losses can be much higher for a partially open valve. Figure 6.18/? gives average losses for three valves 0.34 0.32 0.30 0.28 S 0.26 I 9—1 ^ 0.24 0.22 0.20 0.18 0.16 0.05 0.1 0.2 0.3 0.5 1.0 2.0 3.0 4.0 Legend • Plastic elbow • Metal elbow no. 1 Reynolds number (millions) 6.9 Minor or Local Losses in Pipe Systems 383 Fig. 6.18b Average loss coefficients for partially open valves (see sketches in Fig. 6.17). (a) Fig. 6.19 Performance of butterfly valves: (a) typical geometry (Courtesy of Tyco Engineered Products and Services); (b) loss coefficients for three different manufacturers. (b) as a function of “percentage open,” as defined by the opening-distance ratio h/D (see Fig. 6.17 for the geometries). Again we should warn of a possible uncertainty of ±50 percent. Of all minor losses, valves, because of their complex geometry, are most sensitive to manufacturers’ design details. For more accuracy, the particular design and manufacturer should be consulted . The butterfly valve of Fig. 6.19a is a stem-mounted disk that, when closed, seats against an 0-ring or compliant seal near the pipe surface. A single 90° turn opens the valve completely, hence the design is ideal for controllable quick-opening and quick¬ closing situations such as occur in fire protection and the electric power industry. However, considerable dynamic torque is needed to close these valves, and losses are high when the valves are nearly closed. Figure 6.196 shows butterfly-valve loss coefficienls as a function of fhe opening angle 9 for furbulenf flow condifions (9 = 0 is closed). The losses are huge when the 384 Chapter 6 Viscous Flow in Ducts Fig. 6.20 Resistance coefficients for smooth- walled 45°, 90°, and 180° hends, at Re^ = 200,000, after Ito Source: After H. Ito, “Pressure Losses in Smooth Pipe Bends, ” Journal of Basic Engineering, March 1960, pp. 131-143. d opening is small, and K drops off nearly exponentially with the opening angle. There is a factor of 2 spread among the various manufacturers. Note that K in Fig. b.\9b is, as usual, based on the average pipe velocity V = Q/A, not on the increased velocity of the flow as it passes through the narrow valve passage. A bend or curve in a pipe, as in Fig. 6.20, always induces a loss larger than the simple straight-pipe Moody friction loss, due to flow separation on the curved walls and a swirling secondary flow arising from the centripetal acceleration. The smooth- wall loss coefficients K in Fig. 6.20, from the data of Ito , are for total loss, including Moody friction effects. The separation and secondary flow losses decrease with R/d, while the Moody losses increase because the bend length increases. The curves in Fig. 6.20 thus show a minimum where the two effects cross. Ito gives a curve-fit formula for the 90° bend in turbulent flow: /r\0-&4 /r\~^-^^ 90° bend: 0.388a f-j Reo” ^’ where a = 0.95 + 4.42 f-j >1 (6.80fl) The formula accounts for Reynolds number, which equals 200,000 in Fig. 6.20. Com¬ prehensive reviews of curved-pipe flow, for both laminar and turbulent flow, are given by Berger et al. and for 90° bends by Spedding et al. . As shown in Fig. 6.21, entrance losses are highly dependent on entrance geometry, but exit losses are not. Sharp edges or protrusions in the entrance cause large zones of flow separation and large losses. A little rounding goes a long way, and a well- rounded entrance (r = 0.2d) has a nearly negligible loss K = 0.05. At a submerged exit, on the other hand, the flow simply passes out of the pipe into the large down¬ stream reservoir and loses all its velocity head due to viscous dissipation. Therefore iiT = 1.0 for all submerged exits, no matter how well rounded. If the entrance is from a finite reservoir, it is termed a sudden contraction (SC) between two sizes of pipe. If the exit is to finite-sized pipe, it is termed a sudden expansion (SE). The losses for both are graphed in Fig. 6.22. For the sudden expansion. 6.9 Minor or Local Losses in Pipe Systems 385 Fig. 6.21 Entrance and exit loss coefficients: (a) reentrant inlets; (b) rounded and beveled inlets. Exit losses are ~ 1.0 for all shapes of exit (reentrant, sharp, beveled, or rounded). Source: From ASHRAE Handbook-2012 Fundamentals, ASHRAE, Atlanta, GA, 2012. Fig. 6.22 Sudden expansion and contraction losses. Note that the loss is based on velocity head in the small pipe. 386 Chapter 6 Viscous Flow in Ducts Gradual Expansion— The Diffuser Fig. 6.23 Flow losses in a gradual conical expansion region, as calculated from Gibson’s suggestion [15, 50], Eq. (6.79), for a smooth wall. the shear stress in the comer separated flow, or deadwater region, is negligible, so that a control volume analysis between the expansion section and the end of the separation zone gives a theoretical loss: - _ ^ “ V £>V V^Hlg) (6.77) Note that K is based on the velocity head in the small pipe. Equation (6.77) is in excellent agreement with experiment. For the sudden contraction, however, flow separation in the downstream pipe causes the main stream to contract through a minimum diameter called the vena contracta, as sketched in Fig. 6.22. Because the theory of the vena contracta is not well developed, the loss coefficient in the figure for sudden contraction is experimen¬ tal. It fits the empirical formula K^c “ 0.42 (^1 - ^ (6.78) up to the value dID = 0.76, above which it merges into the sudden-expansion predic¬ tion, Eq. (6.77). As flow enters a gradual expansion or diffuser, such as the conical geometry of Fig. 6.23, the velocity drops and the pressure rises. An efficient diffuser reduces the pumping power required. Head loss can be large, due to flow separation on the walls, if the cone angle is too great. A thinner entrance boundary layer, as in Fig. 6.6, causes 6.9 Minor or Local Losses in Pipe Systems 387 a slightly smaller loss than a fully developed inlet flow. The flow loss is a combina¬ tion of nonideal pressure recovery plus wall friction. Some correlating curves are shown in Fig. 6.23. The loss coefficient K is based on the velocity head in the inlet (small) pipe and depends upon cone angle 29 and the diffuser diameter ratio dxld2. There is scatter in the reported data [15, 16]. The curves in Fig. 6.23 are based on a correlation by A. Ft. Gibson , cited in Ref. 15: for 29 < 45° (6.79) For large angles, 29 > 45°, drop the coefficient (2.61 sin 9), which leaves us with a loss equivalent to the sudden expansion of Eq. (6.77). As seen, the formula is in reasonable agreement with the data from Ref. 16. The minimum loss lies in the region 5° < 26 < 15°, which is the best geometry for an efficient diffuser. For angles less than 5°, the diffuser is too long and has too much friction. Angles greater than 15° cause flow separation, resulting in poor pressure recovery. Professor Gordon Holloway provided the writer a recent example, where an improved diffuser design reduced the power requirement of a wind tunnel by 40 percent (100 hp decrease!). We shall look again at diffusers in Sec. 6.11, using the data of Ref. 14. For a gradual contraction, the loss is very small, as seen from the following experi¬ mental values : Contraction cone angle 29, deg 30 45 60 K for gradual contraction 0.02 0.04 0.07 References 15, 16, 43, and 46 contain additional data on minor losses. EXAMPLE 6.16 Water, p = 1.94 slugs/fP and v = 0.000011 ftVs, is pumped between two reservoirs at 0.2 ft^/s through 400 ft of 2-in-diameter pipe and several minor losses, as shown in Fig. E6.16. The roughness ratio is eld = 0.001. Compute the pump horsepower required. Screwed regular' 90° elbow exit © = 20 ft Sharp entrance Half-open gate valve Open globe valve 400 ft of pipe, d = E6.16 388 Chapter 6 Viscous Flow in Ducts Solution Write the steady flow energy equation between sections 1 and 2, the two reservoir surfaces: - P8 2g - ^ ^ + 12] + hf+ - hp Pg^g' where hp is the head increase across the pump. But since pi = p2 and Vi = 1^2 ~ 0, solve for the pump head: hp = Z2 — Zi + hf + = 120 ft — 20 ft + Now with the flow rate known, calculate Q 0.2 ftVs V =- = 1 - ^ - y = 9.17 ft/s A Now list and sum the minor loss coefficients: 2gW + 2^ (1) Loss K Sharp entrance (Fig. 6.21) 0.5 Open globe valve (2 in, Table 6.5) 6.9 12-in bend (Fig. 6.20) 0.25 Regular 90° elbow (Table 6.5) 0.95 Half-closed gate valve (from Fig. 6.18i?) 3.8 Sharp exit (Fig. 6.21) 1.0 tK = 13.4 Calculate the Reynolds number and pipe friction factor: Rew = — = Vd 9.17(^) = 139,000 V 0.000011 For eld = 0.001, from the Moody chart read/ = 0.0216. Substitute into Eq. (1): hn — 100 ft + (9.17 ft/s) 2 r 0.0216(400) 2_ 12 + 13.4 2(32.2 ft/s^L = 100 ft + 85 ft = 185 ft pump head The pump must provide a power to the water of P = pgQhp = 1.94(32.2) lbf/ft^(185 ft) « 2300 ft ■ Ibf/s The conversion factor is 1 hp = 550 ft • Ibf/s. Therefore 2300 P = - = 4.2 hp Ans. 550 ^ Allowing for an efficiency of 70 to 80 percent, a pump is needed with an input of about 6 hp. Laminar Flow Minor Losses The data in Table 6.5 are for turbulent flow in fittings. If the flow is laminar, a dif¬ ferent form of loss occurs, which is proportional to V, not V^. By analogy with Eqs. (6.12) for Poiseuille flow, the laminar minor loss takes the form ^lam Apioss d fiV 6.10 Multiple-Pipe Systems 389 Laminar minor losses are just beginning to be studied, due to increased interest in micro- and nano-flows in tubes. They can be substantial, comparable to the Poiseuille loss. Professor Bruce Finlayson, of the University of Washington, kindly provided the writer with new data in the following table: Laminar Minor Loss Coefficients in Tube Fittings for 1 < Re^ < 10 Type of fitting ^lam 45° bend, long radius 0.2 90° bend, short radius 0.5 90° bend, long radius 0.36 2:1 pipe contraction 7.3 3:1 pipe contraction 8.6 4:1 pipe contraction 9.0 2:1 pipe expansion 3.1 3:1 pipe expansion 4.1 4:1 pipe expansion 4.5 For the bends in the table, is the excess loss after calculating Poiseuille flow around the centerline of the bend. For the contractions and expansion, is based upon the velocity in the smaller section. 6.10 Multiple-Pipe Systems® Pipes in Series If you can solve the equations for one-pipe systems, you can solve them all; but when systems contain two or more pipes, certain basic rules make the calculations very smooth. Any resemblance between these rules and the rules for handling electric circuits is not coincidental. Figure 6.24 shows three examples of multiple-pipe systems. The first is a set of three (or more) pipes in series. Rule 1 is that the flow rate is the same in all pipes: Qi = Qi = Qs = const (6.80) or Virfi = V2‘^2 = V'srif (6.81) Rule 2 is that the total head loss through the system equals the sum of the head loss in each pipe: = Ahi + A/z2 + A/j3 (6.82) In terms of the friction and minor losses in each pipe, we could rewrite this as _l_ Vj f /3A3 2g\d, and so on for any number of pipes in the series. Since V2 and V3 are proportional to Vi from Eq. (6.81), Eq. (6.83) is of the form 2g \ d2 + 2^3 (6.83) A '^1 ~ (^0 + <^1/1 + <^2/2 + <^3/3) 2g (6.84) ^This section may be omitted without loss of continuity. 390 Chapter 6 Viscous Flow in Ducts (a) @ ^ — ^ Fig. 6.24 Examples of multiple- pipe systems: (a) pipes in series; (b) pipes in parallel; (c) the three- reservoir junction problem. where the ct, are dimensionless constants. If the flow rate is given, we can evaluate the right-hand side and hence the total head loss. If the head loss is given, a little iteration is needed, since /i,/2, and/3 depend on through the Reynolds number. Begin by calculating /i,/2, and/3, assuming fully rough flow, and the solution for Vi will converge with one or two iterations. EXAMPLE 6.17 Given is a three-pipe series system, as in Fig. 6.24a. The total pressure drop is — Pb — 150,000 Pa, and the elevation drop is 2,4 — = 5 m. The pipe data are Pipe L, m d, cm e, mm eld 1 100 8 0.24 0.003 2 150 6 0.12 0.002 3 80 4 0.20 0.005 The fluid is water, p = 1000 kg/m^ and u = 1.02 X 10 m/s. Calculate the flow rate Q in m^/h through the system. 6.10 Multiple-Pipe Systems 391 Solution The total head loss across the system is ~ Za Zb — Pg From the continuity relation (6.84) the velocities are and Neglecting minor losses and substituting into Eq. (6.83), we obtain or Vi 20.3 m = — (1250/i 7900/2 + 32,000/,) 2g (1) This is the form that was hinted at in Eq. (6.84). It seems to be dominated by the third pipe loss 32,000/3. Begin by estimating /,, /2, and from the Moody-chart fully rough regime: /i = 0.0262 /2 = 0.0234 /, = 0.0304 Substitute in Eq. (1) to find Vj ~ 2g(20.3)/(33 + 185 + 973). The first estimate thus is Vi = 0.58 m/s, from which Re, « 45,400 Re2 = 60,500 Res = 90,800 Hence, from the Moody chart, /i = 0.0288 /2 = 0.0260 /, = 0.0314 Substitution into Eq. (1) gives the better estimate Vi = 0.565 m/s Q = \-Kd\Vi = 2.84 X lO'^mVs or Q = 10.2 m^/h Ans. A second iteration gives Q = 10.22 m^/h, a negligible change. Pipes in Parallel The second multiple-pipe system is the parallel flow case shown in Fig. 6.24/. Here the pressure drop is the same in each pipe, and the total flow is the sum of the individual flows: = Ahi = A/t2 = Ah^ Q = Qi + Qi + Qi (6.85fl) (6.85/) If the total head loss is known, it is straightforward to solve for (2, in each pipe and sum them, as will be seen in Example 6.18. The reverse problem, of determining when hf is known, requires iteration. Each pipe is related to hf by the Moody relation hf = f{Lld){V^/2g) = fQ^/C, where C = TT^gcf/^L. Thus each pipe has nearly quadratic nonlinear parallel resistance, and head loss is related to total flow rate by (6.86) 392 Chapter 6 Viscous Flow in Ducts Since the ^ vary with Reynolds number and roughness ratio, one begins Eq. (6.86) by guessing values of^ (fully rough values are recommended) and calculating a first estimate of hf. Then each pipe yields a flow-rate estimate 2, ~ {Cjhf/f)''^ and hence a new Reynolds number and a better estimate of^. Then repeat Eq. (6.86) to convergence. It should be noted that both of these parallel-pipe cases — finding either XQ or hf — are easily solved by Excel if reasonable guesses are given. EXAMPLE 6.18 Assume that the same three pipes in Example 6.17 are now in parallel with the same total head loss of 20.3 m. Compute the total flow rate Q, neglecting minor losses. Solution From Eq. (6.85fl) we can solve for each V separately: y2 y2 y2 20.3 m = 1250/1 = 2500/) = 2000/, (1) 2g 2g 2g Guess fully rough flow in pipe I:/, = 0.0262, Vi = 3.49 m/s; hence Re, = Vidi/u = 273,000. From the Moody chart read / = 0.0267; recompute Vi = 3.46 m/s, Qi = 62.5 m^/h. Next guess for pipe 2: /2 ~ 0.0234, V2 ~ 2.61 m/s; then Re2 = 153,000, and hence /2 = 0.0246, V2 = 2.55 m/s, Q2 = 25.9 mVh. Finally guess for pipe 3: ~ 0.0304, ~ 2.56 m/s; then Re3 = 100,000, and hence /, = 0.0313, V3 = 2.52 m/s, Q3 = 11.4 m^/h. This is satisfactory convergence. The total flow rate is Q = Qi + Q2 + Qi = 62.5 -f 25.9 -f 11.4 = 99.8 mVh Ans. These three pipes carry 10 times more flow in parallel than they do in series. This example may he solved by Excel iteration using the Colebrook-formula procedure outlined in Ex. 6.9. Each pipe is a separate iteration of friction factor, Reynolds number, and flow rate. The pipes are rough, so only one iteration is needed. Here are the Excel results: A B c D E E Rci (e/rf)i Ex. 6.18 — Pipe 1 Vi — m/s Qi - m^/h /i /i-guess 1 313053 0.003 3.991 72.2 0.0267 0.0200 2 271100 0.003 3.457 62.5 0.0267 0.0267 Re2 (e/d)2 Ex. 6.18 — Pipe 2 V2 ~ m/s Qz - m^/h fi /2-guess 1 166021 0.002 2.822 28.7 0.0246 0.0200 2 149739 0.002 2.546 25.9 0.0246 0.0246 Rej (e/rf)3 Ex. 6.18 — Pipe 3 V} — m/s Qi - m^/h fi /3-guess 1 123745 0.005 3.155 14.3 0.0313 0.0200 2 98891 0.005 2.522 11.4 0.0313 0.0313 Thus, as in the hand calculations, the total flow rate = 62.5 -1- 25.9 -1- 11.4 = 99.8 m^/h. Ans. 6.10 Multiple-Pipe Systems 393 Three-Reservoir Junction Consider the third example of a three-reservoir pipe junction, as in Fig. 6.24c. If all flows are considered positive toward the junction, then Qi + Q2 + Q3 = 0 (6.87) which obviously implies that one or two of the flows must be away from the junction. The pressure must change through each pipe so as to give the same static pressure p j at the junction. In other words, let the HGL at the junction have the elevation where py is in gage pressure for simplicity. Then the head loss through each, assuming Pi — Pi — P3 — ^ (gage) at each reservoir surface, must be such that ^hl A/!3 2g di = Zi - hj Vi fiLi 2g d2 = Zi 2g di = Zi - hj (6.88) We guess the position hj and solve Eqs. (6.88) for Vi, V2, and V3 and hence Qi, Q2, and Qi, iterating until the flow rates balance at the junction according to Eq. (6.87). If we guess hj too high, the sum Qi + Q2 + Qi will be negative and the remedy is to reduce hj, and vice versa. EXAMPLE 6.19 Take the same three pipes as in Example 6.17, and assume that they connect three reservoirs at these surface elevations Zi = 20 m Z2— 100 m 23 = 40 m Find the resulting flow rates in each pipe, neglecting minor losses. Solution As a first guess, take hj equal to the middle reservoir height, Zi = hj = AQ m. This saves one calculation (Qi = 0) and enables us to get the lay of the land: Reservoir hj, m Zi - hj, m fi Vi, m/s 0„ m^/h L/di 1 40 -20 0.0267 -3.43 -62.1 1250 2 40 60 0.0241 4.42 45.0 2500 3 40 0 0 0 XQ = -17.1 2000 394 Chapter 6 Viscous Flow in Ducts Pipe Networks Fig. 6.25 Schematic of a piping network. Since the sum of the flow rates toward the junction is negative, we guessed hj too high. Reduce /jy to 30 m and repeat: Reservoir h], m Zi — hj, m fi Vi, m/s Qi, m’/h 1 30 -10 0.0269 -2.42 -43.7 2 30 70 0.0241 4.78 48.6 3 30 10 0.0317 1.76 8.0 22 = 12.9 This is positive and Make one final list: so we can linearly interpolate to get an accurate guess: hj~ 34.3 m. Reservoir hj, m S 1 fi Vi, m/s Qi, m^/h 1 34.3 -14.3 0.0268 -2.90 -52.4 2 34.3 65.7 0.0241 4.63 47.1 3 34.3 5.7 0.0321 1.32 6.0 22 = 0.7 This is close enough; hence we calculate that the flow rate is 52.4 m^/h toward reservoir 3, balanced hy 47.1 m^/h away from reservoir 1 and 6.0 m^/h away from reservoir 3. One further iteration with this problem would give hj = 34.53 m, resulting in Qi = —52.8, Q2 = 47.0, and Qs = 5.8 m^/h, so that XQ = 0 to three-place accuracy. Peda- gogically speaking, we would then be exhausted. The ultimate case of a multipipe system is the piping network illustrated in Fig. 6.25. This might represent a water supply system for an apartment or subdivision 6.11 Experimental Duct Flows: Diffuser Performance 395 or even a city. This network is quite complex algebraically but follows the same basic rules: 1 . The net flow into any junction must be zero. 2. The net pressure change around any closed loop must be zero. In other words, the HGL at each junction must have one and only one elevation. 3. All pressure changes must satisfy the Moody and minor-loss friction correlations. By supplying these rales to each junction and independent loop in the network, one obtains a set of simultaneous equations for the flow rates in each pipe leg and the HGL (or pres¬ sure) at each junction. Solution may then be obtained by numerical iteration, as first developed in a hand calculation technique by Prof. Hardy Cross in 1936 . Computer solution of pipe network problems is now quite common and is covered in at least one specialized text . Network analysis is quite useful for real water distribution systems if well calibrated with the actual system head loss data. The Moody chart is such a great correlation for tubes of any cross section with any roughness or flow rate that we may be deluded into thinking that the world of internal flow prediction is at our feet. Not so. The theory is reliable only for ducts of constant cross section. As soon as the section varies, we must rely principally on experiment to determine the flow properties. As mentioned many times before, experimentation is a vital part of fluid mechanics. Literally thousands of papers in the literature report experimental data for specific internal and external viscous flows. We have already seen several examples: 1. Vortex shedding from a cylinder (Fig. 5.2). 2. Drag of a sphere and a cylinder (Fig. 5.3). 3. Hydraulic model of a dam spillway (Fig. 5.9). 4. Rough-wall pipe flows (Fig. 6.12). 5. Secondary flow in ducts (Fig. 6.16). 6. Minor duct loss coefficients (Sec. 6.9). Chapter 7 will treat a great many more external flow experiments, especially in Sec. 7.6. Here we shall show data for one type of internal flow, the diffuser. Diffuser Performance A diffuser, shown in Fig. 6.26a and b, is an expansion or area increase intended to reduce velocity in order to recover the pressure head of the flow. Rouse and Ince relate that it may have been invented by customers of the early Roman (about 100 A.D.) water supply system, where water flowed continuously and was billed according to pipe size. The ingenious customers discovered that they could increase the flow rate at no extra cost by flaring the outlet section of the pipe. Engineers have always designed diffusers to increase pressure and reduce kinetic energy of ducted flows, but until about 1950, diffuser design was a combination of art, luck, and vast amounts of empiricism. Small changes in design parameters caused large changes in performance. The Bernoulli equation seemed highly suspect as a useful tool. 6.11 Experimental Duct Flows; Diffuser Performance^ ^This section may be omitted without loss of continuity. 396 Chapter 6 Viscous Flow in Ducts Exit {b) (c) Fig. 6.26 Diffuser geometry and typical flow regimes: (a) geometry of a flat- walled diffuser; {b) geometiy of a conical diffuser; (c) flat diffuser stability map. (From Ref. 14, by permission ofCreare, Inc.) Neglecting losses and gravity effects, the incompressible Bernoulli equation predicts that P + = Po = const (6.89) where pq is the stagnation pressure the fluid would achieve if the fluid were slowed to rest (V = 0) without losses. The basic output of a diffuser is the pressure-recovery coefficient Cp, defined as Pe - Pt Pot - Pt (6.90) where subscripfs e and t mean the exit and the throat (or inlet), respectively. Higher Cp means better performance. Consider the flat-walled diffuser in Fig. 6.26a, where section 1 is the inlet and section 2 the exit. Application of Bernoulli’s equation (6.89) to this diffuser predicts that Poi = Fi + ~ Pi iP^i ~ Poi C, /7, frictionless or 2 (6.91) 6.11 Experimental Duct Flows: Diffuser Performance 397 Fig. 6.27 Diffuser performance: (a) ideal pattern with good performance; (b) actual measured pattern with boundary layer separation and resultant poor performance. ih) Meanwhile, steady one-dimensional continuity would require that Q=ViAi = V2A2 (6.92) Combining (6.91) and (6.92), we can write the performance in terms of the area ratio AR = A2/A1, which is a basic parameter in diffuser design: Cp.frictionless = 1 ~ (A-R) ^ (6.93) A typical design would have AR = 5:1, for which Eq. (6.93) predicts Cp = 0.96, or nearly full recovery. But, in fact, measured values of Cp for this area ratio are only as high as 0.86 and can be as low as 0.24. The basic reason for the discrepancy is flow separation, as sketched in Fig. 6.21 b. The increasing pressure in the diffuser is an unfavorable gradient (Sec. 7.5), which causes the viscous boundary layers to break away from the walls and greatly reduces the performance. Computational fluid dynamics (CFD) can now predict this behavior. 398 Chapter 6 Viscous Flow in Ducts As an added complication to boundary layer separation, the flow patterns in a dif¬ fuser are highly variable and were considered mysterious and erratic until 1955, when Kline revealed the structure of these patterns with flow visualization techniques in a simple water channel. A complete stability map of diffuser flow patterns was published in 1962 by Fox and Kline , as shown in Fig. 6.26c. There are four basic regions. Below line aa there is steady viscous flow, no separation, and moderately good perfor¬ mance. Note that even a very short diffuser will separate, or stall, if its half-angle is greater than 10°. Between lines aa and bb is a transitory stall pattern with strongly unsteady flow. Best performance (highest C^) occurs in this region. The third pattern, between bb and cc, is steady bistable stall from one wall only. The stall pattern may flip-flop from one wall to the other, and performance is poor. The fourth pattern, above line cc, is jet flow, where the wall separation is so gross and pervasive that the mainstream ignores the walls and simply passes on through at nearly constant area. Performance is extremely poor in this region. Dimensional analysis of a flat-walled or conical diffuser shows that Cp should depend on the following parameters; 1. Any two of the following geometric parameters: a. Area ratio AR = A2/A1 or {DJUf b. Divergence angle 29 c. Slenderness L/Wi or LID 2. Inlet Reynolds number Re, = V-flVilv or Re, = ViD/v' 3. Inlet Mach number Ma, = Vi/a^ 4. Inlet boundary layer blockage factor B, = A^JAy, where Abl is the wall area blocked, or displaced, by the retarded boundary layer flow in the inlet (typically B, varies from 0.03 to 0.12) A flat-walled diffuser would require an additional shape parameter to describe its cross section; 5. Aspect ratio AS = blW^ Even with this formidable list, we have omitted five possible important effects: inlet turbulence, inlet swirl, inlet profile vorticity, superimposed pulsations, and down¬ stream obstruction, all of which occur in practical machinery applications. The three most important parameters are AR, 9, and B. Typical performance maps for diffusers are shown in Fig. 6.28. For this case of 8 to 9 percent blockage, both the flat-walled and conical types give about the same maximum performance, Cp = 0.70, but at different divergence angles (9° flat versus 4.5° conical). Both types fall far short of the Bernoulli estimates of Cp = 0.93 (flat) and 0.99 (conical), primarily because of the blockage effect. From the data of Ref. 14 we can determine that, in general, performance decreases with blockage and is approximately the same for both flat-walled and conical dif¬ fusers, as shown in Table 6.6. In all cases, the best conical diffuser is 10 to 80 percent longer than the best flat-walled design. Therefore, if length is limited in the design, the flat-walled design will give the better performance depending on duct cross section. 6.11 Experimental Duct Flows: Diffuser Performance 399 Fig. 6.28a Typical performance maps for flat-wall and conical diffusers at similar operating conditions: flat wall. Source: From P. W. Runstadler, Jr., et at, "Dijf user Data Book," Creme Inc. Tech. Note 186, Hanover, NH, 1975., by permis.don of Creare, Inc. AS = 1.0 Ma, =0.2 nj — u.uo = 279,000 ITi (a) The experimental design of a diffuser is an excellent example of a successful attempt to minimize the undesirable effects of adverse pressure gradient and flow separation. Table 6.6 Maximum Diffuser Performance Data Source: From P. W. Runstadler, Jr., et ah, “Diffuser Data Book, ” Creme Inc. Tech. Note 186, Hanover, NH, 1975. Inlet blockage B, Flat-walled Conical Cp,max L/ITi Cp, max L/d 0.02 0.86 18 0.83 20 0.04 0.80 18 0.78 22 0.06 0.75 19 0.74 24 0.08 0.70 20 0.71 26 0.10 0.66 18 0.68 28 0.12 0.63 16 0.65 30 400 Chapter 6 Viscous Flow in Ducts Fig. 6.28b Typical performance maps for flat-wall and conical diffusers at similar operating conditions: conical wall. (From Ref. 14, by permission of Creare, Inc. ) 6.12 Fluid Meters Local Velocity Measurements M, = 0.2 B, =0.09 Re^= 120,000 Conical (b) Almost all practical fluid engineering problems are associated with the need for an accurate flow measurement. There is a need to measure local properties (velocity, pressure, temperature, density, viscosity, turbulent intensity), integrated properties (mass flow and volume flow), and global properties (visualization of the entire flow field). We shall concentrate in this section on velocity and volume flow measurements. We have discussed pressure measurement in Sec. 2.10. Measurement of other ther¬ modynamic properties, such as density, temperature, and viscosity, is beyond the scope of this text and is treated in specialized books such as Refs. 22 and 23. Global visualization techniques were discussed in Sec. 1.11 for low-speed flows, and the special optical techniques used in high-speed flows are treated in Ref. 34 of Chap. 1 . Flow measurement schemes suitable for open-channel and other free-surface flows are treated in Chap. 10. Velocity averaged over a small region, or point, can be measured by several different physical principles, listed in order of increasing complexity and sophistication: 6.12 Fluid Meters 401 1. Trajectory of floats or neutrally buoyant particles. 2. Rotating mechanical devices: a. Cup anemometer. b. Savonius rotor. c. Propeller meter. d. Turbine meter. 3. Pitot-static tube (Fig. 6.30). 4. Electromagnetic current meter. 5. Hot wires and hot Aims. 6. Laser-doppler anemometer (EDA). 7. Particle image velocimetry (PIV). Some of these meters are sketched in Fig. 6.29. Fig. 6.29 Eight common velocity meters: (a) three-cup anemometer; (b) Savonius rotor; (c) turbine mounted in a duct; (d) free-propeller meter; (e) hot-wire anemometer; (/) hot-film anemometer; (g) pitot- static tube; (h) laser-doppler anemometer. (g) Display 402 Chapter 6 Viscous Flow in Ducts Fig. 6.30 Pitot-static tube for combined measurement of static and stagnation pressure in a moving stream. Floats or Buoyant Particles. A simple but effective estimate of flow velocity can be found from visible particles entrained in the flow. Examples include flakes on the surface of a channel flow, small neutrally buoyant spheres mixed with a liquid, or hydrogen bubbles. Sometimes gas flows can be estimated from the motion of entrained dust par¬ ticles. One must establish whether the particle motion truly simulates the fluid motion. Floats are commonly used to track the movement of ocean waters and can be designed to move at the surface, along the bottom, or at any given depth . Many official tidal current charts were obtained by releasing and timing a floating spar attached to a length of string. One can release whole groups of spars to determine a flow pattern. Rotating Sensors. The rotating devices of Fig. 6.29a to d can he used in either gases or liquids, and their rotation rate is approximately proportional to the flow velocity. The cup anemometer (Fig. 6.29fl) and Savonius rotor (Fig. 6.29b) always rotate the same way, regardless of flow direction. They are popular in atmospheric and oceano¬ graphic applications and can be fitted with a direction vane to align themselves with the flow. The ducted-propeller (Fig. 6.29c) and free-propeller (Fig. 6.29(f) meters must be aligned with the flow parallel to their axis of rotation. They can sense reverse flow because they will then rotate in the opposite direction. All these rotating sensors can be attached to counters or sensed hy electromagnetic or slip-ring devices for either a continuous or a digital reading of flow velocity. All have the disadvantage of being relatively large and thus not representing a “point.” Pitot-Static Tube. A slender tube aligned with the flow (Figs. 6.29g and 6.30) can measure local velocity by means of a pressure difference. It has sidewall holes to measure the static pressure in the moving stream and a hole in the front to measure the stagnation pressure pq, where the stream is decelerated to zero velocity. Instead of measuring po or p^ separately, it is customary to measure their difference with, say, a transducer, as in Fig. 6.30. If Re^) > 1000, where D is the prohe diameter, the flow around the probe is nearly frictionless and Bernoulli’s relation, Eq. (3.54), applies with good accuracy. For incompressible flow Ps + ~ Po + hpi^f + Pgza 6.12 Fluid Meters 403 Assuming that the elevation pressure difference pg(Zs — Zo) is negligible, this reduces to y = , iPo - Ps) 1/2 (6.94) This is the Pitot formula, named after the French engineer, Henri de Pitot, who designed the device in 1732. The primary disadvantage of the pitot tube is that it must be aligned with the flow direction, which may be unknown. For yaw angles greater than 5°, there are substantial errors in both the po and measurements, as shown in Fig. 6.30. The pitot-static tube is useful in liquids and gases; for gases a compressibility correction is necessary if the stream Mach number is high (Chap. 9). Because of the slow response of the fluid-fllled tubes leading to the pressure sensors, it is not useful for unsteady flow measurements. It does resemble a point and can be made small enough to measure, for example, blood flow in arteries and veins. It is not suitable for low-velocity measurement in gases because of the small pressure differences developed. For example, if y = 1 ft/s in standard air, from Eq. (6.94) we compute po — p equal to only 0.001 Ibf/ft^ (0.048 Pa). This is beyond the resolution of most pressure gages. Electromagnetic Meter. If a magnetic held is applied across a conducting fluid, the fluid motion will induce a voltage across two electrodes placed in or near the flow. The electrodes can be streamlined or built into the wall, and they cause little or no flow resistance. The output is very strong for highly conducting fluids such as liquid metals. Seawater also gives good output, and electromagnetic current meters are in common use in oceanography. Even low-conductivity fresh water can be measured by amplifying the output and insulating the electrodes. Commercial instruments are available for most liquid flows but are relatively costly. Electromagnetic flowmeters are treated in Ref. 26. Hot-Wire Anemometer. A very flne wire {d = 0.01 mm or less) heated between two small probes, as in Fig. 6.29e, is ideally suited to measure rapidly fluctuating flows such as the turbulent boundary layer. The idea dates back to work by L. V. King in 1914 on heat loss from long, thin cylinders. If electric power is supplied to heat the cylinder, the loss varies with flow velocity across the cylinder according to King ’s law q = fR== a + b(pVT (6.95) where n ~ j at very low Reynolds numbers and equals \ at high Reynolds numbers. The hot wire normally operates in the high-Reynolds-number range but should be calibrated in each situation to find the best-fit a, b, and n. The wire can be operated either at constant current I, so that resistance /? is a measure of V, or at constant resistance R (constant temperature), with I a measure of velocity. In either case, the output is a nonlinear function of V, and the equipment should contain a linearizer to produce convenient velocity data. Many varieties of commercial hot-wire equipment are available, as are do-it-yourself designs . Excellent detailed discussions of the hot wire are given in Ref. 28. Because of its frailty, the hot wire is not suited to liquid flows, whose high density and entrained sediment will knock the wire right off. A more stable yet quite sensitive alternative for liquid flow measurement is the hot-film anemometer (Fig. 6.29/). A thin metallic film, usually platinum, is plated onto a relatively thick support, which 404 Chapter 6 Viscous Flow in Ducts can be a wedge, a cone, or a cylinder. The operation is similar to the hot wire. The cone gives best response but is liable to error when the flow is yawed to its axis. Hot wires can easily be arranged in groups to measure two- and three-dimensional velocity components. Laser-Doppler Anemometer. In the LDA a laser beam provides highly focused, coherent monochromatic light that is passed through the flow. When this light is scattered from a moving particle in the flow, a stationary observer can detect a change, or doppler shift, in the frequency of the scattered light. The shift A/ is proportional to the velocity of the particle. There is essentially zero disturbance of the flow by the laser. Figure 6.29h shows the popular dual-beam mode of the LDA. A focusing device splits the laser into two beams, which cross the flow at an angle 6. Their intersection, which is the measuring volume or resolution of the measurement, resembles an ellip¬ soid about 0.5 mm wide and 0.1 mm in diameter. Particles passing through this measuring volume scatter the beams; they then pass through receiving optics to a photodetector, which converts the light to an electric signal. A signal processor then converts electric frequency to a voltage that can be either displayed or stored. If X. is the wavelength of the laser light, the measured velocity is given by A A/ 2 sin (6»/2) (6.96) Multiple components of velocity can be detected by using more than one photodetec¬ tor and other operating modes. Either liquids or gases can be measured as long as scattering particles are present. In liquids, normal impurities serve as scatterers, but gases may have to be seeded. The particles may be as small as the wavelength of the light. Although the measuring volume is not as small as with a hot wire, the LDA is capable of measuring turbulent fluctuations. The advantages of the LDA are as follows: 1. No disturbance of the flow. 2. High spatial resolution of the flow field. 3. Velocity data that are independent of the fluid thermodynamic properties. 4. An output voltage that is linear with velocity. 5. No need for calibration. The disadvantages are that both the apparatus and the fluid must be transparent to light and that the cost is high (a basic system shown in Fig. 6.29h begins at about $50,000). Once installed, an LDA can map the entire flow field in minutest detail. To truly appreciate the power of the LDA, one should examine, for instance, the amazingly detailed three-dimensional flow profiles measured by Eckardt in a high-speed centrifugal com¬ pressor impeller. Extensive discussions of laser velocimetry are given in Refs. 38 and 39. Particle Image Velocimetry. This popular new idea, called PIV for short, measures not just a single point but instead maps the entire field of flow. An illustration was shown in Fig. 1.1 8fc. The flow is seeded with neutrally buoyant particles. A planar laser light sheet across the flow is pulsed twice and photographed twice. If Ar is the particle displacement vector over a short time At, an estimate of its velocity is V ~ Ar/At. 6.12 Fluid Meters 405 Volume Flow Measurements A dedicated computer applies this formula to a whole cloud of particles and thus maps the flow field. One can also use the data to calculate velocity gradient and vorticity fields. Since the particles all look alike, other cameras may be needed to identify them. Three-dimensional velocities can be measured by two cameras in a stereoscopic arrangement. The PIV method is not limited to stop-action. New high-speed cameras (up to 10,000 frames per second) can record movies of unsteady flow fields. For further details, see the monograph by M. Raffel . EXAMPLE 6.20 The pitot-static tube of Fig. 6.30 uses mercury as a manometer fluid. When it is placed in a water flow, the manometer height reading is h = 8.4 in. Neglecting yaw and other errors, what is the flow velocity V in ft/s? Solution From the two-fluid manometer relation (2.23b), with za — Z2, the pressure difference is related to h by Po- Ps= (.1m - lw)h Taking the specific weights of mercury and water from Table 2.1, we have Po- Ps= (846 - 62.4 Ibf/fP) 12 -ft = 549 Ibf/ft 2 The density of water is 62.4/32.2 = 1.94 slugs/ft^. Introducing these values into the pitot- static formula (6.97), we obtain V = 2(549 Ibf/ft-) 1/2 . 1.94 slugs/fP . 23.8 ft/s Ans. Since this is a low-speed flow, no compressibility correction is needed. It is often desirable to measure the integrated mass, or volume flow, passing through a duct. Accurate measurement of flow is vital in billing customers for a given amount of liquid or gas passing through a duct. The different devices available to make these measurements are discussed in great detail in the ASME text on fluid meters . These devices split into two classes; mechanical instruments and head loss instruments. The mechanical instruments measure actual mass or volume of fluid by trapping it and counting it. The various types of measurement are 1. Mass measurement a. Weighing tanks b. Tilting traps 2. Volume measurement a. Volume tanks b. Reciprocating pistons 406 Chapter 6 Viscous Flow in Ducts c. Rotating slotted rings d. Nutating disc e. Sliding vanes f. Gear or lobed impellers g. Reciprocating bellows h. Sealed-drum compartments The last three of these are suitable for gas flow measurement. The head loss devices obstruct the flow and cause a pressure drop, which is a measure of flux: 1. Bernoulli-type devices a. Thin-plate orifice b. Flow nozzle c. Venturi tube 2. Friction loss devices a. Capillary tube b. Porous plug The friction loss meters cause a large nonrecoverable head loss and obstruct the flow too much to be generally useful. Six other widely used meters operate on different physical principles: 1. Turbine meter 2. Vortex meter 3. Ultrasonic flowmeter 4. Rotameter 5. Coriolis mass flowmeter 6. Laminar flow element Nutating Disc Meter. For measuring liquid volumes, as opposed to volume rates, the most common devices are the nutating disc and the turbine meter. Figure 6.31 shows Fig. 6.31 Cutaway sketch of a nutating disc fluid meter. A: metered- volume chamber; fl: nutating disc; C: rotating spindle; D\ drive magnet; E: magnetic counter sensor. Source: Courtesy of Badger Meter, Inc., Milwaukee, Wisconsin. 6.12 Fluid Meters 407 a cutaway sketch of a nutating disc meter, widely used in both water and gasoline delivery systems. The mechanism is clever and perhaps beyond the writer’s capabil¬ ity to explain. The metering chamber is a slice of a sphere and contains a rotating disc set at an angle to the incoming flow. The fluid causes the disc to nutate (spin eccentrically), and one revolution corresponds to a certain fluid volume passing through. Total volume is obtained by counting the number of revolutions. Turbine Meter. The turbine meter, sometimes called a propeller meter, is a freely rotating propeller that can be installed in a pipeline. A typical design is shown in Fig. 6.32a. There are flow straighteners upstream of the rotor, and the rotation is Fig. 6.32 The turbine meter widely used in the oil and gas industry: (a) basic design; (b) the linearity curve is the measure of variation in the signal output across the 10% to 100% nominal flow range of the meter. (Daniel Measurement and Control, Houston, TX.) Source: (a) Daniel Industries of Fluke Calibration, Houston, TX. Magnetic Flow rate % 408 Chapter 6 Viscous Flow in Ducts Fig. 6.33 A Commercial handheld wind velocity turbine meter. { Courtesy of Nielsen-Kellerman Company.) measured by electric or magnetic pickup of pulses caused by passage of a point on the rotor. The rotor rotation is approximately proportional to the volume flow in the pipe. Like the nutating disc, a major advantage of the turbine meter is that each pulse corresponds to a finite incremental volume of fluid, and the pulses are digital and can be summed easily. Liquid flow turbine meters have as few as two blades and produce a constant number of pulses per unit fluid volume over a 5:1 flow rate range with ±0.25 percent accuracy. Gas meters need many blades to produce sufficient torque and are accurate to ±1 percent. Since turbine meters are very individualistic, flow calibration is an absolute neces¬ sity. A typical liquid meter calibration curve is shown in Fig. 6.32b. Researchers attempting to establish universal calibration curves have met with little practical suc¬ cess as a result of manufacturing variabilities. Turbine meters can also be used in unconfined flow situations, such as winds or ocean currents. They can be compact, even microsize with two or three component directions. Figure 6.33 illustrates a handheld wind velocity meter that uses a seven- bladed turbine with a calibrated digital output. The accuracy of this device is quoted at ±2 percent. Vortex Flowmeters. Recall from Fig. 5.2 that a bluff body placed in a uniform crossflow sheds alternating vortices at a nearly uniform Strouhal number St = fUU, where U is the approach velocity and L is a characteristic body width. Since L and St are constant, this means that the shedding frequency is proportional to velocity: /= (const) (t/) (6.97) The vortex meter introduces a shedding element across a pipe flow and picks up the shedding frequency downstream with a pressure, ultrasonic, or heat transfer type of sensor. A typical design is shown in Fig. 6.34. 6.12 Fluid Meters 409 Fig. 6.34 A vortex flowmeter, f Courtesy ofinvensys p/c.) The advantages of a vortex meter are as follows: 1. Absence of moving parts. 2. Accuracy to ±1 percent over a wide flow rate range (up to 100:1). 3. Ability to handle very hot or very cold fluids. 4. Requirement of only a short pipe length. 5. Calibration insensitive to fluid density or viscosity. For further details see Ref. 40. Ultrasonic Flowmeters. The sound-wave analog of the laser velocimeter of Fig. 6.29h is the ultrasonic flowmeter. Two examples are shown in Fig. 6.35. The pulse-type flowmeter is shown in Fig. 635a. Upstream piezoelectric transducer A is excited with a short sonic pulse that propagates across the flow to downstream transducer B. The arrival at B triggers another pulse to be created at A, resulting in a regular pulse frequency fj^. The same process is duplicated in the reverse direction from B to A, creating frequency fg. The difference “ /s is proportional to the flow rate. Figure 6.35b shows a doppler-type arrangement, where sound waves from transmitter T are scattered by particles or contaminants in the flow to receiver R. Comparison of the two signals reveals a doppler frequency shift that is proportional to the flow rate. Ultrasonic meters are nonintrusive and can be directly attached to pipe flows in the 410 Chapter 6 Viscous Flow in Ducts Fig. 6.35 Ultrasonic flowmeters: (a) pulse type; (b) doppler-shift type {from Ref. 41 J; (c) a portable noninvasive installation (Courtesy of Thermo Polysonics, Houston, TX. ) field (Fig. 6.35c). Their quoted uncertainty of ±1 to 2 percent can rise to ±5 percent or more due to irregularities in velocity profile, fluid temperature, or Reynolds num¬ ber. For further details see Ref. 41. Rotameter. The variable-area transparent rotameter of Fig. 6.36 has a float that, under the action of flow, rises in the vertical tapered tube and takes a certain equilibrium position for any given flow rate. A student exercise for the forces on the float would yield the approximate relation Q = C,A, f 21V„et Y' vAfioatPfluid/ (6.98) where VTnet is the float’s net weight in the fluid, = Atube “ Afloat is the annular area between the float and the tube, and is a dimensionless discharge coeffi¬ cient of order unity, for the annular constricted flow. For slightly tapered tubes, Aa varies nearly linearly with the float position, and the tube may be calibrated and marked with a flow rate scale, as in Fig. 6.36. The rotameter thus provides a readily visible measure of the flow rate. Capacity may be changed by using different-sized floats. Obviously the tube must be vertical, and the device does not give accurate readings for fluids containing high concentrations of bubbles or particles. Coriolis Mass Flowmeter. Most commercial meters measure volume flow, with mass flow then computed by multiplying by the nominal fluid density. An attractive 6.12 Fluid Meters 411 modern alternative is a mass flowmeter, which operates on the principle of the Coriolis acceleration associated with noninertial coordinates [recall Fig. 3.11 and the Coriolis term 2fl X V in Eq. (3.48)]. The output of the meter is directly pro¬ portional to mass flow. Figure 6.37 is a schematic of a Coriolis device, to be inserted into a piping system. The flow enters a loop arrangement, which is electromagnetically vibrated at a high natural frequency (amplitude < 1 mm and frequency > 100 Hz). The Coriolis effect induces a downward force on the loop entrance and an upward force on the loop exit, as shown. The loop twists, and the twist angle can be measured and is proportional to the mass flow through the tube. Accuracy is typically less than 1 percent of full scale. Laminar Flow Element. In many, perhaps most, commercial flowmeters, the flow through the meter is turbulent and the variation of flow rate with pressure drop is nonlinear. In laminar duct flow, however, Q is linearly proportional to Ap, as in Eq. (6.12): Q = [irK^/iSfiL)] Ap. Thus a laminar flow sensing element is attractive, since its calibration will be linear. To ensure laminar flow for what otherwise would be a turbulent condition, all or part of the fluid is directed into small pas¬ sages, each of which has a low (laminar) Reynolds number. A honeycomb is a popular design. Figure 6.38 uses axial flow through a narrow annulus to create laminar flow. The theory again predicts Q^Ap, as in Eq. (6.70). However, the flow is very sensitive to passage size; for example, halving the annulus clearance increases Ap more than eight times. Careful calibration is thus necessary. In Eig. 6.38 the laminar flow concept has been synthesized into a complete mass flow system, with temperature control, dif¬ ferential pressure measurement, and a microprocessor all self-contained. The accuracy of this device is rated at ±0.2 percent. Fig. 6.36 A commercial rotameter. The float rises in the tapered tube to an equilibrium position, which is a measure of the fluid flow rate. ( Courtesy of Blue White Industries, Huntington Beach, CA.) Fig. 6.37 Schematic of a Coriolis mass flowmeter. Flow out 412 Chapter 6 Viscous Flow in Ducts Fig. 6.38 A complete flowmeter system using a laminar flow element (in this case a narrow annulus). The flow rate is linearly proportional to the pressure drop. Source: Courtesy of Martin Girard, DH Instruments, Inc. Bernoulli Obstruction Theory. Consider the generalized flow obstruction shown in Fig. 6.39. The flow in the basic duct of diameter D is forced through an obstruction of diameter d\ the (3 ratio of the device is a key parameter: [3 = ^ (6.99) After leaving the obstruction, the flow may neck down even more through a vena contracta of diameter £>2 < d, as shown. Apply the Bernoulli and continuity equations for incompressible steady frictionless flow to estimate the pressure change: Continuity: Q = ^D^Vi = ^D\V2 Bernoulli: Fo = Fi + ^ Pi 3- ^pVl Eliminating V^, we solve these for V2 or Q in terms of the pressure change pi — pf 2{pt-p2) ^2 P(i d\id^) (6.100) But this is surely inaccurate because we have neglected friction in a duct flow, where we know friction will be very important. Nor do we want to get into the business of measuring vena contracta ratios D2ld for use in (6.100). Therefore we assume that 6.12 Fluid Meters 413 Fig. 6.39 Velocity and pressure change through a generalized Bernoulli obstruction meter. Horizontal D2ID ~ (3 and then calibrate the device to fit the relation Q = A,V, = C,A, 2(/h " P2)lp 1 - /3" 1/2 (6.101) where subscript t denotes the throat of the obstruction. The dimensionless discharge coefficient Q accounts for the discrepancies in the approximate analysis. By dimen¬ sional analysis for a given design we expect Q = fiP, Rcd) where Re^ = ViD V (6.102) The geometric factor involving (5 in (6.101) is called the velocity-of-approach factor: E= (I - (6.103) One can also group Q and E in Eq. (6.101) to form the dimensionless flow coefficient a: Q (1 - a = CjE = (6.104) 414 Chapter 6 Viscous Flow in Ducts Thus Eq. (6.101) can be written in the equivalent form r 2(pi - P2)V'^ Q = qA, — - — V P J Obviously the flow coefficient is correlated in the same manner: a =m Ren) (6.105) (6.106) Occasionally one uses the throat Reynolds number instead of the approach Reynolds number: Red = V,d u Reg /3 (6.107) Since the design parameters are assumed known, the correlation of a from Eq. (6.106) or of Cd from Eq. (6.102) is the desired solution to the fluid metering problem. The mass flow is related to Q by m = pQ (6.108) and is thus correlated by exactly the same formulas. Figure 6.40 shows the three basic devices recommended for use by the International Organization for Standardization (ISO) : the orifice, nozzle, and venturi tube. (a) (b) Fig. 6.40 Standard shapes for the three primary Bernoulli ohstmction- type meters: (a) long-radius nozzle; (b) thin-plate orifice; (c) venturi nozzle. (Based on data from the International Organization for Standardization. } 6.12 Fluid Meters 415 Thin-Plate Orifice. The thin-plate orifice, Fig. 6.40f>, can be made with (3 in the range of 0.2 to 0.8, except that the hole diameter d should not be less than 12.5 mm. To measure pi and p2, three types of tappings are commonly used: 1. Corner taps where the plate meets the pipe wall. 2. D: \p taps: pipe-wall taps at D upstream and \D downstream. 3. Flange taps: 1 in (25 mm) upstream and 1 in (25 mm) downstream of the plate, regardless of the size D. Types 1 and 2 approximate geometric similarity, but since the flange taps 3 do not, they must be correlated separately for every single size of pipe in which a flange-tap plate is used [30, 31]. Figure 6.41 shows the discharge coefficient of an orifice with D: or type 2 taps in the Reynolds number range Re^ = 10“ to lO’ of normal use. Although detailed charts such as Fig. 6.41 are available for designers , the ASME recommends use of the curve-fit formulas developed by the ISO . The basic form of the curve fit is 0 090 Q =m + 91.71/32-5Rea“-’5 + - 0.03310% (6.109) 1 jJ where f{0) = 0.5959 + 0.0312/3^' - 0.184/3® Fig. 6.41 Discharge coefficient for a thin-plate orifice with D: \D taps, plotted from Eqs. (6.109) and (6.110fc). 416 Chapter 6 Viscous Flow in Ducts The correlation factors Fi and F2 vary with tap position: Corner taps: Cj = 0 f 2 = 0 (h.llOa) D- \D taps: Fj = 0.4333 Fj = 0.47 (6.110^) 1 I ^ D > 2.3 in Flange taps: F2 = - ^ — Fj = < O (in) (6.110c) i 0.4333 2.0 <£)<2.3in Note that the flange taps (6.110c), not being geometrically similar, use raw diameter in inches in the formula. The constants will change if other diameter units are used. We cautioned against such dimensional formulas in Example 1.4 and Eq. (5.17) and give Eq. (6.110c) only because flange taps are widely used in the United States. Flow Nozzle. The flow nozzle comes in two types, a long-radius type shown in Fig. 6.40a and a short-radius type (not shown) called the ISA 1932 nozzle [30, 31]. The flow nozzle, with its smooth, rounded entrance convergence, practically elimi¬ nates the vena contracta and gives discharge coefficients near unity. The nonrecover- able loss is still large because there is no diffuser provided for gradual expansion. The ISO recommended correlation for long-radius-nozzle discharge coefficient is 7 106X1/2 /10®\''^ Q = 0.9965 - 0.00653/3'^^ - = 0.9965 - 0.00653 - (6.111) VRezj/ \Red/ The second form is independent of the f3 ratio and is plotted in Fig. 6.42. A similar ISO correlation is recommended for the short-radius ISA 1932 flow nozzle: Q « 0.9900 - 0.2262/3'-^ -f (0.000215 - 0.001125/3 + 0.00249/3''’) - (6.112) \^&dJ Flow nozzles may have /3 values between 0.2 and 0.8. Fig. 6.42 Discharge coefficient for long-radius nozzle and classical Herschel-type venturi. Re^, Re^ 10' 6.12 Fluid Meters 417 Fig. 6.43 Discharge coefficient for a venturi nozzle. 1.00 0.98 0.96 0.94 0.92 Inte St£ 0.316 10^ < rnatio ndard 3< Re^ nal s: 0.775 c2.0; 1.5 X < 10'’ 0.3 0.4 0.5 0.6 P 0.7 0.8 Venturi Meter. The third and final type of obstruction meter is the venturi, named in honor of Giovanni Venturi (1746-1822), an Italian physicist who first tested conical expansions and contractions. The original, or classical, venturi was invented by a U.S. engineer, Clemens Herschel, in 1898. It consisted of a 21° conical contraction, a straight throat of diameter d and length d, then a 7° to 15° conical expansion. The discharge coefficient is near unity, and the nonrecoverable loss is very small. Herschel venturis are seldom used now. The modern venturi nozzle. Fig. 6.40c, consists of an ISA 1932 nozzle entrance and a conical expansion of half-angle no greater than 15°. It is intended to be operated in a narrow Reynolds number range of 1.5 X 10^ to 2 X 10^. Its discharge coefficient, shown in Fig. 6.43, is given by the ISO correlation formula Crf = 0.9858 - 0.196/3'^ (6.113) It is independent of Re^, within the given range. The Herschel venturi discharge varies with Re^, but not with f3, as shown in Fig. 6.42. Both have very low net losses. The choice of meter depends on the loss and the cost and can be illustrated by the following table: Type of meter Net head loss Cost Orifice Lai'ge Small Nozzle Medium Medium Venturi Small Large As so often happens, the product of inefficiency and initial cost is approximately constant. The average nonrecoverable head losses for the three types of meters, expressed as a fraction of the throat velocity head vfl{2g), are shown in Fig. 6.44. The orifice has the greatest loss and the venturi the least, as discussed. The orifice and nozzle 418 Chapter 6 Viscous Flow in Ducts Fig. 6.44 Nonrecoverable head loss in Bernoulli ohstruction meters. (Adapted from Ref. 30.) Th c n-plate rilice Vent m: Flow nozzle / 5° cone a ngle cone ang e 1 1 i 0 I I I I I 0.2 0.3 0.4 0.5 0.6 0.7 0.8 P simulate partially closed valves as in Fig. 6.18/?, while the venturi is a very minor loss. When the loss is given as a fraction of the measured pressure drop, the orifice and nozzle have nearly equal losses, as Example 6.21 will illustrate. The other types of instruments discussed earlier in this section can also serve as flowmeters if properly constructed. For example, a hot wire mounted in a tube can be calibrated to read volume flow rather than point velocity. Such hot-wire meters are commercially available, as are other meters modified to use velocity instruments. For further details see Ref. 30. Compressible Gas Flow Correction Factor. The orifice/nozzle/venturi formulas in this section assume incompressible flow. If the fluid is a gas, and the pressure ratio (p2/ Pi) is not near unity, a compressibility correction is needed. Equation (6.101) is rewrit¬ ten in terms of mass flow and the upstream density pp. . _ /2pi(pi - pi) d m = CaYA,y^ - ^ ^ — where /3 = “ (6.114) The dimensionless expansion factor T is a function of pressure ratio, (3, and the type of meter. Some values are plotted in Eig. 6.45. The orifice, with its strong jet contrac¬ tion, has a different factor from the venturi or the flow nozzle, which are designed to eliminate contraction. 6.12 Fluid Meters 419 Fig. 6.45 Compressible flow expansion factor Y for flowmeters. EXAMPLE 6.21 We want to meter the volume flow of water (p = 1000 kg/m^, v = 1.02 X 10~ mVs) moving through a 200-mm-diameter pipe at an average velocity of 2.0 m/s. If the differen¬ tial pressure gage selected reads accurately at Pi — P2 = 50,000 Pa, what size meter should be selected for installing (a) an orifice with D: \D taps, {b) a long-radius flow nozzle, or (c) a venturi nozzle? What would be the nonrecoverable head loss for each design? Solution Here the unknown is the (3 ratio of the meter. Since the discharge coefficient is a complicated function of /3, iteration will be necessary. We are given D = 0.2 m and Vi = 2.0 m/s. The pipe-approach Reynolds number is thus V (2.0)(0.2) 1.02 X 10“® 392,000 For all three cases [(a) to (c)] the generalized formula (6.105) holds: V"! V, = 4 = a /3^ 2(Pi - Pi) C, (1 - isy (1) where the given data are Vi = 2.0 m/s, p = 1000 kg/m^, and Ap = 50,000 Pa. Inserting these known values into Eq. (1) gives a relation between fi and a: /3" 2(50,000) 1000 1/2 or (2) The unknowns are f3 (or a) and C^. Parts {a) to (c) depend on the particular chart or formula needed for = fcn(Refl_ jS). We can make an initial guess P ~ 0.5 and iterate to convergence. Part (a) For the orifice with D\ jD taps, use Eq. (6.109) or Eig. 6.41. The iterative sequence is Pi « 0.5, Crfi « 0.604, ai « 0.624, Pi « 0.566, C^i « 0.606, ai « 0.640, P, = 0.559 420 Chapter 6 Viscous Flow in Ducts We have converged to three figures. The proper orifice diameter is d = f3D =112 mm Ans. (a) Part (b) For the long-radius flow nozzle, use Eq. (6.1 1 1) or Fig. 6.42. The iterative sequence is /3i « 0.5, Cji « 0.9891, ai « 1.022, /Sj « 0.442, = 0.9896, ctj = 1.009, = 0.445 We have converged to three figures. The proper nozzle diameter is d = I3D = 89 mm Ans. (b) Part (c) For the venturi nozzle, use Eq. (6.1 13) or Fig. 6.43. The iterative sequence is /3i « 0.5, Cai « 0.977, cti « 1.009, « 0.445, = 0.9807, Qj « 1.0004, (3^ = 0.447 We have converged to three figures. The proper venturi diameter is d = I3D = 89 mm Ans. (c) Comments: These meters are of similar size, hut their head losses are not the same. From Fig. 6.44 for the three different shapes we may read the three K factors and compute ^m.orifice 3.5 m Itm.nozzle 3.6 m yejjtyj.; 0.8 m The venturi loss is only about 22 percent of the orifice and nozzle losses. Solution by Excel Iteration for the Flow Nozzle Parts (a, b, c) were solved by hand, but Excel is ideal for these calculations. You may review this procedure from the instructions in Example 6.5. We need five columns: C calculated from Eq. (6. Ill), throat velocity V, calculated from Ap, a as calculated from Eq. (6. 104), and /3 calculated from the velocity ratio (V7V,). The fifth column is an initial guess for j3, which is replaced in its next row by the newly computed /3. Any initial j3 < \ will do. Here we chose /3 = 0.5 as in part (h) for the flow nozzle. Remember to use cell names, not symbols: in row 1, Q = Al, V, = Bl, a = Cl, and /3 = Dl. The process converges rapidly, in only two or three iterations: Cd from a = /3 = Eq.(6.114) V, = a{2Ap/p) Cja - I3''4)''0.5 (V/y,)''0.5 /3-guess A B c D E 1 0.9891 10.216 1.0216 0.4425 0.5000 2 0.9896 10.091 1.0091 0.4452 0.4425 3 0.9895 10.096 1.0096 0.4451 0.4452 4 0.9895 10.096 1.0096 0.4451 0.4451 The final answers for the long-radius flow nozzle are: a = 1.0096 Ci = 0.9895 /3 = 0.4451 Ans. (b) EXAMPLE 6.22 A long-radius nozzle of diameter 6 cm is used to meter airflow in a 10-cm-diameter pipe. Upstream conditions are pi = 200 kPa and Tj = 100°C. If the pressure drop through the nozzle is 60 kPa, estimate the flow rate in mVs. Summary 421 Solution • Assumptions: The pressure drops 30 percent, so we need the compressibility factor Y, and Eq. (6.114) is applicable to this problem. • Approach: Find pi and Q and apply Eq. (6.114) with [3 = 6/10 = 0.6. • Property values: Given pi and Tj, pi = PilRTi = (200,000)/[287(100 + 273)] = 1.87 kg/m^. The downstream pressure is p2 = 200 — 60 = 140 kPa, hence P2lp\ = 0.7. At 100°C, from Table A. 2, the viscosity of air is 2.17 E-5 kg/m-s. • Solution steps: Initially apply Eq. (6.114) by guessing, from Fig. 6.42, that Cj ~ 0.98. From Fig. 6.45, for a nozzle with P2/P1 = 0.7 and /3 = 0.6, read Y ~ 0.80. Then m = Cd YA, 2pi{pi - Pi) 1 - (0.98)(0.80)- (0.06 m)' 2(1.87 kg/m^) (60,000 Pa) 1 - (0.6) kg 1.13 — s Now estimate Re^, putting it in the convenient mass flow form: pVd 4ih 4(1.13 kg/s) Re,i = ^ — = - = - S - « I II e6 p irpd 7r(2. 17 E-5 kg/m — s) (0.06 m) Returning to Fig. 6.42, we could read a slightly better Cd ~ 0.99. Thus our final estimate is til ~ 1.14 kg/s Ans. Comments: Figure 6.45 is not just a “chart” for engineers to use casually. It is based on the compressible flow theory of Chap. 9. There, we may reassign this example as a theory. Summary This chapter has been concerned with internal pipe and duct flows, which are prob¬ ably the most common problems encountered in engineering fluid mechanics. Such flows are very sensitive to the Reynolds number and change from laminar to transi¬ tional to turbulent flow as the Reynolds number increases. The various Reynolds number regimes were outlined, and a semiempirical approach to turbulent flow modeling was presented. The chapter then made a detailed analysis of flow through a straight circular pipe, leading to the famous Moody chart (Fig. 6.13) for the friction factor. Possible uses of the Moody chart were discussed for flow rate and sizing problems, as well as the application of the Moody chart to noncircular ducts using an equivalent duct “diameter.” The addition of minor losses due to valves, elbows, fittings, and other devices was presented in the form of loss coefficients to be incorporated along with Moody-type friction losses. Multiple-pipe systems were discussed briefly and were seen to be quite complex algebraically and appropriate for computer solution. Diffusers are added to ducts to increase pressure recovery at the exit of a system. Their behavior was presented as experimental data, since the theory of real diffusers is still not well developed. The chapter ended with a discussion of flowmeters, espe¬ cially the pitot-static tube and the Bernoulli obstruction type of meter. Flowmeters also require careful experimental calibration. 422 Chapter 6 Viscous Flow in Ducts Problems Most of the problems herein are fairly straightforward. More difficult or open-ended assignments are labeled with an asterisk. Problems labeled with a computer icon may require the use of a computer. The standard end-of-chapter problems P6.1 to P6.163 (categorized in the problem list here) are followed by word problems W6.1 to W6.4, fundamentals of engineering exam problems FE6.1 to FE6.15, comprehensive problems C6.1 to C6.9, and design projects D6.1 and D6.2. Problem Distribution Section Topic Problems 6.1 Reynolds number regimes P6.DP6.5 6.2 Internal and external flow P6.6-P6.8 6.3 Head loss — friction factor P6.9-P6.ll 6.4 Laminar pipe flow P6.12-P6.33 6.5 Turbulence modeling P6.34-P6.40 6.6 Turbulent pipe flow P6.41-P6.62 6.7 Flow rate and sizing problems P6.63-P6.85 6.8 Noncircular ducts P6.86-P6.98 6.9 Minor or local losses P6.99-P6.110 6.10 Series and parallel pipe systems P6.lll-P6.120 6.10 Three-reservoir and pipe network systems P6.121-P6.130 6.11 Diffuser performance P6.131-P6.134 6.12 The pitot-static tube P6.135-P6.139 6.12 Flowmeters: the orifice plate P6.140-P6.148 6.12 Flowmeters: the flow nozzle P6.149-P6.153 6.12 Flowmeters: the venturi meter P6.154-P6.159 6.12 Flowmeters: other designs P6.160-P6.161 6.12 Flowmeters: compressibility correction P6.162-P6.163 Reynolds number regimes P6.1 An engineer claims that the flow of SAE 30W oil, at 20°C, through a 5-cm-diameter smooth pipe at 1 million N/h, is laminar. Do you agree? A million newtons is a lot, so this sounds like an awfully high flow rate. P6.2 The present pumping rate of crude oil through the Alaska Pipeline, with an ID of 48 in, is 550,000 barrels per day (1 barrel = 42 U.S. gallons), (a) Is this a turbulent flow? (b) What would be the maximum rate if the flow were constrained to be laminar? Assume that Alaskan oil fits Pig. A.l of the Appendix at 60°C. P6.3 The Keystone Pipeline in the chapter opener photo has a maximum proposed flow rate of 1.3 million barrels of crude oil per day. Estimate the Reynolds number and whether the flow is laminar. Assume that Keystone crude oil fits Fig. A. 1 of the Appendix at 40°C. P6.4 For flow of SAE 30 oil through a 5-cm-diameter pipe, from Pig. A.l, for what flow rate in m^/h would we expect transi¬ tion to turbulence at (a) 2Q°C and (b) 100°C? P6.5 In flow past a body or wall, early transition to turbulence can be induced by placing a trip wire on the wall across the flow, as in Fig. P6.5. If the trip wire in Fig. P6.5 is placed where the local velocity is U, it will trigger turbulence if Udiv = 850, where d is the wire diameter [3, p. 388]. If the sphere diameter is 20 cm and transition is observed at Re^ = 90,000, what is the diameter of the trip wire in mm? Trip wire d P6.5 Internal and external flow P6.6 For flow of a uniform stream parallel to a sharp flat plate, transition to a turbulent boundary layer on the plate may occur at Re^, = pUx/^ ~ 1 E6, where U is the approach velocity and x is distance along the plate. If U = 2.5 m/s, determine the distance x for the following fluids at 20°C and 1 atm: (a) hydrogen, {b) air, (c) gasoline, {d) water, ie) mercury, and (/) glycerin. P6.7 SAE 10W30 oil at 20°C flows from a tank into a 2-cm- diameter tube 40 cm long. The flow rate is 1 . 1 mVhr. Is the entrance length region a significant part of this tube flow? P6.8 When water at 20°C is in steady turbulent flow through an 8-cm-diameter pipe, the wall shear stress is 72 Pa. What is the axial pressure gradient (dp/dx) if the pipe is (a) horizon¬ tal and (b) vertical with the flow up? Head loss — friction factor P6.9 A light liquid (p ~ 950 kg/m^) flows at an average velocity of 10 m/s through a horizontal smooth tube of diameter 5 cm. The fluid pressure is measured at 1-m intervals along the pipe, as follows: X, m 0 1 2 3 4 5 6 p, kPa 304 273 255 240 226 213 200 Estimate (a) the total head loss, in meters; (b) the wall shear stress in the fully developed section of the pipe; and (c) the overall friction factor. Problems 423 P6.10 Water at 20°C flows through an inclined 8-cm-diameter pipe. At sections A and B the following data are taken: Pa — 186 kPa, Va = 3.2 m/s, za = 24.5 m, and pg = 260 kPa, Vg = 3.2 m/s, zg = 9.1 m. Which way is the flow go¬ ing? What is the head loss in meters? P6.ll Water at 20°C flows upward at 4 m/s in a 6-cm-diameter pipe. The pipe length between points 1 and 2 is 5 m, and point 2 is 3 m higher. A mercury manometer, connected between 1 and 2, has a reading /? = 135 mm, with pi higher. (a) What is the pressure change (pi — P2)? (b) What is the head loss, in meters? (c) Is the manometer reading propor¬ tional to head loss? Explain, (d) What is the friction factor of the flow? In Probs. 6.12 to 6.99, neglect minor losses. Laminar pipe flow — no minor losses P6.12 A 5-mm-diameter capillary tube is used as a viscometer for oils. When the flow rate is 0.071 m^/h, the measured pres¬ sure drop per unit length is 375 kPa/m. Estimate the viscosity of the fluid. Is the flow laminar? Can you also estimate the density of the fluid? P6.13 A soda straw is 20 cm long and 2 mm in diameter. It delivers cold cola, approximated as water at 10°C, at a rate of 3 cmVs. (a) What is the head loss through the straw? What is the axial pressure gradient dpidx if the flow is {b) vertically up or (c) horizontal? Can the human lung deliver this much flow? P6.14 Water at 20°C is to be siphoned through a tube 1 m long and 2 mm in diameter, as in Fig. P6.14. Is there any height H for which the flow might not be laminar? What is the flow rate if // = 50 cm? Neglect the tube curvature. P6.15 Professor Gordon Holloway and his students at the Uni¬ versity of New Brunswick went to a fast-food emporium and tried to drink chocolate shakes (p ~ 1200 kg/m^, p ~ 6 kg/m-s) through fat straws 8 mm in diameter and 30 cm long, (a) Verify that their human lungs, which can develop approximately 3000 Pa of vacuum pressure, would be unable to drink the milkshake through the verti¬ cal straw, (b) A student cut 15 cm from his straw and proceeded to drink happily. What rate of milkshake flow was produced by this strategy? P6.16 Fluid flows steadily, at volume rate Q, through a large pipe and then divides into two small pipes, the larger of which has an inside diameter of 25 mm and carries three times the flow of the smaller pipe. Both small pipes have the same length and pressure drop. If all flows are laminar, estimate the diameter of the smaller pipe. P6.17 A capillary viscometer measures the time required for a spec¬ ified volume V of liquid to flow through a small-bore glass tube, as in Fig. P6. 17. This transit time is then correlated with fluid viscosity. For the system shown, (a) derive an approxi¬ mate formula for the time required, assuming laminar flow with no entrance and exit losses. (b)lf L = 12 cm, 1 = 2 cm, w = 8 cm^, and the fluid is water at 20°C, what capillary diam¬ eter D will result in a transit time ? of 6 seconds? P6.18 SAE 50W oil at 20°C flows from one tank to another through a tube 160 cm long and 5 cm in diameter. Estimate the flow rate in m^/hr if zi = 2 m and Z2 = 0.8 m. P6.19 An oil (SG = 0.9) issues from the pipe in Fig. P6.19 at 2 = 35 ft^/h. What is the kinematic viscosity of the oil in ft^/s? Is the flow laminar? 424 Chapter 6 Viscous Flow in Ducts Dt= \ cm P6.20 The oil tanks in Tinyland are only 160 cm high, and they discharge to the Tinyland oil truck through a smooth tube 4 mm in diameter and 55 cm long. The tube exit is open to the atmosphere and 145 cm below the tank surface. The fluid is medium fuel oil, p = 850 kg/m^ and p, = 0.11 kg/(m ■ s). Estimate the oil flow rate in cmVh. P6.21 In Tinyland, houses are less than a foot high! The rain¬ fall is laminar! The drainpipe in Fig. P6.21 is only 2 mm in diameter, (a) When the gutter is full, what is the rate of draining? (b) The gutter is designed for a sudden rain¬ storm of up to 5 mm per hour. For this condition, what is the maximum roof area that can be drained successfully? (c) What is Re^? P6.23 S AE 1 0 oil at 20°C flows in a vertical pipe of diameter 2.5 cm. It is found that the pressure is constant throughout the fluid. What is the oil flow rate in m^/h? Is the flow up or down? P6.24 Two tanks of water at 20°C are connected by a capillary tube 4 mm in diameter and 3.5 m long. The surface of tank 1 is 30 cm higher than the surface of tank 2. (a) Estimate the flow rate in m^/h. Is the flow laminar? (b) For what tube diameter will Re^ be 500? P6.25 For the conhguration shown in Fig. P6.25, the fluid is ethyl alcohol at 20°C, and the tanks are very wide. Find the flow rate which occurs in m^/h. Is the flow laminar? V 50 cm P6.21 Tinyland governor’s mansion P6.22 A steady push on the piston in Fig. P6.22 causes a flow rate Q = 0.15 cmVs through the needle. The fluid has p = 900 kg/m^ and p = 0.002 kg/(m • s). What force F is required to maintain the flow? P6.25 V - 2 mm 40 t 80 cm 1 m P6.26 Two oil tanks are connected by two 9-m-long pipes, as in Fig. P6.26. Pipe 1 is 5 cm in diameter and is 6 m higher than pipe 2. It is found that the flow rate in pipe 2 is twice as large as the flow in pipe 1 . (a) What is the diameter of pipe 2? (b) Are both pipe flows laminar? (c) What is the flow rate in pipe 2 (mVs)? Neglect minor losses. Problems 425 P6.27 Let us attack Prob. P6.25 in symbolic fashion, using Fig. P6.27. All parameters are constant except the upper tank depth Z(t). Find an expression for the flow rate Q(t) as a function of Z(t). Set up a differential equation, and solve for the time ?o to drain the upper tank completely. Assume quasi-steady laminar flow. P6.29 SAE 30W oil at 20°C flows through a straight pipe 25 m long, with diameter 4 cm. The average velocity is 2 m/s. (a) Is the flow laminar? Calculate (b) the pressure drop and (c) the power required, (d) If the pipe diameter is doubled, for the same average velocity, by what percent does the required power increase? P6.30 SAE 10 oil at 20°C flows through the 4-cm-diameter verti¬ cal pipe of Fig. P6.30. For the mercury manometer reading h = 42 cm shown, (a) calculate the volume flow rate in m^/h and (b) state the direction of flow. -D- V p.fl V Z(t) H 3 m P6.30 P6.27 _ _ L P6.28 For straightening and smoothing an airflow in a 50-cm- diameter duct, the duct is packed with a “honeycomb” of thin straws of length 30 cm and diameter 4 mm, as in Fig. P6.28. The inlet flow is air at 1 10 kPa and 20°C, mov¬ ing at an average velocity of 6 m/s. Estimate the pressure drop across the honeycomb. P6.31 A laminar flow element (LFE) (Meriam Instrument Co.) measures low gas-flow rates with a bundle of capillary tubes or ducts packed inside a large outer tube. Consider oxygen at 20°C and 1 atm flowing at 84 ftVmin in a 4-in- diameter pipe, (a) Is the flow turbulent when approaching the element? (b) If there are 1000 capillary tubes, L = 4 in, select a tube diameter to keep Re^ below 1500 and also to keep the tube pressure drop no greater than 0.5 Ibf/in^. (c) Do the tubes selected in part (b) fit nicely within the approach pipe? 426 Chapter 6 Viscous Flow in Ducts P6.32 SAE 30 oil at 20°C flows in the 3-cm-diameter pipe in Fig. P6.32, which slopes at 37°. For the pressure measurements shown, determine (a) whether the flow is up or down and (b) the flow rate in m^/h. = 180 kPa Estimate (a) the wall shear stress and (b) the velocity u at y = 0.22 in. P6.37 Two infinite plates a distance h apart are parallel to the xz plane with the upper plate moving at speed V, as in Eig. P6.37. There is a fluid of viscosity fi and constant pres¬ sure between the plates. Neglecting gravity and assuming incompressible turbulent flow u(y} between the plates, use the logarithmic law and appropriate boundary conditions to derive a formula for dimensionless wall shear stress versus dimensionless plate velocity. Sketch a typical shape of the profile u(y). P6.33 Water at 20°C is pumped from a reservoir through a verti¬ cal tube 10 ft long and l/16th in in diameter. The pump provides a pressure rise of 1 1 Ibf/in^ to the flow. Neglect entrance losses, (a) Calculate the exit velocity, (b) Approx¬ imately how high will the exit water jet rise? (c) Verify that the flow is laminar. Turbulence modeling P6.34 Derive the time-averaged v-momentum equation (6.21) by direct substitution of Eqs. (6.19) into the momentum equation (6.14). It is convenient to write the convective acceleration as du d ^ d d ^ — (u) + — (uv) + — (uw) dt dx dy dz which is valid because of the continuity relation, Eq. (6.14). P6.35 In the overlap layer of Fig. 6.9a, turbulent shear is large. If we neglect viscosity, we can replace Eq. (6.24) with the approximate velocity-gradient function du — = fcn(y, T^,p) dy Show by dimensional analysis that this leads to the loga¬ rithmic overlap relation (6.28). P6.36 The following turbulent flow velocity data u{y), for air at 75°F and 1 atm near a smooth flat wall were taken in the University of Rhode Island wind tunnel: V, in 0.025 0.035 0.047 0.055 0.065 u, ft/s 51.2 54.2 56.8 57.6 59.1 P6.38 Suppose in Fig. P6.37 that A = 3 cm, the fluid in water at 20°C, and the flow is turbulent, so that the logarithmic law is valid. If the shear stress in the fluid is 15 Pa, what is V in m/s? P6.39 By analogy with laminar shear, t = fi du/dy, T. V. Boussinesq in 1877 postulated that turbulent shear could also be related to the mean velocity gradient Ttmi, = sduldy, where e is called the eddy viscosity and is much larger than fi. If the logarithmic overlap law, Eq. (6.28), is valid with Tturb ~ show that e ~ Kpuy. P6.40 Theodore von Karman in 1930 theorized that turbulent shear could be represented by Tmrb = e duldy, where e = pK^y^\duldy\ is called the mixing-length eddy viscosity and K ~ 0.41 is Karman’ s dimensionless mixing-length constant [2, 3]. Assuming that « T„, near the wall, show that this expression can be integrated to yield the logarithmic overlap law, Eq. (6.28). Turbulent pipe flow — no minor losses P6.41 Two reservoirs, which differ in surface elevation by 40 m, are connected by 350 m of new pipe of diameter 8 cm. If the desired flow rate is at least 130 N/s of water at 20°C, can the pipe material be made of (a) galvanized iron, {b) commercial steel, or (c) cast iron? Neglect minor losses. P6.42 Eluid flows steadily, at volume rate Q, through a large hor¬ izontal pipe and then divides into two small pipes, the larger of which has an inside diameter of 25 mm and car¬ ries three times the flow of the smaller pipe. Both small pipes have the same length and pressure drop. If all flows are turbulent, at Re^ near lO", estimate the diameter of the smaller pipe. Problems 427 P6.43 A reservoir supplies water through 100 m of 30-cm-diameter cast iron pipe to a turbine that extracts 80 hp from the flow. The water then exhausts to the atmosphere. Neglect minor losses, (a) Assuming that/^ 0.019, find the flow rate (which results in a cubic polynomial). Explain why there are two legitimate solutions, (b) For extra credit, solve for the flow rates using the actual friction factors. P6.44 Mercury at 20°C flows through 4 m of 7-mm-diameter glass tuhing at an average velocity of 5 m/s. Estimate the head loss in m and the pressure drop in kPa. P6.45 Oil, SG = 0.88 and = 4 E-5 m^/s, flows at 400 gal/min through a 6-in asphalted cast iron pipe. The pipe is 0.5 mi long and slopes upward at 8° in the flow direction. Com¬ pute the head loss in ft and the pressure change. P6.46 The Keystone Pipeline in the chapter opener photo has a diameter of 36 inches and a design flow rate of 590,000 bar¬ rels per day of cmde oil at 40°C. If the pipe material is new steel, estimate the pump horsepower required per mile of pipe. P6.47 The gutter and smooth drainpipe in Fig. P6.47 remove rain¬ water from the roof of a building. The smooth drainpipe is 7 cm in diameter, (a) When the gutter is full, estimate the rate of draining, (b) The gutter is designed for a sudden rainstorm of up to 5 inches per hour. For this condition, what is the maximum roof area that can be drained successfully? P6.48 Follow up Prob. P6.46 with the following question. If the total Keystone pipeline length, from Alberta to Texas, is 2147 miles, how much flow, in barrels per minute, will result if the total available pumping power is 8,000 hp? P6.49 The tank-pipe system of Fig. P6.49 is to deliver at least 11 mVh of water at 20°C to the reservoir. What is the maximum roughness height e allowable for the pipe? , L = 5 m, = 3 cm V P6.49 P6.50 Ethanol at 20°C flows at 125 U.S. gal/min through a horizontal cast iron pipe with L = 12 m and d = 5 cm. Neglecting en¬ trance effects, estimate (a) the pressure gradient dpidx, {b) the wall shear stress t„, and (c) the percentage reduction in friction factor if the pipe walls are polished to a smooth surface. P6.51 The viscous sublayer (Fig. 6.9) is normally less than 1 percent of the pipe diameter and therefore very difficult to probe with a finite-sized instmment. In an effort to generate a thick sublayer for probing, Pennsylvania State University in 1964 built a pipe with a flow of glycerin. Assume a smooth 12-in-diameter pipe with V = 60 ft/s and glycerin at 20°C. Compute the sublayer thickness in inches and the pumping horsepower required at 75 percent efficiency if L = 40 ft. P6.52 The pipe flow in Fig. P6.52 is driven by pressurized air in the tank. What gage pressure pi is needed to provide a 20°C water flow rate Q = 60 m^/h? 30 m Smooth pipe: Q Open jet 428 Chapter 6 Viscous Flow in Ducts P6.53 Water at 20°C flows by gravity through a smooth pipe from one reservoir to a lower one. The elevation difference is 60 m. The pipe is 360 m long, with a diameter of 12 cm. Calculate the expected flow rate in m^/h. Neglect minor losses. P6.54 A swimming pool Why Yhy h deep is to be emptied by gravity through the long pipe shown in Fig. P6.54. Assuming an aver¬ age pipe friction factor and neglecting minor losses, derive a formula for the time to empty the tank from an initial level h^. P6.60 In the spirit of Haaland’s explicit pipe friction factor approximation, Eq. (6.49), Jeppson proposed the following explicit formula: 1 Vf -2.01ogio (eld_ V3.7 + 5.74 \ ReSV (a) Is this identical to Haaland’s formula with just a simple rearrangement? Explain, (b) Compare leppson’s formula to Haaland’ s for a few representative values of (turbulent) Re^ and eld and their errors compared to the Colebrook formula (6.48). Discuss briefly. P6.61 What level h must be maintained in Eig. P6.61 to deliver a flow rate of 0.015 ft^/s through the j-in commercial steel pipe? V _ ^ P6.55 The reservoirs in Pig. P6.55 contain water at 20°C. If the pipe is smooth with L = 4500 m and d = A cm, what will the flow rate in m^/h be for Az = 100 m? P6.56 The Alaska Pipeline in the chapter opener photo has a de¬ sign flow rate of 4.4 E7 gallons per day of crude oil at 60°C (see Fig. A. 1). (a) Assuming a galvanized-iron wall, esti¬ mate the total pressure drop required for the 800-mile trip. (b) If there are nine equally spaced pumps, estimate the horsepower each pump must deliver. P6.57 Apply the analysis of Prob. P6.54 to the following data. Let IT = 5 m, T = 8 m, /to = 2 m, L = 15 m, D = 5 cm, and £■ = 0. (a) By letting h= 1.5 m and 0.5 m as representative depths, estimate the average friction factor. Then (b) esti¬ mate the time to drain the pool. P6.58 For the system in Prob. 6.53, a pump is used at night to drive water back to the upper reservoir. If the pump deliv¬ ers 15,000 W to the water, estimate the flow rate. P6.59 The following data were obtained for flow of 20°C water at 20 m^/h through a badly corroded 5-cm-diameter pipe that slopes downward at an angle of 8°: pi — 420 kPa, zi = 12 m, P2 = 250 kPa, Z2 = 3 m. Estimate (a) the roughness ratio of the pipe and (b) the percentage change in head loss if the pipe were smooth and the flow rate the same. P6.61 Water at 20“C L = 80 ft D=-m 2 P6.62 Water at 20°C is to be pumped through 2000 ft of pipe from reservoir 1 to 2 at a rate of 3 ft^/s, as shown in Pig. P6.62. If the pipe is cast iron of diameter 6 in and the pump is 75 percent efficient, what horsepower pump is needed? Flow rate and sizing problems P6.63 A tank contains 1 m^ of water at 20°C and has a drawn¬ capillary outlet tube at the bottom, as in Fig. P6.63. Find the outlet volume flux Q in m^/h at this instant. P6.64 For the system in Fig. P6.63, solve for the flow rate in m^/h if the fluid is SAE 10 oil at 20°C. Is the flow laminar or turbulent? Problems 429 V 1 m 1 m 3 L = 80 cm D= 4 cm 1 P6.63 Q P6.73 P6.74 P6.75 P6.65 P6.66 P6.67 P6.68 P6.69 P6.70 P6.71 P6.72 In Prob. P6.63 the initial flow is turbulent. As the water P6.76 drains out of the tank, will the flow revert to laminar mo¬ tion as the tank becomes nearly empty? If so, at what tank depth? Estimate the time, in h, to drain the tank completely. Ethyl alcohol at 20°C flows through a 10-cm horizontal drawn tube 100 m long. The fully developed wall shear stress is 14 Pa. Estimate {a) the pressure drop, {b) the vol¬ ume flow rate, and (c) the velocity u&Xr = 1 cm. A straight 10-cm commercial-steel pipe is 1 km long and is laid on a constant slope of 5°. Water at 20°C flows down¬ ward, due to gravity only. Estimate the flow rate in mVh. What happens if the pipe length is 2 km? The Moody chart cannot find V directly, since V appears in both ordinate and abscissa, (a) Arrange the variables {ht, d, g, L, v) into a single dimensionless group, with hfd in the numerator, denoted as ^ , which equals (fKe//2). (b) Rear¬ range the Colebrook formula (6.48) to solve for Re^ in terms of (c) Eor extra credit, solve Example 6.9 with this new formula. P6.77 Eor Prob. P6.62 suppose the only pump available can deliver 80 hp to the fluid. What is the proper pipe size in inches to maintain the 3 ft^/s flow rate? Ethylene glycol at 20°C flows through 80 m of cast iron pipe of diameter 6 cm. The measured pressure drop is 250 kPa. Neglect minor losses. Using a noniterative formu¬ lation, estimate the flow rate in m^/h. It is desired to solve Prob. 6.62 for the most economical pump and cast iron pipe system. If the pump costs $125 per P6.78 horsepower delivered to the fluid and the pipe costs $7000 per inch of diameter, what are the minimum cost and the pipe and pump size to maintain the 3 ft^/s flow rate? Make P6.79 some simplifying assumptions. Modify Prob. P6.57 by letting the diameter be unknown. Eind the proper pipe diameter for which the pool will drain in about two hours flat. Eor 20°C water flow in a smooth, horizontal 10-cm pipe, with Ap/L = 1000 Pa/m, the writer computed a flow rate of 0.030 m^/s. (a) Verify, or disprove, the writer’s answer. (b) If verified, use the power-law friction factor relation, Eq. (6.41), to estimate the pipe diameter that will triple this flow rate, (c) For extra credit, use the more exact friction factor relation, Eq. (6.38), to solve part (b). Two reservoirs, which differ in surface elevation by 40 m, are connected by a new commercial steel pipe of diameter 8 cm. If the desired flow rate is 200 N/s of water at 20°C, what is the proper length of the pipe? You wish to water your garden with 100 ft of \|-in-diameter hose whose roughness is 0.011 in. What will be the deliv¬ ery, in tf/s, if the gage pressure at the faucet is 60 Ibf/in^? If there is no nozzle (just an open hose exit), what is the maximum horizontal distance the exit jet will carry? The small turbine in Fig. P6.76 extracts 400 W of power from the water flow. Both pipes are wrought iron. Compute the flow rate Q in m^/h. Why are there two solutions? Which is better? V _ Turbine o 10 m D = 6 cm 30 m D = A cm P6.76 Modify Prob. P6.76 into an economic analysis, as follows: Let the 40 m of wrought iron pipe have a uniform diameter d. Let the steady water flow available be 2 = 30 m^/h. The cost of the turbine is $4 per watt developed, and the cost of the piping is $75 per centimeter of diameter. The power generated may be sold for $0.08 per kilowatt-hour. Find the proper pipe diameter for minimum payback time — that is, the minimum time for which the power sales will equal the initial cost of the system. In Fig. P6.78 the connecting pipe is commercial steel 6 cm in diameter. Estimate the flow rate, in m^/h, if the fluid is water at 20°C. Which way is the flow? A garden hose is to be used as the return line in a waterfall display at a mall. In order to select the proper pump, you need to know the roughness height inside the garden hose. Unfortunately, roughness information is not supplied by the hose manufacturer. So you devise a simple experiment 430 Chapter 6 Viscous Flow in Ducts to measure the roughness. The hose is attached to the drain of an above-ground swimming pool, the surface of which is 3.0 m above the hose outlet. You estimate the minor loss coefficient of the entrance region as 0.5, and the drain valve has a minor loss equivalent length of 200 diameters when fully open. Using a bucket and stopwatch, you open the valve and measure the flow rate to be 2.0 X 10”' mVs for a hose that is 10.0 m long and has an inside diameter of 1.50 cm. Estimate the roughness height in mm inside the hose. reservoir. The pipe diameter is increased from 12 cm to provide more flow. If the resultant flow rate is 90 m^/h, esti¬ mate the new pipe size. Noncircular ducts P6.86 SAE 10 oil at 20°C flows at an average velocity of 2 m/s between two smooth parallel horizontal plates 3 cm apart. Estimate (a) the centerline velocity, {b) the head loss per meter, and (c) the pressure drop per meter. P6.87 A commercial steel annulus 40 ft long, with a = 1 in and b = \ in, connects two reservoirs that differ in surface height by 20 ft. Compute the flow rate in ft^/s through the annulus if the fluid is water at 20°C. P6.88 An oil cooler consists of multiple parallel-plate pas¬ sages, as shown in Fig. P6.88. The available pressure drop is 6 kPa, and the fluid is SAE lOW oil at 20°C. If the desired total flow rate is 900 m^/h, estimate the appropriate number of passages. The plate walls are hydraulically smooth. P6.80 The head-versus-flow-rate characteristics of a centrifugal pump are shown in Fig. P6.80. If this pump drives water at 20°C through 120 m of 30-cm-diameter cast iron pipe, what will be the resulting flow rate, in mVs? P6.81 The pump in Fig. P6.80 is used to deliver gasoline at 20°C through 350 m of 30-cm-diameter galvanized iron pipe. Estimate the resulting flow rate, in mVs. (Note that the pump head is now in meters of gasoline.) P6.82 Eluid at 20°C flows through a horizontal galvanized-iron pipe 20 m long and 8 cm in diameter. The wall shear stress is 90 Pa. Calculate the flow rate in m^/h if the fluid is (a) glycerin and (b) water. P6.83 Eor the system of Fig. P6.55, let Az = 80 m and L = 185 m of cast iron pipe. What is the pipe diameter for which the flow rate will be 7 m^/h? P6.84 It is desired to deliver 60 m^/h of water at 20°C through a horizontal asphalted cast iron pipe. Estimate the pipe diam¬ eter that will cause the pressure drop to be exactly 40 kPa per 100 m of pipe length. P6.85 Eor the system in Prob. P6.53, a pump, which delivers 15,000 W to the water, is used at night to refill the upper P6.89 An annulus of narrow clearance causes a very large pressure drop and is useful as an accurate measure¬ ment of viscosity. If a smooth annulus 1 m long with a = 50 mm and b = 49 mm carries an oil flow at 0.001 mVs, what is the oil viscosity if the pressure drop is 250 kPa? P6.90 A rectangular sheet-metal duct is 200 ft long and has a fixed height H = 6 in. The width B, however, may vary from 6 to 36 in. A blower provides a pressure drop of 80 Pa of air at 20°C and 1 atm. What is the optimum width B that will provide the most airflow in ftVs? P6.91 Heat exchangers often consist of many triangular pas¬ sages. Typical is Fig. P6.91, with L = 60 cm and an isos¬ celes-triangle cross section of side length a = 2 cm and included angle /3 = 80°. If the average velocity is V = 2 m/s and the fluid is SAE 10 oil at 20°C, estimate the pressure drop. Problems 431 P6.92 A large room uses a fan to draw in atmospheric air at 20°C through a 30-cm hy 30-cm commercial-steel duct 12 m long, as in Fig. P6.92. Estimate (a) the airflow rate in m^/h if the room pressure is 10 Pa vacuum and (b) the room pressure if the flow rate is 1200 m^/h. Neglect minor losses. P6.93 In Moody’s Example 6.6, the 6-inch diameter, 200-ft-long asphalted cast iron pipe has a pressure drop of about 280 Ihf/tf when the average water velocity is 6 ft/s. Com¬ pare this to an annular cast iron pipe with an inner diameter of 6 in and the same annular average velocity of 6 ft/s. (a) What outer diameter would cause the flow to have the same pressure drop of 280 Ihf/ft^? (b) How do the cross- section areas compare, and why? Use the hydraulic diam¬ eter approximation. P6.94 Air at 20°C flows through a smooth duct of diameter 20 cm at an average velocity of 5 m/s. It then flows into a smooth square duct of side length a. Find the square duct size a for which the pressure drop per meter will he exactly the same as the circular duct. P6.95 Although analytical solutions are available for laminar flow in many duct shapes , what do we do about ducts of arbitrary shape? Bahrami et al. propose that a better approach to the pipe result, /Re = 64, is achieved by replacing the hydraulic diameter with VS, where A is the area of the cross section. Test this idea for the isosceles triangles of Table 6.4. If time is short, at least try 10°, 50°, and 80°. What do you con¬ clude about this idea? P6.96 A fuel cell consists of air (or oxygen) and hydrogen micro ducts, separated by a membrane that promotes pro¬ ton exchange for an electric current, as in Fig. P6.96. Sup¬ pose that the air side, at 20°C and approximately 1 atm, has five 1 mm by 1 mm ducts, each 1 m long. The total flow rate is 1.5 E-4 kg/s. (a) Determine if the flow is laminar or turbulent, {b) Estimate the pressure drop. (Problem cour¬ tesy of Dr. Pezhman Shirvanian.) Hydrogen Air flow flow P6.97 A heat exchanger consists of multiple parallel-plate pas¬ sages, as shown in Fig. P6.97. The available pressure drop is 2 kPa, and the fluid is water at 20°C. If the desired total flow rate is 900 m^/h, estimate the appropriate number of passages. The plate walls are hydraulically smooth. P6.98 A rectangular heat exchanger is to be divided into smaller sections using sheets of commercial steel 0.4 mm thick, as sketched in Fig. P6.98. The flow rate is 20 kg/s of water at 20°C. Basic dimensions are L = 1 m, W = 20 cm, and H = 10 cm. What is the proper number of square sections if the overall pressure drop is to be no more than 1600 Pa? 432 Chapter 6 Viscous Flow in Ducts Minor or local losses P6.99 In Sec. 6.11 it was mentioned that Roman aqueduct cus¬ tomers obtained extra water hy attaching a diffuser to their pipe exits. Fig. P6.99 shows a simulation: a smooth inlet pipe, with or without a 15° conical diffuser expanding to a 5-cm-diameter exit. The pipe entrance is sharp-edged. Cal¬ culate the flow rate (a) without and {b) with the diffuser. 1 5° diffuser P6.102 A 70 percent efficient pump delivers water at 20°C from one reservoir to another 20 ft higher, as in Fig. P6. 102. The piping system consists of 60 ft of galvanized iron 2-in pipe, a reentrant entrance, two screwed 90° long-radius elhows, a screwed-open gate valve, and a sharp exit. What is the input power required in horsepower with and without a 6° well-designed conical expansion added to the exit? The flow rate is 0.4 ftVs. P6.102 P6.103 The reservoirs in Fig. P6.103 are connected hy cast iron pipes joined abruptly, with sharp-edged entrance and exit. Including minor losses, estimate the flow of water at 20°C if the surface of reservoir 1 is 45 ft higher than that of reservoir 2. P6.99 P6.100 Modify Prob. P6.55 as follows: Assume a pump can deliver 3 kW to pump the water back up to reservoir 1 from reservoir 2. Accounting for an open flanged globe valve and sharp-edged entrance and exit, estimate the predicted flow rate, in m^/h. P6.101 In Fig. P6.101 a thick filter is being tested for losses. The flow rate in the pipe is 7 m^/min, and the upstream pressure is 120 kPa. The fluid is air at 20°C. Using the water manom¬ eter reading, estimate the loss coefficient K of the filter. V D = 2 in L = 20 ft \ 1 1 in 2 in - - 1 © \ D = 1 in L = 20 ft P6.103 P6.104 Consider a 20°C flow at 2 m/s through a smooth 3-mm diameter microtube which consists of a straight run of 10 cm, a long radius bend, and another straight run of 10 cm. Compute the total pressure drop if the fluid is (a) water; and (b) ethylene glycol. P6.105 The system in Fig. P6.105 consists of 1200 m of 5 cm cast iron pipe, two 45° and four 90° flanged long-radius elbows, a fully open flanged globe valve, and a sharp exit into a reservoir. If the elevation at point 1 is 400 m, what gage pressure is required at point 1 to deliver 0.005 mVs of water at 20°C into the reservoir? Problems 433 Elevation 500 m P6.106 The water pipe in Fig. P6.106 slopes upward at 30°. The pipe has a 1-in diameter and is smooth. The flanged glohe valve is fully open. If the mercury manometer shows a 7-in deflection, what is the flow rate in ft^/s? P6.108 P6.109 In Fig. P6.109 there are 125 ft of 2-in pipe, 75 ft of 6-in pipe, and 150 ft of 3-in pipe, all cast iron. There are three 90° elhows and an open globe valve, all flanged. If the exit elevation is zero, what horsepower is extracted hy the tur¬ bine when the flow rate is 0. 16 ft^/s of water at 20°C? P6.107 A tank of water 4 m in diameter and 7 m deep is to be drained by a 5-cm-diameter exit pipe at the bottom, as in Fig. P6. 107. In design (1), the pipe extends out for 1 m and into the tank for 10 cm. In design (2), the interior pipe is removed and the entrance beveled. Fig. 6.21, so that A" ~ 0.1 in the entrance. (a) An engineer claims that design (2) will drain 25 percent faster than design (1). Is this claim tme? {b) Estimate the time to drain of design (2), assuming/ — 0.020. (1) P6.107 P6.108 The water pump in Fig. P6.108 maintains a pressure of 6.5 psig at point 1. There is a filter, a half-open disk valve, and two regular screwed elbows. There are 80 ft of 4-in diam¬ eter commercial steel pipe, (a) If the flow rate is 0.4 ftVs, what is the loss coefficient of the filter? (b) If the disk valve is wide open and ^ffluer = 7, what is the resulting flow rate? P6.109 P6.110 In Fig. P6. 1 10 the pipe entrance is sharp-edged. If the flow rate is 0.004 mVs, what power, in W, is extracted hy the turbine? L = 125 m, D = 5 cm P6.110 Series and parallel pipe systems P6.111 For the parallel-pipe system of Fig. P6.111, each pipe is cast iron, and the pressure drop pi — P2 = ^ VoUir?. Com¬ pute the total flow rate between 1 and 2 if the fluid is SAE 10 oil at 20°C. 434 Chapter 6 Viscous Flow in Ducts D = 3 in L = 250 ft P6.111 P6.112 If the two pipes in Fig. P6.111 are instead laid in series with the same total pressure drop of 3 Ihf/in^, what will the flow rate he? The fluid is SAE 10 oil at 20°C. P6.113 The parallel galvanized iron pipe system of Fig. P6.113 delivers water at 20°C with a total flow rate of 0.036 mVs. If the pump is wide open and not running, with a loss coef¬ ficient K = 1.5, determine (a) the flow rate in each pipe and (b) the overall pressure drop. L[ = 60 m, D[ = 5 cm P6.114 A hlower supplies standard air to a plenum that feeds two horizontal square sheet-metal ducts with sharp-edged en¬ trances. One duct is 100 ft long, with a cross-section 6 in hy 6 in. The second duct is 200 ft long. Each duct exhausts to the atmosphere. When the plenum pressure is 5.0 Ihf/ft^ (gage) the volume flow in the longer duct is three times the flow in the shorter duct. Estimate both volume flows and the cross-section size of the longer duct. P6.115 In Fig. P6.1 15 all pipes are 8-cm-diameter cast iron. Deter- '(X mine the flow rate from reservoir 1 if valve C is (a) closed and (b) open, K = 0.5. P6.116 For the series-parallel system of Fig. P6.116, all pipes are 8-cm-diameter asphalted cast iron. If the total pressure drop Pi — P2 = 750 kPa, find the resulting flow rate Q m^/h for water at 20°C. Neglect minor losses. P6.117 A blower delivers air at 3000 m^/h to the duct circuit in Fig. P6.117. Each duct is commercial steel and of square cross section, with side lengths Oi = 03 = 20 cm and 02 = 04 = 12 cm. Assuming sea-level air conditions, esti¬ mate the power required if the blower has an efficiency of 75 percent. Neglect minor losses. P6.115 L = 250 m Blower 40 m P6.117 P6.118 For the piping system of Fig. P6. 1 1 8, all pipes are concrete with a roughness of 0.04 in. Neglecting minor losses, com¬ pute the overall pressure drop pi — p2 in Ibf/in^ if Q =20 ftVs. The fluid is water at 20°C. D = 8 in Problems 435 P6.119 For the piping system of Prob. P6. 1 1 1 , let the fluid be gaso¬ line at 20°C, with both pipes cast iron. If the flow rate in the 2- in pipe {b) is 1.2 ftVmin, estimate the flow rate in the 3- in pipe (a), in ft^/min. P6.120 Three cast iron pipes are laid in parallel with these dimensions: Pipe Length, m Diameter, cm 1 800 12 2 600 8 3 900 10 The total flow rate is 200 mVh of water at 20°C. Determine (a) the flow rate in each pipe and (b) the pressure drop across the system. Three-reservoir and pipe network systems P6.121 Consider the three-reservoir system of Fig. P6.121 with the following data: Li = 95m Lz = 125 m L3 = 160 m Zi = 25 m Z2 = 1 15 m 23 = 85 m All pipes are 28-cm-diameter unfinished concrete (e = 1 mm). Compute the steady flow rate in all pipes for water at 20°C. P6.122 Modify Prob. P6.121 as follows: Reduce the diameter to 15 cm (with £ = 1 mm), and compute the flow rates for water at 20'’C. These flow rates distribute in nearly the same manner as in Prob. P6.121 but are about 5.2 times lower. Can you explain this difference? P6. 123 Modify Prob. P6.121 as follows: All data are the same except that zs is unknown. Find the value of zs for which the flow rate in pipe 3 is 0.2 mVs toward the junction. (This problem requires iteration and is best suited to a computer.) P6.124 The three-reservoir system in Fig. P6. 124 delivers water at 20°C. The system data are as follows: Di = 8 in D2 = 6 in D3 = 9 in L, = 1800 ft L2 = 1200 ft L3 = 1600 ft All pipes are galvanized iron. Compute the flow rate in all pipes. Z2 = 100 ft P6.125 Suppose that the three cast iron pipes in Prob. P6. 120 are instead connected to meet smoothly at a point B, as shown in Fig. P6. 125. The inlet pressures in each pipe are Pi = 200 kPa P2 = 160 kPa p^ = 100 kPa. The fluid is water at 20°C. Neglect minor losses. Estimate the flow rate in each pipe and whether it is toward or away from point B. P6.125 P6.126 Modify Prob. P6.124 as follows: Let all data be the same except that pipe 1 is fitted with a butterfly valve (Fig. 6. 1 9b). Estimate the proper valve opening angle (in degrees) for the flow rate through pipe 1 to be reduced to 1.5 ft^/s to¬ ward reservoir 1. (This problem requires iteration and is best suited to a computer.) P6.127 In the five-pipe horizontal network of Eig. P6.127, as- sume that all pipes have a friction factor/ = 0.025. For the given inlet and exit flow rate of 2 ft^/s of water at 436 Chapter 6 Viscous Flow in Ducts 20°C, determine the flow rate and direction in all pipes. If Pa — 120 Ibf/in^ gage, determine the pressures at points B, C, and D. P6.127 P6.128 Modify Proh. P6.127 as follows: Let the inlet flow rate at A and the exit flow at D be unknown. Let Pa~ Pb— 100 Ibf/in^. Compute the flow rate in all five pipes. P6.129 In Fig. P6.129 all four horizontal cast iron pipes are 45 m long and 8 cm in diameter and meet at junction a, deliver¬ ing water at 20°C. The pressures are known at four points as shown: Pi = 950 kPa p2 = 350 kPa P3 = 675 kPa p4 = 100 kPa Neglecting minor losses, determine the flow rate in each pipe. Pi P6.130 In Fig. P6.130 lengths AB and BD are 2000 and 1500 ft, respectively. The friction factor is 0.022 everywhere, and Pa = 90 Ibf/in^ gage. All pipes have a diameter of 6 in. For water at 20°C, determine the flow rate in all pipes and the pressures at points fi, C, and D. 0.5 ft^/s 0.5 fP/s Diffuser performance P6.131 A water tunnel test section has a 1-m diameter and flow properties V = 20 m/s, p = 100 kPa, and T = 20°C. The boundary layer blockage at the end of the section is 9 percent. If a conical diffuser is to be added at the end of the section to achieve maximum pressure recovery, what should its angle, length, exit diameter, and exit pressure be? P6.132 For Proh. P6.131 suppose we are limited by space to a total diffuser length of 10 m. What should the diffuser angle, exit diameter, and exit pressure be for maximum recovery? P6.133 A wind tunnel test section is 3 ft square with flow proper¬ ties V = 150 ft/s, p = 15 Ibf/in^ absolute, and T = 68°F. Boundary layer blockage at the end of the test section is 8 percent. Find the angle, length, exit height, and exit pres¬ sure of a flat-walled diffuser added onto the section to achieve maximum pressure recovery. P6.134 For Proh. P6.133 suppose we are limited by space to a total diffuser length of 30 ft. What should the diffuser angle, exit height, and exit pressure be for maximum recovery? The pitot-static tube P6.135 An airplane uses a pitot-static tube as a velocimeter. The measurements, with their uncertainties, are a static tem¬ perature of ( — 11 ± 3)°C, a static pressure of 60 ± 2 kPa, and a pressure difference {p„ — p^ = 3200 ± 60 Pa. {a) Estimate the airplane’s velocity and its uncertainty. ib) Is a compressibility correction needed? P6.136 For the pitot-static pressure arrangement of Fig. P6.136, the manometer fluid is (colored) water at 20°C. Estimate (a) the centerline velocity, (b) the pipe volume flow, and (c) the (smooth) wall shear stress. P6.137 For the 20°C water flow of Fig. P6.137, use the pitot-static arrangement to estimate (a) the centerline velocity and (b) the volume flow in the 5-in-diameter smooth pipe. (c) What error in flow rate is caused by neglecting the 1-ft elevation difference? Problems 437 P6.138 An engineer who took college fluid mechanics on a pass-fail basis has placed the static pressure hole far upstream of the stagnation prohe, as in Fig. P6.138, thus contaminating the pitot measurement ridiculously with pipe friction losses. If the pipe flow is air at 20°C and 1 atm and the manometer fluid is Meriam red oil (SG = 0.827), estimate the air centerline velocity for the given manometer reading of 16 cm. Assume a smooth-walled tube. afford a pitot-static probe, but instead inserts a total head prohe and a static pressure probe, as shown in Fig. P6.139, a distance hi apart from each other. Both prohes are in the main free stream of the water tunnel, unaffected hy the thin boundary layers on the sidewalls. The two probes are connected as shown to a U-tube manometer. The densities and vertical distances are shown in Fig. P6. 139. {a) Write an expression for velocity V in terms of the parameters in the problem, ib) Is it critical that distance hi be measured accurately? (c) How does the expression for velocity V dif¬ fer from that which would be obtained if a pitot-static probe had been available and used with the same U-tube manometer? P6.139 -lOm P6.138 P6.139 Professor Walter Tunnel needs to measure the flow velocity in a water tunnel. Due to budgetary restrictions, he cannot Flowmeters: the orifice plate P6.140 Gasoline at 20°C flows at 3 m^/h in a 6-cm-diameter pipe. A 4-cm-diameter thin-plate orifice with corner taps is installed. Estimate the measured pressure drop, in Pa. P6.141 Gasoline at 20°C flows at 105 mVh in a 10-cm-diameter pipe. We wish to meter the flow with a thin-plate ori¬ fice and a differential pressure transducer that reads best at about 55 kPa. What is the proper /3 ratio for the orifice? P6.142 The shower head in Fig. P6.142 delivers water at 50°C. An orifice-type flow reducer is to be installed. The upstream pressure is constant at 400 kPa. What flow rate, in gal/min, results without the reducer? What reducer orifice diameter would decrease the flow by 40 percent? 438 Chapter 6 Viscous Flow in Ducts D = 1.5 cm P6.143 A 10-cm-diameter smooth pipe contains an orifice plate with D: \D taps and fi = 0.5. The measured orifice pressure drop is 75 kPa for water flow at 20°C. Estimate the flow rate, in m^/h. What is the nonrecoverahle head loss? P6.144 Water at 20°C flows through the orifice in Fig. P6.154, which is monitored by a mercury manometer. If rf = 3 cm, (a) what is h when the flow rate is 20 m^/h and (b) what is Q in mVh when h = 5S cm? I Water ^ — 5 cm - ►- d P6.145 The 1-m-diameter tank in Fig. P6.145 is initially filled with gasoline at 20°C. There is a 2-cm-diameter orifice in the bottom. If the orifice is suddenly opened, estimate the time for the fluid level h(t) to drop from 2.0 to 1.6 m. P6.146 A pipe connecting two reservoirs, as in Fig. P6.146, con- tains a thin-plate orifice. For water flow at 20°C, estimate (a) the volume flow through the pipe and (b) the pressure drop across the orifice plate. P6.146 r-r~ lOOm D = 5 cm 20 m 3-cm orifice P6.147 Air flows through a 6-cm-diameter smooth pipe that has a 2-m-long perforated section containing 500 holes (diame¬ ter 1 mm), as in Fig. P6.147. Pressure outside the pipe is sea-level standard air. If pi = 105 kPa and = 1 10 m^/h, estimate p2 and Q2, assuming that the holes are approxi¬ mated by thin-plate orifices. {Hint: A momentum control volume may be very useful.) 500 holes (diameter 1 mm) — © o°OoogOo°oOoO°o 0 oOq 00 0 ooo 0 0^0 0 00 on © 0 - - 2 m - - \ D = 6 cm P6.147 P6.148 A smooth pipe containing ethanol at 20°C flows at 7 m^/h through a Bernoulli obstruction, as in Fig. P6.148. Three piezometer tubes are installed, as shown. If the obstruction is a thin-plate orifice, estimate the piezometer levels (a) /t2 and (b) h^. V -1 m- h(0) = 2 m hit) P6.145 P6.148 Problems 439 Flowmeters: the flow nozzle P6.149 In a laboratory experiment, air at 20°C flows from a large tank through a 2-cm-diameter smooth pipe into a sea-level atmosphere, as in Fig. P6.149. The flow is metered by a long-radius nozzle of 1-cm diameter, using a manometer with Meriam red oil (SG = 0.827). The pipe is 8 m long. The measurements of tank pressure and manometer height are as follows: /’tank, Pa(gage): 60 320 1200 2050 2470 3500 4900 ^^mano’ rniTl- 6 38 160 295 380 575 820 Use these data to calculate the flow rates Q and Reynolds numbers Re^ and make a plot of measured flow rate versus tank pressure. Is the flow laminar or turbulent? Compare the data with theoretical results obtained from the Moody chart, including minor losses. Discuss. P6.150 Gasoline at 20°C flows at 0.06 mVs through a 15-cm pipe and is metered by a 9-cm long-radius flow nozzle (Fig. 6.40a). What is the expected pressure drop across the nozzle? P6.151 An engineer needs to monitor a flow of 20°C gasoline at about 250 ± 25 gal/min through a 4-in-diameter smooth pipe. She can use an orifice plate, a long-radius flow noz¬ zle, or a venturi nozzle, all with 2-in-diameter throats. The only differential pressure gage available is accurate in the range 6 to 10 Ibf/in^. Disregarding flow losses, which device is best? P6.152 Kerosene at 20°C flows at 20 m^/h in an 8-cm-diameter pipe. The flow is to be metered by an ISA 1932 flow nozzle so that the pressure drop is 7000 Pa. What is the proper nozzle diameter? P6.153 Two water tanks, each with base area of 1 ft^, are connected by a 0.5-in-diameter long-radius nozzle as in Fig. P6.153. If /i = 1 ft as shown for t = 0, estimate the time for h(t) to drop to 0.25 ft. Flowmeters: the venturi meter P6.154 Gasoline at 20°C flows through a 6-cm-diameter pipe. It is metered by a modem venturi nozzle with d = 4 cm. The measured pressure drop is 8.5 kPa. Estimate the flow rate in gallons per minute. P6.155 It is desired to meter methanol at 20°C flowing through a 5-in-diameter pipe. The expected flow rate is about 300 gal/min. Two flowmeters are available: a venmri nozzle and a thinplate orifice, each with d = 2in. The differential pres¬ sure gage on hand is most accurate at about 12-15 Ibs/in^. Which meter is better for this job? P6.156 Ethanol at 20°C flows down through a modem venturi noz- zle as in Fig. P6. 156. If the mercury manometer reading is 4 in, as shown, estimate the flow rate, in gal/min. P6.157 P6.158 Modify Prob. P6.156 if the fluid is air at 20°C, entering the venturi at a pressure of 18 Ibf/in^. Should a compressibility correction be used? Water at 20°C flows in a long horizontal commercial steel 6-cm-diameter pipe that contains a classical Herschel venturi with a 4-cm throat. The venturi is connected to a mercury manometer whose reading is h = 40 cm. Estimate 440 Chapter 6 Viscous Flow in Ducts (a) the flow rate, in mVh, and (b) the total pressure difference between points 50 cm upstream and 50 cm downstream of the venturi. P6.159 A modem venturi nozzle is tested in a laboratory flow with water at 20°C. The pipe diameter is 5.5 cm, and the venturi throat diameter is 3.5 cm. The flow rate is measured by a weigh tank and the pressure drop by a water-mercury manometer. The mass flow rate and manometer readings are as follows: w, kg/s 0.95 1.98 2.99 5.06 8.15 /?, mm 3.7 15.9 36.2 102.4 264.4 Use these data to plot a calibration curve of venturi dis¬ charge coefficient versus Reynolds number. Compare with the accepted correlation, Eq. (6.1 14). Flowmeters: other designs P6.160 An instrument popular in the beverage industry is the tar¬ get flowmeter in Fig. P6.160. A small flat disk is mounted in the center of the pipe, supported by a strong but thin rod. (a) Explain how the flowmeter works, (b) If the bending moment M of the rod is measured at the wall, derive a for¬ mula for the estimated velocity of the flow, (c) List a few advantages and disadvantages of such an instrument. P6.161 An instrument popular in the water supply industry, sketched in Fig. P6.161, is the single jet water meter, (a) How does it work? (b) What do you think a typical calibration curve would look like? (c) Can you cite further details, for example, reliability, head loss, cost ? Flowmeters: compressibility correction P6.162 Air flows at high speed through a Herschel venturi moni¬ tored by a mercury manometer, as shown in Fig. P6. 162. The upstream conditions are 150 kPa and 80°C. If h = 37 cm, estimate the mass flow in kg/s. (Hint: The flow is compressible.) P6.163 Modify Prob. P6.162 as follows: Find the manometer reading h for which the mass flow through the venturi is approximately 0.4 kg/s. (Hint: The flow is compressible.) Word Problems W6.1 In fully developed straight-duct flow, the velocity profiles do not change (why?), but the pressure drops along the pipe axis. Thus there is pressure work done on the fluid. If, say, the pipe is insulated from heat loss, where does this energy go? Make a thermodynamic analysis of the pipe flow. W6.2 From the Moody chart (Fig. 6.13), rough surfaces, such as sand grains or ragged machining, do not affect laminar flow. Can you explain why? They do affect turbulent flow. Can you develop, or suggest, an analytical-physical model of turbulent flow near a rough surface that might be used to predict the known increase in pressure drop? W6.3 Differentiation of the laminar pipe flow solution, Eq. (6.40), shows that the fluid shear stress T(r) varies linearly from zero at the axis to at the wall. It is claimed that this Fundamentals of Engineering Exam Problems 441 is also true, at least in the time mean, for fully developed turbulent flow. Can you verify this claim analytically? W6.4 A porous medium consists of many tiny tortuous passages, and Reynolds numbers based on pore size are usually very low, of order unity. In 1856 H. Darcy proposed that the pressure gradient in a porous medium was directly propor¬ tional to the volume-averaged velocity V of the fluid: where K is termed the permeability of the medium. This is now called Darcy’s law of porous flow. Can you make a Fundamentals of Engineering Exam Problems FE6.1 In flow through a straight, smooth pipe, the diameter Reynolds number for transition to turbulence is generally taken to be [a) 1500, (b) 2300, (c) 4000, (d) 250,000, (e) 500,000 FE6.2 Eor flow of water at 20°C through a straight, smooth pipe at 0.06 m^/h, the pipe diameter for which transition to turbu¬ lence occurs is approximately (a) 1.0 cm, ib) 1.5 cm, (c) 2.0 cm, id) 2.5 cm, (e) 3.0 cm EE6.3 Eor flow of oil [/i = 0.1 kg/(m ■ s), SG = 0.9] through a long, straight, smooth 5-cm-diameter pipe at 14 m^/h, the pressure drop per meter is approximately (fl) 2200 Pa, (b) 2500 Pa, (c) 10,000 Pa, (d) 160 Pa, (e) 2800 Pa EE6.4 Eor flow of water at a Reynolds number of 1 .03 E6 through a 5-cm-diameter pipe of roughness height 0.5 mm, the approximate Moody friction factor is (a) 0.012, (b) 0.018, (c) 0.038, (d) 0.049, (e) 0.102 EE6.5 Minor losses through valves, fittings, bends, contractions, and the like are commonly modeled as proportional to (a) total head, (b) static head, (c) velocity head, (d) pressure drop, (e) velocity EE6.6 A smooth 8-cm-diameter pipe, 200 m long, connects two reservoirs, containing water at 20°C, one of which has a surface elevation of 700 m and the other a surface elevation of 560 m. If minor losses are neglected, the expected flow rate through the pipe is (a) 0.048 m^/h, (b) 2.87 m^/h, (c) 134 m^/h, (d) 172 m^/h, (e) 385 m^/h EE6.7 If, in Prob. PE6.6 the pipe is rough and the actual flow rate is 90 m^/h, then the expected average roughness height of the pipe is approximately (a) 1.0 mm, (b) 1.25 mm, (c) 1.5 mm, (d) 1.75 mm, (e) 2.0 mm Poiseuille flow model of porous-media flow that verifies Darcy’s law? Meanwhile, as the Reynolds number in¬ creases, so that > 1, the pressure drop becomes nonlinear, as was shown experimentally by P. H. Eor- scheimer as early as 1782. The flow is still decidedly lami¬ nar, yet the pressure gradient is quadratic: fi Vp = - V — C\|V\|V Darcy-Eorscheimer law K where C is an empirical constant. Can you explain the reason for this nonlinear behavior? EE6.8 Suppose in Prob. PE6.6 the two reservoirs are connected, not by a pipe, but by a sharp-edged orifice of diameter 8 cm. Then the expected flow rate is approximately (fl) 90 mVh, (fc) 579 mVh, (c) 748 mVh, {d) 949 mVh, (e) 1048 mVh EE6.9 Oil [/i = 0.1 kg/(m • s), SG = 0.9] flows through a 50-m- long smooth 8-cm-diameter pipe. The maximum pressure drop for which laminar flow is expected is approximately (fl) 30 kPa, (b) 40 kPa, (c) 50 kPa, (d) 60 kPa, (e) 70 kPa EE6.10Air at 20°C and approximately 1 atm flows through a smooth 30-cm-square duct at 1500 ft^/min. The expected pressure drop per meter of duct length is (fl) 1.0 Pa, (b) 2.0 Pa, (c) 3.0 Pa, (d) 4.0 Pa, (e) 5.0 Pa EE6.11 Water at 20°C flows at 3 mVh through a sharp-edged 3-cm-diameter orifice in a 6-cm-diameter pipe. Estimate the expected pressure drop across the orifice. (fl) 440 Pa, (b) 680 Pa, (c) 875 Pa, (d) 1750 Pa, (e) 1870 Pa EE6.12 Water flows through a straight 10-cm-diameter pipe at a diameter Reynolds number of 250,000. If the pipe rough¬ ness is 0.06 mm, what is the approximate Moody friction factor? (fl) 0.015, (b) 0.017, (c) 0.019, (d) 0.026, (e) 0.032 EE6.13What is the hydraulic diameter of a rectangular air- ventilation duct whose cross section is 1 m by 25 cm? (fl) 25 cm, (b) 40 cm, (c) 50 cm, (d) 75 cm, (e) 100 cm EE6.14 Water at 20°C flows through a pipe at 300 gal/min with a friction head loss of 45 ft. What is the power required to drive this flow? (a)0.16 kW, (b) 1.88 kW, (c) 2.54 kW, (d) 3.41 kW, (e) 4.24 kW EE6.15 Water at 20°C flows at 200 gal/min through a pipe 150 m long and 8 cm in diameter. If the friction head loss is 12 m, what is the Moody friction factor? (fl) 0.010, (b) 0.015, (c) 0.020, (d) 0.025, (e) 0.030 442 Chapter 6 Viscous Flow in Ducts Comprehensive Problems C6.1 A pitot-static probe will be used to measure the velocity distribution in a water tunnel at 20°C. The two pressure lines from the probe will be connected to a U-tube manom¬ eter that uses a liquid of specibc gravity 1.7. The maximum velocity expected in the water tunnel is 2.3 m/s. Your job is to select an appropriate U-tube from a manufacturer that supplies manometers of heights 8, 12, 16, 24, and 36 in. The cost increases significantly with manometer height. Which of these should you purchase? C6.2 A pump delivers a steady flow of water {p, p) from a large tank to two other higher-elevation tanks, as shown in Fig. C6.2. The same pipe of diameter d and roughness e is used throughout. All minor losses except through the valve are neglected, and the partially closed valve has a loss coeffi¬ cient Turbulent flow may be assumed with all kinetic energy flux correction coefficients equal to 1 .06. The pump net head His a. known function of 2^ and hence also of Vji = Ga/Apipe; for example, H = a — BVa, where a and b are constants. Subscript J refers to the junction point at the tee where branch A splits into B and C. Pipe length is much longer than Lg. It is desired to predict the pressure at J, the three pipe velocities and friction factors, and the pump head. Thus there are eight variables: H, Va, Vs, Vc,fA,fB,fc< Pj. Write down the eight equations needed to resolve this problem, but do not solve, since an elaborate iteration pro¬ cedure would be required. C6.3 A small water slide is to be installed inside a swimming pool. See Fig. C6.3. The slide manufacturer recommends a continuous water flow rate Q of 1.39 X 10”^ mVs (about 22 gal/min) down the slide, to ensure that the customers do not burn their bottoms. A pump is to be installed under the slide, with a 5.00-m-long, 4.00-cm-diameter hose supply¬ ing swimming pool water for the slide. The pump is 80 percent efficient and will rest fully submerged 1.00 m below the water surface. The roughness inside the hose is about 0.0080 cm. The hose discharges the water at the top of the slide as a free jet open to the atmosphere. The hose outlet is 4.00 m above the water surface. For fully devel¬ oped turbulent pipe flow, the kinetic energy flux correction factor is about 1.06. Ignore any minor losses here. Assume that p = 998 kg/m^ and v = 1.00 X 10“® m^/s for this water. Find the brake horsepower (that is, the actual shaft power in watts) required to drive the pump. C6.4 Suppose you build a rural house where you need to run a pipe to the nearest water supply, which is fortunately at an elevation of about 1000 m above that of your house. The pipe will be 6.0 km long (the distance to the water sup¬ ply), and the gage pressure at the water supply is 1000 kPa. You require a minimum of 3.0 gal/min of water when the end of your pipe is open to the atmosphere. To minimize cost, you want to buy the smallest-diameter pipe possible. The pipe you will use is extremely smooth, (a) Find the total head loss from the pipe inlet to its exit. Neglect any minor losses due to valves, elbows, entrance lengths, and so on, since the length is so long here and major losses dominate. Assume the outlet of the pipe is open to the atmosphere, (b) Which is more important in this problem, the head loss due to elevation difference or the head loss due to pressure drop in the pipe? (c) Find the minimum required pipe diameter. C6.5 Water at room temperature flows at the same volume flow rate, Q = 9.4 X 10”"^ mVs, through two ducts, one a round pipe and one an annulus. The cross-sectional area A of the two ducts is identical, and all walls are made of commer¬ cial steel. Both ducts are the same length. In the cross sec¬ tions shown in Fig. C6.5 R = 15.0 mm and a = 25.0 mm. Comprehensive Problems 443 C6.3 C6.5 (a) What is the radius b such that the cross-sectional areas of the two ducts are identical? (b) Compare the frictional head loss hf per unit length of pipe for the two cases, assuming fully developed flow. For the annulus, do both a quick estimate (using the hydraulic diameter) and a more accurate estimate (using the effective diameter correction), and compare, (c) If the losses are different for the two cases, explain why. Which duct, if any, is more “efficient”? C6.6 John Laufer (NACA Tech Rep. 1 174, 1954) gave velocity data 20°C airflow in a smooth 24.7-cm-diameterpipe at Re ~ 5 E5: W/«CL 1.0 0.997 0.988 0.959 0.908 0.847 0.818 0.771 0.690 r/R 0.0 0.102 0.206 0.412 0.617 0.784 0.846 0.907 0.963 Source: John Laufer (NASA Tech Rep. 1174, 1954) The centerline velocity uql was 30.5 m/s. Determine (a) the average velocity by numerical integration and (b) the wall shear stress from the log law approximation. Compare with the Moody chart and with Eq. (6.43). C6.7 Consider energy exchange in fully developed laminar flow between parallel plates, as in Eqs. (6.60). Let the pressure drop over a length L be Ap. Calculate the rate of work done by this pressure drop on the fluid in the region (0 < x < L, — h). 16. E. F. Brater, H. W. King, 1. E. Lindell, and C. Y. Wei, Hand¬ book of Hydraulics, 7th ed., McGraw-Hill, New York, 1996. 17. H. Cross, “Analysis of Flow in Networks of Conduits or Con¬ ductors,” Univ. III. Bull. 286, November 1936. 18. P. K. Swamee and A. K. Sharma, Design of Water Supply Pipe Networks, Wiley-Interscience, New York, 2008. 19. D. C. Wilcox, Turbulence Modeling for CFD, 3d ed., DCW Industries, La Canada, CA, 2006. 20. R. W. leppson. Analysis of Flow in Pipe Networks, Butterworth-Heinemann, Woburn, MA, 1976. 21. R. W. Fox and S. 1. Kline, “Flow Regime Data and Design Methods for Curved Subsonic Diffusers,” J. Basic Eng., vol. 84, 1962, pp. 303-312. 22. R. C. Baker, Flow Measurement Handbook: Industrial Designs, Operating Principles, Performance, and Applica¬ tions, Cambridge University Press, New York, 2005. 23. R. W. Miller, Flow Measurement Engineering Handbook, 3d edition, McGraw-Hill, New York, 1997. 24. B. Warren and C. Wunsch (eds.). Evolution of Physical Oceanography, M.I.T. Press, Cambridge, MA, 1981. 25. U.S. Department of Commerce, Tidal Current Tables, National Oceanographic and Atmospheric Administration, Washington, DC, 1971. 26. 1. A. Shercliff, Electromagnetic Flow Measurement, Cambridge University Press, New York, 1962. I. A. Miller, “A Simple Linearized Hot-Wire Anemometer,” J. Fluids Eng., December 1976, pp. 749-752. R. 1. Goldstein (ed.). Fluid Mechanics Measurements, 2d ed.. Hemisphere, New York, 1996. D. Eckardt, “Detailed Flow Investigations within a High Speed Centrifugal Compressor Impeller,” J. Fluids Eng., September 1976, pp. 390^02. H. S. Bean (ed.). Fluid Meters: Their Theory and Applica¬ tion, 6th ed., American Society of Mechanical Engineers, New York, 1971. “Measurement of Fluid Flow by Means of Orifice Plates, Nozzles, and Venturi Tubes Inserted in Circular Cross Section Conduits Running Full,” Int. Organ. Stand. Rep. DIS-5167, Geneva, April 1976. 32. P. Sagaut and C. Meneveau, Large Eddy Simulation for Incompressible Flows: An Introduction, 3d ed.. Springer, New York, 2006. 33. S. E. Haaland, “Simple and Explicit Formulas for the Friction Factor in Turbulent Pipe Flow,” J. Fluids Eng., March 1983, pp. 89-90. 34. R. K. Shah and A. L. London, Laminar Flow Forced Convec¬ tion in Ducts, Academic, New York, 1979. 35. P. L. Skousen, Valve Handbook, 3d ed. McGraw-Hill, New York, 2011. 36. W. Li, W.-X. Chen, and S.-Z. Xie, “Numerical Simulation of Stress-Induced Secondary Flows with Hybrid Finite Analytic Method,” Journal of Hydrodynamics, vol. 14, no. 4, December 2002, pp. 24-30. 37. ASHRAE Handbook — 2012 Fundamentals, ASHRAE, Atlanta, GA, 2012. 38. F. Durst, A. Melling, and 1. H. Whitelaw, Principles and Practice of Laser-Doppler Anemometry, 2d ed.. Academic, New York, 1981. 39. A. P. Lisitsyn et al.. Laser Doppler and Phase Doppler Mea¬ surement Techniques, Springer- Verlag, New York, 2003. 40. 1. E. Amadi-Echendu, H. Zhu, and E. H. Higham, “Analysis of Signals from Vortex Flowmeters,” Flow Measurement and Instrumentation, vol. 4, no. 4, Oct. 1993, pp. 225-231. 41. G. Vass, “Ultrasonic Flowmeter Basics,” Sensors, vol. 14, no. 10, Oct. 1997, pp. 73-78. 42. ASME Fluid Meters Research Committee, “The ISO-ASME Orifice Coefficient Equation,” Mech. Eng. luly 1981, pp. 44-45. 43. R. D. Blevins, Applied Fluid Dynamics Handbook, Van Nostrand Reinhold, New York, 1984. 44. O. C. lones, Ir., and J. C. M. Leung, “An Improvement in the Calculation of Turbulent Friction in Smooth Concentric Annuli,” J. Fluids Eng., December 1981, pp. 615-623. 45. P. R. Bandyopadhyay, “Aspects of the Equilibrium Puff in Transitional Pipe Flow,” J. Fluid Mech., vol. 163, 1986, pp. 439-458. 446 Chapter 6 Viscous Flow in Ducts 46. I. E. Idelchik, Handbook of Hydraulic Resistance, 3d ed., CRC Press, Boca Raton, FL, 1993. 47. S. Klein and W. Beckman, Engineering Equation Solver (EES), University of Wisconsin, Madison, WI, 2014. 48. R. D. Coffield, P. T. McKeown, and R. B. Hammond, “Irrecoverable Pressure Loss Coefficients for Two Elbows in Series with Various Orientation Angles and Separation Distances,” Report WAPD-T-3117, Bettis Atomic Power Laboratory, West Mifflin, PA, 1997. 49. H. Ito, “Pressure Losses in Smooth Pipe Bends,” Journal of Basic Engineering, March 1960, pp. 131-143. 50. A. H. Gibson, “On the Flow of Water through Pipes and Passages,” Proc. Roy. Soc. London, Ser. A, vol. 83, 1910, pp. 366-378. 51. M. Raffel et ah. Particle Image Velocimetry: A Practical Guide, 2d ed.. Springer, New York, 2007. 52. Crane Co., Plow of Fluids through Valves, Fittings, and Pipe, Crane, Stanford, CT, 2009. 53. S. A. Berger, L. Talbot, and L.-S. Yao, “Flow in Curved Pipes,” Annual Review of Fluid Mechanics, vol. 15, 1983, pp. 461-512. 54. P. L. Spedding, E. Benard, and G. M. McNally, “Fluid Flow through 90° Bends,” Developments in Chemical Engineering and Mineral Processing, vol. 12, nos. 1-2, 2004, pp. 107-128. 55. R. R. Kerswell, “Recent Progress in Understanding the Tran¬ sition to Turbulence in a Pipe,” Nonlinearity, vol. 18, 2005, pp. R17-R44. 56. B. J. McKeon et ah, “Friction Factors for Smooth Pipe Flow,” J. Fluid Meek, vol. 511, 2004, pp. 41-44. 57. M. Bahrami, M. M. Yovanovich, and J. R. Culham, “Pressure Drop of Fully-Developed Laminar Flow in Microchannels of Arbitrary Cross-Section,” J. Fluids Engineering, vol. 128, Sept. 2006, pp. 1036-1044. 58. G. S. Larraona, A. Rivas, and J. C. Ramos, “Computa¬ tional Modeling and Simulation of a Single-Jet Water Me¬ ter,” J. Fluids Engineering, vol. 130, May 2008, pp. 0511021-05110212. 59. C. Spiegel, Designing and Building Fuel Cells, McGraw- Hill, New York, 2007. 60. B. A. Finlay son et ah. Microcomponent Flow Characteriza¬ tion, Chap. 8 of Micro Instrumentation, M. V. Koch (Ed.), John Wiley, Hoboken, NJ, 2007. © Solar Impulse \| Revillard \| Rezo This chapter is devoted to lift and drag of various bodies immersed in an approaching stream of fluid. Pictured is the Swiss solar-powered aircraft, Solar Impulse, over the Golden Gate Bridge. Earlier solar aircraft needed to be towed aloft before flying and were not able to fly at night. The Solar Impulse is the first solar airplane to fly day and night, approaching the notion of perpetual flight. The long, high-aspect-ratio wings have more lift, and less drag, than a short wing of the same area. Its first international flight was from Switzerland to Brussels, on May 14, 2011. In the summer of 2013, as shown, it flew from San Francisco to New York City, in five legs. The pilots were Bertrand Piccard and Andre Borschberg. 448 Chapter 7 Flow Past Immersed Bodies 7.1 Reynolds Number and Geometry Effects Motivation. This chapter is devoted to “external” flows around bodies immersed in a fluid stream. Such a flow will have viscous (shear and no-slip) effects near the body surfaces and in its wake, but will typically be nearly inviscid far from the body. These are unconfined boundary layer flows. Chapter 6 considered “internal” flows confined by the walls of a duct. In that case the viscous boundary layers grow from the sidewalls, meet downstream, and fill the entire duct. Viscous shear is the dominant effect. For example, the Moody chart of Fig. 6.13 is essentially a correlation of wall shear stress for long ducts of constant cross section. External flows are unconfined, free to expand no matter how thick the viscous layers grow. Although boundary layer theory (Sec. 7.3) and computational fluid dynamics (CFD) are helpful in understanding external flows, complex body geometries usually require experimental data on the forces and moments caused by the flow. Such immersed-body flows are commonly encountered in engineering studies: aerodynamics (airplanes, rockets, projectiles), hydrodynamics (ships, submarines, torpedos), transportation (automobiles, trucks, cycles), wind engineering (buildings, bridges, water towers, wind turbines), and ocean engineering (buoys, breakwaters, pilings, cables, moored instruments). This chapter provides data and analysis to assist in such studies. The technique of boundary layer (BL) analysis can be used to compute viscous effects near solid walls and to “patch” these onto the outer inviscid motion. This patching is more successful as the body Reynolds number becomes larger, as shown in Fig. 7.1. In Fig. 7.1 a uniform stream U moves parallel to a sharp flat plate of length L. If the Reynolds number ULIv is low (Fig. 7.1a), the viscous region is very broad and extends far ahead and to the sides of the plate. The plate retards the oncoming stream greatly, and small changes in flow parameters cause large changes in the pressure 449 450 Chapter 7 Flow Past Immersed Bodies Fig. 7.1 Comparison of flow past a sharp flat plate at low and high Reynolds numbers: (a) laminar, low-Re flow; (b) high-Re flow. U ■ Re^= 10'^ Small (b) distribution along the plate. Thus, although in principle it should be possible to patch the viscous and inviscid layers in a mathematical analysis, their interaction is strong and nonlinear [1 to 3]. There is no existing simple theory for external flow analysis at Reynolds numbers from 1 to about 1000. Such thick-shear-layer flows are typically studied by experiment or by numerical modeling of the flow field on a computer . A high-Reynolds-number flow (Fig. 7.1b) is much more amenable to boundary layer patching, as first pointed out by Prandtl in 1904. The viscous layers, either laminar or turbulent, are very thin, thinner even than the drawing shows. We define the boundary layer thickness <5 as the locus of points where the velocity u parallel to the plate reaches 99 percent of the external velocity U. As we shall see in Sec. 7.4, the accepted formulas for flat-plate flow, and their approximate ranges, are ' 5.0 laminar 10' < Re, < 10® Re]'' - 0.16 turbulent 10® < Re , 1 Re]'’ (7.1a) (7.1h) 7.1 Reynolds Number and Geometry Effects 451 where Re^^ = Uxlv is called the local Reynolds number of the flow along the plate surface. The turbulent flow formula applies for Re^^ greater than approximately 10®. Some computed values from Eq. (7.1) are Re,, 10' 10^ 10^ 10’ 10 (^/-^)lam 0.050 0.016 0.005 0.022 0.016 0.011 The blanks indicate that the formula is not applicable. In all cases these boundary layers are so thin that their displacement effect on the outer inviscid layer is negligible. Thus the pressure distribution along the plate can be computed from inviscid theory as if the boundary layer were not even there. This external pres¬ sure held then “drives” the boundary layer flow, acting as a forcing function in the momentum equation along the surface. We shall explain this boundary layer theory in Secs. 7.4 and 7.5. For slender bodies, such as plates and airfoils parallel to the oncoming stream, we conclude that this assumption of negligible interaction between the boundary layer and the outer pressure distribution is an excellent approximation. For a blunt-body flow, however, even at very high Reynolds numbers, there is a discrepancy in the viscous-inviscid patching concept. Figure 7.2 shows two Fig. 7.2 Illustration of the strong interaction between viscous and inviscid regions in the rear of blunt-body flow: (a) idealized and definitely false picture of blunt-body flow; (b) actual picture of blunt-body flow. Beautifully behaved 452 Chapter 7 Flow Past Immersed Bodies sketches of flow past a two- or three-dimensional blunt body. In the idealized sketch {1 .2a), there is a thin film of boundary layer about the body and a narrow sheet of viscous wake in the rear. The patching theory would be glorious for this picture, but it is false. In the actual flow (Fig. 1 .2b), the boundary layer is thin on the front, or windward, side of the body, where the pressure decreases along the surface {favorable pressure gradient). But in the rear the boundary layer encounters increasing pressure {adverse pressure gradient) and breaks off, or sepa¬ rates, into a broad, pulsating wake. (See Fig. 5.2a for a photograph of a specific example.) The mainstream is deflected by this wake, so that the external flow is quite different from the prediction from inviscid theory with the addition of a thin boundary layer. The theory of strong interaction between blunt-body viscous and inviscid layers is not well developed. Flows like that of Fig. 1.2b are normally studied experimentally or with CFD . Reference 5 is an example of efforts to improve the theory of sepa¬ rated flows. Reference 6 is another textbook devoted to separated flow. EXAMPLE 7.1 A long, thin, flat plate is placed parallel to a 20-ft/s stream of water at 68°F. At what distance x from the leading edge will the boundary layer thickness be 1 in? Solution • Assumptions: Flat-plate flow, with Eqs. (7.1) applying in their appropriate ranges. • Approach: Guess laminar flow first. If contradictory, try turbulent flow. • Property values: From Table A.l for water at 68°F, i/ ~ 1.082 E-5 fF/s. • Solution step 1: With S = 1 in = 1/12 ft, try laminar flow, Eq. (7.1a): 1/12 ft _ 5 _ X [(20ft/s)x/(1.082E-5fF/s)]‘“ Solve for x = 513 ft Pretty long plate! This does not sound right. Check the local Reynolds number: te _ (20ft/s)(513ft) 1^ 1.082 E-5 fF/s 9.5 E8 (1) This is impossible, since laminar boundary layer flow only persists up to about 10^ (or, with special care to avoid disturbances, up to 3 X 10^). • Solution step 2: Try turbulent flow, Eq. {1 .lb)\ d_ 0.16 1/12 ft _ _ CUB _ {Uxlvf'' “ X [(20ft/s)x/(1.082E-5fF/s)]''’ Solve for x ~ 5.17 ft Ans. Check Re,, = (20 ft/s)(5.17 ft)/(1.082 E-5 fF/s) = 9.6 E6 > lO'. OK, turbulent flow. • Comments: The flow is turbulent, and the inherent ambiguity of the theory is resolved. 7.2 Momentum Integral Estimates 453 7.2 Momentum Integral Estimates Karman’s Analysis of the Flat Plate When we derived the momentum integral relation, Eq. (3.37), and applied it to a flat- plate boundary layer in Example 3.11, we promised to consider it further in Chap. 7. Well, here we are! Let us review the problem, using Eig. 7.3. A shear layer of unknown thickness grows along the sharp flat plate in Eig. 7.3. The no-slip wall condition retards the flow, making it into a rounded profile u(x, y), which merges into the external velocity U = constant at a “thickness” y = A(x). By utilizing the control volume of Fig. 3.11, we found (without making any assumptions about laminar versus turbulent flow) in Example 3.11 that the drag force on the plate is given by the following momentum integral across the exit plane: Dix) pb S(x) u(U — u) dy (7.2) where b is the plate width into the paper and the integration is carried out along a vertical plane x = constant. You should review the momentum integral relation (3.37) and its use in Example 3.11. Equation (7.2) was derived in 1921 by Karman , who wrote it in the convenient form of the momentum thickness 9: D(x) = pbU^e (7.3) Momentum thickness is thus a measure of total plate drag. Karman then noted that the drag also equals the integrated wall shear stress along the plate: D{x) = b T„{x) dx or — = br^ dx Meanwhile, the derivative of Eq. (7.3), with U = constant, is (7.4) dD dO — = pbU^ — dx dx y Fig. 7.3 Growth of a boundary layer on a flat plate. The thickness is exaggerated. x = 0 x = L X 454 Chapter 7 Flow Past Immersed Bodies By comparing this with Eq. (7.4) Karman arrived at what is now called the momentum integral relation for flat-plate boundary layer flow: T„, = pU^ M dx (7.5) It is valid for either laminar or turbulent flat-plate flow. To get a numerical result for laminar flow, Karman assumed that the velocity profiles had an approximately parabolic shape ..2' u{x, y) ~ U ^1 0 < y < 5{x) 2y 5 which makes it possible to estimate both momentum thickness and wall shear: (7.6) e = 2y 6 T = M' F , 15 du dy 2pU (7.7) y = 0 By substituting (7.7) into (7.5) and rearranging, we obtain V 5 d5 ~ 15 — dx U (7.8) where v = pip. We can integrate from 0 to x, assuming that <5 = 0 at x = 0, the leading edge: or 1 2 _ 2^ ~ U 5.5 Ref (7.9) This is the desired thickness estimate. It is all approximate, of course, part of Karman’ s momentum integral theory , but it is startlingly accurate, being only 10 percent higher than the known accepted solution for laminar flat-plate flow, which we gave as Eq. (7.1a). By combining Eqs. (7.9) and (7.7) we also obtain a shear stress estimate along the plate: pf UeJ Ref (7.10) Again this estimate, in spite of the crudeness of the profile assumption [Eq. (7.6)] is only 10 percent higher than the known exact laminar-plate-flow solution Cf = 0.664/Ref, treated in Sec. 7.4. The dimensionless quantity Cf, called the skin friction coefficient, is analogous to the friction factor /in ducts. A boundary layer can be judged as “thin” if, say, the ratio d/x is less than about 0.1. This occurs at 6lx = 0.1 = 5.0/Ref or at Re^ = 2500. Eor Re^, less than 2500 we can estimate that boundary layer theory fails because the thick layer has a significant effect on the outer inviscid flow. The upper limit on Re^, for laminar flow 7.2 Momentum Integral Estimates 455 is about 3 X 10®, where measurements on a smooth flat plate show that the flow undergoes transition to a turbulent boundary layer. From 3 X 10® upward the turbulent Reynolds number may be arbitrarily large, and a practical limit at present is 5 X 10® for oil supertankers. Displacement Thickness Another interesting effect of a boundary layer is its small but finite displacement of the outer streamlines. As shown in Fig. 7.4, outer streamlines must deflect outward a distance J(jr) to satisfy conservation of mass between the inlet and outlet: rh pUb dy fS pub dy d = h + 6 (7.11) The quantity 5 is called the displacement thickness of the boundary layer. To relate it to uiy), cancel p and b from Eq. (7.11), evaluate the left integral, and slyly add and subtract U from the right integrand: or Uh = {U + u — U) dy = U{h + J) + •'o (m — U) dy rS. 6 = 1 - - «y (7.12) Thus the ratio of d/d varies only with the dimensionless velocity profile shape ulU. Introducing our profile approximation (7.6) into (7.12), we obtain by integration this approximate result: 5 1.83 X ~ Re^^ (7.13) These estimates are only 6 percent away from the exact solutions for laminar flat-plate flow given in Sec. 7.4: 5 = 0.3445 = 1.721x/Rey^. Since 5 is much smaller than X for large Re^ and the outer streamline slope VIU is proportional to 5, we conclude that the velocity normal to the wall is much smaller than the velocity parallel to the wall. This is a key assumption in boundary layer theory (Sec. 7.3). We also conclude from the success of these simple parabolic estimates that Karman’s momentum integral theory is effective and useful. Many details of this theory are given in Refs. 1 to 3. Fig. 7.4 Displacement effect of a boundary layer. y = h + 5 1 \ u u u y = h t Outer streamline h 1 ,■/ - ► U '//////////////////////////////////////// o, d Simulated effect 456 Chapter 7 Flow Past Immersed Bodies Part (a) Part (b) 7.3 The Boundary Layer Equations EXAMPLE 7.2 Are low-speed, small-scale air and water boundary layers really thin? Consider flow at U = \ ft/s past a flat plate 1 ft long. Compute the boundary layer thickness at the trailing edge for {a) air and (b) water at 68°F. Solution From Table A.2, 1.61 E-4 fr/s. The trailing-edge Reynolds number thus is UL (1 ft/s) (1ft) Re^ = - = = 6200 V 1.61 E-4 ftVs Since this is less than 10®, the flow is presumed laminar, and since it is greater than 2500, the boundary layer is reasonably thin. From Eq. (7.1a), the predicted laminar thickness is 6 _ 5.0 X ~ VMm S = 0.0634 ft = 0.76 in Arts, (a) ■a,. = 0.0634 or, at X = 1 ft. From Table A.l, i^water ~ 1-08 E-5 fC/s. The trailing-edge Reynolds number is (1 ft/s)(l ft) Re^, = ^ ^ « 92,600 1.08 E-5 ft-/s This again satisfies the laminar and thinness conditions. The boundary layer thickness is d 5.0 X V92,600 = 0.0164 or, at X = 1 ft, S = 0.0164 ft = 0.20 in Ani’. (b) Thus, even at such low velocities and short lengths, both airflows and water flows satisfy the boundary layer approximations. In Chaps. 4 and 6 we learned that there are several dozen known analytical laminar flow solutions [1 to 3]. None are for external flow around immersed bodies, although this is one of the primary applications of fluid mechanics. No exact solutions are known for turbulent flow, whose analysis typically uses empirical modeling laws to relate time-mean variables. There are presently three techniques used to study external flows: (1) numerical (computer) solutions, (2) experimentation, and (3) boundary layer theory. Computational fluid dynamics is now well developed and described in advanced texts such as that by Anderson . Thousands of computer solutions and models have been published; execution times, mesh sizes, and graphical presentations are improving each year. Both laminar and turbulent flow solutions have been published, and turbulence modeling is a current research topic . Except for a brief discussion of computer analysis in Chap. 8, the topic of CFD is beyond our scope here. 7.3 The Boundary Layer Equations 457 Experimentation is the most common method of studying external flows. Chapter 5 outlined the technique of dimensional analysis, and we shall give many nondimen- sional experimental data for external flows in Sec. 7.6. The third tool is boundary layer theory, first formulated by Ludwig Prandtl in 1904. We shall follow Prandtl’s ideas here and make certain order-of-magnitude assumptions to greatly simplify the Navier-Stokes equations (4.38) into boundary layer equations that are solved relatively easily and patched onto the outer inviscid flow field. One of the great achievements of boundary layer theory is its ability to predict the flow separation that occurs in adverse (positive) pressure gradients, as illustrated in Fig. 1 .2b. Before 1904, when Prandtl published his pioneering paper, no one realized that such thin shear layers could cause such a gross effect as flow separation. Even today, however, boundary layer theory cannot accurately predict the behavior of the separated-flow region and its interaction with the outer flow. Modern research [4, 9] has focused on detailed CFD simulations of separated flow, and the resultant wakes, to gain further insight. Derivation for Two-Dimensional Flow We consider only steady two-dimensional incompressible viscous flow with the X direction along the wall and y normal to the wall, as in Fig. 7.3. We neglect gravity, which is important only in boundary layers where fluid buoyancy is dominant [2, sec. 4.14]. From Chap. 4, the complete equations of motion consist of continuity and the x- and y-momentum relations: du - h dx dv dy = 0 du du u - V — dx dy dv dv u - h V - dx dy dp dx dy + /i + M d^u d^u\ dx^ dyV d^V d^v\ dx^ dyV (7.14fl) (7.14^7) (7.14c) These should be solved for u, v, and p subject to typical no-slip, inlet, and exit bound¬ ary conditions, but in fact they are too difficult to handle for most external flows except with CFD. In 1904 Prandtl correctly deduced that a shear layer must be very thin if the Reynolds number is large, so that the following approximations apply: Velocities: Rates of change: Reynolds number: V < u du du dv dv — ^ ^ - dx dy dx dy 1 (7.15fl) (7.15/7) (7.15c) 'For a curved wall, x can represent the arc length along the wall and y can be everywhere normal to x with negligible change in the boundary layer equations as long as the radius of curvature of the wall is large compared with the boundary layer thickness [1 to 3]. 458 Chapter 7 Flow Past Immersed Bodies Our discussion of displacement thickness in the previous section was intended to justify these assumptions. Applying these approximations to Eq. (7.14c) results in a powerful simplification: dx small pi 1 + pi 17 dV dy small dp — = 0 dy dp - + p dy d^V dx' very small + /i d^V dy" small p ~ p(x) only (7.16) In other words, the y-momentum equation can be neglected entirely, and the pressure varies only along the boundary layer, not through it. The pressure gradient term in Eq. (7.14h) is assumed to be known in advance from Bernoulli’s equation applied to the outer inviscid flow: dp dp dx dx (7.17) Presumably we have already made the inviscid analysis and know the distribution of U(x) along the wall (Chap. 8). Meanwhile, one term in Eq. (7.14h) is negligible due to Eqs. (7.15): d^u dx^ d^u dy" (7.18) However, neither term in the continuity relation (7.14fl) can be neglected — another warning that continuity is always a vital part of any fluid flow analysis. The net result is that the three full equations of motion (7.14) are reduced to Prandtl’s two boundary layer equations for two-dimensional incompressible flow: Continuity: Momentum along wall: where dll dV - h — = 0 dx dy du dll dU 1 dT u - h V — « u — + — dx dy dx p dy p — laminar flow dy a - pu'v' turbulent flow dy (7.1%) (7.1%) These are to be solved for u{x, y) and v{x, y), with U{x) assumed to be a known function from the outer inviscid flow analysis. There are two boundary conditions on u and one on v: Aty = 0 (wall): u = v = 0 (no slip) (7.20a) Asy = (5(x) (other stream): u = U{x) (patching) (7.20b) Unlike the Navier-Stokes equations (7.14), which are mathematically elliptic and must be solved simultaneously over the entire flow field, the boundary layer equations (7.19) 7.4 The Flat-Plate Boundary Layer 459 are mathematically parabolic and are solved by beginning at the leading edge and marching downstream as far as you like, stopping at the separation point or earlier if you prefer.^ The boundary layer equations have been solved for scores of interesting cases of internal and external flow for both laminar and turbulent flow, utilizing the inviscid distribution U{x) appropriate to each flow. Full details of boundary layer theory and results and comparison with experiment are given in Refs. 1 to 3. Here we shall confine ourselves primarily to flat-plate solutions (Sec. 7.4). 7.4 The Flat-Plate Boundary Layer The classic and most often used solution of boundary layer theory is for flat-plate flow, as in Fig. 7.3, which can represent either laminar or turbulent flow. Laminar Flow For laminar flow past the plate, the boundary layer equations (7.19) can be solved exactly for u and v, assuming that the free-stream velocity U is constant {dUldx = 0). The solution was given by Prandtl’s student Blasius, in his 1908 dissertation from Gottingen. With an ingenious coordinate transformation, Blasius showed that the dimensionless velocity profile ulU is a function only of the single composite dimen¬ sionless variable (y)[t//(i/x)]^^^; U = f'(r]) 'n = y U vx 1/2 (7.21) where the prime denotes differentiation with respect to rj. Substitution of (7.21) into the boundary layer equations (7.19) reduces the problem, after much algebra, to a single third-order nonlinear ordinary differential equation for / [1-3]: /"' + hr = 0 The boundary conditions (7.20) become Aty = 0: /(0)=/'(0) = 0 Asy— >c>o: /'(oo)— >1.0 (7.22) (7.23fl) {1.23b) This is the Blasius equation, for which accurate solutions have been obtained only by numerical integration. Some tabulated values of the velocity profile shape /'(tj) = uHJ are given in Table 7.1. Since uHJ approaches 1.0 only as y — > oo, it is customary to select the boundary layer thickness 5 as that point where ulU = 0.99. From the table, this occurs at T] = 5.0: vl/2 _^Y' i/xj 5.0 or (7.24) ^For further mathematical details, see Ref. 2, Sec. 2.8. 460 Chapter 7 Flow Past Immersed Bodies Table 7.1 The Blasius Velocity Profile [1 to 3] ylU/(vx)f^ u/U yWKvxif^ ulV 0.0 0.0 2.8 0.81152 0.2 0.06641 3.0 0.84605 0.4 0.13277 3.2 0.87609 0.6 0.19894 3.4 0.90177 0.8 0.26471 3.6 0.92333 1.0 0.32979 3.8 0.94112 1.2 0.39378 4.0 0.95552 1.4 0.45627 4.2 0.96696 1.6 0.51676 4.4 0.97587 1.8 0.57477 4.6 0.98269 2.0 0.62977 4.8 0.98779 2.2 0.68132 5.0 0.99155 2.4 0.72899 oo 1.00000 2.6 0.77246 With the profile known, Blasius, of course, could also compute the wall shear and displacement thickness: 9 ' 0.664 Re'/- 5 X 1.721 Re'/- (7.25) Notice how close these are to our integral estimates, Eqs. (7.9), (7.10), and (7.13). When Cf is converted to dimensional form, we have ^1/2 The wall shear drops off with because of boundary layer growth and varies as velocity to the 1.5 power. This is in contrast to laminar pipe flow, where ^ U and is independent of x. If T„(x) is substituted into Eq. (7.4), we compute the total drag force: D{x) = b Xx)dx = (7.26) •'o The drag increases only as the square root of the plate length. The nondimensional drag coefficient is defined as _ 2Z)(L) pU^bL 1.328 Ref 2cf{L) (7.27) Thus, for laminar plate flow, Co equals twice the value of the skin friction coefficient at the trailing edge. This is the drag on one side of the plate. Karman pointed out that the drag could also be computed from the momentum relation (7.2). In dimensionless form, Eq. (7.2) becomes Co 2 L u U 1 - dy 'o (7.28) 7.4 The Flat-Plate Boundary Layer 461 This can be rewritten in terms of the momentum thickness at the trailing edge: ^ _ 20(L) L Computation of 0 from the profile ulU or from Co gives (7.29) e 0.664 . . ^ . — = - TTT laminar flat plate (7.30) Since 6 is so ill defined, the momentum thickness, being definite, is often used to correlate data taken for a variety of boundary layers under differing conditions. The ratio of displacement to momentum thickness, called the dimensionless-profile shape factor, is also useful in integral theories. For laminar flat-plate flow ^ _ 1.721 T ~ 0.664 2.59 (7.31) A large shape factor then implies that boundary layer separation is about to occur. If we plot the Blasius velocity profile from Table 7.1 in the form of ulU versus yid, we can see why the simple integral theory guess, Eq. (7.6), was such a great success. This is done in Fig. 7.5. The simple parabolic approximation is not far from the true Blasius profile; hence its momentum thickness is within 10 percent of the true value. Also shown in Fig. 7.5 are three typical turbulent flat-plate velocity pro¬ files. Notice how strikingly different in shape they are from the laminar profiles. Fig. 7.5 Comparison of dimensionless laminar and turbulent flat-plate velocity profiles. 6 462 Chapter 7 Flow Past Immersed Bodies Transition to Turbulence Instead of decreasing parabolically to zero, the turbulent profiles are very flat and then drop off sharply at the wall. As you might guess, they follow the logarithmic law shape and thus can be analyzed by momentum integral theory if this shape is properly represented. The laminar flat-plate boundary layer eventually becomes turbulent, but there is no unique value for this change to occur. With care in polishing the wall and keeping the free stream quiet, one can delay the transition Reynolds number to Re^. „ ~ 3 E6 . However, for typical commercial surfaces and gusty free streams, a more realistic value is Re,. u ~ 5 E5. EXAMPLE 7.3 A sharp fiat plate with L = 50 cm and h = 3 m is parallel to a stream of velocity 2.5 m/s. Find the drag on one side of the plate, and the boundary thickness 6 at the trailing edge, for (a) air and {b) water at 20°C and 1 atm. Solution ■ Assumptions: Laminar flat-plate flow, but we should check the Reynolds numbers. ■ Approach: Find the Reynolds number and use the appropriate boundary layer formulas. • Property values: From Table A.2 for air at 20°C, p = 1.2 kg/m^, v = \. 5 E-5 mVs. From Table A.l for water at 20°C, p = 998 kg/m^, v = 1.005 E-6 mVs. • (a) Solution for air: Calculate the Reynolds number at the trailing edge: VL (2.5 m/s)(0.5 m) Re, = - = - - = 83,300 < 5 E5 therefore assuredly laminar i^air 1.5E-5m^/s The appropriate thickness relation is Eq. (7.24): 6 5 5 — = — pr = - py = 0.0173, or = 0.0173(0.5 m) = 0.0087 m Ans. (a) L Ref (83,300)'“ The laminar boundary layer is only 8.7 mm thick. The drag coefficient follows from Eq. (7.27): Cd 1.328 Ref 1.328 (83,300)'“ 0.0046 p , 1.2 kg/m , or Ooneside = bL = (0.0046) - - (2.5 m/s)''(3 m)(0.5 m) « 0.026 N Ans. (a) • Comment (a): This is purely /ncftow drag and is very small for gases at low velocities. • (b) Solution for water: Again calculate the Reynolds number at the trailing edge: UL (2.5 m/s)(0.5 m) r^water 1.005 E-6m^/s 1.24 E6 > 5 E5 therefore it might be turbulent 7.4 The Flat-Plate Boundary Layer 463 This is a quandary. If the plate is rough or encounters disturbances, the flow at the trailing edge will be turbulent. Let us assume a smooth, undisturbed plate, which will remain laminar. Then again the appropriate thickness relation is Eq. (7.24): 6 5 5 - = — TTT = - TTT = 0.00448 or 4=^ = 0.00448(0.5 m) = 0.0022 m Ans. (b) L Ref (L24E6)‘“ This is four times thinner than the air result in part {a), due to the high laminar Reynolds number. Again the drag coefficient follows from Eq. (J.ll): Cn = 1.328 1.328 = 0.0012 p , 998 kg/m , or Ooneside = Co\| U^bL = (0.0012) - Y - (2.5 m/s)^(3 m)(0.5 m) • 5.6 N Ans. (b) • Comment (b): The drag is 215 times larger for water, although Co is lower, reflecting that water is 56 times more viscous and 830 times denser than air. From Eq. (7.26), for the same U and X, the water drag should be (56)^^(830)^^ ~ 215 times higher. Note: If transition to turbulence had occurred at = 5 E5 (at about x = 20 cm), the drag would be about 2.5 times higher, and the trailing edge thickness about four times higher than for fully laminar flow. Turbulent Flow There is no exact theory for turbulent flat-plate flow, although there are many elegant computer solutions of the boundary layer equations using various empirical models for the turbulent eddy viscosity . The most widely accepted result is simply an integral analysis similar to our study of the laminar profile approximation (7.6). We begin with Eq. (7.5), which is valid for laminar or turbulent flow. We write it here for convenient reference: tUx) (7.32) From the definition of Cf, Eq. (7.10), this can be rewritten as 9 = (7.33) Now recall from Fig. 7.5 that the turbulent profiles are nowhere near parabolic. Going back to Fig. 6.10, we see that flat-plate flow is very nearly logarithmic, with a slight outer wake and a thin viscous sublayer. Therefore, just as in turbulent pipe flow, we assume that the logarithmic law (6.28) holds all the way across the boundary layer u 1 yu — ~ — In — -f B 1/2 (7.34) with, as usual, K = 0.41 and B = 5.0. At the outer edge of the boundary layer, y = 5 and u = U, and Eq. (7.34) becomes u 1 5u = - In -f B K V (7.35) 464 Chapter 7 Flow Past Immersed Bodies But the definition of the skin friction coefficient, Eq. (7.10), is such that the following identities hold: u (7.36) Therefore, Eq. (7.35) is a skin friction law for turbulent fiat-plate flow: 9. 1/2 2.44 In Re^ 1/2 -\| + 5.0 (7.37) It is a complicated law, but we can at least solve for a few values and list them: Rea 10“ 10^ lO 10^ c/ 0.00493 0.00315 0.00217 0.00158 Following a suggestion of Prandtl, we can forget the complex log friction law (7.37) and simply fit the numbers in the table to a power-law approximation: 9=0.02Re5"® (7.38) This we shall use as the left-hand side of Eq. (7.33). For the right-hand side, we need an estimate for 9(x) in terms of d(x). If we use the logarithmic law profile (7.34), we shall be up to our hips in logarithmic integrations for the momentum thickness. Instead we follow another suggestion of Prandtl, who pointed out that the turbulent profiles in Fig. 7.5 can be approximated by a one-seventh-power law: (7.39) This is shown as a dashed line in Fig. 7.5. It is an excellent fit to the low-Reynolds- number turbulent data, which were all that were available to Prandtl at the time. With this simple approximation, the momentum thickness (7.28) can easily be evaluated: 9 = (7.40) We accept this result and substitute Eqs. (7.38) and (7.40) into Karman’s momentum law (7.33): .,-0.02 Re,-. 2 Rel'^® = 9.72^ = 9.72 dd dx d{Rss) /i(Re,) Separate the variables and integrate, assuming J = 0 at x = 0: (7.41) Rea =0.16 Ref or 6 ^ 0.16 X ~ Ref (7.42) Thus the thickness of a turbulent boundary layer increases as x^^’, far more rapidly than the laminar increase x^^. Equation (7.42) is the solution to the problem, because 7.4 The Flat-Plate Boundary Layer 465 all other parameters are now available. For example, combining Eqs. (7.42) and (7.38), we obtain the friction variation 9 0.027 Writing this out in dimensional form, we have (7.43) T,v,turb = - ^ - (7.44) X Turbulent plate friction drops slowly with x, increases nearly as p and if', and is rather insensitive to viscosity. We can evaluate the drag coefficient by integrating the wall friction: or Cd D = rL T„b dx 2D pU^bL 1 Cfd Cn 0.031 Ref 7 6 Cf (L) (7.45) Then Cd is only 16 percent greater than the trailing-edge skin friction coefficient [compare with Eq. (7.27) for laminar flow]. The displacement thickness can be estimated from the one-seventh-power law and Eq. (7.12): r4r /,.\l/7- 6 •’0 '-'5 The turbulent flat-plate shape factor is approximately H = = f= 1.3 (7.46) (7.47) These are the basic results of turbulent flat-plate theory. Eigure 7.6 shows flat-plate drag coefficients for both laminar and turbulent flow conditions. The smooth-wall relations (7.27) and (7.45) are shown, along with the effect of wall roughness, which is quite strong. The proper roughness parameter here is xle or Lie, by analogy with the pipe parameter eld. In the fully rough regime, Cd is independent of the Reynolds number, so that the drag varies exactly as if and is independent of p. Reference 2 presents a theory of rough flat-plate flow, and Ref. 1 gives a curve fit for skin friction and drag in the fully rough regime: 9= 2.87 -f 1.58 log -2.5 -2.5 Cd = ( 1.89 + 1.62 log - (7.48fl) {l.A%b) 466 Chapter 7 Flow Past Immersed Bodies Fig. 7.6 Drag coefficient of laminar and turbulent boundary layers on smooth and rough flat plates. This chart is the flat-plate analog of the Moody diagram of Fig. 6.13. 10^ 10® 10’ 10® 10’ Rsl Equation (7.48^;) is plotted to the right of the dashed line in Fig. 7.6. The figure also shows the behavior of the drag coefficient in the transition region 5 X 10^ < Re/_ < 8 X lO’, where the laminar drag at the leading edge is an appreciable fraction of the total drag. Schlichting suggests the following curve fits for these transition drag curves, depending on the Reynolds number Re^ Cn 0.031 1440 Re/. 0.031 8700 Ref Ref Re, •tians where transition begins: Retrans = 5 X 10^ (7.49a) Retrans = 3 X 10® (7.49/?) EXAMPLE 7.4 A hydrofoil 1.2 ft long and 6 ft wide is placed in a seawater flow of 40 ft/s, with p = 1.99 slugs/ft^ and u = 0.000011 ft’/s. (a) Estimate the boundary layer thickness at the end of the plate. Estimate the friction drag for (b) turbulent smooth-wall flow from the leading 7.4 The Flat-Plate Boundary Layer 467 edge, (c) laminar turbulent flow with Re,rans = 5 X 10^, and (d) turbulent rough-wall flow with e = 0.0004 ft. Solution Part (a) The Reynolds number is RSi = UL V (40 ft/s) (1.2 ft) 0.000011 ft^/s = 4.36 X 10® Thus the trailing-edge flow is certainly turbulent. The maximum boundary layer thickness would occur for turbulent flow starting at the leading edge. From Eq. (7.42), or 5{L-) L 0.16 (4.36 X 10®)"’ 0.018 5= 0.018(1.2 ft) = 0.0216 ft Ans. (a) This is 7.5 times thicker than a fully laminar boundary layer at the same Reynolds number. Part (b) Part (c) For fully turbulent smooth-wall flow, the drag coefficient on one side of the plate is, from Eq. (7.45), Cd 0.031 (4.36 X 10®)"’ 0.00349 Then the drag on both sides of the foil is approximately D = 2Co{\pU'^)bL = 2(0.00349)(^)(1.99)(40)’(6.0)(1.2) = 80 Ibf Ans. (b) With a laminar leading edge and Re,rans = 5 X 10®, Eq. (7.49a) applies: 1440 Cd = 0.00349 - 7 = 0.00316 4.36 X 10® The drag can be recomputed for this lower drag coefficient: D = ICoilpU^ybL = 72 Ibf Ans. (c) Part (d) Finally, for the rough wall, we calculate L £ 1.2 ft 0.0004 ft = 3000 From Fig. 7.6 at Re^ = 4.36 X 10®, this condition is just inside the fully rough regime. Equation (7.48/>) applies: Cd = (1.89 -f 1.62 log 3000)”’ ® = 0.00644 and the drag estimate is D = 2CD{\pU'^)bL = 148 Ibf Ans. (d) This small roughness nearly doubles the drag. It is probable that the total hydrofoil drag is still another factor of 2 larger because of trailing-edge flow separation effects. 468 Chapter 7 Flow Past Immersed Bodies 7.5 Boundary Layers with Pressure Gradient^ The flat-plate analysis of the previous section should give us a good feeling for the behavior of both laminar and turbulent boundary layers, except for one important effect: flow separation. Prandtl showed that separation like that in Fig. 1 .2b is caused by excessive momentum loss near the wall in a boundary layer trying to move down¬ stream against increasing pressure, dpidx > 0, which is called an adverse pressure gradient. The opposite case of decreasing pressure, dpidx < 0, is called & favorable gradient, where flow separation can never occur. In a typical immersed-body flow, such as in Fig. 1.2b, the favorable gradient is on the front of the body and the adverse gradient is in the rear, as discussed in detail in Chap. 8. We can explain flow separation with a geometric argument about the second deriv¬ ative of velocity u at the wall. From the momentum equation (1.19b) at the wall, where u = v = 0, we obtain or dr d^u dU dy wall dy wall n2 0 U 1 dp a/ wall M dx dp dx (7.50) for either laminar or turbulent flow. Thus in an adverse gradient the second deriva¬ tive of velocity is positive at the wall; yet it must be negative at the outer layer (y = (5) to merge smoothly with the mainstream flow U(x). It follows that the second derivative must pass through zero somewhere in between, at a point of inflection, and any boundary layer profile in an adverse gradient must exhibit a characteristic S shape. Figure 7.7 illustrates the general case. In a favorable gradient (Fig. 1.1a) the profile is very rounded, there is no point of inflection, there can be no separa¬ tion, and laminar profiles of this type are very resistant to a transition to turbu¬ lence [1 to 3]. In a zero pressure gradient (Fig. 1.1b), such as a flat-plate flow, the point of inflec¬ tion is at the wall itself. There can be no separation, and the flow will undergo transi¬ tion at Re^j no greater than about 3 X 10®, as discussed earlier. In an adverse gradient (Fig. 1.1c to e), a point of inflection (PI) occurs in the boundary layer, its distance from the wall increasing with the strength of the adverse gradient. For a weak gradient (Fig. 1.1c) the flow does not actually sepa¬ rate, but it is vulnerable to transition to turbulence at Re;^ as low as 10® [1, 2]. At a moderate gradient, a critical condition (Fig. 1 .Id) is reached where the wall shear is exactly zero (duldy = 0). This is defined as fhe separation point (t„ = 0), because any sfronger gradient will actually cause backflow at the wall (Fig. 1 .le)\ the boundary layer thickens greatly, and the main flow breaks away, or separates, from the wall (Fig. 1 .2b). The flow profiles of Fig. 7.7 usually occur in sequence as the boundary layer progresses along the wall of a body. For example, in Fig. 1 .2a, a favorable gradient occurs on the front of the body, zero pressure gradient occurs just upstream of the shoulder, and an adverse gradient occurs successively as we move around the rear of the body. ^This section may be omitted without loss of continuity. 7.5 Boundary Layers with Pressure Gradient 469 u PI U U U Favorable (b) Zero gradient: gradient: i^>0 ^ = 0 dx dx <0 dp dx dx Mo separation, Mo sepai'ation PI inside wall PI at wall dp dx >0- PI ,T„, = 0 PI ' Backflow Fig. 7.7 Effect of pressure gradient on boundary layer profiles; PI = point of inflection. (c) Weak adverse (d) Critical adverse {e) Excessive a gradient: gradient: gradient: dU<0 Zero slope Backflow dx at the wall: at the wall: dp Separation Separated flow region No separation, PI in the flow A second practical example is the flow in a duct consisting of a nozzle, throat, and diffuser, as in Fig. 7.8. The nozzle flow is a favorable gradient and never separates, nor does the throat flow where the pressure gradient is approximately zero. But the expanding-area diffuser produces low velocity and increasing pres¬ sure, an adverse gradient. If the diffuser angle is too large, the adverse gradient is excessive, and the boundary layer will separate at one or both walls, with backflow. 470 Chapter 7 Flow Past Immersed Bodies Fig. 7.8 Boundary layer growth and separation in a nozzle-diffuser configuration. Sepai'ation point Nearly inviscid core flow Nozzle-. Throat: Dijfuser: Decreasing Constant Increasing pressure pressure and area pressure and area and area Increasing velocity Velocity constant Decreasing velocity Favorable Zero Adverse gradient gradient gradient (boundary layer thickens) increased losses, and poor pressure recovery. In the diffuser literature this condition is called diffuser stall, a term used also in airfoil aerodynamics (Sec. 7.6) to denote airfoil boundary layer separation. Thus the boundary layer behavior explains why a large-angle diffuser has heavy flow losses (Fig. 6.23) and poor performance (Fig. 6.28). Presently boundary layer theory can compute only up to the separation point, after which it is invalid. Techniques are now developed for analyzing the strong interaction effects caused by separated flows [5, 6]. Laminar Integral Theory' Both laminar and turbulent theories can be developed from Karman’s general two- dimensional boundary layer integral relation [2, 7], which extends Eq. (7.33) to vari¬ able U{x) by integration across the boundary layer: I de e du pu¬ ll dx (7.51) This section may be omitted without loss of continuity. c;\| = 7.5 Boundary Layers with Pressure Gradient 471 where 6(x) is the momentum thickness and H(x) = 5(x)l9{x) is the shape factor. From Eq. (7.17) negative dUldx is equivalent to positive dpidx — that is, an adverse gradient. We can integrate Eq. (7.51) to determine 8{x) for a given U{x) if we correlate Cf and H with the momentum thickness. This has been done by examining typical veloc¬ ity profiles of laminar and turbulent boundary layer flows for various pressure gradi¬ ents. Some examples are given in Fig. 7.9, showing that the shape factor // is a good indicator of the pressure gradient. The higher the H, the stronger the adverse gradient, and separation occurs approximately at f 3.5 laminar flow //= (7.52) [ 2.4 turbulent flow The laminar profiles (Fig. 1.9a) clearly exhibit the S shape and a point of inflection with an adverse gradient. But in the turbulent profiles (Fig. 1.9b) the points of inflec¬ tion are typically buried deep within the thin viscous sublayer, which can hardly be seen on the scale of the figure. There are scores of turbulent theories in the literature, but they are all compli¬ cated algebraically and will be omitted here. The reader is referred to advanced texts [1-3, 9]. y 6 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 y s (a) {b) Fig. 7.9 Velocity profiles with pressure gradient: (a) laminar flow; (b) turbulent flow with adverse gradients. 472 Chapter 7 Flow Past Immersed Bodies For laminar flow, a simple and effective method was developed by Thwaites , who found that Eq. (7.51) can be correlated by a single dimensionless momentum thickness variable A, defined as V dx (7.53) Using a straight-line fit to his correlation, Thwaites was able to integrate Eq. (7.51) in closed form, with the result + 0.451/ if' fdx (7.54) where 9q, is the momentum thickness at x = 0 (usually taken to be zero). Separation (Cf = 0) was found to occur at a particular value of A; Separation; A = -0.09 (7.55) Finally, Thwaites correlated values of the dimensionless shear stress S = T„9l{iiU) with A, and his graphed result can be curve-fitted as follows: 5(A) = — = (A + 0.09)“® (7.56) /if/ This parameter is related to the skin friction by the identity 5 - ic/Ree (7.57) Equations (7.54) to (7.56) constitute a complete theory for the laminar boundary layer with variable U{x), with an accuracy of ±10 percent compared with computer solu¬ tions of the laminar-boundary-layer equations (7.19). Complete details of Thwaites’ s and other laminar theories are given in Ref. 2. As a demonstration of Thwaites’s method, take a flat plate, where U = constant, X = 0, and 9q = 0. Equation (7.54) integrates to or 9^ = 9 _ x 0.45 i/x U 0.671 Ref This is within 1 percent of Blasius’s numerical solution, Eq. (7.30). With X = 0, Eq. (7.56) predicts the flat-plate shear to be (7.58) or — = (0.09)°® = 0.225 /if/ _ 2t^ _ 0.671 pU^ ~ Ref (7.59) This is also within 1 percent of the Blasius result, Eq. (7.25). However, the general accuracy of this method is poorer than 1 percent because Thwaites actually “tuned” his correlation constants to make them agree with exact flat-plate theory. We shall not compute any more boundary layer details here; but as we go along, investigating various immersed-body flows, especially in Chap. 8, we shall 7.5 Boundary Layers with Pressure Gradient 473 use Thwaites’s method to make qualitative assessments of the boundary layer behavior. EXAMPLE 7.5 In 1938 Howarth proposed a linearly decelerating external velocity distribution U{x) = f/o (1) as a theoretical model for laminar-boundary-layer study, {a) Use Thwaites’s method to compute the separation point for 0o = 0, and compare with the exact computer solution JCsep/L = 0.119863 given by H. Wipperman in 1966. (b) Also compute the value of Cf = iTjipU^) at xlL = 0.1. Part (a) Part (b) Solution First note that dUldx = — UgfL = constant: Velocity decreases, pressure increases, and the pressure gradient is adverse throughout. Now integrate Eq. (7.54): (f = 0.45v (7^(1 -xILf Jo (7^1 l---]dx = 0.075^^ (2) Then the dimensionless factor X is given by A = (PdU V dx i/L = -0.075 1 - - 1 (3) From Eq. (7.55) we set this equal to —0.09 for separation: Asep = -0.09 = -0.075 1 - - 1 — = 1 - (2.2)“'® = 0.123 L Arts, (a) This is less than 3 percent higher than Wipperman’ s exact solution, and the computational effort is very modest. To compute at .r/L = 0.1 (just before separation), we first compute X at this point, using Eq. (3): Mx = O.IL) = -0.075[(1 - 0.1)“® - 1] = -0.0661 Then from Eq. (7.56) the shear parameter is S(x = O.IL) = (-0.0661 -f 0.09)°“ = 0.099 = jC/Rep (4) We can compute Rep in terms of Re/, from Eq. (2) or (3): ^ _ 0.0661 _ 0.0661 L^ ~ UUv ~ Re/, X at - = 0.1 L or Rep = 0.257 Re}'^ 474 Chapter 7 Flow Past Immersed Bodies 7.6 Experimental External Flows Drag of Immersed Bodies Substitute into Eq. (4): 0.099 = \c0.251 Rel'^) 0.77 UL Ans. (b) We cannot actually compute Cf without the value of, say, U^iLlv. Boundary layer theory is very interesting and illuminating and gives us a great qualita¬ tive grasp of viscous flow behavior; but, because of flow separation, the theory does not generally allow a quantitative computation of the complete flow field. In particu¬ lar, there is at present no satisfactory theory, except CFD results, for the forces on an arbitrary body immersed in a stream flowing at an arbitrary Reynolds number. There¬ fore, experimentation is the key to treating external flows. Literally thousands of papers in the literature report experimental data on specific external viscous flows. This section gives a brief description of the following external flow problems: 1. Drag of two- and three-dimensional bodies: a. Blunt bodies. b. Streamlined shapes. 2. Performance of lifting bodies: a. Airfoils and aircraft. b. Projectiles and finned bodies. c. Birds and insects. For further reading see the goldmine of data compiled in Hoerner . In later chapters we shall study data on supersonic airfoils (Chap. 9), open-channel friction (Chap. 10), and turbomachinery performance (Chap. 11). Any body of any shape when immersed in a fluid stream will experience forces and moments from the flow. If the body has arbitrary shape and orientation, the flow will exert forces and moments about all three coordinate axes, as shown in Fig. 7.10. It is customary to choose one axis parallel to the free stream and positive downstream. The force on the body along this axis is called drag, and the moment about that axis the rolling moment. The drag is essentially a flow loss and must be overcome if the body is to move against the stream. A second and very important force is perpendicular to the drag and usually per¬ forms a useful job, such as bearing the weight of the body. It is called the lift. The moment about the lift axis is called yaw. The third component, neither a loss nor a gain, is the side force, and about this axis is the pitching moment. To deal with this three-dimensional force-moment situ¬ ation is more properly the role of a textbook on aerodynamics [for example, 13]. We shall limit the discussion here to lift and drag. 7.6 Experimental External Flows 475 Fig. 7.10 Definition of forces and moments on a body immersed in a uniform flow. When the body has symmetry about the lift-drag axis, as with airplanes, ships, and cars moving directly into a stream, the side force, yaw, and roll vanish, and the prob¬ lem reduces to a two-dimensional case: two forces, lift and drag, and one moment, pitch. A final simplification often occurs when the body has two planes of symmetry, as in Fig. 7.11. A wide variety of shapes such as cylinders, wings, and all bodies of revolution satisfy this requirement. If the free stream is parallel to the intersection of these two planes, called the principal chord line of the body, the body experiences drag only, with no lift, side force, or moments.^ This type of degenerate one-force drag data is what is most commonly reported in the literature, but if the free stream is not parallel to the chord line, the body will have an unsymmetric orientation and all three forces and three moments can arise in principle. In low-speed flow past geometrically similar bodies with identical orientation and relative roughness, the drag coefficient should be a function of the body Reynolds number: Co=/(Re) (7.60) Fig. 7.11 Only the drag force occurs if the flow is parallel to both planes of symmetry. ^In bodies with shed vortices, such as the cylinder in Fig. 5.2, there may be oscillating lift, side force, and moments, but their mean value is zero. 476 Chapter 7 Flow Past Immersed Bodies Characteristic Area Friction Drag and Pressure Drag The Reynolds number is based upon the free-stream velocity V and a characteristic length L of the body, usually the chord or body length parallel to the stream: VL Re = — (7.61) V For cylinders, spheres, and disks, the characteristic length is the diameter D. Drag coefficients are defined by using a characteristic area A, which may differ depending on the body shape: (7.62) The factor \ is our traditional tribute to Euler and Bernoulli. The area A is usually one of three types: 1 . Frontal area, the body as seen from the stream; suitable for thick, stubby bodies, such as spheres, cylinders, cars, trucks, missiles, projectiles, and torpedoes. 2. Planform area, the body area as seen from above; suitable for wide, flat bodies such as wings and hydrofoils. 3. Wetted area, customary for surface ships and barges. In using drag or other fluid force data, it is important to note what length and area are being used to scale the measured coefficients. As we have mentioned, the theory of drag is weak and inadequate, except for the flat plate. This is because of flow separation. Boundary layer theory can predict the sepa¬ ration point but cannot accurately estimate the (usually low) pressure distribution in the separated region. The difference between the high pressure in the front stagnation region and the low pressure in the rear separated region causes a large drag contribu¬ tion called pressure drag. This is added to the integrated shear stress or friction drag of the body, which it often exceeds: Cd ~ Q),press + C'/jfric (7.63) The relative contribution of friction and pressure drag depends upon the body’s shape, especially its thickness. Figure 7.12 shows drag data for a streamlined cyl¬ inder of very large depth into the paper. At zero thickness the body is a flat plate and exhibits 100 percent friction drag. At thickness equal to the chord length, simu¬ lating a circular cylinder, the friction drag is only about 3 percent. Friction and pressure drag are about equal at thickness tic = 0.25. Note that in Fig. 1 .\2b looks quite different when based on frontal area instead of planform area, planform being the usual choice for this body shape. The two curves in Fig. 1 .\2h represent exactly the same drag data. 7.6 Experimental External Flows 477 Fig. 7.12 Drag of a streamlined two- dimensional cylinder at Re^ = 10®: (a) effect of thickness ratio on percentage of friction drag; (b) total drag versus thickness when based on two different areas. Figure 7.13 illustrates the dramatic effect of separated flow and the subsequent failure of boundary layer theory. The theoretical inviscid pressure distribution on a circular cylinder (Chap. 8) is shown as the dashed line in Fig. 7.13c: Cp = = 1-4 sin^6l where and V are the pressure and velocity, respectively, in the free stream. The actual laminar and turbulent boundary layer pressure distributions in Fig. 7.13c are startlingly different from those predicted by theory. Laminar flow is very vulnerable to the adverse gradient on the rear of the cylinder, and separation occurs at 0 = 82°, which certainly could not have been predicted from inviscid theory. The broad wake and very low pressure in the separated laminar region cause the large drag Co = 1.2. The turbulent boundary layer in Fig. 7.13b is more resistant, and separation is delayed until 6 = 120°, with a resulting smaller wake, higher pressure on the rear, and 75 percent less drag, Co = 0.3. This explains the sharp drop in drag at transition in Fig. 5.3. 478 Chapter 7 Flow Past Immersed Bodies Fig. 7.13 Flow past a circular cylinder: {a) laminar separation; {b) turbulent separation; (c) theoretical and actual surface pressure distributions. The same sharp difference between vulnerable laminar separation and resistant tur¬ bulent separation can be seen for a sphere in Fig. 7.14. The laminar flow (Fig. 7.14a) separates at about 80°, Cq = 0.5, while the turbulent flow (Fig. lAAb) separates at 120°, Co = 0.2. Here the Reynolds numbers are exactly the same, and the turbulent boundary layer is induced by a patch of sand roughness at the nose of the ball. Golf balls fly in this range of Reynolds numbers, which is why they are deliberately dimpled — to induce a turbulent boundary layer and lower drag. Again we would find the actual pressure distribution on the sphere to be quite different from that predicted by inviscid theory. In general, we cannot overstress the importance of body streamlining to reduce drag at Reynolds numbers above about 100. This is illustrated in Fig. 7.15. The rectangular cylinder (Fig. 7.15a) has separation at all sharp corners and very high drag. Rounding its nose (Fig. 7.151?) reduces drag by about 45 percent, but Co is still high. Streamlin¬ ing its rear to a sharp trailing edge (Fig. 7.15c) reduces its drag another 85 percent to a practical minimum for the given thickness. As a dramatic contrast, the circular cylinder (Fig. 7.15r/) has one-eighth the thickness and one-three-hundredth the cross 7.6 Experimental External Flows 479 Fig. 7.14 Strong differences in laminar and turbulent separation on an 8.5-in bowling ball entering water at 25 ft/s: (a) smooth ball, laminar boundary layer; (b) same entry, turbulent flow induced by patch of nose-sand roughness. (NAVAIR Weapons Division Historical Archives.) Two-Dimensional Bodies Creeping Flow Fig. 7.15 The importance of streamlining in reducing drag of a body (Cd based on frontal area): (a) rectangular cylinder; (b) rounded nose; (c) rounded nose and streamlined sharp trailing edge; (d) circular cylinder with the same drag as case (c). (a) (b) section (c) (Fig. 7.15c), yet it has the same drag. For high-performance vehicles and other moving bodies, the name of the game is drag reduction, for which intense research continues for both aerodynamic and hydrodynamic applications [20, 39]. The drag of some representative wide-span (nearly two-dimensional) bodies is shown versus the Reynolds number in Fig. 7.16a. All bodies have high C^, at very low {creeping flow) Re < 1.0, while they spread apart at high Reynolds numbers according to their degree of streamlining. All values of Co are based on the planform area except the plate normal to the flow. The birds and the sailplane are, of course, not very two- dimensional, having only modest span length. Note that birds are not nearly as effi¬ cient as modern sailplanes or airfoils [14, 15]. In 1851 G. G. Stokes showed that, if the Reynolds number is very small. Re ^ 1, the acceleration terms in the Navier-Stokes equations {1 .\Ab, c) are negligible. The flow is termed creeping flow, or Stokes flow, and is a balance between pressure (c) {d) 480 Chapter 7 Flow Past Immersed Bodies Re (a) Fig. 7.16 Drag coefficients of smooth bodies at low Mach numbers: (a) two-dimensional bodies; (b) three-dimensional bodies. Note the Reynolds number independence of blunt bodies at high Re. (b) gradient and viscous stresses. Continuity and momentum reduce to two linear equa¬ tions for velocity and pressure: Re « 1: V-V = 0 and Vp = /iV^V If the geometry is simple (for example, a sphere or disk), closed-form solutions can be found and the body drag can be computed . Stokes himself provided the sphere drag formula: or ^sphere 37V ^Ud F _ 24 _ 24 ^ pUd/fi Rerf (7.64) This relation is plotted in Fig. 7.16b and is seen to be accurate for about Re^ < 1. 7.6 Experimental External Flows 481 Table 7.2 gives a few data on drag, based on frontal area, of two-dimensional bodies of various cross section, at Re > 10“. The sharp-edged bodies, which tend to cause flow separation regardless of the character of the boundary layer, are insensitive to the Reynolds number. The elliptic cylinders, being smoothly rounded, have the Table 7.2 Drag of Two- Dimensional Bodies at Re S 10“ Cd based Cd based Co based on frontal on frontal on frontal Shape area Shape area Shape area Square cylinder: Half cylinder: Plate: Thin plate normal to a wall: - ► 1.4 Hexagon: Shape Rounded nose section: H Cd based on frontal area L/H: 0.5 1.0 2.0 4.0 6.0 Cd- 1.16 0.90 0.70 0.68 0.64 Flat nose section: L/H: 0.1 1 0.4 0.7 1.2 2.0 2.5 3.0 6.0 H Cd- 1.9 1 2.3 2.7 2.1 1.8 1.4 1.3 0.9 Elliptical cylinder: Laminar Turbulent 1:1 - - o 1.2 0.3 2:1 - - o 0.6 0.2 4:1 - c ^ 0.35 0.15 8:1 - — ► cr ^ 0.25 0.1 482 Chapter 7 Flow Past Immersed Bodies Three-Dimensional Bodies laminar-to-turbulent transition effect of Figs. 7.13 and 7.14 and are therefore quite sensitive to whether the boundary layer is laminar or turbulent. EXAMPLE 7.6 A square 6-in piling is acted on by a water flow of 5 ft/s that is 20 ft deep, as shown in Fig. E7.6. Estimate the maximum bending exerted by the flow on the bottom of the piling. E7.6 Solution Assume seawater with p = 1.99 slugs/ft^ and kinematic viscosity u = 0.000011 fF/s. With a piling width of 0.5 ft, we have Re (5 ft/s) (0.5 ft) ^ ^ = 2.3 X 10- 0.000011 ft-/s This is the range where Table 7.2 applies. The worst case occurs when the flow strikes the flat side of the piling, Co ~ 2.1. The frontal area is A = Lh = (20 ft)(0.5 ft) = 10 ft^. The drag is estimated by F = Cd{\pV^A) «2.1(^)(1.99 slugs/ft^)(5ft/s)^10fF) = 522 Ibf If the flow is uniform, the center of this force should be at approximately middepth. There¬ fore, the bottom bending moment is Mo« FL 2 = 522(10) = 5220 ft • Ibf Ans. According to the flexure formula from strength of materials, the bending stress at the bottom would be S = Mgy I (5220 ft - lb) (0.25 ft) il(0.5 ft)" = 251,000 Ibf/fF = 17401bf/in^ to be multiplied, of course, by the stress concentration factor due to the built-in end conditions. Some drag coefficients of three-dimensional bodies are listed in Table 7.3 and Fig. lAbb. Again we can conclude that sharp edges always cause flow separation and high drag 7.6 Experimental External Flows 483 Table 7.3 Drag of Three-Dimensional Bodies at Re > 10"^ Co based on Body frontal area Body Co based on frontal area Cup: 1.4 0.4 Disk: - 1.17 Parachute (Low porosity): 1.2 Streamlined train (approximately 5 cars): Upright: Cj^A = 0.51 m^; Racing: Cj^A = 0.30 m^ Short cylinder, laminar flow: L 9\ 10° 20° 30° 40° 60° 75° 90° Cd- 0.30 0.40 0.55 0.65 0.80 1.05 1.15 UD-. 1 2 3 5 10 20 1 40 CO C^: 0.64 0.68 0.72 0.74 0.82 0.91 1 0.98 1.20 D Porous pai'abolic dish : ' Porosity: 0 0.1 0.2 0.3 0.4 0.5 ^ — ^D- 1.42 1.33 1.20 1.05 0.95 0.82 0.95 0.92 0.90 0.86 0.83 0.80 Average person: CflA = 9 ft^ :1.2 ft Ff2 Pine and spruce trees : - ► U, m/s: 1 20 30 40 Cd- 1 1.2 + 0.2 1.0 + 0.2 0.7 + 0.2 0.5 + 0.2 Without deflector: 0.96; with deflector: 0.76 Co based on Body Ratio frontal area Body Co based on Ratio frontal area Rectangular plate: Ellipsoid: b/h 1 1.18 Flat-faced cylinder: h 10 1.3 b 20 oo 1.5 2.0 Laminar Turbulent L/d 0.75 1 2 4 0.5 0.47 0.27 0.25 0.2 0.2 0.2 0.13 0.1 0.08 Buoyant rising sphere , Ud 0.5 1 2 4 Cn « 0.95 1.15 0.90 0.85 0.87 0.99 L 135 < Re^ < 1E5 484 Chapter 7 Flow Past Immersed Bodies Buoyant Rising Light Spheres Aerodynamic Forces on Road Vehicles that is insensitive to the Reynolds number. Rounded bodies like the ellipsoid have drag that depends on the point of separation, so both the Reynolds number and the character of the boundary layer are important. Body length will generally decrease pressure drag by making the body relatively more slender, but sooner or later the friction drag will catch up. For the flat-faced cylinder in Table 7.3, pressure drag decreases with Lid but friction increases, so minimum drag occurs at about Ud = 2. The sphere data in Fig. l.lbb are for fixed models in wind tunnels and from falling sphere tests and indicate a drag coefficient of about 0.5 in the range 1 E3 < Re^ < 1 E5. It was pointed out that this is not the case for a freely rising buoyant sphere or bubble. If the sphere is light, ^sphere < 0.8 jOfluid, a wake instability arises in the range 135 < Re^ < 1 E5. The sphere then spirals upward at an angle of about 60° from the horizontal. The drag coefficient is approximately doubled, to an average value Co ~ 0.95, as listed in Table 7.3 . For a heavier body, Pspheie ~ Pfluid. the buoyant sphere rises vertically and the drag coefficient follows the standard curve in Fig. I.l6b. EXAMPLE 7.7 According to Ref. 12, the drag coefficient of a blimp, based on surface area, is approximately 0.006 if Re^ > 10®. A certain blimp is 75 m long and has a surface area of 3400 m^. Esti¬ mate the power required to propel this blimp at 18 m/s at a standard altitude of 1000 m. Solution ■ Assumptions: We hope the Reynolds number will be high enough that the given data are valid. ■ Approach: Determine if Re^ >10® and, if so, compute the drag and the power required. ■ Property values: Table A.6 at z = 1000 m: p = 1.112 kg/m^, T = 282 K, thus p = 1.75 E-5 kg/m ■ s. • Solution steps: Determine the Reynolds number of the blimp; Re, = pUL (1.112 kg/m°)(18 m/s)(75 m) P 1.75 E-5 kg/m ■ s = 8.6 E7 > 10® OK The given drag coefficient is valid. Compute the blimp drag and the power = (drag) X (velocity): p , 1.112kg/m'^ - , F = Co^ U^A^a = (0.006) - - (18 m/s)^ (3400 m^) = 3675 N Power = FV= (3675 N)(18 m/s) = 66,000 W (89 hp) Ans. • Comments: These are nominal estimates. Drag is highly dependent on both shape and Reynolds number, and the coefficient Co = 0.006 has considerable uncertainty. Automobiles and trucks are now the subject of much research on aerodynamic forces, both lift and drag . At least one textbook is devoted to the subject . A very readable description of race car drag is given by Katz . Consumer, manufacturer, and government interest has cycled between high speed/high horsepower and lower 7.6 Experimental External Flows 485 speed/lower drag. Better streamlining of car shapes has resulted over the years in a large decrease in the automobile drag coefficient, as shown in Fig. 7.17a. Modern cars have an average drag coefficient of about 0.25, based on the frontal area. Since the frontal area has also decreased sharply, the actual raw drag force on cars has dropped even more than indicated in Fig. 7.17o. The theoretical minimum shown in the figure, Co ~ 0.15, is about right for a commercial automobile, but lower values are possible for experimental vehicles; see Prob. P7.109. Note that basing Co on the frontal area is awkward, since one would need an accurate drawing of the automobile to estimate its frontal area. For this reason, some technical articles simply report the raw drag in newtons or pound-force, or the product C^A. Many companies and laboratories have automotive wind tunnels, some full- scale and/or with moving floors to approximate actual kinematic similarity. The blunt shapes of most automobiles, together with their proximity to the ground, cause a wide variety of flow and geometric effects. Simple changes in part of the shape can have (a) Fig. 7.17 Aerodynamics of automobiles: (a) the historical trend for drag coefficients (from Ref. 21); (b) effect of bottom rear upsweep angle on drag and downward lift force (from Ref. 25). 486 Chapter 7 Flow Past Immersed Bodies a large influence on aerodynamic forces. Figure 1 Alb shows force data by Bearman et al. for an idealized smooth automobile shape with upsweep in the rear of the bottom section. We see that by simply adding an upsweep angle of 25°, we can qua¬ druple the downward force, gaining tire traction at the expense of doubling the drag. For this study, the effect of a moving floor was small — about a 10 percent increase in both drag and lift compared to a fixed floor. It is difficult to quantify the exact effect of geometric changes on automotive forces, since, for example, changes in a windshield shape might interact with down¬ stream flow over the roof and trunk. Nevertheless, based on correlation of many model and full-scale tests. Ref. 26 proposes a formula for automobile drag that adds separate effects such as front ends, cowls, fenders, windshield, roofs, and rear ends. Figure 7.1 So illustrates the power required to drive a typical tractor-trailer truck. An approximation is that the rolling resistance increases linearly and the drag qua- dratically with speed. The two are about equal at 55 mi/h. Figure 7.18f> shows that air drag can be reduced by attaching a deflector to the top of the cab. When the deflector angle is adjusted to carry the flow smoothly over the top and sides of the trailer, the reduction in Cq is about 20 percent. This type of applied fluids engineering is very important for modern transportation problems . The velocity effect on rolling resistance is mostly due to the engine-transmission-wheel- bearing system. The tires generally have a nearly constant rolling resistance coefficient, where F„. is the resistance force and N is the normal force on the tires . This coefficient is analogous to a solid friction factor but is much smaller: about 0.01 to 0.04 for passenger car tires and 0.006 to 0.01 for truck tires. Progress in computational fluid dynamics means that complicated vehicle flow fields can be predicted fairly well. Reference 42 compares one-equation and two-equation turbulence models with NASA data for a simplified tractor-trailer model. Even with two million mesh points, the predicted vehicle drag is from 20 to 50 percent higher than the measurements. The turbulence models do not reproduce the pressures and wake Fig. 7.18 Drag reduction of a tractor-trailer truck: (a) horsepower required to overcome resistance; (b) deflector added to cab reduces air drag by 20 percent. & 250 I 200 ■3 400 o % 300 o - 350 550 500 150 100 50 0 0 10 20 30 40 50 60 70 80 Vehicle speed, mi/h (a) 7.6 Experimental External Flows 487 structure in the rear of the vehicle. Newer models, such as large eddy simulation (LES) and direct numerical simulation (DNS) will no doubt improve the calculations. EXAMPLE 7.8 A high-speed car with m = 2000 kg, Co = 0.3, and A = 1 deploys a 2-m parachute to slow down from an initial velocity of 100 m/s (Fig. E7.8). Assuming constant Co, brakes free, and no rolling resistance, calculate the distance and velocity of the car after 1, 10, 100, and 1000 s. For air assume p = 1.2 kg/m^, and neglect interference between the wake of the car and the parachute. dp = 2m Vo= too m/s Solution Newton’s law applied in the direction of motion gives dV 1 -^ = -Fc - Fp= --pV {CdcA, + CopAp) where subscript c denotes the car and subscript p the parachute. This is of the form dt m K= ^CoA^ Separate the variables and integrate: 'dV dt K /Q - V - - t m Rearrange and solve for the velocity V\ Vo V = V„-‘ - = -- r {CocAc + Ci}pAp)p K = 1 + {K/m)Vot We can integrate this to find the distance traveled: Vn S = — In (1 + at) a K a = - Vo m Now work out some numbers. From Table 7.3, Cop ~ 1.2; hence CdcA, + CopAp = 0.3(1 m^) -f 1.2 y (2 m)^ = 4.07 m^ Then K i(4.07m2)(1.2kg/m^)(100tn/s) 2000 kg ' Vo = = 0.122 s"' = a (1) (2) Now make a table of the results for V and S from Eqs. (1) and (2): 488 Chapter 7 Flow Past Immersed Bodies Other Methods of Drag Reduction t, s 1 to too 1000 V, m/s 89 45 7.6 0.8 S, m 94 654 2110 3940 Air resistance alone will not stop a body completely. If you don’t apply the brakes, you'll be halfway to the Yukon Territory and still going. Sometimes drag is good, for example, when using a parachute. Do not jump out of an airplane holding a flat plate parallel to your motion (see Prob. P7.81). Mostly, though, drag is bad and should be reduced. The classical method of drag reduction is streamlining (Figs. 7.15 and 7.18). For example, nose fairings and body panels have produced motorcycles that can travel over 200 mi/h. More recent research has uncovered other methods that hold great promise, especially for turbulent flows. 1. Oil pipelines introduce an annular strip of water to reduce the pumping power . The low-viscosity water rides the wall and reduces friction up to 60 percent. 2. Turbulent friction in liquid flows is reduced up to 60 percent by dissolving small amounts of a high-molecular-weight polymer additive . Without changing pumps, the Trans-Alaska Pipeline System (TAPS) increased oil flow 50 percent by injecting small amounts of polymer dissolved in kerosene. 3. Stream-oriented surface vee-groove microriblets can reduce turbulent friction up to 8 percent . Riblet heights are of order 1 mm and were used on the Stars and Stripes yacht hull in the Americas Cup races. Riblets are also effective on aircraft skins. 4. Small, near-wall large-eddy breakup devices (LEBUs) reduce local turbulent friction up to 10 percent . However, one must add these small structures to the surface, and LEBU drag may be significant. 5. Air microbubbles injected at the wall of a water flow create a low-shear bubble blanket . At high void fractions, drag reduction can be 80 percent. 6. Spanwise (transverse) wall oscillation may reduce turbulent friction up to 30 percent . 7. Active flow control, especially of turbulent flows, is the wave of the future, as reviewed in Ref. 47. These methods generally require expenditure of energy but can be worth it. Eor example, tangential blowing at the rear of an auto evokes the Coanda ejfect, in which the separated near-wake flow attaches itself to the body surface and reduces auto drag up to 10 percent. Drag reduction is presently an area of intense and fruitful research and applies to many types of airflows [39, 53] and water flows for both vehicles and conduits. Drag of Surface Ships The drag data given so far, such as Tables 7.2 and 7.3, are for bodies “fully immersed” in a free stream — that is, with no free surface. If, however, the body moves at or near 7.6 Experimental External Flows 489 Body Drag at High Mach Numbers a free liquid surface, wave-making drag becomes important and is dependent on both the Reynolds number and the Fronde number. To move through a water surface, a ship must create waves on both sides. This implies putting energy into the water surface and requires a finite drag force to keep the ship moving, even in a frictionless fluid. The total drag of a ship can then be approximated as the sum of friction drag and wave-making drag; F ^fhc ^wave C)r Cq f^n.fric ^D.wave The friction drag can be estimated by the (turbulent) flat-plate formula, Eq. (7.45), based on the below-water or wetted area of the ship. Reference 27 is an interesting review of both theory and experiment for wake¬ making surface ship drag. Generally speaking, the bow of the ship creates a wave system whose wavelength is related to the ship speed but not necessarily to the ship length. If the stern of the ship is a wave trough, the ship is essentially climbing uphill and has high wave drag. If the stern is a wave crest, the ship is nearly level and has lower drag. The criterion for these two conditions results in certain approximate Froude numbers : ^ y 0.53 highdragifA? = 1, 3, 5, 7, ... ; Fr = — 1= ~ — 1= (7.65) Vn low drag if iV = 2, 4, 6, 8, . . . where V is the ship’s speed, L is the ship’s length along the centerline, and N is the number of half-lengths, from bow to stern, of the drag-making wave system. The wave drag will increase with the Froude number and oscillate between lower drag (Fr ~ 0.38, 0.27, 0.22, . . .) and higher drag (Fr ~ 0.53, 0.31, 0.24, . . .) with neg¬ ligible variation for Fr < 0.2. Thus it is best to design a ship to cruise at N = 2, 4, 6, 8. Shaping the bow and stern can further reduce wave-making drag. Figure 7.19 shows the data of Inui for a model ship. The main hull, curve A, shows peaks and valleys in wave drag at the appropriate Froude numbers > 0.2. Introduction of a bu/b protrusion on the bow, curve B, greatly reduces the drag. Add¬ ing a second bulb to the stern, curve C, is still better, and Inui recommends that the design speed of this two-bulb ship be at A = 4, Fr ~ 0.27, which is a nearly “wave¬ less” condition. In this figure Cq ^^ave i® defined as 2F instead of using the wetted area. The solid curves in Fig. 7.19 are based on potential flow theory for the below-water hull shape. Chapter 8 is an introduction to potential flow theory. Modern computers can be programmed for numerical CFD solutions of potential flow over the hulls of ships, submarines, yachts, and sailboats, including boundary layer effects driven by the potential flow . Thus theoretical prediction of flow past surface ships is now at a fairly high level. See also Ref. 15. All the data presented to this point are for nearly incompressible flows, with Mach numbers assumed less than about 0.3. Beyond this value compressibility can be very important, with C^, = fcn(Re, Ma). As the stream Mach number increases, at some subsonic value < 1 that depends on the body’s bluntness and thickness, the local velocity at some point near the body surface will become sonic. If Ma increases beyond Mac-jt, shock waves form, intensify, and spread, raising surface 490 Chapter 7 Flow Past Immersed Bodies Fig. 7.19 Wave-making drag on a ship model. {After Iniii .) Note: The drag coefficient is dehned as Cdw = 2F/{pV^L^). o A Main hull (without bulb) pressures near the front of the body and therefore increasing the pressure drag. The effect can he dramatic with Co increasing tenfold, and 70 years ago this sharp increase was called the sonic barrier, implying that it could not he surmounted. Of course, it can he — the rise in Co is finite, as supersonic bullets have proved for centuries. Figure 7.20 shows the effect of the Mach number on the drag coefficient of various body shapes tested in air.® We see that compressibility affects blunt bodies earlier, with Macrit equal to 0.4 for cylinders, 0.6 for spheres, and 0.7 for airfoils and pointed projectiles. Also the Reynolds number (laminar versus turbulent boundary layer flow) has a large effect below Macnt for spheres and cylinders but becomes unimportant above Ma ~ 1. In contrast, the effect of the Reynolds number is small for airfoils and projectiles and is not shown in Fig. 7.20. A general statement might divide Reynolds and Mach number effects as follows: Ma < 0.3: Reynolds number important, Mach number unimportant 0.3 < Ma < 1 : both Reynolds and Mach numbers important Ma > 1 .0: Reynolds number unimportant, Mach number important At supersonic speeds, a broad bow shock wave forms in front of the body (see Figs. 9.106 and 9.19), and the drag is mainly due to high shock-induced pressures on the front. Making the bow a sharp point can sharply reduce the drag (Fig. 9.28) but ^There is a slight effect of the specihc-heat ratio k, which would appear if other gases were tested. 7.6 Experimental External Flows 491 Fig. 7.20 Effect of the Mach number on the drag of various body shapes. (Data from Refs. 23 and 29.) Biological Drag Reduction 2.0 1.8 1.6 1.4 1.2 Cd 1.0 0.8 0.6 0.4 0.2 0.0 0.0 1.0 2.0 3.0 4.0 Mach number Cy inder in cross flow: — — Laminar, Re ss 1 E5 Turbulent. Re « 1 E6 — Sphere ^ Laminar, Re ss 1 E5 Turbulent, Re 1 E6 Pointed body of revolution 77 Airfoil M 1 1 1 1 1 1 1 1 1 1 1 1 1 does not eliminate the bow shock. Chapter 9 gives a brief treatment of compressible flow. References 30 and 31 are more advanced textbooks devoted entirely to com¬ pressible flow. A great deal of engineering effort goes into designing immersed bodies to reduce their drag. Most such effort concentrates on rigid-body shapes. A different process occurs in nature, as organisms adapt to survive high winds or currents, as reported in a series of papers by S. Vogel [33, 34]. A good example is a tree, whose flexible structure allows it to reconfigure in high winds and thus reduce drag and damage. Tree root systems have evolved in several ways to resist wind-induced bending moments, and trunk cross sections have become resistant to bending but relatively easy to twist and reconfigure. We saw this in Table 7.3, where tree drag coefficients reduced by 60 percent as wind velocity increased. The shape of the tree changes to offer less resistance. The individual branches and leaves of a tree also curl and cluster to reduce drag. Figure 7.21 shows the results of wind tunnel experiments by Vogel . A tulip tree leaf. Fig. 7.21fl, broad and open in low wind, curls into a conical low-drag shape as wind increases. A compound black walnut leaf group. Fig. 1.2lb, clusters into a low- drag shape at high wind speed. Although drag coefficients were reduced up to 50 percent by flexibility, Vogel points out that rigid structures are sometimes just as effective. An interesting recent symposium was devoted entirely to the solid mechanics and fluid mechanics of biological organisms. 492 Chapter 7 Flow Past Immersed Bodies Fig. 7.21 Biological adaptation to wind forces: (a) a tulip tree leaf curls into a conical shape at high velocity; (&) hlack walnut leaves cluster into a low-drag shape as wind increases. (From Vogel. Ref. 33.) Forces on Lifting Bodies Fig. 7.22 Definition sketch for a lifting vane. Lifting bodies (airfoils, hydrofoils, or vanes) are intended to provide a large force normal to the free stream and as little drag as possible. Conventional design practice has evolved a shape not unlike a bird’s wing — that is, relatively thin (f/c ^ 0.24) with a rounded leading edge and a sharp trailing edge. A typical shape is sketched in Fig. 7.22. For our purposes we consider the body to be symmetric, as in Fig. 7.11, with the free-stream velocity in the vertical plane. If the chord line between the leading and trailing edge is not a line of symmetry, the airfoil is said to be cambered. The camber line is the line midway between the upper and lower surfaces of the vane. The angle between the free stream and the chord line is called the angle of attack a. The lift L and the drag D vary with this angle. The dimensionless forces are defined Planform 7.6 Experimental External Flows 493 with respect to the planform area Ap = be. Lift coefficient: (7.66a) Drag coefficient: (7.66^) Fig. 7.23 Transient stages in the development of lift: (a) start-up: rear stagnation point on the upper surface: no lift; (b) sharp trailing edge induces separation, and a starting vortex forms: slight lift; (c) starting vortex is shed; and streamlines flow smoothly from trailing edge: lift is now 80 percent developed; (d) starting vortex now shed far behind, trailing edge now very smooth: lift fully developed. If the chord length is not constant, as in the tapered wings of modem aircraft, Ap = jc db. For low-speed flow with a given roughness ratio, and Cp should vary with a and the chord Reynolds number: Cl = fia, ReJ or Co = f(a, Re^) where Re^ = Vc/u. The Reynolds numbers are commonly in the turbulent boundary layer range and have a modest effect. The rounded leading edge prevents flow separation there, but the sharp trailing edge causes a tangential wake motion that generates the lift. Figure 7.23 shows what happens when a flow starts up past a lifting vane or an airfoil. Just after start-up in Fig. 7.23a the streamline motion is irrotational and inviscid. The rear stagnation point, assuming a positive angle of attack, is on the upper surface, and there is no lift; but the flow cannot long negotiate the sharp turn at the trailing edge: it separates, and a starting vortex forms in Fig. 7.23b. This starting vortex is shed downstream in Figs. 7.23c and d, and a smooth streamline flow develops over the wing, leaving the foil in a direction approximately parallel to the chord line. Lift at this time is fully developed, and the starting vortex is gone. Should the flow now cease, a stopping vortex of opposite (clockwise) sense will form and be shed. During flight, increases or decreases in lift will cause incremental starting or stopping vortices. Fig. 7.23 Transient stages in the development of lift: (a) start-up: rear stagnation point on the upper surface: no lift; (b) sharp trailing edge induces separation, and a starting vortex forms: slight lift; (c) starting vortex is shed; and streamlines flow smoothly from trailing edge: lift is now 80 percent developed; (d) starting vortex now shed far behind, trailing edge now very smooth: lift fully developed. 494 Chapter 7 Flow Past Immersed Bodies always with the effect of maintaining a smooth parallel flow at the trailing edge. We pursue this idea mathematically in Chap. 8. At a low angle of attack, the rear surfaces have an adverse pressure gradient but not enough to cause significant boundary layer separation. The flow pattern is smooth, as in Fig. 1.23d, and drag is small and lift excellent. As the angle of attack is increased, the upper-surface adverse gradient becomes stronger, and generally a separation bub¬ ble begins to creep forward on the upper surface.^ At a certain angle a = 15 to 20°, the flow is separated completely from the upper surface, as in Fig. 7.24. The airfoil is said to be stalled'. Lift drops off markedly, drag increases markedly, and the foil is no longer flyable. Early airfoils were thin, modeled after birds’ wings. The German engineer Otto Lilienthal (1848-1896) experimented with flat and cambered plates on a rotating arm. He and his brother Gustav flew the world’s first glider in 1891. Horatio Frederick Phillips (1845-1912) built the first wind tunnel in 1884 and measured the lift and drag of cambered vanes. The first theory of lift was proposed by Frederick W. Lanchester shortly afterward. Modern airfoil theory dates from 1905, when the Russian hydro- dynamicist N. E. Joukowsky (1847-1921) developed a circulation theorem (Chap. 8) for computing airfoil lift for arbitrary camber and thickness. With this basic theory, as extended and developed by Prandtl and Karman and their students, it is now pos¬ sible to design a low-speed airfoil to satisfy particular surface pressure distributions and Fig. 7.24 At high angle of attack, smoke flow visualization shows stalled flow on the upper surface of a lifting vane. {National Committee for Fluid Mechanics Films, Education Development Center, Inc., © 1972.) ’For some airfoils the bubble leaps, not creeps, forward, and stall occurs rapidly and dangerously. 7.6 Experimental External Flows 495 boundary layer characteristics. There are whole families of airfoil designs, notably those developed in the United States under the sponsorship of the NACA (now NASA). Extensive theory and data on these airfoils are contained in Ref. 16. We shall discuss this further in Chap. 8. The history of aeronautics is a rich and engaging topic and highly recommended to the reader [43, 44]. Figure 7.25 shows the lift and drag on a symmetric airfoil denoted as the NACA 0009 foil, the last digit indicating the thickness of 9 percent. With no flap extended, this airfoil, as expected, has zero lift at zero angle of attack. Up to about 12° the lift coefficient increases linearly with a slope of 0.1 per degree, or 6.0 per radian. This is in agreement with the theory outlined in Chap. 8: CL,theory ~ 27r smi o + — 1 (7.67) where hic is the maximum camber expressed as a fraction of the chord. The NACA 0009 has zero camber; hence = Itt sin a ~ 0.1 let, where a is in degrees. This is excellent agreement. The drag coefficient of the smooth-model airfoils in Fig. 7.25 is as low as 0.005, which is actually lower than both sides of a flat plate in turbulent flow. This is mis¬ leading inasmuch as a commercial foil will have roughness effects; for example, a paint job will double the drag coefficient. The effect of increasing Reynolds number in Fig. 7.25 is to increase the maximum lift and stall angle (without changing the slope appreciably) and to reduce the drag coefficient. This is a salutary effect since the prototype will probably be at a higher Reynolds number than the model (10^ or more). For takeoff and landing, the lift is greatly increased by deflecting a split flap, as shown in Fig. 7.25. This makes the airfoil unsymmetric (or effectively cambered) Fig. 7.25 Lift and drag of a symmetric NACA 0009 airfoil of infinite span, including effect of a split-flap deflection. Note that roughness can increase Co from too to 300 percent. Cz,» 496 Chapter 7 Flow Past Immersed Bodies and changes the zero-lift point to a = —12°. The drag is also greatly increased by the flap, but the reduction in takeoff and landing distance is worth the extra power needed. A lifting craft cruises at low angle of attack, where the lift is much larger than the drag. Maximum lift-to-drag ratios for the common airfoils lie between 20 and 50. Some airfoils, such as the NACA 6 series, are shaped to provide favorable gradients over much of the upper surface at low angles. Thus separation is small, and transition to turbulence is delayed; the airfoil retains a good length of laminar flow even at high Reynolds numbers. The lift-drag polar plot in Fig. 7.26 shows the NACA 0009 data from Fig. 7.25 and a laminar flow airfoil, NACA 63-009, of the same thickness. The laminar flow airfoil has a low-drag bucket at small angles but also suffers lower stall angle and lower maximum lift coefficient. The drag is 30 percent less in the bucket, but the bucket disappears if there is significant surface roughness. All the data in Figs. 7.25 and 7.26 are for infinite span — that is, a two-dimensional flow pattern about wings without tips. The effect of finite span can be correlated with the dimensionless slenderness, or aspect ratio, denoted (AR): (7.68) where c is the average chord length. Finite-span effects are shown in Fig. 7.27. The lift slope decreases, but the zero-lift angle is the same; and the drag increases, but the zero-lift drag is the same. The theory of finite-span airfoils predicts that the effective angle of attack increases, as in Fig. 7.27, by the amount Aa ~ Cl ttAR (7.69) Fig. 7.26 Lift-drag polar plot for standard (0009) and a laminar flow (63-009) NACA airfoil. 0 0.008 0.024 7.6 Experimental External Flows 497 Fig. 7.27 Effect of finite aspect ratio on lift and drag of an airfoil: (a) effective angle increase; (b) induced drag increase. (a) Cd (b) When applied to Eq. (1 .61), the finite-span lift becomes Cz.« 2tt sin (a + 2h/c) 1 -I- 2/AR (7.70) The associated drag increase is AC^, ~ sin Act ~ Aa, or Cn — Cn Cj ttAR (7.71) where Cooo is the drag of the infinite-span airfoil, as sketched in Fig. 7.25. These correlations are in good agreement with experiments on finite-span wings . The existence of a maximum lift coefficient implies the existence of a minimum speed, or stall speed, for a craft whose lift supports its weight: or L = W= C, AkpylAp) V, = 2iy (7.72) The stall speed of typical aircraft varies between 60 and 200 ft/s, depending on the weight and value of C/^max- The pilot must hold the speed greater than about 1.217 to avoid the instability associated with complete stall. The split flap in Fig. 7.25 is only one of many devices used to secure high lift at low speeds. Figure 1 .2%a shows six such devices whose lift performance is given in Fig. 1 .2%b along with a standard {A) and laminar flow {B) airfoil. The double- slotted flap achieves C^max ~ 3.4, and a combination of this plus a leading-edge slat can achieve C^niax ~ 4.0. These are not scientific curiosities; for instance, the Boeing 727 commercial jet aircraft uses a triple-slotted flap plus a leading-edge slat during landing. A violation of conventional aerodynamic wisdom is that military aircraft are begin¬ ning to fly, briefly, above the stall point. Fighter pilots are learning to make quick 498 Chapter 7 Flow Past Immersed Bodies Fig. 7.28 Performance of airfoils with and without high-lift devices: A = NACA 0009; B = NACA 63-009; C = Kline-Fogleman airfoil {from Ref. 17), D to I shown in (a): (a) types of high-lift devices; (b) lift coefficients for various devices (Ref. 62). The Kline-Fogelman Airfoil A Wing Inspired by the Humpback Whale Plain flap or aileron d Split flap d External airfoil flap c : Slotted flap c r Double- slotted flap :::::::: Leading-edge slat (a) (b) maneuvers in the stalled region as detailed in Ref. 32. Some planes can even fly con¬ tinuously while stalled — the Grumman X-29 experimental aircraft recently set a record by flying at a = 67°. Traditionally, an airfoil is a thin teardrop shape, with a rounded leading edge and a sharp trailing edge, Fig. 7.28a. It provides low drag but stalls at low a ~ 10 to 15°. In 1972 R. F. Kline and F. F. Fogelman designed an airfoil with a sharp leading edge and a rear cut-out . When tested in a wind tunnel. Fig. 1.28b, it did not stall until a ~ 45°, but the drag was very high. Fertis rounded the leading edge and reduced the drag. Finaish and Witherspoon made further improvements, but the drag is still too high for full-scale commercial applications. The KF airfoil, though, is extremely popular for radio-controlled model aircraft. Biologists have long noticed the high maneuverability of the humpback whale when it seeks prey. Unlike most whales, the humpback has tubercles, or bumps, on the leading edge of its flippers. Miklosovic et al. tested this idea, using a standard wing with periodic bumps glued to its leading edge, as in Fig. 7.29. They report a 40 percent increase in stall angle, compared to the same wing without bumps, plus higher lift and higher lift-to-drag ratios. The concept has promise for commercial applications such as wind turbine blades. Flow visualization shows that the bumps create energetic streamwise vortices along the wing surface, helping to delay separation. 7.6 Experimental External Flows 499 Fig. 7.29 New experimental airfoils: plan view of a wing modeled on the humpback whale flipper , Source: D. S. Miklosovic etal, “Leading Edge Tubercles Delay Stall on Humpback Whale, ” Physics of Fluids, vol. 16, no. 5, May 2004, pp. U9-L42. A Combination Car and Airplane Engineers have long dreamed of a viable car that can fly. Hop in at home, drive to the airport, fly somewhere, then drive to the motel. Designer efforts date back to the Glenn Curtiss 1917 Autoplane, with other projects in the 1930s and 1940s. Perhaps the most famous was Moulton Taylor’s Aerocar in 1947. Only five Aerocars were built. The year 2008 seems to have been the Year of the Car-Plane, with at least five different companies working on designs. Engineers can now use lightweight materials, better engines, and guidance systems. The writer’s favorite is the Transition®, made by Terrafugia, Inc., shown in Eig. 7.30. The Transition® has wings that fold out to a span of 27.5 ft. The twin tails do not fold. The front canard wing doubles as a bumper for the highway. The 100-hp engine drives both the rear propeller for flight and also the front wheels for the highway. Gross takeoff weight is 1430 Ibf. Operators need only a Light Sport Airplane license. The Transition® made a successful maiden flight on Mar. 5, 2009. This craft’s data will clearly be useful for setting end-of-chapter problems. Fig. 7.30 The Transition® car-plane in flight on March 23, 2012. It has a gross weight of 1,430 Ibf and a cruise velocity of 105 mi/h. (Image from Terrafugia, Inc., http:// www.terrafugia.com) 500 Chapter 7 Flow Past Immersed Bodies Further information on the performance of lifting craft can be found in Refs. 12, 13, and 16. We discuss this matter again briefly in Chap. 8. EXAMPLE 7.9 An aircraft weighs 75,000 Ihf, has a planform area of 2500 ft", and can deliver a constant thrust of 12,000 Ih. It has an aspect ratio of 7, and ~ 0.02. Neglecting rolling resistance, estimate the takeoff distance at sea level if takeoff speed equals 1.2 times stall speed. Take = 2.0. Solution The stall speed from Eq. (7.72), with sea-level density p = 0.00237 slug/ft^, is 1/2 V, = / 2W 2(75,000) \ ^L,m3ixP^pJ .2.0(0.00237) (2500). = 112.5 ft/s Hence takeoff speed Vg = 1.21^ = 135 ft/s. The drag is estimated from Eq. (7.71) for AR = 7 as Co « 0.02 + — = 0.02 -f 0.0455Ci In A force balance in the direction of takeoff gives Fj = ni— = thrust — drag = T — kV^ k = \CopAp (1) Since we are looking for distance, not time, we introduce dVIdt = V dV/ds into Eq. (1), separate variables, and integrate: or m r° d{V^) 2 I T- kV^ T m In - y = /rr T - kVl 2k k ~ const (2) where Dg = kVg is the takeoff drag. Equation (2) is the desired theoretical relation for takeoff distance. Eor the particular numerical values, take 75,000 m = - = 2329 slugs 32.2 ® ^ _ W _ _ 75,000 _ _ ^ ^ iPVlAp ^(0.00237) (135)^(2500) Co„ = 0.02 -f 0.0455 (Ci„)^ = 0.108 k « I Co^pAp = (I) (0.108) (0.00237) (2500) = 0.319 slug/ft Dg = kVl = 5820 Ibf Then Eq. (2) predicts that 2329 slugs _ 12,000 2(0.319 slug/ft) ” 12,000 - 5820 3650 In 1.94 = 2420 ft Ans. A more exact analysis accounting for variable k gives the same result to within 1 percent. Summary 501 EXAMPLE 7.10 For the aircraft of Example 7.9, if maximum thrust is applied during flight at 6000 m stan¬ dard altitude, estimate the resulting velocity of the plane, in mi/h. Solution ■ Assumptions: Given W = 75,000 Ihf, Ap = 2500 ft^ T = 12,000 Ihf, AR = 7, Cb„ = 0.02. • Approach: Set lift equal to weight and drag equal to thrust and solve for the velocity. • Property values: From Table A.6, at z = 6000 m, p = 0.6596 kg/m^ = 0.00128 slug/ft^. • Solution steps: Write out the formulas for lift and drag. The unknowns will he Q, and V. p , 0.00128 slug/fF , W = 75,000 Ihf = lift = = Cl - - ^^(2500 ft^) T = 12,000 Ihf = drag = 0.02 + d ] 0.00128 slug/ft^ 2 V^(2500 ft-) Some clever manipulation (dividing W hy T) would reveal a quadratic equation for Cl- The final solution is Cl = 0.13 V « 600 ft/s = 410 mi/h Ans. • Comments: These are preliminary design estimates that do not depend on airfoil shape. Summary This chapter has dealt with viscous effects in external flow past bodies immersed in a stream. When the Reynolds number is large, viscous forces are confined to a thin boundary layer and wake in the vicinity of the body. Flow outside these “shear lay¬ ers” is essentially inviscid and can be predicted by potential theory and Bernoulli’s equation. The chapter began with a discussion of the flat-plate boundary layer and the use of momentum integral estimates to predict the wall shear, friction drag, and thickness of such layers. These approximations suggest how to eliminate certain small terms in the Navier-Stokes equations, resulting in Prandtl’s boundary layer equations for laminar and turbulent flow. Section 7.4 then solved the boundary layer equations to give very accurate formulas for flat-plate flow at high Reynolds numbers. Rough-wall effects were included, and Sec. 7.5 gave a brief introduction to pressure gradient effects. An adverse (decelerating) gradient was seen to cause flow separa¬ tion, where the boundary layer breaks away from the surface and forms a broad, low- pressure wake. Boundary layer theory fails in separated flows, which are commonly studied by experiment or CFD. Section 7.6 gave data on drag coefficients of various two- and three-dimensional body shapes. The chapter ended with a brief discussion of lift forces generated by lifting bodies such as airfoils and hydrofoils. Airfoils also suffer flow separation or stall at high angles of incidence. 502 Chapter 7 Flow Past Immersed Bodies Problems Most of the problems herein are fairly straightforward. More diffi¬ cult or open-ended assignments are labeled with an asterisk. Prob¬ lems labeled with a computer icon may require the use of a computer. The standard end-of-chapter problems P7.1 to P7.127 (categorized in the problem list here) are followed by word prob¬ lems W7.1 to W7.12, fundamentals of engineering exam problems FE7.1 to FE7.10, comprehensive problems C7.1 to C7.5, and design project D7.1. Problem Distribution Section Topic Problems 7.1 Reynolds number and geometry P7.1-P7.5 7.2 Momentum integral estimates P7.6-P7.12 7.3 The boundary layer equations P7.13-P7.15 lA Laminar flat-plate flow P7.16-P7.29 lA Turbulent flat-plate flow P7.30-P7.47 7.5 Boundaiy layers with pressure gradient P7.48-P7.50 7.6 Drag of bodies P7.51-P7.114 7.6 Lifting bodies — airfoils P7.115-P7.127 Reynolds number and geometry P7.1 An ideal gas, at 20°C and 1 atm, flows at 12 m/s past a thin flat plate. At a position 60 cm downstream of the leading edge, the boundary layer thickness is 5 mm. Which of the 13 gases in Table A.4 is this likely to be? P7.2 A gas at 20°C and 1 atm flows at 6 ft/s past a thin flat plate. At X = 3 ft, the boundary layer thickness is 0.052 ft. Assuming laminar flow, which of the gases in Table A.4 is this likely to be? P7.3 Equation (1 Ab) assumes that the boundary layer on the plate is turbulent from the leading edge onward. Devise a scheme for determining the boundary layer thickness more accurately when the flow is laminar up to a point RSjr.crit turbulent thereafter. Apply this scheme to computation of the boundary layer thickness at x = 1.5 m in 40 m/s flow of air at 20°C and 1 atm past a flat plate. Compare your result with Eq. {1 Ab). Assume Re.v^crit ~ 1.2 E6. P7.4 A smooth ceramic sphere (SG = 2.6) is immersed in a flow of water at 20°C and 25 cm/s. What is the sphere diameter if it is encountering (a) creeping motion, Re^ = 1 or (b) transition to turbulence, Re^ = 250,000? P7.5 SAE 30 oil at 20°C flows at 1.8 ftVs from a reservoir into a 6-in-diameter pipe. Use flat-plate theory to estimate the position X where the pipe wall boundary layers meet in the center. Compare with Eq. (6.5), and give some explana¬ tions for the discrepancy. Momentum integral estimates P7.6 For the laminar parabolic boundary layer profile of Eq. (7.6), compute the shape factor H and compare with the exact Blasius result, Eq. (7.31). P7.7 Air at 20°C and 1 atm enters a 40-cm-square duct as in Fig. P7.7. Using the “displacement thickness” concept of Fig. 7.4, estimate (a) the mean velocity and (b) the mean pressure in the core of the flow at the position x = 3 m. (c) What is the average gradient, in Pa/m, in this section? P7.8 Air, p = 1.2 kg/m^ and p, = 1.8 E-5 kg/(m ■ s), flows at 10 m/s past a flat plate. At the trailing edge of the plate, the following velocity profile data are measured: y, mm 0 0.5 1.0 2.0 3.0 4.0 5.0 6.0 u, m/s 0 1.75 3.47 6.58 8.70 9.68 10.0 10.0 If the upper surface has an area of 0.6 m^, estimate, using momentum concepts, the friction drag, in N, on the upper surface. P7.9 Repeat the flat-plate momentum analysis of Sec. 7.2 by replacing Eq. (7.6) with the simple but unrealistic linear velocity profile suggested by Schlichting : u y for O^y^c^ Compute momentum-integral estimates of Cf, Olx, 5lx, and H. P7.10 Repeat Prob. P7.9, using a trigonometric profile approxi¬ mation: u . f Try — ~ sin — r U \2d Does this profile satisfy the conditions of laminar flat-plate flow? P7.ll Air at 20°C and 1 atm flows at 2 m/s past a sharp flat plate. Assuming that Roman’s parabolic-profile analysis, Eqs. (7.6-7.10), is accurate, estimate (a) the local velocity u and (b) the local shear stress t at the position (x, y) = (50 cm, 5 mm). Problems 503 P7.12 The velocity profile shape ulU « 1 — exp (-4.605y/(^ is a smooth curve with m = 0 at y = 0 and u = Q.99U at y = ^ and thus would seem to be a reasonable substitute for the parabolic flat-plate profile of Eq. (7.3). Yet when this new profile is used in the integral analysis of Sec. 7.3, we get the lousy result 6lx ~ 9.2/Re]^^, which is 80 percent high. What is the reason for the inaccuracy? [Hint: The answer lies in evaluating the laminar boundary layer momentum equation (7.19fe) at the wall, y = 0.] The boundary layer equations P7.13 Derive modified forms of the laminar boundary layer equations (7.19) for the case of axisymmetric flow along the outside of a circular cylinder of constant radius R, as in Fig. P7.13. Consider the two special cases (a) 6 < R and ib) 6~R. What are the proper boundary conditions? V P7.14 Show that the two-dimensional laminar flow pattern with dpidx = 0 u = (/o(l - e'^) v= Vo < 0 P7.18 Air at 20°C and 1 atm flows at 5 m/s past a flat plate. At jc = 60 cm and y = 2.95 mm, use the Blasius solution. Table 7.1, to find (a) the velocity u; and (b) the wall shear stress, (c) For extra credit, find a Blasius formula for the shear stress away from the wall. P7.19 Air at 20°C and 1 atm flows at 50 ft/s past a thin flat plate whose area (bL) is 24 ft^. If the total friction drag is 0.3 Ibf, what are the length and width of the plate? P7.20 Air at 20°C and 1 atm flows at 20 m/s past the flat plate in Fig. P7.20. A pitot stagnation tube, placed 2 mm from the wall, develops a manometer head h = 16 mm of Me- riam red oil, SG = 0.827. Use this information to esti¬ mate the downstream position x of the pitot tube. Assume laminar flow. P7.21 For the experimental setup of Fig. P7.20, suppose the stream velocity is unknown and the pitot stagnation tube is traversed across the boundary layer of air at 1 atm and 20°C. The manometer fluid is Meriam red oil, and the fol¬ lowing readings are made: is an exact solution to the boundary layer equations (7.19). Find the value of the constant C in terms of the flow param¬ eters. Are the boundary conditions satisfied? What might this flow represent? P7.15 Discuss whether fully developed laminar incompressible flow between parallel plates, Eq. (4.134) and Fig. 4. 14fc, represents an exact solution to the boundary layer equa¬ tions (7.19) and the boundary conditions (7.20). In what sense, if any, are duct flows also boundary layer flows? V, mm 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 h, mm 1.2 4.6 9.8 15.8 21.2 25.3 27.8 29.0 29.7 29.7 Using these data only (not the Blasius theory) estimate (a) the stream velocity, (b) the boundary layer thickness, (c) the wall shear stress, and (d) the total friction drag between the leading edge and the position of the pitot tube. P7.22 In the Blasius equation (7.22), / is a dimensionless plane stream function: Laminar flat-plate flow P7.16 A thin flat plate 55 by 1 10 cm is immersed in a 6-m/s stream of SAE 10 oil at 20°C. Compute the total Ifiction drag if the stream is parallel to (a) the long side and (b) the short side. P7.17 Consider laminar flow past a sharp flat plate of width b and length L. What percentage of the friction drag on the plate is carried by the rear half of the plate? m) ipix, y) Vi^ Values of /are not given in Table 7.1, but one published value is /(2.0) = 0.6500. Consider airflow at 6 m/s, 20°C, and 1 atm past a flat plate. At v = 1 m, estimate (a) the height y; (b) the velocity, and (c) the stream function atr) = 2.0. 504 Chapter 7 Flow Past Immersed Bodies P7.23 Suppose you buy a 4- by 8-ft sheet of plywood and put it on your roof rack. (See Fig. P7.23.) You drive home at 35 mi/h. (a) Assuming the board is perfectly aligned with the airflow, how thick is the boundary layer at the end of the board? (b) Estimate the drag on the sheet of plywood if the boundary layer remains laminar, (c) Esti¬ mate the drag on the sheet of plywood if the boundary layer is turbulent (assume the wood is smooth), and compare the result to that of the laminar boundary layer case. Square duct L = 8 mm d- V=24 m/s 1 m P7.25 P7.23 P7.24 Air at 20°C and 1 atm flows past the flat plate in Fig. P7.24 under laminar conditions. There are two equally spaced pitot stagnation tubes, each placed 2 mm from the wall. The manometer fluid is water at 20°C. If U = 15 m/s and 1 3 2 4 (a) L = 50 cm, determine the values of the manometer read- 1 2 3 4 ings hi and /i2, in mm. - ^ P7.25 Consider the smooth square lO-cm-by-lO-cm duct in Fig. P7.25. The fluid is air at 20°C and 1 atm, flowing at Favg “ 24 m/s. It is desired to increase the pressure drop over the 1-m length by adding sharp 8-mm-long flat plates across the duct, as shown, (a) Estimate the pres¬ sure drop if there are no plates, (b) Estimate how many plates are needed to generate an additional 100 Pa of pressure drop. P7.26 Consider laminar boundary layer flow past the square-plate arrangements in Fig. P7.26. Compared to the friction drag of a single plate 1, how much larger is the drag of four plates together as in configurations (a) and (/))? Explain your results. P7.26 P7.27 Air at 20°C and 1 atm flows at 3 m/s past a sharp flat plate 2 m wide and 1 m long, (a) What is the wall shear stress at the end of the plate? (b) What is the air velocity at a point 4.5 mm normal to the end of the plate? (c) What is the total friction drag on the plate? P7.28 Elow straighteners are arrays of narrow ducts placed in wind tunnels to remove swirl and other in-plane secondary velocities. They can be idealized as square boxes con¬ structed by vertical and horizontal plates, as in Eig. P7.28. The cross section is a by a, and the box length is L. Assum¬ ing laminar flat-plate flow and an array of N X N boxes, derive a formula for (a) the total drag on the bundle of boxes and (b) the effective pressure drop across the bundle. P7.29 Let the flow straighteners in Eig. P7.28 form an array of 20 X 20 boxes of size a = 4 cm and L = 25 cm. If the Problems 505 approach velocity is C/q = 12 m/s and the fluid is sea-level standard air, estimate (a) the total array drag and (b) the pressure drop across the array. Compare with Sec. 6.8. Turbulent flat-plate flow P7.30 In Ref. 56 of Chap. 6, McKeon et al. propose new, more accurate values for the turbulent log-law constants, K = 0.421 and B = 5.62. Use these constants, and the one- seventh power-law, to repeat the analysis that led to the formula for turbulent boundary layer thickness, Eq. (7.42). By what percent is 6/x in your new formula different from that in Eq. (7.42)? Comment. P7.31 The centerboard on a sailboat is 3 ft long parallel to the flow and protrudes 7 ft down below the hull into sea¬ water at 20°C. Using flat-plate theory for a smooth sur¬ face, estimate its drag if the boat moves at 10 knots. Assume Re^. ,;. = 5 E5. P7.32 A flat plate of length L and height S is placed at a wall and is parallel to an approaching boundary layer, as in Fig. P7.32. Assume that the flow over the plate is fully turbulent and that the approaching flow is a one-seventh-power law: u(y) = Uo(^^ Using strip theory, derive a formula for the drag coefficient of this plate. Compare this result with the drag of the same plate immersed in a uniform stream Uq. y P7.33 An alternate analysis of turbulent flat-plate flow was given by Prandtl in 1927, using a wall shear stress formula from pipe flow: T, = 0.0225pU^(j^ Show that this formula can be combined with Eqs. (7.33) and (7.40) to derive the following relations for turbulent flat-plate flow: 6 _ 0.37 _ 0.0577 _ 0.072 ^ “ Rep ■ Ref ^ These formulas are limited to Re^^ between 5 X 10^ and 10^. P7.34 Consider turbulent flow past a sharp, smooth flat plate of width b and length L. What percentage of the friction drag on the plate is carried by the rear half of the plate? P7.35 Water at 20°C flows at 5 m/s past a 2-m-wide sharp flat plate, (a) Estimate the boundary layer thickness at v = 1.2 m. (b) If the total drag (on both sides of the plate) is 310 N, estimate the length of the plate using, for simplicity, Eq. (7.45). P7.36 A ship is 125 m long and has a wetted area of 3500 m^. Its propellers can deliver a maximum power of 1.1 MW to seawater at 20°C. If all drag is due to friction, estimate the maximum ship speed, in kn. P7.37 Air at 20°C and 1 atm flows past a long flat plate, at the end of which is placed a narrow scoop, as shown in Fig. P7.37. (a) Estimate the height h of the scoop if it is to extract 4 kg/s per meter of width into the paper, (b) Find the drag on the plate up to the inlet of the scoop, per meter of width. 30 m/s P7.38 Atmospheric boundary layers are very thick but follow formulas very similar to those of flat-plate theory. Consider wind blowing at 10 m/s at a height of 80 m above a smooth beach. Estimate the wind shear stress, in Pa, on the beach if the air is standard sea-level conditions. What will the wind velocity striking your nose be if (a) you are standing up and your nose is 170 cm off the ground and (b) you are lying on the beach and your nose is 17 cm off the ground? P7.39 A hydrofoil 50 cm long and 4 m wide moves at 28 kn in seawater at 20°C. Using flat-plate theory with Re,j = 5 E5, estimate its drag, in N, for (a) a smooth wall and (b) a rough wall, £ = 0.3 mm. P7.40 Hoerner [12, p. 3.25] states that the drag coefficient of a flag in winds, based on total wetted area 2bL, is approxi¬ mated by Co ~ 0.01 + 0.05L/b, where L is the flag length in the flow direction. Test Reynolds numbers Re^ were 1 E6 or greater, (a) Explain why, for Lib a 1, these drag values are much higher than for a flat plate. Assuming sea-level standard air at 50 mi/h, with area bL = A m^, find (b) the proper flag dimensions for which the total drag is approximately 400 N. 506 Chapter 7 Flow Past Immersed Bodies P7.41 Repeat Prob. P7.20 with the sole change that the pitot probe is now 10 mm from the wall (5 times higher). Show that the flow there cannot possibly be laminar, and use smooth-wall turbulent flow theory to estimate the position X of the probe, in m. P7.42 A light aircraft flies at 30 m/s in air at 20°C and 1 atm. Its wing is an NACA 0009 airfoil, with a chord length of 150 cm and a very wide span (neglect aspect ratio effects). Estimate the drag of this wing, per unit span length, (a) by flat plate theory and (b) using the data from Fig. 7.25 for a = 0°. P7.43 In the flow of air at 20°C and 1 atm past a flat plate in Fig. P7.43, the wall shear is to be determined at position x by a floating element (a small area connected to a strain-gage force measurement). At x = 2 m, the element indicates a shear stress of 2. 1 Pa. Assuming turbulent flow from the lead¬ ing edge, estimate (a) the stream velocity U, (b) the boundary layer thickness <5 at the element, and (c) the boundary layer velocity u, in m/s, at 5 mm above the element. P7.43 P7.44 Extensive measurements of wall shear stress and local velocity for turbulent airflow on the flat surface of the University of Rhode Island wind tunnel have led to the following proposed correlation: py'T„ 0.0207 Thus, if y and u(y) are known at a point in a flat-plate boundary layer, the wall shear may be computed directly. If the answer to part (c) of Prob. P7.43 is u ~ 26.3 m/s, determine the shear stress and compare with Prob. P7.43. Discuss. P7.45 A thin sheet of fiberboard weighs 90 N and lies on a roof¬ top, as shown in Fig. P7.45. Assume ambient air at 20°C and 1 atm. If the coefficient of solid friction between board and roof is cr « 0.12, what wind velocity will generate enough fluid friction to dislodge the board? P7.46 A ship is 150 m long and has a wetted area of 5000 m^. If it is encrusted with barnacles, the ship requires 7000 hp to overcome friction drag when moving in seawater at 1.5 m 2 m P7.45 15 kn and 20°C. What is the average roughness of the barnacles? How fast would the ship move with the same power if the surface were smooth? Neglect wave drag. P7.47 Local boundary layer effects, such as shear stress and heat transfer, are best correlated with local variables, rather using distance x from the leading edge. The momentum thickness 9 is often used as a length scale. Use the analysis of turbulent flat-plate flow to write local wall shear stress in terms of dimensionless 0 and compare with the formula recom¬ mended by Schlichting : Cf ~ 0.033 Re^i”’'^®. Boundary layers with pressure gradient P7.48 In 1957 H. Gdrtler proposed the adverse gradient test cases (1 -f x/L)" and computed separation for laminar flow at n = 1 to be Xsep/L = 0.159. Compare with Thwaites’s method, assum¬ ing 6»o = 0. P7.49 Based strictly on your understanding of flat-plate theory plus adverse and favorable pressure gradients, explain the direction (left or right) for which airflow past the slender airfoil shape in Fig. P7.49 will have lower total (friction -I- pressure) drag. P7.49 P7.50 Consider the flat-walled diffuser in Fig. P7.50, which is similar to that of Fig. 6.26a with constant width b. If x is measured from the inlet and the wall boundary layers are thin, show that the core velocity U{x) in the diffuser is given approximately by 1 -f (2x tan 9)IW Problems 507 where W is the inlet height. Use this velocity distribution with Thwaites’s method to compute the wall angle 0 for which laminar separation will occur in the exit plane when diffuser length L = 2W. Note that the result is independent of the Reynolds number. P7.56 A delivery vehicle carries a long sign on top, as in Fig. P7.56. If the sign is very thin and the vehicle moves at 65 mi/h, (a) estimate the force on the sign with no cross wind and (b) discuss the effect of a crosswind. Constant width b Drag of bodies P7.51 A 2-cm-diameter solid metal sphere falls steadily at about 1 m/s in 20°C fresh water. If we use Table 7.3 for a drag estimate, is the sphere made of steel, aluminum, or copper? P7.52 Clift et al. give the formula F ~ (67r/5)(4 + a/b)iiUb for the drag of a prolate spheroid in creeping motion, as shown in Fig. P7.52. The half-thickness h is 4 mm. If the fluid is SAE 50W oil at 20°C, (a) check that Rej < 1 and (b) estimate the spheroid length if the drag is 0.02 N. 2a P7.52 P7.53 From Table 7.2, the drag coefficient of a wide plate normal to a stream is approximately 2.0. Let the stream conditions he f/oo and poo. If the average pressure on the front of the plate is approximately equal to the free-stream stagnation pressure, what is the average pressure on the rear? P7.54 If a missile takes off vertically from sea level and leaves the atmosphere, it has zero drag when it starts and zero drag when it finishes. It follows that the drag must be a maximum somewhere in between. To simplify the analysis, assume a constant drag coefficient, C/j, and a constant vertical acceleration, a. Let the density variation be modeled by the troposphere relation, Eq. (2.20). Eind an expression for the altitude z where the drag is a maximum. Comment on your result. P7.55 A ship tows a submerged cylinder, which is 1.5 m in diam¬ eter and 22 m long, at 5 m/s in fresh water at 20°C. Esti¬ mate the towing power, in kW, required if the cylinder is (a) parallel and (b) normal to the tow direction. P7.56 P7.57 The main cross-cable between towers of a coastal suspen¬ sion bridge is 60 cm in diameter and 90 m long. Estimate the total drag force on this cable in crosswinds of 50 mi/h. Are these laminar flow conditions? P7.58 Modify Proh. P7.54 to he more realistic by accounting for missile drag during ascent. Assume constant thrust T and missile weight W. Neglect the variation of g with altitude. Solve for the altitude z in the standard atmosphere where the drag is a maximum, for T = 16,000 N, VT = 8000 N, and CoA = 0.4 m^. The writer does not believe an analytic solution is possible. P7.59 loe can pedal his bike at 10 m/s on a straight level road with no wind. The rolling resistance of his bike is 0.80 N • s/m — that is, 0.80 N of force per m/s of speed. The drag area (CoA) of Joe and his bike is 0.422 m^. Joe’s mass is 80 kg and that of the bike is 15 kg. He now encounters a headwind of 5.0 m/s. (a) Develop an equation for the speed at which Joe can pedal into the wind. Hint: A cubic equation for V will result, (b) Solve for V; that is, how fast can Joe ride into the headwind? (c) Why is the result not simply 10 — 5.0 = 5.0 m/s, as one might first suspect? P7.60 A fishnet consists of 1 -mm-diameter strings overlapped and knotted to form 1-by-l-cm squares. Estimate the drag of 1 m^ of such a net when towed normal to its plane at 3 m/s in 20°C seawater. What horsepower is required to tow 400 ft^ of this net? P7.61 A filter may be idealized as an array of cylindrical fibers normal to the flow, as in Eig. P7.6L Assuming that the fibers are uniformly distributed and have drag coeffi¬ cients given by Fig. 7.16a, derive an approximate ex¬ pression for the pressure drop Ap through a filter of thickness L. 508 Chapter 7 Flow Past Immersed Bodies U p + Ap P7.61 Filter section OOOOOOOOOO OOOOOOOOOOO OOOOOOOOOO OOOOOOOOOOO OOOOOOOOOO OOOOOOOOOOO OOOOOOOOOO OOOOOOOOOOO OOOOOOOOOO OOOOOOOOOOO ' Array of cylinders (fibers) P7.62 A sea-level smokestack is 52 m high and has a square cross section. Its supports can withstand a maximum side force of 90 kN. If the stack is to survive 90-mi/h hurricane winds, what is its maximum possible width? P7.63 For those who think electric cars are sissy, Keio University in Japan has tested a 22-ft-long prototype whose eight electric motors generate a total of 590 horsepower. The “Kaz” cruises at 180 mi/h (see Popular Science, August 2001, p. 15). If the drag coefficient is 0.35 and the frontal area is 26 ft^, what per¬ centage of this power is expended against sea-level air drag? P7.64 A parachutist jumps from a plane, using an 8.5-m-diameter chute in the standard atmosphere. The total mass of the chutist and the chute is 90 kg. Assuming an open chute and quasi-steady motion, estimate the time to fall from 2000- to 1000-m altitude. P7.65 As soldiers get higger and packs get heavier, a parachutist and load can weigh as much as 400 Ibf. The standard 28-ft parachute may descend too fast for safety. For heavier loads, the U.S. Army Natick Center has developed a 28-ft, higher-drag, less porous XT- 11 parachute (see ). This parachute has a sea-level descent speed of 16 ft/s with a 400-lbf load, (a) What is the drag coefficient of the XT- 1 1 ? (h) How fast would the stan¬ dard chute descend at sea level with such a load? P7.66 A sphere of density pj and diameter D is dropped from rest in a fluid of density p and viscosity p. Assuming a constant drag coefficient derive a differential equation for the fall velocity V(f) and show that the solution is V = C = 4gD(5 - 1) 3Q„ 3gC45- 1)U« tanh Ct where S = pjp is the specific gravity of the sphere material. P7.67 The Toyota Prius has a drag coefficient of 0.25, a frontal area of 23.4 ft^, and an empty weight of 3042 Ibf. Its rolling resistance coefficient is C„ = 0.03, that is, the rolling resis¬ tance is 3 percent of the normal force on the tires. If rolling freely down a slope of 8° at an altitude of 500 m, calculate its maximum velocity, in mi/h. P7.68 The Mars roving-laboratory parachute, in the Chap. 5 opener photo, is a 5 1 -ft-diameter disk-gap-band chute, with a measured drag coefficient of 1.12 . Mars has very low density, about 2.9 E-5 slug/ft^, and its gravity is only 38 percent of earth gravity. If the mass of payload and chute is 2400 kg, estimate the terminal fall velocity of the parachute. P7.69 Two baseballs of diameter 7.35 cm are connected to a rod 7 mm in diameter and 56 cm long, as in Fig. P7.69. What power, in W, is required to keep the system spinning at 400 r/min? Include the drag of the rod, and assume sea-level standard air. P7.70 The Army’s new ATPS personnel parachute is said to be able to bring a 400-lbf load, trooper plus pack, to ground at 16 ft/s in “mile-high” Denver, Colorado. If we assume that Table 7.3 is valid, what is the approximate diameter of this new parachute? P7.71 The 2013 Toyota Camry has an empty weight of 3190 Ibf, a frontal area of 22.06 tf, and a drag coefficient of 0.28. Its rolling resistance is C„ ~ 0.035. Estimate the maximum velocity, in mi/h, this car can attain when rolling freely at sea level down a 4° slope. P7.72 A settling tank for a municipal water supply is 2.5 m deep, and 20°C water flows through continuously at 35 cm/s. Estimate the minimum length of the tank that will ensure that all sediment (SG = 2.55) will fall to the bottom for particle diameters greater than (a) 1 mm and (b) 100 pm. P7.73 A balloon is 4 m in diameter and contains helium at 1 25 kPa and 15°C. Balloon material and payload weigh 200 N, not including the helium. Estimate (a) the terminal ascent velocity in sea-level standard air, (b) the final standard altitude (neglecting winds) at which the balloon will come to rest, and (c) the minimum diameter (< 4 m) for which the balloon will just barely begin to rise in sea- level standard air. Problems 509 P7.74 It is difficult to define the “frontal area” of a motorcycle due to its complex shape. One then measures the drag area (that is, CdA) in area units. Hoemer reports the drag area of a typical motorcycle, including the (upright) driver, as about 5.5 ft^. Rolling friction is typically about 0.7 Ibf per mi/h of speed. If that is the case, estimate the maximum sea-level speed (in mi/h) of the Harley-Davidson V-Rod’’^'^ cycle, whose liquid-cooled engine produces 1 15 hp. P7.75 The helium-filled balloon in Fig. P7.75 is tethered at 20°C and 1 atm with a string of negligible weight and drag. The diameter is 50 cm, and the balloon material weighs 0.2 N, not including the helium. The helium pressure is 120 kPa. Estimate the tilt angle 0 if the airstream velocity (/ is (a) 5 m/s or (b) 20 m/s. P7.75 P7.76 The movie The World’s Fastest Indian tells the story of Burt Munro, a New Zealander who, in 1967, set a motor¬ cycle record of 201 mi/h on the Bonneville Salt Flats. Using the data of Prob. P7.74, (a) estimate the horsepower needed to drive this fast, (b) What horsepower would have gotten Burt up to 250 mi/h? ¥1.11 To measure the drag of an upright person, without violating human subject protocols, a life-sized mannequin is attached to the end of a 6-m rod and rotated at fl = 80 rev/min, as in Fig. P7.77. The power required to maintain the rotation is 60 kW. By including rod drag power, which is significant, estimate the drag area C^A of the mannequin, in m^. P7.78 On April 24, 2007, a French bullet train set a new speed record, for rail-driven trains, of 357.2 mi/h, beating the old record by 12 percent. Using the data in Table 7.3, estimate the sea-level horsepower required to drive this train at such a speed. P7.79 Assume that a radioactive dust particle approximates a sphere of density 2400 kg/m^. How long, in days, will it take such a particle to settle to sea level from an altitude of 12 km if the particle diameter is {a) 1 /im or (b) 20 /tm? P7.80 A heavy sphere attached to a string should hang at an angle 9 when immersed in a stream of velocity U, as in Fig. P7.80. Derive an expression for 0 as a function of the sphere and flow properties. What is 6 if the sphere is steel (SG = 7.86) of diameter 3 cm and the flow is sea-level standard air at 1/ = 40 m/s? Neglect the string drag. P7.81 A typical U.S. Army parachute has a projected diameter of 28 ft. For a payload mass of 80 kg, (a) what terminal veloc¬ ity will result at 1000-m standard altitude? For the same velocity and net payload, what size drag-producing “chute” is required if one uses a square flat plate held (b) vertically and (cj horizontally? (Neglect the fact that flat shapes are not dynamically stable in free fall.) P7.82 Skydivers, flying over sea-level ground, typically jump at about 8000 ft altitude and free-fall spread-eagled until they open their chutes at about 2000 ft. They take about 10 s to reach terminal velocity. Estimate how many seconds of free-fall they enjoy if (a) they fall spread-eagled or (b) they fall feet first? Assume a total skydiver weight of 220 Ibf. P7.83 A blimp approximates a 4:1 spheroid that is 196 ft long. It is powered by two 150 hp ducted fans. Estimate the maxi¬ mum speed attainable, in mi/h, at an altitude of 8200 ft. P7.84 A Ping-Pong ball weighs 2.6 g and has a diameter of 3.8 cm. It can be supported by an air jet from a vacuum cleaner outlet, as in Fig. P7.84. For sea-level standard air, what jet velocity is required? Mmi P7.84 510 Chapter 7 Flow Past Immersed Bodies P7.85 In this era of expensive fossil fuels, many alternatives have been pursued. One idea from SkySails, Inc., shown in Fig. P7.85, is the assisted propulsion of a ship hy a large tethered kite. The tow force of the kite assists the ship’s propeller and is said to reduce annual fuel consumption hy 10-35 percent. For a typical example, let the ship he 120 m long, with a wetted area of 2800 m^. The kite area is 330 m^ and has a force coefficient of 0.8. The kite cable makes an angle of 25° with the horizontal. Let Fwind = 30 mi/h. Neglect ship wave drag. Estimate the ship speed (a) due to the kite only and (b) if the propeller delivers 1250 hp to the water. [Hint: The kite sees the relative velocity of the wind.] P7.85 Ship propulsion assisted by a large kite. ( Courtesy of SkySails, Inc.) P7.86 Hoerner [Ref. 12, pp. 3-25] states that the drag coeffi¬ cient of a flag of 2: 1 aspect ratio is 0. 1 1 based on plan- form area. The University of Rhode Island has an aluminum flagpole 25 m high and 14 cm in diameter. It flies equal-sized national and state flags together. If the fracture stress of aluminum is 210 MPa, what is the maximum flag size that can be used without breaking the flagpole in hurricane (75 mi/h) winds? (Neglect the drag of the flagpole.) P7.87 A tractor-trailer truck has a drag area C^A = 8 m^ bare and 6.7 m^ with an aerodynamic deflector (Fig. 7.18h). Its roll¬ ing resistance is 50 N for each mile per hour of speed. Calculate the total horsepower required at sea level with and without the deflector if the truck moves at (a) 55 mi/h and (b) 75 mi/h. P7.88 A pickup truck has a clean drag area C^A of 35 ft^. Estimate the horsepower required to drive the truck at 55 mi/h (a) clean and (ft) with the 3- by 6-ft sign in Eig. P7.88 installed if the rolling resistance is 150 Ibf at sea level. 6 ft P7.89 The AMTRAK Acela train passes through Kingston, RI, at 130 mi/h, scaring all the villagers daily. Its total weight is 624 short tons, with a rolling resistance C,.r ~ 0.0024. Esti¬ mate the horsepower required to drive the train this fast. P7.90 In the great hurricane of 1938, winds of 85 mi/h blew over a boxcar in Providence, Rhode Island. The boxcar was 10 ft high, 40 ft long, and 6 ft wide, with a 3-ft clearance above tracks 4.8 ft apart. What wind speed would topple a boxcar weighing 40,000 Ibf? P7.91 A cup anemometer uses two 5-cm-diameter hollow hemi¬ spheres connected to 15-cm rods, as in Pig. P7.91. Rod drag is negligible, and the central bearing has a retarding torque of 0.004 N • m. Making simplifying assumptions to average out the time- varying geometry, estimate and plot the variation of anemometer rotation rate kl with wind velocity U in the range 0 < U <25 m/s for sea-level standard air. D = 5 cm Problems 511 P7.92 A 1500-kg automobile uses its drag area QjA = 0.4 m^, P7.96 plus brakes and a parachute, to slow down from 50 m/s. Its brakes apply 5000 N of resistance. Assume sea-level stan¬ dard air. If the automobile must stop in 8 s, what diameter parachute is appropriate? P7.93 A hot-film probe is mounted on a cone-and-rod system in a sea- level airsheam of 45 m/s, as in Fig. P7.93. Estimate the maxi¬ mum cone vertex angle allowable if the flow-induced bending moment at the root of the rod is not to exceed 30 N ■ cm. Hot film P7.93 P7.94 Baseball drag data from the University of Texas are shown in Fig. P7.94. A baseball weighs approximately 5.12 ounces and has a diameter of 2.91 in. Hall-of-Famer Nolan Ryan, in a 1974 game, threw the fastest pitch ever recorded; 108.1 mi/h. If it is 60 ft from Nolan’s hand to the catcher’s mitt, estimate the sea-level hall velocity which the catcher experiences for (a) a normal hasehall, and (ft) a perfectly smooth hasehall. P7.97 % P7.98 P7.95 An airplane weighing 28 kN, with a drag area C^A « 5 m^, P7.99 lands at sea level at 55 m/s and deploys a drag parachute 3 m in diameter. No other brakes are applied, {a) How long will it take the plane to slow down to 20 m/s? {b) How far will it have traveled in that time? A Savonius rotor (Fig. 6.29b) can be approximated by the two open half-tuhes in Fig. P7.96 mounted on a central axis. If the drag of each tube is similar to that in Table 7.2, derive an approximate formula for the rotation rate H as a function of U, D, L, and the fluid properties (p, fi). P7.96 C! ^ A simple measurement of automohile drag can be found by an unpowered coastdown on a level road with no wind. Assume rolling resistance proportional to velocity. For an automobile of mass 1500 kg and frontal area 2 m^, the following velocity-versus-time data are obtained during a coastdown: t, s 0 10 20 30 40 V, m/s 27.0 24.2 21.8 19.7 17.9 Estimate (a) the rolling resistance and (b) the drag coeffi¬ cient. This problem is well suited for computer analysis but can be done by hand also. A buoyant ball of specific gravity SG < 1 dropped into water at inlet velocity Vq will penetrate a distance h and then pop out again, as in Fig. P7.98. Make a dynamic analysis of this problem, assuming a constant drag coef¬ ficient, and derive an expression for fi as a function of the system properties. How far will a 5-cm-diameter ball with SG = 0.5 and Co ~ 0.47 penetrate if it enters at 10 m/s? Two steel balls (SG = 7.86) are connected by a thin hinged rod of negligible weight and drag, as in Fig. P7.99. A stop prevents the rod from rotating counterclockwise. Estimate the sea-level air velocity U for which the rod will first begin to rotate clockwise. 512 Chapter 7 Flow Past Immersed Bodies P7.98 Diameter P7.102 Sand particles (SG = 2.7), approximately spherical with diameters from 100 to 250 fim, are introduced into an upward-flowing stream of water at 20°C. What is the minimum water velocity that will carry all the sand par¬ ticles upward? P7.103 When immersed in a uniform stream V, a heavy rod hinged at A will hang at Pode ’s angle 9, after an analysis hy L. Pode in 1951 (Fig. P7.103). Assume that the cylinder has normal drag coefficient Con tangential coefficient Cdt that relate the drag forces to Vjv and Vj, respectively. Derive an expression for Pode’s angle as a function of the flow and rod parameters. Find 9 for a steel rod, L = 40 cm, D = 1 cm, hanging in sea-level air at F = 35 m/s. D = 2 cm P7.100 A tractor-trailer truck is coasting freely, with no brakes, down an 8° slope at 1000-m standard altitude. Rolling resistance is 120 N for every m/s of speed. Its frontal area is 9 m^, and the weight is 65 kN. Estimate the ter¬ minal coasting velocity, in mi/h, for (a) no deflector and (b) a deflector installed. P7.101 Icebergs can be driven at substantial speeds by the wind. Let the iceberg be idealized as a large, flat cylinder, D> L, with one-eighth of its bulk exposed, as in Fig. P7.101. Let the seawater be at rest. If the upper and lower drag forces depend on relative velocities between the iceberg and the fluid, derive an approximate expression for the steady ice¬ berg speed V when driven by wind velocity U. u -■ - D»L - «- V L/8 y Iceberg IL /8 A P7.104 The Russian Typhoon-class submarine is 170 m long, with a maximum diameter of 23 m. Its propulsor can deliver up to 80,000 hp to the seawater. Model the submarine as an 8: 1 ellipsoid and estimate the maximum speed, in knots, of this ship. P7.105 A ship 50 m long, with a wetted area of 800 m^, has the hull shape tested in Fig. 7.19. There are no bow or stem bulbs. The total propulsive power available is 1 MW. For sea¬ water at 20°C, plot the ship’s velocity Fkn versus power P for 0 < P < 1 MW. What is the most efficient setting? P7.106 For the kite-assisted ship of Prob. P7.85, again neglect wave drag and let the wind velocity be 30 mi/h. Estimate the kite area that would tow the ship, unaided by the propeller, at a ship speed of 8 knots. P7.107 The largest flag in Rhode Island stands outside Herb Cham¬ bers’ auto dealership, on the edge of Route 1-95 in Provi¬ dence. The flag is 50 ft long, 30 ft wide, weighs 250 Ibf, and takes four strong people to raise it or lower it. Using Prob. P7.40 for input, estimate (a) the wind speed, in mi/h, for which the flag drag is 1000 Ibf and (b) the flag drag when the wind is a low-end category 1 hurricane, 74 mi/h. [Hint: Providence is at sea level.] P7.108 The data in Fig. P7.108 are for the lift and drag of a spinning sphere from Ref. 45. Suppose that a tennis ball iW ~ 0.56 N, D ~ 6.35 cm) is struck at sea level with initial velocity Vq = 30 m/s and “topspin” (front of the ball rotating downward) of 120 r/s. If the initial height of the ball is 1.5 m, estimate the horizontal distance traveled before it strikes the ground. P7.101 Problems 513 mission of the American Society of Mechanical Engineers.) V P7.109 The world record for automobile mileage, 12,665 miles per gallon, was set in 2005 by the PAC-CAR II in Fig. P7.I09, built by students at the Swiss Federal Institute of Technol¬ ogy in Zurich . This little car, with an empty weight of 64 Ibf and a height of only 2.5 ft, traveled a 21 -km course at 30 km/h to set the record. It has a reported drag coefficient P7.109 The world’s best mileage set by PAC-Car II of ETH Zurich. of 0.075 (comparahle to an airfoil), based upon a frontal area of 3 ft^. (a) What is the drag of this little car when on the course? (b) What horsepower is required to pro¬ pel it? (c) Do a bit of research and explain why a value of miles per gallon is completely misleading in this par¬ ticular case. P7.110 A baseball pitcher throws a curveball with an initial ve¬ locity of 65 mi/h and a spin of 6500 r/min about a verti¬ cal axis. A baseball weighs 0.32 Ibf and has a diameter of 2.9 in. Using the data of Fig. P7.108 for turbulent flow, estimate how far such a curveball will have devi¬ ated from its straight-line path when it reaches home plate 60.5 ft away. P7.111 A table tennis ball has a mass of 2.6 g and a diameter of 3.81 cm. It is struck horizontally at an initial velocity of 20 m/s while it is 50 cm above the table, as in Fig. P7. 1 1 1 . For sea-level air, what spin, in r/min, will cause the ball to strike the opposite edge of the table, 4 m away? Make an analytical estimate, using Fig. P7.108, and account for the fact that the ball decelerates during flight. P7.112 A smooth wooden sphere (SG = 0.65) is connected by a thin rigid rod to a hinge in a wind tunnel, as in Fig. P7.1 12. Air at 20°C and 1 atm flows and levitates the sphere. (a) Plot the angle 9 versus sphere diameter d in the range 1 cm < < 15 cm. (b) Comment on the feasibility of this configuration. Neglect rod drag. 514 Chapter 7 Flow Past Immersed Bodies 4 m- 20 m/s 9 / ' / _ 50 cm P7.111 P7.113 An automobile has a mass of 1000 kg and a drag area CoA = 0.7 m^. The rolling resistance of 70 N is approxi¬ mately constant. The car is coasting without brakes at 90 km/h as it begins to climb a hill of 10 percent grade (slope = tan” 0.1 = 5.71°). How far up the hill will the car come to a stop? P7.114 The deep submergence vehicle ALVIN is 23 ft long and 8.5 ft wide. It weighs about 36,000 Ibf in air and ascends (descends) in the seawater due to about 360 Ibf of positive (negative) buoyancy. Noting that the front face of the ship is quite different for ascent and descent, (a) estimate the ve¬ locity for each direction, in meters per minute, (b) How long does it take to ascend from its maximum depth of 4500 m? Lifting bodies — airfoils P7.115 The Cessna Citation executive jet weighs 67 kN and has a wing area of 32 m^. It cruises at 10 km standard altitude with a lift coefficient of 0.21 and a drag coefficient of 0.015. Estimate (a) the cruise speed in mi/h and (b) the horsepower required to maintain cruise velocity. P7.116 An airplane weighs 180 kN and has a wing area of 160 m^ and a mean chord of 4 m. The airfoil properties are given by Fig. 7.25. If the plane is designed to land at Vq = 1.2Vstaii, using a split flap set at 60°, (a) what is the proper landing speed in mi/h? (b) What power is required for take¬ off at the same speed? P7.117 The Transition® auto-car in Fig. 7.30 has a weight of 1200 Ibf, a wingspan of 27.5 ft, and a wing area of 150 ft’, with a symmetrical airfoil, ~ 0.02. Assume that the fuse¬ lage and tail section have a drag-area comparable to the Toyota Prius , C/jA = 6.24 t^. If the pusher propeller provides a thrust of 250 Ibf, how fast, in mi/h, can this car- plane fly at an altitude of 8200 ft? P7.118 Suppose that the airplane of Prob. P7.1 16 is fitted with all the best high-lift devices of Fig. 7.28. What is its minimum stall speed in mi/h? Estimate the stopping distance if the plane lands at Vq = 1.25Vstaii with constant Ci^ = 3.0 and Co = 0.2 and the braking force is 20 percent of the weight on the wheels. P7.119 A transport plane has a mass of 45,000 kg, a wing area of 160 m^, and an aspect ratio of 7. Assume all lift and drag due to the wing alone, with C^oo = 0.020 and = 1.5. If the aircraft flies at 9000 m standard altitude, make a plot of drag (in N) versus speed (from stall to 240 m/s) and determine the optimum cmise velocity (minimum drag per unit speed). P7.120 Show that if Eqs. (7.70) and (7.71) are valid, the maxi¬ mum lift-to-drag ratio occurs when Co = 2Cooo- What are (L/Z))^ax and a for a symmetric wing when AR = 5 and Cfloo = 0.009? P7.121 In gliding (unpowered) flight, the lift and drag are in equi¬ librium with the weight. Show that if there is no wind, the aircraft sinks at an angle tan 0 ~ drag lift For a sailplane of mass 200 kg, wing area 12 m^, and aspect ratio 1 1, with an NACA 0009 airfoil, estimate (a) the stall speed, (b) the minimum gliding angle, and (c) the maxi¬ mum distance it can glide in still air when it is 1200 m above level ground. P7.122 A boat of mass 2500 kg has two hydrofoils, each of chord 30 cm and span 1.5 m, with C^max =1-2 and Cooo = 0.08. Its engine can deliver 130 kW to the water. For seawater at 20°C, estimate (a) the minimum speed for which the foils support the boat and (b) the maximum speed attainable. P7.123 In prewar days there was a controversy, perhaps apocry¬ phal, about whether the bumblebee has a legitimate aero¬ dynamic right to fly. The average bumblebee (Bombus terrestris) weighs 0.88 g, with a wingspan of 1.73 cm and a wing area of 1.26 cm^. It can indeed fly at 10 m/s. Using fixed-wing theory, what is the lift coefficient of the bee at this speed? Is this reasonable for typical airfoils? P7.124 Tbe bumblebee can hover at zero speed by flapping its wings. Using the data of Prob. P7.123, devise a theory for flapping wings where the downstroke approximates a short flat plate normal to the flow (Table 7.3) and the upstroke is feathered at nearly zero drag. How many flaps per second of such a model wing are needed to support the bee’s weight? (Actual mea¬ surements of bees show a flapping rate of 194 Hz.) Fundamentals of Engineering Exam Problems 515 P7.125 The Solar Impulse aircraft in the chapter-opener photo has a wingspan of 208 ft, a wing area of 2140 ft^, and a weight of 1600 kgf. Its propellers deliver an average of 24 hp to the air at a cruising altitude of 8.5 km. Assuming an NACA 0009 airfoil, and neglecting the drag of the fuselage and tail, estimate (a) the wing aspect ratio, {b) the cruise speed, in mi/h, and (c) the wing angle of attack. [Hint: Simplify hy using Fig. 7.25 to estimate lift and drag.] P7.126 Using the data for the Transition® auto-car from Proh. P7.117, and a maximum lift coefficient of 1.3, estimate Word Problems W7.1 How do you recognize a boundary layer? Cite some physi¬ cal properties and some measurements that reveal appro¬ priate characteristics. W7.2 In Chap. 6 the Reynolds number for transition to turbu¬ lence in pipe flow was about Re,j ~ 2300, whereas in flat- plate flow Re^. ~ 1 E6, nearly three orders of magnitude higher. What accounts for the difference? W7.3 Without writing any equations, give a verbal description of boundary layer displacement thickness. W7.4 Describe, in words only, the basic ideas behind the “boundary layer approximations.” W7.5 What is an adverse pressure gradient? Give three examples of flow regimes where such gradients occur. W7.6 What is a. favorable pressure gradient? Give three exam¬ ples of flow regimes where such gradients occur. W7.7 The drag of an airfoil (Fig. 7.12) increases considerably if you turn the sharp edge around 1 80° to face the stream. Can you explain this? Fundamentals of Engineering Exam Problems FE7.1 A smooth 12-cm-diameter sphere is immersed in a stream of 20°C water moving at 6 m/s. The appropriate Reynolds number of this sphere is approximately (a) 2.3 E5, (b) 7.2 E5, (c) 2.3 E6, (d) 1.2 E6, (e) 7.2 E7 rE7.2 If, in Prob. FE7. 1 , the drag coefficient based on frontal area is 0.5, what is the drag force on the sphere? {a) 17 N, (ft) 51 N, (c) 102 N, {d) 130 N, (e) 203 N FE7.3 If, in Prob. FE7. 1 , the drag coefficient based on frontal area is 0.5, at what terminal velocity will an aluminum sphere (SG = 2.7) fall in still water? (fl) 2.3 m/s, (ft) 2.9 m/s, (c) 4.6 m/s, {d) 6.5 m/s, (e) 8.2 m/s FE7.4 For flow of sea-level standard air at 4 m/s parallel to a thin flat plate, estimate the boundary layer thickness at x = 60 cm from the leading edge: (a) 1.0 mm, (ft) 2.6 mm, (c) 5.3 mm, (d) 7.5 mm, (e) 20.2 mm the distance for the vehicle to take off at a speed of 1.2 Vs, all. Note that we have to add the car-body drag to the wing drag. P7.127 The so-called Rocket Man, Yves Rossy, flew across the Alps in 2008, wearing a rocket-propelled wing-suit with the following data: thrust = 200 Ibf, altitude = 8,200 ft, and wingspan = 8 ft ( Further assume a wing area of 12 fP, total weight of 280 Ibf, Cd^ = 0.08 for the wing, and a drag area of 1 .7 ft^ for Rocket Man. Estimate the maximum velocity possible for this condition, in mi/h. W7.8 In Table 7.3, the drag coefficient of a spruce tree decreases sharply with wind velocity. Can you explain this? W7.9 Thrust is required to propel an airplane at a finite forward velocity. Does this imply an energy loss to the system? Explain the concepts of thrust and drag in terms of the first law of thermodynamics. W7.10 How does the concept of drafting, in automobile and bicycle racing, apply to the material studied in this chapter? W7.ll The circular cylinder of Fig. 7. 13 is doubly symmetric and therefore should have no lift. Yet a lift sensor would defi¬ nitely reveal a finite root-mean-square value of lift. Can you explain this behavior? W7.12 Explain in words why a thrown spinning ball moves in a curved trajectory. Give some physical reasons why a side force is developed in addition to the drag. FE7.5 In Prob. EE7.4, for the same flow conditions, what is the wall shear stress at x = 60 cm from the leading edge? (a) 0.053 Pa, (ft) 0.11 Pa, (c) 0.16 Pa, (d) 0.32 Pa, (e) 0.64 Pa FE7.6 Wind at 20°C and 1 atm blows at 75 km/h past a flagpole 18 m high and 20 cm in diameter. The drag coefficient, based on frontal area, is 1.15. Estimate the wind-induced bending moment at the base of the pole. (a) 9.7 kN ■ m, (ft) 15.2 kN ■ m, (c) 19.4 kN ■ m, (d) 30.5 kN ■ m, (e)61.0kN-m FE7.7 Consider wind at 20°C and 1 atm blowing past a chimney 30 m high and 80 cm in diameter. If the chimney may frac¬ ture at a base bending moment of 486 kN-m, and its drag coefficient based on frontal area is 0.5, wbat is the approxi¬ mate maximum allowable wind velocity to avoid fracture? (a) 50 mi/h, (ft) 75 mi/h, (c) 100 mi/h, (d) 125 mi/h, (e) 150 mi/h 516 Chapter 7 Flow Past Immersed Bodies FE7.8 A dust particle of density 2600 kg/m^, small enough to satisfy Stokes’s drag law, settles at 1.5 mm/s in air at 20°C and 1 atm. What is its approximate diameter? (a) 1.8 fim, (b) 2.9 fim, (c) 4.4 fim, (d) 16.8 fim, (e) 234 fim FE7.9 An airplane has a mass of 19,550 kg, a wingspan of 20 m, and an average wing chord of 3 m. When flying in air of density 0.5 kg/m^, its engines provide a thrust of 12 kN Comprehensive Problems C7.1 Jane wants to estimate the drag coefficient of herself on her bicycle. She measures the projected frontal area to he 0.40 m^ and the rolling resistance to he 0.80 N ■ s/m. The mass of the hike is 15 kg, while the mass of Jane is 80 kg. Jane against an overall drag coefficient of 0.025. What is its approximate velocity? (fl) 250 mi/h, (b) 300 mi/h, (c) 350 mi/h, (d) 400 mi/h, (e) 450 mi/h EE7.10For the flight conditions of the airplane in Proh. FE7.9 above, what is its approximate lift coefficient? (fl) 0.1, (b) 0.2, (c) 0.3, (d) 0.4, (e) 0.5 coasts down a long hill that has a constant 4° slope. (See Fig. C7.1.) She reaches a terminal (steady state) speed of 14 m/s down the hill. Estimate the aerodynamic drag coef¬ ficient Co of the rider and bicycle combination. C7.2 Air at 20°C and 1 atm flows at = 5 m/s between long, smooth parallel heat exchanger plates 10 cm apart, as in Eig. C7.2. It is proposed to add a number of widely spaced 1-cm-long interrupter plates to increase the heat transfer, as shown. Although the flow in the channel is turbulent, the boundary layers over the interrupter plates are essentially laminar. Assume all plates are 1 m wide into the paper. Eind (fl) the pressure drop in Pa/m without the small plates present. Then find (b) the number of small plates per meter of channel length that will cause the pressure drop to rise to 10.0 Pa/m. Interrupter plates L = 1 cm C7.3 A new pizza store is planning to open. It will, of course, offer free delivery, and therefore need a small delivery car with a large sign attached. The sign (a flat plate) is 1.5 ft high and 5 ft long. The boss (having no feel for fluid mechanics) mounts the sign bluntly facing the wind. One of his drivers is taking fluid mechanics and tells his boss he can save lots of money by mounting the sign parallel to the wind. (See Pig. C7.3.) (fl) Calculate the drag (in Ibf) on the sign alone at 40 mi/h (58.7 ft/s) in both orientations, (b) Suppose the car without any sign has a drag coefficient of 0.4 and a frontal area of 40 ft^. For F = 40 mi/h, calculate the total drag of the car-sign combination for both orientations, (c) If the car has a rolling resistance of 40 Ibf at 40 mi/h, calculate the horse¬ power required by the engine to drive the car at 40 mi/h in both orientations, (d) Finally, if the engine can deliver 10 hp for 1 h on a gallon of gasoline, calculate the fuel efficiency in mi/gal for both orientations at 40 mi/h. PIZZA C7.2 References 517 C7.4 Consider a pendulum with an unusual bob shape: a hemi- spherical cup of diameter D whose axis is in the plane of oscillation, as in Fig. C7.4. Neglect the mass and drag of C7.5 the rod L. (a) Set up the differential equation for the oscil¬ lation 9(1), including different cup drag (air density p) in each direction, and (b) nondimensionalize this equation, (c) Determine the natural frequency of oscillation for small 6^1 rad. (d) For the special case L = Im, D = 10 cm, m = 50 g, and air at 20°C and 1 atm, with 9{0) = 30°, find (numerically) the time required for the oscillation ampli¬ tude to drop to 1°. Program a method of numerical solution of the Blasius flat- plate relation, Eq. (7.22), subject to the conditions in Eqs. (7.23). You will find that you cannot get started without knowing the initial second derivative /"(O), which lies between 0.2 and 0.5. Devise an iteration scheme that starts at/"(0) « 0.2 and converges to the correct value. Print out ulU =f'(ri) and compare with Table 7.1. Design Project D7.1 It is desired to design a cup anemometer for wind speed, similar to Pig. P7.91, with a more sophisticated approach than the “average-torque” method of Prob. P7.91. The design should achieve an approximately linear relation between wind velocity and rotation rate in the range 20 < U < 40 mi/h, and the anemometer should rotate at about 6 r/s at U = 30 mi/h. All specifications — cup diameter D, rod length L, rod diameter d, the bearing type, and all materials — are to be selected through your analysis. Make suitable as¬ sumptions about the instantaneous drag of the cups and rods at any given angle 9(t) of the system. Compute the instantaneous torque T(t), and find and integrate the instan¬ taneous angular acceleration of the device. Develop a com¬ plete theory for rotation rate versus wind speed in the range 0 < U < 50 mi/h. Try to include actual commercial bear¬ ing friction properties. References 1. Ft. Schlichting and K. Gersten, Boundary Layer Theory, 8th ed.. Springer, New York, 2000. 2. F. M. White, Viscous Fluid Flow, 3d ed., McGraw-Hill, New York, 2005. 3. J. Cousteix, Modeling and Computation of Boundary-Layer Flows, 2d ed., Springer- Verlag, New York, 2005. 4. J. D. Anderson, Computational Fluid Dynamics: An Intro¬ duction, 3d ed.. Springer, New York, 2010. 5. V. V. Sychev et ah. Asymptotic Theory of Separated Flows, Cambridge University Press, New York, 2008. 6. I. J. Sobey, Introduction to Interactive Boundary Layer Theory, Oxford University Press, New York, 2001. 7. T. von Karman, “On Laminar and Turbulent Friction,” Z Angew. Math. Mech., vol. 1, 1921, pp. 235-236. 8. G. B. Schubauer and H. K. Skramstad, “Laminar Boundary Layer Oscillations and Stability of Laminar Flow,” Natl. Bur. Stand. Res. Pap. 1772, April 1943 (see also J. Aero. Sci., vol. 14, 1947, pp. 69-78, and NACA Rep. 909, 1947). 9. P. S. Bernard and 1. M. Wallace, Turbulent Flow: Analysis, Measurement, and Prediction, Wiley, New York, 2002. 10. P. W. Runstadler, Ir., et ah, “Diffuser Data Book,” Create Inc., Tech. Note 186, Hanover, NH, May 1975. 11. B. Thwaites, “Approximate Calculation of the Laminar Boundary Layer,” Aeronanf. Q., vol. 1, 1949, pp. 245-280. 12. S. F. Hoemer, Fluid Dynamic Drag, published by the author. Midland Park, Nl, 1965. 13. 1. D. Anderson, Fundamentals of Aerodynamics, 5th ed., McGraw-Hill, New York, 2010. 14. V. Tucker and G. C. Parrott, “Aerodynamics of Gliding Flight of Falcons and Other Birds,” J. Exp. Biol., vol. 52, 1970, pp. 345-368. 15. E. C. Tupper, Introduction to Naval Architecture, 5th ed., Butterworth-Heinemann, Burlington, MA, 2013. 16. I. H. Abbott and A. E. von Doenhoff, Theory of Wing Sec¬ tions, Dover, New York, 1981. 518 Chapter 7 Flow Past Immersed Bodies 17. R. L. Kline and F. F. Fogelman, “Airfoil for Aircraft,” U. S. Patent 3,706,430, Dec. 19, 1972. 18. A. Azuma, The Biokinetics of Swimming and Flying, AIAA, Reston, VA, 2006. 19. National Committee for Fluid Mechanics Films, Illustrated Experiments in Fluid Mechanics, M.I.T. Press, Cambridge, MA, 1972. 20. D. M. Bushnell and J. Hefner (Eds.), Viscous Drag Reduction in Boundary Layers, American Institute of Aeronautics & Astronautics, Reston, VA, 1990. 21. “Automobile Drag Coefficient,” URL . 22. R. H. Barnard, Road Vehicle Aerodynamic Design, 3d ed., Mechaero Publishing, St. Albans, U.K., 2010. 23. R. D. Blevins, Applied Fluid Dynamics Handbook, BBS, New York, 2009. 24. R. C. Johnson, Jr., G. E. Ramey, and D. S. O'Hagen, “Wind Induced Forces on Trees,” J. Fluids Eng., vol. 104, March 1983, pp. 25-30. 25. P. W. Bearman et ah, “The Effect of a Moving Floor on Wind-Tunnel Simulation of Road Vehicles,” Paper No. 880245, SAE Transactions, J. Passenger Cars, vol. 97, sec. 4, 1988, pp. 4.200-4.214. 26. CRC Handbook of Tables for Applied Engineering Science, 2d ed., CRC Press, Boca Raton, FL, 1973. 27. T. Inui, “Wavemaking Resistance of Ships,” Trans. Soc. Nav. Arch. Marine Engrs., vol. 70, 1962, pp. 283-326. 28. L. Larsson, “CFD in Ship Design — Prospects and Limita¬ tions,” Ship Technology Research, vol. 44, no. 3, July 1997, pp. 133-154. 29. R. L. Street, G. Z. Watters, and J. K. Vennard, Elementary Eluid Mechanics, 7th ed., Wiley, New York, 1995. 30. J. D. Anderson, Jr., Modern Compressible Elow: with His¬ torical Perspective, 3d ed., McGraw-Hill, New York, 2002. 31. J. D. Anderson, Jr., Hypersonic and High Temperature Gas Dynamics, AIAA, Reston, VA, 2000. 32. J. Rom, High Angle of Attack Aerodynamics: Subsonic, Tran¬ sonic, and Supersonic Flows, Springer-Verlag, New York, 2011. 33. S. Vogel, “Drag and Reconfiguration of Broad Leaves in High Winds,” J. Exp. Bot., vol. 40, no. 217, August 1989, pp. 941-948. 34. S. Vogel, Life in Moving Eluids, Princeton University Press 2d ed., Princeton, NJ, 1996. 35. J. A. C. Humphrey (ed.). Proceedings 2d International Sym¬ posium on Mechanics of Plants, Animals, and Their Environ¬ ment, Engineering Eoundation, New York, January 2000. 36. D. D. Joseph, R. Bai, K. P. Chen, and Y. Y. Renardy, “Core- Annular Flows,” Annu. Rev. Eluid Mech., vol. 29, 1997, pp. 65-90. 37. J. W. Hoyt and R. H. J. Sellin, “Scale Effects in Polymer Solution Pipe Flow,” Experiments in Eluids, vol. 15, no. 1, June 1993, pp. 70-74. 38. S. Nakao, “Application of V-Shape Riblets to Pipe Flows,” J. Eluids Eng., vol. 113, December 1991, pp. 587-590. 39. P. Thiede (ed.). Aerodynamic Drag Reduction Technologies, Springer, New York, 2001. 40. C. L. Merkle and S. Deutsch, “Microbubble Drag Reduc¬ tion in Liquid Turbulent Boundary Layers,” Applied Mechanics Reviews, vol. 45, no. 3 part 1, March 1992, pp. 103-127. 41. K. S. Choi and G. E. Kamiadakis, “Mechanisms on Trans¬ verse Motions in Turbulent Wall Elows,” Annual Review of Fluid Mechanics, vol. 35, 2003, pp. 45-62. 42. C. J. Roy, J. Payne, and M. McWherter-Payne, “RANS Simu¬ lations of a Simplified Tractor-Trailer Geometry,” J. Fluids Engineering, vol. 128, Sept. 2006, pp. 1083-1089. 43. Evolution ofElight, Internet URL . 44. J. D. Anderson, Jr., A History of Aerodynamics, Cambridge University Press, New York, 1999. 45. Y. Tsuji, Y. Morikawa, and O. Mizuno, “Experimental Mea¬ surement of the Magnus Force on a Rotating Sphere at Low Reynolds Numbers,” Journal of Eluids Engineering, vol. 107, 1985, pp. 484-488. 46. R. Clift, J. R. Grace, and M. E. Weber, Bubbles, Drops and Particles, Dover, NY, 2005. 47. M. Gad-el-Hak, “Flow Control: The Future,” Journal of Aircraft, vol. 38, no. 3, 2001, pp. 402-418. 48. D. Geropp and H. J. Odenthal, “Drag Reduction of Motor Vehicles by Active Flow Control Using the Coanda Effect,” Experiments in Eluids, vol. 28, no. 1, 2000, pp. 74—85. 49. Z. Zapryanov and S. Tabakova, Dynamics of Bubbles, Drops, and Rigid Particles, Kluwer Academic Pub., New York, 1998. 50. D. G. Karamanev, and L. N. Nikolov, “Ereely Rising Spheres Do Not Obey Newton’s Law for Free Settling,” AIChE Jour¬ nal, vol. 38, no. 1, Nov. 1992, pp. 1843-1846. 51. Katz J., Race-Car Aerodynamics, Robert Bentley Inc., Cambridge, MA, 2003. 52. A. S. Brown, “More than 12,000 Miles to the Gallon,” Mechanical Engineering, January 2006, p. 64. 53. D. M. Bushnell, “Aircraft Drag Reduction: A Review,” Pro¬ ceedings of the Institution of Mechanical Engineers, Part G: Journal of Aerospace Engineering, vol. 217, no. 1, 2003, pp. 1-18. 54. D. B. Spalding, “A Single Formula for the Law of the Wall,” J. Appl. Mechanics, vol. 28, no. 3, 1961, pp. 444^58. 55. D. G. Fertis, “New Airfoil-Design Concept with Improved Aerodynamic Characteristics,” J. Aerospace Engineering, vol. 7, no. 3, July 1994, pp. 328-339. References 519 56. F. Finnish and S. Witherspoon, “Aerodynamic Performance of an Airfoil with Step-Induced Vortex for Lift Augmentation,” J. Aerospace Engineering, vol. 11, no. 1, Jan. 1998, pp. 9-16. 57. D. S. Miklosovic et ah, “Leading Edge Tubercles Delay Stall on Humpback Whale,” Physics of Fluids, vol. 16, no. 5, May 2004, pp. L39-L42. 58. R. McCallen, J. Ross, and F. Browand, The Aerodynamics of Heavy Vehicles: Trucks, Buses, and Trains, Springer-Verlag, New York, 2005. 59. J. R. Cruz et ah, “Wind Tunnel Testing of Various Disk- Gap-Band Parachutes,” AIAA Paper 2003-2129, 17th AIAA Aerodynamic Decelerator Systems Conference, May 2003. 60. B. de Gomars, “Drag of Cones at Zero Incidence,” URL cones. 61. National Highway Traffic Safety Administration, “NHTSA Tire Fuel Efficiency,” Report DOT HS 811 154, August 2009. 62. J. E. Cahill, “Summary of Section Data on Trailing-Edge High-Lift Devices,” National Advisory Committee for Aero¬ nautics, Report 938, 1949. Until they reach shallow water and feel the bottom, ocean waves are almost frictionless. Waves are created by winds, especially storms. Long waves — large distances between crests — travel the fastest and decay the slowest. Short waves decay more quickly but are still nearly friction¬ less. The pictured long waves, breaking on the beach in Narragansett, Rhode Island, might have been formed from a storm off the coast of Africa. The theory of ocean waves is based almost entirely upon frictionless flow. (Photo courtesy of Ellen Emerson White.) 8.1 Introduction and Review Chapter 8 Potential Flow and Computational Fluid Dynamics Motivation. The basic partial differential equations of mass, momentum, and energy were discussed in Chap. 4. A few solutions were then given for incompressible viscous flow in Sec. 4.10. The viscous solutions were limited to simple geometries and uni¬ directional flows, where the difficult nonlinear convective terms were neglected. Potential flows are not limited by such nonlinear terms. Then, in Chap. 7, we found an approxi¬ mation: patching boundary layer flows onto an outer inviscid flow pattern. For more complicated viscous flows, we found no theory or solutions, just experimental data or computer solutions. The purposes of the present chapter are (1) to explore examples of potential theory and (2) to indicate some flows that can be approximated by computational fluid dynamics (CFD). The combination of these two gives us a good picture of incom¬ pressible-flow theory and its relation to experiment. One of the most important appli¬ cations of potential-flow theory is to aerodynamics and marine hydrodynamics. First, however, we will review and extend the concepts of Chap. 4. Figure 8.1 reminds us of the problems to be faced. A free stream approaches two closely spaced bodies, creating an “internal” flow between them and “external” flows above and below them. The fronts of the bodies are regions of favorable gradient (decreasing pressure along the surface), and the boundary layers will be attached and thin: Inviscid theory will give excellent results for the outer flow if Re > 10“. For the internal flow between bodies, the boundary layers will grow and eventually meet, and the inviscid core vanishes. Inviscid theory works well in a “short” duct LID < 10, such as the nozzle of a wind tunnel. For longer ducts we must estimate boundary layer growth and be cautious about using inviscid theory. 521 522 Chapter 8 Potential Flow and Computational Fluid Dynamics Freestream Fig. 8.1 Patching viscous and inviscid flow regions. Potential theory in this chapter does not apply to the boundary layer regions. For the external flows above and below the bodies in Fig. 8.1, inviscid theory should work well for the outer flows, until the surface pressure gradient becomes adverse (increasing pressure) and the boundary layer separates or stalls. After the separation point, boundary layer theory becomes inaccurate, and the outer flow streamlines are deflected and have a strong interaction with the viscous near-wall regions. The theoretical analysis of separated-flow regions is an active research area at present. Review of Velocity Potential Concepts Recall from Sec. 4.9 that if viscous effects are neglected, low-speed flows are irrota- tional, V X V = 0, and the velocity potential (p exists, such that d(p V = V0 or M = — dx d(j) V = — dy (8.1) The continuity equation (4.73), V ■ V = 0, reduces to Laplace’s equation for (p: V (p = - y -F - y -t- - y = 0 d.r” dy^ dz~ and the momentum equation (4.74) reduces to Bernoulli’s equation: dcp pi, - h — H — V' + sz = const where V = \V(p\ dt p 2 ' ' Typical boundary conditions are known free-stream conditions d(p d(p d(p Outer boundaries: Known — , — , — dx dy dz (8.2) (8.3) (8.4) and no velocity normal to the boundary at the body surface: d(p Solid surfaces: — = 0 where n is perpendicular to body (8.5) dn 8.1 Introduction and Review 523 Unlike the no-slip condition in viscous flow, here there is no condition on the tangential surface velocity = d(p/ds, where s is the coordinate along the surface. This velocity is determined as part of the solution to the problem. Occasionally the problem involves a free surface, for which the boundary pressure is known and equal to p^, usually a constant. The Bernoulli equation (8.3) then supplies a relation at the surface between V and the elevation z of the surface. For steady flow. Free surface: = \|V(/)p = const — 2gZsmf (8.6) It should be clear to the reader that this use of Laplace’s equation, with known values of the derivative of (p along the boundaries, is much easier than a direct attack using the fully viscous Navier-Stokes equations. The analysis of Laplace’s equation is very well developed and is termed potential theory, with whole books written about its application to fluid mechanics [1 to 4]. There are many analytical techniques, includ¬ ing superposition of elementary functions, conformal mapping , numerical finite differences , numerical finite elements , numerical boundary elements , and electric or mechanical analogs that are now outdated. Having found (p(x, y, z, t) from such an analysis, we then compute V by direct differentiation in Eq. (8.1), after which we compute p from Eq. (8.3). The procedure is quite straightforward, and many interesting albeit idealized results can be obtained. A beautiful collection of computer¬ generated potential flow sketches is given by Kirchhoff . Review of Stream Function Concepts Recall from Sec. 4.7 that if a flow is described by only two coordinates, the stream function xjj also exists as an alternate approach. Eor plane incompressible flow in xy coordinates, the correct form is dtp dtp U = - V = - dy dx (8.7) The condition of irrotationality reduces to Laplace’s equation for tp also: or dV du 2UJ=0 = - dx dy d_/dtp\ dy\dyj (8.8) The boundary conditions again are known velocity in the stream and no flow through any solid surface: Free stream: Solid surface: dtp dtp Known — , — dx dy -i/Jbody = const (8.9fl) (8.%) Equation (8.9f>) is particularly interesting because any line of constant tpin a. flow can therefore be interpreted as a body shape and may lead to interesting applications. For the applications in this chapter, we may compute either (p or tp or both, and the solution will be an orthogonal flow net as in Fig. 8.2. Once found, either set of lines may be considered the of Thwaites’s laminar boundary method from Eqs. (7.54) and (7.56) reveals that separation does not occur on the front nose of the half-body. Therefore, Fig. 8.9fl is a very realistic picture of streamlines past a half-body nose. In contrast, when applied to the tail, Fig. 8.9c, Thwaites’s method predicts separation at about s/a ~ —2.2, or 9 ~ 110°. Thus, if a half-body is a solid surface. Fig. 8.9c is not realistic and a broad separated wake will form. However, if the half-body tail is a fluid line separating the sink-directed flow from the outer stream, as in Example 8.2, then Fig. 8.9c is quite realistic and useful. Computations for turbulent boundary layer theory would be similar: separation on the tail, no separation on the nose. Fig. 8.9 The Rankine half-body; pattern (c) is not found in a real fluid because of boundary layer separation, (a) UnifoiTn stream plus a source equals a half-body; stagnation point at.ic = —a = —m/U^. (b) Slight adverse gradient for s/a greater than 3.0: no separation. (c) Uniform stream plus a sink equals the rear of a half-body; stagnation point at X = a = m/U^. (d) Strong adverse gradient for s/a > —3.0: separation. t/j (max) = 1 .26 Laminar separation 8.3 Superposition of Plane Flow Solutions 533 EXAMPLE 8.2 An offshore power plant cooling- water intake sucks in 1500 ftVs in water 30 ft deep, as in Fig. E8.2. If the tidal velocity approaching the intake is 0.7 ft/s, (a) how far downstream does the intake effect extend and (b) how much width L of tidal flow is entrained into the intake? Solution Recall from Eq. (8.13) that the sink strength m is related to the volume flow Q and the depth b into the paper: Q 1500 ft7s iTTb 27r(30 ft) Therefore from Fig. 8.9 the desired lengths a and L are = 7.96 ft7s m 7.96 ft7s = 11.4 ft 0.7 ft/s L = 27Ta = 27r(11.4ft) = 71 ft Ans. (a) Arts, (b) Flow Past a Vortex Consider a uniform stream Uoo in the x direction flowing past a vortex of strength K with center at the origin. By superposition the combined stream function is l/^stream "f V^vortex sin 0 K\xi K (8.26) The velocity components are given by \ dip dip K Vr = - — = UooCos 9 Vg= — — = - Coo sm 0 + — (8.27) r dd dr r The streamlines are plotted in Fig. 8.10 by the graphical method, intersecting the circular streamlines of the vortex with the horizontal lines of the uniform stream. Fig. 8.10 Flow of a uniform stream past a vortex constructed by the graphical method. 534 Chapter 8 Potential Flow and Computational Fluid Dynamics By setting = Vg = 0 from (8.27) we find a stagnation point at 9 = 90°, r = a = K/Uao, or (x, y) = (0, a). This is where the counterclockwise vortex velocity Kir exactly cancels the stream velocity C/oo- Prohahly the most interesting thing about this example is that there is a nonzero lift force normal to the stream on the surface of any region enclosing the vortex, but we postpone this discussion until the next section. An Infinite Row of Vortices Consider an infinite row of vortices of equal strength K and equal spacing a, as in Fig. 8.11a. This case is included here to illustrate the interesting concept of a vortex sheet. From Eq. (8.14), the ith vortex in Fig. 8.11a has a stream function tpj = —K In r„ so that the total infinite row has a combined stream function CX t/t = 2 T i=l (b) Fig. 8.11 Superposition of vortices: (c) (a) an infinite row of equal strength; (b) streamline pattern for part (a); (c) vortex sheet: part (b) viewed from afar. y u = -'kKIo u = +'KKIa 8.3 Superposition of Plane Flow Solutions 535 The Vortex Sheet The Doublet It can be shown [2, Sec. 4.51] that this infinite sum of logarithms is equivalent to a closed-form function: i/t = —\K\vi 1 f iTty 27Vx\ - cosh - — cos - 2 ^ a a J (8.28) Since the proof uses the complex variable z = x iy, i = (—1)^^, we are not going to show the details here. The streamlines from Eq. (8.28) are plotted in Fig. 8.11fi, showing what is called a cat’s-eye pattern of enclosed flow cells surrounding the individual vortices. Above the cat’s eyes the flow is entirely to the left, and below the cat’s eyes the flow is to the right. Moreover, these left and right flows are uniform if \|y\| > a, which follows by differentiating Eq. (8.28): u ttK where the plus sign applies below the row and the minus sign above the row. This uniform left and right streaming is sketched in Fig. 8.11c. We stress that this effect is induced by the row of vortices: There is no uniform stream approaching the row in this example. When Fig. 8. life is viewed from afar, the streaming motion is uniform left above and uniform right below, as in Fig. 8.11c, and the vortices are packed so closely together that they are smudged into a continuous vortex sheet. The strength of the sheet is defined as 7 = IttK a (8.29) and in the general case 7 can vary with x. The circulation about any closed curve that encloses a short length dx of the sheet would be, from Eqs. (8.23) and (8.29), IttK dT = ui dx — u„ dx = {ui — m„) dx = - dx = j dx (8.30) where the subscripts / and u stand for lower and upper, respectively. Thus the sheet strength 7 = dTIdx is the circulation per unit length of the sheet. Thus when a vortex sheet is immersed in a uniform stream, 7 is proportional to the lift per unit length of any surface enclosing the sheet. Note that there is no velocity normal to the sheet at the sheet surface. Therefore a vortex sheet can simulate a thin-body shape, like a plate or thin airfoil. This is the basis of the thin airfoil theory mentioned in Sec. 8.7. As we move far away from the source-sink pair of Fig. 8.4, the flow pattern begins to resemble a family of circles tangent to the origin, as in Fig. 8.12. This limit of vanishingly small distance a is called a doublet. To keep the flow strength large enough to exhibit decent velocities as a becomes small, we specify that the product 536 Chapter 8 Potential Flow and Computational Fluid Dynamics Fig. 8.12 A doublet, or source-sink pair, is the limiting case of Fig. 8.4 viewed from afar. Streamlines are circles tangent to the x axis at the origin. This figure was prepared using the contour feature of MATLAB [34, 35]. lam remain constant. Let us call this constant A. Then the stream function of a doublet is 'ip lim a—>^0 2am = A — m tan , + lamy Xy 2 , 2 “ 2 , 2 X + y X + y (8.31) We have used the fact that tan” a ~ a as a becomes small. The quantity A is called the strength of the doublet. Equation (8.31) can be rearranged to yield x^ + so that, as advertised, the streamlines are circles tangent to the origin with centers on the y axis. This pattern is sketched in Fig. 8.12. Although the author has in the past laboriously sketched streamlines by hand, this is no longer necessary. Figure 8.12 was computer-drawn, using the contour feature of the student version of MATLAB . Simply set up a grid of points, spell out the stream function, and call for a contour. For Fig. 8.12, the actual statements were [X, Y] = meshgrid (-1 : .02 : 1); PSI = -Y. / (X. ''2 + Y. ''2); contour (X, Y, PSI, 100) This would produce 100 contour lines of tp from Eq. (8.31), with A = 1 for conve¬ nience. The plot would include grid lines, scale markings, and a surrounding box, and 8.4 Plane Flow Past Closed-Body Shapes 537 8.4 Plane Flow Past Closed-Body Shapes The Rankine Oval the circles might look a bit elliptical. These blemishes can be eliminated with three statements of cosmetic improvement: axis square grid off axis off The final plot, Fig. 8.12, has no markings but the streamlines themselves. MATLAB is thus a recommended tool and, in addition, has scores of other uses. All this chap¬ ter’s problem assignments that call for “sketch the streamlines/potential lines” can be completed using this contour feature. For further details, consult Ref. 34. In a similar manner the velocity potential of a doublet is found by taking the limit of Eq. (8.15) as a^O and 2am = A: or ^doublet Ax 2 , 2 X + y (8.32) The potential lines are circles tangent to the origin with centers on the x axis. Simply turn Fig. 8.12 clockwise 90° to visualize the (p lines, which are everywhere normal to the streamlines. The doublet functions can also be written in polar coordinates: 11;= - A sin 0 . The circulating streamlines inside the oval are uninteresting and not usually shown. The oval is the line ip = 0. There are stagnation points at the front and rear, x = ±L, and points of maximum velocity and minimum pressure at the shoulders, y = ±h, of the oval. All these 538 Chapter 8 Potential Flow and Computational Fluid Dynamics Fig. 8.13 Flow past a Rankine oval: {a) uniform stream plus a source- sink pair; (&) oval shape and streamlines for = 1.0. Flow Past a Circular Cylinder with Circulation ■ - ► ' AIA ^1 -m — - " " '/tv Source - ► Sink (a) Shoulder {b) parameters are a function of the basic dimensionless parameter m/(Uoca), which we can determine from Eq. (8.34): h hla L ( 2m — = cot - — = 1 H - ) a 2ml(Uaoa) a \ UooaJ (8.35) ttmax _ j ^ 2ml(^U^Q2} Uoo 1 + }^la' As we increase ml{UooO) from zero to large values, the oval shape increases in size and thickness from a flat plate of length 2a to a huge, nearly circular cylinder. This is shown in Table 8.1. In the limit as m/(Uooa) —■ oo, LIh 1.0 and 2.0, which is equivalent to flow past a circular cylinder. All the Rankine ovals except very thin ones have a large adverse pressure gradient on their leeward surface. Thus boundary layer separation will occur in the rear with a broad wake flow, and the inviscid pattern is unrealistic in that region. From Table 8.1 at large source strength the Rankine oval becomes a large circle, much greater in diameter than the source-sink spacing 2a. Viewed on the scale of the 8.4 Plane Flow Past Closed-Body Shapes 539 Table 8.1 Rankine Oval Parameters from Eq. (8.30) ml(U,oa) h/a L/a L/h 0.0 0.0 1.0 OO 1.0 0.01 0.031 1.010 32.79 1.020 0.1 0.263 1.095 4.169 1.187 1.0 1.307 1.732 1.326 1.739 10.0 4.435 4.583 1.033 1.968 100.0 14.130 14.177 1.003 1.997 00 oo OO 1.000 2.000 cylinder, this is equivalent to a uniform stream plus a doublet. We also throw in a vortex at the doublet center, which does not change the shape of the cylinder. Thus the stream function for flow past a circular cylinder with circulation, centered at the origin, is a uniform stream plus a doublet plus a vortex: A sin 9 ip = U^r sm 9 - A'ln r + const (8.36) The doublet strength A has units of velocity times length squared. For convenience, let A = where a is a length, and let the arbitrary constant in Eq. (8.36) equal K In a. Then the stream function becomes (8.37) The streamlines are plotted in Fig. 8.14 for four different values of the dimension¬ less vortex strength KI{UooO). For all cases the line 'ip = 0 corresponds to the circle r = a — that is, the shape of the cylindrical body. As circulation F = IttK increases, the velocity becomes faster and faster below the cylinder and slower and slower above it. The velocity components in the flow are given by Vr 1 dip r 39 Uoc cos 9\ 1 Ve= - dip dr — sin 9{ 1 (8.38) The velocity at the cylinder surface r = a is purely tangential, as expected: K vpr = a) = 0 Vg{r = a) = —2Uoc sin 0 + — (8.39) For small K, two stagnation points appear on the surface at angles 9^ where Vg = 0; or, from Eq. (8.39), sin = K 2 co^ (8.40) Eigure 8.14a is for K = 0, 9^ = 0 and 180°, or doubly symmetric inviscid flow past a cylinder with no circulation. Eigure 8.14h is for K/(Uaoa) = i, 9^ = 30 and 150°; and Eig. 8.14c is the limiting case where the two stagnation points meet at the top, K/iU^a) = 2,9, = 90°. 540 Chapter 8 Potential Flow and Computational Fluid Dynamics Fig. 8.14 Flow past a circular cylinder with circulation for values of KHU^a) of (fl) 0, (b) 1.0, (c) 2.0, and (d) 3.0. (c) (d) The Kutta-Joukowski Lift Theorem For K > 2Uooa, Eq. (8.40) is invalid, and the single stagnation point is above the cylinder, as in Fig. 8.14t/, at a point y = h given by ^ = i[^ + ,3= - 4)'“] /3 > 2 In Fig. %.\Ad, KI{UooO) = 3.0, and hla = 2.6. The cylinder flows with circulation. Figs. 8.14h to d, develop an inviscid downward lift normal to the free stream, called the Magnus-Robins force. This lift is proportional to stream velocity and vortex strength. Its discovery, by experiment, has long been attributed to the German physicist Gustav Magnus, who observed it in 1853. It is now known [40, 45] that the brilliant British engineer Benjamin Robins first reported a lift force on a spinning body in 1761. We see from the streamline patterns that the veloc¬ ity on top of the cylinder is less, and, thus, from Bernoulli’s equation, the pressure is higher. On the bottom, we see tightly packed streamlines, high velocity, and low pressure; viscosity is neglected. Inviscid theory predicts this force. The surface velocity is given by Eq. (8.39). From Bernoulli’s equation (8.3), neglecting gravity, the surface pressure is given by 1 2 1 K Foo + - pUoo = Ps + If P\ -2C/oe sm 61 -f — or Fi = Foo + iPU «(1 - 4 sin^ 9 -f 4/3 sin 0 - 0^) (8.41) 8.4 Plane Flow Past Closed-Body Shapes 541 where (3 = K/(Uaoa) and p^o is the free-stream pressure. If b is the cylinder depth into the paper, the drag D is the integral over the surface of the horizontal component of pressure force: D = - Itt (Ps — Poo) cos 9 ba dd where p^ — Poo is substituted from Eq. (8.41). But the integral of cos 9 times any power of sin 9 over a full cycle 2tt is identically zero. Thus we obtain the (perhaps surprising) result ^(cylinder with circulation) = 0 (8.42) This is a special case of d’Alembert’s paradox, mentioned in Sec. 1.2: According to inviscid theory, the drag of any body of any shape immersed in a uniform stream is identically zero. D’Alembert published this result in 1752 and pointed out himself that it did not square with the facts for real fluid flows. This unfortunate paradox caused everyone to over¬ react and reject all inviscid theory until 1904, when Prandtl first pointed out the profound effect of the thin viscous boundary layer on the flow pattern in the rear, as in Fig. 1.2b, for example. The lift force L normal to the stream, taken positive upward, is given by summa¬ tion of vertical pressure forces: L = - iPs ~ Poo) sin 9 ba d9 -'o Since the integral over 27r of any odd power of sin 9 is zero, only the third term in the parentheses in Eq. (8.41) contributes to the lift: L = 1 put 4K aU at ba l-K sim 9 d9 = —pUoo{2.TvK)b or (8.43) Notice that the lift seems independent of the radius a of the cylinder. Actually, though, as we shall see in Sec. 8.7, the circulation T depends on the body size and orientation through a physical requirement. Equation (8.43) was generalized by W. M. Kutta in 1902 and independently by N. Joukowski in 1906 as follows: According to inviscid theory, the lift per unit depth of any cylinder of any shape immersed in a uniform stream equals pUooT, where E is the total net circulation contained within the body shape. The direction of the lift is 90° from the stream direction, rotating opposite to the circulation. 542 Chapter 8 Potential Flow and Computational Fluid Dynamics The problem in airfoil analysis, Sec. 8.7, is thus to determine the circulation T as a function of airfoil shape and orientation. Lift and Drag of Rotating Cylinders^ The flows in Fig. 8.14 are mathematical: a doublet plus a vortex plus a uniform stream. The physical realization could be a rotating cylinder in a free stream. The no- slip condition would cause the fluid in contact with the cylinder to move tangen¬ tially at velocity ve = au), setting up a net circulation T. Measurement of forces on a spinning cylinder is very difficult, and no reliable drag data are known to the author. However, Tokumaru and Dimotakis used a clever auxiliary scheme to measure lift forces at Re^, = 3800. Figure 8.15 shows lift and drag coefficients, based on frontal area {lab), for a rotating cylinder at Re^ = 3800. The drag curve is from CFD calculations . Reported CFD drag results, from several different authors, are quite controversial because they do not agree, even qualitatively. The writer feels that Ref. 41 gives the most reliable results. Note that the experimental C/, increases to a value of 15.3 at auj/Uoo = 10. This contradicts an early theory of Prandtl, in 1926, that the maximum possible value of would be 47r ~ 12.6, corresponding to the flow conditions in Fig. 8.14c. The inviscid theory for lift would be: L InpUn^Kb InVg, (8.44) where Vg^ = Kla is the peripheral speed of the cylinder. Figure 8.15 shows that the theoretical lift from Eq. (8.44) is much too high, but the measured lift is quite respectable, much larger in fact than a typical airfoil of the same chord length, as in Fig. 7.25. Thus rotating cylinders have practical possibilities. The Flettner rotor ship built in Germany in 1924 employed rotating vertical cylinders that developed a thrust due to any winds blowing past the ship. The Flettner design did not gain any popularity, but such inventions may be more attractive in this era of high energy costs. Theory, = rrauj/t/c Fig. 8.15 Drag and lift of a rotating cylinder of large aspect ratio at Re^ = 3800, after Tokumaru and Dimotakis and Sengupta et al. . 0 2 4 6 a Lt^/Uoo 8 10 ^The writer is indebted to Prof. T. K. Sengupta of I.I.T. Kanpur for data and discussion for this subsection. 8.4 Plane Flow Past Closed-Body Shapes 543 EXAMPLE 8.3 The experimental Flettner rotor sailboat at the University of Rhode Island is shown in Fig. E8.3. The rotor is 2.5 ft in diameter and 10 ft long and rotates at 220 r/min. It is driven by a small lawnmower engine. If the wind is a steady 10 kn and boat relative motion is neglected, what is the maximum thrust expected for the rotor? Assume standard air and water density. Solution Convert the rotation rate to w = 27r(220)/60 = 23.04 rad/s. The wind velocity is 10 kn = 16.88 ft/s, so the velocity ratio is au) _ (1.25 ft) (23.04 rad/s) ~ 16.88 ft/s Using Fig. 8.15, we read Co ~ 0.7 and Ci ~ 2.5. From Table A.6, standard air density in BG units is 0.00238 slug/tf. Then the estimated rotor lift and drag are 1 1 L = C^-pUi 2ba = (2.5)-( 0.00238 D = Co-pUi 2ba = (0.7)-( 0.00238 ^ slugV _ ft ft^ / slug 16.88- 2(10 ft) (1.25 ft) = 21.21bf 16.88- I 2(10 ft) (1.25 ft) =5.91bf The maximum thrust available to the sailboat is the resultant of these two: F = [(21.2)^ -f (5.9)^] = 22.0 Ibf Ans. Note that water density did not enter into this calculation, which is a force due to air. If aligned along the boat’s keel, this thrust will drive the boat through the water at a speed of about 4 kn. E8.3 (Courtesy of R. C. Lessmann, University of Rhode Island.) 544 Chapter 8 Potential Flow and Computational Fluid Dynamics The Kelvin Oval Potential Flow Analogs Fig. 8.16 Kelvin oval body shapes as a function of the vortex strength parameter KI{lJ^a)\ outer streamlines not shown. Comment: For the sake of a numerical example, we have done something improper here. We have used data for Re^ = 3800 to estimate forces when the rotor Re^ ~ 260,000. Do not do this in your real job after you graduate! A family of body shapes taller than they are wide can be formed by letting a uniform stream flow normal to a vortex pair. If Uao is to the right, the negative vortex —K is placed at y = +a and the counterclockwise vortex -\-K placed at y = —a, as in Fig. 8.16. The combined stream function is 1 + (y + a)^ 'iP=U^y--K\n (8.45) 2 jC + {y - a) The body shape is the line 'ip = Q, and some of these shapes are shown in Fig. 8.16. For K/(Uoca) > 10 the shape is within 1 percent of a Rankine oval (Fig. 8.13) turned 90°, but for small K/(Uooa) the waist becomes pinched in, and a figure-eight shape occurs at 0.5. For Kl(Uooa) < 0.5 the stream blasts right between the vortices and isolates two more or less circular body shapes, one surrounding each vortex. A closed body of practically any shape can be constructed by proper superposition of sources, sinks, and vortices. See the advanced work in Refs. 1 to 3 for further details. A summary of elementary potential flows is given in Table 8.2. For complicated potential flow geometries, one can resort to other methods than super¬ position of sources, sinks, and vortices. A variety of devices simulate solutions to Laplace’s equation. From 1897 to 1900 Hele-Shaw developed a technique whereby laminar flow between very closely spaced parallel plates simulated potential flow when viewed v 8.4 Plane Flow Past Closed-Body Shapes 545 Table 8.2 Summary of Plane Incompressible Potential Flows Type of flow Potential functions Remarks Stream \U rj) = Uy 0) or sink {m < 0) tp = mO - r See Fig. 8.12 Rankine oval ^ = Ur sin 9 + m{9i — ^2) See Fig. 8.13 Cylinder with circulation ip = U sin 6 r K\n- See Fig. 8.14 v r J a from above the plates. Obstructions simulate body shapes, and dye streaks represent the streamlines. The Hele-Shaw apparatus makes an excellent laboratory demonstra¬ tion of potential flow [10, pp. 197-198, 219-220]. Figure 8.17a illustrates Hele-Shaw (potential) flow through an array of cylinders, a flow pattern that would be difficult to analyze just using Laplace’s equation. However beautiful this array pattern may be, it is not a good approximation to real (laminar viscous) array flow. Figure 8.177> shows experimental streakline patterns for a similar staggered-array flow at Re ~ 6400. We see that the interacting wakes of the real flow (Fig. 8.177>) cause intensive mixing and transverse motion, not the smooth streaming passage of the potential flow model (Fig. 8.17a). The moral is that this is an internal flow with multiple bodies and, therefore, not a good candidate for a realistic potential flow model. Other flow-mapping techniques are discussed in Ref. 8. Electromagnetic fields also satisfy Laplace’s equation, with voltage analogous to velocity potential and current lines analogous to streamlines. At one time commercial analog field plotters were available, using thin conducting paper cut to the shape of the flow geometry. Potential lines (voltage contours) were plotted by probing the paper with a potentiometer pointer. Hand- sketching “curvilinear square” techniques were also popular. The avail¬ ability and the simplicity of computer potential flow methods [5 to 7] have made analog models obsolete. EXAMPLE 8.4 A Kelvin oval from Fig. 8.16 has KI{U^a) = 1.0. Compute the velocity at the top shoulder of the oval in terms of U^. Solution We must locate the shoulder y = h from Eq. (8.45) for tp = 0 and then compute the velocity by differentiation. Atip = 0 and y = h and x = 0, Eq. (8.45) becomes h K h/a + 1 — = - In - a a hla — 1 With KI(U^a) = 1.0 and the initial guess hla ~ 1.5 from Fig. 8.16, we iterate and find the location hla = 1.5434. 546 Chapter 8 Potential Flow and Computational Fluid Dynamics By inspection w = 0 at the shoulder because the streamline is horizontal. Therefore the shoulder velocity is, from Eq. (8.45), d-lp K K u = — = U^ + - dy y=i, h — a h + a Introducing K = and h = 1.5434a, we obtain “shoulder = (^ao(l-0 + 1.84 - 0.39) = 2A5U^ Ans. Because they are short-waisted compared with a circular cylinder, all the Kelvin ovals have shoulder velocity greater than the cylinder result 2.0U^ from Eq. (8.39). Fig. 8.17 Elow past a staggered array of cylinders: (a) potential flow model using the Hele-Shaw apparatus (TQ Education and Training Ltd.)’, (b) experimental streaklines for actual staggered- array flow at Reo « 6400. (From Ref. 36, Courtesy of Jack Hoyt, with the permission of the American Society of Mechanical Engineers.) 8.5 Other Plane Potential Flows 547 8.5 Other Plane Potential Flows^ References 2 to 4 treat many other potential flows of interest in addition to the cases presented in Secs. 8.3 and 8.4. In principle, any plane potential flow can be solved by the method of conformal mapping, by using the complex variable z = X + iy i = ( - 1 ) It turns out that any arbitrary analytic function of this complex variable z has the remarkable property that both its real and its imaginary parts are solutions of Laplace’s equation. If f{z) =f{x + iy) = ffx,y) + if2{x,y) then dx^ a/ dx^ dy^ (8.46) We shall assign the proof of this as Prob. W8.4. Even more remarkable if you have never seen it before is that lines of constant fi will be everywhere perpendicular to lines of constant /2: dy dx /i = C {dy/dx)f^^c (8.47) This is true for totally arbitrary /(z) as long as this function is analytic; that is, it must have a unique derivative dfidz at every point in the region. The net result of Eqs. (8.46) and (8.47) is that the functions fi and/2 can be interpreted to be the potential lines and streamlines of an inviscid flow. By long custom we let the real part of fig) be the velocity potential and the imaginary part be the stream function: f{z) = (p{x, y) + iip{x, y) (8.48) We try various functions /(z) and see whether any interesting flow pattern results. Of course, most of them have already been found, and we simply report on them here. We shall not go into the details, but there are excellent treatments of this complex- variable technique on both an introductory and a more advanced [2, 3] level. The method is less important now because of the popularity of computer techniques. As a simple example, consider the linear function f{z) = t/ooZ = t/ooX + iUooy It follows from Eq. (8.48) that (p = Ua^x and t/ = Uooy, which, we recall from Eq. (8.12), represents a uniform stream in the x direction. Once you get used to the complex variable, the solution practically falls in your lap. To find the velocities, you may either separate a, there is an adverse pressure gradient along the wall, and boundary layer theory should be used to predict separation. As mentioned in conjunction with the Kutta-Joukowski lift theorem, Eq. (8.43), the problem in airfoil theory is to determine the net circulation E as a function of airfoil shape and free-stream angle of attack a. Even if the airfoil shape and free-stream angle of attack are specified, the potential flow theory solution is nonunique: An infinite family of solutions can be found cor¬ responding to different values of circulation E. Eour examples of this nonuniqueness were shown for the cylinder flows in Fig. 8.14. The same is true of the airfoil, and Fig. 8.22 shows three mathematically acceptable “solutions” to a given airfoil flow for small (Fig. 8.22a), large (Fig. 8.22f>), and medium (Fig. 8.22c) net circulation. ^This section may be omitted without loss of continuity. 8.7 Airfoil Theory 555 You can guess which case best simulates a real airfoil from the earlier discussion of transient lift development in Fig. 7.23. It is the case (Fig. 8.22c) where the upper and lower flows meet and leave the trailing edge smoothly. If the trailing edge is rounded slightly, there will be a stagnation point there. If the trailing edge is sharp, approxi¬ mating most airfoil designs, the upper- and lower-surface flow velocities will be equal as they meet and leave the airfoil. This statement of the physically proper value of F is generally attributed to W. M. Kutta, hence the name Kutta condition, although some texts give credit to Joukowski and/or Chaplygin. All airfoil theories use the Kutta condition, which is in good agree¬ ment with experiment. It turns out that the correct circulation F^utta depends on flow velocity, angle of attack, and airfoil shape. Potential Theory for Thick Cambered Airfoils The theory of thick cambered airfoils is covered in advanced texts [for example, 2 to 4] ; Ref. 1 3 has a thorough and comprehensive review of both in viscid and viscous aspects of airfoil behavior. Basically the theory uses a complex-variable mapping that transforms the flow about a cylinder with circulation in Fig. 8.14 into flow about a foil shape with circula¬ tion. The circulation is then adjusted to match the Kutta condition of smooth exit flow from the trailing edge. Regardless of the exact airfoil shape, the inviscid mapping theory predicts that the correct circulation for any thick cambered airfoil is rKutta = ttCI/oo (^1 + si” + /3) (8.57) where j3 = tan” (2/i/C) and h is the maximum camber, or maximum deviation of the airfoil midline from its chord line, as in Fig. 8.24a. The lift coefficient of the infinite-span airfoil is thus jOf/0 2am = X We leave the proof of this limit as a problem , / m m\ X cos 0 tPdoubiet = lim - h - = - r — (8.79) ^source ^sink/ V The streamlines and potential lines are shown in Fig. 8.29. Unlike the plane doublet flow of Fig. 8.12, neither set of lines represents perfect circles. A sin^ 0 - m cos 6»si„k) = - - - (8.78) . The point-doublet velocity potential is 8.8 Axisymmetric Potential Flow 565 Uniform Stream plus a Point Source Uniform Stream plus a Point Doublet Fig. 8.30 Streamlines for a Rankine half-body of revolution. By combining Eqs. (8.75) and (8.77), we obtain the stream function for a uniform stream plus a point source at the origin; 'ip = sin^ 9 + m cos 9 (8.80) From Eq. (8.73) the velocity components are, by differentiation, m F’r = Uoo cos 0 + -y Itg = - f/oo sin 0 (8.81) r Setting these equal to zero reveals a stagnation point at 9 = 180° and r = a = as shown in Fig. 8.30. If we let m = the stream function can be rewritten as tp 1 / r Y 2 - y = COS 9 - — ) sin^ 9 (8.82) 2\aJ The stream surface that passes through the stagnation point (r, 9) = (a, tt) has the value tp = —U^a^ and forms a half-body of revolution enclosing the point source, as shown in Fig. 8.30. This half-body can be used to simulate a pitot tube. Far down¬ stream the half-body approaches the constant radius R = 2a about the x axis. The maximum velocity and minimum pressure along the half-body surface occur at 9 = 70.5°, r = aVY, U, = 1.155C/oo. Downstream of this point there is an adverse gradient as U, slowly decelerates to [/oo, but boundary layer theory indicates no flow separation. Thus Eq. (8.82) is a very realistic simulation of a real half-body flow. But when the uniform stream is added to a sink to form a half-body rear surface, similar to Fig. 8.9c, separation is predicted and the rear inviscid pattern is not realistic. From Eqs. (8.75) and (8.78), combination of a uniform stream and a point doublet at the origin gives tp j. 2 sin^ 9 + (8.83) Examination of this relation reveals that the stream surface tp sphere of radius 0 corresponds to the (8.84) 566 Chapter Potential Flow and Computational Fluid Dynamics The Concept of Hydrodynamic Mass Fig. 8.31 Streamlines and potential lines for inviscid flow past a sphere. This is exactly analogous to the cylinder flow of Fig. 8.14a formed by combining a uniform stream and a line doublet. Letting A = \Uaoa^ for convenience, we rewrite Eq. (8.83) as T-^ = -sm^0K-7 (8.85) 2^00^ \a ry The streamlines for this sphere flow are plotted in Fig. 8.31. By differentiation from Eq. (8.73) the velocity components are Vr = t/oo COS '^6 = “2 0^2 H — (8.86) We see that the radial velocity vanishes at the sphere surface r = a, as expected. There is a stagnation point at the front (a, tt) and the rear (a, 0) of the sphere. The maximum velocity occurs at the shoulder (a, ±y7r), where D,. = 0 and Vg = +1.5 Loo. The surface velocity distribution is K = -Vg\r=a = lUoo sin 6 (8.87) Note the similarity to the cylinder surface velocity equal to 2Loo sin 0 from Eq. (8.39) with zero circulation. Equation (8.87) predicts, as expected, an adverse pressure gradient on the rear (9 < 90°) of the sphere. If we use this distribution with laminar boundary layer theory [for example, 15, p. 294], separation is computed to occur at about 9 = 76°, so that in the actual flow pattern of Fig. 7.14 a broad wake forms in the rear. This wake interacts with the free stream and causes Eq. (8.87) to be inaccurate even in the front of the sphere. The measured maximum surface velocity is equal only to about 1.3 Loo and occurs at about 9 = 107° (see Ref. 15, Sec. 4.10.4, for further details). When a body moves through a fluid, it must push a finite mass of fluid out of the way. If the body is accelerated, the surrounding fluid must also be accelerated. The body behaves as if it were heavier by an amount called the hydrodynamic mass (also l.SFoo Potential lines 8.8 Axisymmetric Potential Flow 567 called the added or virtual mass) of the fluid. If the instantaneous body velocity is U(f), the summation of forces must include this effect: ^ t/U 2 F = (m + m,,) - (8.88) dt where mi,, the hydrodynamic mass, is a function of body shape, the direction of motion, and (to a lesser extent) flow parameters such as the Reynolds number. According to potential theory [2, Sec. 6.4; 3, Sec. 9.22], m^ depends only on the shape and direction of motion and can be computed by summing the total kinetic energy of the fluid relative to the body and setting this equal to an equivalent body energy: RE fluid \dm Vfei = \mi,U^ (8.89) The integration of fluid kinetic energy can also be accomplished by a body-surface integral involving the velocity potential [16, Sec. 11]. Consider the previous example of a sphere immersed in a uniform stream. By subtracting out the stream velocity we can replot the flow as in Fig. 8.32, showing the streamlines relative to the moving sphere. Note the similarity to the doublet flow in Fig. 8.29. The relative velocity components are found by subtracting U from Eqs. (8.86): Ua^ cos 6 Ua^ sin 6 Fig. 8.32 Potential flow streamlines relative to a moving sphere. Compare with Figs. 8.29 and 8.31. Fluid particle: df(KE) = ^dmV^ 568 Chapter Potential Flow and Computational Fluid Dynamics 8.9 Numerical Analysis The element of fluid mass, in spherical polar coordinates, is dm = p{27vr sin 9)r dr dO When dm and are substituted into Eq. (8.89), the integral can be evaluated: KEnuid = or m;,(sphere) = Ipna^ (8.90) Thus, according to potential theory, the hydrodynamic mass of a sphere equals one- half of its displaced mass, independent of the direction of motion. A similar result for a cylinder moving normal to its axis can be computed from Eqs. (8.38) after subtracting out the stream velocity. The result is m;, (cylinder) = pna^L (8.91) for a cylinder of length L, assuming two-dimensional motion. The cylinder’s hydro- dynamic mass equals its displaced mass. Tables of hydrodynamic mass for various body shapes and directions of motion are given by Patton . See also Ref. 21. When potential flow involves complicated geometries or unusual stream conditions, the classical superposition scheme of Secs. 8.3 and 8.4 becomes less attractive. Con¬ formal mapping of body shapes, by using the complex-variable technique of Sec. 8.5, is no longer popular. Numerical analysis is the appropriate modern approach, and at least three different approaches are in use: 1. The finite element method (FEM) [6, 19] 2. The finite difference method (EDM) [5, 20, 23-26], or its close sibling, the finite volume method . 3. a. Integral methods with distributed singularities b. The boundary element method (BEM) [7, 38] Methods 3a and 3b are closely related, having first been developed on an ad hoc basis by aerodynamicists in the 1960s and then generalized into a multipurpose applied mechanics technique in the 1970s . Methods 1 (or FEM) and 2 (or EDM), though strikingly different in concept, are comparable in scope, mesh size, and general accuracy. We concentrate here on the latter method for illustration purposes. All three of these methods — FEM, EDM, and BEM — are popular in present-day computational fluid dynamics. Although simplified online CFD codes are available — sometimes free — the writer believes that CFD, as a serious method of flow analysis, should wait until one studies the professional software available. This subject is more appropriate for advanced electives or graduate school. The discussion here will be brief and descriptive, with only nominal illustrations for an EDM method. The Finite Element Method The finite element method is applicable to all types of linear and nonlinear partial differential equations in physics and engineering. The computational domain is divided 8.9 Numerical Analysis 569 The Finite Difference Method into small regions, usually triangular or quadrilateral. These regions are delineated with a finite number of nodes where the field variables — temperature, velocity, pres¬ sure, stream function, and so on — are to be calculated. The solution in each region is approximated by an algebraic combination of local nodal values. Then the approxi¬ mate functions are integrated over the region, and their error is minimized, often by using a weighting function. This process yields a set of N algebraic equations for the N unknown nodal values. The nodal equations are solved simultaneously, by matrix inversion or iteration. For further details see Ref. 6 or 19. Although textbooks on numerical analysis [5, 20] apply finite difference techniques to many different problems, here we concentrate on potential flow. The idea of FDM is to approximate the partial derivatives in a physical equation by “differences” between nodal values spaced a finite distance apart — a sort of numerical calculus. The basic partial differential equation is thus replaced by a set of algebraic equations for the nodal values. For potential (inviscid) flow, these algebraic equations are linear, but they are generally nonlinear for viscous flows. The solution for nodal values is obtained by iteration or matrix inversion. Nodal spacings need not be equal. Here we illustrate the two-dimensional Laplace equation, choosing for convenience the stream-function form (8.92) subject to known values of tp along any body surface and known values of d'lp/dx and d'lp/dy in the free stream. Our finite difference technique divides the flow field into equally spaced nodes, as shown in Fig. 8.33. To economize on the use of parentheses or functional notation. The Finite Difference Method Av t-Uj o -o Fig. 8.33 Definition sketch for a two-dimensional rectangular finite difference grid. 570 Chapter Potential Flow and Computational Fluid Dynamics subscripts i and j denote the position of an arbitrary, equally spaced node, and ipi j denotes the value of the stream function at that node: '0ij = 'ipixo + i Ax, yo + j Ay) Thus 'ipi+ij is just to the right of t/;,-,-, and 'tpij+i is just above. An algebraic approximation for the derivative dip/dx is dip tp(x + Ax, y) — ip{x, y) dx Ax A similar approximation for the second derivative is d^ip dx^ J_ Ax ipix + Ax, y) — ipix, y) ipix, y) — ip{x — Ax, y) Ax Ax The subscript notation makes these expressions more compact: dip 1 dx Ax i+ Ij '^u j) (8.93) d^ip dx^ 1 Ax^ - ^ipij + ipi-ij) These formulas are exact in the calculus limit as Ax — > 0, but in numerical analysis we keep Ax and Ay finite, hence the term finite differences. In an exactly similar manner we can derive the equivalent difference expressions for the y direction: dip dy J_ iti. J+i d^ip 1 lly^ “ V ~ (8.94) The use of subscript notation allows these expressions to be programmed directly into a scientific computer language. When (8.93) and (8.94) are substituted into Laplace’s equation (8.92), the result is the algebraic formula 2(1 + f3)ipij = ipi-ij + ipi+ij + f3(ipij-i + ipij+i) (8.95) where (3 = (Ax/Ay)^ depends on the mesh size selected. This finite difference model of Laplace’s equation states that every nodal stream-function value ipjj is a linear combination of its four nearest neighbors. The most commonly programmed case is a square mesh (/3 = 1), for which Eq. (8.95) reduces to 'ipi.j ^ + ipij-i + ipi+uj + ipi-i.j) (8.96) Thus, for a square mesh, each nodal value equals the arithmetic average of the four neighbors shown in Fig. 8.33. The formula is easily remembered and easily pro¬ grammed. The formula is applied in iterative fashion sweeping over each of the 8.9 Numerical Analysis 571 internal nodes (I, J), with known values of P specified at each of the surrounding boundary nodes. Any initial guesses can be specified for the internal nodes, and the iteration process will converge to the final algebraic solution in a finite number of sweeps. The numerical error, compared with the exact solution of Laplace’s equation, is proportional to the square of the mesh size. The Boundary Element Method A relatively new technique for numerical solution of partial differential equations is the boundary element method (BEM). Reference 7 is an introductory textbook outlin¬ ing the concepts of BEM. There are no interior elements. Rather, all nodes are placed on the boundary of the domain, as in Eig. 8.34. The “element” is a small piece of the boundary surrounding the node. The “strength” of the element can be either constant or variable. Eor plane potential flow, the method takes advantage of the particular solution 1 1 -lb = — In - (8.97) Itt r which satisfies Laplace’s equation, = 0. Each element i is assumed to have a different strength t/;,. Then r represents the distance from that element to any other point in the flow field. Summing all these elemental effects, with proper boundary conditions, will give the total solution to the potential flow problem. At each element of the boundary, we typically know either the value of -ip or the value of d-ip/dn, where n is normal to the boundary. (Mixed combinations of -ip and dtpldn are also possible but are not discussed here.) The correct strengths t/;, are such that these boundary conditions are satisfied at every element. Summing these effects over N elements requires integration by parts plus a careful evaluation of the (singular) effect of element i upon itself. The mathematical details are given in Ref. 7. The result is a set of N algebraic equations for the unknown boundary values. In the case of elements of constant strength, the final expression is ]_ 2 N / 'IPi + S'/) J=1 k d-ib \ -^ds . dn J i= ItoN (8.98) The integrals, which involve the logarithmic particular solution ip from Eq. (8.97), are evaluated numerically for each element. Fig. 8.34 Boundary elements of constant strength in plane potential flow. n Element j 572 Chapter 8 Potential Flow and Computational Fluid Dynamics Viscous Flow Computer Models One-Dimensional Unsteady Flow Fig. 8.35 An equally spaced finite difference mesh for one-dimensional viscous flow [Eq. (8.99)]. Reference 7 is a general introduction to boundary elements, while Ref. 38 empha¬ sizes programming methods. Meanwhile, research continues. Dargush and Grigoriev have developed a multilevel boundary element method for steady Stokes or creeping flows (see Sec. 7.6) in irregular geometries. Their scheme avoids the heavy memory and CPU-time requirements of most boundary element methods. They esti¬ mate that CPU time is reduced by a factor of 700,000 and required memory is reduced by a factor of 16,000. Our previous finite difference model of Laplace’s equation, as in Eq. (8.96), was very well behaved and converged nicely with or without overrelaxation. Much more care is needed to model the full Navier-Stokes equations. The challenges are quite differ¬ ent, and they have been met to a large extent, so there are now many textbooks [5, 20, 23 to 27] on (fully viscous) CFD. This is not a textbook on CFD, but we will address some of the issues in this section. We consider a simplified problem, showing that even a single viscous term introduces new effects and possible instabilities. Recall (or review) Prob. P4.85, where a wall moves and drives a viscous fluid parallel to itself. Gravity is neglected. Let the wall be the plane y = 0, moving at a speed Uoit), as in Fig. 8.35. A uniform vertical grid, of spacing Ay, has nodes n at which the local velocity is to be calculated, where superscript j denotes the time-step y'Af. The wall is n = 1. If m = u(y, i) only and V = w = 0, continuity, V • V = 0, is satisfied, and we need only solve the x-momentum Navier-Stokes equation: du d^u St dy^ (8.99) where u = pip. Utilizing the same finite difference approximations as in Eq. (8.93), we may model Eq. (8.99) algebraically as a forward time difference and a central spatial difference: - u,i At Mi+ 1 - luj + “n-1 Ay^ i «+ 1 Ay 0 Ay Q n-1 Ay Ay u=U„ Wall n = \ 8.9 Numerical Analysis 573 Steady Two-Dimensional Laminar Flow Rearrange and find that we can solve explicitly for at the next time-step j + 1: ... i/Af ui ~ (1 - 2cr) M,' + a{ui-i + + cr = (8.100) A/ Thus u at node n at the next time-step y + 1 is a weighted average of three previous values, similar to the “four-nearest-neighbors” average in the laplacian model of Eq. (8.96). Since the new velocity is calculated immediately, Eq. (8.100) is called an explicit model. It differs from the well-behaved laplacian model, however, because it may be unstable. The weighting coefficients in Eq. (8.100) must all be positive to avoid divergence. Now cr is positive, but (1 — 2cr) may not be. Therefore, our explicit viscous flow model has a stability requirement: i/Af 2 (8.101) Normally one would first set up the mesh size Ay in Fig. 8.35, after which Eq. (8.101) would limit the time-step At. The solutions for nodal values would then be stable, but not necessarily that accurate. The mesh sizes Ay and At could be reduced to increase accuracy, similar to the case of the potential flow laplacian model (8.96). For example, to solve Prob. P4.85 numerically, one sets up a mesh with plenty of nodes (30 or more Ay within the expected viscous layer); selects At according to Eq. (8.101); and sets two boundary conditions® for all j: = Uq sin LVt and % = 0, where N is the outermost node. For initial conditions, perhaps assume the fluid initially at rest: m = 0 for 2 < n < N — 1. Sweeping the nodes 2 < n < N — 1 using Eq. (8.100) (an Excel spreadsheet is excellent for this), one generates numerical values of m/, for as long as one desires. After an initial transient, the final “steady” fluid oscillation will approach the classical solution in viscous flow textbooks . Try Prob. P8.115 to demonstrate this. The previous example, unsteady one-dimensional flow, had only one viscous term and no convective accelerations. Let us look briefly at incompressible two-dimensional steady flow, which has four of each type of term, plus a nontrivial continuity equation: du dv Continuity: - 1 - = 0 (8.102fl) dx dy du du I dp ( d^u d^u\ X momentum: u - h v — = - h i/ — ^ H - ^ (8.102h) dx dy p dx \dx^ 5yV dv dv I dp ( d^V d^v\ y momentum: u - h v — = - h H ^ H - ^ (8.102c) dx dy p dy 5yV These equations, to be solved for (w, v, p) as functions of (x, y), are familiar to us from analytical solutions in Chaps. 4 and 6. However, to a numerical analyst, they are odd, because there is no pressure equation — that is, a differential equation for which the dominant derivatives involve p. This situation has led to several different Tinite differences are not analytical; one must set Uq and u) equal to numerical values. 574 Chapter Potential Flow and Computational Fluid Dynamics “pressure adjustment” schemes in the literature [20, 23 to 27], most of which manip¬ ulate the continuity equation to insert a pressure correction. A second difficulty in Eqs. (8.102& and c) is the presence of nonlinear convective accelerations such as u(duldx), which creates asymmetry in viscous flows. Early aftempfs, which modeled such ferms wifh a cenfral difference, led to numerical insta¬ bility. The remedy is to relate convection finite differences solely to the upwind flow enfering the cell, ignoring the downwind cell. Eor example, the derivative duldx could be modeled, for a given cell, as (Uupwind ~ Such improvements have made fully viscous CFD an effective tool, with various commercial user-friendly codes available. For details beyond our scope, see Refs. 20 and 23 to 27. Mesh generation and gridding have also become quite refined in modern CFD. Figure 8.36 illustrates a CFD solution of two-dimensional flow pasf an NACA {a) Fig. 8.36 CFD results for water flow past an NACA 66(MOD) hydrofoil (from Ref. 28, with permission of the American Society of Mechanical Engineers): (a) C gridding, 262 by 91 nodes; (b) surface pressures at a = 1°. 0.6 0.5 0.4 0.3 0.2 0.1 0.0 -0.1 -0.2 Expt. Comp. I I I I I I 0.0 0.5 1.0 xtC (b) 8.9 Numerical Analysis 575 Commercial CFD Codes 66(MOD) hydrofoil . The gridding in Fig. 8.36fl is of the C type, which wraps around the leading edge and trails off behind the foil, thus capturing the important near-wall and wake details without wasting nodes in front or to the sides. The grid size is 262 by 91. The CFD model for this hydrofoil flow is also quite sophisticated: a full Navier-Stokes solver with turbulence modeling and allowance for cavitation bubble formation when surface pressures drop below the local vapor pressure. Figure 8.36h compares computed and experimental surface pressure coefficients for an angle of attack of 1°. The dimensionless pressure coefficient is defined as Cp = (^surface ^ Poo)/(pVl^/2). The agreement is excellent, as indeed it is also for cases where the hydrofoil cavitates . Clearly, when properly implemented for the proper flow cases, CFD can be an extremely effective tool for engineers. The arrival of the third millennium has seen an enormous emphasis on computer applications in nearly every field, fluid mechanics being a prime example. It is now possible, at least for moderately complex geometries and flow patterns, to model on a computer, approximately, the equations of motion of fluid flow, with dedicated CFD textbooks available [5, 20, 23 to 27]. The flow region is broken into a fine grid of elements and nodes, which algebraically simulate the basic partial differential equations of flow. While simple two-dimensional flow simulations have long been reported and can be programmed as student exercises, three-dimensional flows, involving thousands or even millions of grid points, are now solvable with the modem supercomputer. Although elementary computer modeling was treated briefly here, the general topic of CFD is essentially for advanced study or professional practice. The big change over the past decade is that engineers, rather than laboriously programming CFD problems themselves, can now take advantage of any of several commercial CFD codes. These extensive software packages allow engineers to construct a geometry and boundary conditions to simulate a given viscous flow problem. The software then grids the flow region and attempts to compute flow properties at each grid element. The convenience is great; the danger is also great. That is, computations are not merely automatic, like when using a hand calculator, but rather require care and concern from the user. Convergence and accuracy are real problems for the modeler. Use of the codes requires some art and experience. In particular, when the flow Reynolds number. Re = pVUp, goes from moderate (laminar flow) to high (turbulent flow), the accuracy of the simulation is no longer assured in any real sense. The reason is that turbulent flows are not completely resolved by the full equations of motion, and one resorts to using approximate turbulence models. Turbulence models are developed for particular geometries and flow condi¬ tions and may be inaccurate or unrealistic for others. This is discussed by Freitas , who compared eight different commercial code calculations (FLOW-3D, FLOTRAN, STAR-CD, N3S, CFD-ACE, FLUENT, CFDS-FLOW3D, and NISA/3D-FLUID) with experimental results for five benchmark flow experiments. Calculations were made by the vendors themselves. Ereitas concludes that commercial codes, though promising in general, can be inaccurate for certain laminar and turbulent flow situations. Recent modifications to the standard turbulence models have improved their accuracy and reliability, as shown by Elkhoury . 576 Chapter 8 Potential Flow and Computational Fluid Dynamics Fig. 8.37 Flow over a surface- mounted cube creates a complex and perhaps unexpected pattern: {a) experimental oil-streak visualization of surface flow at Re = 40,000 (based on cube height) { Courtesy of Robert Martinuzzi with the permission of the American Society of Mechanical Engineers); (b) computational large-eddy simulation of the surface flow in (a) (from Ref. 32, courtesy of Kishan Shah, Stanford University); and (c) a side view of the flow in (a) visualized by smoke generation and a laser light sheet ( Courtesy of Robert Martinuzzi with the permission of the American Society of Mechanical Engineers). Problems 577 An example of erratic CFD results has already been mentioned here, namely, the drag and lift of a rotating cylinder. Fig. 8.15. Perhaps because the flow itself is physi¬ cally unstable [41, 44], results computed by different workers are strikingly different: Some predicted forces are high, some low, some increase, some decrease. In spite of this warning to treat CFD codes with care, one should also realize that the results of a given CFD simulation can be spectacular. Figure 8.37 illustrates tur¬ bulent flow past a cube mounted on the floor of a channel whose clearance is twice the cube height. Compare Fig. 8.37a, a top view of the experimental surface flow as visualized by oil streaks, with Fig. 8.37h, a CFD supercomputer result using the method of large-eddy simulation [32, 33]. The agreement is remarkable. The C-shaped flow pattern in front of the cube is caused by formation of a horseshoe vortex, as seen in a side view of the experiment in Fig. 8.37c. Horseshoe vortices commonly result when surface shear flows meet an obstacle. We conclude that CFD has a tre¬ mendous potential for flow prediction. Summary This chapter has analyzed a highly idealized but very useful type of flow: inviscid, incompressible, irrotational flow, for which Laplace’s equation holds for the veloc¬ ity potential (8.1) and for the plane stream function (8.7). The mathematics is well developed, and solutions of potential flows can be obtained for practically any body shape. Some solution techniques outlined here are (1) superposition of elementary line or point solutions in both plane and axisymmetric flow, (2) the analytic functions of a complex variable, and (3) numerical analysis on a computer. Potential theory is especially useful and accurate for thin bodies such as airfoils. The only require¬ ment is that the boundary layer be thin — in other words, that the Reynolds number be large. For blunt bodies or highly divergent flows, potential theory serves as a first approx¬ imation, to be used as input to a boundary layer analysis. The reader should consult the advanced texts [for example, 2 to 4, 11 to 13] for further applications of potential theory. Section 8.9 discussed computational methods for viscous (nonpotential) flows. Problems Most of the problems herein ai'e fairly straightforward. More difficult or open-ended assignments are labeled with an asterisk. Problems la¬ beled with a computer icon may require the use of a computer. The standard end-of-chapter problems P8.1 to P8.115 (categorized in the problem list here) are followed by word problems W8. 1 to W8.7, com¬ prehensive problems C8.1 to C8.7, and design projects D8.1 to D8.3. Problem Distribution Section Topic Problems 8.1 Introduction and review P8.1-P8.7 8.2 Elementary plane flow solutions P8.8-P8.17 8.3 Superposition of plane flows P8.18-P8.34 8.4 Plane flow past closed-body shapes P8.35-P8.59 8.5 The complex potential P8.60-P8.71 8.6 Images P8.72-P8.79 8.7 Airfoil theory: two-dimensional P8.80-P8.84 8.7 Airfoil theory: finite-span wings P8.85-P8.90 8.8 Axisymmetric potential flow P8.91-P8.103 OO bo Hydrodynamic mass P8.104-P8.105 8.9 Numerical methods P8.106-P8.115 Introduction and review P8.1 Prove that the streamlines ijj (r, 9) in polar coordinates from Eqs. (8.10) are orthogonal to the potential lines (p(r, 9). P8.2 The steady plane flow in Fig. P8.2 has the polar velocity components Vg = Dr and v,. = 0. Determine the circula¬ tion r around the path shown. 578 Chapter 8 Potential Flow and Computational Fluid Dynamics P8.3 Using cartesian coordinates, show that each velocity com¬ ponent {u, V, w) of a potential flow satisfies Laplace’s equation separately. P8.4 Is the function 1/r a legitimate velocity potential in plane polar coordinates? If so, what is the associated stream func¬ tion Tpir, 0)1 P8.5 A proposed harmonic function F(x, y, z) is given by F = 2)^ + - Axz+ f(y) (a) If possible, find a function /(y) for which the laplacian of F is zero. If you do indeed solve part (a), can your final function F serve as (b) a velocity potential or (c) a stream function? P8.6 An incompressible plane flow has the velocity potential (p = 2Kxy, where 5 is a constant. Find the stream function of this flow, sketch a few streamlines, and interpret the flow pattern. P8.7 Consider a flow with constant density and viscosity. If the flow possesses a velocity potential as defined by Eq. (8.1), show that it exactly satisfies the full Navier-Stokes equa¬ tions (4.38). If this is so, why for inviscid theory do we back away from the full Navier-Stokes equations? Elementary plane flow solutions P8.8 For the velocity distribution u = —By, v = +Bx, w = 0, evaluate the circulation F about the rectangular closed curve defined by (x, y) = (1,1), (3,1), (3,2), and (1,2). Inter¬ pret your result, especially vis-a -vis the velocity potential. P8.9 Consider the two-dimensional flow ii = —Ax, v = Ay, where A is a constant. Evaluate the circulation F around the rectangular closed curve defined by (x, y) = (1, 1), (4, 1), (4, 3), and (1,3). Interpret your result, especially vis-a -vis the velocity potential. P8.10 A two-dimensional Rankine half-body, 8 cm thick, is placed in a water tunnel at 20°C. The water pressure far upstream along the body centerline is 105 kPa. What is the nose radius of the half-body? At what tunnel flow velocity will cavitation bubbles begin to form on the sur¬ face of the body? P8.ll A power plant discharges cooling water through the mani¬ fold in Fig. P8. 1 1 , which is 55 cm in diameter and 8 m high and is perforated with 25,000 holes 1 cm in diameter. Does this manifold simulate a line source? If so, what is the equivalent source strength ml V _ — O 0 o o — ► ^ _ O 0 o O 0 o o ^ _ O 0 o o o o o _ ^ ^ _ o o o O 0 o o _ ^ _ o o o O 0 o o _ ^ ^ _ o o o O 0 o o o o o P8.ll P8.12 Consider the flow due to a vortex of strength K at the ori¬ gin. Evaluate the circulation from Eq. (8.23) about the clockwise path from (r, 9) = {a, 0) to (2a, 0) to {2a, 37r/2) to (a, 37r/2) and back to (a, 0). Interpret the result. P8.13 Starting at the stagnation point in Fig. 8.6, the fluid accel¬ eration along the half-body surface rises to a maximum and eventually drops off to zero far downstream, (a) Does this maximum occur at the point in Fig. 8.6 where (/max ~ 1.26(7? (b) If not, does the maximum acceleration occur before or after that point? Explain. P8.14 A tornado may be modeled as the circulating flow shown in Eig. P8.14, with v,. = = 0 and Vg(r) such that utr r < R - r > R r Determine whether this flow pattern is irrotational in either the inner or outer region. Using the r-momentum equation (D.5) of App. D, determine the pressure distribution p{r) in the tornado, assuming p = p^ as r — » oo. Eind the location and magnitude of the lowest pressure. Problems 579 P8.15 Hurricane Sandy, which hit the New Jersey coast on Oct. 29, 2012, was extremely broad, with wind velocities of 40 mi/h at 400 miles from its center. Its maximum velocity was 90 mi/h. Using the model of Fig. P8.14, at 20°C with a pressure of 100 kPa far from the center, estimate (a) the radius R of maximum velocity, in mi; and (b) the pressure atr = R. P8.16 Air flows at 1.2 m/s along a flat surface when it encounters a jet of air issuing from the horizontal wall at point A, as in Fig. P8.16. The jet volume flow is 0.4 mVs per unit depth into the paper. If the jet is approximated as an inviscid line source, (a) locate the stagnation point S on the wall. (b) How far vertically will the jet flow extend into the stream? 1.2 m/s • - ► S 0.4 m^/(s • m) P8.22 Consider inviscid stagnation flow, tp = Kxy (see Fig. 8. 19fc), superimposed with a source at the origin of strength m. Plot the resulting streamlines in the upper half-plane, using the length scale a = {m/K)^'^. Give a physical interpretation of the flow pattern. P8.23 Sources of strength m = 10 mVs are placed at points A and B in Fig. P8.23. At what height h should source B be placed so that the net induced horizontal velocity component at the origin is 8 m/s to the left? P8.23 2 m I I I I A ^ P8.16 P8.17 Find the position (x, y) on the upper surface of the half-body in Fig. 8.9a for which the local velocity equals the uniform stream velocity. What should be the pressure at this point? P8.24 Line sources of equal strength m = Ua, where U is a reference velocity, are placed at (x, y) — (0, a) and (0, —a). Sketch the stream and potential lines in the upper half plane. Is y = 0 a “wall”? If so, sketch the pressure coefficient Cp = P - Po \pU^ Superposition of plane flows P8.18 Plot the streamlines and potential lines of the flow due to a line source of strength m at (a, 0) plus a source 3m at (—a, 0). What is the flow pattern viewed from afar? P8.19 Plot the streamlines and potential lines of the flow due to a line source of strength 3m at (a, 0) plus a sink —m at (— fl, 0). What is the pattern viewed from afar? P8.20 Plot the streamlines of the flow due to a line vortex -I- A" at (0, -l-a) and a vortex —K at (0, —a). What is the pattern viewed from afar? P8.21 At point A in Fig. P8.21 is a clockwise line vortex of strength A" = 12 m^/s. At point B is a line source of strength m = 25 mVs. Determine the resultant velocity induced by these two at point C. y B 3 m c 4 m A along the wall, where po is the pressure at (0, 0). Find the minimum pressure point and indicate where flow separa¬ tion might occur in the boundary layer. P8.25 Let the vortex/sink flow of Eq. (8.16) simulate a tornado as in Fig. P8.25. Suppose that the circulation about the tornado is r = 8500 m^/s and that the pressure at r = 40 m is 2200 Pa less than the far-field pressure. Assuming inviscid flow at sea- level density, estimate (a) the appropriate sink strength — m, (b) the pressure at r = 15 m, and (c) the angle (3 at which the streamlines cross the circle at r = 40 m (see Fig. P8.25). P8.26 A coastal power plant takes in cooling water through a vertical perforated manifold, as in Fig. P8.26. The total volume flow intake is 110 mVs. Currents of 25 cm/s flow past the manifold, as shown. Estimate (a) how far down¬ stream and (b) how far normal to the paper the effects of the intake are felt in the ambient 8-m-deep waters. P8.21 580 Chapter 8 Potential Flow and Computational Fluid Dynamics P8.26 P8.27 Water at 20°C flows past a half-body as shown in Fig. P8.27. Measured pressures at points A and B are 160 kPa and 90 kPa, respectively, with uncertainties of 3 kPa each. Estimate the stream velocity and its uncertainty. P8.28 Sources of equal strength m are placed at the four symmet¬ ric positions (x, y) = (a, a), (—a, a), (—a, —a), and (a, —a). Sketch the streamline and potential line patterns. Do any plane “walls” appear? P8.27 P8.29 A uniform water stream, = 20 m/s and p = 998 kg/m^, combines with a source at the origin to form a half-body. At (x, y) = (0, 1.2 m), the pressure is 12.5 kPa less than p^. (a) Is this point outside the body? Estimate (b) the appropriate source strength m and (c) the pressure at the nose of the body. P8.30 A tornado is simulated by a line sink m = — 1000 mVs plus a line vortex K = 1600 m^/s. Find the angle between any streamline and a radial line, and show that it is independent of both r and 6. If this tornado forms in sea-level standard air, at what radius will the local pressure be equivalent to 29 inHg? P8.31 A Rankine half-body is formed as shown in Fig. P8.3 1 . For the stream velocity and body dimension shown, compute (a) the source strength m in mVs, (b) the distance a, (c) the distance h, and (d) the total velocity at point A. A (0, 3 m) 1 1 1 h \ 7 m/s , (4m, 0) Source P8.31 P8.32 Line sources nii and m2 are near point A, as in Fig. P8.32. If mi = 30 mV2, find the value of m2 for which the resultant velocity at point A is exactly vertical. 3 m 4 m P8.32 I 4m I 3m I P8.33 Sketch the streamlines, especially the body shape, due to equal line sources +m at (0, +a) and (0, —a) plus a uni¬ form stream t/^ = rna. P8.34 Consider three equally spaced sources of strength m placed at (x, y) = ( + a, 0), (0, 0), and (—a, 0). Sketch the resulting streamlines, noting the position of any stagna¬ tion points. What would the pattern look like from afar? Plane flow past closed-body shapes P8.35 A uniform stream, U^o = 4 m/s, approaches a Rankine oval as in Fig. 8.13, with a = 50 cm. Find the strength m of the source-sink pair, in mVs, which will cause the total length of the oval to be 250 cm. What is the maximum width of this oval? P8.36 When a line source-sink pair with m = 2 mVs is combined with a uniform stream, it forms a Rankine oval whose min¬ imum dimension is 40 cm. If a = 15 cm, what are the stream velocity and the velocity at the shoulder? What is the maximum dimension? P8.37 A Rankine oval 2 m long and 1 m high is immersed in a stream Uoo = 10 m/s, as in Fig. P8.37. Estimate (a) the velocity at point A and (b) the location of point B where a particle approaching the stagnation point achieves its max¬ imum deceleration. A P8.37 P8.38 Consider potential flow of a uniform stream in the x direc¬ tion plus two equal sources, one at (x, y) = (0, -fa) and the other at (x, y) = (0, —a). Sketch your ideas of the body contours that would arise if the sources were (a) very weak and (b) very strong. Problems 581 P8.39 A large Rankine oval, with a = 1 m and h = 1 m, is immersed in 20°C water flowing at 10 m/s. The upstream pressure on the oval centerline is 200 kPa. Calculate (a) the value of m\ and {b) the pressure on the top of the oval (analogous to point A in Fig. P8.37). P8.40 Modify the Rankine oval in Fig. P8.37 so that the stream velocity and body length are the same hut the thickness is unknown (not 1 m). The fluid is water at 30°C and the pres¬ sure far upstream along the body centerline is 108 kPa. Find the body thickness for which cavitation will occur at point A. P8.41 A Kelvin oval is formed by a line-vortex pair with K = 9 mVs, a = 1 m, and (/ = 10 m/s. What are the height, width, and shoulder velocity of this oval? P8.42 The vertical keel of a sailboat approximates a Rankine oval 125 cm long and 30 cm thick. The boat sails in seawater in standard atmosphere at 14 knots, parallel to the keel. At a section 2 m below the surface, estimate the lowest pressure on the surface of the keel. P8.43 Water at 20°C flows past a 1-m-diameter circular cylinder. The upstream centerline pressure is 128,500 Pa. If the low¬ est pressure on the cylinder surface is exactly the vapor pressure, estimate, by potential theory, the stream velocity. P8.44 Suppose that circulation is added to the cylinder flow of Prob. P8.43 sufficient to place the stagnation points at 9 equal to 35° and 145°. What is the required vortex strength K in m^/s? Compute the resulting pressure and surface velocity at (a) the stagnation points and (b) the upper and lower shoulders. What will the lift per meter of cylinder width be? P8.45 If circulation K is added to the cylinder flow in Prob. P8.43, (a) for what value of K will the flow begin to cavitate at the surface? (b) Where on the surface will cavitation begin? (c) For this condition, where will the stagnation points lie? P8.46 A cylinder is formed by bolting two semicylindrical chan¬ nels together on the inside, as shown in Fig. P8.46. There are 10 bolts per meter of width on each side, and the inside pressure is 50 kPa (gage). Using potential theory for the outside pressure, compute the tension force in each bolt if the fluid outside is sea-level air. P8.48 Wind at Uoo and flows past a Quonset hut which is a half-cylinder of radius a and length L (Fig. P8.48). The in¬ ternal pressure is p,-. Using inviscid theory, derive an ex¬ pression for the upward force on the hut due to the difference between p, and p^. P8.48 I I P8.49 In strong winds the force in Prob. P8.48 can be quite large. Suppose that a hole is introduced in the hut roof at point A to make p, equal to the surface pressure there. At what angle 9 should hole A be placed to make the net wind force zero? P8.50 It is desired to simulate flow past a two-dimensional ridge or bump by using a streamline that passes above the flow over a cylinder, as in Fig. P8.50. The bump is to be a/2 high, where a is the cylinder radius. What is the elevation h of this streamline? What is Umax on the bump compared with stream velocity (/? P8.50 D = 2m P8.47 A circular cylinder is fitted with two surface-mounted pres¬ sure sensors, to measure p„ at 0 = 180° and pj at 0 = 105°. The intention is to use the cylinder as a stream velocimeter. Using inviscid theory, derive a formula for estimating I/„ in terms of p^, pj, p, and the cylinder radius a. P8.51 A hole is placed in the front of a cylinder to measure the stream velocity of sea-level fresh water. The measured pressure at the hole is 2840 Ibf/ft^. If the hole is misaligned by 12° from the stream, and misinterpreted as stagnation pressure, what is the error in velocity? P8.52 The Flettner rotor sailboat in Fig. E8.3 has a water drag coefficient of 0.006 based on a wetted area of 45 ft^. If the rotor spins at 220 r/min, find the maximum boat velocity that can be achieved in a 15-mi/h wind. What is the opti¬ mum angle between the boat and the wind? P8.53 Modify Prob. P8.52 as follows. For the same sailboat data, find the wind velocity, in mi/h, that will drive the boat at an optimum speed of 8 kn parallel to its keel. 582 Chapter 8 Potential Flow and Computational Fluid Dynamics P8.54 P8.55 P8.56 P8.57 P8.58 P8.59 The original Flettner rotor ship was approximately 100 ft long, displaced 800 tons, and had a wetted area of 3500 ft^. As sketched in Fig. P8.54, it had two rotors 50 ft high and 9 ft in diameter rotating at 750 r/min, which is far outside the range of Fig. 8.15. The measured lift and drag coefficients for each rotor were about 10 and 4, respectively. If the ship is moored and subjected to a crosswind of 25 ft/s, as in Fig. P8.54, what will the wind force parallel and normal to the ship centerline be? Estimate the power required to drive the rotors. Assume that the Flettner rotor ship of Fig. P8.54 has a wa¬ ter resistance coefficient of 0.005. How fast will the ship sail in seawater at 20°C in a 20-ft/s wind if the keel aligns itself with the resultant force on the rotors? [Hint: This is a problem in relative velocities.] P8.54 11111^°° A proposed free-stream velocimeter would use a cylinder with pressure taps at 6 = 180° and at 150°. The pressure difference would be a measure of stream velocity U^. How¬ ever, the cylinder must be aligned so that one tap exactly faces the free stream. Let the misalignment angle be 5\ that is, the two taps are at (180° -I- 5) and (150° -I- (5). Make a plot of the percentage error in velocity measurement in the range —20° < 5 < -1-20° and comment on the idea. In principle, it is possible to use rotating cylinders as air¬ craft wings. Consider a cylinder 30 cm in diameter, rotat¬ ing at 2400 r/min. It is to lift a 55-kN airplane cruising at 100 m/s. What should the cylinder length be? How much power is required to maintain this speed? Neglect end ef¬ fects on the rotating wing. Plot the streamlines due to the combined flow of a line sink — m at the origin plus line sources +m at (a, 0) and (4a, 0). [Hint: A cylinder of radius 2a will appear.] The Transition® car-plane in Fig. 7.30 has a gross weight of 1430 Ibf. Suppose we replace the wing with a 1-ft-diameter rotating cylinder 20 ft long, (a) What rotation rate from Fig. 8.15, in r/min, would lift the plane at a take-off speed of 55 mi/h? (b) Estimate the cylinder drag at this rotation rate. Neglect fuselage lift and cylinder end effects. The complex potential P8.60 One of the comer flow patterns of Fig. 8. 1 8 is given by the cartesian stream function 'ip = A(3yx^ — y^). Which one? Can the correspondence be proved from Eq. (8.53)? P8.61 Plot the streamlines of Eq. (8.53) in the upper right quad¬ rant for M = 4. How does the velocity increase with x out¬ ward along the x axis from the origin? For what comer angle and value of n would this increase be linear in x? For what corner angle and n would the increase be as x^? P8.62 Combine stagnation flow. Fig. 8.19h, with a source at the origin: fiz) = Az^ + m\nz Plot the streamlines for m = Al}, where L is a length scale. Interpret. P8.63 The superposition in Prob. P8.62 leads to stagnation flow near a curved bump, in contrast to the flat wall of Fig. 8.19/>. Determine the maximum height H of the bump as a function of the constants A and m. P8.64 Consider the polar-coordinate stream function ip = sin(1.2 0), with B equal, for convenience, to 1.0 ft°'Vs. (a) Plot the streamline tp = 0 in the upper half plane. (b) Plot the streamline ip = l.O and interpret the flow pattern, (c) Find the locus of points above ip = 0 for which the resultant velocity = 1.2 ft/s. P8.65 Potential flow past a wedge of half-angle 0 leads to an im¬ portant application of laminar boundary layer theory called the Falkner-Skan flows [15, pp. 239-245]. Let x denote dis¬ tance along the wedge wall, as in Fig. P8.65, and let 0= 10°. Use Eq. (8.53) to find the variation of surface velocity U(x) along the wall. Is the pressure gradient adverse or favorable? P8.65 P8.66 The inviscid velocity along the wedge in Prob. P8.65 has the analytic form U{x} = ex'", where m = n — I and n is the exponent in Eq. (8.53). Show that, for any C and n, computation of the boundary layer by Thwaites’s method, Eqs. (7.53) and (7.54), leads to a unique value of the Thwaites parameter A. Thus wedge flows are called similar [15, p. 241]. P8.67 Investigate the complex potential function /(z) = Uooiz + aVz) and interpret the flow pattern. P8.68 Investigate the complex potential function y(z) = U^z + m In [(z -f a)/(z— a)] and interpret the flow pattern. Problems 583 P8.69 Investigate the complex potential /(z) = A cosh [7r(z/a)], and plot the streamlines inside the region shown in Fig. P8.69. What hyphenated word (originally French) might describe such a flow pattern? y = a {ip = Q) P8.69 Plot the streamlines inside this region P8.73 Set up an image system to compute the flow of a source at unequal distances from two walls, as in Fig. P8.73. Find the point of maximum velocity on the y axis. J’ P8.70 Show that the complex potential / = ?7oo{z + coth [7r(z/a)] ) represents flow past an oval shape placed mid¬ way between two parallel walls y = ±^fl. What is a prac¬ tical application? P8.71 Figure P8.71 shows the streamlines and potential lines of flow over a thin-plate weir as computed by the complex potential method. Compare qualitatively with Fig. 10.16a. State the proper boundary conditions at all boundaries. The velocity potential has equally spaced values. Why do the flow-net “squares” become smaller in the overflow jet? P8.73 P8.74 A positive line vortex K is trapped in a comer, as in Fig. P8.74. Compute the total induced velocity vector at point fl, (x, y) = (2a, a), and compare with the induced velocity when no walls are present. 2a ■0 P8.71 Images P8.72 Use the method of images to constmct the flow pattern for a source +m near two walls, as shown in Fig. P8.72. Sketch the velocity distribution along the lower wall (y = 0). Is there any danger of flow separation along this wall? y P8.74 a 2a P8.75 Using the four-source image pattern needed to constmct the flow near a comer in Fig. P8.72, find the value of the source strength m that will induce a wall velocity of 4.0 m/s at the point (x, y) = (a, 0) just below the source shown, if a = 50 cm. P8.76 Use the method of images to approximate the flow pattern past a cylinder a distance 4a from a single wall, as in Fig. P8.76. To illustrate the effect of the wall, compute the velocities at corresponding points A, B, C, and D, com¬ paring with a cylinder flow in an infinite expanse of fluid. P8.76 D C 584 Chapter 8 Potential Flow and Computational Fluid Dynamics P8.77 Discuss how the flow pattern of Prob. P8.58 might be inter¬ preted to be an image system construction for circular walls. Why are there two images instead of one? P8.78 Indicate the system of images needed to construct the flow of a uniform stream past a Rankine half-body constrained between two parallel walls, as in Fig. P8.78. For the par¬ ticular dimensions shown in this figure, estimate the posi¬ tion of the nose of the resulting half-body. P8.78 P8.79 Explain the system of images needed to simulate the flow of a line source placed unsymmetrically between two par¬ allel walls as in Fig. P8.79. Compute the velocity on the lower wall atx = a. How many images are needed to esti¬ mate this velocity within 1 percent? drag coefficient Further assume sufficient thrust to balance whatever drag is calculated, (a) Find an algebraic expression for the best cruise velocity 14, which occurs when the ratio of drag to speed is a minimum, (b) Apply your formula to the data in Prob. P7. 119 for which a labori¬ ous graphing procedure gave an answer 14 ~ 180 m/s. P8.82 The ultralight plane Gossamer Condor in 1977 was the first to complete the Kremer Prize figure-eight course under hu¬ man power. Its wingspan was 29 m, with Cav = 2.3 m and a total mass of 95 kg. The drag coefficient was approximately 0.05. The pilot was able to deliver \| hp to propel the plane. Assuming two-dimensional flow at sea level, estimate (a) the cruise speed attained, (b) the lift coefficient, and (c) the horsepower required to achieve a speed of 15 kn. P8.83 The world’s largest airplane, the Airbus A380, has a maxi¬ mum weight of 1,200,000 Ibf, wing area of 9100 ft^, wing¬ span of 262 ft, and Cco = 0.026. When cruising at maximum weight at 35,000 ft, the four engines each pro¬ vide 70,000 Ibf of thrust. Assuming all lift and drag are due to the wing, estimate the cruise velocity, in mi/h. P8.84 Reference 12 contains in viscid theory calculations for the upper and lower surface velocity distributions V{x} over an airfoil, where x is the chordwise coordinate. A typical re¬ sult for small angle of attack is as follows: P8.79 y Airfoil theory: two-dimensional xfc F/l/„(upper) V/F^ilower) 0.0 0.0 0.0 0.025 0.97 0.82 0.05 1.23 0.98 0.1 1.28 1.05 0.2 1.29 1.13 0.3 1.29 1.16 0.4 1.24 1.16 0.6 1.14 1.08 0.8 0.99 0.95 1.0 0.82 0.82 P8.80 The beautiful expression for lift of a two-dimensional air¬ foil, Eq. (8.59), arose from applying the Joukowski trans¬ formation, ^ = z -f a^/z, where z =x + iy and C ~ V The constant a is a length scale. The theory transforms a certain circle in the z plane into an airfoil in the ^ plane. Taking a = \ unit for convenience, show that {a) a circle with center at the origin and radius > 1 will become an el¬ lipse in the ^ plane and (h) a circle with center at .tr = — £" 1, y = 0, and radius (1 -I- £■) will become an airfoil shape in the ^ plane. [Hint: The Excel spreadsheet is excellent for solving this problem.] P8.81 Given an airplane of weight W, wing area A, aspect ratio AR, and flying at an altitude where the density is p. Assume all drag and lift is due to the wing, which has an infinite-span Use these data, plus Bernoulli’s equation, to estimate (a) the lift coefficient and (b) the angle of attack if the airfoil is symmetric. Airfoil theory: finite-span wings P8.85 A wing of 2 percent camber, 5-in chord, and 30-in span is tested at a certain angle of attack in a wind tunnel with sea- level standard air at 200 ft/s and is found to have lift of 30 Ibf and drag of 1 .5 Ibf. Estimate from wing theory (a) the angle of attack, (b) the minimum drag of the wing and the angle of at¬ tack at which it occurs, and (c) the maximum lift-to-drag ratio. P8.86 An airplane has a mass of 20,000 kg and flies at 175 m/s at 5000-m standard altitude. Its rectangular wing has a 3-m chord Problems 585 and a symmetric airfoil at 2.5° angle of attack. Estimate (a) the wing span, (b) the aspect ratio, and (c) the induced drag. P8.87 A freshwater boat of mass 400 kg is supported by a rectan¬ gular hydrofoil of aspect ratio 8, 2 percent camber, and 12 percent thickness. If the boat travels at 7 m/s and a = 2.5°, estimate (a) the chord length, (b) the power required if Cd^ = 0.01, and (c) the top speed if the boat is refitted with an engine that delivers 20 hp to the water. P8.88 The Boeing 787-8 Dreamliner has a maximum weight of 502,500 Ibf, a wingspan of 197 ft, a wing area of 3501 ft^, and cruises at 567 mi/h at 35,000 ft altitude. When cruis¬ ing, its overall drag coefficient is about 0.027. Estimate (a) the aspect ratio, (b) the lift coefficient, (c) the cmise Mach number, and (d) the engine thrust needed when cruising. P8.89 The Beechcraft T-34C aircraft has a gross weight of 5500 Ibf and a wing area of 60 ft^ and flies at 322 mi/h at 10,000-ft standard altitude. It is driven by a propeller that delivers 300 hp to the air. Assume for this problem that its airfoil is the NACA 2412 section described in Figs. 8.23 and 8.24, and neglect all drag except the wing. What is the appropriate aspect ratio for the wing? P8.90 NASA is developing a swing-wing airplane called the Bird of Prey . As shown in Fig. P8.90, the wings pivot like a pocketknife blade: forward (a), straight (b), or backward (c). Discuss a possible advantage for each of these wing positions. If you can’t think of one, read the article and report to the class. Axisymmetric potential flow P8.91 If (p(r, 0) in axisymmetric flow is defined by Eq. (8.72) and the coordinates are given in Fig. 8.28, determine what par¬ tial differential equation is satisfied by (p. P8.92 A point source with volume flow 2 = 30 mVs is immersed in a uniform stream of speed 4 m/s. A Rankine half-body of revolution results. Compute (a) the distance from source to the stagnation point and (b) the two points (r, 6) on the body surface where the local velocity equals 4.5 m/s. P8.93 The Rankine half-body of revolution (Fig. 8.30) could sim¬ ulate the shape of a pitot-static tube (Fig. 6.30). According to inviscid theory, how far downstream from the nose should the static pressure holes be placed so that the local velocity is within ±0.5 percent of (7cc? Compare your an¬ swer with the recommendation x ~ 8D in Fig. 6.30. P8.94 Determine whether the Stokes streamlines from Eq. (8.73) are everywhere orthogonal to the Stokes potential lines from Eq. (8.74), as is the case for Cartesian and plane polar coordinates. P8.95 Show that the axisymmetric potential flow formed by su¬ perposition of a point source +m at (x, y) = (—a, 0), a point sink —ma.t(+a, 0), and a stream (/„ in the x direction forms a Rankine body of revolution as in Fig. P8.95. Find analytic expressions for determining the length 2L and maximum diameter 2R of the body in terms of m, Uo^, and a. P8.96 Consider inviscid flow along the streamline approaching the front stagnation point of a sphere, as in Fig. 8.31. Find (a) the maximum fluid deceleration along this streamline and (b) its position. P8.97 The Rankine body of revolution in Fig. P8.97 is 60 cm long and 30 cm in diameter. When it is immersed in the low- pressure water tunnel as shown, cavitation may appear at point A. Compute the stream velocity U, neglecting surface wave formation, for which cavitation occurs. P8.97 P8.98 We have studied the point source (sink) and the line source (sink) of infinite depth into the paper. Does it make any sense to define a finite-length line sink (source) as in 586 Chapter 8 Potential Flow and Computational Fluid Dynamics Fig. P8.98? If so, how would you establish the mathemati¬ cal properties of such a finite line sink? When combined with a uniform stream and a point source of equivalent strength as in Fig. P8.98, should a closed-body shape be formed? Make a guess and sketch some of these possible shapes for various values of the dimensionless parameter m/{U^l}). Point y Line sink of source \| total strength P8.98 0 L P8.99 Consider air flowing past a hemisphere resting on a flat surface, as in Fig. P8.99. If the internal pressure is p„ find an expression for the pressure force on the hemisphere. By analogy with Prob. P8.49, at what point A on the hemi¬ sphere should a hole be cut so that the pressure force will be zero according to inviscid theory? P8.100 A 1-m-diameter sphere is being towed at speed V in fresh water at 20°C as shown in Fig. P8.100. Assuming inviscid theory with an undistorted free surface, estimate the speed V in m/s at which cavitation will first appear on the sphere surface. Where will cavitation appear? For this condition, what will be the pressure at point A on the sphere, which is 45° up from the direction of travel? P8.101 Consider a steel sphere (SG = 7.85) of diameter 2 cm, dropped from rest in water at 20°C. Assume a constant drag coefficient = 0.47. Accounting for the sphere’s hydro- dynamic mass, estimate (a) its terminal velocity and (h) the time to reach 99 percent of terminal velocity. Compare these to the results when hydrodynamic mass is neglected, terminal ~ 1-95 m/s and ~ 0.605 s, and discuss. P8.102 A golf ball weighs 0.102 Ibf and has a diameter of 1.7 in. A professional golfer strikes the ball at an initial velocity of 250 ft/s, an upward angle of 20°, and a backspin (front of the ball rotating upward). Assume that the lift coefficient on the ball (based on frontal area) follows Fig. P7.108. If the ground is level and drag is neglected, make a simple analysis to predict the impact point (d) without spin and {V) with backspin of 7500 r/min. P8.103 Consider inviscid flow past a sphere, as in Fig. 8.31. Find (a) the point on the front surface where the fluid acceleration Umax is maximum and (b) the magnitude of (c) If the stream velocity is 1 m/s, find the sphere diameter for which flmax is 10 times the acceleration of gravity. Comment. Hydrodynamic mass P8.104 Consider a cylinder of radius a moving at speed U^o through a still fluid, as in Fig. P8. 104. Plot the streamlines relative to the cylinder by modifying Eq. (8.32) to give the relative flow with K = 0. Integrate to find the total relative kinetic energy, and verify the hydrodynamic mass of a cylinder from Eq. (8.91). Still P8.104 P8.105 A 22-cm-diameter solid aluminum sphere (SG = 2.7) is ac¬ celerating at 12 m/s^ in water at 20°C. (a) According to poten¬ tial theory, what is the hydrodynamic mass of the sphere? (b) Estimate the force being applied to the sphere at this instant. Numerical methods P8.106 Laplace’s equation in plane polar coordinates, Eq. (8.11), is complicated by the variable radius. Consider the finite difference mesh in Eig. P8.106, with nodes (i, j) equally spaced AF and Ar apart. Derive a finite difference model for Eq. (8.1 1) similar to the cartesian expression (8.96). Problems 587 P8.107 SAE 10W30 oil at 20°C is at rest near a wall when the wall suddenly begins moving at a constant 1 m/s. (a) Use Ay = 1 cm and Af = 0.2 s and check the stability criterion (8.101). {b) Carry out Eq. (8. 100) to f = 2 s and report the velocity u at y = 4 cm. P8.108 Consider two-dimensional potential flow into a step con¬ 's^ traction as in Eig. P8.108. The inlet velocity Ui = 1 m/s, and the outlet velocity U2 is uniform. The nodes (;, j) are labeled in the figure. Set up the complete finite difference algebraic relations for all nodes. Solve, if possible, on a computer and plot the streamlines in the flow. Ui P8.108 P8.109 Consider inviscid flow through a two-dimensional 90° bend with a contraction, as in Fig. P8.109. Assume uniform flow at the entrance and exit. Make a finite difference computer analy¬ sis for small grid size (at least 150 nodes), determine the dimen¬ sionless pressure distribution along the walls, and sketch the streamlines. (You may use either square or rectangular grids.) 6 m -10m- 10 m -5 m- - Vl = 10 m/s 16 m -15 m- P8.110 For fully developed laminar incompressible flow through a straight noncircular duct, as in Sec. 6.8, the Navier-Stokes equations (4.38) reduce to d^u d^u 1 dp — y -I - y = - = const < 0 dz M dx where (y, z) is the plane of the duct cross section and v is along the duct axis. Gravity is neglected. Using a non¬ square rectangular grid (Ax, Ay), develop a finite differ¬ ence model for this equation, and indicate how it may be applied to solve for flow in a rectangular duct of side lengths a and b. P8.111 Solve Prob. P8.110 numerically for a rectangular duct of side length b by 2b, using at least 100 nodal points. Evalu¬ ate the volume flow rate and the friction factor, and com¬ pare with the results in Table 6.4: b'^ f dp\ 62.19 where Z);, = 4A/P = Ab/3 for this case. Comment on the possible truncation errors of your model. P8.112 In CFD textbooks [5, 23-27], one often replaces the left- hand sides of Eqs. (8.IO2/1 and c) with the following two expressions, respectively: d 2 ^ ^ ^9 — (m ) -I - (vu) and — (uv) H - {v ) dx dy dx dy Are these equivalent expressions, or are they merely sim¬ plified approximations? Either way, why might these forms be better for finite difference purposes? P8.113 Formulate a numerical model for Eq. (8.99), which has no instability, by evaluating the second derivative at the next time step, / + 1. Solve for the center velocity at the next time step and comment on the result. This is called an im¬ plicit model and requires iteration. P8.114 If your institution has an online potential flow boundary element computer code, consider flow past a symmetric airfoil, as in Fig. P8.114. The basic shape of an NACA symmetric airfoil is defined by the function ^ « 1.4845C‘'^ - 0.63C - 1.758C^ ^max -f 1.4215C^-0.5075C‘ where ^ = xlC and the maximum thickness t,^^ occurs at C = 0.3. Use this shape as part of the lower boundary for zero angle of attack. Let the thickness be fairly large, say, fmax = 0.12, 0.15, or 0.18. Choose a generous number of nodes (£60), and calculate and plot the velocity distribu¬ tion V7(7„ along the airfoil surface. Compare with the P8.109 588 Chapter 8 Potential Flow and Computational Fluid Dynamics L[ x = 0 x = C Lj P8.114 Word Problems W8.1 What simplihcations have been made, in the potential flow theory of this chapter, which result in the elimination of the Reynolds number, Froude number, and Mach number as important parameters? W8.2 In this chapter we superimpose many basic solutions, a concept associated with linear equations. Yet Bernoulli’s equation (8.3) is nonlinear, being proportional to the square of the velocity. How, then, do we justify the use of superposition in inviscid flow analysis? W8.3 Give a physical explanation of circulation F as it relates to the lift force on an immersed body. If the line integral de- hned by Eq. (8.23) is zero, it means that the integrand is a perfect differential — but of what variable? theoretical results in Ref. 12 for NACA 0012, 0015, or 0018 airfoils. If time permits, investigate the effect of the boundary lengths Li, L2, and L3, which can initially be set equal to the chord length C. P8.115 Use the explicit method of Eq. (8.100) to solve Prob. P4.85 numerically for SAE 30 oil at 20°C with (/q = 1 m/s and ijO = M rad/s, where M is the number of letters in your surname. (This author will solve the problem for M = 5.) When steady oscillation is reached, plot the oil velocity versus time at y = 2 cm. W8.4 Give a simple proof of Eq. (8.46) — namely, that both the real and imaginary parts of a function /(z) are laplacian if z = X + iy. What is the secret of this remarkable behavior? W8.5 Eigure 8.18 contains five body corners. Without carrying out any calculations, explain physically what the value of the inviscid fluid velocity must be at each of these five comers. Is any flow separation expected? W8.6 Explain the Kutta condition physically. Why is it necessary? W8.7 We have briefly outlined finite difference and boundary element methods for potential flow but have neglected the finite element technique. Do some reading and write a brief essay on the use of the finite element method for potential flow problems. Comprehensive Problems C8.I Did you know that you can solve simple fluid mechanics problems with Microsoft Excel? The successive relaxation technique for solving the Laplace equation for potential flow problems is easily set up on a spreadsheet, since the stream function at each interior cell is simply the average of its four neighbors. As an example, solve for the irrotational potential flow through a contraction, as given in Fig. C8.1. Note: To avoid the “circular reference” error, you must turn on the iteration option. Use the help index for more informa¬ tion. For full credit, attach a printout of your spreadsheet, with stream function converged and the value of the stream function at each node displayed to four digits of accuracy. C8.2 Use an explicit method, similar to but not identical to Eq. (8.100), to solve the case of SAE 30 oil at 20°C starting from rest near a fixed wall. Far from the wall, the oil ac¬ celerates linearly; that is, = at, where a = 9 m/s^. At f = Is, determine {a) the oil velocity at y = 1 cm and (b) the instantaneous boundary layer thickness (where u ~ 0.99 Moo)- Hint: There is a nonzero pressure gradient in the outer (nearly shear-free) stream, n = N, which must be in¬ cluded in Eq. (8.99) and your explicit model. tp = 5 tp = 4 tp = 3 tp = 2 tp = 1 tp = 0 Wall, tp = 5 / Inlet Wall, tp = 0 tp = 3.333 Outlet tp= 1.667 Wall, tp = 0 C8.1 C8.3 Consider plane inviscid flow through a symmetric diffuser, '0^ as in Fig. C8.3. Only the upper half is shown. The flow is to expand from inlet half-width h to exit half-width 2h, as shown. The expansion angle 0 is 18.5° (L ~ 3h). Set up a nonsquare potential flow mesh for this problem, and Design Projects 589 calculate and plot {a) the velocity distribution and (jb) the pressure coefficient along the centerline. Assume uniform inlet and exit flows. C8.3 C8.4 Use potential flow to approximate the flow of air being sucked up into a vacuum cleaner through a two- dimensional slit attachment, as in Fig. C8.4. In the xy plane through the centerline of the attachment, model the flow as a line sink of strength ( — m), with its axis in the z direction at height a above the floor, (a) Sketch the streamlines and locate any stagnation points in the flow. (b) Find the magnitude of velocity V(x) along the floor in terms of the parameters a and m. (c) Let the pressure far away be p^, where velocity is zero. Define a velocity scale U = m/a. Determine the variation of dimensionless pressure coefficient, Cp = {p — p^)l{pU^I2), along the floor, {d) The vacuum cleaner is most effective where Cp is a minimum — that is, where velocity is maximum. Find the locations of minimum pressure coefficient along the X axis, (e) At which points along the x axis do you expect the vacuum cleaner to work most effectively? Is it best at X = 0 directly beneath the slit, or at some other x location along the floor? Conduct a scientific experiment at home with a vacuum cleaner and some small pieces of dust or dirt to test your prediction. Report your results and dis¬ cuss the agreement with prediction. Give reasons for any disagreements. Design Projects D8.1 In 1927, Theodore von Karman developed a scheme to use a uniform stream, plus a row of sources and sinks, to generate an arbitrary closed-body shape. A schematic of the idea is sketched in Fig. D8.1. The body is symmetric and at zero an¬ gle of attack. A total of N sources and sinks are distributed along the axis within the body, with strengths m, at positions C8.4 C8.5 Consider a three-dimensional, incompressible, irrotational flow. Use the following two methods to prove that the viscous term in the Navier-Stokes equation is identically zero: (a) using vector notation; and (b) expanding out the scalar terms and substituting terms from the definition of irrotationality. C8.6 Find, either on-line or in Ref. 12, lift-drag data for the NACA 4412 airfoil, (a) Draw the polar lift-drag plot and compare qualitatively with Fig. 7.26. (b) Find the maxi¬ mum value of the lift-to-drag ratio, (t) Demonstrate a straight-line construction on the polar plot that will imme¬ diately yield the maximum LID in {b). (d) If an aircraft could use this two-dimensional wing in actual flight (no induced drag) and had a perfect pilot, estimate how far (in miles) this aircraft could glide to a sea-level runway if it lost power at 25,000 ft altitude. C8.7 Find a formula for the stream function for flow of a doublet of strength A a distance a from a wall, as in Fig. C8.7. (a) Sketch the streamlines, (b) Are there any stagnation points? (c) Find the maximum velocity along the wall and its position. A x„ for i = 1 to N. The object is to find the correct distribution of strengths that approximates a given body shape y(x) at a fi¬ nite number of surface locations and then to compute the ap¬ proximate surface velocity and pressure. The technique should work for either two-dimensional bodies (distributed line sources) or bodies of revolution (distributed point sources). 590 Chapter 8 Potential Flow and Computational Fluid Dynamics D8.1 For our body shape let us select the NACA 0018 airfoil, given by the formula in Prob. P8.114 with “ 0.18. Develop the ideas stated here into N simultaneous alge¬ braic equations that can be used to solve for the N unknown line source/sink strengths. Then program your equations for a computer, with N S 20; solve for m,-; compute the surface velocities; and compare with the theoretical References 1. J. Wermer, Potential Theory, Springer- Verlag, New York, 2008. 2. J. M. Robertson, Hydrodynamics in Theory and Application, Prentice-Hall, Englewood Cliffs, NJ, 1965. 3. L. M. Milne-Thomson, Theoretical Hydrodynamics, 4th ed., Dover, New York, 1996. 4. D. H. Armitage and S. J. Gardiner, Classical Potential The¬ ory, Springer, New York, 2013. 5. J. Tu, G. H. Yeoh, and C. Liu, Computational Fluid Dynam¬ ics: A Practical Approach, 2d ed., Elsevier Science, New York, 2012. 6. O. C. Zienkiewicz, R. L. Taylor, and P. Nithiarasu, The Finite Element Method for Fluid Dynamics, vol. 3, 6th ed., Butterworth- Heinemann, Burlington, MA, 2005. 7. G. Beer, I. Smith, and C. Duenser, The Boundary Element Method with Programming: For Engineers and Scientists, Springer- Verlag, New York, 2010. 8. A. D. Moore, “Eields from Eluid Plow Mappers,” J. Appl. Phys., vol. 20, 1949, pp. 790-804. 9. H. J. S. Hele-Shaw, “Investigation of the Nature of the Sur¬ face Resistance of Water and of Streamline Motion under Certain Experimental Conditions,” Trans. Inst. Nav. Archit., vol. 40, 1898, p. 25. 10. S. W. Churchill, Viscous Flows: The Practical Use of Theory, Butterworth, Stoneham, MA, 1988. 11. J. D. Anderson, Jr., Fundamentals of Aerodynamics, 5th ed., McGraw-Hill, New York, 2010. 12. I. H. Abbott and A. E. von Doenhoff, Theory of Wing Sec¬ tions, Dover, New York, 1981. 13. P. O. Smetana, Introductory Aerodynamics and Hydrody¬ namics of Wings and Bodies: A Software-Based Approach, AIAA, Reston, VA, 1997. velocities for this shape in Ref. 12. Your goal should be to achieve accuracy within ± 1 percent of the classic results. If necessary, you should adjust N and the locations of the sources. D8.2 Modify Prob. D8.1 to solve for the point-source distribu¬ tion that approximates an “0018” body-of-revolution shape. Since no theoretical results are published, simply make sure that your results converge to ± 1 percent. D8.3 Consider water at 20°C flowing at 12 m/s in a water chan¬ nel. A Rankine oval cylinder, 40 cm long, is to be placed parallel to the flow, where the water static pressure is 120 kPa. The oval’s thickness is a design parameter. Prepare a plot of the minimum pressure on the oval’s surface as a function of body thickness. Especially note the thicknesses where {a) the local pressure is 50 kPa and {b) cavitation first occurs on the surface. 14. L. Prandtl, “Applications of Modern Hydrodynamics to Aeronautics,” VACA Rep. 116, 1921. 15. P. M. White, Viscous Fluid Flow, 3d ed., McGraw-Hill, New York, 2005. 16. C. S. Yih, Fluid Mechanics, McGraw-Hill, New York, 1969. 17. K. T. Patton, “Tables of Hydrodynamic Mass Pactors for Translational Motion,” ASME Winter Annual Meeting, Paper 65-WA/UNT-2, 1965. 18. J. L. Hess and A. M. O. Smith, “Calculation of Nonlifting Potential Flow about Arbitrary Three-Dimensional Bodies,” J. Ship Res., vol. 8, 1964, pp. 22-44. 19. K. H. Huebner, The Finite Element Method for Engineers, 4th ed., Wiley, New York, 2001. 20. J. C. Tannehill, D. A. Anderson, and R. H. Fletcher, Compu¬ tational Fluid Mechanics and Heat Transfer, 3d ed., Taylor and Francis, Bristol, PA, 2011. 21. J. N. Newman, Marine Hydrodynamics, M.I.T. Press, Cambridge, MA, 1977. 22. P. T. Tokumaru and P. E. Dimotakis, “The Lift of a Cylinder Executing Rotary Motions in a Uniform Flow,” J. Fluid Me¬ chanics, vol. 255, 1993, pp. 1-10. 23. J. H. Ferziger and M. Peric, Computational Methods for Fluid Dynamics, 3d ed. Springer- Verlag, New York, 2002. 24. P. J. Roache, Fundamentals of Computational Fluid Dynam¬ ics, Hermosa Pub., Albuquerque, NM, 1998. 25. B. A. Finlayson, Introduction to Chemical Engineering Com¬ puting, Wiley, New York, 2012. 26. B. Andersson, Computational Fluid Dynamics, Cambridge University Press, New York, 2012. 27. H. Versteeg and W. Malalasekera, Computational Fluid Dynamics: The Finite Volume Method, 2d ed., Prentice-Hall, Upper Saddle River, NJ, 2007. References 591 28. M. Deshpande, J. Feng, and C. L. Merkle, “Numerical Mod¬ eling of the Thermodynamic Effects of Cavitation,” J. Fluids Eng., June 1997, pp. 420^27. 29. P. A. Durbin and R. B. A. Pettersson, Statistical Theory and Modeling for Turbulent Flows, Wiley, New York, 2001. 30. C. J. Freitas, “Perspective: Selected Benchmarks from Com¬ mercial CFD Codes,” J. Fluids Eng., vol. 117, June 1995, pp. 208-218. 31. R. Martinuzzi and C. Tropea, “The Flow around Surface- Mounted, Prismatic Obstacles in a Fully Developed Channel Flow,” J. Fluids Eng., vol. 1 15, March 1993, pp. 85-92. 32. K. B. Shah and J. H. Ferzier, “Fluid Mechanicians View of Wind Engineering: Large Eddy Simulation of Flow Past a Cubic Obstacle,” J. Wind Engineering and Industrial Aerody¬ namics, vol. 67-68, 1997, pp. 221-224. 33. P. Sagaut, Large Eddy Simulation for Incompressible Flows: An Introduction, 3rd ed.. Springer, New York, 2010. 34. W. J. Palm, Introduction to MATLAB 7 for Engineers, 3d ed. McGraw-Hill, New York, 2010. 35. A. Gilat, MATLAB: An Introduction with Applications, 4th ed., Wiley, New York, 2010. 36. J. W. Hoyt and R. H. J. Sellin, “Flow over Tube Banks — A Visualization Study,” J. Fluids Eng., vol. 119, June 1997, pp. 480-483. 37. S. Douglass, “Switchblade FightevBombei,” Popular Science, Nov. 2000, pp. 52-55. 38. G. Beer, 1. Smith, and C. Duenser, The Boundary Element Method with Programming: For Engineers and Scientists, Springer, New York, 2010. 39. J. D. Anderson, A History of Aerodynamics and Its Impact on Elying Machines, Cambridge University Press, Cambridge, UK, 1999. 40. B. Robins, Mathematical Tracts 1 & 2,I. Nourse, London, 1761. 41. T. K. Sengupta, A. Kasliwal, S. De, and M. Nair, “Temporal Flow Instability for Magnus-Robins Effect at High Rotation Rates,” J. Eluids and Structures, vol. 17, 2003, pp. 941-953. 42. G. F. Dargush and M. M. Grigoriev, “Fast and Accurate Solutions of Steady Stokes Flows Using Multilevel Boundary Element Methods,” J. Fluids Eng., vol. 127, July 2005, pp. 640-646. 43. R. H. Kirchhoff, Potential Plows: Computer Graphic Solu¬ tions, Marcel Dekker, New York, 2001. 44. H. Werle, “Hydrodynamic Visualization of the Flow around a Streamlined Cylinder with Suction: Cousteau-Malavard Turbine Sail Model,” Le Recherche Aerospatiale, vol. 4, 1984, pp. 29-38. 45. T. K. Sengupta and S. R. Talla, “Robins-Magnus Effect: A Continuing Saga,” Current Science, vol. 86, no. 7, 2004, pp. 1033-1036. 46. P. R. Spalart, “Airplane Trailing Vortices,” Annual Review Pluid Mechanics, vol. 30, 1998, pp. 107-138. 47. M. Elkhoury, “Assessment and Modification of One-Equation Models of Turbulence for Wall-Bounded Flows,” J. Fluids Eng., vol. 129, July 2007, pp. 921-928. Ever since the demise of the Concorde, engineers have been working on the design of an overland supersonic airplane. For such airplanes to he practical, sonic hooms must be reduced to an acceptable level. Theory, though useful, cannot solve this problem without extensive testing. Shown here is Boeing’s Lynx design, being tested in NASA’s Glenn Research Center in Cleveland. Sensors capture both the forces on the plane and the pressures far from the vehicle. The goal is to generate sonic booms so low that they barely register on the ground. [Photo courtesy of NASA] 592 9.1 Introduction; Review of Thermodynamics Chapter 9 Compressible Flow Motivation. All eight of our previous chapters have been concerned with “low-speed” or “incompressible” flow, where the fluid velocity is much less than its speed of sound. In fact, we did not even develop an expression for the speed of sound of a fluid. That is done in this chapter. When a fluid moves at speeds comparable to its speed of sound, density changes become significant and the flow is termed compressible. Such flows are difficult to obtain in liquids, since high pressures on the order of 1000 atm are needed to generate sonic velocities. In gases, however, a pressure ratio of only 2:1 will likely cause sonic flow. Thus compressible gas flow is quite common, and this subject is often called gas dynamics. The most important parameter is the Mach number. Probably the two most important and distinctive effects of compressibility on flow are (1) choking, wherein the duct flow rate is sharply limited by the sonic condition, and (2) shock waves, which are nearly discontinuous property changes in a supersonic flow. The purpose of this chapter is to explain such striking phenomena and to famil¬ iarize the reader with engineering calculations of compressible flow. Speaking of calculations, the present chapter can use the help of Excel. Com¬ pressible flow analysis is filled with scores of complicated algebraic equations, many of which are difficult to manipulate or invert. Consequently, for nearly a century, compressible flow textbooks have relied on extensive tables of Mach number rela¬ tions (see App. B) for numerical work. With Excel, however, any equation in this chapter can be typed into a cell and iterated to solve for any variables — see part (b) of Example 9.13 for an especially intricate example. With such a tool, App. B serves only as a backup, for initial estimates, and may soon vanish from textbooks. We took a brief look in Chap. 4 [Eqs. (4.13) to (4.17)] to see when we might safely neglect the compressibility inherent in every real fluid. We found that the proper criterion for a nearly incompressible flow was a small Mach number V Ma = — ^ 1 a 593 594 Chapter 9 Compressible Flow The Mach Number where V is the flow velocity and a is the speed of sound of the fluid. Under small Mach number conditions, changes in fluid density are everywhere small in the flow field. The energy equation becomes uncoupled, and temperature effects can be either ignored or put aside for later study. The equation of state degenerates into the simple statement that density is nearly constant. This means that an incompressible flow requires only a momentum and continuity analysis, as we showed with many examples in Chaps. 7 and 8. This chapter treats compressible flows, which have Mach numbers greater than about 0.3 and thus exhibit nonnegligible density changes. If the density change is significant, it follows from the equation of state that the temperature and pressure changes are also substantial. Large temperature changes imply that the energy equation can no longer be neglected. Therefore, the work is doubled from two basic equations to four 1. Continuity equation 2. Momentum equation 3. Energy equation 4. Equation of state to be solved simultaneously for four unknowns: pressure, density, temperature, and flow velocity (p, p, T, V). Thus the general theory of compressible flow is quite complicated, and we try here to make further simplifications, especially by assuming a reversible adiabatic or isentropic flow. We note in passing that at least two flow patterns depend strongly on very small density differences, acoustics, and natural convection. Acoustics [7, 9] is the study of sound wave propagation, which is accompanied by extremely small changes in den¬ sity, pressure, and temperature. Natural convection is the gentle circulating pattern set up by buoyancy forces in a fluid stratified by uneven heating or uneven concentration of dissolved materials. Here we are concerned only with steady compressible flow where the fluid velocity is of magnitude comparable to that of the speed of sound. The Mach number is the dominant parameter in compressible flow analysis, with dif¬ ferent effects depending on its magnitude. Aerodynamicists especially make a distinc¬ tion between the various ranges of Mach number, and the following rough classifications are commonly used: Ma < 0.3: 0.3 < Ma < 0.8: 0.8 < Ma < 1.2: incompressible flow, where density effects are negligible. subsonic flow, where density effects are important but no shock waves appear. transonic flow, where shock waves first appear, dividing subsonic and supersonic regions of the flow. Powered flight in the transonic region is difficult because of the mixed character of the flow field. 1.2 < Ma < 3.0: supersonic flow, where shock waves are present but there are no subsonic regions. 3.0 < Ma: hypersonic flow , where shock waves and other flow changes are especially strong. 9. 1 Introduction; Review of Thermodynamics 595 The Specific-Heat Ratio The Perfect Gas The numerical values listed are only rough guides. These five categories of flow are appropriate to external high-speed aerodynamics. For internal (duct) flows, the most important question is simply whether the flow is subsonic (Ma < 1) or supersonic (Ma 1), because the effect of area changes reverses, as we show in Sec. 9.4. Since supersonic flow effects may go against intuition, you should study these differences carefully. In addition to geometry and Mach number, compressible flow calculations also depend on a second dimensionless parameter, the specific-heat ratio of the gas: Earlier, in Chaps. 1 and 4, we used the same symbol k to denote the thermal conduc¬ tivity of a fluid. We apologize for the duplication; thermal conductivity does not appear in these later chapters of the text. Recall from Fig. 1.4 that k for the common gases decreases slowly with tempera¬ ture and lies between 1.0 and 1.7. Variations in k have only a slight effect on com¬ pressible flow computations, and air, k ~ 1.40, is the dominant fluid of interest. Therefore, although we assign some problems involving other gases like steam and CO2 and helium, the compressible flow tables in App. B are based solely on the single value k = 1.40 for air. This text contains only a single chapter on compressible flow, but, as usual, whole books have been written on the subject. Here we list only certain recent or classic texts. References 1 to 4 are introductory or intermediate treatments, while Refs. 5 to 10 are advanced books. One can also become specialized within this specialty of compressible flow. Reference 11 concerns hypersonic fiow — that is, at very high Mach numbers. Reference 12 explains the exciting new technique of direct simulation of gas flows with a molecular dynamics model. Compressible flow is also well suited for computational fluid dynamics (CFD), as described in Ref. 13. Finally, a short, thoroughly readable (no calculus) Ref. 14 describes the principles and promise of high-speed (supersonic) flight. From time to time we shall defer some specialized topic to these other texts. In principle, compressible flow calculations can be made for any fluid equation of state, and we shall assign a few problems involving the steam tables , the gas tables , and liquids [Eq. (1.19)]. But in fact most elementary treatments are confined to the perfect gas with constant specific heats: Cp p = pRT R = Cp — Cl, = const k = — = const (9.2) Cp, For all real gases, Cp, c^, and k vary with temperature but only moderately; for exam¬ ple, Cp of air increases 30 percent as temperature increases from 0 to 5000°F. Since we rarely deal with such large temperature changes, it is quite reasonable to assume constant specific heats. 596 Chapter 9 Compressible Flow Recall from Sec. 1.8 that the gas constant is related to a universal constant A divided by the gas molecular weight; (9.3) A = 49,720 ft-lbf/(lbmol ■ °R) = 8314 J/(kmol ■ K) where For air, M = 28.97, and we shall adopt the following property values for air through¬ out this chapter: R = 1716 ft^/(s^ • °R) = 287 m^/(s^ ■ K) k= 1.400 = 4293 ft^/(s^ • °R) = 718 m^/(s^ • K) (9.4) k - 1 = 6009 ft^/(s^ • °R) = 1005 mV(s^ ■ K) Experimental values of k for eight common gases were shown in Fig. 1.4. From this figure and the molecular weight, the other properties can be computed, as in Eqs. (9.4). The changes in the internal energy u and enthalpy of a perfect gas are computed for constant specific heats as M2 - Ml = c„(7’2 - Ti) h2- hi = Cp{T2 - Ti) (9.5) Eor variable specific heats one must integrate u = and h = Jc^dT or use the gas tables . Most modern thermodynamics texts now contain software for evaluat¬ ing properties of nonideal gases . Isentropic Process The isentropic approximation is common in compressible flow theory. We compute the entropy change from the first and second laws of thermodynamics for a pure substance [17 or 18]: dp P Tds = dh — (9.6) Introducing dh = CpdT for a perfect gas and solving for ds, we substitute pT = pIR from the perfect-gas law and obtain f2 dT ds = c„ - R • 1 P y 1 ^ J If Cp is variable, the gas tables will be needed, but for constant Cp we obtain the analytic results (9.8) Equations (9.8) are used to compute the entropy change across a shock wave (Sec. 9.5), which is an irreversible process. 9. 1 Introduction; Review of Thermodynamics 597 For isentropic flow, we set S2 = and obtain these interesting power-law relations for an isentropic perfect gas: (9.9) These relations are used in Sec. 9.3. EXAMPLE 9.1 Argon flows through a mhe such that its initial condition is pi = 1.7 MPa and pi = 18 kg/m^ and its final condition is p2 = 248 kPa and T2 = 400 K. Estimate (a) the initial tem¬ perature, (b) the final density, (c) the change in enthalpy, and (d) the change in entropy of the gas. Solution From Table A.4 for argon, R = 208 mV(s^ ■ K) and k = 1.67. Therefore estimate its specific heat at constant pressure from Eq. (9.4): kR 1.67(208) , , — ^-^;6^«519m^/(s^.K) The initial temperature and final density are estimated from the ideal-gas law, Eq. (9.2): 1.7 E6N/m" piR (18kg/mO[208mV(s -K)] = 454 K P2 = P2 248 E3 N/m" T2R (400 K)[208 m^/(s^ ■ K)] From Eq. (9.5) the enthalpy change is = 2.98 kg/m^ Ans. (a) Ans. (b) hi - hi = CplTi - Ti) = 519(400 - 454) « -28,000 J/kg (orm^/s^) Ans. (c) The argon temperature and enthalpy decrease as we move down the tube. Actually, there may not be any external cooling; that is, the fluid enthalpy may be converted by friction to increased kinetic energy (Sec. 9.7). Finally, the entropy change is computed from Eq. (9.8): ^2 — ^1 = Cp In - R In P2 Pi 400 0.248 E6 = 519 In - 208 In - 454 1.7 E6 = -66 -f 400 ^ 334 mV(s^ ■ K) Ans. (d) The fluid entropy has increased. If there is no heat transfer, this indicates an irreversible process. Note that entropy has the same units as the gas constant and specific heat. This problem is not just arbitrary numbers. It correctly simulates the behavior of argon moving subsonically through a tube with large frictional effects (Sec. 9.7). 598 Chapter 9 Compressible Flow 9.2 The Speed of Sound The so-called speed of sound is the rate of propagation of a pressure pulse of infini¬ tesimal strength through a still fluid. It is a thermodynamic property of a fluid. Let us analyze it hy first considering a pulse of finite strength, as in Fig. 9.1. In Fig. 9.1a the pulse, or pressure wave, moves at speed C toward the still fluid (p, p, T, V = 0) at the left, leaving behind at the right a fluid of increased properties (p + Ap, p + Ap, T + AT) and a fluid velocity AV toward the left following the wave but much slower. We can determine these effects by making a control volume analysis across the wave. To avoid the unsteady terms that would be necessary in Fig. 9.1a, we adopt instead the control volume of Fig. 9. lb, which moves at wave speed C to the left. The wave appears fixed from this viewpoint, and the fluid appears to have velocity C on the left and C — AV on the right. The thermodynamic properties p, p, and T are not affected by this change of viewpoint. The flow in Fig. 9.\b is steady and one-dimensional across the wave. The continu¬ ity equation is thus, from Eq. (3.24), pAC = {p + Ap)(A){C - AV) Ap or AV=C - ^ (9.10) p+ Ap This proves our contention that the induced fluid velocity on the right is much smaller than the wave speed C. In the limit of infinitesimal wave strength (sound wave) this speed is itself infinitesimal. Notice that there are no velocity gradients on either side of the wave. Therefore, even if fluid viscosity is large, frictional effects are confined to the interior of the wave. Advanced texts [for example, 9] show that the thickness of pressure waves in P P T V=0 P P T Fig. 9.1 Control volume analysis of y a finite-strength pressure wave: (a) control volume fixed to still fluid at left; (b) control volume moving left at wave speed C. C - p + Ap p + Ap T+ AT AV Moving wave of frontal area A (a) Friction and heat transfer effects are confined to wave interior p + Ap p + Ap T+ AT V=C-AV - ► Fixed wave ib) 9.2 The Speed of Sound 599 gases is of order 10 ° ft at atmospheric pressure. Thus we can safely neglect friction and apply the one-dimensional momentum equation (3.40) across the wave: or bright ( ^out ^in ) pA- {p + Ap)A = ipAC)(C - AV- C) Again the area cancels, and we can solve for the pressure change: Ap = pC AV (9.11) (9.12) If the wave strength is very small, the pressure change is small. Finally, combine Eqs. (9.10) and (9.12) to give an expression for the wave speed: , Ap Ap 1 -f P (9.13) The larger the strength Ap/p of the wave, the faster the wave speed; that is, powerful explosion waves move much more quickly than sound waves. In the limit of infini¬ tesimal strength Ap — ■ 0, we have what is defined to be the speed of sound a of a fluid: 2 a = dp dp (9.14) But the evaluation of the derivative requires knowledge of the thermodynamic process undergone by the fluid as the wave passes. Sir Isaac Newton in 1686 made a famous error by deriving a formula for sound speed that was equivalent to assuming an iso¬ thermal process, the result being 20 percent too low for air, for example. He rational¬ ized the discrepancy as being due to the “crassitude” (dust particles and so on) in the air; the error is certainly understandable when we reflect that it was made 180 years before the proper basis was laid for the second law of thermodynamics. We now see that the correct process must be adiabatic because there are no tem¬ perature gradients except inside the wave itself. For vanishing-strength sound waves we therefore have an infinitesimal adiabatic or isentropic process. The correct expres¬ sion for the sound speed is ( 9p ( 1 X./2 a - \ — \dp J - V dp j (9.15) for any fluid, gas or liquid. Even a solid has a sound speed. Eor a perfect gas, from Eq. (9.2) or (9.9), we deduce that the speed of sound is a = vl/2 — ) = {kRT^'^ PJ (9.16) The speed of sound increases as the square root of the absolute temperature. Eor air, with k = 1.4, an easily memorized dimensional formula is a(ft/s) = 49[r(°R)]''^ a(m/s) = 20[7’(K)]'^ (9.17) 600 Chapter 9 Compressible Flow Table 9.1 Sound Speed of Various Materials at 60°F (15.5°C) and 1 atm Material a, ft/s a, m/s Gases: Hi 4,246 1,294 He 3,281 1,000 Air 1,117 340 Ar 1,040 317 CO2 873 266 CH4 607 185 297 91 Liquids: Glycerin 6,100 1,860 Water 4,890 1,490 Mercury 4,760 1,450 Ethyl alcohol 3,940 1,200 Solids: Aluminum 16,900 5,150 Steel 16,600 5,060 Hickory 13,200 4,020 Ice 10,500 3,200 Plane waves. Solids also have a shear-wave speed. At sea-level standard temperature, 60°F = 520°R, a = 1117 ft/s. This decreases in the upper atmosphere, which is cooler; at 50,000-ft standard altitude, T = — 69.7°F = 389.9°R and a = 49(389.9)'^ = 968 ft/s, or 13 percent less. Some representative values of sound speed in various materials are given in Table 9.1. For liquids and solids it is common to define the bulk modulus K of the material: dp K = -r— dr (9.18) In terms of bulk modulus, then, a = For example, at standard conditions, the bulk modulus of liquid carbon tetrachloride is 1.32 GPa absolute, and its density is 1590 kg/m^. Its speed of sound is therefore a = (1.3 E9 Pa/1590 = 911 m/s = 2980 ft/s. Steel has a bulk modulus of about 2 Ell Pa and water about 2.2 E9 Pa (see Table A. 3), or 90 times less than steel. For solids, it is sometimes assumed that the bulk modulus is approximately equiva¬ lent to Young’s modulus of elasticity E, but in fact their ratio depends on Poisson’s ratio cr: ^ = 3(1 - 2(t) (9.19) A The two are equal for cr = \|, which is approximately the case for many common metals such as steel and aluminum. EXAMPLE 9.2 Estimate the speed of sound of carbon monoxide at 200-kPa pressure and 300°C in m/s. Solution From Table A.4, for CO, the molecular weight is 28.01 and k ~ 1.40. Thus from Eq. (9.3) Rco = 8314/28.01 = 297 mV(s^ ■ K), and the given temperature is 300^ -f 273 = 573 K. Thus from Eq. (9.16) we estimate flco = (k/?r)‘“ = [1.40(297)(573)]‘“ = 488 m/s Ans. 9.3 Adiabatic and Isentropic As mentioned in Sec. 9.1, the isentropic approximation greatly simplifies a compress- Steady Flow ible flow calculation. So does the assumption of adiabatic flow, even if nonisentropic. Consider high-speed flow of a gas past an insulated wall, as in Fig. 9.2. There is no shaft work delivered to any part of the fluid. Therefore, every streamtube in the flow satisfies the steady flow energy equation in the form of Eq. (3.70): hi + iVi + gZi = h2 + ^Vl + gZ2 - q + (9.20) where point 1 is upstream of point 2. You may wish to review the details of Eq. (3.70) and its development. We saw in Example 3.20 that potential energy changes of a 9.3 Adiabatic and Isentropic Steady Flow 601 Fig. 9.2 Velocity and stagnation enthalpy distributions near an insulated wall in a typical high¬ speed gas flow. gas are extremely small compared with kinetic energy and enthalpy terms. We shall neglect the terms gzi and gZ2 in all gas dynamic analyses. Inside the thermal and velocity boundary layers in Fig. 9.2 the heat transfer and viscous work terms q and w„ are not zero. But outside the boundary layer q and w„ are zero by definition, so that the outer flow satisfies the simple relation /ti + ^Vi = h2 + ^^2 = const (9.21) The constant in Eq. (9.21) is equal to the maximum enthalpy that the fluid would achieve if brought to rest adiabatically. We call this value Hq, the stagnation enthalpy of the flow. Thus we rewrite Eq. (9.21) in the form h + = Hq = const (9.22) This should hold for steady adiabatic flow of any compressible fluid outside the boundary layer. The wall in Eig. 9.2 could be either the surface of an immersed body or the wall of a duct. We have shown the details of Eig. 9.2; typically the thermal layer thickness dj is greater than the velocity layer thickness dy because most gases have a dimensionless Prandtl number Pr less than unity (see, for exam¬ ple, Ref. 19, Sec. 4-3.2). Note that the stagnation enthalpy varies inside the thermal boundary layer, but its average value is the same as that at the outer layer due to the insulated wall. For nonperfect gases we may have to use the steam tables or the gas tables to implement Eq. (9.22). But for a perfect gas h = CpT, and Eq. (9.22) becomes CpT + iy" = CpTo (9.23) This establishes the stagnation temperature Tq of an adiabatic perfect-gas flow — that is, the temperature it achieves when decelerated to rest adiabatically. An alternate interpretation of Eq. (9.22) occurs when the enthalpy and temperature drop to (absolute) zero, so the velocity achieves a maximum value: V„,ax = = (2cpTof^ (9.24) No higher flow velocity can occur unless additional energy is added to the fluid through shaft work or heat transfer (Sec. 9.8). 602 Chapter 9 Compressible Flow Mach Number Relations Isentropic Pressure and Density Relations Fig. 9.3 Adiabatic (TITq and a/oo) and isentropic (p/po and p/po) properties versus Mach number for k = 1.4. The dimensionless form of Eq. (9.23) brings in the Mach number Ma as a param¬ eter, by using Eq. (9.16) for the speed of sound of a perfect gas. Divide through by CpT to obtain 2cpT T But, from the perfect-gas law, CpT = {kRI{k — 1)]?’ becomes (9.25) a^l{k - 1), so that Eq. (9.25) ^ ^ {k - l)y^ To 2a^ T or (9.26) This relation is plotted in Eig. 9.3 versus the Mach number for k the temperature has dropped to ^ Tq. Since a 7^^^, the ratio aja is the square root of (9.26): Go a \ + - {k- l)Ma^ 1/2 1.4. At Ma = 5 (9.27) Equation (9.27) is also plotted in Fig. 9.3. At Ma = 5 the speed of sound has dropped to 41 percent of the stagnation value. Note that Eqs. (9.26) and (9.27) require only adiabatic flow and hold even in the presence of irreversibilities such as friction losses or shock waves. Mach number 9.3 Adiabatic and Isentropic Steady Flow 603 Relationship to Bernoulli’s Equation Critical Values at the Sonic Point If the flow is also isentropic, then for a perfect gas the pressure and density ratios can be computed from Eq. (9.9) as a power of the temperature ratio: (9.28fl) (9.28/7) These relations are also plotted in Fig. 9.3; at Ma = 5 the density is 1.13 percent of its stagnation value, and the pressure is only 0.19 percent of stagnation pressure. The quantities po and po ^re the isentropic stagnation pressure and density, respectively that is, the pressure and density that the flow would achieve if brought isentropically to rest. In an adiabatic nonisentropic flow pq and po retain their local meaning, but they vary throughout the flow as the entropy changes due to friction or shock waves. The quantities Hq, Tq, and oq are constant in an adiabatic nonisen¬ tropic flow (see Sec. 9.7 for further details). \ 1 21 1 + -{k - l)Ma^ po_^Toy‘~'^ _ r 1 7I 1 + -{k - l)Ma- The isentropic assumptions (9.28) are effective, but are they realistic? Yes. To see why, take the differential of Eq. (9.22): Adiabatic: dh + V dV = 0 (9.29) Meanwhile, from Eq. (9.6), if = 0 (isentropic process), dp dh = — (9.30) P Combining (9.29) and (9.30), we find that an isentropic streamtube flow must be — +VdV=0 (9.31) P But this is exactly the Bernoulli relation, Eq. (3.54), for steady frictionless flow with negligible gravity terms. Thus we see that the isentropic flow assumption is equivalent to use of the Bernoulli or streamline form of the frictionless momentum equation. The stagnation values (gq, Tq, po, Po) are useful reference conditions in a compressible flow, but of comparable usefulness are the conditions where the flow is sonic, Ma = 1.0. These sonic, or critical, properties are denoted by asterisks: p, p, a, and T. They are certain ratios of the stagnation properties as given by Eqs. (9.26) to (9.28) when Ma = 1.0; for k = \A p Po 2 \kl{k-l) k+ T 2 To~ k + 1 = 0.5283 0.8333 p Po a Go 2 k + 1 ^ + 1 \ !/(<:- 1) j = 0.6339 1/2 I = 0.9129 (9.32) 604 Chapter 9 Compressible Flow In all isentropic flow, all critical properties are constant; in adiabatic nonisentropic flow, a and T are constant, but p and p may vary. The critical velocity V equals the sonic sound speed a by definition and is often used as a reference velocity in isentropic or adiabatic flow; V = a (9.33) The usefulness of these critical values will become clearer as we study compressible duct flow with friction or heat transfer later in this chapter. Some Useful Numbers for Air Since the great bulk of our practical calculations are for air, k = 1.4, the stagnation property ratios p/po and so on from Eqs. (9.26) to (9.28) are tabulated for this value in Table B.l. The increments in Mach number are rather coarse in this table because the values are meant as only a guide; these equations are now a trivial matter to manipulate on a hand calculator. Thirty years ago every text had extensive compress¬ ible flow tables with Mach number spacings of about 0.01, so that accurate values could be interpolated. Even today, reference books are available [20, 21, 29] with tables and charts and computer programs for a wide variety of compressible flow situ¬ ations. Reference 22 contains formulas and charts applying to the thermodynamics of real (nonperfect) gas flows. For k = 1.4, the following numerical versions of the isentropic and adiabatic flow formulas are obtained: — = 1 4- 0.2 Ma^ — = (1 -f 0.2 Ma^)^-^ T p (9.34) — = (1 + 0.2 Ma^)^'^ P Or, if we are given the properties, it is equally easy to solve for the Mach number (again with k = 1.4): (9.35) Note that these isentropic flow formulas serve as the equivalent of the friction¬ less adiabatic momentum and energy equations. They relate velocity to physical properties for a perfect gas, but they are not the “solution” to a gas dynamics problem. The complete solution is not obtained until the continuity equation has also been satisfied, for either one-dimensional (Sec. 9.4) or multidimensional (Sec. 9.9) flow. One final note: These isentropic-ratio-versus-Mach-number formulas are seduc¬ tive, tempting one to solve all problems by jumping right into the tables. Actually, many problems involving (dimensional) velocity and temperature can be solved more easily from the original raw dimensional energy equation (9.23) plus the perfect-gas law (9.2), as the next example will illustrate. 9.3 Adiabatic and Isentropic Steady Flow 605 EXAMPLE 9.3 Air flows adiabatically through a duct. At point 1 the velocity is 240 m/s, with Ti = 320 K and Pi = 170 kPa. Compute (a) Tq, (b) p^, {c) Pq, (d) Ma, (e) and (f) F . At point 2 further downstream V2 = 290 m/s and p2 = 135 kPa. (g) What is the stagnation pressure Pq2^ Solution • Assumptions: Let air be approximated as an ideal gas with constant k. The flow is adiabatic but not isentropic. Isentropic formulas are used only to compute local po and Pq, which vary. • Approach: Use adiabatic and isentropic formulas to find the various properties. • Ideal gas parameters: For air, R = 287 mV(s^ ■ K), k = 1.40, and Cp = 1005 m^/(s^ • K). • Solution steps (a, b, c, d): With Ti, pi, and Vi known, other properties at point 1 follow: Vi (240 m/s)^ Tm = Ti + — = 320 + - ^ - r — ^ - = 320 + 29 = 349 K Ans. (a) 2cp 2[1005m^/(s"-K)] Once the Mach number is found from Eq. (9.35), local stagnation pressure and density follow: „2x3.5 Poi = Pi(l + 0.2 Maf)^'^ = (170kPa)[l + 0.2(0.67)^]^'= = 230 kPa Poi Poi 230,000 N/m" RToi 287mV(s^-K) N ■ s^/m kg ■ = 2.29 - 5 — = 2.29 Ans. (b) Ans. id) Ans. (c) • Comment: Note that we used dimensional (non-Mach-number) formulas where convenient. • Solution steps (e, f): Both Fnax and V are directly related to stagnation temperature from Eqs. (9.24) and (9.33): = VlCpTo = V2 1005 m^/(s^ ■ K) = 837 V = (349 K) = 342 Ans. (e) Ans. if) • At point 2 downstream, the temperature is unknown, but since the flow is adiabatic, the stagnation temperature is constant: Tgi = T112 = 349 K. Thus, from Eq. (9.23), Ti — Tta 2Cp = 349 - (290 m/s)^ 2[1005m^/(s^-K)] = 307 K Hence, from Eq. (9.28a), the isentropic stagnation pressure at point 2 is = (135 kPa) /349 V307 Ky Pm = Pi = 211 kPa Ans. (g) 606 Chapter 9 Compressible Flow 9.4 Isentropic Flow with Area Changes Fig. 9.4 Compressible flow through a duct: (a) real-fluid velocity profile; (b) one-dimensional approximation. • Comments: Part (g), a ratio-type ideal-gas formula, is more direct than finding the Mach number, which turns out to be Ma2 = 0.83, and using the Mach number formula, Eq. (9.34) for po2- Note that /Jo2 is 8 percent less than pm. The flow is nonisentropic: Entropy rises downstream, and stagnation pressure and density drop, due in this case to frictional losses. By combining the isentropic and/or adiabatic flow relations with the equation of con¬ tinuity we can study practical compressible flow problems. This section treats the one-dimensional flow approximation. Figure 9.4 illustrates the one-dimensional flow assumption. A real flow. Fig. 9.4a, has no slip at the walls and a velocity profile V(x, y) that varies across the duct sec¬ tion (compare with Fig. 7.8). If, however, the area change is small and the wall radius of curvature large — « 1 b(x) ^ R{x) (9.36) dx then the flow is approximately one-dimensional, as in Fig. 9Ab, with V ~ V{x) reacting to area change A(x). Compressible flow nozzles and diffusers do not always satisfy conditions (9.36), but we use the one-dimensional theory anyway because of its simplicity. For steady one-dimensional flow the equation of continuity is, from Eq. (3.24), p{x)V{x)A{x) = m = const (9.37) Before applying this to duct theory, we can learn a lot from the differential form of Eq. (9.37): dp dV dA — + — — p V A 0 (9.38) The differential forms of the frictionless momentum equation (9.31) and the sound- speed relation (9.15) are recalled here for convenience: (a) (b) 9.4 Isentropic Flow with Area Changes 607 Duct geometry Subsonic Ma < 1 Supersonic Ma > 1 Fig. 9.5 Effect of Mach number on property changes with area change in duct flow. dV0 Supersonic diffuser Momentum — + VdV = 0 P (9.39) Sound speed: dp = dp Now eliminate dp and dp between Eqs. (9.38) and (9.39) to obtain the following relation between velocity change and area change in isentropic duct flow: dV dA 1 dp — = - 5 - = - ^ (9.40) y A Ma^ - 1 Inspection of this equation, without actually solving it, reveals a fascinating aspect of compressible flow: Property changes are of opposite sign for subsonic and supersonic flow because of the term Ma^ — 1. There are four combinations of area change and Mach number, summarized in Fig. 9.5. From earlier chapters we are used to subsonic behavior (Ma < 1): When area increases, velocity decreases and pressure increases, which is denoted a subsonic dif¬ fuser. But in supersonic flow (Ma > 1), the velocity actually increases when the area increases, a supersonic nozzle. The same opposing behavior occurs for an area decrease, which speeds up a subsonic flow and slows down a supersonic flow. What about the sonic point Ma = 1 ? Since infinite acceleration is physically impos¬ sible, Fq. (9.40) indicates that dV can be finite only when dA = 0 — that is, a minimum area (throat) or a maximum area (bulge). In Fig. 9.6 we patch together a throat section and a bulge section, using the rules from Fig. 9.5. The throat or converging-diverging section can smoothly accelerate a subsonic flow through sonic to supersonic flow, as in Fig. 9.6a. This is the only way a supersonic flow can be created by expanding the gas from a stagnant reservoir. The bulge section fails; the bulge Mach number moves away from a sonic condition rather than toward it. Although supersonic flow downstream of a nozzle requires a sonic throat, the opposite is not true: A compressible gas can pass through a throat section without becoming sonic. 608 Chapter 9 Compressible Flow Fig. 9.6 From Eq. (9.40), in flow through a throat (a) the fluid can accelerate smoothly through sonic and supersonic flow. In flow through the bulge {b) the flow at the bulge cannot be sonic on physical grounds. Perfect-Gas Area Change A max Subsonic Ma = 1 Supersonic («) (b) We can use the perfect-gas and isentropic flow relations to convert the continuity relation (9.37) into an algebraic expression involving only area and Mach number, as follows. Equate the mass flow at any section to the mass flow under sonic conditions (which may not actually occur in the duct): or pVA = pVA ^ _ p^V^ A ~ p V (9.41) Both terms on the right are functions only of Mach number for isentropic flow. From Eqs. (9.28) and (9.32) p ^ p ^ P Po P k + 1 From Eqs. (9.26) and (9.32) we obtain 1 9 1 -f - (A: - l)Ma^ !/(<:- 1) y V V TJ ,1/2 f 2 r 1 9I U + 1 1 + - {k- l)Ma^ L 2 J 1/2 (9.42) (9.43) Combining Eqs. (9.41) to (9.43), we get the desired result: A 1 ■ 1 + \{k — 1 ) Ma^ ■ (l/2)(/r+ !)/(- 1) A Ma \(k + 1) J (9.44) For k = 1.4, Eq. (9.44) takes the numerical form A 1 (l-f0.2Ma^)^ — = - ^ (9.45) A Ma 1.728 which is plotted in Fig. 9.7. Equations (9.45) and (9.34) enable us to solve any one-dimensional isentropic airflow problem given, say, the shape of the duct A(x) and the stagnation conditions and assuming that there are no shock waves in the duct. 9.4 Isentropic Flow with Area Changes 609 Fig. 9.7 Area ratio and fluid properties versus Mach number for isentropic flow of a perfect gas with k = 1.4. Figure 9.7 shows that the minimum area that can occur in a given isentropic duct flow is the sonic, or critical, throat area. All other duct sections must have A greater than A. In many flows a critical sonic throat is not actually present, and the flow in the duct is either entirely subsonic or, more rarely, entirely supersonic. Choking From Eq. (9.41) the inverse ratio A/A equals pVI{pV), the mass flow per unit area at any section compared with the critical mass flow per unit area. From Fig. 9.7 this inverse ratio rises from zero at Ma = 0 to unity at Ma = 1 and back down to zero at large Ma. Thus, for given stagnation conditions, the maximum possible mass flow passes through a duct when its throat is at the critical or sonic condition. The duct is then said to be choked and can carry no additional mass flow unless the throat is wid¬ ened. If the throat is constricted further, the mass flow through the duct must decrease. From Eqs. (9.32) and (9.33) the maximum mass flow is / 2 / Ik / j \(i/2){k+mk-i) = Apo(RTof^ (9.46fl) Eor ^ = 1.4 this reduces to m max 0.6847APo(^7’o)'^^ = 0.6847po A (RToy^ (9.461?) For isentropic flow through a duct, the maximum mass flow possible is proportional to the throat area and stagnation pressure and inversely proportional to the square root of the stagnation temperature. These are somewhat abstract facts, so let us illustrate with some examples. 610 Chapter 9 Compressible Flow The Local Mass Flow Function Equations (9.46) give the maximum mass flow, which occurs at the choking condition (sonic exit). They can be modified to predict the actual (nonmaximum) mass flow at any section where local area A and pressure p are known. The algebra is convoluted, so here we give only the final result, expressed in dimensionless form: Mass flow function m "s/RTq A po (9.47) We stress that p and A in this relation are the local values at position x. As p/p^ falls, this function rises rapidly and then levels out at the maximum of Eqs. (9.46). A few values may be tabulated here for A: = 1.4: P/Pa 1.0 1 0.98 1 0.95 1 0.9 0.8 1 0.7 1 0.6 1 <0.5283 Function 1 0.0 0.1978 0.3076 0.4226 0.5607 0.6383 0.6769 0.6847 Equation (9.47) is handy if stagnation conditions are known and the flow is not choked. When A/A is known and the Mach number is unknown, no algebraic solution of Eq. (9.44) is known to the writer. One could interpolate in Table B.l or simply iterate Eq. (9.44) with a calculator. But Excel can iterate Eq. (9.44) for subsonic flow in its direct form: Subsonic flow: Ma A A 1 + 0.5 (/t - l)Ma^ 0.5{k + 1) 0.5(r+ !)/(<:- 1) (9.48) Make a subsonic guess for Ma on the right side and then replace it with the value calculated on the left side. Eor example, suppose A/A = 2.035, corresponding to Ma = 0.300. A poor guess of Ma = 0.5 in Eq. (9.44) leads to a better Ma = 0.329, then 0.303, then 0.300. Eor supersonic flow, iteration of Eq. (9.44) diverges. Instead, simply try different Mach numbers in Eq. (9.44) until the proper area is achieved. For example, suppose A/A = 3.183, corresponding to Ma = 2.70. A poor guess of Ma = 2.4 yields A/A = 2.403, 24 percent low. Improve the guess to Ma = 2.8 to give A/A = 3.500, or 10 percent high. Interpolate to Ma = 2.72, A/A = 3.244, 2 percent high. Finally settle on Ma = 2.70, correct. These calculations simply require that you retype your guess for Ma, check the error, and convergence only depends upon your cleverness. Note that two solutions are possible for a given A/A, one subsonic and one super¬ sonic. The proper solution cannot be selected without further information, such as known pressure or temperature at the given duct section. EXAMPLE 9.4 Air flows isentropically through a duct. At section 1 the area is 0.05 and Vi = 180 m/s. Pi = 500 kPa, and Ti = 470 K. Compute (a) Tq, (b) Mai, (c) Po> ™d (d) both A and m. If at section 2 the area is 0.036 m^, compute Ma2 and p2 if the flow is (e) subsonic or (f) supersonic. Assume k = 1.4. The author is indebted to Georges Aigret, of Chimay, Belgium, for suggesting this useful function. 9.4 Isentropic Flow with Area Changes 611 Part (a) Part (b) Part (c) Part (d) Part (e) Solution A general sketch of the problem is shown in Fig. E9.4. With Vi and Ti known, the energy equation (9.23) gives vi Jo = + — = 470 + 2Cp (180)^ 2(1005) = 486 K Ans. (a) The local sound speed Oi = \/kRTi = [(1.4)(287)(470)]^^ = 435 m/s. Hence Mai Cl\ 180 435 0.414 Ans. (b) With Mai known, the stagnation pressure follows from Eq. (9.34): Po = Pi(l + 0.2 Mai)^'^ = (500kPa)[l + 0.2(0.414)^]^'’ = 563 kPa Ans. (c) Similarly, from Eq. (9.45), the critical sonic throat area is Ai (1 + 0.2 Ma?)^ [1 + 0.2(0.414)"]^ A ~ 1.728 Mai ~ 1.728(0.414) or A = ^1 1.547 0.05 m^ 1.547 0.0323 m^ Ans. (d) This throat must actually be present in the duct if the flow is to become supersonic. We now know A. So to compute the mass flow we can use Eqs. (9.46), which remain valid, based on the numerical value of A, whether or not a throat actually exists: m = 0.6847 PqA vm = 0.6847 (563,000)(0.0323) V(287)(486) = 33.4 kg/s Ans. (d) Or we could fare equally well with our new “local mass flow” formula, Eq. (9.47), using, say, the pressure and area at section 1. Given pdpt, = 500/563 = 0.889, Eq. (9.47) yields V287(486) 563,000(0.05) ^0 4 ^ (0.889)^^''^[1 -(0.889)°'‘"'‘] =0.444 kg m=33.4 — Ans. (d) Eor subsonic flow upstream of the throat at section 2E, the area ratio is A2/A = 0.036/0.0323 = 1.115, corresponding to the left side of Fig. 9.7 or the subsonic numbers 612 Chapter 9 Compressible Flow in Table B.l, neither of which is very accurate. Guess Ma2 at section 2E, from Fig. 9.7, at about 0.70. Enter this guess into Eq. (9.48) and repeat. The Excel table is: Ma - guess Ma — Eq. (9.48) A/A 0.700 0.687 1.115 0.687 0.680 1.115 0.680 0.677 1.115 0.677 0.675 1.115 0.675 0.674 1.115 0.674 0.674 1.115 The (slowly) converged subsonic Mach number is Maj = 0.674 The pressure is given by the isentropic relation Pi = Po [1 + 0.2(0.674)^]^'^ 563 kPa 1.356 = 415 kPa Ans. (e) Ans. (e) Part (e) does not require a throat, sonic or otherwise; the flow could simply be contracting subsonically from Ai to A^. Part (f) For supersonic flow at section IF, again the area ratio is 0.036/0.0323 = 1.115. On the right side of Fig. 9.7 we estimate Ma2 —1.5. The table from Eq. (9.44) is Ma - guess AJA - Eq. (9.44) AM 1.5000 1.1762 1.1150 1.4000 1.1149 1.1150 1.4001 1.1150 1.1150 We were lucky that this Mach number is easy to guess: Ma2 = 1.4001 Ans. (/) Again the pressure is given by the isentropic relation at this new Mach number: Pi = _ Po _ [1 + 0.2(1.4001)^]^'^ 563 kPa 3.183 = 177 kPa Ans. if) Note that the supersonic-flow pressure level is much less than p2 in part (e), and a sonic throat must have occurred between sections 1 and 2F. EXAMPLE 9.5 It is desired to expand air from po = 200 kPa and Tg = 500 K through a throat to an exit Mach number of 2.5. If the desired mass flow is 3 kg/s, compute (a) the throat area and the exit (b) pressure, (c) temperature, (d) velocity, and (e) area, assuming isentropic flow, with k = 1.4. 9.5 The Normal Shock Wave 613 9.5 The Normal Shock Wave Solution The throat area follows from Eq. (9.47), because the throat flow must be sonic to produce a supersonic exit: or 0.6847po 3.0[287(500)]''^ 0.6847(200,000) 0.00830 m^ 4 -Dthroat = 10.3 cm Ans. (a) With the exit Mach number known, the isentropic flow relations give the pressure and temperature: Pe = Po 200,000 [1 + 0.2(2.5)^]^ T, T = p 17.08 500 = 11,700 Pa = 222 K 1 + 0.2(2.5)^ 2.25 The exit velocity follows from the known Mach number and temperature: P, = Ma,{kRT,)^'^ = 2.5[1.4(287)(222)]‘'^ = 2.5(299 m/s) = 747 m/s Ans. (b) Ans. (c) Ans. (d) The exit area follows from the known throat area and exit Mach number and Eq. (9.45): A, [1 + 0.2(2.5)^f — = - = 2.64 A 1.728(2.5) or A, = 2.64A = 2.64(0.0083 m^) = 0.0219 m^ = or Dg = 16.7 cm Ans. (e) One point might be noted: The computation of the throat area A did not depend in any way on the numerical value of the exit Mach number. The exit was supersonic; therefore the throat is sonic and choked, and no further information is needed. Shock waves are nearly discontinuous changes in a supersonic flow. They can occur due to a higher downstream pressure, a sudden change in flow direction, blockage by a downstream body, or the result of an explosion. The simplest algebraically is a one-dimensional change, or normal shock wave, shown in Fig. 9.8. We select a control volume just before and after the wave. The analysis is identical to that of Fig. 9.1; that is, a shock wave is a fixed strong pressure wave. To compute all property changes rather than just the wave speed, we use all our basic one-dimensional steady flow relations, letting section 1 be upstream and section 2 be downstream: Continuity: PiVi = P2V2 = G = const (9.49fl) Momentum: P\ ~ Pi ~ Pi^i ~ (9.49/7) Energy: hi + = h2 + ^Vl = pQ = const (9.49c) Pi P'’ Perfect gas: (9.49r/) PiTi P2T2 Constant Cp. h = CpT k = const (9.49e) 614 Chapter 9 Compressible Flow Fig. 9.8 Flow through a fixed normal shock wave. Note that we have canceled out the areas Ai ~ A2, which is justified even in a vari¬ able duct section because of the thinness of the wave. The first successful analyses of these normal shock relations are credited to W. J. M. Rankine (1870) and A. Hugoniot (1887), hence the modern term Rankine-Hugoniot relations. If we assume that the upstream conditions (pi, Vi, pi, hi, Ti) are known, Eqs. (9.49) are five alge¬ braic relations in the five unknowns (p2, Vi, Pi, h2, 7’2). Because of the velocity- squared term, two solutions are found, and the correct one is determined from the second law of thermodynamics, which requires that J2 > Si- The velocities Vi and V2 can be eliminated from Eqs. (9.49a) to (9.49c) to obtain the Rankine-Hugoniot relation: hi - hi = ^ (P2 - Pi) ( — + - ) (9.50) ^ \Pi PiJ This contains only thermodynamic properties and is independent of the equation of state. Introducing the perfect-gas law h = CpT = kpl{{k — l)p], we can rewrite this as Pi _l + Ppilpi g _ h + I Pi (3 + P2/P1 k - 1 (9.51) We can compare this with the isentropic flow relation for a very weak pressure wave in a perfect gas: fh Pi Pi\ Pi) Uk (9.52) Also, the actual change in entropy across the shock can be computed from the perfect- gas relation: Si - Si = In Pi \Pi (9.53) Assuming a given wave strength P2/P1, we can compute the density ratio and the entropy change and list them as follows for k = 1.4: 9.5 The Normal Shock Wave 615 Pi Pi Pi/Pi 2 - Si Eq. (9.51) Isentropic Cv 0.5 0.6154 0.6095 -0.0134 0.9 0.9275 0.9275 -0.00005 1.0 1.0 1.0 0.0 1.1 1.00704 1.00705 0.00004 1.5 1.3333 1.3359 0.0027 2.0 1.6250 1.6407 0.0134 We see that the entropy change is negative if the pressure decreases across the shock, which violates the second law. Thus a rarefaction shock is impossible in a perfect gas.^ We see also that weak shock waves {p2lp\ — 2.0) are very nearly isentropic. Mach Number Relations For a perfect gas all the property ratios across the normal shock are unique functions of k and the upstream Mach number Mai. For example, if we eliminate p2 and V2 from Eqs. (9.49a) to (9.49c) and introduce h = kpl{{k — l)p], we obtain P2 Pi 1 r 2pivt ^ + 1 1. Pi -{k-\) (9.54) But for a perfect gas p{V\lpi = kV\/{kRTi) = k Ma?, so that Eq. (9.54) is equivalent to a = ^[2»Mai-(k-,)) (S.SS) Erom this equation we see that, for any k, p2 > P\ only if Mai > 10. Thus for flow through a normal shock wave, the upstream Mach number must be supersonic to satisfy the second law of thermodynamics. What about the downstream Mach number? Erom the perfect-gas identity pV^ = kp Ma^, we can rewrite Eq. (9.49h) as P2 1 + A: Mai — = - y (9.56) Pi I + kMal which relates the pressure ratio to both Mach numbers. By equating Eqs. (9.55) and (9.56) we can solve for Ma? (k - 1) Mai -t- 2 2k Mai - (k - 1 ) (9.57) Since Mui must be supersonic, this equation predicts for all A: > 1 that Ma2 must be subsonic. Thus a normal shock wave decelerates a flow almost discontinuously from supersonic to subsonic conditions. ^This is true also for most real gases; see Ref. 9, Sec. 7.3. 616 Chapter 9 Compressible Flow Further manipulation of the basic relations (9.49) for a perfect gas gives addi¬ tional equations relating the change in properties across a normal shock wave in a perfect gas: T2 Ti fh _ (k + l)Ma^ _ pi~ (k- l)Ma? + 2~ V2 [2 + (k- l)Ma?] 2A:Mai - - 1) (k + l)^Ma? (9.58) Tqi — Tqi Pai P02 {k + l)Ma? k/{k-l) k + i Pm Poi .2 4- (A: - l)Ma^ 2kMaj - (k- 1)_ Of additional interest is the fact that the critical, or sonic, throat area A in a duct increases across a normal shock: A\| A Ma2 Maj 2 + {k- \) (9.59) 2 + {k- \) Ma^J All these relations are given in Table B.2 and plotted versus upstream Mach number Mai in Fig- 9.9 for k = 1.4. We see that pressure increases greatly while temperature and density increase moderately. The effective throat area A increases slowly at hrst and then rapidly. The failure of students to account for this change in A is a common source of error in shock calculations. The stagnation temperature remains the same, but the stagnation pressure and den¬ sity decrease in the same ratio; in other words, the flow across the shock is adiabatic but nonisentropic. Other basic principles governing the behavior of shock waves can 9.5 The Normal Shock Wave 617 be summarized as follows: 1. The upstream flow is supersonic, and the downstream flow is subsonic. 2. For perfect gases (and also for real fluids except under bizarre thermodynamic conditions) rarefaction shocks are impossible, and only a compression shock can exist. 3. The entropy increases across a shock with consequent decreases in stagnation pressure and stagnation density and an increase in the effective sonic throat area. 4. Weak shock waves are very nearly isentropic. Normal shock waves form in ducts under transient conditions, such as in shock tubes, and in steady flow for certain ranges of the downstream pressure. Figure 9.10fl shows a normal shock in a supersonic nozzle. Flow is from left to right. The oblique wave pattern to the left is formed by roughness elements on the nozzle walls and indicates Fig. 9.10 Normal shocks form in both internal and external flows. (a) Normal shock in a duct; note the Mach wave pattern to the left (upstream), indicating supersonic flow. ( Courtesy of U.S. Air Force Arnold Engineering Development Center.) (b) Supersonic flow past a blunt body creates a normal shock at the nose; the apparent shock thickness and body-comer curvature are optical distortions. (Courtesy of U.S. Army Ballistic Research Laboratory, Aberdeen Proving Ground.) 618 Chapter 9 Compressible Flow that the upstream flow is supersonic. Note the absence of these Mach waves (see Sec. 9.10) in the subsonic flow downstream. Normal shock waves occur not only in supersonic duct flows but also in a variety of supersonic external flows. An example is the supersonic flow past a blunt body shown in Fig. 9.10b. The bow shock is curved, with a portion in front of the body that is essentially normal to the oncoming flow. This normal portion of the bow shock satisfies the property change conditions just as outlined in this section. The flow inside the shock near the body nose is thus subsonic and at relatively high temperature Ti, and convective heat transfer is especially high in this region. Each nonnormal portion of the bow shock in Fig. 9.1 Oh satisfies the oblique shock relations to be outlined in Sec. 9.9. Note also the oblique recompression shock on the sides of the body. What has happened is that the subsonic nose flow has accelerated around the corners back to supersonic flow at low pressure, which must then pass through the second shock to match the higher downstream pressure conditions. Note the fine-grained turbulent wake structure in the rear of the body in Fig. 9.10h. The turbulent boundary layer along the sides of the body is also clearly visible. The analysis of a complex multidimensional supersonic flow such as in Fig. 9.10 is beyond the scope of this book. For further information see, for example. Ref. 9, Chap. 9, or Ref. 5, Chap. 16. Moving Normal Shocks The preceding analysis of the fixed shock applies equally well to the moving shock if we reverse the transformation used in Fig. 9.1. To make the upstream conditions simulate a still fluid, we move the shock of Fig. 9.8 to the left at speed Vi, that is, we fix our coordinates to a control volume moving with the shock. The downstream flow then appears to move to the left at a slower speed Vi — V2 following the shock. The thermodynamic properties are not changed by this transformation, so that all our Eqs. (9.50) to (9.59) are still valid. EXAMPLE 9.6 Air flows from a reservoir where p = 300 kPa and T = 500 K through a throat to section 1 in Fig. E9.6, where there is a normal shock wave. Compute {a) pi, {b) p2, (c) po2, (d) A , ie) Pm, (f) Af, (g) Pi, and (h) Toi. Solution ■ System sketch: This is shown in Fig. E9.6. Between sections 1 and 2 is a normal shock. ■ Assumptions: Isentropic flow before and after the shock. Lower po and po after the shock. • Approach: After first noting that the throat is sonic, work your way from 1 to 2 to 3. • Property values: For air, R = 287 m^/(s^ ■ K), k = 1.40, and Cp = 1005 mV(s^ • K). The inlet stagnation pressure of 300 kPa is constant up to point 1. • Solution step (a): A shock wave cannot exist unless Mai is supersonic. Therefore the throat is sonic and choked: = A = 1 m^. The area ratio gives Mai from Eq. (9.45) for k = 1.4: A^ 2^_ 1 (l+0.2Maf)^ A't 1 m^ Mai 1.728 solve for Mai = 2.1972 9.5 The Normal Shock Wave 619 Such four-decimal-place accuracy might require iteration or the use of Excel. Linear inter¬ polation in Table B.l would give Mai ~ 2.194, quite good also. The pressure at section 1 then follows from the isentropic relation, Eq. (9.28): Pi Pm 300 kPa (1 -f 0.2Ma?)^'^ [1 -f 0.2(2.194)^]^-’ = 28.2 kPa Ans. (a) Steps (b, c, d): The pressure p2 is found from the normal shock Eq. (9.55) or Table B.2: Pi o 28.2 kPa T Pi = -^[2kMa?-(k-l)] = ———[2(1.4)(2.194)" -(1.4-1)] = 154 kPa Ans. (b) k + I (1.4 -1-1) Similarly, for Mai ~ 2.20, Table B.2 gives pailpm ~ 0.628 (Excel gives 0.6294) and AyA^ ~ 1.592 (Excel gives 1.5888). Thus, to good accuracy, Po2 ~ 0.628poi = 0.628(300 kPa) ~ 188 kPa Ans. (c) AJ = 1.59A = 1.59(1.0 m^) « 1.59 m^ Ans. (d) • Comment: To calculate A’^ directly, without Table B.2, you would need to pause and calculate Ma2 ~ 0.547 from Eq. (9.57), since Eq. (9.59) involves both Mai and Ma2. • Step (e, f): The flow from 2 to 3 is isentropic (but at higher entropy than upstream of the shock); therefore • Steps Next, Excel Po3 = Po2 ~ 188 kPa A = A « 1.59 m^ Ans. (e) Ans. (f) (g, h): The flow is adiabatic throughout, so the stagnation temperature is constant: To3 = Tq2 = Toi — 500 K Ans. (h) the area ratio, using the new sonic area, gives the Mach number at section 3: A3 3m^ . 1 (l-f0.2Mai)^ A? 1.59 m^ = 1.89 = Mai 1.728 solve for Mai ~ 0.33 would yield Mai “ 0.327. Finally, with po2 known, Eq. (9.28) yields py Pi = P02 188 kPa (1 -f 0.2 Ma^)^'^ [1 -f 0.2(0.33)^]^ 174 kPa Au.v. (g) • Comments: Excel would give p^ = 175 kPa, so we see that Table B.2 is satisfactory for this type of problem. A duct flow with a normal shock wave requires straightforward application of algebraic perfect-gas relations, coupled with a little thought as to which formula is appropriate for the given property. EXAMPLE 9.7 An explosion in air, k = 1.4, creates a spherical shock wave propagating radially into still air at standard conditions. At the instant shown in Fig. E9.7, the pressure just inside the shock is 200 Ibf/in^ absolute. Estimate (a) the shock speed C and (b) the air velocity V just inside the shock. 620 Chapter 9 Compressible Flow Solution Part (a) In spite of the spherical geometry, the flow across the shock moves normal to the spherical wave front; hence the normal shock relations (9.50) to (9.59) apply. Fixing our control volume to the moving shock, we find that the proper conditions to use in Fig. 9.8 are C= Vi Pi = 14.7 Ibf/in^ absolute Ti = 520“R V= Vi - V2 P2 = 200 Ibf/in^ absolute The speed of sound outside the shock is ai ~ 49Ti^ the known pressure ratio across the shock: P2 200 Ibf/in^ absolute Pi 14.7 Ibf/in^ absolute From Eq. (9.55) or Table B.2 1 9 13.61 = — (2.8 Maf - 0.4) 01 2.4 Then, by definition of the Mach number, C= Vi = Mai fli = 3.436(1117 ft/s) = 3840 ft/s Ans. (a) Part (b) To hnd V2, we need the temperature or sound speed inside the shock. Since Mai is known, from Eq. (9.58) or Table B.2 for Mai = 3.436 we compute T2/T1 = 3.228. Then T2 = 3.228ri = 3.228(520°R) = 1679“R At such a high temperature we should account for non-perfect-gas effects or at least use the gas tables , but we won’t. Here just estimate from the perfect-gas energy equation (9.23) that F\| = 2cp(ri - T2) + Vj = 2(6010)(520 - 1679) -f (3840)^ = 815,000 or V2 « 903 ft/s Notice that we did this without bothering to compute Ma2, which equals 0.454, or 02 ~ 49T = 2000 ft/s. = 1117 ft/s. We can find Mai from = 13.61 Mai = 3.436 9.6 Operation of Converging and Diverging Nozzles 621 9.6 Operation of Converging and Diverging Nozzles Converging Nozzle Finally, the air velocity behind the shock is F = Vi - 1/2 = 3840 - 903 « 2940 ft/s Ans. (b) Thus a powerful explosion creates a brief but intense blast wind as it passes.^ By combining the isentropic flow and normal shock relations plus the concept of sonic throat choking, we can outline the characteristics of converging and diverging nozzles. First consider the converging nozzle sketched in Fig. 9.1 la. There is an upstream reservoir at stagnation pressure p^. The flow is induced by lowering the downstream outside, or back, pressure pi, helow p^, resulting in the sequence of states a to e shown in Figs. 9.1 Ifo and c. For a moderate drop in to states a and b, the throat pressure is higher than the critical value p that would make the throat sonic. The flow in the nozzle is subsonic throughout, and the jet exit pressure p^ equals the back pressure p^. The mass flow is predicted hy subsonic isentropic theory and is less than the critical value as shown in Fig. 9.1 Ic. For condition c, the back pressure exactly equals the critical pressure p of the throat. The throat becomes sonic, the jet exit flow is sonic, p^ = p/,, and the mass flow equals its maximum value from Eqs. (9.46). The flow upstream of the throat is subsonic everywhere and predicted by isentropic theory based on the local area ratio A(x)/A and Table B.l. Finally, if p^, is lowered further to conditions d or e below p, the nozzle can¬ not respond further because it is choked at its maximum throat mass flow. The throat remains sonic with p^ = p, and the nozzle pressure distribution is the same as in state c, as sketched in Fig. 9.\\b. The exit jet expands supersonically so that the jet pressure can be reduced from p down to p^. The jet structure is complex and multidimensional and is not shown here. Being supersonic, the jet cannot send any signal upstream to influence the choked flow conditions in the nozzle. If the stagnation plenum chamber is large or supplemented by a compressor, and if the discharge chamber is larger or supplemented by a vacuum pump, the converging nozzle flow will be steady or nearly so. Otherwise the nozzle will be blowing down, with po decreasing and p^ increasing, and the flow states will be changing from, say, state e backward to state a. Blowdown calculations are usually made by a quasi-steady analysis based on isentropic steady flow theory for the instantaneous pressures po(t) and pk{t). ^This is the principle of the shock tube wind tunnel, in which a controlled explosion creates a brief flow at very high Mach number, with data taken by fast-response instruments. See, for example. Ref. 5. 622 Chapter 9 Compressible Flow Fig. 9.11 Operation of a converging nozzle: (a) nozzle geometry showing characteristic pressures; (b) pressure distribution caused by various back pressures; (c) mass flow versus back pressure. e d c (c) EXAMPLE 9.8 A converging nozzle has a throat area of 6 cm^ and stagnation air conditions of 120 kPa and 400 K. Compute the exit pressure and mass flow if the back pressure is (a) 90 kPa and (b) 45 kPa. Assume k = 1.4. Solution From Eq. (9.32) for fc = 1.4 the critical (sonic) throat pressure is — = 0.5283 or p = (0.5283)(120 kPa) = 63.4 kPa Po If the back pressure is less than this amount, the nozzle flow is choked. 9.6 Operation of Converging and Diverging Nozzles 623 Part (a) Part (b) Converging-Diverging Nozzle For Pj = 90 kPa > p, the flow is subsonic, not choked. The exit pressure isp^ = p^. The throat Mach number is found from the isentropic relation (9.35) or Table B.l: = 0.4283 Ma, = 0.654 To find the mass flow, we could proceed with a serial attack on Ma,., T^, a^, V^, and p^, hence to compute pgA^Vg. However, since the local pressure is known, this part is ideally suited for the dimensionless mass flow function in Eq. (9.47). With pjpo = 90/120 = 0.75, compute mVafo Apo 2(1.4) 0.4 (0.75)^''"'[1 - (0.75)°"''"] 0.6052 hence m = 0.6052- (0.0006)(120,000) \/287(400) 0.129 kg/s Ans. (a) for Pe= Pb = 90 kPa Ans. (a) For ph = 45 kPa < /?, the flow is choked, similar to condition d in Fig. 9.1 IZ?. The exit pres¬ sure is sonic: Pe ^ p = 63.4 kPa The (choked) mass flow is a maximum from Eq. {9A6h)\ Jll 0.6U7poA, 0.6847( 120,000) (0.0006) {RToV [287(400)] Ans. (b) = 0.145 kg/s Ans. (b) Any back pressure less than 63.4 kPa would cause this same choked mass flow. Note that the 50 percent increase in exit Mach number, from 0.654 to 1.0, has increased the mass flow only 12 percent, from 0.128 to 0.145 kg/s. Now consider the converging-diverging nozzle sketched in Fig. 9.12a. If the back pressure pi, is low enough, there will be supersonic flow in the diverging portion and a variety of shock wave conditions may occur, which are sketched in Fig. 9.12h. Let the back pressure be gradually decreased. For curves A and B in Fig. 9.12h the back pressure is not low enough to induce sonic flow in the throat, and the flow in the nozzle is subsonic throughout. The pres¬ sure distribution is computed from subsonic isentropic area-change relations, such as in Table B.l. The exit pressure = py, and the jet is subsonic. For curve C the area ratio AJA, exactly equals the critical ratio AJA for a subsonic Ma^ in Table B.l. The throat becomes sonic, and the mass flux reaches a maximum in Fig. 9.12c. The remainder of the nozzle flow is subsonic, including the exit jet, and p^ = pb. Now jump for a moment to curve H. Here py is such that pylpo exactly corresponds to the critical area ratio AJA for a supersonic Ma^ in Table B.l. The diverging flow is entirely supersonic, including the jet flow, and p^ = py. This is called the design pressure ratio of the nozzle and is the back pressure suitable for operating a super¬ sonic wind tunnel or an efficient rocket exhaust. 624 Chapter 9 Compressible Flow Fig. 9.12 Operation of a converging-diverging nozzle: (a) nozzle geometry with possible flow configurations; (b) pressure distribution caused by various back pressures; (c) mass flow versus back pressure. (c) Now back up and suppose that pf, lies between curves C and H, which is impos¬ sible according to purely isentropic flow calculations. Then back pressures D to F occur in Fig. 9.12b. The throat remains choked at the sonic value, and we can match Pe ~ Pb by placing a normal shock at just the right place in the diverging section to cause a subsonic dijfuser flow back to the back-pressure condition. The mass flow remains at maximum in Fig. 9.12c. At back pressure F the required normal shock stands in the duct exit. At back pressure G no single normal shock can do the job, and so the flow compresses outside the exit in a complex series of oblique shocks until it matches pf,. Finally, at back pressure I, pf, is lower than the design pressure H, but the nozzle is choked and cannot respond. The exit flow expands in a complex series of supersonic wave motions until it matches the low back pressure. See, Ref. 7, Sec. 5.4, for further details of these off-design jet flow configurations. 9.6 Operation of Converging and Diverging Nozzles 625 Note that for pi, less than back pressure C, there is supersonic flow in the nozzle, and the throat can receive no signal from the exit behavior. The flow remains choked, and the throat has no idea what the exit conditions are. Note also that the normal shock-patching idea is idealized. Downstream of the shock, the nozzle flow has an adverse pressure gradient, usually leading to wall bound¬ ary layer separation. Blockage by the greatly thickened separated layer interacts strongly with the core flow (recall Fig. 6.27) and usually induces a series of weak two-dimensional compression shocks rather than a single one-dimensional normal shock (see. Ref. 9, pp. 292 and 293, for further details). EXAMPLE 9.9 A converging-diverging nozzle (Fig. 9.12fl) has a throat area of 0.002 and an exit area of 0.008 m^. Air stagnation conditions are po = 1000 kPa and Tq = 500 K. Compute the exit pressure and mass flow for (a) design condition and the exit pressure and mass flow if (b) Pt ~ 300 kPa and (c) pi, ~ 900 kPa. Assume k = 1.4. Solution Part (a) The design condition corresponds to supersonic isentropic flow at the given area ratio AJA, = 0.008/0.002 = 4.0. We can find the design Mach number by iteration of the area ratio formula (9.45): Ma^desig,^ 2.95 The design pressure ratio follows from Eq. (9.34): — = [1 + 0.2(2.95)^]^'^ = 34.1 Pe 1000 kPa Pe.design I 29.3 kPa Since the throat is clearly sonic at design conditions, Eq. (9.46fo) applies: 0.6847poA, 0.6847(10® Pa) (0.002 m^) = (Pro)>« " [287(500)]‘« = 3.61 kg/s Part (b) Eor Pf, = 300 kPa we are definitely far below the subsonic isentropic condition C in Fig. 9.\2b, but we may even be below condition F with a normal shock in the exit — that is, in condition G, where oblique shocks occur outside the exit plane. If it is condition G, then p^ = Pe,design = 29.3 kPa because no shock has yet occurred. To find out, compute condition F by assuming an exit normal shock with Mai = 2.95 — that is, the design Mach number just upstream of the shock. From Eq. (9.55) Ans. (a) Ans. (a) P2 Pi ^[2.8(2.95)^ 0.4] = 9.99 or P2 = 9.99pi = 9.99p,4esigi. = 293 kPa 626 Chapter 9 Compressible Flow Since this is less than the given pj, = 300 kPa, there is a normal shock just upstream of the exit plane (condition E). The exit flow is subsonic and equals the back pressure: Pe — Pb — 300 kPa Also m = niniax = 3.61 kg/s The throat is still sonic and choked at its maximum mass flow. Part (c) Finally, for pj = 900 kPa, which is up near condition C, we compute Ma,, and p^ for condition C as a comparison. Again AJA, = 4.0 for this condition, with a subsonic Ma,, estimated from Eq. (9.48): Ans. (b) Ans. (b) Ma,(C) « 0.147 (exact = 0.14655) Then the isentropic exit pressure ratio for this condition is or — = [1 + 0.2(0.147)^]^'^ Pe 1.0152 P. 1000 1.0152 = 985 kPa The given back pressure of 900 kPa is less than this value, corresponding roughly to condi¬ tion D in Fig. 9.12h. Thus for this case there is a normal shock just downstream of the throat, and the throat is choked: Pe = Pb = 900 kPa rh = = 3.61 kg/s Ans. (c) For this large exit area ratio, the exit pressure would have to be larger than 985 kPa to cause a subsonic flow in the throat and a mass flow less than maximum. 9.7 Compressible Duct Flow with Friction^' 1. Steady one-dimensional adiabatic flow. 2. Perfect gas with constant speciflc heats. 3. Constant- area straight duct. 4. Negligible shaft work and potential energy changes. 5. Wall shear stress correlated by a Darcy friction factor. In effect, we are studying a Moody-type pipe friction problem but with large changes in kinetic energy, enthalpy, and pressure in the flow. This type of duct flow — constant area, constant stagnation enthalpy, constant mass flow, but variable momentum (due to friction) — is often termed F anno flow, after Gino '^This section may be omitted without loss of continuity. Section 9.4 showed the effect of area change on a compressible flow while neglecting friction and heat transfer. We could now add friction and heat transfer to the area change and consider coupled effects, which is done in advanced texts [for example. Ref. 5, Chap. 8]. Instead, as an elementary introduction, this section treats only the effect of friction, neglecting area change and heat transfer. The basic assumptions are 9.7 Compressible Duct Flow with Friction 627 Fig. 9.13 Elemental control volume for flow in a constant-area duct with friction. Control volume T X X + dx Fanno, an Italian engineer born in 1 882, who first studied this flow. For a given mass flow and stagnation enthalpy, a plot of enthalpy versus entropy for all possible flow states, subsonic or supersonic, is called a Fanno line. See Probs. P9.94 and P9.111 for examples of a Fanno line. Consider the elemental duct control volume of area A and length dx in Fig. 9.13. The area is constant, but other flow properties ip, p, T, h, V) may vary with x. Appli¬ cation of the three conservation laws to this control volume gives three differential equations: Continuity: pv m — = G = const A or X momentum: dp P + dV V = 0 pA — {p F dp)A — T^.TzDdx = m(V + dV — V) or dp + Ar^dx D + pVdV = 0 (9.60fl) (9.601?) Energy: h + = ha = CpTo = CpT + or CpdT + V dV = 0 (9.60c) Since these three equations have five unknowns — p, p, T, V, and — we need two additional relations. One is the perfect-gas law: dp dp dT p = pRT or — = — -f — (9.61) ‘ ^ P p T To eliminate as an unknown, it is assumed that wall shear is correlated by a local Darcy friction factor/ T^. = kfpV^ = lfkp (9.62) where the last form follows from the perfect-gas speed-of-sound expression = kpip. In practice, / can be related to the local Reynolds number and wall roughness from, say, the Moody chart, Fig. 6.13. 628 Chapter 9 Compressible Flow Adiabatic Flow Equations (9.60) and (9.61) are first-order differential equations and can be inte¬ grated, by using friction factor data, from any inlet section 1, where p^, T^, Vi, and so on are known, to determine pix), T{x), and other properties along the duct. It is practically impossible to eliminate all but one variable to give, say, a single differential equation for p{x), but all equations can be written in terms of the Mach number Ma(x) and the friction factor, by using this definition of Mach number: = Ma^ kRT or 2dV 2dMa dT - — - -1- — y Ma r (9.63) By eliminating variables between Eqs. (9.60) to (9.63), we obtain the working relations dp , _ , 1 -F (k - l)Ma^ Jx — = — kMa T f — P 2(1 - Ma^) D (9.64a) dp k Ma^ dx dV p ~ 2(1 - Ma^)^ D ~ V (9.64b) dpo dpo 1 2 dir - = - = — k Ma f — Po po 2 D (9.64c) dT k(k-l)Ma\dx T 2(1 - Ma^) ^ D (9.64^0 dMa^ 1 + \|(k - l)Ma2 Jx T = kMa , f Ma^ 1 - Ma^ D (9.64e) All these except dpc/po have the factor 1 — Ma^ in the denominator, so that, like the area change formulas in Fig. 9.5, subsonic and supersonic flow have opposite effects: Property Subsonic Supersonic P Decreases Increases P Decreases Increases V Increases Decreases Pa. Pa Decreases Decreases T Decreases Increases Ma Increases Decreases Entropy Increases Increases We have added to this list that entropy must increase along the duct for either subsonic or supersonic flow as a consequence of the second law for adiabatic flow. For the same reason, stagnation pressure and density must both decrease. The key parameter in this discussion is the Mach number. Whether the inlet flow is subsonic or supersonic, the duct Mach number always tends downstream toward Ma = 1 because this is the path along which the entropy increases. If the pressure and density are computed from Eqs. (9.64fl) and (9.64b) and the entropy from Eq. (9.53), the result can be plotted in Fig. 9.14 versus Mach number for k = 1.4. 9.7 Compressible Duct Flow with Friction 629 Fig. 9.14 Adiabatic frictional flow in a constant-area duct always approaches Ma = 1 to satisfy the second law of thermodynamics. The computed curve is independent of the value of the friction factor. The maximum entropy occurs at Ma = 1, so the second law requires that the duct flow properties continually approach the sonic point. Since po ^nd po continually decrease along the duct due to the frictional (nonisentropic) losses, they are not useful as reference properties. Instead, the sonic properties p, p, T, pj, and are the appropriate constant reference quantities in adiabatic duct flow. The theory then com¬ putes the ratios p/p, T/T, and so forth as a function of local Mach number and the integrated friction effect. To derive working formulas, we first attack Eq. (9.64e), which relates the Mach number to friction. Separate the variables and integrate: d.x D _ 1 - Ma^ _ +\|(k- l)Ma^] dMa^ (9.65) The upper limit is the sonic point, whether or not it is actually reached in the duct flow. The lower limit is arbitrarily placed at the position x = 0, where the Mach number is Ma. The result of the integration is ft 1 - Ma^ k+l (k+UMa- — = - ^ + - — In ^ - — r (9.66) D kMa^ 2k 2+(A:-l)Ma- where / is the average friction factor between 0 and L. In practice, an average /is always assumed, and no attempt is made to account for the slight changes in Reynolds number along the duct. For noncircular ducts, D is replaced by the hydraulic diameter Dh = (4 X area)/perimeter as in Eq. 6.56. 630 Chapter 9 Compressible Flow Equation (9.66) is tabulated versus Mach number in Table B.3. The length L is the length of duct required to develop a duct flow from Mach number Ma to the sonic point. Many problems involve short ducts that never become sonic, for which the solution uses the differences in the tabulated “maximum,” or sonic, length. For exam¬ ple, the length AL required to develop from Ma; to Ma2 is given by This avoids the need for separate tabulations for short ducts. It is recommended that the friction factor / he estimated from the Moody chart (Fig. 6.13) for the average Reynolds number and wall roughness ratio of the duct. Available data on duct friction for compressible flow show good agreement with the Moody chart for subsonic flow, but the measured data in supersonic duct flow are up to 50 percent less than the equivalent Moody friction factor. EXAMPLE 9.10 Air flows subsonically in an adiabatic 2-cm-diameter duct. Tbe average friction factor is 0.024. Wbat length of duct is necessary to accelerate tbe flow from Mai “ 0-1 to Ma2 = 0.5? What additional length will accelerate it to Ma3 = 1.0? Assume k = 1.4. Solution Equation (9.67) applies with values of fLID computed from Eq. (9.66) or read from Table B.3: - AL _ 0.024 AL ^ D ~ 0.02 m D D Thus Ma = 0.1 = 66.9216 - 1.0691 = 65.8525 65.8525(0.02 m) AL = — — — = 55 m Ans. (a) 0.024 The additional length AL' to go from Ma = 0.5 to Ma = 1.0 is taken directly from Table B.2: AL' ^ D AL' = L^a=o.5 ./T; D = 1.0691 Ma = 0.5 1.0691(0.02 m) 0.024 = 0.9 m Ans. (b) This is typical of these calculations: It takes 55 m to accelerate up to Ma = 0.5 and then only 0.9 m more to get all the way up to the sonic point. Formulas for other flow properties along the duct can he derived from Eqs. (9.64). Equation (9.64e) can he used to eliminate / dxlD from each of the other relations, giving, for example, dpip as a function only of Ma and d Ma^/Ma^. For convenience in tabulating the results, each expression is then integrated all the way from (p, Ma) 9.7 Compressible Duct Flow with Friction 631 to the sonic point {p, 1.0). The integrated results are 1 p'f Ma P _V _ 1 p V Ma T a" k + 1 11/2 2 + (k- l)Ma^J 2 + {k- l)Ma^ k + 1 4 1 2-\|l/2 PO _ pQ p%~~pi a 1 Ma 2 2+ {k- l)Ma^ 2 + {k- l)Ma^ /c + 1 (l/2)(ifc+ \)l(k- 1) (9.68fl) (9.68h) (9.68c) (9.68t/) All these ratios are also tabulated in Table B.3. For finding changes between points Ma; and Ma2 that are not sonic, products of these ratios are used. For example, P2lP1 1 El Pi Pi since p is a constant reference value for the flow. (9.69) EXAMPLE 9.11 For the duct flow of Example 9.10 assume that, at Mai = 0.1, we have pi = 600 kPa and Ti = 450 K. At section 2 farther downstream, Ma2 = 0.5. Compute (a) p2, (b) T2, (c) V2, and (d) po2- Solution As preliminary information we can compute Vi and poi from the given data: Vi = Maifli = 0.1[(1.4)(287)(450)]‘® = 0.1(425 m/s) = 42.5 m/s Poi =Pi(l + 0.2 Mai)^-^ = (600kPa)[l + 0.2(0. 1)^]^'^ = 604 kPa Now enter Table B.3 or Eqs. (9.68) to find the following property ratios: Section Ma pip jyy VIV Po/Pt 1 0.1 10.9435 1.1976 0.1094 5.8218 2 0.5 2.1381 1.1429 0.5345 1.3399 Use these ratios to compute all properties downstream: pip 2.1381 10:9435^"^'^^ TIT 1.1429 T2 = Ti— - = (450 K) _ _ = 429 K "TIT 1.1976 V2IV 0.5345 m F2 = Vi— - = (42.5 m/s) - = 208 — VIV 0.1094 s P02/P0 1.3399 ■ ''"Poilpl = '^^"^5:8218 = Ans. (a) Ans. (b) Ans. (c) Ans. (d) 632 Chapter 9 Compressible Flow Choking Due to Friction Fig. 9.15 Behavior of duct flow with a nominal supersonic inlet condition Ma = 3.0: (a) LID £ 26, flow is supersonic throughout duct; {b) UD = 40 > LID, normal shock at Ma = 2.0 with subsonic flow then accelerating to sonic exit point; (c) UD = 53, shock must now occur at Ma = 2.5; (d) LID > 63, flow must be entirely subsonic and choked at exit. Note the 77 percent reduction in stagnation pressure due to friction. The formulas are seduc¬ tive, so check your work by other means. For example, check po2 = + 0.2 Ma\|)^'^. The theory here predicts that for adiabatic frictional flow in a constant-area duct, no matter what the inlet Mach number Maj is, the flow downstream tends toward the sonic point. There is a certain duct length L(Mai) for which the exit Mach number will be exactly unity. The duct is then choked. But what if the actual length L is greater than the predicted “maximum” length L1 Then the flow conditions must change, and there are two classifications. Subsonic Inlet. If L > L(Mai), the flow slows down until an inlet Mach number Ma2 is reached such that L = L(Ma2). The exit flow is sonic, and the mass flow has been reduced by frictional choking. Further increases in duct length will continue to decrease the inlet Ma and mass flow. Supersonic Inlet. From Table B.3 we see that friction has a very large effect on supersonic duct flow. Even an infinite inlet Mach number will be reduced to sonic conditions in only 41 diameters for / = 0.02. Some typical numerical values are shown in Fig. 9.15, assuming an inlet Ma = 3.0 and / = 0.02. For this condition L = 26 diameters. If L is increased beyond 2bD, the flow will not choke but a normal shock will form at just the right place for the subsequent subsonic frictional flow to 0 10 20 30 40 50 60 £ D 9.7 Compressible Duct Flow with Friction 633 Part (a) Part (b) become sonic exactly at the exit. Figure 9.15 shows two examples, for LID = 40 and 53. As the length increases, the required normal shock moves upstream until, for Fig. 9.15, the shock is at the inlet for LID = 63. Further increase in L causes the shock to move upstream of the inlet into the supersonic nozzle feeding the duct. Yet the mass flow is still the same as for the very short duct, because presumably the feed nozzle still has a sonic throat. Eventually, a very long duct will cause the feed-nozzle throat to become choked, thus reducing the duct mass flow. Thus supersonic friction changes the flow pattern if L > L but does not choke the flow until L is much larger than L. EXAMPLE 9.12 Air enters a 3-cm-diameter duct at po = 200 kPa, Tq = 500 K, and Vi = 100 m/s. The friction factor is 0.02. Compute {a) the maximum duct length for these conditions, (&) the mass flow if the duct length is 15 m, and (c) the reduced mass flow if L = 30 m. Solution First compute fvf i(100m(s)^ Ti = To - — = 500 - - = 500 - 5 = 495 K Cp 1005 m^/s^ ■ K fli = {kRTi)^''^ « 20(495)“ = 445 m/s Thus Fi 100 Mai = — = - = 0.225 fli 445 For this Mai, from Eq. (9.66) or interpolation in Table B.3, fL D = 11.0 The maximum duct length possible for these inlet conditions is (fLID)D 11.0(0.03 m) L = f 0.02 = 16.5 m Ans. (a) The given L = 15 m is less than L, and so the duct is not choked and the mass flow follows from the inlet conditions: Poi Poi 200,000 Pa Pi RTo 287(500 K) Poi 1.394 = 1.394 kg/m^ [1 -f 0.2(0.225)^]^ 1.0255 = 1.359 kg/m^ whence m = PiAVi = ( 1.359 kg/m^) = 0.0961 kg/s -(0.03 m)^ (100 m/s) Ans. (b) 634 Chapter 9 Compressible Flow Part (c) Since L = 30 m is greater than L, the duct must choke back until L = L, corresponding to a lower inlet Map L = L = 30 m fL 0.02(30 m) - = - - ^ = 20.0 D 0.03 m Although it is difficult to interpolate in the coarse Table B.3 for JL/D = 20, it is a simple matter for Excel to iterate to find this subsonic Mach number. Program Eq. (9.66) into a cell, guess a subsonic Mach number, and calculate fL/D. Adjust Ma until you approximate fUD = 20. The writer’s effort took five guesses, as in the following Excel table: A B Mai (/L/Z> )i-Eq. (9.66) 1 0.2 14.533 2 0.15 27.932 3 0.17 21.115 4 0.175 19.772 5 0.1741 20.005 An accurate solution for Ma is thus Machoked ~ 0.174 (23 percent less) To T = ^ l,new 1 + 0.2(0.174)^ ' = 497 K ai,new « 20(497 K)‘“ = 446 m/s Fi,„ew = Mai ai = 0.174(446) = 77.6 m/s Poi Pi, new [1 + 0.2(0.174)^]^ TT = 1.373 kg/m" ninew = PiAVi = 1.373 (0.03)" (77.6) = 0.0753 kg/s (22 percent less) Ans. (c) Minor Losses in Compressible Flow For incompressible pipe flow, as in Eq. (6.78), the loss coefficient K is the ratio of pressure head loss (Ap/pg) to the velocity head {V^/2g) in the pipe. This is inappropriate in compress¬ ible pipe flow, where p and V are not constant. Benedict suggests that the static pressure loss (pi — Pj) be related to downstream conditions and a static loss coefficient K^: K, = 2(pi - Pi) Pi^i (9.70) Benedict gives examples of compressible losses in sudden contractions and expansions. If data are unavailable, a first approximation would be to use ~ K from Sec. 6.9. Isothermal Flow with Friction: Long Pipelines The adiabatic frictional flow assumption is appropriate to high-speed flow in short ducts. For flow in long ducts, such as natural gas pipelines, the gas state more closely 9.7 Compressible Duct Flow with Friction 635 approximates an isothermal flow. The analysis is the same except that the isoenergetic energy equation (9.60c) is replaced by the simple relation T = const dT = 0 Again it is possible to write all property changes in terms of the Mach number. Inte¬ gration of the Mach number-friction relation yields y^max D I - kMa^ kMa^ -f In (k Ma^) (9.71) which is the isothermal analog of Eq. (9.66) for adiabatic flow. This friction relation has the interesting result that L^ax becomes zero not at the sonic point but at Macut = = 0.845 if A: = 1.4. The inlet flow, whether subsonic or supersonic, tends downstream toward this limiting Mach number 1/A:^^. If the tube length L is greater than L^ax from Eq. (9.71), a subsonic flow will choke back to a smaller Maj and mass flow and a supersonic flow will experience a normal shock adjustment similar to Eig. 9.15. The exit isothermal choked flow is not sonic, and so the use of the asterisk is inappropriate. Let p' , p' , and V’ represent properties at the choking point L = Lmax- Then the isothermal analysis leads to the following Mach number relations for the flow properties: P_ ^ 1 p' Ma y' p — = Mak P 1/2 (9.72) The complete analysis and some examples are given in advanced texts [for example. Ref. 5, Sec. 6.4]. Mass Flow for a Given Pressure Drop An interesting by-product of the isothermal analysis is an explicit relation between the pressure drop and duct mass flow. This common problem requires numerical iteration for adiabatic flow, as outlined here. In isothermal flow, we may substitute dV/V = —dp/p and = G^I[pl{RT)f' in Eq. (9.63) to obtain 2p dp dx +/ — G^RT D 2 dp P = 0 Since G^RT is constant for isothermal flow, this may be integrated in closed form between (x, p) = (0, pi) and (L, p2y. G^ Pi pl RT[fL/D + 2ln(pi/p2)] (9.73) Thus mass flow follows directly from the known end pressures, without any use of Mach numbers or tables. The writer does not know of any direct analogy to Eq. (9.73) for adiabatic flow. However, a useful adiabatic relation, involving velocities instead of pressures, is ^2 ^ «o[l - iVl/Vlf] ' A/L/D -1 (A 4- 1) In (Vz/yi) (9.74) 636 Chapter 9 Compressible Flow where Qq = {kRTQf'^ is the stagnation speed of sound, constant for adiabatic flow. This may be combined with continuity for constant duct area V1IV2 = Pilpx, plus the following combination of adiabatic energy and the perfect-gas relation: Yi ^ P2T\ ^ Pi V2 Pi T2 Pi If we are given the end pressures, neither Vi nor V2 will likely be known in advance. We suggest only the following simple procedure: Begin with gq ~ fli and the bracketed term in Eq. (9.75) approximately equal to 1.0. Solve Eq. (9.75) for a first estimate of V1/V2, and use this value in Eq. (9.74) to get a better estimate of Vi. Use Vi to improve your estimate of aq, and repeat the procedure. The process should converge in a few iterations. Equations (9.73) and (9.74) have one flaw: With the Mach number eliminated, the frictional choking phenomenon is not directly evident. Therefore, assuming a subsonic inlet flow, one should check the exit Mach number Ma2 to ensure that it is not greater than 1/A:^^ for isothermal flow or greater than 1.0 for adiabatic flow. We illustrate both adiabatic and isothermal flow with the following example. 24 - jk- i)vi 2d - (k- 1)V2 (9.75) EXAMPLE 9.13 Air enters a pipe of 1-cm diameter and 1.2-m length at pi = 220 kPa and U = 300 K. If / = 0.025 and the exit pressure is p2 = 140 kPa, estimate the mass flow for (a) isothermal flow and (b) adiabatic flow. Solution Part (a) For isothermal flow Eq. (9.73) applies without iteration: fL Pi (0.025) (1.2 m) 220 — -f 2 In — = + 2 In - = 3.904 D P2 0.01 m 140 G^ = (220,000 Pa)^ - ( 140,000 Pa)^ = 85,700 or G = 293 kg/(s ■ mU 287 m^/(s^ ■ K)(3.904) Since A = (7r/4)(0.01 m)^ = 7.85 E-5 m^, the isothermal mass flow estimate is m = GA = (293)(7.85 E-5) « 0.0230 kg/s Ans. (a) Check that the exit Mach number is not choked: or P2 _ 140,000 RT ~ (287)(300) 1.626 kg/m^ 293 1.626 = 180 m/s Ma2 V2 _ 180 _ 180 Vkl& [1.4(287)(300)]‘'^ 347 This is well below choking, and the isothermal solution is accurate. Part (b) For adiabatic flow, we can iterate by hand, in the time-honored fashion, using Eqs. (9.74) and (9.75), plus the definition of stagnation speed of sound, Oo = (kRT„Y^- A few years ago the writer would have done just that, laboriously. Flowever, these equations can be iterated and manipulated by Excel. First list the given data and requirements: 9.8 Frictionless Duct Flow with Fleat Transfer 637 9.8 Frictionless Duct Flow with Heat Transfer® /AL (0.025) (1.2) k= 1.4; Pi = 220,000 Pa; p2 = 140,000 Pa; Fi = 300 K; - = ^ 3.0,p = „ Now use Excel to apply Eqs. (9.66), for fLID, and (9.68a), for p/p, to points 1 and 2 in the pipe. We are iterating to find the inlet Mach number for which (a) f\L/D = 3.0 and {b) p = p. Even with Excel doing all the work, the iteration is exasperating. We guess Mai and acfjust Ma2 until A(fL/D) = 3.0, after which we check for equal values of p. The tabulated results are below. It took four guesses of Mai to arrive at p ~ 67,900 Pa. A B c D E F G H I Mai Pl/p Ma2 (JL/Dh Pi/P A(fUD) P P 1 0.2 14.533 5.455 0.221 11.533 4.944 3.000 40327 28317 2 0.3 5.299 3.619 0.401 2.299 2.692 3.000 60789 51999 3 0.34 3.752 3.185 0.546 0.752 1.950 3.000 69068 71802 4 0.3343 3.936 3.241 0.518 0.935 2.062 3.001 67884 67886 The mass flow follows from the inlet Mach number and density and is quite close to part (a): Ft = MaiVkRTi = (0.3343)\/l.4(287)(300) = 116 m/s Pi = PiKRTi) = (220,000)/[287(300)] = 2.56 kg/m^ m = PiAV^ = (2.56)(7r/4)(0.01)^(116) = 0.0233 kg/s Ans. (b) Heat addition or removal has an interesting effect on a compressible flow. Advanced texts [for example, Ref. 5, Chap. 8] consider the combined effect of heat transfer coupled with friction and area change in a duct. Here we confine the analysis to heat transfer with no friction in a constant-area duct. This type of duct flow — constant area, constant momentum, constant mass flow, but variable stagnation enthalpy (due to heat transfer) — is often termed Rayleigh flow after John William Strutt, Lord Rayleigh (1842-1919), a famous physicist and engi¬ neer. For a given mass flow and momentum, a plot of enthalpy versus entropy for all possible flow states, subsonic or supersonic, forms a Rayleigh line. See Probs. P9.110 and P9. Ill for examples of a Rayleigh line. Consider the elemental duct control volume in Fig. 9.16. Between sections 1 and 2 an amount of heat (50 is added (or removed) to each incremental mass dm passing through. With no friction or area change, the control volume conservation relations are quite simple: Continuity: PiFi = P2V2 = G = const {9.16a) X momentum: P,-P2 = G{V2 - Fi) {9.16b) Energy: Q = m(h2 + 2VI - hi - \|Fi) or q — . — ^ — «02 "01 (9.76c) m dm ^This section may be omitted without loss of continuity. 638 Chapter 9 Compressible Flow Control volume A2=A Vi.Pi, Ti, Fq) Fig. 9.16 Elemental control volume for frictionless flow in a constant- area duct with heat transfer. The length of the element is indeterminate in this simplified theory. © © The heat transfer results in a change in stagnation enthalpy of the flow. We shall not specify exactly how the heat is transferred — combustion, nuclear reaction, evaporation, condensa¬ tion, or wall heat exchange — ^but simply that it happened in amount q between 1 and 2. We remark, however, that wall heat exchange is not a good candidate for the theory because wall convection is inevitably coupled with wall friction, which we neglected. To complete the analysis, we use the perfect-gas and Mach number relations: Pi Pi hoi hoi — Cp{To2 Toi) PiTi piTi (9.77) For a given heat transfer q = dQ/dm or, equivalently, a given change /jo2 “ /Jou Eqs. (9.76) and (9.77) can be solved algebraically for the property ratios P2/P1, Ma2/Mai, and so on between inlet and outlet. Note that because the heat transfer allows the entropy to either increase or decrease, the second law imposes no restrictions on these solutions. Before writing down these property ratio functions, we illustrate the effect of heat transfer in Fig. 9.17, which shows Tq and T versus Mach number in the duct. Heating - 1 /I Fq (max) at Ma =1.0 Fig. 9.17 Effect of heat transfer on Mach number. 0 0.5 1 1.5 Mach number 2 2.5 9.8 Frictionless Duct Flow with Fleat Transfer 639 Mach Number Relations increases Tq, and cooling decreases it. The maximum possible Tq occurs at Ma = 1.0, and we see that heating, whether the inlet is subsonic or supersonic, drives the duct Mach number toward unity. This is analogous to the effect of friction in the previous section. The temperature of a perfect gas increases from Ma = 0 up to Ma = 1/^^^ and then decreases. Thus there is a peculiar — or at least unexpected — region where heating (increasing Tq) actually decreases the gas temperature, the difference being reflected in a large increase of the gas kinetic energy. For A: = 1.4 this peculiar area lies between Ma = 0.845 and Ma = 1.0 (interesting but not very useful information). The complete list of the effects of simple Tq change on duct flow properties is as follows; Heating Cooling Subsonic Supersonic Subsonic Supersonic To Increases Increases Decreases Decreases Ma Increases Decreases Decreases Increases P Decreases Increases Increases Decreases P Decreases Increases Increases Decreases V Increases Decreases Decreases Increases Pa Decreases Decreases Increases Increases s Increases Increases Decreases Decreases T Increases t Decreases ^Increases up to Ma = and decreases thereafter. tDecreases up to Ma = 1/t:^ and increases thereafter. Probably the most significant item on this list is the stagnation pressure Pq, which always decreases during heating whether the flow is subsonic or supersonic. Thus heating does increase the Mach number of a flow but entails a loss in effective pressure recovery. Equations (9.76) and (9.77) can be rearranged in terms of the Mach number and the results tabulated. For convenience, we specify that the outlet section is sonic, Ma = 1, with reference properties T, 7, p, p, V, and p. The inlet is assumed to be at arbitrary Mach number Ma. Equations (9.76) and (9.77) then take the fol¬ lowing form: To (yt + l)Ma^ [2 -f (yt - l)Ma^] T~ (I + kMaY T _ {k + ifMa^ T ~ {I + kMaY El P% p _ k + \ p 1 + k MsE V _ p _ jk + l)Ma^ V ~ p ~ 1 + kMsE k + 1 1 + yt Ma^ 2 + {k- l)Ma^ k + 1 (9.78fl) (9.78^7) (9.78c) (9.78f/) (9.78e) 640 Chapter 9 Compressible Flow These formulas are all tabulated versus Mach number in Table B.4. The tables are very convenient if inlet properties Maj, V-i, and the like are given but are somewhat cumber¬ some if the given information centers on Tm and Tqi- Let us illustrate with an example. EXAMPLE 9.14 A fuel-air mixture, approximated as air with k = 1.4, enters a duct combustion chamber at V-i = 75 m/s. Pi = 150 kPa, and T = 300 K. The heat addition by combustion is 900 kJ/kg of mixture. Compute (a) the exit properties V2, P2, and T2 and {b) the total heat addition that would have caused a sonic exit flow. Solution Part (a) First compute Lqi = + V\l{2Cj,) = 300 + (75)^/[2(1005)] = 303 K. Then compute the change in stagnation temperature of the gas: 9 = Cp(T(y2 — Tqi) q 900,000 J/kg or Tq2 = Toi -k — = 303 K -k - ^ = 1 199 K Cp 1005 J/(kg • K) We have enough information to compute the initial Mach number: Oi = \/kRTi = [1.4(287)(300)]''^ = 347 m/s ^ ^ ^ ^ For this Mach number, use Eq. (9.78a) or Table B.4 to find the sonic value 7^: At Mai = 0.216: ^ = 0.1992 or T = « 1521 K T% 0.1992 Then the stagnation temperature ratio at section 2 is Tq2IT% = 1199/1521 = 0.788, which corresponds in Table B.4 to a Mach number Ma2 = 0.573. Now use Eqs. (9.78) at Mai und Ma2 to tabulate the desired property ratios. Section Ma V/V pip jyy 1 0.216 0.1051 2.2528 0.2368 2 0.573 0.5398 1.6442 0.8876 The exit properties are computed by using these ratios to find state 2 from state 1: F,/V 0.5398 V2 = Vi - = (75 m/s) - = 385 m/s Ans. (a) Vi/V 0.1051 pjp 1.6442 P2 = Pi^ = (150 kPa)-— = 109 kPa Ans. (a) Pi/p 2.2528 T2IT 0.8876 T2 = - = (300 K) - = 1124 K Ans. (a) 2 1 p /pK 0.2368 Part (b) The maximum allowable heat addition would drive the exit Mach number to unity: To2 = T%= 1521 K ?max = Cp{Tl - Tqi) = 1005 J/(kg-K) « 1.22 E6 J/kg Ans. (b) 9.8 Frictionless Duct Flow with Fleat Transfer 641 Choking Effects Due to Simple Heating Equation (9.78a) and Table B.4 indicate that the maximum possible stagnation tem¬ perature in simple heating corresponds to 7^, or the sonic exit Mach number. Thus, for given inlet conditions, only a certain maximum amount of heat can be added to the flow — for example, 1.22 MJ/kg in Example 9.14. Eor a subsonic inlet there is no theoretical limit on heat addition: The flow chokes more and more as we add more heat, with the inlet velocity approaching zero. For supersonic flow, even if Maj is infinite, there is a finite ratio Tq-^IT^ = 0.4898 for k = 1.4. Thus if heat is added without limit to a supersonic flow, a normal shock wave adjustment is required to accommodate the required property changes. In subsonic flow there is no theoretical limit to the amount of cooling allowed: The exit flow just becomes slower and slower, and the temperature approaches zero. In supersonic flow only a finite amount of cooling can be allowed before the exit flow approaches infinite Mach number, with 7o2/7’g = 0.4898 and the exit temperature equal to zero. There are very few practical applications for supersonic cooling. EXAMPLE 9.15 What happens to the inlet flow in Example 9.14 if the heat addition is increased to 1400 kj/kg and the inlet pressure and stagnation temperature are fixed? What will be the subsequent decrease in mass flow? Solution For q = 1400 kJ/kg, the exit will be choked at the stagnation temperature: 7^ = Toi + — = 303 + Cp 1.4 E6 J/kg 1005 J/ (kg- K) « 1696 K This is higher than the value = 1521 K in Example 9.14, so we know that condition 1 will have to choke down to a lower Mach number. The proper value is found from the ratio Toi/Tq = 303/1696 = 0.1787. Erom Table B.4 or Eq. (9.78a) for this condition, we read the new, lowered entrance Mach number: Mai „ew ~ 0.203. With Toi and Pi known, the other inlet properties follow from this Mach number: Ti =■ 303 - = 301 K 1 + 0.2 Maf 1 + 0.2(0.203)^ ai = \/kRTi = [1.4(287)(301)]‘“ = 348 m/s Vi = Maid = (0.203) (348 m/s) = 71 m/s Pi 150,000 Pi RT, (287)(301) = 1.74 kg/m" Finally, the new lowered mass flow per unit area is = PiVi = (1.74kg/m^)(71 m/s) = 123 kg/(s ■ m^) This is 7 percent less than in Example 9.14, due to choking by excess heat addition. 642 Chapter 9 Compressible Flow Relationship to the Normal Shock Wave 9.9 Mach Waves and Oblique Shock Waves Mach Waves The normal shock wave relations of Sec. 9.5 actually lurk within the simple heating relations as a special case. From Table B.4 or Fig. 9.17 we see that for a given stag¬ nation temperature less than two flow states satisfy the simple heating relations, one subsonic and the other supersonic. These two states have (1) the same value of Tq, (2) the same mass flow per unit area, and (3) the same value of p -f pV^. Therefore these two states are exactly equivalent to the conditions on each side of a normal shock wave. The second law would again require that the upstream flow Maj be supersonic. To illustrate this point, take Maj = 3.0 and from Table B.4 read Tqi/T^ = 0.6540 and pi/p = 0.1765. Now, for the same value Tq2IT% = 0.6540, use Table B.4 or Eq. (9.78a) to compute Ma2 = 0.4752 and pnip = 1.8235. The value of Ma2 is exactly what we read in the shock table. Table B.2, as the downstream Mach number when Mai = 3-0- Ths pressure ratio for these two states is pilpx = ip2lp)l{p\lp) = 1.8235/0.1765 = 10.33, which again is just what we read in Table B.2 for Maj = 3.0. This illustration is meant only to show the physical background of the simple heating relations; it would be silly to make a practice of computing normal shock waves in this manner. Up to this point we have considered only one-dimensional compressible flow theories. This illustrated many important effects, but a one-dimensional world completely loses sight of the wave motions that are so characteristic of supersonic flow. The only “wave motion” we could muster in a one-dimensional theory was the normal shock wave, which amounted only to a flow discontinuity in the duct. When we add a second dimension to the flow, wave motions immediately become apparent if the flow is supersonic. Figure 9.18 shows a celebrated graphical construc¬ tion that appears in every fluid mechanics textbook and was first presented by Ernst Mach in 1887. The figure shows the pattern of pressure disturbances (sound waves) sent out by a small particle moving at speed U through a still fluid whose sound velocity is a. As the particle moves, it continually crashes against fluid particles and sends out spherical sound waves emanating from every point along its path. A few of these spherical disturbance fronts are shown in Eig. 9.18. The behavior of these fronts is quite different according to whether the particle speed is subsonic or supersonic. In Eig. 9.18a, the particle moves subsonically, U < a, Ma = Ula < 1. The spheri¬ cal disturbances move out in all directions and do not catch up with one another. They move well out in front of the particle also, because they travel a distance a dt during the time interval in which the particle has moved only U dt. Therefore a subsonic body motion makes its presence felt everywhere in the flow field: You can “hear” or “feel” the pressure rise of an oncoming body before it reaches you. This is apparently why that pigeon in the road, without turning around to look at you, takes to the air and avoids being hit by your car. At sonic speed, U = a, Eig. 9.18/>, the pressure disturbances move at exactly the speed of the particle and thus pile up on the left at the position of the particle into a 9.9 Mach Waves and Oblique Shock Waves 643 Fig. 9.18 Wave patterns set up by a particle moving at speed U into still fluid of sound velocity a\ (a) subsonic, (b) sonic, and (c) supersonic motion. sort of “front locus,” which is now called a Mach wave, after Ernst Mach. No distur¬ bance reaches beyond the particle. If you are stationed to the left of the particle, you cannot “hear” the oncoming motion. If the particle blew its horn, you couldn’t hear that either: A sonic car can sneak up on a pigeon. In supersonic motion, U > a, the lack of advance warning is even more pronounced. The disturbance spheres cannot catch up with the fast-moving particle that created them. They all trail behind the particle and are tangent to a conical locus called the Mach cone. From the geometry of Fig. 9.18c the angle of the Mach cone is seen to be , a 3 1 , a 1 1 u = sin”^ — — = sin”^ — = sin“ - (9.79) ^ U 5t U Ma The higher the particle Mach number, the more slender the Mach cone; for example, is 30° at Ma = 2.0 and 11.5° at Ma = 5.0. For the limiting case of sonic flow, Ma = 1, /i = 90°; the Mach cone becomes a plane front moving with the particle, in agreement with Fig. 9.18/?. You cannot “hear” the disturbance caused by the supersonic particle in Fig. 9.18c until you are in the zone of action inside the Mach cone. No warning can reach your ears if you are in the zone of silence outside the cone. Thus an observer on the ground beneath a supersonic airplane does not hear the sonic boom of the passing cone until the plane is well past. 644 Chapter 9 Compressible Flow Fig. 9.19 The wave pattern around a model of the X-15 fighter, moving at about Ma = 1.7. The heavy lines are oblique shock waves, caused by sharp turns, and the light lines are Mach waves, caused by gentle turns. [Photo Courtesy of NASA] The Mach wave need not be a cone: Similar waves are formed by a small disturbance of any shape moving supersonically with respect to the ambient fluid. For example, the “particle” in Fig. 9.18c could be the leading edge of a sharp flat plate, which would form a Mach wedge of exactly the same angle fi. Mach waves are formed by small roughnesses or boundary layer irregularities in a supersonic wind tunnel or at the surface of a supersonic body. Look again at Fig. 9.10: Mach waves are clearly visible along the body surface downstream of the recompression shock, especially at the rear comer. Their angle is about 30°, indicating a Mach number of about 2.0 along this surface. A more complicated system of waves, seen in Fig. 9.19, emanates from a model of the supersonic X-15 fighter plane, catapulted at Ma —1.7 through a wind tunnel. The Mach and shock waves are visualized by the knife-edge schlieren photographic technique . Note the supersonic turbulent wake. EXAMPLE 9.16 An observer on the ground does not hear the sonic boom caused by an airplane moving at 5-km altitude until it is 9 km past her. What is the approximate Mach number of the plane? Assume a small disturbance, and neglect the variation of sound speed with altitude. Solution A finite disturbance like an airplane will create a finite-strength oblique shock wave whose angle will be somewhat larger than the Mach wave angle fi and will curve downward due to the variation in atmospheric sound speed. If we neglect these effects, the altitude and distance are a measure of fi, as seen in Fig. E9. 16. 9.9 Mach Waves and Oblique Shock Waves 645 The Oblique Shock Wave Fig. 9.20 Geometry of flow through an oblique shock wave. E9.16 9 km Thus, tan fj, = Hence, from Eq. (9.79), 5 km 9 km 0.5556 or ji = 29.05° Ma = CSC /i = 2.06 Ans. Figures 9.10 and 9.19 and our earlier discussion all indicate that a shock wave can form at an oblique angle to the oncoming supersonic stream. Such a wave will deflect the stream through an angle 6, unlike the normal shock wave, for which the down¬ stream flow is in the same direction. In essence, an oblique shock is caused by the necessity for a supersonic stream to turn through such an angle. Examples could be a finite wedge at the leading edge of a body and a ramp in the wall of a supersonic wind tunnel. The flow geometry of an oblique shock is shown in Fig. 9.20. As for the normal shock of Fig. 9.8, state 1 denotes the upstream conditions and state 2 is downstream. The shock angle has an arbitrary value /3, and the downstream flow V2 turns at an angle 6 which is a function of j3 and state 1 conditions. The upstream flow is always supersonic, but the downstream Mach number Ma2 = V2la2 may be subsonic, sonic, or supersonic, depending on the conditions. Oblique shock wave 646 Chapter 9 Compressible Flow It is convenient to analyze the flow by breaking it up into normal and tangential components with respect to the wave, as shown in Fig. 9.20. For a thin control volume just encompassing the wave, we can then derive the following integral relations. canceling out A; = A2 on each side of the wave: Continuity: = P2K2 (9.80fl) Normal momentum: Pi ~ Pi ~ Pi^ni ~ Pi^ni (9.80f>) Tangential momentum: 0 = PiF„i(V,2 “ ^ti) (9.80c) Energy: h, + + kvfi = K i9Md) We see from Eq. (9.80c) that there is no change in tangential velocity across an oblique shock: Vii = Ki = = const (9.81) Thus tangential velocity has as its only effect the addition of a constant kinetic energy to each side of the energy equation (9.80(f)- We conclude that Eqs. (9.80) are identical to the normal shock relations (9.49), with Vi and V2 replaced by the normal components V„\ and V„2- All the various relations from Sec. 9.5 can be used to com¬ pute properties of an oblique shock wave. The trick is to use the “normal” Mach numbers in place of Mai and Ma2: Ma„i — ^ = Mai sin B fli Ma,j2 V„2 - = Ma2 sin {(B — 0) 0-2 (9.82) Then, for a perfect gas with constant specific heats, the property ratios across the oblique shock are the analogs of Eqs. (9.55) to (9.58) with Mai replaced by Ma„i: - ^^[2/tMa?sin^/3 - (A: - 1)] Pi Pi tan /3 {k + l)Matsin^/3 V„i Pi tan {(3 -6) {k - l)Ma? sin^/3 + 2 V, n2 T2 , 22 , 2A:Maf sin^/3 - (yt - 1) ^ = [2 + {k - f3] ^ , 2-2^ Ti (k + 1) Mafsm p Tm — Tn P02 (yt -f l)Ma?sin^/3 m-i) ^ + 1 Poi .2 -f (A: - l)Ma?sin^/3, . 2yt Ma? sin^ yS - (yt - 1). ll{k- 1) 2 “ l)Ma„i -f 2 Mat, 2 = 2ytMa,-;i - (A: - 1) (9.83fl) (9.831?) (9.83c) (9.83(/) (9.83c) (9.83/) All these are tabulated in the normal shock Table B.2. If you wondered why that table listed the Mach numbers as Ma„i and Ma„2, it should be clear now that the table is also valid for the oblique shock wave. 9.9 Mach Waves and Oblique Shock Waves 647 Fig. 9.21 The oblique shock polar hodograph, showing double solutions (strong and weak) for small deflection angle and no solutions at all for large deflection. Thinking all this over, we realize with hindsight that an oblique shock wave is the flow pattern one would observe by running along a normal shock wave (Fig. 9.8) at a constant tangential speed V,. Thus the normal and oblique shocks are related by a galilean, or inertial, velocity transformation and therefore satisfy the same basic equations. If we continue with this run-along-the-shock analogy, we find that the deflection angle 6 increases with speed V, up to a maximum and then decreases. From the geometry of Fig. 9.20, the deflection angle is given by -1 Vt Vt 6 = tan - tan - (9.84) V„2 V„i If we differentiate ff with respect to V, and set the result equal to zero, we find that the maximum deflection occurs when V,/V„i = We can substitute this back into Eq. (9.84) to compute = tan-' - tan"' r”''" r = ^ (9.85) For example, if Ma„i = 3.0, from Table B.2 we find that V„i/F„2 = 3.8571, the square root of which is 1.9640. Then Eq. (9.85) predicts a maximum deflection of tan" 1.9640 — tan" (1/1.9640) = 36.03°. The deflection is quite limited even for infinite Ma,.,: From Table B.2 for this case V„JV„j = 6.0, and we compute from Eq. (9.85) that = 45.58°. This limited-deflection idea and other facts become more evident if we plot some of the solutions of Eqs. (9.83). Eor given values of Vi and a^, assuming as usual that k = 1.4, we can plot all possible solutions for V2 downstream of the shock. Figure 9.21 does this in velocity-component coordinates I4 and Vy, with x parallel to Vi. Such a plot is called a hodograph. The heavy dark line that looks like a fat airfoil is the locus, or shock polar, of all physically possible solutions for the given Maj. The two dashed-line fishtails are solutions that increase V2; they are physically impossible because they violate the second law. Examining the shock polar in Fig. 9.21, we see that a given deflection line of small angle 9 crosses the polar at two possible solutions: the strong shock, which greatly decelerates the flow, and the weak shock, which causes a much milder deceleration. 648 Chapter 9 Compressible Flow The flow downstream of the strong shock is always subsonic, while that of the weak shock is usually supersonic but occasionally subsonic if the deflection is large. Both types of shock occur in practice. The weak shock is more prevalent, but the strong shock will occur if there is a blockage or high-pressure condition downstream. Since the shock polar is only of flnite size, there is a maximum deflection shown in Fig. 9.21, that just grazes the upper edge of the polar curve. This verifies the kinematic discussion that led to Eq. (9.85). What happens if a supersonic flow is forced to deflect through an angle greater than The answer is illustrated in Fig. 9.22 for flow past a wedge-shaped body. In Fig. 9.22fl the wedge half-angle 6 is less than 9^^, and thus an oblique shock forms at the nose of wave angle f3 just sufficient to cause the oncoming supersonic stream to deflect through the wedge angle 9. Except for the usually small effect of boundary layer growth (see, for example. Ref. 19, Sec. 7-5.2), the Mach number Ma2 is constant along the wedge surface and is given by the solution of Eqs. (9.83). The pressure, density, and temperature along the surface are also nearly constant, as pre¬ dicted by Eqs. (9.83). When the flow reaches the corner of the wedge, it expands to higher Mach number and forms a wake (not shown) similar to that in Eig. 9.10. In Eig. 9.22b the wedge half-angle is greater than an attached oblique shock is impossible. The flow cannot deflect at once through the entire angle yet somehow the flow must get around the wedge. A detached curve shock wave forms in front of the body, discontinuously deflecting the flow through angles smaller than 9^^. The flow then curves, expands, and deflects subsonically around the wedge, becoming sonic and then supersonic as it passes the corner region. The flow just inside each point on the curved shock exactly satisfies the oblique shock relations (9.83) for Ma, > 1 Fig. 9.22 Supersonic flow past a wedge: (a) small wedge angle, attached oblique shock forms; (b) large wedge angle, attached shock not possible, broad curved detached shock forms. 9.9 Mach Waves and Oblique Shock Waves 649 that particular value of /3 and the given Maj. Every condition along the curved shock is a point on the shock polar of Fig. 9.21. Points near the front of the wedge are in the strong shock family, and points aft of the sonic line are in the weak shock family. The analysis of detached shock waves is extremely complex , and experimentation is usually needed, such as the shadowgraph optical technique of Fig. 9.10. The complete family of oblique shock solutions can be plotted or computed from Eqs. (9.83). For a given k, the wave angle /3 varies with Maj and 6, from Eq. (9.83h). By using a trigonometric identity for tan (/3 — 9) this can be rewritten in the more convenient form 2 cot (3 (Mai sin^ /3 — 1 ) tan 9 = - - Ma? (k + cos 2/3) + 2 (9.86) All possible solutions of Eq. (9.86) for k = 1.4 are shown in Fig. 9.23. For deflections 9 < there are two solutions: a weak shock (small (3) and a strong shock (large /3), as expected. All points along the dash-dot line for 9^^ satisfy Eq. (9.85). A dashed line has been added to show where Ma2 is exactly sonic. We see that there is a narrow region near maximum deflection where the weak shock downstream flow is subsonic. For zero deflections (0=0) the weak shock family satisfies the wave angle relation f3 = 11 = sin-' ^ (9.87) Mai Thus weak shocks of vanishing deflection are equivalent to Mach waves. Meanwhile the strong shocks all converge at zero deflection to the normal shock condition /3 = 90°. Fig. 9.23 Oblique shock deflection versus wave angle for various upstream Mach numbers, k = 1.4: dash-dot curve, locus of 0,^^, divides strong (right) from weak (left) shocks; dashed curve, locus of sonic points, divides subsonic Ma2 (right) from supersonic Ma2 (left). k = 1.4 Ma^ = oo-v.^ 10^ li /(' 6 /] /; 3 ] 1 2.5 1 i i 2 i 1.8 1 ' i 1.6 iv 1. 4^ \ \ \ V 1.2 ' .\ 0° 30° 60° 90° Wave angle 650 Chapter 9 Compressible Flow Two additional oblique shock charts are given in App. B for k = 1.4, where Fig. B.l gives the downstream Mach number Ma2 and Fig. B.2 the pressure ratio P2lp\, each plotted as a function of Mai and 9. Additional graphs, tables, and computer programs are given in Refs. 20 and 21. Very Weak Shock Waves For any finite 9 the wave angle (3 for a weak shock is greater than the Mach angle /i. For small 9 Eq. (9.86) can be expanded in a power series in tan 9 with the following linearized result for the wave angle: /c + 1 sin p = sin u H - tan 9 + 4 cos p, + 0(tan^ 9) + (9.88) For Mai between 1 .4 and 20.0 and deflections less than 6° this relation predicts f3 to within 1° for a weak shock. For larger deflections it can be used as a useful initial guess for iterative solution of Eq. (9.86). Other property changes across the oblique shock can also be expanded in a power series for small deflection angles. Of particular interest is the pressure change from Eq. (9.83a), for which the linearized result for a weak shock is P2 - Pi Pi kMaf (Ma? 1) 1/2 tan 0 + ■■■ + ©(tan^ 9) + (9.89) The differential form of this relation is used in the next section to develop a theory for supersonic expansion turns. Figure 9.24 shows the exact weak shock pressure jump Fig. 9.24 Pressure jump across a weak oblique shock wave from Eq. (9.83a) for k = 1.4. For very small deflections Eq. (9.89) applies. 9.9 Mach Waves and Oblique Shock Waves 651 computed from Eq. (9.83fl). At very small deflections the curves are linear with slopes given by Eq. (9.89). Einally, it is educational to examine the entropy change across a very weak shock. Using the same power series expansion technique, we can obtain the following result for small flow deflections: Si - l)Ma? 12(Mat - 1) 3/2 tan^ 0 + • • • + 0(tam 6) + (9.90) The entropy change is cubic in the deflection angle 9. Thus weak shock waves are very nearly isentropic, a fact that is also used in the next section. Mai = 2.0 Pj = 10 Ibf/in^ E9.17 EXAMPLE 9.17 Air at Ma = 2.0 and p = 10 Ibf/in^ absolute is forced to turn through 10° by a ramp at the body surface. A weak oblique shock forms as in Fig. E9.17. For k = \A compute from exact oblique shock theory (a) the wave angle (3, (b) Ma.2, and (c) p2. Also use the linearized theory to estimate (d) (3 and (e) p2. Solution using Excel With Mai = 2.0 and 0 = 10° known, we can estimate /3 ~ 40° ± 2° from Fig. 9.23. For more accuracy, we can set an Excel iteration, using improved guesses for /3. Begin with a guess of 40° and zero in on the correct wave angle. The writer’s guesses are as follows: A B /3 - guess e - Eq. (9.86) 1 40.00 10.623 2 38.00 8.767 3 39.00 9.710 4 39.30 9.987 5 39.32 10.006 The iteration converges to (3 = 39.32° Ans. (a) The normal Mach number upstream is thus Ma„i = Mai sin /3 = 2.0 sin 39.32° = 1.267 With Ma„i we can use the normal shock relations (Table B.2) or Eig. 9.9 or Eqs. (9.56) to (9.58) to compute Ma„2 = 0.8031 — = 1.707 Pi Thus the downstream Mach number and pressure are Ma2 Ma„2 sin {(3 - 0) 0.8031 sin (39.32° - 10°) 1.64 P2 = (10 Ibf/in^ absolute) (1.707) = 17.07 Ibf/in^ absolute Ans. (b) Ans. (c) 652 Chapter 9 Compressible Flow 9.10 Prandtl-Meyer Expansion Waves The Prandtl-Meyer Perfect-Gas Function Notice that the computed pressure ratio agrees with Figs. 9.24 and B.2. For the linearized theory the Mach angle is [i = sin” (1/2.0) = 30°. Equation (9.88) then estimates that or sin P ~ sin 30° + 2.4 tan 10° 4 cos 30° 0.622 P « 38.5° Equation (9.89) estimates that or P2 Pi « 1 1.4(2)^ tan 10° (22 _ 1)1/2 1.57 P2 ~ 1.57(10 Ibf/in^ absolute) = 15.7 Ibf/in^ absolute Ans. (d) Ans. (e) These are reasonable estimates in spite of the fact that 10° is really not a “small” flow deflection. The oblique shock solution of Sec. 9.9 is for a finite compressive deflection 0 that obstructs a supersonic flow and thus decreases its Mach number and velocity. The present section treats gradual changes in flow angle that are primarily expansive; they widen the flow area and increase the Mach number and velocity. The property changes accumulate in infinitesimal increments, and the linearized relations (9.88) and (9.89) are used. The local flow deflections are infinitesimal, so the flow is nearly isentropic according to Eq. (9.90). Figure 9.25 shows four examples, one of which (Fig. 9.25c) fails the test for gradual changes. The gradual compression of Fig. 9.25a is essentially isentropic, with a smooth increase in pressure along the surface, but the Mach angle increases along the surface and the waves tend to coalesce farther out into an oblique shock wave. The gradual expansion of Fig. 9.25b causes a smooth isentropic increase of Mach number and velocity along the surface, with diverging Mach waves formed. The sudden compression of Fig. 9.25c cannot be accomplished by Mach waves: An oblique shock forms, and the flow is nonisentropic. This could be what you would see if you looked at Fig. 9.25a from far away. Finally, the sudden expansion of Fig. 9.25(7 is isentropic and forms a fan of centered Mach waves emanating from the corner. Note that the flow on any streamline passing through the fan changes smoothly to higher Mach number and velocity. In the limit as we near the corner the flow expands almost discontinuously at the surface. The cases in Figs. 9.25a, b, and d can all be handled by the Prandtl-Meyer supersonic wave theory of this section, first formulated by Ludwig Prandtl and his student Theodor Meyer in 1907 to 1908. Note that none of this discussion makes sense if the upstream Mach number is subsonic, since Mach wave and shock wave patterns cannot exist in subsonic flow. Consider a small, nearly infinitesimal flow deflection dO such as occurs between the first two Mach waves in Fig. 9.25a. From Eqs. (9.88) and (9.89) we have, in 9.10 Prandtl-Meyer Expansion Waves 653 Fig. 9.25 Some examples of supersonic expansion and compression: (a) gradual isentropic compression on a concave surface, Mach waves coalesce farther out to form oblique shock; (b) gradual isentropic expansion on convex surface, Mach waves diverge; (c) sudden compression, nonisentropic shock forms; (d) sudden expansion, centered isentropic fan of Mach waves forms. (a) Oblique (b) Mach (c) id) the limit. /3 ~ /i = sin Ma (9.91fl) dp A:Ma^ (Ma^ - 1) 1/2 dB (9.9lb) Since the flow is nearly isentropic, we have the frictionless differential momentum equation for a perfect gas: , dV dp = -pVdV = -kp Ma^ — (9.92) Combining Eqs. (9.91a) and (9.92) to eliminate dp, we obtain a relation between turning angle and velocity change: dB = (9.93) This can be integrated into a functional relation for finite turning angles if we can relate V to Ma. We do this from the definition of Mach number: or y = Maa dV d Ma da y Ma a (9.94) 654 Chapter 9 Compressible Flow Finally, we can eliminate data because the flow is isentropic and hence Qq is constant for a perfect gas: or a = ao[l + \(k - l)Ma^]“^'^ da —\{k— l)Ma(iMa a ~ I +\{k- l)Ma^ (9.95) Eliminating dVIV and dala from Eqs. (9.93) to (9.95), we obtain a relation solely between turning angle and Mach number: (Ma^ - 1)'^^ t/Ma I +\(k- l)Ma^ Ma (9.96) Before integrating this expression, we note that the primary application is to expan¬ sions: increasing Ma and decreasing 9. Therefore, for convenience, we deflne the Prandtl-Meyer angle a;(Ma), which increases when 0 decreases and is zero at the sonic point: duj = -dO 00 = 0 at Ma = 1 (9.97) Thus we integrate Eq. (9.96) from the sonic point to any value of Ma: duo = -'o (Ma^ - 1) 1/2 dMa 1 1 + j(/t - l)Ma Ma (9.98) The integrals are evaluated in closed form, with the result, in radians. ,/Ma- - 1 a;(Ma) = /if'^-tan'M - - — 1/2 tan”' (Ma“ - 1)''^ (9.99) where K = k + 1 k - 1 This is the Prandtl-Meyer supersonic expansion function, which is plotted in Fig. 9.26 and tabulated in Table B.5 for k = lA, K = 6. The angle to changes rapidly at first and then levels off at high Mach number to a limiting value as Ma — > 0°: ^^max = f - 1) = 130.45° if k= 1.4 (9.100) Thus a supersonic flow can expand only through a finite turning angle before it reaches inflnite Mach number, maximum velocity, and zero temperature. Gradual expansion or compression between finite Mach numbers Ma; and Ma2, neither of which is unity, is computed hy relating the turning angle Au; to the differ¬ ence in Prandtl-Meyer angles for the two conditions AaJi^2 = t.^(Ma2) — a;(Mai) (9.101) The change \lo may be either positive (expansion) or negative (compression) as long as the end conditions lie in the supersonic range. Let us illustrate with an example. 9.10 Prandtl-Meyer Expansion Waves 655 Fig. 9.26 The Prandtl-Meyer supersonic expansion from Eq. (9.99) for /t = 1.4. EXAMPLE 9.18 Air (k = 1.4) flows at Mai = 3.0 and pi = 200 kPa. Compute the final downstream Mach number and pressure for (a) an expansion turn of 20° and (h) a gradual compression turn of 20°. Solution using Excel Part (a) The isentropic stagnation pressure is Po = Pi[l + 0.2(3.0)^]^'^ = 7347 kPa and this will he the same at the downstream point. For Mai = 3.0 we find from Table B.5 or Eq. (9.99) that UJi = 49.757°. The flow expands to a new condition such that cjj = Wi + AtJ = 49.757° -f 20° = 69.757° Inversion of Eq. (9.99), to find Ma when u) is given, requires iteration, and Excel is well suited for this job. Hard to read, but Fig. 9.26 indicates a; ~ 4. Make a guess of u; = 4 and program Eq. (9.99) into an Excel cell. The writer’s improved guesses are shown. A B Ma - guess u> - Eq. (9.99) 1 4.00 65.78 2 4.20 68.33 3 4.30 69.54 4 4.32 69.78 656 Chapter 9 Compressible Flow Part (b) Application to Supersonic Airfoils The iteration converges to Ma.2 = 4.32 The isentropic pressure at this new condition is P2 = _ Po _ [1 + 0.2(4.32)^]^'’ 7347 230.1 31.9 kPa Ans. (a) Ans. (a) The flow compresses to a lower Prandtl-Meyer angle: ^2 = 49.757° - 20° = 29.757° Again from Eq. (9.99), Table B.5, or Excel we compute that Maj = 2.125 P2 Po [1 + 0.2(2.125)^]^ 7347 9.51 = 773 kPa Ans. (b) Ans. (b) Similarly, we compute density and temperature changes by noticing that Tq and po are constant for isentropic flow. The oblique shock and Prandtl-Meyer expansion theories can be used to patch together a number of interesting and practical supersonic flow fields. This marriage, called shock expansion theory, is limited by two conditions: (1) Except in rare instances the flow must be supersonic throughout, and (2) the wave pattern must not suffer interference from waves formed in other parts of the flow field. A very successful application of shock expansion theory is to supersonic airfoils. Figure 9.27 shows two examples, a flat plate and a diamond-shaped foil. In contrast to subsonic flow designs (Fig. 8.21), these airfoils must have sharp leading edges, which form attached oblique shocks or expansion fans. Rounded supersonic leading edges would cause detached bow shocks, as in Fig. 9.19 or 9.22b, greatly increasing the drag and lowering the lift. In applying shock expansion theory, one examines each surface turning angle to see whether it is an expansion (“opening up”) or compression (obstruction) to the surface flow. Figure 9.21a shows a flat-plate foil at an angle of attack. There is a leading-edge shock on the lower edge with flow deflection 0 = a, while the upper edge has an expansion fan with increasing Prandtl-Meyer angle Aui = a. We compute P3 with expansion theory and p2 with oblique shock theory. The force on the plate is thus F = (p2 — P3)Cb, where C is the chord length and b the span width (assuming no wingtip effects). This force is normal to the plate, and thus the lift force normal to the stream is L = F cos a, and the drag parallel to the stream is D = F sin a. The dimensionless coefficients and have the same definitions as in low-speed flow, Eqs. (7.66), except that the perfect-gas law identity = ^kp Ma^ is very useful here; Cl L \kpao Ma^ bC Cd D ^2kpoo Mai bC (9.102) 9.10 Prandtl-Meyer Expansion Waves 657 Fig. 9.27 Supersonic airfoils: (a) flat plate, higher pressure on lower surface, drag due to small downstream component of net pressure force; {b) diamond foil, higher pressures on both lower surfaces, additional drag due to body thickness. The typical supersonic lift coefficient is much smaller than the subsonic value ~ lira, hut the lift can he very large because of the large value of at supersonic speeds. At the trailing edge in Fig. 921a, a shock and fan appear in reversed positions and bend the two flows back so fhaf fhey are parallel in the wake and have the same pressure. They do not have quite the same velocity because of the unequal shock strengths on the upper and lower surfaces; hence a vortex sheet trails behind the wing. This is very interest¬ ing, but in the theory you ignore the trailing-edge pattern entirely, since it does not affect the surface pressures; The supersonic surface flow cannof “hear” fhe wake disturbances. The diamond foil in Fig. 921b adds two more wave patterns to the flow. At this particular a less than the diamond half-angle, there are leading-edge shocks on both surfaces, the upper shock being much weaker. Then there are expansion fans on each shoulder of the diamond: The Prandtl-Meyer angle change AtJ equals the sum of the leading-edge and trailing-edge diamond half-angles. Finally, the trailing-edge pattern is similar to that of the flat plate (921a) and can be ignored in the calculation. Both lower-surface pressures p2 and p,^ are greater than their upper counterparts, and the lift is nearly that of the flat plate. There is an additional drag due to thickness because P4 and p^ on the trailing surfaces are lower than their counterparts p2 and p^. The diamond drag is greater than the flat-plate drag, but this must be endured in practice to achieve a wing structure strong enough to support these forces. 658 Chapter 9 Compressible Flow The theory sketched in Fig. 9.27 is in good agreement with measured supersonic lift and drag as long as the Reynolds number is not too small (thick boundary layers) and the Mach number not too large (hypersonic flow). It turns out that for large Rem¬ and moderate supersonic Ma„ the boundary layers are thin and separation seldom occurs, so that the shock expansion theory, although frictionless, is quite successful. Let us look now at an example. EXAMPLE 9.19 A flat-plate airfoil with C = 2 m is immersed at a = 8° in a stream with Ma = 2.5 and = 100 kPa. Compute (a) Ci and (b) C/j, and compare with low-speed airfoils. Compute (c) lift and (d) drag in newtons per unit span width. Solution Instead of using a lot of space outlining the detailed oblique shock and Prandtl-Meyer expansion computations, we list all pertinent results in Fig. E9.19 on the upper and lower surfaces. Using the theories of Secs. 9.9 and 9.10, you should verify every single one of the calculations in Fig. E9.19 to make sure that all details of shock expansion theory are well understood. Ma =2.5 oo = 100 kPa Pol = 1709 kPa u} =39.124° IX Do not compute E9.19 The important final results are p2 and p^, from which the total force per unit width on the plate is F = {P2- Pi)bC = (165.7 - 56.85)(kPa)(l m)(2 m) = 218 kN The lift and drag per meter width are thus L = Fcos 8° = 216 kN D = F sin 8° = 30 kN Ans. (c) Ans. (d) These are very large forces for only 2 m^ of wing area. Erom Eq. (9.102) the lift coefficient is 216 kN = 0.246 Ans. (a) 9.10 Prandtl-Meyer Expansion Waves 659 The comparable low-speed coefficient from Eq. (8.67) is Ci = 27r sin 8° = 0.874, which is 3.5 times larger. From Eq. (9.102) the drag coefficient is Cn = 30 kN \|(1.4)(100kPa)(2.5)^(2m^) = 0.035 Ans. (b) From Fig. 7.25 for the NACA 0009 airfoil, Co at a = 8° is about 0.009, or about 4 times smaller. Notice that this supersonic theory predicts a finite drag in spite of assuming frictionless flow with infinite wing aspect ratio. This is called wave drag, and we see that the d’Alembert paradox of zero body drag does not occur in supersonic flow. Thin-Airfoil Theory In spite of the simplicity of the flat-plate geometry, the calculations in Example 9.19 were laborious. In 1925 Ackeret developed simple yet effective expressions for the lift, drag, and center of pressure of supersonic airfoils, assuming small thickness and angle of attack. The theory is based on the linearized expression (9.89), where tan 9 ~ surface deflection relative to the free stream and condition 1 is the free stream, Maj = Ma„. For the flat-plate airfoil, the total force F is based on Pi - P3 Poo Pi - Poo _ P3 ~ Poo Poo Poo kMaj, (Mai [a - (-a)] (9.103) Substitution into Eq. (9.102) gives the linearized lift coefficient for a supersonic flat- plate airfoil: , ^ (Pi - Pi)bC ^ Aa \kp^MalbC (MaL-1)^'" (9.104) Computations for diamond and other finite-thickness airfoils show no first-order effect of thickness on lift. Therefore, Eq. (9.104) is valid for any sharp-edged supersonic thin airfoil at a small angle of attack. The flat-plate drag coefficient is 4a^ Co = Cz,tan a = ClO = ^ (9.105) However, the thicker airfoils have additional thickness drag. Let the chord line of the airfoil be the x axis, and let the upper-surface shape be denoted by y^ix) and the lower profile by yi(x). Then fhe complefe Ackeref drag theory (see Ref. 5, Sec. 14.6, for details) shows that the additional drag depends on the mean square of the slopes of the upper and lower surfaces, defined by ^ 1 C J fi ax 0 (9.106) 660 Chapter 9 Compressible Flow The final expression for drag [5, p. 442] is 4 Cn (Mai - 1) 1/2 + \iy? + y'l^) (9.107) These are all in reasonable agreement with more exact computations, and their extreme simplicity makes them attractive alternatives to the laborious but accurate shock expansion theory. Consider the following example. EXAMPLE 9.20 Repeat parts (a) and (ft) of Example 9.19, using the linearized Ackeret theory. Solution From Eqs. (9.104) and (9.105) we have, for Ma,, = 2.5 and a = 8° = 0.1396 rad. 4(0.1396) (2.5^ - 1)'^ 0.244 Cd 4(0.1396)^ (2.5^ - 1)'^ 0.034 Ans. These are less than 3 percent lower than the more exact computations of Example 9.19. A further result of the Ackeret linearized theory is an expression for the position Xcp of the center of pressure (CP) of the force distribution on the wing: Xf’p S,j St — = 0.5 + — - ^ (9.108) C 2aC^ where A,, is the cross-sectional area between the upper surface and the chord and Si is the area between the chord and the lower surface. For a symmetric airfoil {Si = Su) we obtain xcp at the half-chord point, in contrast with the low-speed airfoil result, where xcp is at the quarter-chord. The difference in difficulty between the simple Ackeret theory and shock expansion theory is even greater for a thick airfoil, as the following example shows. EXAMPLE 9.21 By analogy with Example 9.19 analyze a diamond, or double- wedge, airfoil of 2° half-angle and C = 2 m at a = 8° and Ma„ = 2.5. Compute and Co by (a) shock expansion theory and (ft) Ackeret theory. Pinpoint the difference from Example 9.19. Solution Part (a) Again we omit the details of shock expansion theory and simply list the properties computed on each of the four airfoil surfaces in Fig. E9.21. Assume p„ = 100 kPa. There are both a force F normal to the chord line and a force P parallel to the chord. For the normal force the pressure difference on the front half isp2— P3 = 1 86.4 — 65 .9 = 1 20.5 kPa, and on the rear half it is p4 — Ps = 146.9 — 48.8 = 98.1 kPa. The average pressure difference is \|(120.5 -I- 98.1) = 109.3 kPa, so that the normal force is F= (109.3 kPa) (2 m^) = 218.6 kN 9.10 Prandtl-Meyer Expansion Waves 661 For the chordwise force P the pressure difference on the top half is P3 — Ps = 65.9 — 48.8 = 17.1 kPa, and on the bottom half it is p2 — Pn = 186.4 — 146.9 = 39.5 kPa. The average dif¬ ference is j(17.1 + 39.5) = 28.3 kPa, which when multiplied by the frontal area (maximum thickness times 1-m width) gives P = (28.3 kPa) (0.07 m)(l m) = 2.0 kN Both F and P have components in the lift and drag directions. The lift force normal to the free stream is L = F cos 8° - P sin 8“ = 216.2 kN and D = F sin 8° + P cos 8° = 32.4 kN For computing the coefficients, the denominator of Eq. (9.102) is the same as in Example 9.19: \kpcc Ma^hC = j(1.4)(100 kPa)(2. 5)^(2 m^) = 875 kN. Thus, finally, shock expansion theory predicts 216.2 kN 875 kN 0.247 32.4 kN 875 kN 0.0370 Arts, (a) Part (b) Meanwhile, by Ackeret theory, Q. is the same as in Example 9.20: 4(0.1396) (2.5^ - 1)‘® 0.244 Ans. (b) This is 1 percent less than the shock expansion result above. For the drag we need the mean-square slopes from Eq. (9.106): = yl^ = tan^ 2° = 0.00122 Then Eq. (9.107) predicts this linearized result: Cd = (2.5^ - !)'■ •[(0.1396)^ -f 1(0.00122 -f 0.00122)] = 0.0362 Ans. (b) This is 2 percent lower than shock expansion theory predicts. We could judge Ackeret theory to be “satisfactory.” Ackeret theory predicts p2 = 167 kPa (—11 percent), p^ = 60 kPa (—9 percent), p^ = 140 kPa (—5 percent), and ps = 33 kPa (—6 percent). 662 Chapter 9 Compressible Flow Three-Dimensional Supersonic Flow Fig. 9.28 Shadowgraph of flow past an 8° half-angle cone at Ma. = 2.0. The turbulent boundary layer is clearly visible. The Mach lines curve slightly, and the Mach number varies from 1.98 just inside the shock to 1.90 at the surface. {Courtesy ofU.S. Army Ballistic Research Laboratory, Aberdeen Proving Ground.) We have gone about as far as we can go in an introductory treatment of compressible flow. Of course, there is much more, and you are invited to study further in the refer¬ ences at the end of the chapter. Three-dimensional supersonic flows are highly complex, especially if they concern blunt bodies, which therefore contain embedded regions of subsonic and transonic flow, as in Fig. 9.10. Some flows, however, yield to accurate theoretical treatment such as flow past a cone at zero incidence, as shown in Fig. 9.28. The exact theory of cone flow is discussed in advanced texts [for example. Ref. 5, Chap. 17], and extensive tables of such solutions have been published . There are similarities between cone flow and the wedge flows illustrated in Fig. 9.22: an attached oblique shock, a thin turbulent boundary layer, and an expansion fan at the rear corner. How¬ ever, the conical shock deflects the flow through an angle less than the cone half-angle, unlike the wedge shock. As in the wedge flow, there is a maximum cone angle above which the shock must detach, as in Fig. 9.22b. For A: = 1.4 and Ma^o = oo, the maxi¬ mum cone half-angle for an attached shock is about 57°, compared with the maximum wedge angle of 45.6° (see Ref. 25). The use of computational fluid dynamics (CFD) is now very popular and successful in compressible flow studies . For example, a supersonic cone flow such as Fig. 9.28, even at an angle of attack, can be solved by numerical simulation of the full three-dimensional (viscous) Navier-Stokes equations . 9.10 Prandtl-Meyer Expansion Waves 663 Fig. 9.29 Wind tunnel test of the Cobra P-530 supersonic interceptor. The surface flow patterns are visualized by the smearing of oil droplets. ( Courtesy of Northrop Grumman. ) New Trends in Aeronautics For more complicated body shapes one usually resorts to experimentation in a supersonic wind tunnel. Figure 9.29 shows a wind tunnel study of supersonic flow past a model of an interceptor aircraft. The many junctions and wingtips and shape changes make theoretical analysis very difficult. Here the surface flow patterns, which indicate boundary layer development and regions of flow separation, have been visualized hy the smearing of oil drops placed on the model surface before the test. As we shall see in the next chapter, there is an interesting analogy between gas dynamic shock waves and the surface water waves that form in an open-channel flow. Chapter 11 of Ref. 9 explains how a water channel can be used in an inexpensive simulation of supersonic flow experiments. The previous edition of this text discussed NASA’s proposed hypersonic scramjet aircraft, the X-43A , which set a new world speed record, in 2004, of Mach 9.6, or nearly 7000 miles per hour. This is hardly a design for a hypersonic airliner, though, since it has to be launched at high altitude from a B-52 bomber. Also discussed earlier was the Air Force X-35 Joint Strike Fighter, whose wind tunnel test is shown here in Figure 9.19. This design is now operational, designated as the F-35, seen in Fig. 9.30, and it has been ordered by the U.S. military and also by Australia and seven NATO countries. A special version, for the U.S. Marine Corps, takes off and lands vertically. It reaches a speed of Mach 1.6 at 40,000 ft altitude. Its shortcomings are the present poor world economy and the fact that the price of one F-35 has risen to 220 million dollars. 664 Chapter 9 Compressible Flow Fig. 9.30 The F-35 Joint Strike Fighter is planned to become the standard supersonic hghter plane for the countries aligned with the United States. [Lockheed Martin photo provided by F-35 Lightning II Program Office.] Summary This chapter briefly introduced a very broad subject, compressible flow, sometimes called gas dynamics. The primary parameter is the Mach number Ma = Via, which is large and causes the fluid density to vary significantly. This means that the continu¬ ity and momentum equations must be coupled to the energy relation and the equation of state to solve for the four unknowns ip, p, T, V). The chapter reviewed the thermodynamic properties of an ideal gas and derived a formula for the speed of sound of a fluid. The analysis was then simplified to one-dimensional steady adiabatic flow without shaft work, for which the stagnation enthalpy of the gas is constant. A further simplification to isentropic flow enables formulas to be derived for high-speed gas flow in a variable-area duct. This reveals the phenomenon of sonic-flow choking (maximum mass flow) in the throat of a nozzle. Problems 665 At supersonic velocities there is the possibility of a normal shock wave, where the gas discontinuously reverts to subsonic conditions. The normal shock explains the effect of back pressure on the performance of converging-diverging nozzles. To illustrate nonisentropic flow conditions, the chapter briefly focused on constant- area duct flow with friction and with heat transfer, both of which lead to choking of the exit flow. The chapter ended with a discussion of two-dimensional supersonic flow, where oblique shock waves and Prandtl-Meyer (isentropic) expansion waves appear. With a proper combination of shocks and expansions one can analyze supersonic airfoils. Problems Most of the problems herein are fairly straightforward. More difficult or open-ended assignments are labeled with an asterisk. Problems la¬ beled with a computer icon may require the use of a computer. The standard end-of-chapter problems P9.1 to P9.157 (categorized in the problem list here) ai'e followed by word problems W9.1 to W9.8, fundamentals of engineering exam problems FE9.1 to FE9.10, com¬ prehensive problems C9. 1 to C9.8, and design projects D9. 1 and D9.2. Problem Distribution Section Topic Problems 9.1 Introduction P9.1-P9.9 9.2 The speed of sound P9.10-P9.18 9.3 Adiabatic and isentropic flow P9.19-P9.33 9.4 Isentropic flow with area changes P9.34-P9.53 9.5 The normal shock wave P9.54-P9.62 9.6 Converging and diverging nozzles P9.63-P9.85 9.7 Duct flow with friction P9.86-P9.106 9.8 Frictionless duct flow with heat transfer P9.107-P9.115 9.9 Mach waves P9.116-P9.121 9.9 The oblique shock wave P9.122-P9.139 9.10 Prandtl-Meyer expansion waves P9.140-P9.148 9.10 Supersonic airfoils P9.149-P9.157 Introduction P9.1 An ideal gas flows adiabatically through a duct. At sec¬ tion l,pi = 140 kPa, Ti = 260°C, and Vi = 75 m/s. Farther downstream, p2 = 30 kPa and T2 = 207°C. Calculate V2 in m/s and ^2 “ ■ti in J/(kg • K) if the gas is (a) air, k = 1.4, and (b) argon, k = 1.67. P9.2 Solve Prob. P9.1 if the gas is steam. Use two approaches: (a) an ideal gas from Table A.4 and (b) real gas data from the steam tables . P9.3 If 8 kg of oxygen in a closed tank at 200“C and 300 kPa is heated until the pressure rises to 400 kPa, calculate (a) the new temperature, (b) the total heat transfer, and (c) the change in entropy. P9.4 Consider steady adiabatic airflow in a duct. At section B, the pressure is 600 kPa and the temperature is 177°C. At section D, the density is 1.13 kg/m^ and the temperature is 156°C. (a) Find the entropy change, if any. (b) Which way is the air flowing? P9.5 Steam enters a nozzle at 377°C, 1.6 MPa, and a steady speed of 200 m/s and accelerates isentropically until it exits at saturation conditions. Estimate the exit velocity and temperature. P9.6 Methane, approximated as a perfect gas, is compressed adiabatically from 101 kPa and 20°C to 300 kPa. Estimate (a) the final temperature, and (b) the final density. P9.7 Air flows through a variable-area duct. At section 1, Ai = 20 cm^. Pi = 300 kPa, pi = 1.75 kg/m^, and Vi = 122.5 m/s. At section 2, the area is exactly the same, but the density is much lower: P2 = 0.266 kg/m^ and 72 = 281 K. There is no transfer of work or heat. Assume one-dimensional steady flow. (a) How can you reconcile these differences? (b) Find the mass flow at section 2. Calculate (c) V2, (d) p2, and (e) ^2 “ ■Ji- [Hint: This problem requires the continuity equation.] P9.8 Atmospheric air at 20°C enters and fills an insulated tank that is initially evacuated. Using a control volume analysis from Eq. (3.67), compute the tank air temperature when it is full. P9.9 Liquid hydrogen and oxygen are burned in a combustion chamber and fed through a rocket nozzle that exhausts at kexit = 1600 m/s to an ambient pressure of 54 kPa. The nozzle exit diameter is 45 cm, and the jet exit density is 0.15 kg/m^. If the exhaust gas has a molecular weight of 18, estimate (a) the exit gas temperature, (b) the mass flow, and (c) the thrust developed by the rocket. The speed of sound P9.10 A certain aircraft flies at 609 mi/h at standard sea level. (a) What is its Mach number? (b) If it flies at the same Mach number at 34,000 ft altitude, how much slower (or faster) is it flying, in mi/h? 666 Chapter 9 Compressible Flow P9.ll At 300°C and 1 atm, estimate the speed of sound of (a) nitrogen, (b) hydrogen, (c) helium, (d) steam, and (e) {k « 1.06). P9.12 Assume that water follows Eq. (1.19) with n~l and B ~ 3000. Compute the bulk modulus (in kPa) and the speed of sound (in m/s) at {a) 1 atm and {b) 1100 atm (the deepest part of the ocean), (c) Compute the speed of sound at 20°C and 9000 atm and compare with the measured value of 2650 m/s (A. H. Smith and A. W. Lawson, J. Chem. Phys., vol. 22, 1954, p. 351). P9.13 Consider steam at 500 K and 200 kPa. Estimate its speed of sound by two different methods: (a) assuming an ideal gas from Table B.4, or (b) using finite differences for isentro- pic densities between 210 kPa and 190 kPa. P9.14 Benzene, listed in Table A. 3, has a measured density of 57.75 Ibm/ft^ at a pressure of 700 bar. Use this data to esti¬ mate the speed of sound of benzene. P9.15 The pressure-density relation for ethanol is approximated by Eq. (1.19) with B = 1600 and n = 7. Use this relation to estimate the speed of sound of ethanol at 2000 atmospheres. P9.16 A weak pressure pulse Ap propagates through still air. Discuss the type of reflected pulse that occurs and the boundary conditions that must be satisfied when the wave strikes normal to, and is reflected from, (a) a solid wall and (b) a free liquid surface. P9.17 A submarine at a depth of 800 m sends a sonar signal and receives the reflected wave back from a similar submerged object in 15 s. Using Prob. P9.12 as a guide, estimate the distance to the other object. P9.18 Race cars at the Indianapolis Speedway average speeds of 185 mi/h. After determining the altitude of Indianapolis, find the Mach number of these cars and estimate whether compressibility might affect their aerodynamics. Adiabatic and isentropic flow P9.19 In 1976, the SR-71A, flying at 20 km standard altitude, set a jet-powered aircraft speed record of 3326 km/h. Estimate the temperature, in °C, at its front stagnation point. At what Mach number would it have a front stagnation-point tem¬ perature of 500°C? P9.20 Air flows isentropically in a channel. Properties at sec¬ tion 1 are Vi = 250 m/s, T, = 330 K, and pi = 80 kPa. At section 2 downstream, the temperature has dropped to 0°C. Find (a) the pressure, (b) velocity, and (c) Mach number at section 2. P9.21 N2O expands isentropically through a duct from pi = 200 kPa and Tj = 250°C to a downstream section where P2 = 26 kPa and V2 = 594 m/s. Compute (a) T2', (b) Ma2; (c) r„; (rf)p„; (e)Fi;and(/-)Mai. P9.22 Given the pitot stagnation temperature and pressure and the static pressure measurements in Fig. P9.22, estimate the air velocity V, assuming (a) incompressible flow and {b) com¬ pressible flow. P9.22 P9.23 A gas, assumed ideal, flows isentropically from point 1, where the velocity is negligible, the pressure is 200 kPa, and the temperature is 300°C, to point 2, where the pressure is 40 kPa. What is the Mach number Ma2 if the gas is (fl) air, {b) argon, or (cj CH4? {d) Can you tell, without calculating, which gas will be the coldest at point 2? P9.24 For low-speed (nearly incompressible) gas flow, the stagna¬ tion pressure can be computed from Bernoulli’s equation; 1 , Po= P + ^PV- (a) For higher subsonic speeds, show that the isentropic relation (9.28fl) can be expanded in a power series as follows: 1 1 2 2 - k , Po ~ p P ^py ( 1 “f — Ma -t- - Ma ^ ^ 2^^ V 4 24 (b) Suppose that a pitot-static tube in air measures the pres¬ sure difference Po — p and uses the Bernoulli relation, with stagnation density, to estimate the gas velocity. At what Mach number will the en'or be 4 percent? P9.25 If it is known that the air velocity in the duct is 750 ft/s, use the mercury manometer measurement in Fig. P9.25 to esti¬ mate the static pressure in the duct in Ibf/in^ absolute. Problems 667 P9.26 Show that for isentropic flow of a perfect gas if a pitot- static prohe measures po, p, and Tq, the gas velocity can he calculated from 2CpTo 1 - \PoJ What would he a source of error if a shock wave were formed in front of the prohe? P9.27 A pitot tube, mounted on an airplane flying at 8000 m stan¬ dard altitude, reads a stagnation pressure of 57 kPa. Estimate the plane’s (a) velocity and (b) Mach number. P9.28 Air flows isentropically through a duct. At section 1, the pressure and temperature are 250 kPa and 125°C, and the velocity is 200 m/s. At section 2, the area is 0.25 m^ and the Mach number is 2.0. Determine (a) Maj; (b) T2, (c) V2, and () the mass flow. P9.29 Steam from a large tank, where T = 400°C and p = I MPa, expands isentropically through a nozzle until, at a section of 2-cm diameter, the pressure is 500 kPa. Using the steam tables , estimate (a) the temperature, {b) the velocity, and (c) the mass flow at this section. Is the flow subsonic? P9.30 When does the incompressible-flow assumption begin to fail for pressures? Construct a graph of p^lp for incom¬ pressible flow of a perfect gas as compared to Eq. (9.28a). Plot both versus Mach number for 0 S Ma £ 0.6 and decide for yourself where the deviation is too great. P9.31 Air flows adiabatically through a duct. At one section Vi = 400 ft/s, Ti = 200°F, and pi = 35 Ibf/in^ absolute, while farther downstream P2 = 1100 ft/s and P2 “ 18 Ibf/in^ absolute. Compute (a) Ma2, (b) and (c) /^o24'oi- P9.32 The large compressed-air tank in Fig. P9.32 exhausts from a nozzle at an exit velocity of 235 m/s. The mercury ma¬ nometer reads h = 30 cm. Assuming isentropic flow, com¬ pute the pressure (a) in the tank and (b) in the atmosphere, (c) What is the exit Mach number? P9.33 Air flows isentropically from a reservoir, where p = 300 kPa and T = 500 K, to section 1 in a duct, where Ai = 0.2 m^ and Vi = 550 m/s. Compute (a) Mai, ib) Ti, (c) pi, (d) m, and (e) A. Is the flow choked? Isentropic flow with area changes P9.34 Air in a large tank, at 300°C and 400 kPa, flows through a converging-diverging nozzle with throat diameter 2 cm. It exits smoothly at a Mach number of 2.8. According to one¬ dimensional isentropic theory, what is (a) the exit diame¬ ter, and (b) the mass flow? P9.35 Helium, at Tq = 400 K, enters a nozzle isentropically. At section 1, where Ai = 0.1 m^, a pitot-static awangement (see Fig. P9.25) measures stagnation pressure of 150 kPa and static pressure of 123 kPa. Estimate (a) Mai, {b) mass flow rii, (c) Ty and {d) A. P9.36 An air tank of volume 1.5 m^ is initially at 800 kPa and 20°C. At f = 0, it begins exhausting through a converging nozzle to sea-level conditions. The throat area is 0.75 cm^. Estimate (a) the initial mass flow in kg/s, (b) the time required to blow down to 500 kPa, and (c) the time at which the nozzle ceases being choked. P9.37 Make an exact control volume analysis of the blowdown process in Fig. P9.37, assuming an insulated tank with negligible kinetic and potential energy within. Assume critical flow at the exit, and show that both po and To decrease during blowdown. Set up first-order differential equations for Po(0 and Tq(1), and reduce and solve as far as you can. Insulated tank P9.38 Prob. P9.37 makes an ideal senior project or combined laboratory and computer problem, as described in Ref. 27, Sec. 8.6. In Bober and Kenyon’s lab experiment, the tank had a volume of 0.0352 ft^ and was initially filled with air at 50 Ib/in^ gage and 72°F. Atmospheric pressure was 14.5 Vol'vc? absolute, and the nozzle exit diameter was 0.05 in. After 2 s of blowdown, the measured tank pres¬ sure was 20 Ib/in^ gage and the tank temperature was — 5°F. Compare these values with the theoretical analysis of Prob. P9.37. 668 Chapter 9 Compressible Flow P9.39 Consider isentropic flow in a channel of varying area, from section 1 to section 2. We know that Mai = 2.0 and desire that the velocity ratio be 1.2. Estimate (a) Ma.2 and (b) A2/A1. (c) Sketch what this channel looks like. For example, does it converge or diverge? Is there a throat? P9.40 Steam, in a tank at 300 kPa and 600 K, discharges isentro- pically to a low-pressure atmosphere through a converging nozzle with exit area 5 cm^. (a) Using an ideal gas approximation from Table B.4, estimate the mass flow. (b) Without actual calculations, indicate how you would use real properties of steam to And the mass flow. P9.41 Air, with a stagnation pressure of 100 kPa, flows through the nozzle in Fig. P9.41, which is 2 m long and has an area variation approximated by A « 20 - 20x -f lOx^ with A in cm^ and x in m. It is desired to plot the complete family of isentropic pressures p(x) in this nozzle, for the range of inlet pressures 1 < p(0) <100 kPa. Indicate which inlet pressures are not physically possible and discuss briefly. If your computer has an online graphics routine, plot at least 15 pressure profiles; otherwise just hit the highlights and explain. P9.41 ^ P9.42 A bicycle tire is filled with air at an absolute pressure of 169.12 kPa, and the temperature inside is 30.0°C. Suppose the valve breaks, and air starts to exhaust out of the tire into the atmosphere (p„ = 100 kPa absolute and = 20.0°C). The valve exit is 2.00 mm in diameter and is the smallest cross-sectional area of the entire system. Fric¬ tional losses can be ignored here; one-dimensional isentro¬ pic flow is a reasonable assumption, (a) Find the Mach number, velocity, and temperature at the exit plane of the valve (initially), (b) Find the initial mass flow rate out of the tire, (c) Estimate the velocity at the exit plane using the incompressible Bernoulli equation. How well does this estimate agree with the “exact” answer of part (a)? Explain. P9.43 Air flows isentropically through a variable-area duct. At section 1, Ai= 20 cm^, pi= 300 kPa, pi= 1.75 kg/m^, and Mai= 0.25. At section 2, the area is exactly the same, but the flow is much faster. Compute (a) V2, (b) Ma2, (c) T2, and (d) the mass flow, (e) Is there a sonic throat between sections 1 and 2? If so, find its area. P9.44 In Prob. P3.34 we knew nothing about compressible flow at the time, so we merely assumed exit conditions p2 and T2 and computed V2 an application of the continuity equa¬ tion. Suppose that the throat diameter is 3 in. For the given stagnation conditions in the rocket chamber in Fig. P3.34 and assuming k = 1.4 and a molecular weight of 26, com¬ pute the actual exit velocity, pressure, and temperature ac¬ cording to one-dimensional theory. If p^ = 14.7 Ibf/in^ absolute, compute the thrust from the analysis of Prob. P3.68. This thrust is entirely independent of the stagnation temperature (check this by changing Tq to 2000°R if you like). Why? P9.45 It is desired to have an isentropic airflow achieve a veloc¬ ity of 550 m/s at a 6-cm-diameter section where the pres¬ sure is 87 kPa and the density 1.3 kg/m^. (a) Is a sonic throat needed? (b) If so, estimate its diameter, and com¬ pute (c) the stagnation temperature and (d) the mass flow. P9.46 A one-dimensional isentropic airflow has the following properties at one section where the area is 53 cm^: p = 12 kPa, p = 0.182 kg/m^, and V = 760 m/s. Determine (fl) the throat area, (b) the stagnation temperature, and (c) the mass flow. P9.47 In wind tunnel testing near Mach 1 , a small area decrease caused by model blockage can be important. Suppose the test section area is 1 w?, with unblocked test conditions Ma = 1.10 and T = 20°C. What model area will first cause the test section to choke? If the model cross section is 0.004 m^ (0.4 percent blockage), what percentage change in test section velocity results? P9.48 A force F = 1100 N pushes a piston of diameter 12 cm through an insulated cylinder containing air at 20°C, as in Fig. P9.48. The exit diameter is 3 mm, and Pa = 1 atm. Estimate {a) (b) Vp, and (c) rhg. Insulated P9.48 Problems 669 P9.49 Consider the venturi nozzle of Fig. 6.40c, with D = 5 cm and d = 3 cm. Stagnation temperature is 300 K, and the upstream velocity Vi = 72 m/s. If the throat pressure is 124 kPa, estimate, with isentropic flow theory, (a) pi, (b) Ma2, and (c) the mass flow. P9.50 Methane is stored in a tank at 120 kPa and 330 K. It dis¬ charges to a second tank through a converging nozzle whose exit area is 5 cm^. What is the initial mass flow rate if the second tank has a pressure of (a) 70 kPa or (b) 40 kPa? P9.51 The scramjet engine is supersonic throughout. A sketch is shown in Fig. C9.8. Test the following design. The flow enters at Ma = 7 and air properties for 10,000 m altitude. Inlet area is 1 m^, the minimum area is 0.1 m^, and the exit area is 0.8 m^. If there is no combustion, (a) will the flow still he supersonic in the throat? Also, determine (b) the exit Mach number, (c) exit velocity, and (d) exit pressure. P9.52 A converging-diverging nozzle exits smoothly to sea- level standard atmosphere. It is supplied hy a 40-m^ tank initially at 800 kPa and 100°C. Assuming isentropic flow in the nozzle, estimate (a) the throat area and (b) the tank pressure after 10 s of operation. The exit area is 10 cm^. P9.53 Air flows steadily from a reservoir at 20°C through a nozzle of exit area 20 cm^ and strikes a vertical plate as in Fig. P9.53. The flow is subsonic throughout. A force of 135 N is required to hold the plate stationary. Compute (a) 14, (b) Ma,,, and (c) Po if Pa ~ 101 kPa. Plate 135 N The normal shock wave P9.54 The airflow in Prob. P9.46 undergoes a normal shock just past the section where data was given. Determine the (a) Mach number, (b) pressure, and (c) velocity just down¬ stream of the shock. P9.55 Air, supplied by a reservoir at 450 kPa, flows through a converging-diverging nozzle whose throat area is 12 cm^. A normal shock stands where Ai = 20 cm^. (a) Compute the pressure just downstream of this shock. Still farther downstream, at A3 = 30 cm^, estimate (b) p^, (c) A, and id) Maj. P9.56 Air from a reservoir at 20°C and 500 kPa flows through a duct and forms a normal shock downstream of a throat of area 10 cm^. By an odd coincidence it is found that the stagnation pressure downstream of this shock exactly equals the throat pressure. What is the area where the shock wave stands? P9.57 Air flows from a tank through a nozzle into the standard atmosphere, as in Fig. P9.57. A normal shock stands in the exit of the nozzle, as shown. Estimate (a) the pressure in the tank and {b) the mass flow. P9.58 Downstream of a normal shock wave, in airflow, the condi¬ tions are T2 = 603 K, 1^2 = 222 m/s, and p2 = 900 kPa. Estimate the following conditions just upstream of the shock: (a) Maj; (b) Tp, (c) pp, (d) p^p, and (e) T^i- P9.59 Air, at stagnation conditions of 450 K and 250 kPa, flows through a nozzle. At section 1, where the area is 15 cm^, there is a normal shock wave. If the mass flow is 0.4 kg/s, estimate (a) the Mach number and (b) the stagnation pres¬ sure just downstream of the shock. P9.60 When a pitot tube such as in Fig. 6.30 is placed in a supersonic flow, a normal shock will stand in front of the probe. Suppose the probe reads po = 190 kPa and p = 150 kPa. If the stagnation temperature is 400 K, estimate the (supersonic) Mach number and velocity upstream of the shock. P9.61 Air flows from a large tank, where T = 376 K and p = 360 kPa, to a design condition where the pressure is 9800 Pa. The mass flow is 0.9 kg/s. However, there is a normal shock in the exit plane just after this condition is reached. Estimate (a) the throat area and, just downstream of the shock, {b) the Mach number, (c) the temperature, and {d) the pressure. P9.62 An atomic explosion propagates into still air at 14.7 Ibf/in^ absolute and 520°R. The pressure just inside the shock is 5000 Ibf/in^ absolute. Assuming k= 1.4, what are the speed C of the shock and the velocity V just inside the shock? Converging and diverging nozzles P9.63 Sea-level standard air is sucked into a vacuum tank through a nozzle, as in Fig. P9.63. A normal shock stands where the nozzle area is 2 cm^, as shown. Estimate (a) the pressure in the tank and (ft) the mass flow. 670 Chapter 9 Compressible Flow P9.70 P9.64 Air, from a reservoir at 350 K and 500 kPa, flows through a converging-diverging nozzle. The throat area is 3 cm^. A normal shock appears, for which the downstream Mach number is 0.6405. (a) What is the area where the shock appears? Calculate (b) the pressure and (c) the temperature downstream of the shock. P9.65 Air flows through a converging-diverging nozzle between two large reservoirs, as shown in Fig. P9.65. A mercury manometer between the throat and the downstream reser¬ voir reads h = 15 cm. Estimate the downstream reservoir pressure. Is there a normal shock in the flow? If so, does it stand in the exit plane or farther upstream? P9.71 P9.72 P9.73 P9.74 P9.66 P9.67 P9.68 P9.69 P9.65 In Prob. P9.65 what would be the mercury manometer reading h if the nozzle were operating exactly at supersonic design conditions? A supply tank at 500 kPa and 400 K feeds air to a converging- diverging nozzle whose throat area is 9 cm^. The exit area is 46 cm^. State the conditions in the nozzle if the pressure outside the exit plane is (a) 400 kPa, (b) 120 kPa, and (c) 9 kPa. (d) In each of these cases, find the mass flow. Air in a tank at 120 kPa and 300 K exhausts to the atmo¬ sphere through a 5-cm^-throat converging nozzle at a rate of 0. 12 kg/s. What is the atmospheric pressure? What is the maximum mass flow possible at low atmospheric pressure? With reference to Prob. P3.68, show that the thrust of a rocket engine exhausting into a vacuum is given by F = PoA,(l + kMai) 1 -f k - 1 -Ma; P9.75 P9.76 where = exit area Ma^ = exit Mach number Po = stagnation pressure in combustion chamber Note that stagnation temperature does not enter into the thrust. Air, with p^ = 500 kPa and = 600 K, flows through a converging-diverging nozzle. The exit area is 51.2 cm^, and mass flow is 0.825 kg/s. What is the highest possible back pressure that will still maintain supersonic flow inside the diverging section? A converging-diverging nozzle has a throat area of 10 cm^ and an exit area of 28.96 cm^. A normal shock stands in the exit when the back pressure is sea-level standard. If the upstream tank temperature is 400 K, estimate (a) the tank pressure and (b) the mass flow. A large tank at 500 K and 165 kPa feeds air to a converging nozzle. The back pressure outside the nozzle exit is sea- level standard. What is the appropriate exit diameter if the desired mass flow is 72 kg/h? Air flows isentropically in a converging-diverging nozzle with a throat area of 3 cm^. At section 1 , the pres¬ sure is 101 kPa, the temperature is 300 K, and the veloc¬ ity is 868 m/s. (a) Is the nozzle choked? Determine (b) Ai and (c) the mass flow. Suppose, without changing stagnation conditions or Ai, the (flexible) throat is reduced to 2 cm^. Assuming shock-free flow, will there be any change in the gas properties at section 1? If so, compute new pi, Fj, and Tj and explain. Use your strategic ideas, from part (b) of Prob. P9.40, to actually carry out the calculations for mass flow of steam, with Po = 300 kPa and Tg = 600 K, discharging through a converging nozzle of choked exit area 5 cm^. A double-tank system in Fig. P9.75 has two identical con¬ verging nozzles of 1-in^ throat area. Tank 1 is very large, and tank 2 is small enough to be in steady-flow equilib¬ rium with the jet from tank 1. Nozzle flow is isentropic, but entropy changes between 1 and 3 due to jet dissipa¬ tion in tank 2. Compute the mass flow. (If you give up. Ref. 9, pp. 288-290, has a good discussion.) P9.75 Air © © L - ► 100 Ibf/in^ abs 520“ R 10 Ibf/in^ abs A large reservoir at 20°C and 800 kPa is used to fill a small insulated tank through a converging-diverging nozzle with 1-cm^ throat area and 1.66-cm^ exit area. The small tank has a volume of 1 m^ and is initially at 20°C and 100 kPa. Problems 671 Estimate the elapsed time when {a) shock waves begin to appear inside the nozzle and {b) the mass flow begins to drop below its maximum value. P9.77 A perfect gas (not air) expands isentropically through a supersonic nozzle with an exit area 5 times its throat area. The exit Mach number is 3.8. What is the specific-heat ratio of the gas? What might this gas be? If po = 300 kPa, what is the exit pressure of the gas? P9.78 The orientation of a hole can make a difference. Consider holes A and B in Fig. P9.78, which are identical but reversed. For the given air properties on either side, compute the mass flow through each hole and explain why they are different. P9.79 A large tank, at 400 kPa and 450 K, supplies air to a converging-diverging nozzle of throat area 4 cm^ and exit area 5 cm^. For what range of back pressures will the flow {a) be entirely subsonic; (b) have a shock wave inside the nozzle; (c) have oblique shocks outside the exit; and {d) have supersonic expansion waves outside the exit? P9.80 A sea-level automobile tire is initially at 32 Ibf/in^ gage pressure and 75°F. When it is punctured with a hole that re¬ sembles a converging nozzle, its pressure drops to 15 Ibf/in^ gage in 12 min. Estimate the size of the hole, in thousandths of an inch. The tire volume is 2.5 ft^. P9.81 Air, at = 160 Ibf/in^ and = 300°F, flows isentropi¬ cally through a converging-diverging nozzle. At section 1 , where Aj = 288 in^, the velocity is Vi = 2068 ft/s. Calcu¬ late (a) Mai, {b)A, (c) pi, and {d) the mass flow, in slug/s. P9.82 Air at 500 K flows through a converging-diverging nozzle with throat area of 1 cm^ and exit area of 2.7 cm^. When the mass flow is 1 82.2 kg/h, a pitot-static probe placed in the exit plane reads po = 250.6 kPa and p = 240.1 kPa. Estimate the exit velocity. Is there a normal shock wave in the duct? If so, compute the Mach number just downstream of this shock. P9.83 When operating at design conditions (smooth exit to sea- level pressure), a rocket engine has a thrust of 1 million Ibf. The chamber pressure and temperature are 600 Ibf/in^ absolute and 4000°R, respectively. The exhaust gases approximate k = 1.38 with a molecular weight of 26. Esti¬ mate (a) the exit Mach number and {b) the throat diameter. P9.84 Air flows through a duct as in Fig. P9.84, where Ai = 24 cm^, A2 = 18 cm^, and A3 = 32 cm^. A normal shock stands at section 2. Compute {a) the mass flow, {b) the Mach number, and (c) the stagnation pressure at section 3. Ti = 30°C P9.84 P9.85 A typical carbon dioxide tank for a paintball gun holds about 12 oz of liquid CO2. The tank is filled no more than one-third with liquid, which, at room temperature, main¬ tains the gaseous phase at about 850 psia. (a) If a valve is opened that simulates a converging nozzle with an exit diameter of 0.050 in, what mass flow and exit velocity results? (b) Repeat the calculations for helium. Duct flow with friction P9.86 Air enters a 3-cm-diameter pipe 15 m long at Vi = 73 m/s. Pi = 550 kPa, and Ti = 60°C. The friction factor is 0.018. Compute 1^2^ P2< T2, and po2 at the end of the pipe. How much additional pipe length would cause the exit flow to be sonic? P9.87 Problem C6.9 gives data for a proposed Alaska-to-Canada natural gas (assume CH4) pipeline. If the design flow rate is 890 kg/s and the entrance conditions are 2500 Ibf/in^ and 140°F, determine the maximum length of adiabatic pipe before choking occurs. P9.88 Air flows adiabatically, with / = 0.024, down a long 6-cm-diameter pipe. At section 1, conditions are Ti = 300 K, Pi = 400 kPa, and Vi = 104 m/s. At section 2, 1^2 = 233 m/s. (a) How far downstream is section 2? Estimate (b) Ma2, (c) p2, and (d) T2. P9.89 Carbon dioxide flows through an insulated pipe 25 m long and 8 cm in diameter. The friction factor is 0.025. At the entrance, p = 300 kPa and T = 400 K. The mass flow is 1.5 kg/s. Estimate the pressure drop by (a) compressible and (&) incompressible (Sec. 6.6) flow theory, (c) For what pipe length will the exit flow be choked? 672 Chapter 9 Compressible Flow P9.90 Air flows through a rough pipe 120 ft long and 3 in in diameter. Entrance conditions are p = 90 Ibf/in^, T = 68°F, and V = 225 ft/s. The flow chokes at the end of the pipe. (a) What is the average friction factor? (b) What is the pressure at the end of the pipe? P9.91 Air flows steadily from a tank through the pipe in Fig. P9.91. There is a converging nozzle on the end. If the mass flow is 3 kg/s and the nozzle is choked, estimate (a) the Mach number at section 1 and (b) the pressure inside the tank. P9.92 Air enters a 5-cm-diameter pipe at 380 kPa, 3.3 kg/m^, and 120 m/s. The friction factor is 0.017. Find the pipe length for which the velocity (a) doubles, (b) triples, and (c) quadruples. P9.93 Air flows adiabatically in a 3-cm-diameter duct, with / = 0.018. At the entrance, Ti = 323 K, pi = 200 kPa, and Vi = 72 m/s. (a) What is the mass flow? (b) For what tube length will the flow choke? (c) If the tube length is increased to 112 m, with the same inlet pressure and temperature, what will be the new mass flow? P9.94 Compressible pipe flow with friction. Sec. 9.7, assumes constant stagnation enthalpy and mass flow but variable momentum. Such a flow is often called Fanno flow, and a line representing all possible property changes on a tem¬ perature-entropy chart is called a Fanno line. Assuming a perfect gas with k = \. A and the data of Prob. P9.86, draw a Fanno curve of the flow for a range of velocities from very low (Ma ^ 1) to very high (Ma ^ 1). Com¬ ment on the meaning of the maximum-entropy point on this curve. P9.95 Helium (Table A.4) enters a 5-cm-diameter pipe at pi = 550 kPa, Vi = 312 m/s, and Ti = 40°C. The friction factor is 0.025. If the flow is choked, determine (a) the length of the duct and (b) the exit pressure. P9.96 Methane (CH4) flows through an insulated 15-cm-diameter pipe with / = 0.023. Entrance conditions are 600 kPa, 100°C, and a mass flow of 5 kg/s. What lengths of pipe will (a) choke the flow, (b) raise the velocity by 50 percent, or (c) decrease the pressure by 50 percent? P9.97 By making a few algebraic substitutions, show that Eq. (9.74) may be written in the density form P\ = pl + 2k ft k + \ D -f 2 In Why is this formula awkward if one is trying to solve for the mass flow when the pressures are given at sections 1 and 2? P9.98 Compressible laminar flow,/— 64/Re, may occur in capil- lary tubes. Consider air, at stagnation conditions of 100°C and 200 kPa, entering a tube 3 cm long and 0.1 mm in diameter. If the receiver pressure is near vacuum, estimate (a) the average Reynolds number, {b) the Mach number at the entrance, and (c) the mass flow in kg/h. P9.99 A compressor forces air through a smooth pipe 20 m long and 4 cm in diameter, as in Fig. P9.99. The air leaves at 101 kPa and 200°C. The compressor data for pressure rise versus mass flow are shown in the figure. Using the Moody chart to estimate /, compute the resulting mass flow. D = 4 cm P9.99 P9.100 Natural gas, approximated as CH4, flows through a Sched¬ ule 40 six-inch pipe from Providence to Narragansett, RI, a distance of 3 1 miles. Gas companies use the barg as a pres¬ sure unit, meaning a bar of pressure gage, above ambient pressure. Assuming isothermal flow at 68°F, with / = 0.019, estimate the mass flow if the pressure is 5 bargs in Providence and 1 barg in Narragansett. P9.101 How do the compressible pipe flow formulas behave for small pressure drops? Let air at 20°C enter a tube of diam¬ eter 1 cm and length 3 m. If / = 0.028 with pi = 102 kPa and P2 = 100 kPa, estimate the mass flow in kg/h for (a) isothermal flow, (b) adiabatic flow, and (c) incompress¬ ible flow (Chap. 6) at the entrance density. P9.102 Air at 550 kPa and 100°C enters a smooth 1-m-long pipe and then passes through a second smooth pipe to a 30-kPa reservoir, as in Fig. P9. 102. Using the Moody chart to com¬ pute /, estimate the mass flow through this system. Is the flow choked? Problems 673 : 30 kPa P9.102 P9.103 Natural gas, with k ~ 1.3 and a molecular weight of 16, is to he pumped through 100 km of 81-cm-diameter pipeline. The downstream pressure is 150 kPa. If the gas enters at 60°C, the mass flow is 20 kg/s, and f = 0.024, estimate the required entrance pressure for (a) isothermal flow and (b) adiabatic flow. P9.104 A tank of oxygen (Table A.4) at 20°C is to supply an astro- naut through an umhilical tube 12 m long and 1.5 cm in diameter. The exit pressure in the tube is 40 kPa. If the desired mass flow is 90 kg/h and / = 0.025, what should be the pressure in the tank? P9.105 Modify Prob. P9.87 as follows: The pipeline will not be allowed to choke. It will have pumping stations about every 200 miles, (a) Find the length of pipe for which the pressure has dropped to 2000 Ibf/in^. (&) What is the temperature at that point? P9.106 Air, from a 3 cubic meter tank initially at 300 kPa and 200°C, blows down adiabatically through a smooth pipe 1 cm in diameter and 2.5 m long. Estimate the time required to reduce the tank pressure to 200 kPa. For simplicity, assume constant tank temperature and/— 0.020. t = 0: 200° C 300 kPa 3 (1) P9.106 (2) = 100 kPa P9.109 A jet engine at 7000-m altitude takes in 45 kg/s of air and adds 550 kj/kg in the combustion chamber. The chamber cross section is 0.5 m^, and the air enters the chamber at 80 kPa and 5°C. After combustion the air expands through an isentropic converging nozzle to exit at atmospheric pres¬ sure. Estimate (a) the nozzle throat diameter, {b) the nozzle exit velocity, and (c) the thrust produced by the engine. P9.110 Compressible pipe flow with heat addition. Sec. 9.8, assumes constant momenrnm (p -I- pV^) and constant mass flow but variable stagnation enthalpy. Such a flow is often called Rayleigh flow, and a line representing all possible property changes on a temperamre-entropy chart is called a Rayleigh line. Assuming air passing through the flow state pi = 548 kPa, Ti = 588 K, Vi = 266 m/s, andA = 1 m^ draw a Rayleigh curve of the flow for a range of velocities from very low (Ma ^ 1) to very high (Ma 1). Comment on the meaning of the maximum-entropy point on this curve. P9.111 Add to your Rayleigh line of Prob. P9.110 a Fanno line (see Prob. P9.94) for stagnation enthalpy equal to the value associated with state 1 in Prob. P9. 110. The two curves will intersect at state 1, which is subsonic, and at a certain state 2, which is supersonic. Interpret these two states vis-a-vis Table B.2. P9.112 Air enters a duct at Vi = 144 m/s, pi = 200 kPa, and Tj = 323 K. Assuming frictionless heat addition, estimate (a) the heat addition needed to raise the velocity to 372 m/s; and (b) the pressure at this new section 2. P9.113 Air enters a constant-area duct at pi = 90 kPa, Vi = 520 m/s, and T, = 558°C. It is then cooled with negligible friction until it exits at P2 = 160 kPa. Estimate (a) V2, (b) T2, and (c) the total amount of cooling in kJ/kg. P9.114 The scramjet of Fig. C9.8 operates with supersonic flow throughout. Assume that the heat addition of 500 kJ/kg, between sections 2 and 3, is frictionless and at constant area of 0.2 m^. Given Ma2 = 4.0, p2 = 260 kPa, and T2 = 420 K. Assume airflow at fc = 1.40. At the combustion section exit, find {a) Ma3, (b) p3, and (c) T^. P9.115 Air enters a 5-cm-diameter pipe at 380 kPa, 3.3 kg/m^, and 120 m/s. Assume frictionless flow with heat addition. Find the amount of heat addition for which the velocity (a) doubles, (b) triples, and (c) quadruples. Frictionless flow with heat transfer P9.107 A fuel-air mixture, assumed equivalent to air, enters a duct combustion chamber at Vi = 104 m/s and Ti = 300 K. What amount of heat addition in kJ/kg will cause the exit flow to be choked? What will be the exit Mach number and temperature if 504 kJ/kg are added during combustion? P9.108 What happens to the inlet flow of Prob. P9.107 if the com¬ bustion yields 1500 kJ/kg heat addition and poi and Toi remain the same? How much is the mass flow reduced? Mach waves P9.116 An observer at sea level does not hear an aircraft flying at 12,000-ft standard altitude until it is 5 (statute) mi past her. Estimate the aircraft speed in ft/s. P9.117 A tiny scratch in the side of a supersonic wind tunnel creates a very weak wave of angle 17°, as shown in Fig. P9.1 17, after which a normal shock occurs. The air tem¬ perature in region (1) is 250 K. Estimate the temperature in region (2). 674 Chapter 9 Compressible Flow © Shock © P9.121 A thermistor probe, in the shape of a needle parallel to the flow, reads a static temperature of — 25°C when inserted into a supersonic airstream. A conical distur¬ bance cone of half-angle 17° is created. Estimate (a) the Mach number, (b) the velocity, and (c) the stagnation temperature of the stream. 17° P9.117 P9.118 A particle moving at uniform velocity in sea-level stan¬ dard air creates the two disturbance spheres shown in Fig. P9.118. Compute the particle velocity and Mach number. The oblique shock wave P9.122 Supersonic air takes a 5° compression turn, as in Fig. P9.122. Compute the downstream pressure and Mach number and the wave angle, and compare with small- disturbance theory. P9.119 The particle in Fig. P9.119 is moving supersonically in sea-level standard air. From the two given disturbance spheres, compute the particle Mach number, velocity, and Mach angle. P9.120 The particle in Fig. P9.120 is moving in sea-level stan¬ dard air. From the two disturbance spheres shown, estimate (a) the position of the particle at this instant and (b) the temperature in °C at the front stagnation point of the particle. P9.122 P9.123 The 10° deflection in Example 9.17 caused a final Mach number of 1.641 and a pressure ratio of 1.707. Compare this with the case of the flow passing through two 5° deflec¬ tions. Comment on the results and why they might be higher or lower in the second case. P9.124 When a sea-level flow approaches a ramp of angle 20°, an oblique shock wave forms as in Figure P9.124. Calculate (a) Mai, () P2, (c) T2, and (d) ¥2- P9.125 We saw in the text that, for k = 1.40, the maximum possi¬ ble deflection caused by an oblique shock wave occurs at infinite approach Mach number and is = 45.58°. Assuming an ideal gas, what is 6^^ for (a) argon and (b) carbon dioxide? Problems 675 P9.126 Airflow at Ma = 2.8, p = 80 kPa, and T = 280 K under¬ goes a 15° compression turn. Find the downstream values of (a) Mach number, (b) pressure, and (c) temperature. P9.127 Do the Mach waves upstream of an oblique shock wave intersect with the shock? Assuming supersonic down¬ stream flow, do the downstream Mach waves intersect the shock? Show that for small deflections the shock wave angle fi lies halfway between /ij and /i2 + 0 for any Mach number. P9.128 Air flows past a two-dimensional wedge-nosed body as in Fig. P9.128. Determine the wedge half-angle d for which the horizontal component of the total pressure force on the nose is 35 kN/m of depth into the paper. P9.129 Air flows at supersonic speed toward a compression ramp, as in Fig. P9. 129. A scratch on the wall at point a creates a wave of 30° angle, while the oblique shock created has a 50° angle. What is (a) the ramp angle 0 and (b) the wave angle (j) caused by a scratch at h? a P9.130 A supersonic airflow, at a temperature of 300 K, strikes a wedge and is deflected 12°. If the resulting shock wave is attached, and the temperature after the shock is 450 K, (a) estimate the approach Mach number and wave angle. (b) Why are there two solutions? P9.131 The following formula has been suggested as an alternate to Eq. (9.86) to relate upstream Mach number to the oblique shock wave angle /3 and turning angle 9: , 1 (k -I- 1) sin B sin 9 sin^/3 = - r -b Mai 2 cos (/3 - 9) Can you prove or disprove this relation? If not, try a few nu¬ merical values and compare with the results from Eq. (9.86). P9.132 Air flows at Ma = 3 and p = 10 Ibf/in^ absolute toward a wedge of 16° angle at zero incidence in Eig. P9.132. If the pointed edge is forward, what will be the pressure at point A? If the blunt edge is forward, what will be the pressure at point S? Ma = 3 p = 10 Ibf/in^ abs P9.132 P9.133 Air flows supersonically toward the double-wedge system in Fig. P9.133. The (x, y) coordinates of the tips are given. The shock wave of the forward wedge strikes the tip of the aft wedge. Both wedges have 15° deflection angles. What is the free-stream Mach number? P9.134 When an oblique shock strikes a solid wall, it reflects as a shock of sufficient strength to cause the exit flow Ma3 to be parallel to the wall, as in Fig. P9. 134. For airflow with Ma; = 2.5 and pi = 100 kPa, compute Ma3, p^, and the angle (p. P9.134 P9.135 A bend in the bottom of a supersonic duct flow induces a shock wave that reflects from the upper wall, as in Fig. P9.135. Compute the Mach number and pressure in region 3. 676 Chapter 9 Compressible Flow P9.135 P9.136 Figure P9.136 is a special application of Prob. P9.135. With careful design, one can orient the bend on the lower wall so that the reflected wave is exactly canceled by the return bend, as shown. This is a method of reducing the Mach num¬ ber in a channel (a supersonic diffuser). If the bend angle is (p= 10°, find (a) the downstream width h and (b) the down¬ stream Mach number. Assume a weak shock wave. P9.136 P9.137 A 6° half-angle wedge creates the reflected shock system in Fig. P9.137. If Ma3 = 2.5, find (a) Mai and (b) the angle a. P9.137 P9.138 The supersonic nozzle of Fig. P9.138 is overexpanded (case G of Fig. 9.12b) with AJA, = 3.0 and a stagnation pressure of 350 kPa. If the jet edge makes a 4° angle with the nozzle centerline, what is the back pressure in kPa? P9.139 Airflow at Ma = 2.2 takes a compression turn of 12° and then another turn of angle 0 in Fig. P9.139. What is the maximum value of 6 for the second shock to be attached? Will the two shocks intersect for any 9 less than ©max? P9.139 Prandtl-Meyer expansion waves P9.140 The solution to Prob. P9.122 is Ma2 = 2.750 and p2 = 145.5 kPa. Compare these results with an isentropic com¬ pression turn of 5°, using Prandtl-Meyer theory. P9.141 Supersonic airflow takes a 5° expansion turn, as in Fig. P9.141. Compute the downstream Mach number and pressure, and compare with small-disturbance theory. P9.141 P9.142 A supersonic airflow at Mai = 3.2 andpi = 50 kPa under¬ goes a compression shock followed by an isentropic expan¬ sion turn. The flow deflection is 30° for each turn. Compute Ma2 and p2 if (a) the shock is followed by the expansion and (b) the expansion is followed by the shock. P9.143 Airflow at Ma = 3.4 and 300 K encounters a 28° oblique shock turn. What subsequent isentropic expansion turn will bring the temperature back to 300 K? Problems 677 P9.144 The 10° deflection in Example 9.17 caused the Mach number to drop to 1.64. (a) What turn angle will create a Prandtl-Meyer fan and bring the Mach number back up to 2.0? (b) What will be the final pressure? P9.145 Air at Mai = 2.0 and pi = 100 kPa undergoes an isentropic expansion to a downstream pressure of 50 kPa. What is the desired turn angle in degrees? P9.146 Air flows supersonically over a surface that changes direc¬ tion twice, as in Fig. P9.146. Calculate (a) Ma2 and (b) p^. P9.146 P9.147 A converging-diverging nozzle with a 4: 1 exit-area ratio and po = 500 kPa operates in an underexpanded condition (case I of Fig. 9.12b) as in Fig. P9.147. The receiver pres¬ sure is = 10 kPa, which is less than the exit pressure, so that expansion waves form outside the exit. For the given conditions, what will the Mach number Ma2 and the angle (p of the edge of the jet be? Assume k = 1 .4 as usual. P9.147 n = 10 kPa ^ a Jet edge Jet edge P9.148 Air flows supersonically over a circular-arc surface as in Fig. P9. 148. Estimate (a) the Mach number Ma2 and (b) the pressure p2 as the flow leaves the circular surface. Supersonic airfoils P9.149 Air flows at Ma^j, = 3.0 past a doubly symmetric diamond airfoil whose front and rear included angles are both 24°. For zero angle of attack, compute the drag coefficient obtained using shock-expansion theory and compare with Ackeret theory. P9.150 A flat-plate airfoil with C = 1.2 m is to have a lift of 30 kN/m when flying at 5000-m standard altitude with U„= 641 m/s. Using Ackeret theory, estimate (a) the angle of attack and (b) the drag force in N/m. P9.151 Air flows at Ma = 2.5 past a half-wedge airfoil whose angles are 4°, as in Fig. P9.151. Compute the lift and drag coefficient at a. equal to (a) 0° and (b) 6°. Ma„„ = 2.5 P9.151 P9.152 The X-43 model A scramjet aircraft in Fig. C9.8 is small W = 3000 Ibf, and unmanned, only 12.33 ft long and 5.5 ft wide. The aerodynamics of a slender arrowhead-shaped hypersonic vehicle is beyond our scope. Instead, let us assume it is a flat plate airfoil of area 2.0 m^. Let Ma = 7 at 12,000 m standard altitude. Estimate the drag, by shock- expansion theory. Hint: Use Ackeret theory to estimate the angle of attack. P9.153 A supersonic transport has a mass of 65 Mg and cruises at 11-km standard altitude at a Mach number of 2.25. If the angle of attack is 2° and its wings can be approximated by flat plates, estimate (a) the required wing area in m^ and (b) the thrust required in N. P9.154 The F-22 supersonic fighter cruises at 11,000 m altitude, with a weight of 50,000 Ibf and thrust of 10,000 Ibf. Its wing area is 840 ft^. Assume the wing is a 6-percent-thick diamond shape and provides all lift and thrust. Use Ackeret theory to estimate the resulting Mach number. P9.155 The F-35 airplane in Fig. 9.30 has a wingspan of 10 m and a wing area of 41.8 m^. It cruises at about 10 km altitude with a gross weight of about 200 kN. At that altitude, the engine develops a thrust of about 50 kN. Assume the wing has a symmetric diamond airfoil with a thickness of 8 per¬ cent, and accounts for all lift and drag. Estimate the cruise Mach number of the airplane. For extra credit, explain why there are two solutions. P9.156 Consider a flat-plate airfoil at an angle of attack of 6°. The Mach number is Max. = 3.2 and the stream pressure p„ is unspecified. Calculate the predicted lift and drag coefficients by (a) shock-expansion theory and (b) Ackeret theory. P9.157 The Ackeret airfoil theory of Eq. (9. 104) is meant for ?nod- erate supersonic speeds, 1.2 < Ma < 4. How does it fare for hypersonic speeds? To illustrate, calculate {a) Ci and (b) Co for a flat-plate airfoil at a = 5° and Ma = 8.0, using shock-expansion theory, and compare with Ackeret theory. Comment. 678 Chapter 9 Compressible Flow Word Problems W9.1 Notice from Table 9.1 that (a) water and mercury and (b) aluminum and steel have nearly the same speeds of sound, yet the second of each pair of materials is much denser. Can you account for this oddity? Can molecular theory explain it? W9.2 When an object approaches you at Ma = 0.8, you can hear it, according to Fig. 9.18a. But would there be a Doppler shift? For example, would a musical tone seem to you to have a higher or a lower pitch? W9.3 The subject of this chapter is commonly called gas dynam¬ ics. But can liquids not perform in this manner? Using water as an example, make a rule-of-thumb estimate of the pressure level needed to drive a water flow at velocities comparable to the sound speed. W9.4 Suppose a gas is driven at compressible subsonic speeds by a large pressure drop, pi to p2- Describe its behavior on an appropriately labeled Mollier chart for (a) frictionless Fundamentals of Engineering Exam Problems One-dimensional compressible flow problems have become quite popular on the FE Exam, especially in the afternoon sessions. In the following problems, assume one-dimensional flow of ideal air, R = 287 J/(kg ■ K) and k = 1.4. FE9.1 Eor steady isentropic flow, if the absolute temperature in¬ creases 50 percent, by what ratio does the static pressure increase? (a) 1.12, (b) 1.22, (c) 2.25, (d) 2.76, (e) 4.13 FE9.2 Eor steady isentropic flow, if the density doubles, by what ratio does the static pressure increase? (a) 1.22, (b) 1.32, (c) 1.44, (d) 2.64, (e) 5.66 EE9.3 A large tank, at 500 K and 200 kPa, supplies isentropic airflow to a nozzle. At section 1, the pressure is only 120 kPa. What is the Mach number at this section? (a) 0.63, (b) 0.78, (c) 0.89, (d) 1.00, (e) 1.83 EE9.4 In Prob. FE9.3 what is the temperature at section 1? (a) 300 K, (b) 408 K, (c) 417 K, (d) 432 K, (e) 500 K EE9.5 In Prob. PE9.3, if the area at section 1 is 0.15 m^, what is the mass flow? (a) 38.1 kg/s, (b) 53.6 kg/s, (c) 57.8 kg/s, (d) 67.8 kg/s, (e) 77.2 kg/s EE9.6 Eor steady isentropic flow, what is the maximum possible mass flow through the duct in Eig. EE9.6? (a) 9.5 kg/s, (b) 15.1 kg/s, (c) 26.2 kg/s, (d) 30.3 kg/s, (e) 52.4 kg/s flow in a converging nozzle and (b) flow with friction in a long duct. W9.5 Describe physically what the “speed of sound” represents. What kind of pressure changes occur in air sound waves during ordinary conversation? W9.6 Give a physical description of the phenomenon of choking in a converging-nozzle gas flow. Could choking happen even if wall friction were not negligible? W9.7 Shock waves are treated as discontinuities here, but they actually have a very small finite thickness. After giving it some thought, sketch your idea of the distribution of gas velocity, pressure, temperature, and entropy through the inside of a shock wave. W9.8 Describe how an observer, running along a normal shock wave at finite speed V, will see what appears to be an oblique shock wave. Is there any limit to the running speed? EE9.7 If the exit Mach number in Eig. EE9.6 is 2.2, what is the exit area? (a) 0.10 m^ (fe) 0.12 m^ fc) 0.15 m^ {d) 0.18 m^ (e) 0.22 m^ EE9.8 If there are no shock waves and the pressure at one duct section in Eig. FE9.6 is 55.5 kPa, what is the velocity at that section? (a) 166 m/s, (/>) 232 m/s, (c) 554 m/s, {d) 706 m/s, (e) 774 m/s EE9.9 If, in Eig. EE9.6, there is a normal shock wave at a section where the area is 0.07 m^, what is the air density just up¬ stream of that shock? (a) 0.48 kg/m^ {b) 0.78 kg/m^ (c) 1.35 kg/m^ (if) 1.61 kg/m^, (e) 2.61 kg/m^ EE9.10 In Prob. PE9.9, what is the Mach number just downstream of the shock wave? (a) 0.42, (b) 0.55, (c) 0.63, (d) 1.00, (e) 1.76 Comprehensive Problems 679 Comprehensive Problems C9.1 The converging-diverging nozzle sketched in Fig. C9.1 is designed to have a Mach number of 2.00 at the exit plane (assuming the flow remains nearly isentropic). The flow trav¬ els from tank a to tank b, where tank a is much larger than tank b. (a) Find the area at the exit and the back pressure Pj that will allow the system to operate at design conditions. (b) As time goes on, the back pressure will grow, since the second tank slowly fills up with more air. Since tank a is huge, the flow in the nozzle will remain the same, however, until a normal shock wave appears at the exit plane. At what back pressure will this occur? (c) If tank b is held at constant temperature, T = 20°C, estimate how long it will take for the flow to go from design conditions to the condition of part (b) — that is, with a shock wave at the exit plane. C9.2 Two large air tanks, one at 400 K and 300 kPa and the other at 300 K and 100 kPa, are connected by a straight tube 6 m long and 5 cm in diameter. The average friction factor is 0.0225. Assuming adiabatic flow, estimate the mass flow through the tube. C9.3 Figure C9.3 shows the exit of a converging-diverging noz¬ zle, where an oblique shock pattern is formed. In the exit plane, which has an area of 15 cm^, the air pressure is 16 kPa and the temperature is 250 K. lust outside the exit shock, which makes an angle of 50° with the exit plane, the tem¬ perature is 430 K. Estimate (a) the mass flow, (b) the throat area, (c) the turning angle of the exit flow, and, in the tank supplying the air, (d) the pressure and (e) the temperature. C9.4 The properties of a dense gas (high pressure and low temperature) are often approximated by van der Waals’s equation of state [17, 18]: pRT where constants Oi and bi can be found from the critical temperature and pressure 97 Oi = — - ^ = 9.0 X 10^ Ibf • ft^/slug- 64p, for air, and RT bi = — - = 0.65 ftVslug iPc for air. Find an analytic expression for the speed of sound of a van der Waals gas. Assuming k = 1.4, compute the speed of sound of air in ft/s at — 100°F and 20 atm for (a) a perfect gas and (fe) a van der Waals gas. What percentage higher density does the van der Waals rela¬ tion predict? C9.5 Consider one-dimensional steady flow of a nonideal gas, steam, in a converging nozzle. Stagnation conditions are Po = 100 kPa and Tq = 200°C. The nozzle exit diameter is 2 cm. {a) If the nozzle exit pressure is 70 kPa, calculate the mass flow and the exit temperature for real steam from the steam tables. (As a first estimate, assume steam to be an ideal gas from Table A.4.) Is the flow choked? {b) Find the nozzle exit pressure and mass flow for which the steam flow is choked, using the steam tables. C9.6 Extend Prob. C9.5 as follows: Let the nozzle be converging- diverging, with an exit diameter of 3 cm. Assume isentro¬ pic flow. Find the exit Mach number, pressure, and temperature for an ideal gas from Table A.4. Does the mass flow agree with the value of 0.0452 kg/s in Prob. C9.5? C9.7 Professor Gordon Holloway and his student, Jason Settle, of the University of New Brunswick obtained the following tabulated data for blow-down airflow through a converging- diverging nozzle similar in shape to Fig. P3.22. The supply tank pressure and temperature were 29 psig and 74°F, respectively. Atmospheric pressure was 14.7 psia. Wall pressures and centerline stagnation pressures were measured 680 Chapter 9 Compressible Flow in the expansion section, which was a frustrum of a cone. The nozzle throat is at x = 0. jc(cm) 0 1.5 3 4.5 6 7.5 9 Diameter (cm) 1.00 1.098 1.195 1.293 1.390 1.488 1.585 FwallCpsig) 7.7 -2.6 -4.9 -7.3 -6.5 -10.4 -7.4 /^Stagnation (psi§) 29 26.5 22.5 18 16.5 14 10 Use the stagnation pressure data to estimate the local Mach number. Compare the measured Mach numbers and wall pressures with the predictions of one-dimensional theory. For X > 9 cm, the stagnation pressure data was not thought by Holloway and Settle to be a valid measure of Mach number. What is the probable reason? C9.8 Engineers call the supersonic combustion in a scramjet engine almost miraculous, “like lighting a match in a hur¬ ricane.” Figure C9.8 is a crude idealization of the engine. Air enters, bums fuel in the narrow section, then exits, all at supersonic speeds. There are no shock waves. Assume ar¬ eas of 1 m^ at sections 1 and 4 and 0.2 m^ at sections 2 and 3. Let the entrance conditions be Mai = 6, at 10,000 m standard altitude. Assume isentropic flow from 1 to 2, frictionless heat transfer from 2 to 3 with Q = 500 kJ/kg, and isentropic flow from 3 to 4. Calculate the exit condi¬ tions and the thrust produced. Design Projects D9.1 It is desired to select a rectangular wing for a fighter aircraft. The plane must be able (a) to take off and land on a 4500-ft-long sea-level runway and (b) to cruise supersonically at Ma = 2.3 at 28,000-ft altitude. For simplicity, assume a wing with zero sweepback. Let the aircraft maximum weight equal (30 -I- m)(1000) Ibf, where n is the number of letters in your surname. Let the available sea-level maximum thrust be one-third of the maximum weight, decreasing at altitude proportional to ambient density. Making suitable assumptions about the effect of finite aspect ratio on wing lift and drag for both subsonic and supersonic flight, select a wing of mini¬ mum area sufficient to perform these takeoff/landing and cruise requirements. Some thought should be given to analyzing the wingtips and wing roots in supersonic flight, where Mach cones form and the flow is not two- dimensional. If no satisfactory solution is possible, grad¬ ually increase the available thrust to converge to an acceptable design. D9.2 Consider supersonic flow of air at sea-level conditions past a wedge of half-angle 0, as shown in Fig. D9.2. Assume that the pressure on the back of the wedge equals the fluid pressure as it exits the Prandtl-Meyer fan. (a) Suppose Ma = 3.0. For what angle 0 will the super¬ sonic wave drag coefficient C^, based on frontal area, be exactly 0.5? (b) Suppose that 0 = 20°. Is there a free- stream Mach number for which the wave drag coeffi¬ cient Co, based on frontal area, will be exactly 0.5? (c) Investigate the percentage increase in Co from (a) and (b) due to including boundary layer friction drag in the calculation. D9.2 References 681 References 1. J. E. A. John and T. G. Keith, Gas Dynamics, 3d ed., Pearson Education, Upper Saddle River, NJ, 2005. 2. B. K. Hodge and K. Koenig, Compressible Fluid Dynamics: With Personal Computer Applications, Pearson Prentice-Hall, Upper Saddle River, NJ, 1995. 3. R. D. Zucker and O. Biblarz, Fundamentals of Gas Dynamics, 2d ed., Wiley, New York, 2002. 4. J. D. Anderson, Modern Compressible Flow: with Historical Perspective, 3d ed., McGraw-Hill, New York, 2002. 5. A. H. Shapiro, The Dynamics and Thermodynamics of Compressible Fluid Flow, 2 vols., Wiley, New York, 1953. 6. C. Cercignani, Rarefied Gas Dynamics, Cambridge University Press, New York, 2000. 7. H. W. Liepmann and A. Roshko, Elements of Gas Dynamics, Dover, New York, 2001. 8. I. Straskraba, Introduction to the Mathematical Theory of Compressible Flow, Oxford University Press, New York, 2004. 9. P. A. Thompson, Compressible Fluid Dynamics, McGraw-Hill, New York, 1972. 10. P. H. Oosthuizen and W. E. Carscallen, Compressible Fluid Flow, McGraw-Hill, New York, 2003. 11. J. D. Anderson, Hypersonic and High Temperature Gas Dynamics, 2d ed., AIAA, Reston, VA, 2006. 12. G. A. Bird, Molecular Gas Dynamics and the Direct Simula¬ tion of Gas Flows, Clarendon Press, Oxford, 1994. 13. D. D. Knight, Elements of Numerical Methods for Com¬ pressible Flows, Cambridge University Press, New York, 2012. 14. L. W. Reithmaier, Mach 1 and Beyond: The Illustrated Guide to High-Speed Flight, McGraw-Hill, 1994. 15. W. T. Parry, ASME International Steam Tables for Industrial Use, 2d ed., ASME, New York, 2009. 16. J. H. Keenan et ah. Gas Tables: International Version, Krieger Publishing, Melbourne, FL, 1992. 17. Y. A. Cengel and M. A. Boles, Thermodynamics: An Engi¬ neering Approach, 7th ed., McGraw-Hill, New York, 2010. 18. M. J. Moran and H. A. Shapiro, Fundamentals of Engineer¬ ing Thermodynamics, 7th ed., Wiley, New York, 2010. 19. F. M. White, Viscous Eluid Flow, 3d ed., McGraw-Hill, New York, 2005. 20. J. Palmer, K. Ramsden, and E. Goodger, Compressible Flow Tables for Engineers: With Appropriate Computer Programs, Scholium Inti., Port Washington, NY, 1989. 21. M. R. Lindeburg, Consolidated Gas Dynamics Tables, Professional Publications, Inc., Belmont, CA, 1994. 22. A. M. Shektman, Gasdynamic Functions of Real Gases, Taylor and Francis, New York, 1991. 23. J. H. Keenan and E. P. Neumann, “Measurements of Friction in a Pipe for Subsonic and Supersonic Flow of Air,” J. Applied Mechanics, vol. 13, no. 2, 1946, p. A-91. 24. R. P. Benedict, Fundamentals of Pipe Flow, John Wiley, New York, 1980. 25. J. L. Sims, Tables for Supersonic Flow around Right Circular Cones at Zero Angle of Attack, NASA SP-3004, 1964 (see also NASA SP-3007). 26. J. L. Thomas, “Reynolds Number Effects on Supersonic Asymmetrical Flows over a Cone,” J. Aircraft, vol. 30, no. 4, 1993, pp. 488-495. 27. W. Bober and R. A. Kenyon, Fluid Mechanics, Wiley, New York, 1980. 28. J. Ackeret, “Air Forces on Airfoils Moving Faster than Sound Velocity,” AACA Tech. Memo. 317, 1925. 29. W. B. Brower, Theory, Tables and Data for Compressible Flow, Taylor & Francis, New York, 1990. 30. M. Belfiore, “The Hypersonic Age is Near,” Popular Science, January 2008, pp. 36-41. 31. G. S. Settles, Schlieren and Shadowgraph Techniques: Visu¬ alizing Phenomena in Transparent Media, Springer-Verlag, Berlin, 2001. In March of 2010, the normally placid Saugatucket River, in South Kingstown, Rhode Island, was flooded by heavy rains. Instead of its usual gentle trickle over the Main Street dam, the flow rate was enormously increased and flooded the medical building in the background, ruin¬ ing their offices and X-ray machines. The open-channel flow analysis methods in this chapter can handle both the trickle and the flood. [Photo courtesy of Independent Newspapers} 682 10.1 Introduction Chapter 10 Open- Channel Flow Motivation. An open-channel flow denotes a flow with a free surface touching an atmosphere, like a river or a canal or a flume. Closed-duct flows (Chap. 6) are full of fluid, either liquid or gas, have no free surface within, and are driven by a pressure gradient along the duct axis. The open-channel flows here are driven by gravity alone, and the pressure gradient at the atmospheric interface is negligible. The basic force balance in an open channel is between gravity and friction. Open-channel flows are an especially important mode of fluid mechanics for civil and environmental engineers. One needs to predict the flow rates and water depths that result from a given channel geometry, whether natural or artificial, and a given wet-surface roughness. Water is almost always the relevant fluid, and the channel size is usually large. Thus open-channel flows are generally turbulent, three-dimensional, sometimes unsteady, and often quite complex. This chapter presents some simple engineering theories and experimental correlations for steady flow in straight channels of regular geometry. We can borrow and use some concepts from duct flow analysis: hydraulic radius, friction factor, and head losses. Simply stated, open-channel flow is the flow of a liquid in a conduit with a free surface. There are many practical examples, both artificial (flumes, spillways, canals, weirs, drainage ditches, culverts) and natural (streams, rivers, estuaries, floodplains). This chapter introduces the elementary analysis of such flows, which are dominated by the effects of gravity. The presence of the free surface, which is essentially at atmospheric pressure, both helps and hurts the analysis. It helps because the pressure can be taken as constant along the free surface, which therefore is equivalent to the hydraulic grade line (HGL) of the flow. Unlike flow in closed ducts, the pressure gradient is not a direct factor in open-channel flow, where the balance of forces is confined to gravity and friction. But the free surface complicates the analysis because its shape is a priori unknown: 'Surface tension is rarely important because open channels are normally quite large and have a very large Weher number. Surface tension affects small models of large channels. 683 684 Chapter 10 Open-Channel Flow The One-Dimensional Approximation The depth profile changes with conditions and must be computed as part of the prob¬ lem, especially in unsteady problems involving wave motion. Before proceeding, we remark, as usual, that whole books have been written on open-channel hydraulics [1 to 7, 32]. There are also specialized texts devoted to wave motion [8 to 10] and to engineering aspects of coastal free-surface flows [11 to 13]. This chapter is only an introduction to broader and more detailed treatments. The writer recommends, as an occasional break from free-surface flow analysis. Ref. 31, which is an enchanting and spectacular gallery of ocean wave photographs. An open channel always has two sides and a bottom, where the flow satisfies the no-slip condition. Therefore, even a straight channel has a three-dimensional velocity distribution. Some measurements of straight-channel velocity contours are shown in Fig. 10.1. The profiles are quite complex, with maximum velocity typically occurring in the midplane about 20 percent below the surface. In very broad shallow channels the maximum velocity is near the surface, and the velocity profile is nearly logarithmic from the bottom to the free surface, as in Eq. (6.62). In noncircular channels there are also secondary motions similar to Fig. 6.16 for closed-duct flows. If the channel curves or meanders, the secondary motion intensifies due to centrifugal effects, with high velocity occurring near the outer radius of the bend. Curved natural channels are subject to strong bottom erosion and deposition effects. With the advent of the supercomputer, it is possible to make numerical simulations of complex flow patterns such as those in Fig. 10.1 [27, 28]. However, the practical engineering approach, used here, is to make a one-dimensional flow approximation, as in Fig. 10.2. Since the liquid density is nearly constant, the steady flow continuity equation reduces to constant-volume flow Q along the channel Q = V{x)A{x) = const (10.1) where V is average velocity and A the local cross-sectional area, as sketched in Fig. 10.2. A second one-dimensional relation between velocity and channel geometry is the energy equation, including friction losses. If points 1 (upstream) and 2 (downstream) are on the free surface, P\ = P2 — Pa^ and we have, for steady flow, y 2 y 2 -^ + z, = ^ + Z2 + hf (10.2) 2g 2g where z denotes the total elevation of the free surface, which includes the water depth y (see Fig. 10.2fl) plus the height of the (sloping) bottom. The friction head loss hf is analogous to head loss in duct flow from Eq. (6.10): X2 — X. Vh AA hf~f — — - — Dh = hydraulic diameter = — (10.3) where / is the average friction factor (Fig. 6.13) between sections 1 and 2. Since channels are irregular in shape, their “size” is taken to be the hydraulic radius: (10.4) 10.1 Introduction 685 Fig. 10.1 Measured isovelocity contours in typical straight open- channel flows. (From Ref. 2.) Source: From V. T. Chow, Open Channel Hydraulics, Blackburn Press, Caldwell, NJ, 2009. Fig. 10.2 Geometry and notation for open-channel flow: (a) side view; (b) cross section. All these parameters are constant in uniform flow. Triangular channel Narrow rectangular section 686 Chapter 10 Open-Channel Flow Flow Classification by Depth Variation Fig. 10.3 Open-channel flow classified hy regions of rapidly varying flow (RVF), gradually varying flow (GVF), and uniform flow depth profiles. The local Reynolds number of the channel would be Re = VRi,lu, which is usually highly turbulent (>1 E5). The only commonly occurring laminar channel flows are the thin sheets that form as rainwater drains from crowned streets and airport runways. The wetted perimeter P (see Fig. \0.2b) includes the sides and bottom of the chan¬ nel but not the free surface and, of course, not the parts of the sides above the water level. For example, if a rectangular channel is b wide and h high and contains water to depth y, its wetted perimeter is P = b + 2y not 2b + 2h. Although the Moody chart (Fig. 6.13) would give a good estimate of the friction factor in channel flow, in practice it is seldom used. An alternative correlation due to Robert Manning, discussed in Sec. 10.2, is the formula of choice in open-channel hydraulics. The most common method of classifying open-channel flows is by the rate of change of the free-surface depth. The simplest and most widely analyzed case is uniform flow, where the depth and area (hence the velocity in steady flow) remain constant. Uniform flow conditions are approximated by long, straight runs of constant-slope and constant- area channel. A channel in uniform flow is said to be moving at its normal depth y„, which is an important design parameter. If the channel slope or cross section changes or there is an obstruction in the flow, then the depth changes and the flow is said to be varied. The flow is gradually vary¬ ing if the one-dimensional approximation is valid and rapidly varying if not. Some examples of this method of classification are shown in Fig. 10.3. The classes can be summarized as follows: 1. Uniform flow (constant depth and slope) 2. Varied flow: a. Gradually varied (one-dimensional) b. Rapidly varied (multidimensional) GVF 10.1 Introduction 687 Typically uniform flow is separated from rapidly varying flow by a region of gradu¬ ally varied flow. Gradually varied flow can be analyzed by a first-order differential equation (Sec. 10.6), but rapidly varying flow usually requires experimentation or three-dimensional computational fluid dynamics [14, 27, 28]. Flow Classification by Fronde Number Surface Wave Speed A second and very useful classification of open-channel flow is hy the dimensionless Froude number, Fr, which is the ratio of channel velocity to the speed of propagation of a small-disturhance wave in the channel. For a rectangular or very wide constant- depth channel, this takes the form flow velocity Fr = surface wave speed V (10.5) where y is the water depth. The flow behaves differently depending on these three flow regimes: Fr < 1.0 subcritical flow Fr = 1.0 critical flow (10.6) Fr > 1.0 supercritical flow The Froude number for irregular channels is defined in Sec. 10.4. As mentioned in Sec. 9.10, there is a strong analogy here with the three compressible flow regimes of the Mach number: subsonic (Ma < 1), sonic (Ma = 1), and supersonic (Ma > 1). We shall use the analogy in Secs. 10.4 and 10.5. The analogy is pursued in Ref. 21. The Froude number denominator (gy)^'^ is the speed of an infinitesimal shallow- water surface wave. We can derive this with reference to Fig. 10.4a, which shows a wave of height ^ propagating at speed c into still liquid. To achieve a steady flow inertial frame of reference, we fix the coordinates on the wave as in Fig. 10.4h, so that the still water moves to the right at velocity c. Figure 10.4 is exactly analogous to Fig. 9.1, which analyzed the speed of sound in a fluid. It can be used to analyze tidal bores. Fig. P10.86, which are described by Chanson . c Fig. 10.4 Analysis of a small surface wave propagating into still shallow water; (a) moving wave, nonsteady frame; (b) fixed wave, inertial frame of reference. Still j, water - 5V (a) Fixed wave Sy (b) 688 Chapter 10 Open-Channel Flow For the control volume of Fig. IQAb, the one-dimensional continuity relation is, for channel width b. or pcby = p{c — ^)(y + Sy)b 6V = dy c - ^ y + dy (10.7) This is analogous to Eq. (9.10); the velocity change SV induced by a surface wave is small if the wave is “weak,” ^ y. If we neglect bottom friction in the short dis¬ tance across the wave in Fig. 10.4fc, the momentum relation is a balance between the net hydrostatic pressure force and momentum: -hpsb[{y + 6yf - y^] = pcby{c - 5V - c) g 1 + dy = c dV (10.8) This is analogous to Eq. (9.12). By eliminating JV between Eqs. (10.7) and (10.8) we obtain the desired expression for wave propagation speed: c = gy\ 1+^ y 1 + y (10.9) The “stronger” the wave height dy, the faster the wave speed c, by analogy with Eq. (9.13). In the limit of an infinitesimal wave height dy ^ 0, the speed becomes c6 = gy (10.10) This is the surface-wave equivalent of fluid sound speed a, and thus the Eroude number in channel flow Fr = V/cq is the analog of the Mach number. For y = 1 m, Co = 3.1 m/s. As in gas dynamics, a channel flow can accelerate from subcritical to critical to supercritical flow and then return to subcritical flow through a sort of normal shock called a hydraulic jump (Sec. 10.5). This is illustrated in Fig. 10.5. The flow upstream of the sluice gate is subcritical. It then accelerates to critical and supercritical flow as it passes under the gate, which serves as a sort of “nozzle.” Further downstream the Fig. 10.5 Flow under a sluice gate accelerates from subcritical to critical to supercritical flow and then jumps back to subcritical flow. X Subcritical Sluice ' gate Hydraulic jump \b^g) J Supercritical Subcritical / 10.2 Uniform Flow; The Chezy Formula 689 10.2 Uniform Flow; The Chezy Formula flow “shocks” back to subcritical flow because the downstream “receiver” height is too high to maintain supercritical flow. Note the similarity with the nozzle gas flows of Fig. 9.12. The critical depth is sketched as a dashed line in Fig. 10.5 for reference. Like the normal depth y„, y^ is an important parameter in characterizing open-channel flow (see Sec. 10.4). An excellent discussion of the various regimes of open-channel flow is given in Ref. 15. Uniform flow can occur in long, straight runs of constant slope and constant channel cross section. The water depth is constant at y = y„, and the velocity is constant at V = Vq. Let the slope be = tan 0, where 6 is the angle the bottom makes with the horizontal, considered positive for downhill flow. Then Eq. (10.2), with Vi = V2 = Vq, becomes hf = Zi - Z2 = SqL (10.11) where L is the horizontal distance between sections 1 and 2. The head loss thus bal¬ ances the loss in height of the channel. The flow is essentially fully developed, so the Darcy-Weisbach relation, Eq. (6.10), holds hf=f—iz^ Dh = ^Rh (10.12) Dh 2g with D), = 4A/P used to accommodate noncircular channels. The geometry and notation for open-channel flow analysis are shown in Eig. 10.2. By combining Eqs. (10.11) and (10.12) we obtain an expression for flow velocity in uniform channel flow; Vo 1/2 r>l/2ol/2 ■ff/i Jo (10.13) For a given channel shape and bottom roughness, the quantity (8g//)^^ is constant and can be denoted by C. Equation (10.13) becomes To = C(R„5o)''' Q = CA(R,SQy'^ r,,, = pgR^So (10.14) These are called the Chezy formulas, first developed by the Erench engineer Antoine Chezy in conjunction with his experiments on the Seine River and the Courpalet Canal in 1769. The quantity C, called the Chezy coefficient, varies from about 60 ft'^^/s for small, rough channels to 160 ft^^^/s for large, smooth channels (30 to 90 m'^^/s in SI units). Over the past century a great deal of hydraulics research has been devoted to the correlation of the Chezy coefficient with the roughness, shape, and slope of vari¬ ous open channels. Correlations are due to Ganguillet and Kutter in 1869, Manning in 1889, Bazin in 1897, and Powell in 1950 . All these formulations are discussed in delicious detail in Ref. 2, Chap. 5. Here we confine our treatment to Manning’s correlation, the most popular. 690 Chapter 10 Open-Channel Flow EXAMPLE 10.1 A straight rectangular channel is 6 ft wide and 3 ft deep and laid on a slope of 2°. The friction factor is 0.022. Estimate the uniform flow rate in cubic feet per second. Solution • System sketch: The channel cross section is shown in Fig. ElO.l. • Assumptions: Steady, uniform channel flow with 0 = 2°. • Approach: Evaluate the Chezy formula, Eq. (10.13) or (10.14). • Property values: Please note that there are no fluid physical properties involved in the Chezy formula. Can you explain this? ■ Solution step: Simply evaluate each term in the Chezy formula, Eq. (10.13): C = 8(32.2 ft/s^) ft ^ ^ = 108 — 0.022 s 1/2 A = by = (6ft)(3ft) = 18 Rh 18 ft- (3 + 6 + 3) ft = 1.5 ft So = tan(0) = tan(2°) / fti«\ Then Q = CAR]^^So^ = ( 108— j(18ft-)(1.5ft)'“(tan 2°)^'^ « 450 ftVs Ans. • Comments: Uniform flow estimates are straightforward if the geometry is simple. Results are independent of water density and viscosity because the flow is fully rough and driven by grav¬ ity. Note the high flow rate, larger than some rivers. Two degrees is a substantial channel slope. The Manning Roughness Correlation The most fundamentally sound approach to the Chezy formula is to use Eq. (10.13) with / estimated from the Moody friction factor chart. Fig. 6.13. Indeed, the open- channel research establishment strongly recommends use of the friction factor in all calculations. Since typical channels are large and rough, we would generally use the fully rough turbulent flow limit of Eq. (6.48) /- (2.0 log 14^y^ (10.15) where e is the roughness height, with typical values listed in Table 10.1. In spite of the attractiveness of this friction factor approach, most engineers prefer to use a simple (dimensional) correlation published in 1891 by Robert Manning , an Irish engineer. In tests with real channels, Manning found that the Chezy coeffi¬ cient C increased approximately as the sixth root of the channel size. He proposed the simple formula C = (10.16) where « is a roughness parameter. Since the formula is clearly not dimensionally consistent, it requires a conversion factor a that changes with the system of units used: a = 1.0 SI units a = 1.486 BG units (10.17) 10.2 Uniform Flow; The Chezy Formula 691 Table 10.1 Experimental Values of Manning’s n Factor Average roughness height e n ft mm Artificial lined channels: Glass 0.010 ± 0.002 0.0011 0.3 Brass 0.011 ± 0.002 0.0019 0.6 Steel, smooth 0.012 ± 0.002 0.0032 1.0 Painted 0.014 ± 0.003 0.0080 2.4 Riveted 0.015 ± 0.002 0.012 3.7 Cast iron 0.013 ± 0.003 0.0051 1.6 Concrete, finished 0.012 ± 0.002 0.0032 1.0 Unfinished 0.014 ± 0.002 0.0080 2.4 Planed wood 0.012 ± 0.002 0.0032 1.0 Clay tile 0.014 ± 0.003 0.0080 2.4 Brickwork 0.015 ± 0.002 0.012 3.7 Asphalt 0.016 ± 0.003 0.018 5.4 Corrugated metal 0.022 ± 0.005 0.12 37 Rubble masonry 0.025 ± 0.005 0.26 80 Excavated earth channels: Clean 0.022 ± 0.004 0.12 37 Gravelly 0.025 ± 0.005 0.26 80 Weedy 0.030 ± 0.005 0.8 240 Stony, cobbles 0.035 ± 0.010 1.5 500 Natural channels: Clean and straight 0.030 ± 0.005 0.8 240 Sluggish, deep pools 0.040 ± 0.010 3 900 Major rivers 0.035 ± 0.010 1.5 500 Floodplains: Pasture, farmland 0.035 ± 0.010 1.5 500 Light brush 0.05 ± 0.02 6 2000 Heavy brush 0.075 ± 0.025 15 5000 Trees 0.15 ± 0.05 ? 7 A more complete list is given in Ref. 2, pp. 110-113. Recall that we warned about this awkwardness in Example 1.4. You may verify that a is the cube root of the conversion factor between the meter and your chosen length scale: In BG units, a = (3.2808 ft/m)'^ = 1.486.^ The Manning formula for uniform flow velocity is thus Vo(ni/s) Vo (ft/s) « ^^[R„(ft)]“5i'- (10.18) ^An interesting discussion of the history and “dimensionality” of Manning’s formula is given in Ref. 2, pp. 98-99. 692 Chapter 10 Open-Channel Flow The channel slope is dimensionless, and n is taken to be the same in both systems. The volume flow rate simply multiplies this result by the area: Normal Depth Estimates Uniform flow: e = VoA « a n ARrsl,'^ (10.19) Experimental values of n (and the corresponding roughness height) are listed in Table 10.1 for various channel surfaces. There is a factor-of-15 variation from a smooth glass surface (n ~ 0.01) to a tree-lined floodplain (n ~ 0.15). Due to the irregularity of typical channel shapes and roughness, the scatter bands in Table 10.1 should be taken seriously. For routine calculations, always use the average roughness in Table 10.1. Since Manning’s sixth-root size variation is not exact, real channels can have a variable n depending on the water depth. The Mississippi River near Memphis, Tennessee, has n ~ 0.032 at 40-ft flood depth, 0.030 at normal 20-ft depth, and 0.040 at 5-ft low-stage depth. Seasonal vegetative growth and factors such as bottom erosion can also affect the value of n. Even nearly identical man-made channels can vary. Brater et al. report that U.S. Bureau of Reclamation tests, on large concrete-lined canals, yielded values of n ranging from 0.012 to 0.017. EXAMPLE 10.2 Engineers find that the most efficient rectangular channel (maximum uniform flow for a given area) flows at a depth equal to half the bottom width. Consider a rectangular brickwork channel laid on a slope of 0.006. What is the best bottom width for a flow rate of 100 ft^/s? Solution ■ Assumptions: Uniform flow in a straight channel of constant of slope 5 = 0.006. • Approach: Use the Manning formula in English units, Eq. (10.19), to predict the flow rate. • Property values: For brickwork, from Table 10.1, the roughness factor n ~ 0.015. • Solution: For bottom width b, take the water depth to be y = bH. Equation (10.19) becomes A = by = b(b/2) = R, = - = by b^/2 Q = -ARf lA86fb'^\/'b ARfS^'^ = - — - (0.006)'“ = 100 ; n 0.015 V 2 A4y s b + 2y b + 2{bl2) Clean this up: b’^^ = 65.7 solve for b ~ 4.8 ft Ans. • Comments: The Manning approach is simple and effective. The Moody friction factor method, Eq. (10.14), requires laborious iteration and leads to a result b ~ 4.81 ft. With water depth y known, the computation of Q is quite straightforward. However, if Q is given, the computation of the normal depth y„ may require iteration. Since the normal depth is a characteristic flow parameter, this is an important type of problem. 10.2 Uniform Flow; The Chezy Formula 693 The normal depth, y„, is the depth, in uniform flow, of the water in a straight, constant-area, constant-slope channel. It varies with the flow rate and is a useful reference depth, calculated by solving Eq. (10.19) when Q is given. EXAMPLE 10.3 The asphalt-lined trapezoidal channel in Fig. E10.3 carries 300 ft^/s of water under uniform flow conditions when S = 0.0015. What is the normal depth y„? E10.3 Note: See Fig. 10.7 for generalized trapezoid notation. Solution using Excel From Table 10.1, for asphalt, n ~ 0.016. The area and hydraulic radius are functions of y,„ which is unknown: bg = 6 ft + 2y„ cot 50° A = \|(6 -I- bo)y„ = 6y„ -I- yl cot 50° P = 6 -f 2W = 6 -f 2y„ CSC 50° From Manning’s formula (10.19) with a known Q = 300 frVs, we have 1.49 , /6v„ -f cot 50°Y^^ 300 = iTTlTT (6y,. + yl cot 50°) ^ (0.0015)'^ 0.016 V 6 -f 2y„ CSC 50 / or (6y„ -f yl cot 50°)^'^ = 83.2(6 -f 2y„ esc 50°)^'^ (1) One can iterate Eq. (1) by hand laboriously and eventually find y„ ~ 4.6 ft. However, it is a good candidate for iteration in Excel. One could iterate the Chezy formula directly and find the value of y„ that gives a flow rate of 300 ft^/s. The writer chose to deal with the rearranged version, Eq. (1), guessing values of y„ until the left-hand side equals the right-hand side. The writer’s first guess was y„ = 10 ft, which was much too deep. Eour more guesses achieved quite accurate convergence, in the following table: yn Eq. (1) LHS Eq. (1) RHS LHS - RHS A B C D 1 10 3952.01 840.49 3111.52 2 5 700.86 593.54 107.32 3 4 418.74 538.00 -119.26 4 4.6 576.83 571.65 5.18 5 4.578 570.45 570.43 0.02 At 300 ftVs, the normal depth for this channel is y„ = 4.578 ft Ans. 694 Chapter 10 Open-Channel Flow Uniform Flow in a Partly Full Circular Pipe Fig. 10.6 Uniform flow in a partly full circular channel: (a) geometry; (b) velocity and flow rate versus depth. For later use, we might also list the other properties of this channel: b„ = 13.68 ft; P = 17.95 ft; A = 45.05 ft^; R/, = 2.51 ft Hand calculation is too cumbersome for open-channel problems where the depth is unknown. Consider the partially full pipe of Fig. 10.6a in uniform flow. The maximum velocity and flow rate actually occur before the pipe is completely full. In terms of the pipe radius R and the angle 0 up to the free surface, the geometric properties are sin 20 \ p = 2Re Ri ' sin 2d\ Rh = - 1 - V 2 J h 2 \ V 26 y The Manning formulas (10.19) predict a uniform flow as follows: Uo a sin 2ff 26 2/3 cl/2 5o Q = 0 sin 29 (10.20) For a given n and slope ^o, we may plot these two relations versus yID in Fig. lO.hfo. There are two different maxima, as follows: ^^2/3^ 1/2 n ° at 6 = 128.73° and at 6 = 151.21° and (10.21) y D ib) 10.3 Efficient Uniform-Flow Channels 695 10.3 Efficient Uniform-Flow Channels Fig. 10.7 Geometry of a trapezoidal channel section. As shown in Fig. I0.6b, the maximum velocity is 14 percent more than the velocity when running full, and similarly the maximum discharge is 8 percent more. Since real pipes running nearly full tend to have somewhat unstable flow, these differences are not that significant. The engineering design of an open channel has many parameters. If the channel surface can erode or scour, a low-velocity design might be sought. A dirt channel could be planted with grass to minimize erosion. For nonerodible surfaces, construc¬ tion and lining costs might dominate, suggesting a cross section of minimum wetted perimeter. Nonerodible channels can be designed for maximum flow. The simplicity of Manning’s formulation (10.19) enables us to analyze channel flows to determine the most efficient low-resistance sections for given conditions. The most common problem is that of maximizing R;, for a given flow area and discharge. Since /?,, = A/P, maximizing /?;, for given A is the same as minimizing the wetted perimeter P. There is no general solution for arbitrary cross sections, but an analysis of the trapezoid section will show the basic results. Consider the generalized symmetric trapezoid of angle 6 in Fig. 10.7. For a given side angle 0, the flow area is A = by + (3y^ (3 = cot 0 (10.22) The wetted perimeter is P = b + 2W=b + 2y(l + I3Y^ (10.23) Eliminating b between (10.22) and (10.23) gives P = j - (3y + 2yil + I3Y^ (10.24) To minimize P, evaluate dPIdy for constant A and f3 and set equal to zero. The result is A = y\2{l + - (3] P = 4yil + I3Y^ - 2f3y R, = h (10.25) The last result is very interesting: For any angle 0, the most efficient cross section for uniform flow occurs when the hydraulic radius is half the depth. Since a rectangle is a trapezoid with (3 = 0, the most efficient rectangular section is such that A = 2y^ P = Ay Rh ~ \y b = 2y (10.26) 696 Chapter 10 Open-Channel Flow Best Trapezoid Angle Part (a) Part (b) To find the correct depth y, these relations must be solved in conjunction with Manning’s flow rate formula (10.19) for the given discharge Q. Equations (10.25) are valid for any value of (5. What is the best value of /3 for a given depth and area? To answer this question, evaluate dPIdjd from Eq. (10.24) with A and y held constant. The result is 2(3 = {\+f3Y^ (3 = cot9=^ or B= 60° (10.27) Thus the maximum-flow trapezoid section is half a hexagon. Similar calculations with a circular channel section running partially full show best efficiency for a semicircle, y = ^D. In fact, the semicircle is the best of all possible channel sections (minimum wetted perimeter for a given flow area). The percentage improvement over, say, half a hexagon is very slight, however. EXAMPLE 10.4 (a) What are the best dimensions y and b for a rectangular brick channel designed to carry 5 mVs of water in uniform flow with = 0.001? (b) Compare results with a half-hexagon and semicircle. Solution From Eq. (10.26), A = 2y^ and Ri, = \|y. Manning’s formula (10.19) in SI units gives, with n ~ 0.015 from Table 10.1, Q = — ARrsl'^ or 5mVs = (0.001)'" n 0.015 V2 / which can be solved for y'" = 1.882 m'" y = 1.27 m Ans. The proper area and width are A = 2y" = 3.21 m" b = - = 2.53 m Ans. y It is constructive to see what flow rate a half-hexagon and semicircle would carry for the same area of 3.214 m". For the half-hexagon (FIH), with f3 = 1/3'" = 0.577, Eq. (10.25) predicts >1 = + 0.577")'" - 0.577] = 1.732yHH = 3.214 or .Vhh = 1.362 m, whence Rh = \y = 0.681 m. The half-hexagon flow rate is thus Q = ^^^^(3.214)(0.681)"'"(0.001)‘'" = 5.25 m"/s or about 5 percent more than that for the rectangle. 10.4 Specific Energy; Critical Depth 697 10.4 Specific Energy; Critical Depth Fig. 10.8 Illustration of a specific energy curve. The curve for each flow rate Q has a minimum energy corresponding to critical flow. For energy greater than minimum, there are two alternate flow states, one subcritical and one supercritical. For a semicircle, A = 3.214 w? = ttDVS, or D = 2.861 m, whence P = jTtD = 4.494 m and Rf, = A/P = 3.214/4.494 = 0.715 m. The semicircle flow rate will thus be Q = ^^^(3.214)(0.715)^\0.001)'“ = 5.42 mVs or about 8 percent more than that of the rectangle and 3 percent more than that of the half-hexagon. The total head of any incompressible flow is the sum of its velocity head aV^/(2g), pressure head p/j, and potential head z- For open-channel flow, surface pressure is everywhere atmospheric, so channel energy is a balance between velocity and elevation head only. Since the flow is turbulent, we assume that a ~ 1 — recall Eq. (3.77). The final result is the quantity called specific energy E, as suggested by Bakhmeteff in 1913: E = y + — (10.28) 2g where y is the water depth. It is seen from Fig. 10.8 that E is the height of the energy grade line (EGL) above the channel bottom. For a given flow rate, there are usually two states possible, called alternate states, for the same specific energy. There is also a minimum energy, Emin, which corresponds to a Froude number of unity. y 698 Chapter 10 Open-Channel Flow Rectangular Channels The Water Channel Compressible Flow Analogy Consider the possible states at a given location. Let q = Qlb = Vy be the discharge per unit width of a rectangular channel. Then, with q constant, Eq. (10.28) becomes Q E = y + ^ q = f (10.29) 2gy b Figure 10.8 is a plot of y versus E for constant q from Eq. (10.29). There is a mini¬ mum value of E at a certain value of y called the critical depth. By setting dEldy = 0 at constant q, we find that occurs at ,2\ 1/3 ^■■'7) ■ Q^\ b^gJ 1/3 (10.30) The associated minimum energy is £mi„ = EiyJ = iy, (10.31) The depth y^ corresponds to channel velocity equal to the shallow-water wave propagation speed Cq from Eq. (10.10). To see this, rewrite Eq. (10.30) as q" = gyl = (gyM = vlyl (10.32) By comparison it follows that the critical channel velocity is W = (gyy'^ = Co Fr = 1 (10.33) For E < no solution exists in Fig. 10.8, and thus such a flow is impossible physi¬ cally. For E > E^in two solutions are possible: (1) large depth with V < V„ called subcritical, and (2) small depth with V > V^, called supercritical. In subcritical flow, disturbances can propagate upstream because wave speed Cq > V. In supercritical flow, waves are swept downstream: Upstream is a zone of silence, and a small obstruc¬ tion in the flow will create a wedge-shaped wave exactly analogous to the Mach waves in Fig. 9.18c. The angle of these waves must be . -1 Co a = sin — = sm ^ V V = sin 'l — (10.34) The wave angle and the depth can thus be used as a simple measurement of super¬ critical flow velocity. Note from Fig. 10.8 that small changes in E near E^j„ cause a large change in the depth y, by analogy with small changes in duct area near the sonic point in Fig. 9.7. Thus critical flow is neutrally stable and is often accompanied by waves and undula¬ tions in the free surface. Channel designers should avoid long runs of near-critical flow. The simple water wave expression of Eq. (10.34), sin /i = 1/Fr, can be extended into a general, often qualitative, analogy between gas dynamics and water channel flow [21, Chap. 11]. The analogy is derived from small-disturbance theory and has the results Froude number is analogous to Mach number. Water depth is analogous to gas density. Water depth squared is analogous to gas pressure. Hydraulic jumps are analogous to gas shock waves. 10.4 Specific Energy; Critical Depth 699 Both experiments and CFD results have exploited this analogy [37, 38]. A further result is that small-disturbance channel flow is equivalent to a fictitious gas with a specific heat ratio k = 2.0. However, for larger disturbances, such as shock waves, the numerical results differ and, oddly, are more accurate for ^ = 1 .4 than for k = 2.0. A few examples are assigned here as Probs. P10.85, P10.88, and P10.91. EXAMPLE 10.5 A wide rectangular clean-earth channel has a flow rate q = 50 ftV(s ■ ft), (a) What is the critical depth? (b) What type of flow exists if y = 3 ft? Solution Part (a) The critical depth is independent of channel roughness and simply follows from Eq. (10.30): 50- = 4.27 ft Vgy V32.2/ Part (b) If the actual depth is 3 ft, which is less than y^, the flow must be supercritical. Ans. (a) Ans. (b) Nonrectangular Channels If the channel width varies with y, the specific energy must be written in the form The critical point of minimum energy occurs where dE/dy A = A(y), Eq. (10.35) yields, for E = E^^^, dA gA^ dy ~ (10.35) 0 at constant Q. Since (10.36) But dA = boi dy, where bg is the channel width at the free surface. Therefore, Eq. (10.36) is equivalent to (10.37fl) (10.37^) Eor a given channel shape A(y) and bo(y) and a given Q, Eqs. (10.37) have to be solved by trial and error or by Excel iteration to find the critical area A„ from which Vc can be computed. 700 Chapter 10 Open-Channel Flow Critical Uniform Flow: The Critical Slope By comparing the actual depth and velocity with the critical values, we can deter¬ mine the local flow condition. y > y < 14: subcritical flow (Fr < 1) y = y„ V = Vc'. critical flow (Fr = 1) y < yc, V > Vy. supercritical flow (Fr < 1) Note that I4 is equal to the speed of propagation c of a shallow-water wave in the channel and is dependent upon the depth, as in Fig. 10.4fl. For a rectangular channel, c = (gyY'^. If a critical channel flow is also moving uniformly (at constant depth), it must correspond to a critical slope S^, with y„ = y^. This condition is analyzed by equating Eq. (10.37a) to the Chezy (or Manning) formula: <3 2= = ^ bo = C^A^RhS, = or Sr = a^bXc aIrTSc n^g ^ ^hc ^0 /F 8 fto (10.38) where O? equals 1.0 for SI units and 2.208 for BG units. Equation (10.38) is valid for any channel shape. For a wide rectangular channel, S> y^, the formula reduces to n^g f Wide rectangular channel: ~ - ~ — ® aYr 8 This is a special case, a reference point. In most channel flows y„ ¥= y^. For fully rough turbulent flow, the critical slope varies between 0.002 and 0.008. EXAMPLE 10.6 The 50° triangular channel in Fig. E10.6 has a flow rate Q = 16 mVs. Compute (a) y^,, (b) y, and (c) 5, if u = 0.018. Solution Part (a) This is an easy cross section because all geometric quantities can he written directly in terms of depth y: P = 2y CSC 50° A = cot 50° Fa = \|y cos 50° ho = 2y cot 50° (1) The critical flow condition satisfies Eq. (10.37a): = feoG" or g(yl cot 50°)^ = (2y^ cot 50°)G^ / 2Q^ 2(16)' j \g cot^ 50°/ .9.81(0.839)'. Ans. (a) E10.6 10.4 Specific Energy; Critical Depth 701 Part (b) With known, from Eqs. (1) we compute = 6.18 m, Ri,^ = 0.760 m, = 4.70 m^, and f>0c = 3.97 m. The critical velocity from Eq. (10.37fc) is Q 16 mVs V, = — = - y = 3.41 m/s A, 4.70 m^ Ans. (b) Part (c) With n = 0.018, we compute from Eq. (10.38) a critical slope: gn^P 9.81(0.018)-(6.18) 5, = , ^ ^ ^ = 0.00542 c^R^bo 1.0(0.760)‘'^(3.97) Ans. (c) Frictionless Flow over a Bump Fig. 10.9 Erictionless two- dimensional flow over a bump: (a) definition sketch showing Froude number dependence; (b) specific energy plot showing bump size and water depths. A rough analogy to compressible gas flow in a nozzle (Fig. 9.12) is open-channel flow over a bump, as in Fig. 10.9fl. The behavior of the free surface is sharply different according to whether the approach flow is subcritical or supercritical. The height of the bump also can change the character of the results. For frictionless two-dimensional flow, sections 1 and 2 in Fig. 10.9fl are related by continuity and momentum; F? Vl Viyi = V2T2 — + yi = — + yi + ^8 2g Eliminating V2 between these two gives a cubic polynomial equation for the water depth y2 over the bump: yl - ■E2T2 + = 0 where £2 = — + yi - Ah (10.39) 2g 2g . Supercritical approach flow V ■ Subcritical approach Bump : Ah (a) Water depth {b) energy 702 Chapter 10 Open-Channel Flow This equation has one negative and two positive solutions if \h is not too large. Its behavior is illustrated in Fig. 10. 9h and depends on whether condition 1 is on the upper or lower leg of the energy curve. The specific energy £2 i® exactly A/t less than the approach energy Ei, and point 2 will lie on the same leg of the curve as El- A subcritical approach, Frj < 1, will cause the water level to decrease at the bump. Supercritical approach flow, Frj > 1, causes a water level increase over the bump. If the bump height reaches ~ Ei — E^, as illustrated in Fig. 10. 9h, the flow at the crest will be exactly critical (Fr = 1). If A/; > Ah^^, there are no physically correct solutions to Eq. (10.39). That is, a bump too large will “choke” the channel and cause frictional effects, typically a hydraulic jump (Sec. 10.5). These bump arguments are reversed if the channel has a depression (Ah < 0): Subcritical approach flow will cause a water level rise and supercritical flow a fall in depth. Point 2 will be \Ah\ to the right of point 1, and critical flow cannot occur. EXAMPLE 10.7 Water flow in a wide channel approaches a 10-cm-high bump at 1.5 m/s and a depth of 1 m. Estimate (a) the water depth ^2 over the bump and (b) the bump height that will cause the crest flow to be critical. Solution Part (a) First check the approach Froude number, assuming Co = V^: Vi 1.5 m/s Fri = — = — - . . - = 0.479 (subcritical) V(9.81 m/s^)(1.0 m) For subcritical approach flow, if Ah is not too large, we expect a depression in the water level over the bump and a higher subcritical Froude number at the crest. With Ah = 0.1 m, the specific energy levels must be Vi (1.5)^ £1 = — -f yi = - -f 1.0 = 1.115 m £2 = £1 - A/i = 1.015 m 2g 2(9.81) This physical situation is shown on a specific energy plot in Fig. E10.7. With yi in meters, Eq. (10.39) takes on the numerical values yz - 1.015y2 + 0.115 = 0 There are three real roots: yz ~ +0.859 m, +0.451 m, and —0.296 m. The third (negative) solution is physically impossible. The second (smaller) solution is the supercritical condition for £2 and is not possible for this subcritical bump. The first solution is correct: yzfsubcritical) « 0.859 m Ans. (a) The surface level has dropped by yi — yz — Ah = 1.0 — 0.859 — 0.1 = 0.041 m. The crest velocity is V2 = Viyi/yz = 1.745 m/s. The Froude number at the crest is F12 = 0.601. Flow downstream of the bump is subcritical. These flow conditions are shown in Fig. E10.7. 10.4 Specific Energy; Critical Depth 703 E10.7 1.2 Part (b) For critical flow in a wide channel, with q = Vy = \.5 m7s, from Eq. (10.31), E^rnm = Ec = -yc = -\ — 1 =“ - T = 0.918 m 2-^" 2\gJ 2L9.81m/sM Therefore, the maximum height for frictionless flow over this particular bump is ^^max ^ El — £'2,min = 1.115 — 0.918 = 0.197 m Ans. (b) For this bump, the solution of Eq. (10.39) is y2 — yc ~ 0.612 m, and the Froude number is unity at the crest. At critical flow the surface level has dropped by Vi — y2 ~ = 0.191 m. Flow under a Sluice Gate A sluice gate is a bottom opening in a wall, as sketched in Fig. lO.lOa, commonly used in control of rivers and channel flows. If the flow is allowed free discharge through the gap, as in Fig. 10.10a, the flow smoothly accelerates from subcritical (upstream) to critical (near the gap) to supercritical (downstream). The gate is then Fig. 10.10 Flow under a sluice gate passes through critical flow: (a) free discharge with vena contracta; {b) specific energy for free discharge; (c) dissipative flow under a drowned gate. 704 Chapter 10 Open-Channel Flow 10.5 The Hydraulic Jump Fig. 10.11 Hydraulic jump in a laboratory open channel. Note the extreme, dissipative turbulence in the downstream flow. Data are given in Prob. P10.94. (Courtesy of Prof. Hubert Chanson, University of Queensland.) analogous to a converging-diverging nozzle in gas dynamics, as in Fig. 9.12, operating at its design condition (similar to point H in Fig. 9.\2b). For free discharge, friction may be neglected, and since there is no bump (A/t = 0), Eq. (10.39) applies with Ey = £2- . (Vl \ 2 Viyl + ^7^2 + ^ = 0 (10-40) Given subcritical upstream flow (Fi, yi), this cubic equation has only one positive real solution: supercritical flow at the same specific energy, as in Fig. lO.lOfo. The flow rate varies with the ratio y2iyu we ask, as a problem exercise, to show that the flow rate is a maximum when y2/yi = \|- The free discharge. Fig. 10.10a, contracts to a depth 3^2 about 40 percent less than the gate’s gap height, as shown. This is similar to a free orifice discharge, as in Fig. 6.39. If H is the height of the gate gap and b is the gap width into the paper, we can approximate the flow rate by orifice theory: Q = C,HbVWy where Q “ (10.41) VI + O.olHIyi in the range H/yy < 0.5. Thus a continuous variation in flow rate is accomplished by raising the gate. If the tailwater is high, as in Fig. 10.10c, free discharge is not possible. The sluice gate is said to be drowned or partially drowned. There will be energy dissipation in the exit flow, probably in the form of a drowned hydraulic jump, and the downstream flow will return to subcritical. Equations (10.40) and (10.41) do not apply to this situ¬ ation, and experimental discharge correlations are necessary [3, 19]. See Prob. P10.77. In open-channel flow a supercritical flow can change quickly back to a subcritical flow by passing through a hydraulic jump, as in Fig. 10.5. The upstream flow is fast and shallow, and the downstream flow is slow and deep, analogous to the normal shock wave of Fig. 9.8. Unlike the infinitesimally thin normal shock, the hydraulic jump is quite thick, ranging in length from 4 to 6 times the downstream depth y2 . Being extremely turbulent and agitated, the hydraulic jump is a very effective energy dissipator and is a feature of stilling-basin and spillway applications . Figure 10.11 shows the jump formed in a laboratory open channel. It is very important 10.5 The Hydraulic Jump 705 Fig. 10.12 Classification of hydraulic jumps: (a) Fr = 1.0 to 1.7: undular Jump; (b) Fr = 1.7 to 2.5: weak jump; (c) Fr = 2.5 to 4.5: oscillating jump; (d) Fr = 4.5 to 9.0: steady jump; (e) Fr > 9.0: strong jump. Source: Adapted from U.S. Bureau of Reclamation, “Research studies on stilling Basins, Energy Dissipators, and Associated Appurtenances, ” Hydraulic Lab, Rep., Hyd-399, June I, 1995. Classification © (a) (b) ie) that such jumps be located on specially designed aprons; otherwise the channel bottom will be badly scoured by the agitation. Jumps also mix fluids very effectively and have application to sewage and water treatment designs. The principal parameter affecting hydraulic jump performance is the upstream Froude number Fri = The Reynolds number and channel geometry have only a secondary effect. As detailed in Ref. 20, the following ranges of operation can be outlined, as illustrated in Fig. 10.12: Fri < 1.0: Jump impossible, violates second law of thermodynamics. Fri = 1.0 to 1.7: Standing-wave or undular jump about 4y2 long; low dissipation, less than 5 percent. 706 Chapter 10 Open-Channel Flow Fri = 1.7 to 2.5: Fri = 2.5 to 4.5: Fri = 4.5 to 9.0: Fri > 9.0: Smooth surface rise with small rollers, known as a weak jump; dissipation 5 to 15 percent. Unstable, oscillating jump; each irregular pulsation creates a large wave that can travel downstream for miles, damaging earth banks and other structures. Not recommended for design conditions. Dissipation 15 to 45 percent. Stable, well-balanced, steady jump; best performance and action, insensitive to downstream conditions. Best design range. Dissipation 45 to 70 percent. Rough, somewhat intermittent strong jump, but good performance. Dissipation 70 to 85 percent. Further details can be found in Ref. 20 and Ref. 2, Chap. 15. Theory for a Horizontal Jump A jump that occurs on a steep channel slope can be affected by the difference in water-weight components along the flow. The effect is small, however, so the classic theory assumes that the jump occurs on a horizontal bottom. You will be pleased to know that we have already analyzed this problem in Sec. 10.1. A hydraulic jump is exactly equivalent to the strong fixed wave in Fig. 10.4h, where the change in depth dy is not neglected. If Vi and yi upstream are known, V2 and y2 are computed by applying continuity and momentum across the wave, as in Eqs. (10.7) and (10.8). Equation (10.9) is therefore the correct solution for a jump if we interpret C and y in Eig. lOAb as upstream conditions V-[ and yi, respectively, with C — dV and y + ^ being the downstream conditions V2 and y2, respectively, as in Eig. 10.12h. Equation (10.9) becomes Vi = kgyiriir] + 1) (10.42) where ij = y2ly\- Introducing the Eroude number Erj = Vil{gy{}^'^ and solving this quadratic equation for 77, we obtain — = -1 + (1 -t- 8Frl)'^- .Vi (10.43) With y2 thus known, V2 follows from the wide-channel continuity relation: V'lFi V2 = yi (10.44) Einally, we can evaluate the dissipation head loss across the jump from the steady flow energy equation: Introducing y2 and V2 from Eqs. (10.43) and (10.44), we find after considerable alge¬ braic manipulation that (yi - .vi)^ 4yiy2 (10.45) 10.5 The Hydraulic Jump 707 Part (a) Part (b) Part (c) Part (d) Part (e) Equation (10.45) shows that the dissipation loss is positive only if > yi, which is a requirement of the second law of thermodynamics. Equation (10.43) then requires that Fri > 1.0; that is, the upstream flow must be supercritical. Finally, Eq. (10.44) shows that < Vi and the downstream flow is subcritical. All these results agree with our previous experience analyzing the normal shock wave. The present theory is for hydraulic jumps in wide or rectangular horizontal channels. For the theory of prismatic or sloping channels see advanced texts [for example, 2, Chaps. 15 and 16]. EXAMPLE 10.8 Water flows in a wide channel at q = 10 mV(s ■ m) and = 1.25 m. If the flow undergoes a hydraulic jump, compute (a) y2, (b) V2, (c) Fr2, (d) hf, (e) the percentage dissipation, (f) the power dissipated per unit width, and (g) the temperature rise due to dissipation if Cp = 4200 J/(kg ■ K). Solution The upstream velocity is p, =- = yi q lOmV(s-m) 1.25 m = 8.0 m/s The upstream Froude number is therefore 8.0 Fr = ‘ (gyi)'“ [9.81(1.25)]'" = 2.285 From Fig. 10.12 this is a weak jump. The depth y2 is obtained from Eq. (10.43): — = -1 + [1 + 8(2.285)"]'" = 5.54 yi or y2 = \|yi(5.54) = 2(1.25)(5.54) = 3.46 m From Eq. (10.44) the downstream velocity is Fiji 8.0(1.25) V2 = ^ = 2.89 m/s y2 3.46 The downstream Froude number is Fr2 = F2 2.89 = 0.496 (gyz)”^ [9.81(3.46)]' As expected, Fr2 is subcritical. From Eq. (10.45) the dissipation loss is hf = (3.46 - 1.25)" = 0.625 m ^ 4(3.46)(1.25) The percentage dissipation relates hf to upstream energy: vi (8.0)" £■1 = yi + — = 1-25 + ^ ^ 2g 2(9.81) = 4.51 m Ans. (a) Ans. (b) Ans. (c) Ans. (d) 708 Chapter 10 Open-Channel Flow Part (f) Part (g) 10.6 Gradually Varied Flow^ Basic Differential Equation hf 100(0.625) Hence Percentage loss = (100) — = - - = 14 percent Ans. (e) The power dissipated per unit width is Power = pgqhf= (9800 N/m^) 10 mV(s • m) = 61.3kW/m Ans. (f) Finally, the mass flow rate is rk = pq = (1000 kg/m^)[10 mV(s • m)] = 10,000 kg/(s • m), and the temperature rise from the steady flow energy equation is Power dissipated = mCp AT or 61,300 W/m= [ 10,000 kg/(s ■ m)][4200 J/(kg ■ K)]Ar from which AT = 0.0015 K Ans. {g) The dissipation is large, but the temperature rise is negligible. In practical channel flows both the bottom slope and the water depth change with position, as in Fig. 10.3. An approximate analysis is possible if the flow is gradually varied, as is the case if the slopes are small and changes not too sudden. The basic assumptions are 1. Slowly changing bottom slope. 2. Slowly changing water depth (no hydraulic jumps). 3. Slowly changing cross section. 4. One-dimensional velocity distribution. 5. Pressure distribution approximately hydrostatic. The flow then satisfies the continuity relation (10.1) plus the energy equation with bottom friction losses included. The two unknowns for steady flow are velocity Vix) and water depth y{x), where x is distance along the channel. Consider the length of channel dx illustrated in Fig. 10.13. All the terms that enter the steady flow energy equation are shown, and the balance between x and x + dxis /y2\ - \- y + Sodx = S dx - h d I — ] + y + dy 2g 2g \2gJ or dx dx\2g J = So-S (10.46) where Sq is the slope of the channel bottom (positive as shown in Fig. 10.13) and S is the slope of the EGL (which drops due to wall friction losses). ^This section may be omitted without loss of continuity. 10.6 Gradually Varied Flow 709 Fig. 10.13 Energy balance between two sections in a gradually varied open-channel flow. Horizontal To eliminate the velocity derivative, differentiate the continuity relation: V — dx (10.47) But dA = bd dy, where bo is the channel width at the surface. Eliminating dV/dx between Eqs. (10.46) and (10.47), we obtain v\ gA = Sq-S (10.48) Finally, recall from Eq. (10.37) that V^bol{gA) is the square of the Froude number of the local channel flow. The final desired form of the gradually varied flow equation is dy So — S dx 1 - Fr^ (10.49) This equation changes sign according to whether the Froude number is subcritical or supercritical and is analogous to the one-dimensional gas dynamic area-change formula (9.40). The numerator of Eq. (10.49) changes sign according to whether So is greater or less than S, which is the slope equivalent to uniform flow at the same discharge Q\ JLYl - Dh 2g ~ Rf,C- ~ a^Rf (10.50) where C is the Chezy coefficient. The behavior of Eq. (10.49) thus depends on the relative magnitude of the local bottom slope Soix), compared with (1) uniform flow. 710 Chapter 10 Open-Channel Flow Classification of Solutions Fig. 10.14 Gradually varied flow for five classes of channel slope, showing the 12 basic solution y = y„, and (2) critical flow, y = y^. As in Eq. (10.38), the dimensional parameter equals 1.0 for SI units and 2.208 for BG units. It is customary to compare the actual channel slope Sq with the critical slope for the same Q from Eq. (10.38). There are five classes for Sq, giving rise to 12 distinct types of solution curves, all of which are illustrated in Fig. 10.14: (a) (b) (c) Adverse (^) 5o < 0 = imaginary A-2 curves. 10.6 Gradually Varied Flow 711 Numerical Solution Slope class Slope notation Depth class Solution curves > Sc Steep Jc > Jn S-1, S-2, S-3 II Critical yc = Jn C-1, C-3 Sq < Sc Mild yc < Vn M-1, M-2, M-3 So = 0 Horizontal y„ = °° H-2, H-3 < 0 Adverse yn = imaginary A-2, A-3 The solution letters S, C, M, H, and A obviously denote the names of the five types of slopes. The numbers 1, 2, 3 relate to the position of the initial point on the solution curve with respect to the normal depth y„ and the critical depth y^.. In type 1 solutions, the initial point is above both y„ and y„ and in all cases the water depth solution y{x) becomes even deeper and farther away from y„ and y^. In type 2 solutions, the initial point lies between y„ and y^, and if there is no change in or roughness, the solution tends asymptotically toward the lower of y„ or y^. In type 3 cases, the initial point lies below both y„ and y„ and the solution curve tends asymptotically toward the lower of these. Figure 10.14 shows the basic character of the local solutions, but in practice, of course, Sq varies with x, and the overall solution patches together the various cases to form a continuous depth profile y{x) compatible with a given initial condition and a given discharge Q. There is a fine discussion of various composite solutions in Ref. 2, Chap. 9; see also Ref. 22, Sec. 12.7. The basic relation for gradually varied fiow, Eq. (10.49), is a first-order ordinary dif¬ ferential equation that can be easily solved numerically. For a given constant-volume fiow rate Q, it may be written in the form dy So - nQ-/iaWRD -=-5 - ^ (10.51) dx 1 - eV(M ) subject to an initial condition y = yo at x = xq. It is assumed that the bottom slope Soix) and the cross-sectional shape parameters {bo, P, A) are known everywhere along the channel. Then one may solve Eq. (10.51) for local water depth y{x) by any stan¬ dard numerical method. The author uses an Excel spreadsheet for a personal computer. Step sizes Ax may be selected so that each change Ay is limited to no greater than, say, 1 percent. The solution curves are generally well behaved unless there are dis¬ continuous changes in channel parameters. Note that if one approaches the critical depth y,;, the denominator of Eq. (10.51) approaches zero, so small step sizes are required. It helps physically to know what type of solution curve (M-1, S-2, or the like) you are proceeding along, but this is not mathematically necessary. EXAMPLE 10.9 Let us extend the data of Example 10.5 to compute a portion of the profile shape. Given is a wide channel with n = 0.022, Sq = 0.0048, and q = 50 ft^/(s ■ ft). If yo = 3 ft at x = 0, how far along the channel x = L does it take the depth to rise to y^ = 4 ft? Is the 4-ft depth position upstream or downstream in Fig. E10.9fl? 712 Chapter 10 Open-Channel Flow Solution In Example 10.5 we computed = 4.27 ft. Since our initial depth y = 3 ft is less than y^, we know the flow is supercritical. Let us also compute the normal depth for the given slope 5o hy setting q = 50 ftV(s • ft) in the Chezy formula (10.19) with R/, — y„: q = ^ [y„(l ft) ]yf (0.0048)'" = 50 ftV(s ■ ft) E10.9a Solve for: Thus both y(0) = 3 ft and y(L) = 4 ft are less than y„, which is less than y^, so we must he on an S-3 curve, as in Fig. 10.14fl. For a wide channel, Eq. (10.51) reduces to dy _ So- n^q^jaV^'^) dx 1 - ^"/(gy) 0.0048 - (0.022)"(50)"/(2.208y“’') 1 - (50)"/(32.2y") withy(O) = 3 ft The initial slope is y'(0) = 0.00494, and a step size A.r = 5 ft would cause a change Ay « (0.00494)(5 ft) « 0.025 ft, less than 1 percent. We therefore integrate numerically with E10.9b X 230 ft 10.6 Gradually Varied Flow 713 Ax = 5 ft to determine when the depth y = 4 ft is achieved. Tabulate some values: X, ft 0 50 100 150 200 230 3.00 3.25 3.48 3.70 3.90 4.00 The water depth, still supercritical, reaches y = 4 ft at X ~ 230 ft downstream Ans. We verify from Fig. 10.14a that water depth does increase downstream on an S-3 curve. The solution curve y(x) is shown as the bold line in Fig. E10.9Z). For little extra effort we can investigate the entire family of S-3 solution curves for this problem. Figure E10.9fc also shows what happens if the initial depth is varied from 0.5 to 3.5 ft in increments of 0.5 ft. All S-3 solutions smoothly rise and asymptotically approach the uniform flow condition y = y„ = 4.14 ft. Approximate Solution for Irregular Channels Direct numerical solution of Eq. (10.51) is appropriate when we have analytical formulas for the channel variations A(x), 5'o(x), n(x), ho(x), and Rh{x). For natural channels, however, cross sections are often highly irregular, and data can he sparse and unevenly spaced. For such cases, civil engineers use an approximate method to estimate gradual flow changes. Write Eq. (10.46) in finite-difference form between two depths y and y -I- Ay: Ax ~ E{y -F Ay) - E{y) (5o ^ave) where £ = y -I- 1^ (10.52) Average values of velocity, slope, and hydraulic radius are estimated between the two sections. For example, 1 1 nV lyg Kvg = A[V(y) + V(y + Ay)]; = -[£,,(y) + Rhiy + Ay)]; ^avg = 2 A z a Rh,iyg Again, computation can proceed either upstream or downstream, using small values of Ay. Further details of such computations are given in Chap. 10 of Ref. 2. EXAMPLE 10.10 Repeat Example 10.9 using the approximate method of Eq. (10.52) with a 0.25-foot increment in Ay. Find the distance required for y to rise from 3 ft to 4 ft. Solution Recall from Example 10.9 that n = 0.022, So = 0.0048, and q = 50 ftV(s-ft). Note that Ri, = y for a wide channel. Make a table with y varying from 3.0 to 4.0 ft in increments of 0.25 ft, computing V = qly, E = y + V^/(2g), and Savg = [n^V'^/(2.208y''^^)]avg: 714 Chapter 10 Open-Channel Flow Some Dlustrative Composite-Flow Transitions 3-, ft V (ft/s) = 50/y E = y + VVilg) S 5avg Aa: = A£/(S„ - S),v8 X = SAjc 3.0 16.67 7.313 0.01407 _ _ 0 3.25 15.38 6.925 0.01078 0.01243 51 51 3.5 14.29 6.669 0.00842 0.00960 53 104 3.75 13.33 6.511 0.00669 0.00756 57 161 4.0 ft 12.50 ft/s 6.426 ft 0.00539 0.00604 69 ft 230 ft Comment: The accuracy is excellent, giving the same result, x = 230 ft, as the Excel spreadsheet numerical integration in Example 10.9. Much of this accuracy is due to the smooth, slowly varying nature of the prohle. Less precision is expected when the channel is irregular and given as uneven cross sections. The solution curves in Fig. 10.14 are somewhat simplistic, since they postulate constant-hottom slopes. In practice, channel slopes can vary greatly, = So{x), and the solution curves can cross between two regimes. Other parameter changes, such as A(x), and n{x), can cause interesting composite-flow profiles. Some examples are shown in Fig. 10.15."^ Figure 10.15a shows a transition from a mild slope to a steep slope in a constant- width channel. The initial M-2 curve must change to an S-2 curve farther down the steep slope. The only way this can happen physically is for the solution curve to pass smoothly through the critical depth, as shown. The critical point is mathematically singular [2, Sec. 9.6], and the flow near this point is generally rapidly, not gradually, varied. The flow pattern, accelerating from subcritical to supercritical, is similar to a converging-diverging nozzle in gas dynamics. Other scenarios for Fig. 10.15a are impossible. For example, the upstream curve cannot be M-1, for the break in slope would cause an S-1 curve that would move away from uniform steep flow. Figure 10.151; shows a mild slope that suddenly changes to an even milder slope. The approach flow is assumed uniform, and the break in slope makes its presence known upstream. The water depth moves smoothly along an M- 1 curve until it exactly merges, at the break point, with a uniform flow at the new (milder) depth y„2. Figure 10.15c shows a steep slope that suddenly changes to a less steep slope. Note for both slopes that y„ 14) approach flow, the break in slope cannot make its presence known upstream. Thus, not until the break point does an S-3 curve form, and then this profile proceeds smoothly to uniform flow at the new (higher) normal depth. Figure 10.15t/ shows a steep slope that suddenly changes to a mild slope. Various cases may occur, possibly beyond the ability of this author to describe. The two cases shown depend on the relative magnitude of the mild slope. If the downstream depth y„2 is shallow, an M-3 curve will start at the break and develop until the local supercritical flow is just sufficient to form a hydraulic jump up to the new normal depth. As y„2 increases, the jump moves upstream until, for the “high” case shown, it forms on the steep side, followed by an S-1 curve that merges into normal depth y„2 at the break point. The author is indebted to Prof. Bruce Larock for clarification of these transition profiles. 10.6 Gradually Varied Flow 715 Fig. 10.15 Some examples of composite-flow transition profiles. M-2 ~ J’c yn2 yn2 yc ynl Vc yn2 5-1 Uniform flow yn2, high yn2, low yc yn - ■>’c - - - M-2 Critical flow (e) Free overfall 716 Chapter 10 Open-Channel Flow 10.7 Flow Measurement and Control by Weirs Fig. 10.16 Flow over wide, well- ventilated weirs: (a) sharp-crested; (b) broad-crested. Figure 10.15e illustrates a. free overfall with a mild slope. This acts as a control section to the upstream flow, which then forms an M-2 curve and accelerates to criti¬ cal flow near the overfall. The falling stream will be supercritical. The overfall “controls” the water depths upstream and can serve as an initial condition for com¬ putation of y(x). This is the type of flow that occurs in a weir or waterfall. Sec. 10.7. The examples in Fig. 10.15 show that changing conditions in open-channel flow can result in complex flow patterns. Many more examples of composite-flow profiles are given in Ref. 2, pp. 229-233. A weir, of which the ordinary dam is an example, is a channel obstruction over which the flow must deflect. For simple geometries the channel discharge Q correlates with gravity and with the blockage height H to which the upstream flow is backed up above the weir elevation (see Fig. 10.16). Thus a weir is a simple but effective open-channel flowmeter. We used a weir as an example of dimensional analysis in Prob. P5.32. Figure 10.16 shows two common weirs, sharp-crested and broad-crested, assumed to be very wide. In both cases the flow upstream is subcritical, accelerates to critical near the top of the weir, and spills over into a supercritical nappe. For both weirs the discharge q per unit width is proportional to but with somewhat different coefficients. The short-crested (or thin-plate) weir nappe should be ventilated to the ib) 10.7 Flow Measurement and Control by Weirs 717 Analysis of Sharp-Crested Weirs Analysis of Broad-Crested Weirs atmosphere; that is, it should spring clear of the weir crest. Unventilated or drowned nappes are more difficult to correlate and depend on tailwater conditions. (The spill¬ way of Fig. 10.11 is a sort of unventilated weir.) A very complete discussion of weirs, including other designs such as the polygonal “Crump” weir and various contracting flumes, is given in the text by Ackers et al. . See Prob. P10.122. It is possible to analyze weir flow by inviscid potential theory with an unknown (but solvable) free surface, as in Fig. P8.71. Here, however, we simply use one¬ dimensional flow theory plus dimensional analysis to develop suitable weir flow rate correlations. A very early theoretical approach is credited to J. Weisbach in 1855. The velocity head at any point 2 above the weir crest is assumed to equal the total head upstream; in other words, Bernoulli’s equation is used with no losses: y2 — + N- h^ — + H or VzW = V2gh + 2g 2g where h is the vertical distance down to point 2, as shown in Fig. I0.l6a. If we accept for the moment, without proof, that the flow over the crest draws down to /z^in ~ H/3, the volume flow q = Q/b over the crest is approximately q = V2dh ~ crest rH {2gh + Vif^dh ■'HI3 2 3 V2^ „ + UY” _ f « + Yl) ^gJ V3 2gJ 2\3/2-\| Normally the upstream velocity head Vil{2g) is neglected, so this expression reduces to Sharp-crested theory: q = 0.81(\|)(2g)"^//^'^ (10.53) This formula is functionally correct, but the coefficient 0.81 is too high and should be replaced by an experimentally determined discharge coefficient. The broad-crested weir of Fig. 10.16f> can be analyzed more accurately because it creates a short run of nearly one-dimensional critical flow, as shown. Bernoulli’s equation from upstream to the weir crest yields y? — + Y + + T + yc 2g 2g gy^ from Eq. (10.33). Thus we can - ^ ^ 1” If the crest is very wide into the paper, = solve for 2H Vi - + - ■ 3 3g 718 Chapter 10 Open-Channel Flow This result was used without proof to derive Eq. (10.53). Finally, the flow rate follows from wide-channel critical flow, Eq. (10.32): Broad-crested theory: q ^g) (10.54) Again we may usually neglect the upstream velocity head v\l{2g). The coefficient 1/V3 ~ 0.577 is about right, but experimental data are preferred. Experimental Weir Discharge Coefficients Theoretical weir flow formulas may be modified experimentally as follows: Eliminate the numerical coefficients \| and V2, for which there is much sentimental attachment in the literature, and reduce the formula to Q weir CjbVg 2gJ (10.55) where b is the crest width and Q is a dimensionless, experimentally determined weir discharge coefficient, which may vary with the weir geometry, Reynolds number, and Weber number. Many data for many different weirs have been reported in the litera¬ ture, as detailed in Ref. 23. An accurate (±2 percent) composite correlation for wide ventilated sharp crests is recommended as follows : Wide sharp-crested weir: Q 0.564 -f 0.0846 — Y for — < 2 Y (10.56) The Reynolds numbers ViHIu for these data vary from 1 E4 to 2 E6, but the formula should apply to higher Re, such as large dams on rivers. The broad-crested weir of Fig. 10.16/? is considerably more sensitive to geometric parameters, including the surface roughness e of the crest. If the leading-edge nose is rounded, R/L > 0.05, available data [23, Chap. 7] may be correlated as follows: d/LY^ Round-nosed broad-crested weir: ~ 0.544 1 - i HILJ where — ~ 0.001 + 0.2V£/L (10.57) The chief effect is due to turbulent boundary layer displacement-thickness growth 5 on the crest as compared to upstream head H. The formula is limited to HIL < 0.7, dL < 0.002, and V^H/iy > 3 E5. If the nose is round, there is no significant effect of weir height Y, at least if HIY < 2.4. If the broad-crested weir has a sharp leading edge, thus commonly called a rect¬ angular weir, the discharge may depend on the weir height Y. However, in a certain range of weir height and length, is nearly constant: Sharp-nosed ~ 0.462 for broad-crested weir: and 0.08 < — < 0.33 L 0.22 < — < 0.56 y (10.58) 10.7 Flow Measurement and Control by Weirs 719 Surface roughness is not a significant factor here. For HIL < 0.08 there is large scatter (±10 percent) in the data. For HIL > 0.33 and HIY > 0.56, Cj increases up to 10 percent due to each parameter, and complex charts are needed for the discharge coefficient [19, Chap. 5]. EXAMPLE 10.11 A weir in a horizontal channel is 1 m high and 4 m wide. The water depth upstream is 1.6 m. Estimate the discharge if the weir is {a) sharp-crested and {b) round-nosed with an unfinished concrete broad crest 1.2 m long. Neglect V\l{2g). Solution Part (a) We are given T = 1 m and H + Y ~ 1.6 m, hence H ~ 0.6 m. Since H b,v/e assume that the weir is “wide.” For a sharp crest, Eq. (10.56) applies: 0.6 m Cj « 0.564 -f 0.0846 - « 0.615 1 m Part (b) Then the discharge is given hy the basic correlation, Eq. (10.55): Q = CabVgH^'- = (0.615)(4 m)\/(9.81 m/s^)(0.6 m)^® « 3.58 mVs Ans. (a) We check that HIY = 0.6 < 2.0 for Eq. (10.56) to be valid. From continuity, Vi = Ql(byi) = 3.58/[(4.0)(1.6)] = 0.56 m/s, giving a Reynolds number ViHIv ~ 3.4 E5. For a round-nosed broad-crested weir, Eq. (10.57) applies. For an unfinished concrete surface, read e ~ 2.4 mm from Table 10.1. Then the displacement thickness is 6 , _ /0.0024 mV'^ — « 0.001 -f 0.2V^ = 0.001 -f 0.2 - « 0.00994 L V l-2m / Then Eq. (10.57) predicts the discharge coefficient: Q 0.544 0.00994 Y” 0.6 m/1.2 m/ 0.528 The estimated flow rate is thus Q = CabVgH^'^ = 0.528(4 m)V(9.81 m^/s)(0.6 m)^'^ « 3.07 m^/s Ans. (b) Check that HIL = 0.5 < 0.7 as required. The approach Reynolds number is ViHIu ~ 2.9 E5, just barely below the recommended limit in Eq. (10.57). Since Vi ~ 0.5 m/s, V^il{2g) ~ 0.012 m, so the error in taking total head equal to 0.6 m is about 2 percent. We could correct this for upstream velocity head if desired. Other Thin-Plate Weir Designs Weirs are often used for flow measurement and control of artificial channels. The two most common shapes are a rectangle and a V notch, as shown in Table 10.2. All should be fully ventilated and not drowned. Table 10.2fl shows a full-width rectangle, which will have slight end-boundary- layer effects but no end contractions. For a thin-plate design, the top is approximately 720 Chapter 10 Open-Channel Flow Table 10.2 Thin-Plate Weirs for Flow Measurement Thin-plate weir Flow-rate correlation (a) Full-width rectangle. (b) Rectangle with side contractions. 0.564 -I- 0.0846 y j Q « 0.581 (b - H < 0.57 Q » 0.44 tan \| 20° < e < 100° sharp-crested, and Eq. (10.56) should give adequate accuracy, as shown in the table. Since the overfall spans the entire channel, artificial ventilation may be needed, such as holes in the channel walls. Table 10.2h shows a partial-width rectangle, b < L, which will cause the sides of the overfall to contract inward and reduce the flow rate. An adequate contraction cor¬ rection [23, 24] is to reduce the effective weir width by 0.1//, as shown in the table. It seems, however, that this type of weir is rather sensitive to small effects, such as plate thickness and sidewall boundary layer growth. Small heads (// < 75 mm) and small slot widths (b < 30 cm) are not recommended. See Refs. 23 and 24 for further details. The V notch, in Table 10.2c, is intrinsically interesting in that its overfall has only one length scale, H — there is no separate “width.” The discharge will thus be propor¬ tional to rather than a power of \|. Application of Bernoulli’s equation to the triangular opening, in the spirit of Eq. (10.52), leads to the following ideal flow rate for a V notch: 8V2 e , Gideal — 15 2 ^ ^ V notch: (10.59) 10.7 Flow Measurement and Control by Weirs 721 Backwater Curves where 0 is the total included angle of the notch. The actual measured flow is about 40 percent less than this, due to contraction similar to a thin-plate orifice. In terms of an experimental discharge coefficient, the recommended formula is Q Cvnotch = Qtan-g'V^^ Q=0.44 for 20° < 6» < 100° (10.60) for heads H > 50 mm. For smaller heads, both Reynolds number and Weber number effects may be important, and a recommended correction is Low heads, H < 50 mm: Cd.v notch ~ 0-44 + 0.9 (Re We)"® (10.61) where Re = and We = pgH^IY, with Y being the coefficient of surface tension. Liquids other than water may be used with this formula, as long as Re > 300/tan and We > 300. A number of other thin-plate weir designs — trapezoidal, parabolic, circular arc, and U-shaped — are discussed in Ref. 25, which also contains considerable data on broad- crested weirs. See also Refs. 29 and 30. EXAMPLE 10.12 A V notch weir is to be designed to meter an irrigation channel flow. For ease in reading the upstream water-level gage, a reading H S: 30 cm is desired for the design flow rate of 150 m^/h. What is the appropriate angle 9 for the V notch? Solution • Assumptions: Steady flow, negligible Weber number effect because H > 50 mm. • Approach: Equation (10.60) applies with, we hope, a notch angle 20° < 0 < 100°. • Property values: If surface tension is neglected, no fluid properties are needed. Why? • Solution: Apply Equation (10.60) to the known flow rate and solve for 6: 150 mVh 3600 s/h 0.0417 — > Qtan s 0.44 tan 1/2 I (0.3 m)^® Solve for tan < 0.613 or 6» < 63° Ans. ■ Comments: An angle of 63° will create an upstream head of 30 cm. Any angle less than that will create an even larger head. Weir formulas depend primarily on gravity and geometry. Eluid properties such as (p, p, Y) enter only as slight modifications or as correction factors. A weir is a flow barrier that not only alters the local flow over the weir but also modifies the flow depth distribution far upstream. Any strong barrier in an open- channel flow creates a backwater curve, which can be computed by the gradually varied flow theory of Sec. 10.6. If Q is known, the weir formula, Eq. (10.55), determines H and hence the water depth just upstream of the weir, y = H + Y, where 722 Chapter 10 Open-Channel Flow Y is the weir height. We then compute y{x) upstream of the weir from Eq. (10.51), following in this case an M-1 curve (Fig. 10.14c). Such a barrier, where the water depth correlates with the flow rate, is called a channel control point. These are the starting points for numerical analysis of floodwater profiles in rivers . EXAMPLE 10.13 A rectangular channel 8 m wide, with a flow rate of 30 mVs, encounters a 4-m-high sharp- edged dam, as shown in Fig. El 0.1 3a. Determine the water depth 2 km upstream if the channel slope is Sq = 0.0004 and n = 0.025. H \ ^ (From weir theory) Solution First determine the head H produced hy the dam, using sharp-crested full-width weir theory, Eq. (10.56): Q = 30mVs = = fo.564 -f 0.0846—) (8 m)(9.81 V 4 my Since the term 0.0846///4 in parentheses is small, we may proceed hy iteration to the solution H ~ 1 .59 m. Then our initial condition atx = 0, just upstream of the dam, is y(0) = Y + H = 4 + 1.59 = 5.59 m. Compare this to the critical depth from Eq. (10.30): (30 mVs)^ WgJ .(8m)^(9.81 m/s^). Since y(0) is greater than y^, the flow upstream is suhcritical. Finally, for reference purposes, estimate the normal depth from the Chezy equation (10.19): 2 = 30 mVs = Q: 9/, 1/9 1.0 - by = - (8 m)y„ 0.025 8y„ 8 + 2y„ (0.0004)' Summary 723 By trial and error, solve for y„ ~ 3.20 m. If there are no changes in channel width or slope, the water depth far upstream of the dam will approach this value. All these reference values y(0), yc, andy„ are shown in Fig. ElO. 13fc. Since y(0) > y„ > y^, the solution will be an M-1 curve as computed from gradually varied theory, Eq. (10.51), for a rectangular channel with the given input data: dy So- n^Q^Kc^A^R f) dx 1 - a=1.0 A = 8y n = 0.025 Rk 8y 8 + 2y bo = 8 Beginning with y = 5.59 m at x = 0, we integrate backward to x = —2000 m. For the Runge- Kutta method, four-figure accuracy is achieved for Ax = — 100 m. The complete solution curve is shown in Fig. E10.13b. The desired solution value is At X = —2000 m: y ~ 5.00 m Ans. 6 -2000 -1500 -1000 -500 0 E10.13b m Thus, even 2 km upstream, the dam has produced a “backwater” that is 1.8 m above tbe normal depth that would occur without a dam. For this example, a near-normal depth of, say, 10 cm greater than y„, or y ~ 3.3 m, would not be achieved until x = — 13,400 m. Backwater curves are quite far-reaching upstream, especially in flood stages. Summary This chapter has introduced open-channel flow analysis, limited to steady, one¬ dimensional flow conditions. The basic analysis combines the continuity equation with the extended Bernoulli equation including friction losses. Open-channel flows are classified either by depth variation or by Froude number, the latter being analogous to the Mach number in compressible duct flow (Chap. 9). Flow at constant slope and depth is called uniform flow and satisfies the classical Chezy equation (10.19). Straight prismatic channels can be optimized to find the cross section that gives maximum flow rate with minimum friction losses. As the slope and flow velocity increase, the channel reaches a critical condition of Froude number unity, where velocity equals the speed of a small-amplitude surface wave in the channel. Every channel has a critical slope that varies with the flow rate and roughness. If the flow becomes supercritical (Fr > 1), it may undergo a hydraulic jump to a greater depth and lower (subcritical) velocity, analogous to a normal shock wave. 724 Chapter 10 Open-Channel Flow The analysis of gradually varied flow leads to a differential equation (10.51) that can be solved by numerical methods. The chapter ends with a discussion of the flow over a dam or weir, where the total flow rate can be correlated with upstream water depth. Problems Most of the problems herein are fairly straightforward. More difficult or open-ended assignments are labeled with an asterisk. Problems labeled with a computer icon may require the use of a computer. The standard end-of-chapter problems PI 0.1 to PI 0.128 (categorized in the problem list here) are followed by word problems WlO.l to W10.13, fundamentals of engineering exam problems FElO.l to FE10.7, comprehensive problems ClO.l to C10.7, and design projects DlO.l and D10.2. Problem Distribution Section Topic Problems 10.1 Introduction: Froude number, wave speed PlO.l-PlO.lO 10.2 Uniform flow: the Chezy formula PlO.l DP10.36 10.3 Efficient uniform-flow channels P10.37-P10.47 10.4 Specific energy: critical depth P10.48-P10.58 10.4 Flow over a bump P10.59-P10.68 10.4 Sluice gate flow P10.69-P10.78 10.5 The hydraulic jump P10.79-P10.96 10.6 Gradually varied flow P10.97-P10.112 10.7 Weirs and flumes P10.113-P10.123 10.7 Backwater curves P10.124-P10.128 Introduction: Fronde nnmber, wave speed PlO.l The formula for shallow-water wave propagation speed, Eq. (10.9) or (10.10), is independent of the physical prop¬ erties of the liquid, like density, viscosity, or surface ten¬ sion. Does this mean that waves propagate at the same speed in water, mercury, gasoline, and glycerin? Explain. PlO.l Water at 20°C flows in a 30-cm-wide rectangular channel at a depth of 10 cm and a flow rate of 80,000 cmVs. Esti¬ mate (a) the Eroude number and (b) the Reynolds number. P10.3 Narragansett Bay is approximately 21 (statute) mi long and has an average depth of 42 ft. Tidal charts for the area indi¬ cate a time delay of 30 min between high tide at the mouth of the bay (Newport, Rhode Island) and its head (Provi¬ dence, Rhode Island). Is this delay correlated with the propagation of a shallow-water tidal crest wave through the bay? Explain. P10.4 The water flow in Fig. P10.4 has a free surface in three places. Does it qualify as an open-channel flow? Explain. What does the dashed line represent? P10.5 Water flows down a rectangular channel that is 4 ft wide and 2 ft deep. The flow rate is 20,000 gal/min. Estimate the Froude number of the flow. P10.6 Pebbles dropped successively at the same point, into a water channel flow of depth 42 cm, create two circular ripples, as in Fig. PI 0.6. From this information estimate (a) the Froude number and (b) the stream velocity. P10.6 P10.7 Pebbles dropped successively at the same point, into a water channel flow of depth 65 cm, create two circular ripples, as in Fig. P10.7. From this information estimate (a) the Froude number and (b) the stream velocity. P10.8 An earthquake near the Kenai Peninsula, Alaska, creates a single “tidal” wave (called a tsunami) that propagates southward across the Pacific Ocean. If the average ocean depth is 4 km and seawater density is 1025 kg/m^, estimate the time of arrival of this tsunami in Hilo, Hawaii. Problems 725 P10.9 Equation (10.10) is for a single disturbance wave. For periodic small-amplitude surface waves of wavelength X and period T, inviscid theory [8 to 10] predicts a wave propagation speed 2 M^.27iy = where y is the water depth and surface tension is neglected, (a) Determine if this expression is affected by the Reynolds number, Froude number, or Weber number. Derive the limiting values of this expression for (b)y and (c) y > X. (d ) For what ratio y/X is the wave speed within 1 percent of limit (f)? PIO.IO If surface tension Y is included in the analysis of Proh. P10.9, the resulting wave speed is [8 to 10] cl tanh 27iy (a) Determine if this expression is affected by the Reynolds number, Froude number, or Weber number. Derive the limiting values of this expression for ((2) y ^ X and (c) y ^ X. (d) Finally, determine the wavelength Xcrit for a mini¬ mum value of Co, assuming that y ^ X. Uniform flow: the Chezy formula PlO.ll A rectangular channel is 2 m wide and contains water 3 m deep. If the slope is 0.85° and the lining is corrugated metal, estimate the discharge for uniform flow. P10.12 (a) For laminar draining of a wide, thin sheet of water on pavement sloped at angle 0, as in Fig. P4.36, show that the flow rate is given by pebh^ sin 9 Q = — - 3/x where b is the sheet width and h its depth, (b) By (some¬ what laborious) comparison with Eq. (10.13), show that this expression is compatible with a friction factor / = 24/Re, where Re = P10.13 A large pond drains down an asphalt rectangular channel that is 2 ft wide. The channel slope is 0.8 degrees. If the flow is uniform, at a depth of 21 in, estimate the time to drain 1 acre-foot of water. P10.14 The Chezy formula (10.18) is independent of fluid density and viscosity. Does this mean that water, mercury, alcohol, and SAE 30 oil will all flow down a given open channel at the same rate? Explain. P10.15 The painted-steel channel of Fig. P10.15 is designed, with¬ out the barrier, for a flow rate of 6 mVs at a normal depth of 1 m. Determine (a) the design slope of the channel and (b) the reduction in total flow rate if the proposed painted- steel central harrier is installed. P10.15 Proposed barrier V / r 1 m 1 - 3 m - ► P10.16 Water flows in a brickwork rectangular channel 2 m wide, on a slope of 5 m/km. {a) Find the flow rate when the normal depth is 50 cm. (jb) If the normal depth remains 50 cm, find the channel width which will triple the flow rate. Comment on this result. P10.17 The trapezoidal channel of Fig. P10.17 is made of brick¬ work and slopes at 1:500. Determine the flow rate if the normal depth is 80 cm. P10.17 2 m P10.18 A V-shaped painted steel channel, similar to Fig. E10.6, has an included angle of 90°. If the slope, in uniform flow, is 3 m per km, estimate (a) the flow rate, in mVs and (b) the average wall shear stress. Take y = 2 m. P10.19 Modify Prob. P10.18, the 90° V channel, to let the surface be clean earth, which erodes if the average velocity ex¬ ceeds 6 ft/s. Find the maximum depth that avoids erosion. The slope is still 3 m per km. P10.20 An unfinished concrete sewer pipe, of diameter 4 ft, is flowing half-full at 39,500 U.S. gallons per minute. If this is the normal depth, what is the pipe slope, in degrees? P10.21 An engineer makes careful measurements with a weir (see Sec. 10.7) that monitors a rectangular unfinished concrete channel laid on a slope of 1°. She finds, perhaps with sur¬ prise, that when the water depth doubles from 2 ft 2 inches to 4 ft 4 inches, the normal flow rate more than doubles, from 200 to 500 tf/s. (a) Is this plausible? (b) If so, estimate the channel width. P10.22 For more than a century, woodsmen harvested trees in Skowhegan, ME, elevation 171 ft, and floated the logs down the Kennebec River to Bath, ME, elevation 62 ft, a distance of 72 miles. The river has an average depth of 14 ft and an average width of 400 ft. Assuming uniform flow and a stony bottom, estimate the travel time required for this trip. 726 Chapter 10 Open-Channel Flow P10.23 It is desired to excavate a clean-earth channel as a trapezoi¬ dal cross section with 0 = 60° (see Fig. 10.7). The expected flow rate is 500 ft^/s, and the slope is 8 ft per mile. The uniform flow depth is planned, for efficient performance, such that the flow cross section is half a hexagon. What is the appropriate bottom width of the channel? P10.24 A rectangular channel, laid out on a 0.5° slope, delivers a flow rate of 5000 gal/min in uniform flow when the depth is 1 ft and the width is 3 ft. (a) Estimate the value of Manning’s factor n. (b) What water depth will triple the flow rate? P10.25 The equilateral-triangle channel in Fig. P10.25 has con¬ stant slope So and constant Manning factor n.li y = all, find an analytic expression for the flow rate Q. where pj ~ 2400 kg/m^ is the density of sand. If the slope of the channel in Fig. P10.17 is 1:900 and n ~ 0.014, deter¬ mine the maximum water depth to keep from eroding particles of 1-mm diameter. P10.30 A clay tile V-shaped channel, with an included angle of 90°, is 1 km long and is laid out on a 1:400 slope. When running at a depth of 2 m, the upstream end is suddenly closed while the lower end continues to drain. Assuming quasi-steady normal discharge, find the time for the chan¬ nel depth to drop to 20 cm. P10.31 An unfinished-concrete 6-ft-diameter sewer pipe flows half full. What is the appropriate slope to deliver 50,000 gal/min of water in uniform flow? P10.32 Does half a V-shaped channel perform as well as a full V-shaped channel? The answer to Proh. 10.18 is Q = 12.4 mVs. (Do not reveal this to your friends still working on P10.18.) For the painted-steel half-V in Fig. P10.32, at the same slope of 3:1000, find the flow area that gives the same Q and compare with P10.18. P10.32 P10.26 In the spirit of Fig. 10. 6b, analyze a rectangular channel in uniform flow with constant area A = by, constant slope, hut varying width b and depth y. Plot the resulting flow rate Q, normalized hy its maximum value in the range 0.2 < b/y < 4.0, and comment on whether it is crucial for discharge efficiency to have the channel flow at a depth exactly equal to half the channel width. P10.27 A circular corrugated-metal water channel has a slope of 1:800 and a diameter of 6 ft. (a) Estimate the normal dis¬ charge, in gal/min, when the water depth is 4 ft. (b) Eor this condition, calculate the average wall shear stress. P10.28 A new, finished-concrete trapezoidal channel, similar to Eig. 10.7, has /> = 8 ft, y„ = 5 ft, and 9 = 50°. For this depth, the discharge is 500 ft^/s. (a) What is the slope of the channel? (b) As years pass, the channel corrodes and n doubles. What will be the new normal depth for the same flow rate? P10.29 Suppose that the trapezoidal channel of Fig. PI 0.1 7 con- tains sand and silt that we wish not to erode. According to an empirical correlation by A. Shields in 1936, the average wall shear stress T^nt required to erode sand particles of diameter dp is approximated by "^cril (.Ps P)S « 0.5 P10.33 Five sewer pipes, each a 2-m-diameter clay tile pipe running half full on a slope of 0.25°, empty into a single asphalt pipe, also laid out at 0.25°. If the large pipe is also to run half full, what should be its diameter? P10.34 A brick rectangular channel with So = 0.002 is designed to carry 230 ft^/s of water in uniform flow. There is an argu¬ ment over whether the channel width should be 4 or 8 ft. Which design needs fewer bricks? By what percentage? P10.35 In flood stage a natural channel often consists of a deep main channel plus two floodplains, as in Fig. P10.35. The floodplains are often shallow and rough. If the channel has the same slope everywhere, how would you analyze this situation for the discharge? Suppose that Vi = 20 ft, y2 = 5 ft, = 40 ft, ^2 = 100 ft, Ml = 0.020, and M2 = 0.040, with a slope of 0.0002. Estimate the discharge in ftVs. Problems 727 P10.36 The Blackstone River in northern Rhode Island normally flows at about 25 m^/s and resembles Fig. PI 0.35 with a clean-earth center channel, ~ 2Q m and yi « 3 m. The bed slope is about 2 ft/mi. The sides are heavy brush with ^2 ~ 150 m. During Hurricane Carol in 1954, a record flow rate of 1000 m^/s was estimated. Use this information to estimate the maximum flood depth y2 during this event. Efficient uniform-flow channels P10.37 A triangular channel (see Fig. E10.6) is to he constructed of corrugated metal and will carry 8 mVs on a slope of 0.005. The supply of sheet metal is limited, so the engineers want to minimize the channel surface. What are (a) the best included angle 0 for the channel, (b) the normal depth for part (a), and (c) the wetted perimeter for part (b)7 P10.38 For the half-Vee channel in Fig. P10.32, let the interior angle of the Vee be 0. For a given value of area, slope, and n, find the value of 0 for which the flow rate is a maximum. To avoid cumbersome algebra, simply plot Q versus 0 for constant A. P10.39 A trapezoidal channel has n = 0.022 and Sq = 0.0003 and is made in the shape of a half-hexagon for maximum efficiency. What should the length of the side of the hexa¬ gon be if the channel is to carry 225 ftVs of water? What is the discharge of a semicircular channel of the same cross- sectional area and the same Sq and n? P10.40 Using the geometry of Fig. 10.6a, prove that the most effi¬ cient circular open channel (maximum hydraulic radius for a given flow area) is a semicircle. P10.41 Determine the most efficient value of 0 for the V-shaped channel of Fig. P10.41. P10.41 P10.42 It is desired to deliver 30,000 gal/min of water in a brick¬ work channel laid on a slope of 1:100. Which would require fewer bricks, in uniform flow: (a) a V channel with 0 = 45°, as in Fig. P10.41, or (b) an efficient rectangular channel with b = 2y? P10.43 Determine the most efficient dimensions for a clay tile rectangular channel to carry 110,000 gal/min on a slope of 0.002. P10.44 What are the most efficient dimensions for a half-hexagon cast iron channel to carry 15,000 gal/min on a slope of 0.16°? P10.45 Calculus tells us that the most efficient wall angle for a V-shaped channel (Fig. P10.41) is P = 45°. It yields the highest normal flow rate for a given area. But is this a sharp or a flat maximum? For a flow area of 1 m^ and an unfin¬ ished-concrete channel with a slope of 0.004, plot the nor¬ mal flow rate Q, in mVs, versus angle for the range 30° £ 0 < 60° and comment. P10.46 It is suggested that a channel that reduces erosion has a para- bolic shape, as in Fig. P10.46. Formulas for area and perim¬ eter of the parabolic cross section are as follows [7, p. 36]: Vl + o? + - ln(Q: -f Vl + a^) OL 4 hg where a = - b For uniform flow conditions, determine the most efficient ratio hgib for this channel (minimum perimeter for a given constant area). P10.46 P10.47 Calculus tells us that the most efficient water depth for a rectangular channel (such as Fig. ElO.l) is y/b = 1/2. It yields the highest normal flow rate for a given area. But is this a sharp or a flat maximum? Eor a flow area of 1 m^ and a clay tile channel with a slope of 0.006, plot the normal flow rate Q, in m^/s, versus y/b for the range 0.3 £ y/b ^ 0.7 and comment. Specific energy: critical depth P10.48 A wide, clean-earth river has a flow rate g = 150 ftV(s • ft). What is the critical depth? If the actual depth is 12 ft, what is the Eroude number of the river? Compute the critical slope by (a) Manning’s formula and (b) the Moody chart. P10.49 Eind the critical depth of the brick channel in Prob. P10.34 for both the 4- and 8-ft widths. Are the normal flows sub- critical or supercritical? P10.50 A pencil point piercing the surface of a rectangular channel flow creates a wedgelike 25° half-angle wave, as in Pig. P10.50. If the channel surface is painted steel and the depth is 35 cm, determine (a) the Eroude number, (b) the critical depth, and (t ) the critical slope for uniform flow. 728 Chapter 10 Open-Channel Flow P10.50 P10.51 An unhnished concrete duct, of diameter 1.5 m, is flowing half-full at 8.0 m^/s. (a) Is this a critical flow? If not, what is (b) the critical flow rate, (c) the critical slope, and (d) the Froude number? (e) If the flow is uniform, what is the slope of the duct? P10.52 Water flows full in an asphalt half-hexagon channel of bottom width W. The flow rate is 12 mVs. Estimate W if the Froude number is exactly 0.60. P10.53 For the river flow of Prob. P10.48, find the depth y2 that has the same specific energy as the given depth Vi = 12 ft. These are called conjugate depths. What is Fr2? P10.54 A clay tile V-shaped channel has an included angle of 70° and carries 8.5 mVs. Compute {a) the critical depth, {b) the critical velocity, and (c) the critical slope for uniform flow. P10.55 A trapezoidal channel resembles Fig. 10.7 with b = I m and 0 = 50°. The water depth is 2 m, and the flow rate is 32 mVs. If you stick your fingernail in the surface, as in Fig. P10.50, what half-angle wave might appear? P10.56 A 4-ft-diameter finished-concrete sewer pipe is half full of water, {a) In the spirit of Fig. 10.4a, estimate the speed of propagation of a small-amplitude wave propagating along the channel, (b) If the water is flowing at 14,000 gal/min, calculate the Froude number. P10.57 Consider the V-shaped channel of arbitrary angle in Fig. P10.41. If the depth is y, (a) find an analytic expression for the propagation speed Cq of a small-disturbance wave along this channel. [Hint: Eliminate flow rate from the analyses in Sec. 10.4.] If 9 = 45° and the depth is 1 m, determine (b) the propagation speed and (c) the flow rate if the channel is running at a Eroude number of 1/3. P10.58 Eor a half-hexagon channel running full, find an analytic expression for the propagation speed of a small-disturbance wave travelling along this channel. Denote the bottom width as b and use Fig. 10.7 as a guide. Flow over a bump P10.59 Uniform water flow in a wide brick channel of slope 0.02° moves over a 10-cm bump as in Fig. P10.59. A slight depression in the water surface results. If the minimum water depth over the bump is 50 cm, compute (a) the velocity over the bump and (b) the flow rate per meter of width. P10.59 P10.60 Water, flowing in a rectangular channel 2 m wide, encoun¬ ters a bottom bump 10 cm high. The approach depth is 60 cm, and the flow rate 4.8 mVs. Determine (a) the water depth, (b) velocity, and (c) Froude number above the bump. Hint: The change in water depth is rather slight, only about 8 cm. P10.61 Modify Prob. PI 0.59 as follows: Again assuming uniform subcritical approach flow (Vi, yi), find (a) the flow rate and (b) y2 for which the flow at the crest of the bump is exactly critical (Fr2 = 1.0). P10.62 Consider the flow in a wide channel over a bump, as in Fig. P10.62. One can estimate the water depth change or transition with frictionless flow. Use continuity and the Bernoulli equation to show that dy dh/d.x dx ~ 1 - V^Hgy) Is the drawdown of the water surface realistic in Fig. P10.62? Explain under what conditions the surface might rise above its upstream position Vq. P10.63 In Fig. PI 0.62 let Vq “ 1 rn^s andyo = 1 m. If the maximum bump height is 15 cm, estimate (a) the Froude number over the top of the bump and (b) the maximum depression in the water surface. ^ Vo ^ V(x) P10.62 P10.64 For the rectangular channel in Prob. P10.60, the Froude number over the bump is about 1.37, which is 17 percent less than the approach value. For the same entrance condi¬ tions, find the bump height Ah that causes the bump Froude number to be 1 .00. Problems 729 P10.65 Program and solve the differential equation of “frictionless flow over a bump,” from Prob. P10.62, for entrance condi¬ tions 14 = 1 rn/s and Vq = 1 m. Let the bump have the convenient shape h = 0.5/!niax[l ~ cos (27rx/L)], which simulates Fig. P10.62. Let L = 3 m, and generate a numerical solution for y{x) in the bump region 0 < x < L. If you have time for only one case, use /t^ax =15 cm (Prob. P10.63), for which the maximum Froude number is 0.425. If more time is available, it is instructive to examine a com¬ plete family of surface profiles for ~ 1 cm up to 35 cm (which is the solution of Prob. P10.64). P10.66 In Fig. PI0.62, let I4 = 5.5 m/s and = 90 cm. (a) Will the water rise or fall over the bump? (b) For a bump height of 30 cm, determine the Froude number over the bump, (c) Find the bump height that will cause critical flow over the bump. P10.67 Modify Prob. P10.63 so that the 15-cm change in bottom level is a depression, not a bump. Estimate (a) the Froude number above the depression and (b) the maximum change in water depth. P10.68 Modify Prob. P10.65 to have a supercritical approach condition Vg = 6 m/s and vg = 1 m. If you have time for only one case, use = 35 cm (Prob. P10.66), for which the maximum Froude number is 1 .47. If more time is avail¬ able, it is instructive to examine a complete family of sur¬ face profiles for 1 cm < < 52 cm (which is the solution to Prob. P10.67). P10.71 In Fig. P10.69 letyi = 95 cm andy2 = 50 cm. Estimate the flow rate per unit width if the upstream kinetic energy is (a) neglected and (b) included. P10.72 Water approaches the wide sluice gate of Eig. P10.72 at Vi = 0.2 m/s and yi = 1 m. Accounting for upstream kinetic energy, estimate at the outlet, section 2, the (a) depth, (b) velocity, and (c) Eroude number. P10.72 P10.73 In Eig. PI 0.69, let yi = 6 ft and the gate width h = 8 ft. Find the gate opening H that would allow a free-discharge flow of 30,000 gal/min under the gate. P10.74 With respect to Fig. P10.69, show that, for frictionless flow, the upstream velocity may be related to the water levels by „ /2g(yi - yi) Sluice gate flow P10.69 Given is the flow of a channel of large width b under a sluice gate, as in Fig. PI 0.69. Assuming frictionless steady flow with negligible upstream kinetic energy, derive a for¬ mula for the dimensionless flow ratio Q^/(yib^g) as a func¬ tion of the ratio y2lyi- Show by differentiation that the maximum flow rate occurs at y2 = 2yi/3. Gate >2 where K = yi/y2- P10.75 A tank of water 1 m deep, 3 m long, and 4 m wide into the paper has a closed sluice gate on the right side, as in Fig. P10.75. At f = 0 the gate is opened to a gap of 10 cm. Assuming quasi¬ steady sluice gate theory, estimate the time required for the water level to drop to 50 cm. Assume free outflow. V 1 m Gate raised to / a gap of 10 cm Gate closed P10.69 P10.75 P10.70 A periodic and spectacular water release, in China’s Henan province, flows through a giant sluice gate. Assume that the gate is 23 m wide, and its opening is 8 m high. The water depth far upstream is 32 m. Assuming free discharge, estimate the volume flow rate through the gate. P10.76 Figure P10.76 shows a horizontal flow of water through a sluice gate, a hydraulic jump, and over a 6-ft sharp-crested weir. Channel, gate, jump, and weir are all 8 ft wide unfin¬ ished concrete. Determine (a) the flow rate in ft^/s and (b) the normal depth. 730 Chapter 10 Open-Channel Flow P10.77 Equation (10.41) for sluice gate discharge is for free outflow. If the outflow is drowned, as in Fig. 10.10c, there is dissipation, and Cj drops sharply, as shown in Fig. P10.77, taken from Ref. 2. Use this data to restudy Proh. 10.73, with H = 9 in. Plot the estimated flow rate, in gal/min, versus y2 in the range 0.5 ft < y2 < 5 ft. .Vl H P10.77 (From Ref. 2, p. 509.) P10.78 In Fig. P10.69, free discharge, a gate opening of 0.72 ft will allow a flow rate of 30,000 gal/min. Recall yi = 6 ft and the gate width ft = 8 ft. Suppose that the gate is drowned (Fig. P10.77), with y2 = 4 ft. What gate opening would then he required? The hydraulic jump P10.79 Show that the Froude number downstream of a hydraulic jump will be given by Fr2 = 8''^Fri/[(l -f 8 Fr?)‘“ - 1]^'^ Does the formula remain correct if we reverse subscripts 1 and 2? Why? P10.80 Water flowing in a wide channel 25 cm deep suddenly jumps to a depth of 1 m. Estimate (a) the downstream Froude number; (b) the flow rate per unit width; (c) the critical depth; and (d) the percentage of dissipation. P10.81 Water flows in a wide channel at ^ = 25 ftV(s ■ ft), yi = 1 ft, and then undergoes a hydraulic jump. Compute y2, V2, Fr2, hf, the percentage of dissipation, and the horsepower dissi¬ pated per unit width. What is the critical depth? P10.82 Downstream of a wide hydraulic jump the flow is 4 ft deep and has a Froude number of 0.5. Estimate (a) yj, (b) Vi, (c) Eri, (d) the percentage of dissipation, and (e) y^. P10.83 A wide-channel flow undergoes a hydraulic jump from 40 to 140 cm. Estimate (a) Vi, (b) V2, (c) the critical depth, in cm, and (d) the percentage of dissipation. P10.84 Consider the flow under the sluice gate of Fig. P10.84. If yi = 10 ft and all losses are neglected except the dissipa¬ tion in the jump, calculate y2 and y3 and the percentage of dissipation, and sketch the flow to scale with the EGL included. The channel is horizontal and wide. V — Jump Vi = 2 ft/s t M ^ ^3 ■ L t P10.85 The analogy between a hydraulic jump and a normal shock equates Mach number and Froude number, air density and water depth, air pressure and the square of the water depth. Test this analogy for Mai = Ffi “ 4.0 and comment on the results. P10.86 A bore is a hydraulic jump that propagates upstream into a still or slower-moving fluid, as in Fig. P10.86, on the See-Selune channel, near Mont Saint Michel in northwest France. The bore is moving at about 10 ft/s and is about one foot high. Estimate (a) the depth of the water in this area and (b) the velocity induced by the wave. P10.86 Tidal bore on the See-Selune river channel in northwest Erance. ( Courtesy of Prof. Hubert Chanson, University of Queensland.) Problems 731 P10.87 A tidal bore may occur when the ocean tide enters an estuary against an oncoming river discharge, such as on the Severn River in England. Suppose that the tidal hore is 10 ft deep and propagates at 13 mi/h upstream into a river that is 7 ft deep. Estimate the river current in kn. P10.88 Consider supercritical flow, Eri > 1, down a shallow flat water channel toward a wedge of included angle 29, as in Fig. P10.88. By the compressible flow analogy, hydraulic jumps should form, similar to the shock waves in Fig. P9.132fl. Using an approach similar to Fig. 9.20, develop and explain the equations that could be used to find the wave angle (3 and Fr2. P10.88 Frj > 1 P10.92 A circular hydraulic jump in a kitchen sink. (Courtesy of Prof. Hubert Chanson, University of Queensland.) P10.89 Water 30 cm deep is in uniform flow down a 1° unfinished concrete slope when a hydraulic jump occurs, as in Fig. P10.89. If the channel is very wide, estimate the water depth y2 downstream of the jump. For this problem, assume that two-dimensional jump the¬ ory is valid. If the water depth outside the jump is 4 mm, the radius at which the jump appears is R = 3 cm, and the faucet flow rate is 100 cm^/s, find the conditions just upstream of the jump. P10.93 Water in a horizontal channel accelerates smoothly over a hump and then undergoes a hydraulic jump, as in Fig. P10.93. If yi = 1 m and y^ = 40 cm, estimate (a) Vi, (b) V}, (c) y^, and (d) the hump height h. © © Jump P10.90 For the gate/jump/weir system sketched in Fig. P10.76, the flow rate was determined to he 379 ftVs. Determine (a) the water depths y2 and y^, and (b) the Froude numbers Fr2 and Fr3 before and after the hydraulic jump. P10.91 Follow up Prob. P10.88 numerically with flow down a shallow, flat water channel 1 cm deep at an average veloc¬ ity of 0.94 m/s. The wedge half-angle 9 is 20°. Calculate (a) I3- (b) Fr2; and (c) y2- P10.92 A familiar sight is the circular hydraulic jump formed by a faucet jet falling onto a flat sink surface, as in Fig. P10.92. Because of the shallow depths, this jump is strongly depen¬ dent on bottom friction, viscosity, and surface tension . It is also unstable and can form remarkable noncircular shapes, as shown in the website . P10.93 P10.94 In Fig. 10.11, the upstream flow is only 2.65 cm deep. The channel is 50 cm wide, and the flow rate is 0.0359 mVs. Determine (a) the upstream Froude number, (b) the down¬ stream velocity, (c) the downstream depth, and (d) the percent dissipation. P10.95 A 10-cm-high bump in a wide horizontal water channel creates a hydraulic jump just upstream and the flow pattern in Fig. P10.95. Neglecting losses except in the jump, for the case y^ = 30 cm, estimate (a) V4, (b) y4, (c) Vi, and {d)yi. 732 Chapter 10 Open-Channel Flow Bump: h= 10 cm P10.95 P10.96 For the circular hydraulic jump in Fig. P10.92, the water depths before and after the jump are 2 mm and 4 mm, respectively. Assume that two-dimensional jump theory is valid. If the faucet flow rate is 150 cm^/s, estimate the radius R at which the jump will appear. Gradually varied flow P10.97 A hrickwork rectangular channel 4 m wide is flowing at 8.0 mVs on a slope of 0.1°. Is this a mild, critical, or steep slope? What type of gradually varied solution curve are we on if the local water depth is (a) 1 m, (b) 1 .5 m, and (c) 2 m? P10.98 A gravelly earth wide channel is flowing at 10 mVs per meter of width on a slope of 0.75°. Is this a mild, critical, or steep slope? What type of gradually varied solution curve are we on if the local water depth is (a) 1 m, (b) 2 m, or (c) 3 m? P10.99 A clay tile V-shaped channel of included angle 60° is flowing at 1.98 mVs on a slope of 0.33°. Is this a mild, critical, or steep slope? What type of gradually varied solution curve are we on if the local water depth is (a) 1 m, (b) 2 m, or (c) 3 m? PlO.lOO If bottom friction is included in the sluice gate flow of Prob. P10.84, the depths (yi, y2, Vs) will vary with x. Sketch the type and shape of gradually varied solution curve in each region (1,2, 3), and show the regions of rapidly varied flow. PlO.lOl Consider the gradual change from the profile beginning at point a in Fig. PlO.lOl on a mild slope 5oi to a mild but steeper slope So2 downstream. Sketch and label the curve y{x) expected. P10.102 The wide-channel flow in Fig. P10.102 changes from a steep slope to one even steeper. Beginning at points a and b, sketch and label the water surface profiles expected for gradually varied flow. P10.102 P10.103 A gravelly rectangular channel, 7 m wide and 2 m deep, is flowing at 75 mVs on a slope of 0.013. (a) Is this on a mild, critical, or steep curve? (b) Approximately how many meters downstream will the gradually varied solu¬ tion reach the normal depth? P10.104 The rectangular-channel flow in Fig. P10.104 expands to a cross section 50 percent wider. Beginning at points a and b, sketch and label the water surface profiles expected for gradually varied flow. P10.104 105 In Prob. P10.84 the frictionless solution is y2 = 0.82 ft, which we denote as x = 0 just downstream of the gate. If the channel is horizontal with n = 0.018 and there is no hydraulic jump, compute from gradually varied theory the downstream distance where y = 2.0 ft. 106 A rectangular channel with n = 0.018 and a constant slope of 0.0025 increases its width linearly from b to 2b over a distance L, as in Fig. P10.106. (a) Determine the variation y(x) along the channel if = 4 m, L = 250 m, the initial depth is y(0) = 1.05 m, and the flow rate PlO.lOl Mild but steeper Problems 733 is 7 mVs. (b) Then, if your computer program is running well, determine the initial depth y(0) for which the exit flow will be exactly critical. P10.107 A clean-earth wide-channel flow is climbing an adverse slope with Sq = —0.002. If the flow rate is ^ = 4.5 m^/(s ■ m), use gradually varied theory to compute the distance for the depth to drop from 3.0 to 2.0 m. P10.108 Water flows at 1.5 m^/s along a straight, riveted-steel 90° V-shaped channel (see Fig. P10.41, 9 = 45°). At sec¬ tion 1, the water depth is 1.0 m. (a) As we proceed downstream, will the water depth rise or fall? Explain. (b) Depending upon your answer to part (a), calculate, in one numerical swoop, from gradually varied theory, the distance downstream for which the depth rises (or falls) 0.1 m. P10.109 Figure P10.109 illustrates a free overfall or dropdown flow pattern, where a channel flow accelerates down a slope and falls freely over an abrupt edge. As shown, the flow reaches critical just before the overfall. Between and the edge the flow is rapidly varied and does not satisfy gradually varied theory. Suppose that the flow rate is ^ = 1.3 mV(s • m) and the surface is unfinished concrete. Use Eq. (10.51) to estimate the water depth 300 m upstream as shown. / r? PlO.llO We assumed frictionless flow in solving the bump case, Prob. P10.65, for which V2 = 1-21 m/s andy2 = 0.826 m over the crest when = 15 cm, Vi = 1 m/s, and yi = 1 m. However, if the bump is long and rough, fric¬ tion may be important. Repeat Prob. P10.65 for the same bump shape, h = 0.5/!max[l ~ cos (Ittx/L)], to compute conditions (a) at the crest and (b) at the end of the bump, X = L. Let =15 cm and L = 100 m, and assume a clean-earth surface. P10.111 The Rolling Dam on the Blackstone River has a weedy bottom and an average flow rate of 900 ftVs. Assume the river upstream is 150 ft wide and slopes at 10 ft per statute mile. The water depth just upstream of the dam is 7.7 ft. Calculate the water depth one mile upstream {a) for the given initial depth, 7.7 ft; and (b) if flashboards on the dam raise this depth to 10.7 ft. P10.112 The clean-earth channel in Pig. P10.112 is 6 m wide and slopes at 0.3°. Water flows at 30 mVs in the channel and enters a reservoir so that the channel depth is 3 m just before the entry. Assuming gradually varied flow, how far is the distance L to a point in the channel where y = 2 m? What type of curve is the water surface? V Reservoir Weirs and flumes P10.113 Figure PlO.l 13 shows a channel contraction section often called a venturi flume [23, p. 167] because measurements of yi and y2 can be used to meter the flow rate. Show that if losses are neglected and the flow is one-dimensional and subcritical, the flow rate is given by r 2g(yi-y2) 1^” Vuibhl) - ll(b\y\). Apply this to the special case = 3 m, 1)2 = 2 m, and yi = 1.9 m. {a) Find the flow rate if y2 = 1-5 m. {b) Also find the depth y2 for which the flow becomes critical in the throat. P10.109 734 Chapter 10 Open-Channel Flow — Side view P10.113 P10.114 For the gate/jump/weir system sketched in Fig. P10.76, the flow rate was determined to he 379 ftVs. Determine the water depth y4 just upstream of the weir. P10.115 Gradually varied theory, Eq. (10.49), neglects the effect of width changes, dbldx, assuming that they are small. But they are not small for a short, sharp contraction such as the venturi flume in Fig. P10.113. Show that, for a rectangular section with b = b(x), Eq. (10.49) should he modified as follows: dy So- S+ [Vy(gb)]{db/dx) dx 1 - Fr^ Investigate a criterion for reducing this relation to Eq. ( 1 0.49). P10.116 A Cipolletti weir, popular in irrigation systems, is trapezoi¬ dal, with sides sloped at 1 :4 horizontal to vertical, as in Eig. PlO.l 16. The following are flow-rate values, from the U.S. Dept, of Agriculture, for a few different system parameters: P10.116 HJt 0.8 1.0 1.35 1.5 6, ft 1.5 2.0 2.5 3.5 Q, gal/min 1620 3030 5920 9740 Source: U.S. Dept of Agriculture. Use this data to correlate a Cipolletti weir formula with a reasonably constant weir coefficient. P10.117 A popular flow-measurement device in agriculture is the Par shall flume , Eig. PlO.l 17, named after its inven¬ tor, Ralph L. Parshall, who developed it in 1922 for the U.S. Bureau of Reclamation. The suhcritical approach flow is driven, hy a steep constriction, to go critical (y = yfl and then supercritical. It gives a constant reading H for a wide range of tail waters. Derive a formula for estimating Q from measurement of H and knowledge of constriction width b. Neglect the entrance velocity head. P10.117 The Parshall flume P10.118 Using a Bemoulli-type analysis similar to Fig. 10.16a, show that the theoretical discharge of the V-shaped weir in Fig. PlO.l 18 is given hy Q = 0.7542g‘'^ tan a P10.118 P10.119 Data by A. T. Lenz for water at 20°C (reported in Ref. 23) show a signiflcant increase of discharge coefficient of V-notch weirs (Fig. PlO.l 18) at low heads. Fora = 20°, some measured values are as follows: HJt 0.2 0.4 0.6 0.8 1.0 Q 0.499 0.470 0.461 0.456 0.452 Determine if these data can be correlated with the Reynolds and Weber numbers vis-a-vis Eq. (10.61). If not, suggest another correlation. Problems 735 P10.120 The rectangular channel in Fig. P10.120 contains a V-notch weir as shown. The intent is to meter flow rates between 2.0 and 6.0 mVs with an upstream hook gage set to measure water depths between 2.0 and 2.75 m. What are the most appropriate values for the notch height Y and the notch half-angle ct? P10.121 Water flow in a rectangular channel is to be metered by a thin-plate weir with side contractions, as in Table I0.2b, with L = 6 ft and Y = 1 ft. It is desired to measure flow rates between 1500 and 3000 gal/min with only a 6-in change in upstream water depth. What is the most appro¬ priate length for the weir width i? P10.122 In 1952 E. S. Crump developed the triangular weir shape shown in Fig. P10.122 [23, Chap. 4]. The front slope is 1:2 to avoid sediment deposition, and the rear slope is 1:5 to maintain a stable tailwater flow. The beauty of the design is that it has a unique discharge correlation up to near-drowning conditions, H2IH1 £ 0.75: / y2 \3/2 where ~ 0.63 and k/, ~ 0.3 mm The term kj, is a low-head loss factor. Suppose that the weir is 3 m wide and has a crest height T = 50 cm. If the water depth upstream is 65 cm, estimate the flow rate in gal/min. P10.122 The Crump weir [23, Chap. 4] P10.123 Water in a 20-ft-wide rectangular channel, flowing at 120 ft^/s and a depth of 10 ft, is to be metered by a rectangular weir with side contractions, as in Table 10.2h. Suggest some appropriate design values of b, Y, and H to match the table conditions for this weir. Backwater curves P10.124 Water flows at 600 ftVs in a rectangular channel 22 ft wide with n ~ 0.024 and a slope of 0.1°. A dam increases the depth to 15 ft, as in Fig. P10.124. Using gradually varied theory, estimate the distance L upstream at which the water depth will be 10 ft. What type of solution curve are we on? What should be the water depth asymptotically far upstream? Backwater curve PIO, til PIO PIO PIO 125 The Tupperware dam on the Blackstone River is 12 ft high, 100 ft wide, and sharp-edged. It creates a backwater similar to Fig. P10.124. Assume that the river is a weedy-earth rectan¬ gular channel 100 ft wide with a flow rate of 800 ft^/s. Esti¬ mate the water depth 2 mi upstream of the dam if Sq = 0.001 . 126 Suppose that the rectangular channel of Fig. PIO. 120 is made of riveted steel and carries a flow of 8 m^/s on a slope of 0.15°. If the V-notch weir has a = 30° and Y = 50 cm, estimate, from gradually varied theory, the water depth 100 m upstream. 127 A clean-earth river is 50 ft wide and averages 600 ftVs. It contains a dam that increases the water depth to 8 ft, to provide head for a hydropower plant. The bed slope is 0.0025. (a) What is the normal depth of this river? {b) Engineers propose putting dashboards on the dam to raise the water level to 10 ft. Residents a half mile upstream are worried about flooding above their present water depth of about 2.2 ft. Using Eq. (10.52) in one big half-mile step, estimate the new water depth upstream. 128 A rectangular channel 4 m wide is blocked by a broad-crested weir 2 m high, as in Fig. PIO. 128. The channel is horizontal for 200 m upstream and then slopes at 0.7° as shown. The flow rate is 12 m^/s, and n = 0.03. Compute the water depth y at 300 m upstream from gradually varied theory. y(x) P10.128 736 Chapter 10 Open-Channel Flow Word Problems WlO.l Free-surface problems are driven by gravity. Why do so many of the formulas in this chapter contain the square root of the acceleration of gravity? W10.2 Explain why the flow under a sluice gate. Fig. 10.10, either is or is not analogous to compressible gas flow through a converging-diverging nozzle. Fig. 9.12. W10.3 In uniform open-channel flow, what is the balance of forces? Can you use such a force balance to derive the Chezy equation (10.13)? W10.4 A shallow-water wave propagates at the speed Co ~ igy) What makes it propagate? That is, what is the balance of forces in such wave motion? In which direction does such a wave propagate? W10.5 Why is the Manning friction correlation, Eq. (10.16), used almost universally by hydraulics engineers, instead of the Moody friction factor? W10.6 During horizontal channel flow over a bump, is the spe¬ cific energy constant? Explain. W10.7 Cite some similarities, and perhaps some dissimilarities, between a hydraulic jump and a gas dynamic normal shock wave. Fundamentals of Engineering Exam Problems The FE Exam is fairly light on open-channel problems in the general (morning) session, but this subject plays a big part in the specialized civil engineering (afternoon) exam. FElO.l Consider a rectangular channel 3 m wide laid on a 1° slope. If the water depth is 2 m, the hydraulic radius is (a) 0.43 m, (b) 0.6 m, (c) 0.86 m, (d) 1.0 m, (e) 1.2 m FE10.2 Eor the channel of Prob. FElO.l, the most efficient water depth (best flow for a given slope and resistance) is (a) 1 m, (b) 1.5 m, (c) 2 m, (d) 2.5 m, (e) 3 m FE10.3 If the channel of Prob. EElO.l is built of rubble cement (Manning’s n ~ 0.020), what is the uniform flow rate when the water depth is 2 m? (a) 6 mVs, (b) 18 mVs, (c) 36 mVs, (d) 40 mVs, (e) 53 mVs Comprehensive Problems ClO.l Eebmary 1998 saw the failure of the earthen dam impound¬ ing California Jim’s Pond in southern Rhode Island. The resulting flood raised temporary havoc in the nearby vil¬ lage of Peace Dale. The pond is 17 acres in area and 15 ft deep and was full from heavy rains. The breach in the dam was 22 ft wide and 15 ft deep. Estimate the time required for the pond to drain to a depth of 2 ft. W10.8 Give three examples of rapidly varied flow. For each case, cite reasons why it does not satisfy one or more of the five basic assumptions of gradually varied flow theory. W10.9 Is a free overfall. Fig. 10.15e, similar to a weir? Could it be calibrated versus flow rate in the same manner as a weir? Explain. WIO.IO Cite some similarities, and perhaps some dissimilarities, between a weir and a Bernoulli obstruction flowmeter from Sec. 6.12. WlO.ll Is a bump, Eig. 10.9fl, similar to a weir? If not, when does a bump become large enough, or sharp enough, to be a weir? W10.12 After doing some reading and/or thinking, explain the design and operation of a long-throated flume. W10.13 Describe the design and operation of a critical-depth flume. What are its advantages compared to the venturi flume of Prob. P 1 0. 1 1 3 ? FE10.4 For the channel of Prob. FElO.l, if the water depth is 2 m and the uniform flow rate is 24 mVs, what is the approxi¬ mate value of Manning’s roughness factor n? (a) 0.015, (b) 0.020, (c) 0.025, (d) 0.030, (e) 0.035 FE10.5 For the channel of Prob. FElO.l, if Manning’s roughness factor n ~ 0.020 and Q ~ 29 mVs, what is the normal depth y„? (a) 1 m, (b) 1.5 m, (c) 2 m, (d) 2.5 m, (e) 3 m FE10.6 Eor the channel of Prob. EElO.l, if g ~ 24 mVs, what is the critical depth yfl (a) 1.0 m, (b) 1.26 m, (c) 1.5 m, (d) 1.87 m, (e) 2.0 m FE10.7 Eor the channel of Prob. EElO.l, if g ~ 24 mVs and the depth is 2 m, what is the Froude number of the flow? (a) 0.50, (b) 0.77, (c) 0.90, (d) 1.00, (e) 1.11 C10.2 A circular, unfinished concrete drainpipe is laid on a slope of 0.0025 and is planned to carry from 50 to 300 ftVs of runoff water. Design constraints are that (1) the water depth should be no more than three-fourths of the diame¬ ter and (2) the flow should always be subcritical. What is the appropriate pipe diameter to satisfy these require¬ ments? If no commercial pipe is exactly this calculated Comprehensive Problems 737 size, should you buy the next smallest or the next largest pipe? C10.3 Extend Prob. P10.72, whose solution was V2 ~ 4.33 m/s. (a) Use gradually varied theory to estimate the water depth 10 m downstream at section (3) for the 5° unfinished con¬ crete slope shown in Fig. P10.72. {b) Repeat your calcula¬ tion for an upward (adverse) slope of 5°. (c) When you find that part ib) is impossible with gradually varied theory, explain why and repeat for an adverse slope of 1°. C10.4 It is desired to meter an asphalt rectangular channel of width 1.5 m, which is designed for uniform flow at a depth of 70 cm and a slope of 0.0036. The vertical sides of the channel are 1 .2 m high. Consider using a thin-plate rectan¬ gular weir, either full or partial width (Table 10.2a,/)) for this purpose. Sturm [7, p. 51] recommends, for accurate correlation, that such a weir have T S 9 cm and HIY £ 2.0. Determine the feasibility of installing such a weir that will be accurate and yet not cause the water to overflow the sides of the channel. C10.5 Figure CIO. 5 shows a hydraulic model of a compound weir, one that combines two different shapes, (a) Other than measurement, for which it might be poor, what could be the engineering reason for such a weir? (b) For the prototype river, assume that both sections have sides at a 70° angle to the vertical, with the bottom section having a base width of 2 m and the upper section having a base width of 4.5 m, including the cut-out portion. The heights of lower and upper horizontal sections are 1 m and 2 m, respectively. Use engineering estimates and make a plot of upstream water depth versus Petaluma River flow rate in the range 0 to 4 m^/s. (c) For what river flow rate will the water overflow the top of the dam? C10.6 Figure Cl 0.6 shows a horizontal flow of water through a sluice gate, a hydraulic jump, and over a 6-ft sharp-crested weir. Channel, gate. Jump, and weir are all 8 ft wide unfin¬ ished concrete. Determine (a) the flow rate, (b) the normal depth, (c) y2, (d) Vs, and (e) 3)4. C10.5 ( Courtesy of the U.S. Army Corps of Engineers Waterways Experiment Station.) 738 Chapter 10 Open-Channel Flow C10.7 Consider the V-shaped channel in Fig. C10.7, with an arbitrary angle 0. Make a continuity and momentum anal¬ ysis of a small disturbance dy < y, as in Fig. 10.4. Show that the wave propagation speed in this channel is inde¬ pendent of 0 and does not equal the wide-channel result Co = fey)"". C10.7 Design Projects DlO.l A straight weedy-earth channel has the trapezoidal shape of Fig. 10.7, with h = 4 m and 0 = 35°. The channel has a con¬ stant bottom slope of 0.001. The flow rate varies seasonally from 5 up to 10 m"/s. It is desired to place a sharp-edged weir across the channel so that the water depth 1 km upstream remains at 2.0 m ± 10 percent throughout the year. Investi¬ gate the possibility of accomplishing this with a full-width weir; if successful, determine the proper weir height Y. If unsuccessful, try other alternatives, such as (a) a full-width broad-crested weir or (b) a weir with side contractions or (c) a V-notch weir. Whatever your final design, cite the seasonal variation of normal depths and critical depths for comparison with the desired year-round depth of 2 m. D10.2 The Caroselli Dam on the Pawcatuck River is 10 ft high, 90 ft wide, and sharp edged. The Coakley Company References 1. B. A. Bakhmeteff, Hydraulics of Open Channels, McGraw- Hill, New York, 1932. 2. V. T. Chow, Open Channel Hydraulics, Blackburn Press, Caldwell, NJ, 2009. 3. M. H. Chaudhry, Open Channel Flow, 2d ed.. Springer, New York, 2007. 4. R. Srivastava, Flow Through Open Channels, Oxford University Press, New York, 2008. 5. H. Chanson, The Hydraulics of Open Channel Flow, 2d ed., Elsevier, New York, 2004. 6. J. O. Akan, Open Channel Hydraulics, Butterworth-Heinemann, Woburn, MA, 2006. 7. T. W. Sturm, Open Channel Hydraulics, McGraw-Hill, New York, 2001. 8. J. Pedlosky, Waves in the Ocean and Atmosphere: Introduc¬ tion to Wave Dynamics, Springer, New York, 2003. 9. L. H. Holthuijsen, Waves in Oceanic and Coastal Waters, Cambridge University Press, New York, 2007. 10. M. K. Ochi, Ocean Waves: The Stochastic Approach, Cambridge University Press, London, 2008. uses this head to generate hydropower electricity and wants more head. They ask the town for permission to raise the dam higher. The river above the dam may be approximated as rectangular, 90 ft wide, sloping upstream at 12 ft per statute mile, and with a stony, cobbled bed. The average flow rate is 400 ftVs, with a 30-year predicted flood rate of 1200 ftVs. The river sides are steep until 1 mi upstream, where there are low-lying residences. The town council agrees the dam may be heightened if the new river level near these houses, during the 30-year flood, is no more than 3 ft higher than the present level during average flow con¬ ditions. You, as project engineer, have to predict how high the dam crest can be raised and still meet this requirement. 11. G. Masselink, M. Hughes, and J. Knight, Introduction to Coastal Processes and Geomorphology, 2d ed., Routledge, New York, 2011. 12. M. B. Abbott and W. A. Price, Coastal, Estuarial, and Harbor Engineers Reference Book, Taylor & Francis, New York, 1994. 13. P. D. Komar, Beach Processes and Sedimentation, 2d ed., Pearson Education, Upper Saddle River, NJ, 1998. 14. W. Yue, C.-L. Lin, and V. C. Patel, “Large Eddy Simulation of Turbulent Open Channel Elow with Eree Surface Simulated by Level Set Method,” Physics of Fluids, vol. 17, no. 2, Eeb. 2005, pp. 1-12. 15. J. M. Robertson and H. Rouse, “The Eour Regimes of Open Channel Elow,” Civ. Eng., vol. 11, no. 3, March 1941, pp. 169-171. 16. R. W. Powell, “Resistance to Plow in Rough Channels,” Trans. Am. Geophys. Union, vol. 31, no. 4, August 1950, pp. 575-582. 17. R. Manning, “On the Plow of Water in Open Channels and Pipes,” Trans. I.C.E. Ireland, vol. 20, 1891, PP- 161-207. References 739 18. “Friction Factors in Open Channels, Report of the Committee on Hydromechanics,” ASCE J. Hydraul. Div., March 1963, pp. 97-143. 19. E. F. Brater, H. W. King, J. E. Lindell, and C. Y. Wei, Hand¬ book of Hydraulics, 7th ed., McGraw-Hill, New York, 1996. 20. U.S. Bureau of Reclamation, “Research Studies on Stilling Basins, Energy Dissipators, and Associated Appurtenances,” Hydraulic Lab. Rep. Hyd-399, June 1, 1955. 21. P. A. Thompson, Compressible-Fluid Dynamics, McGraw- Hill, New York, 1972. 22. R. M. Olson and S. J. Wright, Essentials of Engineering Fluid Mechanics, 5th ed.. Harper & Row, New York, 1990. 23. P. Ackers et ah, Weirs and Flumes for Flow Measurement, Wiley, New York, 1978. 24. M. G. Bos, J. A. Replogle, and A. J. Clemmens, Flow Measuring Flumes for Open Channel Systems, American Soc. Agricul¬ tural and Biological Engineers, St. Joseph, MI, 1991. 25. M. G. Bos, Long-Throated Plumes and Broad-Crested Weirs, Springer- Verlag, New York, 1984. 26. D. H. Hoggan, Computer-Assisted Floodplain Hydrology and Hydraulics, 2d ed., McGraw-Hill, New York, 1996. 27. R. Jeppson, Open Channel Flow: Numerical Methods and Computer Applications, CRC Press, Boca Raton, FL, 2010. 28. R. Szymkiewicz, Numerical Modeling in Open Channel Hydraulics, Springer, New York, 2010. 29. R. Baban, Design of Diversion Weirs: Small Scale Irrigation in Hot Climates, Wiley, New York, 1995. 30. H. Chanson, Hydraulic Design of Stepped Cascades, Channels, Weirs, and Spillways, Pergamon Press, New York, 1994. 31. D. Kampion and A. Brewer, The Book of Waves: Form and Beauty on the Ocean, 3d ed., Rowman and Littlefield, Lanham, MD, 1997. 32. L. Mays, Water Resources Engineering, Wiley, New York, 2005. 33. D. K. Walkowiak (ed.), Isco Open Channel Flow Measure¬ ment Handbook, 5th ed.. Teledyne Isco, Inc., Lincoln, NE, 2006. 34. H. Chanson, “Photographic Observations of Tidal Bores (Mascarets) in Prance,” Hydraulic Model Report CH71/08, The University of Queensland, 2008, 104 pages. 35. E. J. Watson, “The Spread of a Liquid Jet over a Horizontal Plane,” J. Fluid Mechanics, vol. 20, 1964, pp. 481^99. 36. Z. Arendze and B. W. Skews, “Experimental and Numerical Study of the Hydraulic Analogy to Supersonic Flow,” South African Institution of Mechanical Engineering R&D Journal, vol. 24, 2008, pp. 9-15. 37. T. J. Mueller and W. L. Oberkampf, “Hydraulic Analog for the Expansion Deflection Nozzle,” AIAlA Journal, vol. 5, 1967, pp. 1200-1202. Wind turbines will play a large role in our energy future. The photo shows a 100 kW HAWT, installed in 2011 at the Fisherman’s Memorial State Camp Ground in Narragansett, Rhode Island. It is programmed to generate 100 kW in winds from 13 to 25 m/s and supplies half the electricity needed for the camp’s 18,000 annual visitors. Wind energy is good, but expensive. This turbine cost more, just to install, than it will recover in power savings over its 20-year life span. [Photo courtesy of F. M. White} 740 11.1 Introduction and Classification Classification of Pumps Chapter 11 Turbomachinery Motivation. The most common practical engineering application for fluid mechanics is the design of fluid machinery. The most numerous types are machines that add energy to the fluid (the pump family), but also important are those that extract energy (turbines). Both types are usually connected to a rotating shaft, hence the name turbomachinery. The purpose of this chapter is to make elementary engineering estimates of the performance of fluid machines. The emphasis will be on nearly incompressible flow: liquids or low-velocity gases. Basic flow principles are discussed, but not the detailed construction of the machines. Turbomachines divide naturally into those that add energy (pumps) and those that extract energy (turbines). The prefix turbo- is a Latin word meaning “spin” or “whirl,” appropriate for rotating devices. The pump is the oldest fluid energy transfer device known. At least two designs date before Christ: (1) the undershot-bucket waterwheels, or norias, used in Asia and Africa (1000 b.c.) and (2) Archimedes’ screw pump (250 b.c.), still being manufac¬ tured today to handle solid-liquid mixtures. Paddlewheel turbines were used by the Romans in 70 b.c., and Babylonian windmills date back to 700 b.c. . Machines that deliver liquids are simply called pumps, but if gases are involved, three different terms are in use, depending on the pressure rise achieved. If the pres¬ sure rise is very small (a few inches of water), a gas pump is called a. fan; up to 1 atm, it is usually called a blower; and above 1 atm it is commonly termed a compressor. There are two basic types of pumps: positive-displacement and dynamic or momentum- change pumps. There are several billion of each type in use in the world today. Positive-displacement pumps (PDFs) force the fluid along by volume changes. A cavity opens, and the fluid is admitted through an inlet. The cavity then closes, and the fluid is squeezed through an outlet. The mammalian heart is a good example, and 741 742 Chapter 11 Turbomachinery many mechanical designs are in wide use. References 35-38 give a summary of PDFs. A brief classification of PDP designs is as follows: A. Reciprocating 1. Piston or plunger 2. Diaphragm B. Rotary 1. Single rotor a. Sliding vane b. Flexible tube or lining c. Screw d. Peristaltic (wave contraction) 2. Multiple rotors a. Gear b. Lobe c. Screw d. Circumferential piston All PDFs deliver a pulsating or periodic flow as the cavity volume opens, traps, and squeezes the fluid. Their great advantage is the delivery of any fluid regardless of its viscosity. Figure 11.1 shows schematics of the operating principles of seven of these PDFs. It is rare for such devices to be run backward, so to speak, as turbines or energy extractors, the steam engine (reciprocating piston) being a classic exception. Since PDFs compress mechanically against a cavity filled with liquid, a common feature is that they develop immense pressures if the outlet is shut down for any reason. Sturdy construction is required, and complete shutoff would cause damage if pressure relief valves were not used. Dynamic pumps simply add momentum to the fluid by means of fast-moving blades or vanes or certain special designs. There is no closed volume: The fluid increases momentum while moving through open passages and then converts its high velocity to a pressure increase by exiting into a diffuser section. Dynamic pumps can be classified as follows: A. Rotary 1. Centrifugal or radial exit flow 2. Axial flow 3. Mixed flow (between radial and axial) B. Special designs 1. Jet pump or ejector (see Fig. P3.36) 2. Electromagnetic pumps for liquid metals 3. Fluid-actuated: gas lift or hydraulic ram We shall concentrate in this chapter on the rotary designs, sometimes called rotody- namic pumps. Other designs of both PDP and dynamic pumps are discussed in spe¬ cialized texts [for example, 3, 31]. 11.1 Introduction and Classification 743 Fig. 11.1 Schematic design of positive-displacement pumps: (a) reciprocating piston or plunger, (&) external gear pump, (c) double¬ screw pump, {d) sliding vane, (e) three-lobe pump, if) double circumferential piston, (g) flexible- tube squeegee. Dynamic pumps generally provide a higher flow rate than PDFs and a much steadier discharge hut are ineffective in handling high-viscosity liquids. Dynamic pumps also generally need priming; if they are filled with gas, they cannot suck up a liquid from helow into their inlet. The PDP, on the other hand, is self-priming for most applications. A dynamic pump can provide very high flow rates (up to 300,000 gal/min) but usually with moderate pressure rises (a few atmospheres). In contrast, 744 Chapter 11 Turbomachinery Fig. 11.2 Comparison of performance curves of typical dynamic and positive-displacement pumps at constant speed. 11.2 The Centrifugal Pump a PDF can operate up to very high pressures (300 atm) but typically produces low flow rates (100 gal/min). The relative performance (Ap versus Q) is quite different for the two types of pump, as shown in Fig. 11.2. At constant shaft rotation speed, the PDP produces nearly constant flow rate and virtually unlimited pressure rise, with little effect of viscosity. The flow rate of a PDP cannot be varied except by changing the displace¬ ment or the speed. The reliable constant-speed discharge from PDFs has led to their wide use in metering flows . The dynamic pump, by contrast in Fig. 11.2, provides a continuous constant-speed variation of performance, from near-maximum Ap at zero flow (shutoff conditions) to zero Ap at maximum flow rate. High-viscosity fluids sharply degrade the perfor¬ mance of a dynamic pump. As usual — and for the last time in this text — we remind the reader that this is merely an introductory chapter. Many books are devoted solely to turbomachines: generalized treatments [2 to 7], texts specializing in pumps [8 to 16, 30, 31], fans [17 to 20], compressors [21 to 23], gas turbines [24 to 26], hydropower [27, 28, 29, 32], and PDFs [35 to 38]. There are several useful handbooks [30 to 32], and at least two undergraduate textbooks [33, 34] have a comprehensive discussion of turbomachines. The reader is referred to these sources for further details. Let us begin our brief look at rotodynamic machines by examining the characteristics of the centrifugal pump. As sketched in Fig. 11.3, this pump consists of an impeller rotating within a casing. Fluid enters axially through the eye of the casing, is caught up in the impeller blades, and is whirled tangentially and radially outward until it leaves through all circumferential parts of the impeller into the diffuser part of the casing. The fluid gains both velocity and pressure while passing through the impeller. The doughnut- shaped diffuser, or scroll, section of the casing decelerates the flow and further increases the pressure. The impeller blades are usually backward-curved, as in Fig. 11.3, but there are also radial and forward-curved blade designs, which slightly change the output pressure. 11.2 The Centrifugal Pump 745 Fig. 11.3 Cutaway schematic of a typical centrifugal pump. The blades may be open (separated from the front casing only by a narrow clearance) or closed (shrouded from the casing on both sides by an impeller wall). The diffuser may be vaneless, as in Fig. 11.3, or fitted with fixed vanes to help guide the flow toward the exit. Basic Output Parameters Assuming steady flow, the pump basically increases the Bernoulli head of the flow between point 1, the eye, and point 2, the exit. From Eq. (3.73), neglecting viscous work and heat transfer, this change is denoted by H: H = 2g = K - hf (11.1) where h,. is the pump head supplied and hf the losses. The net head H is a primary output parameter for any turbomachine. Since Eq. (11.1) is for incompressible flow, it must be modified for gas compressors with large density changes. Usually V2 and Vi are about the same, Z2 ~ Zi is no more than a meter or so, and the net pump head is essentially equal to the change in pressure head: Pi ~ Pi H^— - — P8 P8 (11.2) The power delivered to the fluid simply equals the specific weight times the discharge times the net head change: P. = P8QH (11.3) This is traditionally called the water horsepower. The power required to drive the pump is the brake horsepower bhp = cjr (11.4) where u is the shaft angular velocity and T the shaft torque. If there were no losses, and brake horsepower would be equal, but of course P„ is actually less, and the efficiency rj of the pump is defined as P. PSQH bhp loT (11.5) ^Conversion factors may be needed: 1 hp = 550 ft • Ibf/s = 746 W. 746 Chapter 11 Turbomachinery Elementary Pump Theory The chief aim of the pump designer is to make r] as high as possible over as broad a range of discharge Q as possible. The efficiency is basically composed of three parts; volumetric, hydraulic, and mechanical. The volumetric efficiency is r]v Q Q + Ql (11.6) where is the loss of fluid due to leakage in the impeller casing clearances. The hydraulic efficiency is hf %=1-^ (11.7) K where hf has three parts: (1) shock loss at the eye due to imperfect match between inlet flow and the blade entrances, (2) friction losses in the blade passages, and (3) circulation loss due to imperfect match at the exit side of the blades. Finally, the mechanical efficiency is = 1-^ (11.8) where Pf is the power loss due to mechanical friction in the bearings, packing glands, and other contact points in the machine. By definition, the total efficiency is simply the product of its three parts: V^ri^Vhrtm (11-9) The designer has to work in all three areas to improve the pump. You may have thought that Eqs. (11.1) to (11.9) were formulas from pump theory. Not so; they are merely definitions of performance parameters and cannot be used in any predictive mode. To actually predict the head, power, efficiency, and flow rate of a pump, two theoretical approaches are possible: (1) simple one-dimensional flow formulas and (2) complex computer models that account for viscosity and three- dimensionality. Many of the best design improvements still come from testing and experience, and pump research remains a very active field . The last 10 years have seen considerable advances in computational fluid dynamics (CFD) modeling of flow in turbomachines , and at least eight commercial turbulent flow three- dimensional CFD codes are now available. To construct an elementary theory of pump performance, we assume one¬ dimensional flow and combine idealized fluid velocity vectors through the impeller with the angular momentum theorem for a control volume, Eq. (3.59). The idealized velocity diagrams are shown in Fig. 11.4. The fluid is assumed to enter the impeller at r = rj with velocity component Wi tangent to the blade angle Pi plus circumferential speed = utri matching the tip speed of the impeller. Its absolute entrance velocity is thus the vector sum of Wi and Mj, shown as Fj. Similarly, the flow exits at r = r2 with component W2 parallel to the blade angle P2 plus tip speed U2 = CL)r2, with resultant velocity ^2- 11.2 The Centrifugal Pump 747 Fig. 11.4 Inlet and exit velocity diagrams for an idealized pump impeller. We applied the angular momentum theorem to a turbomachine in Example 3.18 (Fig. 3.15) and arrived at a result for the applied torque T: T = pQirjVn - r,Va) (11.10) where Va and Vt2 are the absolute circumferential velocity components of the flow. The power delivered to the fluid is thus or Pu — — pQiuiVa miKi) H=-^ = -{U2V,2 - UiVn) PgQ 8 These are the Euler turbomachine equations, showing that the torque, power, and ideal head are functions only of the rotor-tip velocities Ui2 and the absolute fluid tangential velocities V,i 2, independent of the axial velocities (if any) through the machine. Additional insight is gained by rewriting these relations in another form. From the geometry of Fig. 11.4 — 2uw cos /3 w cos (3 = u — V, or uV, = + u^ - w^) (11.12) Substituting this into Eq. (11.11) gives H = ^[(Vl - Vj) + (ul - u\) - iwl - w?)] 2g (11.13) 748 Chapter 11 Turbomachinery Vi 90° 1(1 Ell.la Thus the ideal head relates to the absolute plus the relative kinetic energy change of the fluid minus the rotor-tip kinetic energy change. Finally, substituting for H from its definition in Eq. (11.1) and rearranging, we obtain the classic relation p - h z H - = const (11.14) Pg 2g 2g This is the Bernoulli equation in rotating coordinates and applies to either two- or three-dimensional ideal incompressible flow. For a centrifugal pump, the power can be related to the radial velocity V,, = V, tan a and the continuity relation Pw = PQiUlVnl cot a2 - UiV^i COtOi) (11.15) where V.2 Q 2nr2b2 and Q 2TTribi and where b^ and b2 are the blade widths at inlet and exit. With the pump parameters ri, r2, (3\, (32, and uu known, Eq. (11.11) or Eq. (11.15) is used to compute idealized power and head versus discharge. The “design” flow rate Q is commonly estimated by assuming that the flow enters exactly normal to the impeller: ai=90° ^1 = ^1 (11.16) We can expect this simple analysis to yield estimates within ±25 percent for the head, water horsepower, and discharge of a pump. Let us illustrate with an example. EXAMPLE 11.1 Given are the following data for a commercial centrifugal water pump: ri = 4 in, r2 = 1 in, /3i = 30°, (32 = 20°, speed = 1440 r/min. Estimate (a) the design point discharge, {b) the water horsepower, and (c) the head if bi = b2 = 1.75 in. Solution Part (a) The angular velocity is cu = 27r r/s = 27r(1440/60) = 150.8 rad/s. Thus the tip speeds are Ml = u)ri = 150.8(4/12) = 50.3 ft/s and U2 = uJr2 = 150.8(7/12) = 88.0 ft/s. From the inlet velocity diagram. Fig. El 1.1a, with Qi = 90° for design point, we compute V„i = Ml tan 30° = 29.0 ft/s 30° whence the discharge is Q = 27rri/)iT„i = (27r)(^^ ftj(^^ ftj(^29.0 ^ , /1728 = (8.87 ftVs)(60 s/min)( ^^gal/tf 1 = 3980 gal/ min Ans. (a) (The actual pump produces about 3500 gal/min.) 11.2 The Centrifugal Pump 749 Part (b) The outlet radial velocity follows from Q: Q V,.2 = 8.87 ftVs iTVrjbn 27r(i^ft) (if ft) = 16.6 ft/s Ell.lb This enables us to construct the outlet velocity diagram as in Fig. El l.li, given f)2 ~ 20°. The tangential component is Va = U2- V„2 cot P2 = 88.0 - 16.6 cot 20° = 42.4 ft/s -1 16-6 ,0 a, = tan - = 21.4° 42.4 The power is then computed from Eq. (11.11) with F,i = 0 at the design point: P„ = pQu2V,2 = (1.94 slugs/ft^)(8.87ftVs)(88.0ft/s)(42.4 ft/s) 64,100 ft ■ Ibf/s 550ft-lbf/(s-hp) 117hp Alls, (b) (The actual pump delivers about 125 water horsepower, requiring 147 bhp at 85 percent efficiency.) Part (c) Einally , the head is estimated from Eq. ( 1 1 . 1 1 ) : Py, _ 64,100 ft -Ibf/s PgQ (62.41bf/ft^)(8.87frVs) 116 ft Ans. (c) (The actual pump develops about 140-ft head.) Improved methods for obtaining closer estimates are given in advanced references [for example, 7, 8, and 31]. Effect of Blade Angle on Pump Head The simple theory just discussed can be used to predict an important blade-angle effect. If we neglect inlet angular momentum, the theoretical water horsepower is P. = pQiiiy,2 (11-17) Q where Va = M2 “ Vni cot (32 V„2 = - — 27rr202 Then the theoretical head from Eq. (11.11) becomes u\ U2 cot (32 Q g 2'Kr2b2g (11.18) The head varies linearly with discharge Q, having a shutoff value u^/g, where U2 is the exit blade-tip speed. The slope is negative if (32 < 90° (backward-curved blades) and positive for (32 > 90° (forward-curved blades). This effect is shown in Fig. 11.5 and is accurate only at low flow rates. The measured shutoff head of centrifugal pumps is only about 60 percent of the theoretical value Hq = LO^r\lg. With the advent of the laser-doppler anemometer, researchers can now make detailed three-dimensional flow measurements inside pumps and can even animate the data into a movie . 750 Chapter 11 Turbomachinery Fig. 11.5 Theoretical effect of blade exit angle on pump head versus discharge. 11.3 Pump Performance Curves and Similarity Rules Unstable: Can cause pump surge Discharge Q — ► The positive slope condition in Fig. 11.5 can be unstable and can cause pump surge, an oscillatory condition where the pump “hunts” for the proper operating point. Surge may cause only rough operation in a liquid pump, but it can be a major prob¬ lem in gas compressor operation. For this reason a backward-curved or radial blade design is generally preferred. A survey of the problem of pump stability is given by Greitzer . Since the theory of the previous section is rather qualitative, the only solid indicator of a pump’s performance lies in extensive testing. For the moment let us discuss the centrifugal pump in particular. The general principles and the presentation of data are exactly the same for mixed flow and axial flow pumps and compressors. Performance charts are almost always plotted for constant shaft rotation speed n (in r/min usually). The basic independent variable is taken to be discharge Q (in gal/min usually for liquids and ft^/min for gases). The dependent variables, or “output,” are taken to be head H (pressure rise Ap for gases), brake horsepower (bhp), and effi¬ ciency T]. Figure 11.6 shows typical performance curves for a centrifugal pump. The head is approximately constant at low discharge and then drops to zero at Q = Qmax- At this speed and impeller size, the pump cannot deliver any more fluid than The posi¬ tive slope part of the head is shown dashed; as mentioned earlier, this region can be unstable and can cause hunting for the operating point. The efficiency rj is always zero at no flow and at Qmax. and it reaches a maxi¬ mum, perhaps 80 to 90 percent, at about 0.6(2max- This is the design flow rate Q or best efficiency point (BEP), rj = tjmax- The head and horsepower at BEP will be termed H and P (or bhp), respectively. It is desirable that the efficiency curve be flat near rjmm, so that a wide range of efficient operation is achieved. However, some designs simply do not achieve flat efficiency curves. Note that rj is not independent of H and P but rather is calculated from the relation in Eq. (11.5), rj = pgQH/P. As shown in Fig. 11.6, the horsepower required to drive the pump typically rises monotonically with the flow rate. Sometimes there is a large power rise beyond the 11.3 Pump Performance Curves and Similarity Rules 751 Fig. 11.6 Typical centrifugal pump performance curves at constant impeller rotation speed. The units are arbitrary. Measured Performance Curves Positive slope may be unstable for certain Flow rate Q BEP, especially for radial-tipped and forward-curved blades. This is considered undesirable because a much larger motor is then needed to provide high flow rates. Backward-curved blades typically have their horsepower level off above BEP (“nonoverloading” type of curve). Figure 11.7 shows actual performance data for a commercial centrifugal pump. Figure 11.7fl is for a basic casing size with three different impeller diameters. The head curves H{Q) are shown, but the horsepower and efficiency curves have to be inferred from the contour plots. Maximum discharges are not shown, being far outside the normal operating range near the BEP. Everything is plotted raw, of course [feet, horsepower, gallons per minute (1 U.S. gal = 231 in^)] since it is to be used directly by designers. Figure W.lb is the same pump design with a 20 percent larger casing, a lower speed, and three larger impeller diameters. Comparing the two pumps may be a little confusing: The larger pump produces exactly the same discharge but only half the horsepower and half the head. This will be readily understood from the scaling or similarity laws we are about to formulate. A point often overlooked is that raw curves like Fig. 1 1.7 are strictly applicable to a fluid of a certain density and viscosity, in this case water. If the pump were used to deliver, say, mercury, the brake horsepower would be about 13 times higher while Q, H, and rj would be about the same. But in that case H should be interpreted as feet of mercury, not feet of water. If the pump were used for SAE 30 oil, all data would change (brake horsepower, Q, H, and rj) due to the large change in viscosity (Reynolds number). Again this should become clear with the similarity rules. 752 Chapter 11 Turbomachinery Fig. 11.7 Measured-performance curves for two models of a centrifugal water pump: (a) basic casing with three impeller sizes; (b) 20 percent larger casing with three larger impellers at slower speed. ( Courtesy of Ingersoll-Rand Corporation, Cameron Pump Division.) Net Positive-Suction Head 12 16 20 24 U.S. gallons per minute X 1000 (b) 28 50 40 'n X in 30 0. Z 20 25 20 15 10 in 0. Z In the top of Fig. 11.7 is plotted the net positive-suction head (NPSH), which is the head required at the pump inlet to keep the liquid from cavitating or boiling. The pump inlet or suction side is the low-pressure point where cavitation will hrst occur. The NPSH is dehned as Pi yf Pv NPSH = PP -y -L - Piy Pg 2g pg (11.19) 11.3 Pump Performance Curves and Similarity Rules 753 Deviations from Ideal Pump Theory where and V, are the pressure and velocity at the pump inlet and p^, is the vapor pressure of the liquid. Given the left-hand side, NPSH, from the pump performance curve, we must ensure that the right-hand side is equal or greater in the actual system to avoid cavitation. If the pump inlet is placed at a height Z, above a reservoir whose free surface is at pressure p^, we can use Bernoulli’s equation to rewrite NPSH as NPSH = -- Z, — (11.20) P8 Pg where hfj is the friction head loss between the reservoir and the pump inlet. Knowing Pa and hfi, we can set the pump at a height Z, that will keep the right-hand side greater than the “required” NPSH plotted in Fig. 11.7. If cavitation does occur, there will be pump noise and vibration, pitting damage to the impeller, and a sharp dropoff in pump head and discharge. In some liquids this deterioration starts before actual boiling, as dissolved gases and light hydrocarbons are liberated. The actual pump head data in Fig. 11.7 differ considerably from ideal theory, Eq. (11.18). Take, for example, the 36.75-in-diameter pump at 1 170 r/min in Fig. 11.7fl. The theoretical shutoff head is iJri [1170(27r/60) rad/s]^[(36.75/2)/(12) ft]^ //o(ideal) = - - - - = 1093 ft g 32.2 ft/s^ From Fig. 11.7a, at Q = 0, we read the actual shutoff head to be only 670 ft, or 61 percent of the theoretical value (see Prob. PI 1.24). This is a sharp dropoff and is indicative of nonrecoverable losses of three types: 1. Impeller recirculation loss, significant only at low flow rates. 2. Friction losses on the blade and passage surfaces, which increase monotonically with the flow rate. 3. “Shock” loss due to mismatch between the blade angles and the inlet flow direction, especially significant at high flow rates. These are complicated three-dimensional flow effects and hence are difficult to predict. Although, as mentioned, numerical (CFD) techniques are becoming more important , modern performance prediction is still a blend of experience, empirical correla¬ tions, idealized theory, and CFD modifications . EXAMPLE 11.2 The 32-in pump of Fig. 11.7a is to pump 24,000 gal/min of water at 1170 r/min from a reservoir whose surface is at 14.7 Ibf/in^ absolute. If head loss from reservoir to pump inlet is 6 ft, where should the pump inlet be placed to avoid cavitation for water at (a) 60°F, pu = 0.26 Ibf/in^ absolute, SG =1.0 and (b) 200°F, = 1 1 .52 Ibf/in^ absolute, SG = 0.9635? 754 Chapter 11 Turbomachinery Part (a) Part (b) Dimensionless Pump Performance Solution For either case read from Fig. 1 1 .7a at 24,000 gal/min that the required NPSH is 40 ft. For this case pg = 62.4 Ibf/ft^. From Eq. (11 .20) it is necessary that or NPSH : Pa - Pv P8 Zi hf. (14.7 - 0.26 lbf/in")( 144 inW) 40 ft < - Zi - 6.0 62.4 Ibf/ft^ or Z, < 27.3 - 40 = -12.7 ft Ans. (a) The pump must be placed at least 12.7 ft below the reservoir surface to avoid cavitation. For this case pg = 62.4(0.9635) = 60.1 Ibf/tf. Equation (11.20) applies again with the higher (14.7 - 1 1.52 Ibf/in^) (144 inW) 40 ft < - - 60.1 Ibf/ft^ -z, - 6.0 or Z,- < 1.6 - 40 = -38.4 ft Ans. (b) The pump must now be placed at least 38.4 ft below the reservoir surface. These are unusually stringent conditions because a large, high-discharge pump requires a large NPSH. For a given pump design, the output variables H and brake horsepower should be dependent on discharge Q, impeller diameter D, and shaft speed n, at least. Other possible parameters are the fluid density p, viscosity /i, and surface roughness e. Thus the performance curves in Fig. 11.7 are equivalent to the following assumed functional relations:^ gH = fi(Q,D,n,p,p,s) hhp = f2{Q,D,n,p,p,e) (11-21) This is a straightforward application of dimensional analysis principles from Chap. 5. As a matter of fact, it was given as an exercise (Example 5.3). For each function in Eq. (11.21) there are seven variables and three primary dimensions (M, L, and 70; hence we expect 7 — 3 = 4 dimensionless pi groups, and that is what we get. You can verify as an exercise that appropriate dimensionless forms for Eqs. (11.21) are gH 1 ' Q pnD^ e 2r»2 n D ' KtiD^ M ' D, bhp / ' Q pnD^ e pn^D^ M ’ H, (11.22) ^We adopt gH as a vaiiable instead of H for dimensional reasons. 11.3 Pump Performance Curves and Similarity Rules 755 The quantities pnD^/fi and elD are recognized as the Reynolds number and roughness ratio, respectively. Three new pump parameters have arisen: Capacity coefficient Cq = fiD'^ Head coefficient Power coefficient Cp gH bhp pr?D^ (11.23) Note that only the power coefficient contains fluid density, the parameters Cq and being kinematic types. Figure 11.7 gives no warning of viscous or roughness effects. The Reynolds numbers are from 0.8 to 1.5 X 10^, or fully turbulent flow in all passages probably. The roughness is not given and varies greatly among commercial pumps. But at such high Reynolds numbers we expect more or less the same percentage effect on all these pumps. Therefore it is common to assume that the Reynolds number and the roughness ratio have a constant effect, so that Eqs. (11.23) reduce to, approximately, C„^Ch{Cq) Cp^CpiCg) (11.24) For geometrically similar pumps, we expect head and power coefficients to be (nearly) unique functions of the capacity coefficient. We have to watch out that the pumps are geometrically similar or nearly so because (1) manufacturers put different-sized impel¬ lers in the same casing, thus violating geometric similarity, and (2) large pumps have smaller ratios of roughness and clearances to impeller diameter than small pumps. In addition, the more viscous liquids will have significant Reynolds number effects; for example, a factor-of-3 or more viscosity increase causes a clearly visible effect on Cfj and Cp. The efficiency rj is already dimensionless and is uniquely related to the other three. It varies with Cq also: ChCq ri(CQ) (11.25) We can test Eqs. (11.24) and (11.25) from the data of Fig. 11.7. The impeller diam¬ eters of 32 and 38 in are approximately 20 percent different in size, and so their ratio of impeller to casing size is the same. The parameters Cq, C^, and Cp are computed with n in r/s, Q in ft^/s (gal/min X 2.23 X 10”^), H and D in ft, g = 32.2 ft/s^, and brake horsepower in horsepower times 550 ft • lbf/(s ■ hp). The nondimensional data are then plotted in Fig. 11.8. A dimensionless suction head coefficient is also defined: ChS ~ g(NPSH) 2n2 n D ~ Ci]s{Cq) (11.26) 756 Chapter 11 Turbomachinery Fig. 11.8 Nondimensional plot of the pump performance data from Fig. 1 1.7. These numbers are not representative of other pump designs. The coefficients Cp and Cus are seen to correlate almost perfectly into a single func¬ tion of Cq, while rj and data deviate by a few percent. The last two parameters are more sensitive to slight discrepancies in model similarity; since the larger pump has smaller roughness and clearance ratios and a 40 percent larger Reynolds number, it develops slightly more head and is more efficient. The overall effect is a resounding victory for dimensional analysis. The best efficiency point in Fig. 11.8 is approximately 0.115 Cf = 0.65 r/max“ 0.88 (11.27) Ch = 5.0 Chs = 0.37 These values can be used to estimate the BEP performance of any size pump in this geometrically similar family. In like manner, the shutoff head is C//(0) ~ 6.0, and by extrapolation the shutoff power is C/.(0) ~ 0.25 and the maximum discharge is CQmax~ 0.23. Note, however, that Fig. 11.8 gives no reliable information about, say, the 28- or 35-in impellers in Fig. 11.7, which have a different impeller-to-casing-size ratio and thus must be correlated separately. By comparing values of n^D^, nD^, and for two pumps in Fig. 11.7, we can see readily why the large pump had the same discharge but less power and head: 11.3 Pump Performance Curves and Similarity Rules 757 Part (a) Part (b) Part (c) Part (d) D, ft II, r/s Discharge nD\ ffVs Head n^D^/g, ft Power pn^D’/SSO, hp Fig. 11. 7.3 32/12 1170/60 370 84 3527 Fig. 11.7/) 38/12 710/60 376 44 1861 Ratio - ■ — 1.02 0.52 0.53 Discharge goes as iiD^, which is about the same for both pumps. Head goes as and power as for the same p (water), and these are about half as much for the larger pump. The NPSH goes as and is also half as much for the 38-in pump. EXAMPLE 11.3 A pump from the family of Fig. 11.8 has Z) = 21 in and n = 1500 r/min. Estimate (a) dis¬ charge, (b) head, (c) pressure rise, and (d) brake horsepower of this pump for water at 60°F and best efficiency. Solution In BG units take D = 21/12 = 1.75 ft and n = 1500/60 = 25 r/s. At 60°F, p of water is 1.94 slugs/ft^. The BEP parameters are known from Fig. 11.8 or Eqs. (11.27). The BEP discharge is thus Q = CqmD^ = 0.115(25 r/s) (1. 75 ft)^ = (15.4 ftVs) (^448.8 = 6900 gal/min Ani’. (a) Similarly, the BEP head is = - g 5.0(25)^(1.75)^ 32.2 300-ft water Ans. (b) Since we are not given elevation or velocity head changes across the pump, we neglect them and estimate ^p « pgH = 1.94(32.2)(300) = 18,600 Ibf/ft^ = 1291bf/in^ Finally, the BEP power is P = Cp,pn^D^ = 0.65(1.94)(25)^(1.75)^ 323,000 ft ■ Ibf/s 550 = 590 hp Ans. (c) Ans. (d) EXAMPLE 11.4 We want to build a pump from the family of Pig. 11.8, which delivers 3000 gal/min water at 1200 r/min at best efficiency. Estimate (a) the impeller diameter, (b) the maximum discharge, (c) the shutoff head, and (d) the NPSH at best efficiency. 758 Chapter 11 Turbomachinery Similarity Rules Solution Part (a) Part (b) Part (c) Part (d) 3000 gal/min = 6.68 ft^/s and 1200 r/min = 20 r/s. At BEP we have Q = CqmD^ = 6.68 ftVs = (0.115)(20)D^ r 6.68 1'^ D = - = 1.43 ft = 17.1 in Lo.115(20)J The maximum Q is related to Q by a ratio of capacity coefficients: 0^2, max 3000(0.23) QmiDL = 6000 gal/min Cg. 0.115 From Fig. 1 1.8 we estimated the shutoff head coefficient to be 6.0. Thus mo) Ch{GWd^ 6.0(20)^(1.43)^ = 152 ft g 32.2 Finally, from Eq. (1 1.27), the NPSH at BEP is approximately CHsnW 0.37(20)^(1.43)^ NPSH = — - = = 9.4 ft g 32.2 Ans. (a) Ans. (b) Ans. (c) Ans. (d) Since this is a small pump, it will be less efficient than the pumps in Fig. 1 1.8, probably about 85 percent maximum. The success of Fig. 11.8 in correlating pump data leads to simple rules for comparing pump performance. If pump 1 and pump 2 are from the same geometric family and are operating at homologous points (the same dimensionless position on a chart such as Fig. 11.8), their flow rates, heads, and powers will be related as follows: Qi nXoJ Hi \nJ\Dj Pi Pi\nJ\Dj (11.28) These are the similarity rules, which can be used to estimate the effect of changing the fluid, speed, or size on any dynamic turbomachine — pump or turbine — within a geometrically similar family. A graphic display of these rules is given in Fig. 11.9, showing the effect of speed and diameter changes on pump performance. In Fig. 1 1 .9a the size is held constant and the speed is varied 20 percent, while Fig. \\.9b shows a 20 percent size change at constant speed. The curves are plotted to scale but with arbitrary units. The speed effect (Fig. 11.9a) is substantial, but the size effect (Fig. 1 1.9/7) is even more dramatic, especially for power, which varies as D^. Generally we see that a given pump family can be adjusted in size and speed to fit a variety of system characteristics. Strictly speaking, we would expect for perfect similarity that r)i = 772, but we have seen that larger pumps are more efficient, having a higher Reynolds number and lower roughness and clearance ratios. Two empirical correlations are recommended for 11.3 Pump Performance Curves and Similarity Rules 759 Fig. 11.9 Effect of changes in size and speed on homologous pump performance: (a) 20 percent change in speed at constant size; (b) 20 percent change in size at constant speed. Source: Courtesy of Vickers Inc., PDN/PACE Division. Effect of Viscosity D = 10 = constant Q ib) maximum efficiency. One, developed by Moody for turbines but also used for pumps, is a size effect. The other, suggested by Anderson from thousands of pump tests, is a flow rate effect: Size changes : Flow rate changes : 1 - r]2 1 - r]i 0.94 - r/2 0.94 - r]i {11. 29a) {n.29b) Anderson’s formula (11.29/?) makes the practical observation that even an infinitely large pump will have losses. He thus proposes a maximum possible efficiency of 94 percent, rather than 100 percent. Anderson recommends that the same formula be used for turbines if the constant 0.94 is replaced by 0.95. The formulas in Eq. (11.29) assume the same value of surface roughness for both machines — one could micropo¬ lish a small pump and achieve the efficiency of a larger machine. Centrifugal pumps are often used to pump oils and other viscous liquids up to 1000 times the viscosity of water. But the Reynolds numbers become low turbulent or even laminar, with a strong effect on performance. Figure 11.10 shows typical test curves of head and brake horsepower versus discharge. High viscosity causes a dramatic drop in head and discharge and increases in power requirements. The efficiency also drops substantially according to the following typical results: Atwater 1.0 10.0 100 1000 85 76 52 11 760 Chapter 11 Turbomachinery Fig. 11.10 Effect of viscosity on centrifugal pump performance. 11.4 Mixed- and Axial-Flow Pumps: The Specific Speed Part (a) Beyond about 300/i„ater the deterioration in performance is so great that a positive- displacement pump is recommended. We have seen from the previous section that the modern centrifugal pump is a for¬ midable device, able to deliver very high heads and reasonable flow rates with excel¬ lent efficiency. It can match many system requirements. But basically the centrifugal pump is a high-head, low-flow machine, whereas there are many applications requir¬ ing low head and high discharge. To see that the centrifugal design is not convenient for such systems, consider the following example. EXAMPLE 11.5 We want to use a centrifugal pump from the family of Fig. 11.8 to deliver 100,000 gal/min of water at 60°F with a head of 25 ft. What should be (a) the pump size and speed and (b) brake horsepower, assuming operation at best efficiency? Solution Enter the known head and discharge into the BEP parameters from Eq. (1 1.27): // = 25 ft = 32.2 Q = 100,000 gal/min = 222.8 ftVs = CqmD^ = O.llSnD^ The two unknowns are n and D. The algebra is quite simple, so we don’t really need Excel. Solve for n in the Q equation and substitute into the H equation: 222.8 1937 _5.0£)Yl937Y 0.115 ~ ’ ~ 32.2 V dV 582,840 or: O' = 23,314 Solve for D = 12.4 ft n = 1.03 r/s = 62 r/min Ans. (a) 11.4 Mixed- and Axial-Flow Pumps: The Specific Speed 761 Part (b) The most efficient horsepower is then, from Eq. (1 1.27), bhp ~ Cfpn^D^ 0.65(1.94)(1.03)^(12.4)^ 550 720 hp Ans. (b) The solution to Example 11.5 is mathematically correct but results in a grotesque pump: an impeller more than 12 ft in diameter, rotating so slowly one can visualize oxen walking in a circle turning the shaft. Other dynamic pump designs provide low head and high discharge. For example, there is a type of 38-in, 710 r/min pump, with the same input parameters as Fig. W.lb, which will deliver the 25-ft head and 100,000 gal/min flow rate called for in Example 11.5. This is done by allowing the flow to pass through the impeller with an axial-flow component and less centrifugal component. The passages can be opened up to the increased flow rate with very little size increase, but the drop in radial outlet velocity decreases the head produced. These are the mixed-flow (part radial, part axial) and axial-flow (propeller-type) families of dynamic pump. Some vane designs are sketched in Fig. 11.11, which introduces an interesting new “design” parameter, the specific speed or N^. Ns r/min (gal/min)i'V(//, (a) Fig. 11.11 (a) Optimum efficiency and (b) vane design of dynamic pump families as a function of specific speed. Specific speed Low High 500 1000 2000 4000 5000 10,000-15,000 (b) 762 Chapter 11 Turbomachinery The Specific Speed Suction Specific Speed Most pump applications involve a known head and discharge for the particular system, plus a speed range dictated by electric motor speeds or cavitation requirements. The designer then selects the best size and shape (centrifugal, mixed, axial) for the pump. To help this selection, we need a dimensionless parameter involving speed, discharge, and head but not size. This is accomplished by eliminating the diameter between Cq and C[], applying the result only to the BEP. This ratio is called the specific speed and has both a dimensionless form and a somewhat lazy, practical form: Rigorous form: K = ^1/2 ^3/4 C// (11.30fl) Lazy but common: (r/min) (gal/min) W)F (11.30^) In other words, practicing engineers do not bother to change n to revolutions per second or Q to cubic feet per second or to include gravity with head, although the latter would be necessary for, say, a pump on the moon. The conversion factor is N, = 17,182iV; Note that is applied only to BEP; thus a single number characterizes an entire family of pumps. Eor example, the family of Fig. 11.8 has NJ ~ (0.1 15)'^^/(5.0)^^'' = 0.1014, Nj = 1740, regardless of size or speed. It turns out that the specific speed is directly related to the most efficient pump design, as shown in Fig. 11.11. Low means low Q and high H, hence a centrifugal pump, and large implies an axial pump. The centrifugal pump is best for between 500 and 4000, the mixed-flow pump for Ns between 4000 and 10,000, and the axial-flow pump for A(, above 10,000. Note the changes in impeller shape as A(, increases. If we use NPSH rather than H in Eq. (1 1.30), the result is called suction-specific speed: Rigorous: (11.31fl) Lazy: (r/min) (gal/min) [NPSH (ft)]^''' (11.31fi) where NPSH denotes the available suction head of the system. Data from Wislicenus show that a given pump is in danger of inlet cavitation if Nss > 0.47 Nss s 8100 In the absence of test data, this relation can be used, given n and Q, to estimate the minimum required NPSH. 11.4 Mixed- and Axial-Flow Pumps: The Specific Speed 763 Axial-Flow Pump Theory Fig. 11.12 Analysis of an axial-flow pump: (a) basic geometry; (b) stator blades and exit velocity diagram; (f) rotor blades and exit velocity diagram. A multistage axial-flow geometry is shown in Fig. 11.12a. The fluid essentially passes almost axially through alternate rows of fixed stator blades and moving rotor blades. The incompressible flow assumption is frequently used even for gases because the pressure rise per stage is usually small. The simplified vector diagram analysis assumes that the flow is one-dimensional and leaves each blade row at a relative velocity exactly parallel to the exit blade angle. Figure 11.12/? shows the stator blades and their exit velocity diagram. Since the stator is fixed, ideally the absolute velocity Vi is parallel to the trailing edge of the blade. After vectorially subtracting the rotor tangential velocity u from Vj, we obtain the veloc¬ ity Wi relative to the rotor, which ideally should be parallel to the rotor leading edge. Figure 11.12c shows the rotor blades and their exit velocity diagram. Here the relative velocity W2 is parallel to the blade trailing edge, while the absolute velocity V2 should be designed to smoothly enter the next row of stator blades. Stator (a) (c) 764 Chapter 11 Turbomachinery Performance of an Axial-Flow Pump Pump Performance versus Specific Speed The theoretical power and head are given by Euler’s turbine relation (11.11). Since there is no radial flow, the inlet and exit rotor speeds are equal, Mj = U2, and one¬ dimensional continuity requires that the axial-velocity component remain constant: Q Vnl = V„2 = V„ = — = const A From the geometry of the velocity diagrams, the normal velocity (or volume flow) can be directly related to the blade rotational speed u: u = = V„i(cotai -I- cot/3i) = V„2{cota2 + cot/32) (11.32) Thus the flow rate can be predicted from the rotational speed and the blade angles. Meanwhile, since V,\ = y„i cot and Va = u — V„2 cot (32, Euler’s relation (11.11) for the pump head becomes gH = MV„(cota2 — cotOi) = — My„(cotQ;i 4- cot/32) (11.33) the preferred form because it relates to the blade angles Oj and (32- The shutoff or no-flow head is seen to be Hq = u^lg, just as in Eq. (11.18) for a centrifugal pump. The blade-angle parameter cot + cot (32 can be designed to be negative, zero, or positive, corresponding to a rising, flat, or falling head curve, as in Fig. 11.5. Strictly speaking, Eq. (11.33) applies only to a single streamtube of radius r, but it is a good approximation for very short blades if r denotes the average radius. For long blades it is customary to sum Eq. (11.33) in radial strips over the blade area. Such complexity may not be warranted since theory, being idealized, neglects losses and usually predicts the head and power larger than those in actual pump performance. At high specific speeds, the most efficient choice is an axial-flow, or propeller, pump, which develops high flow rate and low head. A typical dimensionless chart for a propeller pump is shown in Fig. 11.13. Note, as expected, the higher Cq and lower Cf] compared with Fig. 11.8. The head curve drops sharply with discharge, so a large system head change will cause a mild flow change. The power curve drops with head also, which means a possible overloading condition if the system discharge should suddenly decrease. Finally, the efficiency curve is rather narrow and triangular, as opposed to the broad, parabolic-shaped centrifugal pump efficiency (Fig. 11.8). By inspection of Fig. 11.13, Cq ~ 0.55, Cfj ~ 1.07, Cp ~ 0.70, and ~ 0.84. From this we compute NJ ~ (0.55)'^V(1.07)^^' = 0.705, = 12,000. The relatively low efficiency is due to small pump size: d = 14 in, n = 690 r/min, Q = 4400 gal/min. A repetition of Example 11.5 using Fig. 11.13 would show that this propeller pump family can provide a 25-ft head and 100,000 gal/min discharge if Z) = 46 in and n = 430 r/min, with bhp = 750; this is a much more reasonable design solution, with improvements still possible at larger-A^ conditions. Specific speed is such an effective parameter that it is used as an indicator of both performance and efficiency. Figure 11.14 shows a correlation of the optimum effi¬ ciency of a pump as a function of the specific speed and capacity. Because the 11.4 Mixed- and Axial-Flow Pumps: The Specific Speed 765 Fig. 11.13 Dimensionless performance curves for a typical axial-flow pump, = 12,000. Constructed from data given by Stepanoff for a 14-in pump at 690 r/min. o - Ch - Cp 1.0 0 0.2 0.4 0.6 0.8 n dimensional parameter 2 is a rough measure of both size and Reynolds number, rj increases with Q. When this type of correlation was first published by Wislicenus in 1947, it became known as the pump curve, a challenge to all manufacturers. We can check that the pumps of Figs. 11.7 and 11.13 fit the correlation very well. Fig. 11.14 Optimum efficiency of pumps versus capacity and specific speed. (Adapted from Refs. 4 and 31.) Ns 766 Chapter 11 Turbomachinery Fig. 11.15 Effect of specific speed on pump performance curves. 0 1 2 _e Q _e Q _e Q Figure 11.15 shows the effect of specific speed on the shape of the pump performance curves, normalized with respect to the BEP point. The numerical values shown are representative but somewhat qualitative. The high-specific-speed pumps (N^ ~ 10,000) have head and power curves that drop sharply with discharge, implying overload or start¬ up problems at low flow. Their efficiency curve is very narrow. A low-specific-speed pump (N^ = 600) has a broad efficiency curve, a rising power curve, and a head curve that “droops” at shutoff, implying possible surge or hunting problems. The Free Propeller The propeller-style pump of Fig. 11.12 is enclosed in a duct and captures all the approach flow. In contrast, the/ree propeller, for either aircraft or marine applications, acts in an unbounded fluid and thus is much less effective. The analog of propeller- pump pressure rise is the free propeller thrust per unit area (ttD^/T) swept out by the blades. In a customary dimensional analysis, thrust T and power required P are func¬ tions of fluid density p, rotation rate n (rev/s), forward velocity V, and propeller diameter D. Viscosity effects are small and neglected. You might enjoy analyzing this as a Chap. 5 assignment. The NACA (now the NASA) chose {p, n, D) as repeating variables, and the results are the accepted parameters: T Ct = thrust coefficient = — ^ — 7 pr?D^ fcn(7), J = advance ratio = — riD Cp = power coefficient = — 5 — 7 pn^D^ fcn(7), T] = efficiency VT P JCt (11.34) The advance ratio, J, which compares forward velocity to a measure proportional to blade tip speed, has a strong effect upon thrust and power. Figure 11.16 shows performance data for a propeller used on the Cessna 172 air¬ craft. The thrust and power coefficients are small, of 0(0.05), and are multiplied by 10 for plotting convenience. Maximum efficiency is 83 percent at / = 0.7, where Ct = 0.040 and Cp = 0.034. There are several engineering methods for designing propellers. These theories are described in specialized texts, both for marine and aircraft propellers. 11.5 Matching Pumps to System Characteristics 767 Fig. 11.16 Performance data for a free propeller used on the Cessna 172 aircraft. Compare to Fig. 11.13 for a (ducted) propeller pump. The thrust and power coefficients are much smaller for the free propeller. Computational Fluid Dynamics 11.5 Matching Pumps to System Characteristics VHnD) The design of turbomachinery has traditionally been highly experimental, with simple theories, such as in Sec. 11.2, only able to predict trends. Dimensionless correlations, such as Fig. 11.15, are useful but require extensive experimentation. Consider that flow in a pump is three-dimensional; unsteady (both periodic and turbulent); and involves flow separation, recirculation in the impeller, unsteady blade wakes passing through the diffuser, and blade roots, tips, and clearances. It is no wonder that one¬ dimensional theory cannot give Arm quantitative predictions. Modern computer analysis can give realistic results and is becoming a useful tool for turbomachinery designers. A good example is Ref. 56, reporting combined experi¬ mental and computational results for a centrifugal pump diffuser. A photograph of the device is shown in Fig. 11.17a. It is made of clear Perspex, so that laser measure¬ ments of particle tracking velocimetry (LPTV) and doppler anemometry (LDA) could be taken throughout the system. The data were compared with a CFD simulation of the impeller and diffuser, using the grids shown in Fig. \\.\lb. The computations used a turbulence formulation called the k-e model, popular in commercial CFD codes (see Sec. 8.9). Results were good but not excellent. The CFD model predicted velocity and pressure data adequately up until flow separation, after which it was only qualitative. Clearly, CFD is developing a significant role in turbomachinery design [42, 45]. The ultimate test of a pump is its match with the operating system characteristics. Physically, the system head must match the head produced by the pump, and this intersection should occur in the region of best efficiency. 768 Chapter 11 Turbomachinery Fig. 11.17 Turbomachinery design now involves both experimentation and computational fluid dynamics (CFD): (a) a centrifugal impeller and diffuser; (b) a three-dimensional CFD model grid for this system. Sources: (a) courtesy ofK. Eisele et al, "Flow Analysis in a Pump Diffuser: Part 1, Measurements: Parti, CFD, ” Journal of Fluids Fng. Vol. 119, December 1997, pp. 967-984/American Society of Mechanical Engineers (b) From K. Eisele et al ., "Row Analysis in a Pump Diffuser: Part I Measurements; Part 2, CFD, ” J. Fluids Eng., vol. 119, December 1997, pp. 967-984. by permission of the American Society of Mechanical Engineers. The system head will probably contain a static elevation change Z2 ~ Zi plus fric¬ tion losses in pipes and fittings: 11.5 Matching Pumps to System Characteristics 769 Fig. 11.18 Illustration of pump operating points for three types of system head curves. Pump ViQ) Laminar friction ' Static head Pump Qi Qi Qi Operating points where S K denotes minor losses and V is the flow velocity in the principal pipe. Since V is proportional to the pump discharge Q, the equation represents a system head curve H^Q). Three examples are shown in Fig. 11.18: a static head = a, static head plus laminar friction = a + bQ, and static head plus turbulent friction = a + cQ^. The intersection of the system curve with the pump performance curve H{Q) defines the operating point. In Fig. 11.18 the laminar friction operating point is at maximum efficiency while the turbulent and static curves are off design. This may be unavoidable if system variables change, but the pump should be changed in size or speed if its operating point is consistently off design. Of course, a perfect match may not be possible because commercial pumps have only certain discrete sizes and speeds. Let us illustrate these concepts with an example. EXAMPLE 11.6 We want to use the 32-in pump of Fig. 11.7a at 1170 r/min to pump water at 60°F from one reservoir to another 120 ft higher through 1500 ft of 16-in-lD pipe with friction factor / = 0.030. (a) What will the operating point and efficiency he? (b) To what speed should the pump he changed to operate at the BEP? Solution Part (a) For reservoirs the initial and final velocities are zero; thus the system head is H, = Z2- Zi V^fL 2g D 120 ft -f 0.030(1500 ft) 2^ P From continuity in the pipe, V = Q/A = 2/[j7r(jf ft)^], and so we suhstitute for V to get //,= 120 -f 0.2692^ ginfL/s (1) 770 Chapter 11 Turbomachinery Since Fig. 1 1.7fl uses thousands of gallons per minute for the abscissa, we convert Q in Eq. (1) to this unit: H, = 120 + 1.3352^ Q in 10^ gal/min (2) We can plot Eq. (2) on Fig. 1 1.7a and see where it intersects the 32-in pump head curve, as in Fig. El 1.6. A graphical solution gives approximately // « 430 ft e « 15,000 gal/min H The efficiency is about 82 percent, slightly off design. An analytic solution is possible if we fit the pump head curve to a parabola, which is very accurate: f/pump « 490 - 0.26G^ Q in 10^ gal/min (3) Equations (2) and (3) must match at the operating point: 490 - 0.26G^ = 120 + 1.335G- or 490 - 120 0.26 + 1.335 232 G = 15.2 X 10^ gal/min = 15,200 gal/min Ans. (a) H = 490 - 0.26(15.2)^ = 430 ft Ans. (a) Part (b) To move the operating point to BEP, we change n, which changes both Q ^ n and H oc Erom Eig. 11.7a, at BEP, H ~ 386 ft; thus for any n, H = 386(w/l 170)^. Also read Q ~ 20 X 10^ gal/min; thus for any n, Q = 20(w/1170). Match H to the system characteristics, Eq. (2): « 120 + 1.3351 20 Ans. (b) which gives n"<0 . Thus it is impossible to operate at maximum efficiency with this particular system and pump. 11.5 Matching Pumps to System Characteristics 771 Pumps Combined in Parallel Fig. 11.19 Performance and operating points of two pumps operating singly and (a) in parallel or (b) in series. Source: Copyright United Technologies Corporation 2008. Used with permission. If a pump provides the right head but too little discharge, a possible remedy is to combine two similar pumps in parallel, sharing the same suction and inlet conditions. A parallel arrangement is also used if delivery demand varies, so that one pump is used at low flow and the second pump is started up for higher discharges. Both pumps should have check valves to avoid backflow when one is shut down. The two pumps in parallel need not be identical. Physically, their flow rates will sum for the same head, as illustrated in Fig. 11.19a. If pump A has more head than pump B, pump B cannot be added in until the operating head is below the shutoff head of pump B. Since the system curve rises with Q, the combined delivery Qa+b will be less than the separate operating discharges Qa + Qb but certainly greater than H Pump A Combined H ib) 772 Chapter 11 Turbomachinery Pumps Combined in Series Multistage Pumps Compressors either one. For a very flat (static) curve two similar pumps in parallel will deliver nearly twice the flow. The combined brake horsepower is found by adding brake horsepower for each of pumps A and B at the same head as the operating point. The combined efficiency equals pg{Q^+g)(HA^B)/(55Q bhp^+5). If pumps A and B are not identical, as in Fig. 11.19fl, pump B should not be run and cannot even be started up if the operating point is above its shutoff head. If a pump provides the right discharge but too little head, consider adding a similar pump in series, with the output of pump B fed directly into the suction side of pump A. As sketched in Fig. 11.19h, the physical principle for summing in series is that the two heads add at the same flow rate to give the combined performance curve. The two need not be identical at all, since they merely handle the same discharge; they may even have different speeds, although normally both are driven by the same shaft. The need for a series arrangement implies that the system curve is steep, that is, it requires higher head than either pump A or B can provide. The combined operating point head will be more than either A or B separately but not as great as their sum. The combined power is the sum of brake horsepower for A and B at the operating point flow rate. The combined efficiency is Pg{QA+B){Hj^ + B) 550 bhp^+B similar to parallel pumps. Whether pumps are used in series or in parallel, the arrangement will be uneco¬ nomical unless both pumps are operating near their best efficiency. For very high heads in continuous operation, the solution is a multistage pump, with the exit of one impeller feeding directly into the eye of the next. Centrifugal, mixed- flow, and axial-flow pumps have all been grouped in as many as 50 stages, with heads up to 8000 ft of water and pressure rises up to 5000 Ibf/in^ absolute. Figure 1 1.20 shows a section of a seven-stage centrifugal propane compressor that develops 300 Ibf/in^ rise at 40,000 ftVmin and 35,000 bhp. Most of the discussion in this chapter concerns incompressible flow — that is, negli¬ gible change in fluid density. Even the pump of Fig. 11.7, which can produce 600 ft of head at 1170 r/min, will increase standard air pressure only by 46 Ibf/ft^, about a 2 percent change in density. The picture changes at higher speeds, Ap oc and multiple stages, where very large changes in pressure and density are achieved. Such devices are called compressors, as in Fig. 1 1 .20. The concept of static head, H = Ap/pg, becomes inappropriate, since p varies. Compressor performance is measured by (1) the pressure ratio across the stage polpi and (2) the change in stagnation enthalpy (hQ2 — hQi), where Hq = h + (see Sec. 9.3). Combining m stages in series results in Ffina/Finitiai ~ (PilPiT- As density increases, less area is needed: note the decrease in impeller size from right to left in Fig. 11.20. Compressors may be either of the centrifugal or axial-flow type [21 to 23]. 11.5 Matching Pumps to System Characteristics 773 Fig. 11.20 Cross section of a seven-stage centrifugal propane compressor that delivers 40,000 ft^/min at 35,000 bhp and a pressure rise of 300 Ibf/in^. Note the second inlet at stage 5 and the varying impeller designs. Source: Courtesy of DeLavai-Stork V.OF., Centrifugal Compressor Division. Compressor efficiency, from inlet condition 1 to final outlet /, is defined by the change in gas enthalpy, assuming an adiabatic process: t?comp hf hfji hof ~ hoi ^01 7q/' “ Tqi Compressor efficiencies are similar to hydraulic machines (??max ~ 70 to 80 percent), but the mass flow range is more limited: on the low side by compressor surge, where blade stall and vibration occur, and on the high side by choking (Sec. 9.4), where the Mach number reaches 1.0 somewhere in the system. Compressor mass flow is nor¬ mally plotted using the same type of dimensionless function formulated in Eq. (9.47): m(/?7’o)^^/(Z)^Po)> which will reach a maximum when choking occurs. For further details, see Refs. 21 to 23. 774 Chapter 11 Turbomachinery EXAMPLE 11.7 Investigate extending Example 11.6 by using two 32-in pumps in parallel to deliver more flow. Is this efficient? Solution Since the pumps are identical, each delivers at the same 1170 r/min speed. The system curve is the same, and the balance-of-head relation becomes H = 490 - 0.26(^Qf = 120 -f 1.3352^ , 490 - 120 or 2 = - Q = 16,300 gal/ mm Ans. 1.335 + 0.065 This is only 7 percent more than a single pump. Each pump delivers = 8130 gal/min, for which the efficiency is only 60 percent. The total brake horsepower required is 3200, whereas a single pump used only 2000 bhp. This is a poor design. EXAMPLE 11.8 Suppose the elevation change in Example 11.6 is raised from 120 to 500 ft, greater than a single 32-in pump can supply. Investigate using two 32-in pumps in series at 1170 r/min. Solution Since the pumps are identical, the total head is twice as much and the constant 120 in the system head curve is replaced by 500. The balance of heads becomes H = 2(490 - 0.262^) = 500 -f 1.3352" , 980 - 500 , or 2 = — — — Q = 16.1 X 10^ gal/mm Ans. 1.335 -f 0.52 ^ ^ The operating head is 500 -I- 1.335(16.1)^ = 845 ft, or 97 percent more than that for a single pump in Example 1 1.5. Each pump is operating at 16.1 X 10'^ gal/min, which from Fig. 1 1.7a is 83 percent efficient, a pretty good match to the system. To pump at this operating point requires 4100 bhp, or about 2050 bhp for each pump. Gas Turbines Some modern devices contain both pumps and turbines. A classic case is the gas turbine, which combines a compressor, a combustion chamber, a turbine, and, often, a fan. Gas turbines are used to drive aircraft, helicopters. Army tanks, and small electric power plants. They have a higher power- to- weight ratio than reciprocating engines, but they spin at very high speeds and require high-temperature materials and thus are costly. The compressor raises the inlet air to pressures as much as 30 to 40 times higher, before entering the combustion chamber. The heated air then passes through a turbine, which drives the compressor. The airflow then exits to provide the thrust and is generally a supersonic flow. 11.6 Turbines 775 Fig. 11.21 Cutaway view of a Pratt & Whitney 6000 turbofan aircraft engine. ( Copyright United Technologies Corporation 2008. Used with permission.) 11.6 Turbines Reaction Turbines Fan LPC HPC Combustor HPT LPT Exhaust The example illustrated in Fig. 11.21 is a Pratt & Whitney 6000 turbofan aircraft engine. The large entrance fan greatly increases the airflow into the engine, some of which bypasses the compressors. The central flow enters a low-pressure (LPC) and a high-pressure (HPC) compressor and thence into the combustor. After combustion, the hot, high-velocity gases pass through a high-pressure turbine (HPT), which drives the HPC, and a low-pressure turbine (LPT), which separately drives both the LPC and the fan. The exhaust gases then create the thrust in the usual momentum-exchange manner. The engine shown, designed for shorter airline flights, has a maximum thrust of 24,000 Ibf. A turbine extracts energy from a fluid that possesses high head, but it is fatuous to say a turbine is a pump run backward. Basically there are two types, reaction and impulse, the difference lying in the manner of head conversion. In the reaction turbine, the fluid fills the blade passages, and the head change or pressure drop occurs within the impeller. Reaction designs are of the radial-flow, mixed-flow, and axial-flow types and are essentially dynamic devices designed to admit the high-energy fluid and extract its momentum. An impulse turbine first converts the high head through a nozzle into a high-velocity jet, which then strikes the blades at one position as they pass by. The impeller passages are not fluid-filled, and the jet flow past the blades is essentially at constant pressure. Reaction turbines are smaller because fluid fills all the blades at one time. Reaction turbines are low-head, high-flow devices. The flow is opposite that in a pump, entering at the larger-diameter section and discharging through the eye after giving up most of its energy to the impeller. Early designs were very inefficient because they lacked stationary guide vanes at the entrance to direct the flow smoothly into the impeller passages. The first efficient inward-flow turbine was built in 1849 by James B. Francis, a U.S. engineer, and all radial- or mixed-flow designs are now called Francis turbines. At still lower heads, a turbine can be designed more compactly with purely axial flow and is termed a propeller turbine . The propeller may be either fixed- blade or adjustable (Kaplan type), the latter being complicated mechanically but much 776 Chapter 11 Turbomachinery Fig. 11.22 Reaction turbines: {a) Francis, radial type; {b) Francis mixed-flow; (c) propeller axial- flow; {d) performance curves for a Francis turbine, n = 600 r/min, D = 2.25 ft, N,p = 29. Idealized Radial Turbine Theory Power Specific Speed Cp (d) more efficient at low-power settings. Figure 11.22 shows sketches of runner designs for Francis radial, Francis mixed-flow, and propeller-type turbines. The Euler turbomachine formulas (11.11) also apply to energy-extracting machines if we reverse the flow direction and reshape the blades. Figure 11.23 shows a radial turbine runner. Again assume one-dimensional frictionless flow through the blades. Adjustable inlet guide vanes are absolutely necessary for good efficiency. They bring the inlet flow to the blades at angle 0:2 ^nd absolute velocity V’2 for minimum “shock” or directional-mismatch loss. After vectorially adding in the runner tip speed U2 = u;r2, the outer blade angle should be set at angle (32 to accommodate the relative velocity W2, as shown in the figure. (See Fig. 11.4 for the analogous radial pump velocity diagrams.) Application of the angular momentum control volume theorem, Eq. (3.59), to Fig. 11.23 (see Example 3.18 for a similar case) yields an idealized formula for the power P extracted by the runner: P = ioT = pu)Q{r2V,2 - riV,i) = pQ(u2V2 cos 0:2 - u^Vi cos a^) (11.35) where y,2 and Vfi are the absolute inlet and outlet circumferential velocity components of the flow. Note that Eq. (11.35) is identical to Eq. (11.11) for a radial pump, except that the blade shapes are different. The absolute inlet normal velocity V„2 = ^2 sin 0:2 is proportional to the flow rate Q. If the flow rate changes and the runner speed U2 is constant, the vanes must be adjusted to a new angle 02 so that W2 still follows the blade surface. Thus adjustable inlet vanes are very important to avoid shock loss. Turbine parameters are similar to those of a pump, but the dependent variable is the output brake horsepower, which depends on the inlet flow rate Q, available head H, 11.6 Turbines 111 Fig. 11.23 Inlet and outlet velocity diagrams for an idealized radial- flow reaction turbine runner. Adjustable guide vane impeller speed n, and diameter D. The efficiency is the output brake horsepower divided by the available water horsepower pgQH. The dimensionless forms are Cq, C[], and Cp, defined just as for a pump, Eqs. (11.23). If we neglect Reynolds number and roughness effects, the functional relationships are written with Cp as the indepen¬ dent variable: gH Q bhp CH = ^=C„iCp) Cq = ^=Cq(Cp) r] = ^- = riiCp) (11.36) n D tiD PgQH bhp where Cp = — ^ j pn^D^ Figure ll.22d shows typical performance curves for a small Francis radial turbine. The maximum efficiency point is called the normal power, and the values for this particular turbine are Tjmax = 0.89 Cp = 2.70 Cq = 0.34 Ch = 9.03 A parameter that compares the output power with the available head, independent of size, is found by eliminating the diameter between C^ and Cp. It is called the power specific speed: Rigorous form: Lazy but common: C%'‘- n(bhp)'^^ (r/min)(bhp)''- [//(ft)]"'" (11.37a) (11.37Z7) 778 Chapter 11 Turbomachinery Impulse Turbines For water, p = 1.94 slugs/ft^ and N^p = 2133N'sp. The various turbine designs divide up nicely according to the range of power specific speed, as follows: Turbine type N,p range Ch range Impulse 1-10 15-50 Francis 10-110 5-25 Propeller: Water 100-250 1-4 Gas, steam 25-300 10-80 Note that N^p, like for pumps, is defined only with respect to the BEP and has a single value for a given turbine family. In Fig. 1 1.22^, N^p = 273.3(2.70)'^^/(9.03)^^'^ = 29, regardless of size. Like pumps, turbines of large size are generally more efficient, and Eqs. (11.29) can be used as an estimate when data are lacking. The design of a complete large-scale power-generating turbine system is a major engineering project, involving inlet and outlet ducts, trash racks, guide vanes, wicket gates, spiral cases, generator with cooling coils, bearings and transmission gears, runner blades, draft tubes, and automatic controls. Some typical large-scale reaction turbine designs are shown in Fig. 11.24. The reversible pump-and-turbine design of Fig. 11.24(7 requires special care for adjustable guide vanes to be efficient for flow in either direction. The largest (1000-MW) hydropower designs are awesome when viewed on a human scale, as shown in Fig. 11.25. The economic advantages of small-scale model testing are evident from this photograph of the Francis turbine units at Grand Coulee Dam. For high head and relatively low power (that is, low N^p) not only would a reaction turbine require too high a speed but also the high pressure in the runner would require a massive casing thickness. The impulse turbine of Fig. 11.26 is ideal for this situa¬ tion. Since N^p is low, n will be low and the high pressure is confined to the small nozzle, which converts the head to an atmospheric pressure jet of high velocity V,. The jet strikes the buckets and imparts a momentum change similar to that in our control volume analysis for a moving vane in Example 3.9 or Prob. P3.51. The buckets have an elliptical split-cup shape, as in Fig. 11.26fi. They are named Pelton wheels, after Lester A. Pelton (1829-1908), who produced the first efficient design. In Example 3.10 we found that the force per unit mass flow on a single moving vane, or in this case a single Pelton bucket, was (V,- — m)(1 — cos /3), where u is the vane velocity and (3 is the exit angle of the jet. For a single vane, as in Example 3.10, the mass flow would be pAj{Vj — u), but for a Pelton wheel, where buckets keep entering the jet and capture all the flow, the mass flow would be pQ = pAjVj. An alternative analysis uses the Euler turbomachine equation (11.11) and the velocity diagram of Fig. 11.26c. Noting that Ui = U2 = u, we substitute the absolute exit and inlet tangential velocities into the turbine power relation: or P = pQ{uiV,i - U2Va) = pQ{uVj - u[u + (Vj - m)cos/3]} P = pQu{Vj — m)(1 — cos (3) (11.38) 11.6 Turbines 779 Fig. 11.24 Large-scale turbine designs depend on available head and flow rate and operating conditions: (a) Francis (radial); (b) Kaplan (propeller); (c) bulb mounting with propeller runner; (d) reversible pump turbine with radial runner. (Courtesy of Voith Siemens Hydro Power.) 780 Chapter 11 Turbomachinery Fig. 11.25 Interior view of the 1.1 -million hp (820-MW) turbine units on the Grand Coulee Dam of the Columbia River, showing the spiral case, the outer fixed vanes (“stay ring”), and the inner adjustable vanes (“wicket gates”). (Courtesy ofVoith Siemens Hydro Power.) Fig. 11.26 Impulse turbine: (a) side view of wheel and jet; (b) top view of bucket; (c) typical velocity diagram. (a) (c) 11.6 Turbines 781 where u = limr is the bucket linear velocity and r is the pitch radius, or distance to the jet centerline. A bucket angle /3 = 180° gives maximum power but is physically impractical. In practice, /3 ~ 165°, or 1 — cos (3 ~ 1.966 or only 2 percent less than maximum power. From Eq. (1 1.38) the theoretical power of an impulse turbine is parabolic in bucket speed u and is maximum when dPIdu = 0, or u = 27mr = jVj (11.39) For a perfect nozzle, the entire available head would be converted to jet velocity Vj = (2g//)^^. Actually, since there are 2 to 8 percent nozzle losses, a velocity coefficient C„ is used: Vj = C^{2gHy'^ 0.92 < C„ < 0.98 (11.40) By combining Eqs. (11.36) and (11.40), the theoretical impulse turbine efficiency becomes 77 = 2(1 — cos (3) 8000 ^0.485 where is fhe dimensional specific speed, Eq. (11.30i). Use this correlation to find the appropriate size for a fan that delivers 24,000 ft^/min of air at sea-level conditions when running at 1620 r/min with a pressure rise of 2 inches of water. Hint: Express the fan head in feet of air, not feet of water. P11.54 It is desired to pump 50 ft^/s of water at a speed of 22 r/s, against a head of 80 ft. (a) What type of pump would you recommend? Estimate (b) the required impeller diameter and (c) the brake horsepower. P11.55 Suppose that the axial-flow pump of Fig. 11.13, with D = 18 in, runs at 1800 r/min. (a) Could it efficiently pump 25,000 gal/min of water? (b) If so, what head would result? (c) If a head of 120 ft is desired, what values of D and n would be better? P11.56 Determine if the Bell and Gossett pump of Prob. PI 1.8 (a) fits the three correlations in Figs. P11.46, P11.49, and PI 1.50. (b) If so, use these correlations to find the flow rate and horsepower that would result if the pump is scaled up to D = 24 in but still runs at 1750 r/min. P11.57 Performance data for a 21 -in-diameter air blower running at 3550 r/min are as follows: A/7, in H2O 29 30 28 21 10 Q, fT/min 500 1000 2000 3000 4000 bhp 6 8 12 18 25 Note the fictitious expression of pressure rise in terms of water rather than air. What is the specific speed? How does fhe perfoiTnance compare with Fig. 1 1.8? What are C^, C%, and CP 798 Chapter 11 Turbomachinery P11.58 Aircraft propeller specialists claim that dimensionless propeller data, when plotted as (Cj/fi) versus (CpIJ^), form a nearly straight line, y = mx + b. (a) Test this hypothesis for the data of Fig. 11.16, in the high efficiency range J = VI(nD) equal to 0.6, 0.7, and 0.8. (b) If successful, try this straight line to predict the rotation rate, in r/min, for a propeller with Z) = 5 ft, P = 30 hp, T = 95 Ibf, and V = 95 mi/h, for sea level standard conditions. Comment. P11.59 Suppose it is desired to deliver 700 ftVmin of propane gas (molecular weight = 44.06) at 1 atm and 20°C with a single-stage pressure rise of 8.0 in H2O. Determine the appropriate size and speed for using the pump families of (a) Prob. P11.57 and (b) Fig. 11.13. Which is the better design? P11.60 Performance curves for a certain free propeller, compara¬ ble to Fig. 11.16, can be plotted as shown in Fig. PI 1.60, for thrust T versus speed V for constant power P. (a) What is striking, at least to the writer, about these curves? (b) Can you deduce this behavior by rearranging, or replotting, the data of Fig. 11.16? 1400 1000 800 600 400 200 0 250 hp — 350 hp — 450 hp “ -- ^ •% ^ ^ ^ ^ ^ ^ ^ Assume the same rotation rate, 1170 r/min, and estimate the flow rate this pump will provide to deliver water from a reservoir, through 900 ft of 12-in pipe, to a point 150 ft above the reservoir surface. Assume a friction factor /= 0.019. P11.64 A leaf blower is essentially a centrifugal impeller exiting to a tube. Suppose that the tube is smooth PVC pipe, 4 ft long, with a diameter of 2.5 in. The desired exit velocity is 73 mi/h in sea-level standard air. If we use the pump family of Eqs. (11.27) to drive the blower, what approximate (a) diameter and (b) rotation speed are appropriate? (c) Is this a good design? ^11.65 An 11.5-in-diameter centrifugal pump, running at 1750 r/min, delivers 850 gal/min and a head of 105 ft at best efficiency (82 percent), (a) Can this pump oper¬ ate efficiently when delivering water at 20°C through 200 m of 10-cm-diameter smooth pipe? Neglect minor losses, (b) If your answer to (a) is negative, can the speed n be changed to operate efficiently? (c) If your answer to (b) is also negative, can the impeller diame¬ ter be changed to operate efficiently and still run at 1750 rev/min? P11.66 It is proposed to run the pump of Prob. PI 1.35 at 880 r/min to pump water at 20°C through the system in Fig. PI 1.66. The pipe is 20-cm-diameter commercial steel. What flow rate in ftVmin will result? Is this an efficient application? 50 100 150 Speed, mi/h 200 250 P11.60 P11.61 A mine ventilation fan, running at 295 r/min, delivers 500 m^/s of sea-level air with a pressure rise of 1100 Pa. Is this fan axial, centrifugal, or mixed? Estimate its diam¬ eter in ft. If the flow rate is increased 50 percent for the same diameter, by what percentage will the pressure rise change? P11.62 The actual mine ventilation fan discussed in Prob. PI 1.61 had a diameter of 20 ft [Ref. 20, p. 339]. What would be the proper diameter for the pump family of Fig. 1 1.14 to pro¬ vide 500 m^/s at 295 r/min and BEP? What would be the resulting pressure rise in Pa? Matching pumps to system characteristics P11.63 A good curve-fit to the head vs. flow for the 32-in pump in Pig. 11.7a is H (in ft) « 500 - (2.9E-7) Q in gal/min V 4 m V 3 m ■ 20 m ■ - 12 m ■ P11.66 P11.67 The pump of Prob. PI 1.35, running at 880 r/min, is to pump water at 20°C through 75 m of horizontal galvanized iron pipe. All other system losses are neglected. Determine the flow rate and input power for (a) pipe diameter = 20 cm and (b) the pipe diameter found to yield maximum pump efficiency. P11.68 A popular small aircraft cruises at 230 km/h at 8500 ft altitude. It weighs 2200 Ibf, has a 180-hp engine, a 76-in-diameter propeller, and a drag-area C^A ~ 5.6 ft^. The propeller data in Fig. PI 1.68 is proposed to drive this aircraft. Estimate the required rotation rate, in r/min, and power delivered, in hp. [NOTE: Simply use the coefficient pairs. The actual advance ratio is too high.] Problems 799 60% 50% 40% 30% 20% 10% 0 P11.69 The pump of Prob. PI 1.38, running at 3500 r/min, is used to deliver water at 20°C through 600 ft of cast iron pipe to an elevation 100 ft higher. Determine (a) the proper pipe diameter for BEP operation and (b) the flow rate that results if the pipe diameter is 3 in. P11.70 The pump of Prob. PI 1.28, operating at 2134 r/min, is used with 20°C water in the system of Fig. PI 1.70. (a) If it is operating at BEP, what is the proper elevation Z2? (b) If Z2 = 225 ft, what is the flow rate if = 8 in.? P11.71 The pump of Prob. PI 1.38, running at 3500 r/min, delivers water at 20°C through 7200 ft of horizontal 5-in-diameter commercial steel pipe. There are a sharp entrance, sharp exit, four 90° elbows, and a gate valve. Estimate (a) the flow rate if the valve is wide open and (b) the valve closing percentage that causes the pump to operate at BEP. (c) If the latter condition holds continuously for 1 year, estimate the energy cost at 10 0/kWh. P11.72 Performance data for a small commercial pump are as follows: Q, gal/min 0 10 20 30 40 50 60 70 HJt 75 75 74 72 68 62 47 24 This pump supplies 20°C water to a horizontal \|-in-diameter garden hose (e ~ 0.01 in) that is 50 ft long. Estimate (a) the flow rate and (b) the hose diameter that would cause the pump to operate at BEP. P11.73 The Bell and Gossett pump of Prob. PI 1.8, running under the same conditions, delivers water at 20°C through a long, smooth, 8-in-diameter pipe. Neglect minor losses. How long is the pipe? Pumps in parallel or series P11.74 The 32-in pump in Fig. 11.7a is used at 1170 r/min in a system whose head curve is (ft) = 100 -I- 1.52^, with Q in thousands of gallons of water per minute. Find the dis¬ charge and brake horsepower required for (a) one pump, (b) two pumps in parallel, and (c) two pumps in series. Which configuration is best? P11.75 Two 35-in pumps from Fig. 1 1 Jb are installed in parallel for the system of Fig. PI 1.75. Neglect minor losses. For water at 20°C, estimate the flow rate and power required if (a) both pumps are running and (b) one pump is shut off and isolated. P11.75 P11.76 Two 32-in pumps from Fig. 1 1.7a are combined in parallel to deliver water at 60°F through 1500 ft of horizontal pipe. If/ = 0.025, what pipe diameter will ensure a flow rate of 35,000 gal/min for n = 1 170 r/min? P11.77 Two pumps of the type tested in Prob. PI 1.22 are to be used at 2 140 r/min to pump water at 20°C vertically upward through 100 m of commercial steel pipe. Should they be in series or in parallel? What is the proper pipe diameter for most efficient operation? 800 Chapter 11 Turbomachinery P11.78 Consider the axial-flow pump of Fig. 11.13, running at 4200 r/min, with an impeller diameter of 36 in. The fluid is propane gas (molecular weight 44.06). (a) How many pumps in series are needed to increase the gas pressure from 1 atm to 2 atm? (b) Estimate the mass flow of gas. P11.79 Two 32-in pumps from Fig. 11.7fl are to be used in series at 1170 r/min to lift water through 500 ft of vertical cast iron pipe. What should the pipe diameter be for most effi¬ cient operation? Neglect minor losses. P11.80 Determine if either (a) the smallest or (b) the largest of the seven Taco pumps in Fig. PI 1.24, running in series at 1160 r/min, can efficiently pump water at 20°C through 1 km of horizontal 12-cm-diameter commercial steel pipe. P11.81 Reconsider the system of Fig. P6.62. Use the Byron Jackson pump of Prob. PI 1.28 running at 2134 r/min, no scaling, to drive the flow. Determine the resulting flow rate between the reservoirs. What is the pump efficiency? Pump instability P11.82 The S-shaped head-versus-flow curve in Fig. PI 1.82 occurs in some axial-flow pumps. Explain how a fairly flat system loss curve might cause instabilities in the operation of the pump. How might we avoid instability? H P11.82 0 Q P11.83 The low-shutoff head-versus-flow curve in Eig. PI 1.83 occurs in some centrifugal pumps. Explain how a fairly flat system loss curve might cause instabilities in the operation of the pump. What additional vexation occurs when two of these pumps are in parallel? How might we avoid instability? P11.83 0 Q P11.85 For a high-flow site with a head of 45 ft, it is desired to design a single 7-ft-diameter turbine that develops 4000 bhp at a speed of 360 r/min and 88-percent efficiency. It is decided first to test a geometrically similar model of diameter 1 ft, running at 1180 r/min. (a) What likely type of turbine is in the prototype? What are the appropriate (b) head and (c) flow rate for the model test? (d) Estimate the power expected to be delivered by the model turbine. P11.86 The Tupperware hydroelectric plant on the Blackstone River has four 36-in-diameter turbines, each providing 447 kW at 200 r/min and 205 ftVs for a head of 30 ft. What type of turbine are these? How does their performance compare with Fig. 1 1.22? P11.87 An idealized radial turbine is shown in Fig. PI 1.87. The absolute flow enters at 30° and leaves radially inward. The flow rate is 3.5 m^/s of water at 20°C. The blade thickness is con¬ stant at 10 cm. Compute the theoretical power developed. Reaction and impulse turbines P11.84 Turbines are to be installed where the net head is 400 ft and the flow rate 250,000 gal/min. Discuss the type, number, and size of turbine that might be selected if the generator selected is (a) 48-pole, 60-cycle (n =150 r/min) and (b) 8-pole (n = 900 r/min). Why are at least two turbines desirable from a planning point of view? P11.88 Performance data for a very small (D = 8.25 cm) model water turbine, operating with an available head of 49 ft, are as follows: Q, m^/h 18.7 18.7 18.5 18.3 17.6 16.7 15.1 11.5 RPM 0 500 1000 1500 2000 2500 3000 3500 V 0 14% 27% 38% 50% 65% 61% 11% Problems 801 (a) What type of turbine is this likely to be? (b) What is so different about these data compared to the dimensionless performance plot in Fig. 11.22d? Suppose it is desired to use a geometrically similar turbine to serve where the available head and flow are 150 ft and 6.7 ftVs, respectively. Estimate the most efficient (c) turbine diameter, (d) rotation speed, and (e) horsepower. P11.89 A Pelton wheel of 12-ft pitch diameter operates under a net head of 2000 ft. Estimate the speed, power output, and flow rate for best efficiency if the nozzle exit diameter is 4 in. P11.90 An idealized radial turbine is shown in Fig. PI 1.90. The abso¬ lute flow enters at 25° with the blade angles as shown. The flow rate is 8 m^/s of water at 20°C. The blade thickness is constant at 20 cm. Compute the theoretical power developed. P11.91 The flow through an axial-flow turbine can be idealized by modifying the stator-rotor diagrams of Fig. 1 1 . 1 2 for energy absorption. Sketch a suitable blade and flow arrangement and the associated velocity vector diagrams. P11.92 A dam on a river is being sited for a hydraulic turbine. The flow rate is 1500 m^/h, the available head is 24 m, and the turbine speed is to be 480 r/min. Discuss the estimated turbine size and feasibility for (a) a Francis turbine and (b) a Pelton wheel. P11.93 Figure PI 1.93 shows a cutaway of a cross-flow or “Banki” turbine , which resembles a squirrel cage with slotted curved blades. The flow enters at about 2 o’clock and passes through the center and then again through the blades, leaving at about 8 o’clock. Report to the class on the operation and advantages of this design, including idealized velocity vector diagrams. P11.94 A simple cross-flow turbine, Fig. PI 1.93, was constructed and tested at the University of Rhode Island. The blades were made of PVC pipe cut lengthwise into three 120°-arc pieces. When it was tested in water at a head of 5.3 ft and a flow rate of 630 gal/min, the measured power output was 0.6 hp. Estimate (a) the efficiency and (b) the power specific speed if n = 200 r/min. P11.95 One can make a theoretical estimate of the proper diameter for a penstock in an impulse turbine installation, as in Fig. PI 1.95. Let L and H be known, and let the turbine performance be idealized by Eqs. (11.38) and (11.39). Account for friction loss hf in the penstock, but neglect minor losses. Show that (a) the maximum power is generated when hf = ///3, {b) the optimum jet velocity is (4g///3)^^, and (c) the best nozzle diameter is Dj = [D^/(2fL)]^'', where /is the pipe friction factor. P11.96 Apply the results of Prob. PI 1 .95 to determine the optimum (a) penstock diameter and (b) nozzle diameter for a head of 330 m and a flow rate of 5400 m^/h with a cast iron penstock of length 600 m. 802 Chapter 11 Turbomachinery P11.97 Consider the following nonoptimum version of Prob. PI 1.95: // = 450 m, L = 5 km, D = 1.2 m, Dj = 20 cm. The penstock is concrete, e = I mm. The impulse wheel diameter is 3.2 m. Estimate (a) the power generated by the wheel at 80 percent efficiency and (b) the best speed of the wheel in r/min. Neglect minor losses. P11.98 Francis and Kaplan turbines are often provided with draft tubes, which lead the exit flow into the tailwater region, as in Fig. PI 1.98. Explain at least two advantages in using a draft tube. P11.99 Turbines can also cavitate when the pressure at point 1 in Fig. PI 1.98 drops too low. With NPSH defined by Eq. (1 1.20), the empirical criterion given by Wislicenus for cavitation is (r/min) (gaPmin)''^ [NPSH (ft)]^'' 11,000 Use this criterion to compute how high zi — Z2, the impeller eye in Fig. PI 1.98, can be placed for a Francis turbine with a head of 300 ft, = 40, and Pa = 14 Ibf/in^ absolute before cavitation occurs in 60°F water. Wind turbines Pit. 100 The manufacturer of the wind turbine in the chapter- opener photo claims that it develops exactly 100 kW at a wind speed of 15 m/s. Compare this with an estimate from the correlations in Fig. 1 1.32. Pll.lOl A Darrieus VAWT in operation in Lumsden, Saskatchewan, that is 32 ft high and 20 ft in diameter sweeps out an area of 432 tf. Estimate (a) the maximum power and (h) the rotor speed if it is operating in 16 mi/h winds. P11.102 An American 6-ft-diameter multiblade HAWT is used to pump water to a height of 10 ft through 3-in-diameter cast iron pipe. If the winds are 12 mi/h, estimate the rate of water flow in gal/min. P11.103 Only a mile from the wind turbine in the chapter-opener photo is a 100-ft-high, 23-ft-diameter HAWT, in Fig. PI 1 . 103. It is rated at 10 kW and provides one-half of the electricity for the Salty Brine State Beach bathhouse. From the data in Fig. 1 1.32, at a wind velocity of 20 mi/h, estimate (a) the maximum power developed, and (fe) the rotation speed, in r/min. P11.103 [Rhode Island Department of Environmental Management] P11.104 The controversial Cape Cod Wind Project proposes 130 large wind turbines in Nantucket Sound, intended to pro¬ vide 75 percent of the electric power needs of Cape Cod and the Islands. The turbine diameter is 328 ft. For an average wind velocity of 14 mi/h, what are the best rota¬ tion rate and total power output estimates for {a) a HAWT and (b) a VAWT? Problems 803 P11.105 In 2007, a wind-powered-vehicle contest, held in North Holland , was won with a design by students at the University of Stuttgart. A schematic of the winning three- wheeler is shown in Fig. P11.105. It is powered by a shrouded wind turbine, not a propeller, and, unlike a sail¬ boat, can move directly into the wind, (a) How does it work? (b) What if the wind is off to the side? (c) Cite some design questions you might have. P11.106 Analyze the wind-powered-vehicle of Fig. P11.105 with the following data: turbine diameter D = 6 ft, power coef¬ ficient (Fig. 11.32) = 0.3, vehicle C^A = 4.5 ft^, and tur¬ bine rotation 240 r/min. The vehicle moves directly into a head wind, W = 25 mi/h. The wind backward thrust on the turbine is approximately T ~ C7-(p/2) where is the air velocity relative to the turbine, and Ct ~ 0.7. Eighty percent of the turbine power is deliv¬ ered by gears to the wheels, to propel the vehicle. Estimate the sea-level vehicle velocity V, in mi/h. P11.107 Figure 1 1.32 showed the typical power performance of a wind turbine. The wind also causes a thrust force that must be resisted by the structure. The thrust coefficient Cf of a wind turbine may be defined as follows: Thrust force T ^ {p/2)AV^ (p/2)[{7T/4)D-]V^ Values of Cyfor a typical horizontal-axis wind turbine are shown in Fig. PI 1.107. The abscissa is the same as in Fig. 11.32. Consider the turbine of Prob. PI 1.103. If the wind is 20 mi/h and the rotation rate 115 r/min, estimate the bending moment about the tower base. LJrIVi P11.107 Thrust coefficient for a typical HAWT. P11.108 To avoid the bulky tower and impeller and generator in the HAWT of the chapter-opener photo, we could instead build a number of Darrieus turbines of height 4 m and diameter 3 m. (a) How many of these would we need to match the HAWT’s 100 kW output for 15 m/s wind speed and maximum power? (b) How fast would they rotate? Assume the area swept out by a Darrieus turbine is two- thirds the height times the diameter. P11.105 804 Chapter 11 Turbomachinery Word Problems Wll.l We know that an enclosed rotating bladed impeller will impart energy to a fluid, usually in the form of a pressure rise, but how does it actually happen? Discuss, with sketches, the physical mechanisms through which an impeller actually transfers energy to a fluid. W11.2 Dynamic pumps (as opposed to PDFs) have difficulty mov¬ ing highly viscous fluids. Lobanoff and Ross suggest the following rule of thumb: D (in) > 0.01 5 iVi^a,er, where D is the diameter of the discharge pipe. For example, SAE 30W oil SOOi^water) should require at least a 4.5-in outlet. Can you explain some reasons for this limitation? W11.3 The concept of NPSH dictates that liquid dynamic pumps should generally be immersed below the surface. Can you explain this? What is the effect of increasing the liquid temperature? W11.4 For nondimensional fan performance, Wallis suggests that the head coefficient should be replaced by VTP/(pn^D~), where FTP is the fan total pressure change. Explain the usefulness of this modification. W11.5 Performance data for centrifugal pumps, even if well scaled geometrically, show a decrease in efficiency with decreasing impeller size. Discuss some physical reasons why this is so. W11.6 Consider a dimensionless pump performance chart such as Fig. 11.8. What additional dimensionless parameters might modify or even destroy the similarity indicated in such data? W11.7 One parameter not discussed in this text is the number of blades on an impeller. Do some reading on this subject, and report to the class about its effect on pump performance. W11.8 Explain why some pump performance curves may lead to unstable operating conditions. W11.9 Why are Erancis and Kaplan turbines generally consid¬ ered unsuitable for hydropower sites where the available head exceeds 1000 ft? Wll.lO Do some reading on the performance of the free propeller that is used on small, low-speed aircraft. What dimension¬ less parameters are typically reported for the data? How do the performance and efficiency compare with those for the axial-flow pump? Comprehensive Problems Cll.l The net head of a little aquarium pump is given by the man¬ ufacturer as a function of volume flow rate as listed below: e. m^/s H, mHjO 0 1.10 1.0 E-6 1.00 2.0 E-6 0.80 3.0 E-6 0.60 4.0 E-6 0.35 5.0 E-6 0.0 What is the maximum achievable flow rate if you use this pump to move water from the lower reservoir to the upper reservoir as shown in Fig. Cll.l? Note: The tubing is smooth with an inner diameter of 5.0 mm and a total length of 29.8 m. The water is at room temperature and pressure. Minor losses in the system can be neglected. 0.80 m Cll.l Cll.l Reconsider Prob. P6.62 as an exercise in pump selection. Select an impeller size and rotational speed from the Byron Jackson pump family of Prob. PI 1.28 to deliver a flow rate of 3 ftVs to the system of Fig. P6.62 at minimum input power. Calculate the horsepower required. C11.3 Reconsider Prob. P6.77 as an exercise in turbine selection. Select an impeller size and rotational speed from the Fran¬ cis turbine family of Fig. l\.22d to deliver maximum power generated by the turbine. Calculate the turbine power output and remark on the practicality of your design. C11.4 The system of Fig. Cl 1.4 is designed to deliver water at 20°C from a sea-level reservoir to another through new cast iron pipe of diameter 38 cm. Minor losses are XKi = 0.5 before the pump entrance and 2^2 = 7.2 after the pump exit, {a) Select a pump from either Fig. 1 1 .7fl or 1 1 .7h, run¬ ning at the given speeds, that can perform this task at max¬ imum efficiency. Determine fb) the resulting flow rate, (c) the brake horsepower, and {d) whether the pump as presently situated is safe from cavitation. C11.5 For the 41.5-in water pump of Fig. 11.77i, at 710 r/min and 22,000 gal/min, estimate the efficiency by (a) reading it directly from Fig. \ l.lb\ and fb) reading H and blip and then calculating efficiency from Eq. (11.5). Compare your results. C11.6 An interesting turbomachine is the fluid coupling of Fig. Cl 1.6, which circulates fluid from a primary pump rotor and thus turns a secondary turbine on a separate shaft. Both rotors have radial blades. Couplings are common in Comprehensive Problems 805 C11.4 all types of vehicle and machine transmissions and drives. The slip of the coupling is defined as the dimensionless dif¬ ference between shaft rotation rates, ^ = 1 — u)jLOp. For a given volume of fluid, the torque T transmitted is a function of s, p, ujp, and impeller diameter D. (a) Nondimensional- ize this function into two pi groups, with one pi propor¬ tional to T. Tests on a 1-ft-diameter coupling at 2500 r/min, filled with hydraulic fluid of density 56 Ibm/ft^, yield the following torque versus slip data: Slip, s 0% 5% 10% 15% 20% Torque T, ft-lbf 0 90 275 440 580 (b) If this coupling is mn at 3600 r/min, at what slip value will it transmit a torque of 900 ft-lbf? (c) What is the proper diam¬ eter for a geometrically similar coupling to mn at 3000 r/min and 5 percent slip and transmit 600 ft-lbf of torque? C11.7 Report to the class on the Cordier method for optimal design of turbomachinery. The method is related to, and greatly expanded from, Prob. PI 1.46 and uses both software and charts to develop an efficient design for any given pump or compressor application. C11.8 A pump-turbine is a reversible device that uses a reservoir to generate power in the daytime and then pumps water back up to the reservoir at night. Let us reset Prob. P6.62 as a pump-turbine. Recall that Az = 120 ft, and the water flows through 2000 ft of 6-in-diameter cast iron pipe. For simplicity, assume that the pump operates at BEP (92%) with Hp = 200 ft and the turbine operates at BEP (89%) with H, = 100 ft. Neglect minor losses. Estimate {a) the input power, in watts, required by the pump; and {b) the power, in watts, generated by the turbine. Eor further tech¬ nical reading, consult the URL www.usbr.gov/pmts/ hydraulics_lab/pubs/EM/EM39.pdf. 806 Chapter 11 Turbomachinery Design Project Dll.l To save on electricity costs, a town water supply system uses gravity-driven flow from five large storage tanks dur¬ ing the day and then refills these tanks from 10 p.m. to 6 A.M. at a cheaper night rate of 7 0/kWh. The total resup¬ ply needed each night varies from 5 E5 to 2 E6 gal, with no more than 5 E5 gallons to any one tank. Tank elevations vary from 40 to 100 ft. A single constant-speed pump, drawing from a large groundwater aquifer and valved into five different cast iron tank supply lines, does this job. Distances from the pump to the five tanks vary more or less evenly from 1 to 3 mi. Each line averages one elbow every 100 ft and has four butterfly valves that can be controlled at any desirable angle. Select a suitable pump family from References 1. D. G. Wilson, “Turbomachinery — From Paddle Wheels to 18. Turbojets,” Mech. Eng., vol. 104, Oct. 1982, pp. 28-40. 2. D. Japikse and N. C. Baines, /wtroc/ncrioM to rMrtomac/iinery, 19. Concepts ETl Inc., Hanover, NH, 1997. 3. E. S. Logan and R. Roy (eds.), //anr/lJooko/rMrtomac/imery, 20. 2d ed., Marcel Dekker, New York, 2003. 4. G. F. Wislicenus, Fluid Mechanics of Turbomachinery, 21. 2d ed., McGraw-Hill, New York, 1965. 5. S.E.T)ixona.nAC.Yia[\,FluidMechanicsandThermodynamics 22. of Turbomachinery, 7th ed., Butterworth-Heinemann, Burlington, MA, 2013. 23. 6. W. W. Peng, Fundamentals of Turbomachinery, Wiley, New York, 2007. 7. S. A. Korpela, Principles of Turbomachinery , Wiley, 24. New York, 2011. 8. A. J. Stepanoff, Centrifugal and Axial Flow Pumps, 2d ed., Wiley, New York, 1957. 25. 9. J. Tuzson, Centrifugal Pump Design, Wiley, New York, 2000. 10. P. Girdhar and O. Moniz, Practical Centrifugal Pumps, 26. Elsevier, New York, 2004. 11. L. Bachus and A. Custodio, Know and Understand 27. Centrifugal Pumps, Elsevier, New York, 2003. 12. J. F. Giilich, Centrifugal Pumps, Springer, New York, 2010. 28. 13. R. K. Turton, Rotodynamic Pump Design, Cambridge University Press, Cambridge, UK, 2005. 29. 14. I. J. Karassik and T. McGuire, Centrifugal Pumps, 2d ed.. Springer- Verlag, New York, 1996. 15. V. L. Lobanoff and R. R. Ross, Centrifugal Pumps: Design 30. and Application, 2d ed., Elsevier, New York, 1992. 16. H. L. Stewart, Pumps, 5th ed. Macmillan, New York, 1991. 31. 17. A. B. McKenzie, Axial Flow Fans and Compressors: Aerodynamic Design and Performance, Ashgate Publishing, 32. Brookfield, VT, 1997. one of the six data sets in this chapter: Pigs. 11.8, P11.24, and P11.34 plus Probs. P11.28, P11.35, and P11.38. Assume ideal similarity (no Reynolds number or pump roughness effects). The goal is to determine pump and pipeline sizes that achieve minimum total cost over a 5-year period. Some suggested cost data are (a) Pump and motor: $2500 plus $1500 per inch of pipe size (b) Valves: $100 plus $100 per inch of pipe size (c) Pipelines: 500 per inch of diameter per foot of length Since the flow and elevation parameters vary considerably, a random daily variation within the specified ranges might give a realistic approach. A. J. Wennerstrom, Design of Highly Loaded Axial-Flow Fans and Compressors, Concepts ETl Inc., Hanover, NH, 2001. F. P. Bleier, Fan Handbook: Selection, Application, and Design, McGraw-Hill, New York, 1997. R. A. Wallis, Axial Flow Fans and Ducts, Wiley, New York, 1983. H. P. Bloch, A Practical Guide to Compressor Technology, 2d ed., McGraw-Hill, New York, 2006. P. C. Hanlon, Compressor Handbook, McGraw-Hill, New York, 2001. Ronald H. Aungier, Axial-Flow Compressors: A Strategy for Aerodynamic Design and Analysis, ASME Press, New York, 2003. H. I. H. Saravanamuttoo, G. F. C. Rogers, H. Cohen, and Paul Straznicky, Gas Turbine Theory, 6th ed., Pearson Education Canada, Don Mills, Ontario, 2008. P. P. Walsh and P. Fletcher, Gas Turbine Performance, ASME Press, New York, 2004. M. P. Boyce, Gas Turbine Engineering Handbook, 4th ed.. Gulf Professional Publishing, Burlington, MA, 2011. Fluid Machinery Group, Institution of Mechanical Engineers, Hydropower, Wiley, New York, 2005. Jeremy Thake, The Micro-Hydro Pelton Turbine Manual, Intermediate Technology Pub., Colchester, Essex, UK, 2000. L. Rodriguez and T. Sanchez, Designing and Building Mini and Micro Hydro Power Schemes, Practical Action Publish¬ ing, Warwickshire, UK, 2011. Hydraulic Institute, Hydraulic Institute Pump Standards Complete, 4th ed.. New York, 1994. P. Cooper, J. Messina, C. Heald, and I. J. Karassik (eds.). Pump Handbook, 4th ed., McGraw-Hill, New York, 2008. J. S. Gulliver and R. E. A. Arndt, Hydropower Engineering Handbook, McGraw-Hill, New York, 1990. References 807 33. R. L. Daugherty, J. B. Franzini, and E. J. Finnemore, Fluid 49. Mechanics and Engineering Applications, 9th ed., McGraw- Hill, New York, 1997. 50. 34. R. H. Sabersky, E. M. Gates, A. J. Acosta, and E. G. Hauptmann, Fluid Flow: A First Course in Fluid Mechanics, 4th ed., Pearson Education, Upper Saddle River, NJ, 1994. 51. 35. J. P. Poynton, Metering Pumps, Marcel Dekker, New York, 1983. 52. 36. Hydraulic Institute, Reciprocating Pump Test Standard, New York, 1994. 37. T. L. Henshaw, Reciprocating Pumps, Wiley, New York, 53. mi. 38. J. E. Miller, The Reciprocating Pump: Theory, Design and 54. Use, Wiley, New York, 1987. 39. D. G. Wilson and T. Korakianitis, The Design of High Efficiency Turbomachinery and Gas Turbines, 2d ed., Pearson 55. Education, Upper Saddle River, NJ, 1998. 40. S. O. Kraus et ah, “Periodic Velocity Measurements in a Wide and Large Radius Ratio Automotive Torque Converter 56. at the Pump/Turbine Interface,” J. Fluids Engineering, vol. 127, no. 2, 2005, pp. 308-316. 41. E. M. Greitzer, “The Stability of Pumping Systems: The 1980 57. Freeman Scholar Lecture,” J. Eluids Eng., vol. 103, June 1981, pp. 193-242. 58. 42. R. Elder et al. (eds.). Advances of CFD in Eluid Machinery Design, Wiley, New York, 2003. 59. 43. L. F. Moody, “The Propeller Type Turbine,” ASCE Trans., vol. 89, 1926, p. 628. 44. H. H. Anderson, “Prediction of Head, Quantity, and Efficiency 60. in Pumps — The Area-Ratio Principle,” in Performance Pre¬ diction of Centrifugal Pumps and Compressors, vol. 100127, 61. ASME Symp., New York, 1980, pp. 201-21 1. 45. M. Schobeiri, Turbomachinery Plow Physics and Dynamic 62. Performance, Springer, New York, 2004. 46. D. J. Mahoney (ed.). Proceedings of the 1997 International Conference on Hydropower, ASCE, Reston, VA, 1997. 63. 47. G. W. Koeppl, Putnam’s Power from the Wind, 2d ed.. Van Nostrand Reinhold, New York, 1982. 48. P. Jain, Wind Energy Engineenwg, McGraw-Hill, New York, 64. 2010. D. Wood, Small Wind Turbines: Analysis, Design, and Application, Springer, New York, 2011. E. Hau, Wind Turbines: Eundamentals, Technologies, Application, Economics, 2d ed.. Springer- Verlag, New York, 2005. R. Harrison, E. Hau, and H. Snel, Large Wind Turbines, Wiley, New York, 2000. R. H. Aungier, Turbine Aerodynamics: Axial-Plow and Radial-Flow Turbine Design and Analysis, ASME Press, New York, 2006. M. L. Robinson, “The Darrieus Wind Turbine for Electrical Power Generation,” Aeronanf. J., June J981, pp. 244-255. D. F. Wame and P. G. Calnan, “Generation of Electricity from the Wind,” /EE Rev., vol. 124, no. HR, November 1977, pp. 963-985. L. A. Haimerl, “The Crossflow Turbine,” Waterpower, January J960, pp. 5-13; see also ASME Symp. Small Hydro- power Eluid Mach., vol. 1, 1980, and vol. 2, 1982. K. Eisele et al., “Elow Analysis in a Pump Diffuser: Part 1, Measurements; Part 2, CFD,” J. Fluids Eng., vol. 119, December 1997, pp. 968-984. D. Japikse and N. C. Baines, Turbomachinery Diffuser Design Technology, Concepts ETJ Inc., Hanover, NH, 1998. B. Massey and J. Ward-Smith, Mechanics of Fluids, 7th ed.. Nelson Thornes Publishing, Cheltenham, UK, 1998. American Wind Energy Association, “Global Wind Energy Market Report,” URL: . J. Carlton, Marine Propellers and Propulsion, 3d ed., Butterworth-Heinemann, New York, 2012. M. Hollmann, Modern Propeller and Duct Design, Aircraft Designs, Inc., Monterey, CA, 1993. C. L. Archer and M. Z. Jacobson, “Evaluation of Global Wind Power,” J. Geophys. Res. -Atm., vol. 110, 2005, doi: 10. 1029/2004 JD005462. M. Farinas and A. Garon, “Application of DOE for Optimal Turbomachinery Design,” Paper AJAA-2004-2139, AJAA Fluid Dynamics Conference, Portland, OR, June 2004. C. Crain, “Running Against the Wind,” Popular Science, March 2009, pp. 69-70. Fig. A.l Absolute viscosity of common fluids at 1 atm. Appendix A Physical Properties of Fluids Temperature, °C 808 Physical Properties of Fluids 809 Fig. A.2 Kinematic viscosity of common fluids at 1 atm. Temperature, °C 810 Appendix A Table A.l Viscosity and Density of Water at 1 atm T.°C p, kg/m^ (i, N ■ sW V, mVs r,°F p, slug/fP p, Ib ■ s/ft^ u, fP/s 0 1000 1.788 E-3 1.788 E-6 32 1.940 3.73 E-5 1.925 E-5 10 1000 1.307 E-3 1.307 E-6 50 1.940 2.73 E-5 1.407 E-5 20 998 1.003 E-3 1.005 E-6 68 1.937 2.09 E-5 1.082 E-5 30 996 0.799 E-3 0.802 E-6 86 1.932 1.67 E-5 0.864 E-5 40 992 0.657 E-3 0.662 E-6 104 1.925 1.37 E-5 0.713 E-5 50 988 0.548 E-3 0.555 E-6 122 1.917 1.14 E-5 0.597 E-5 60 983 0.467 E-3 0.475 E-6 140 1.908 0.975 E-5 0.511 E-5 70 978 0.405 E-3 0.414 E-6 158 1.897 0.846 E-5 0.446 E-5 80 972 0.355 E-3 0.365 E-6 176 1.886 0.741 E-5 0.393 E-5 90 965 0.316 E-3 0.327 E-6 194 1.873 0.660 E-5 0.352 E-5 100 958 0.283 E-3 0.295 E-6 212 1.859 0.591 E-5 0.318 E-5 Suggested curve fits for water in the range 0 ^ T < 100°C: p(kg/m^) = 1000 - 0.0178 I rX - 4°C\|'^ ± 0.2% In — « -1.704 - 5.306Z + 7.003z^ Mo 273 K z = Mo = 1-788 E-3kg/(m-s) Table A.2 Viscosity and Density of Air at 1 atm p, kg/m^ p, N • s/m^ V, m^/s r,°F p, slug/fP p, lb • s/fP u, fP/s -40 1.52 1.51 E-5 0.99 E-5 -40 2.94 E-3 3.16E-7 1.07 E-4 0 1.29 1.71 E-5 1.33 E-5 32 2.51 E-3 3.58 E-7 1.43 E-4 20 1.20 1.80 E-5 1.50 E-5 68 2.34 E-3 3.76 E-7 1.61 E-4 50 1.09 1.95 E-5 1.79 E-5 122 2.12 E-3 4.08 E-7 1.93 E-4 100 0.946 2.17 E-5 2.30 E-5 212 1.84 E-3 4.54 E-7 2.47 E-4 150 0.835 2.38 E-5 2.85 E-5 302 1.62 E-3 4.97 E-7 3.07 E-4 200 0.746 2.57 E-5 3.45 E-5 392 1.45 E-3 5.37 E-7 3.71 E-4 250 0.675 2.75 E-5 4.08 E-5 482 1.31 E-3 5.75 E-7 4.39 E-4 300 0.616 2.93 E-5 4.75 E-5 572 1.20 E-3 6.11 E-7 5.12 E-4 400 0.525 3.25 E-5 6.20 E-5 752 1.02 E-3 6.79 E-7 6.67 E-4 500 0.457 3.55 E-5 7.77 E-5 932 0.89 E-3 7.41 E-7 8.37 E-4 Suggested curve fits for air: Power law: Sutherland law: M = ^ /J,i,« 287J/(kg-K) K1 M _ / ry-’ Mo \ To/ ^ f rVyro + Mo \ToJ [t+sJ 5air = 110.4K with Tq = 273 K, /j,q = 1.71 E— 5 kg/(m ■ s), and T in kelvins. Physical Properties of Fluids 811 Table A.3 Properties of Common Liquids at 1 atm and 20°C (68°F) Table A.4 Properties of Common Gases at 1 atm and 20°C (68°F) Liquid p, kg/m^ p, kg/(m • s) Y, N/m p„ nW Bulk modulus A, N/m^ Viscosity parameter Ct Ammonia 608 2.20 E-4 2.13 E-2 9.10 E+5 1.82 E+9 1.05 Benzene 881 6.51 E-4 2.88 E-2 1.01 E+4 1.47 E+9 4.34 Carbon tetrachloride 1590 9.67 E-4 2.70 E-2 1.20 E+4 1.32 E+9 4.45 Ethanol 789 1.20 E-3 2.28 E-2 5.7 E+3 1.09 E+9 5.72 Ethylene glycol 1117 2.14 E-2 4.84 E-2 1.2 E+1 3.05 E+9 11.7 Freon 12 1327 2.62 E-4 — — 7.95 E+8 1.76 Gasoline 680 2.92 E-4 2.16 E-2 5.51 E+4 1.3 E+9 3.68 Glycerin 1260 1.49 6.33 E-2 1.4 E-2 4.35 E+9 28.0 Kerosene 804 1.92 E-3 2.8 E-2 3.11 E+3 1.41 E+9 5.56 Mercury 13,550 1.56 E-3 4.84 E-1 1.1 E-3 2.85 E+10 1.07 Methanol 791 5.98 E-4 2.25 E-2 1.34 E+4 1.03 E+9 4.63 SAE low oil 870 1.04 E-1 3.6 E-2 — 1.31 E+9 15.7 SAE 10W30 oil 876 1.7 E-1 — — — 14.0 SAE 30W oil 891 2.9 E-1 3.5 E-2 — 1.38 E+9 18.3 SAE SOW oil 902 8.6 E-1 — — — 20.2 Water 998 1.00 E-3 7.28 E-2 2.34 E+3 2.19E+9 Table A.l Seawater (30%o) 1025 1.07 E-3 7.28 E-2 2.34 E+3 2.33 E+9 7.28 ln contact with air. fThe viscosity-temperature variation of these liquids may be fitted to the empirical expression 293 K with accuracy of ±6 percent in the range 0 < T ^ 100°C. ^Representative values. The SAE oil classifications allow a viscosity variation of up to ±50 percent, especially at lower temperatures. Gas Molecular weight R, mV(s^ ■ K) pg, nW p, N ■ s/m^ u, m^/s Specific-heat ratio Power-law exponent n Hi 2.016 4124 0.822 9.05 E-6 1.08 E-04 1.41 0.68 He 4.003 2077 1.63 1.97 E-5 1.18 E-04 1.66 0.67 HjO 18.02 461 7.35 1.02 E-5 1.36 E-05 1.33 1.15 Ar 39.944 208 16.3 2.24 E-5 1.35 E-05 1.67 0.72 Dry air 28.96 287 11.8 1.80 E-5 1.49 E-05 1.40 0.67 COi 44.01 189 17.9 1.48 E-5 8.09 E-06 1.30 0.79 CO 28.01 297 11.4 1.82 E-5 1.56 E-05 1.40 0.71 N2 28.02 297 11.4 1.76 E-5 1.51 E-05 1.40 0.67 02 32.00 260 13.1 2.00 E-5 1.50 E-05 1.40 0.69 NO 30.01 277 12.1 1.90 E-5 1.52 E-05 1.40 0.78 N2O 44.02 189 17.9 1.45 E-5 7.93 E-06 1.31 0.89 CI2 70.91 117 28.9 1.03 E-5 3.49 E-06 1.34 1.00 CH4 16.04 518 6.54 1.34 E-5 2.01 E-05 1.32 0.87 The power-law curve fit, Eq. (1.27), ~ (77293)", fits these gases to within ±4 percent in the range 250 < r ^ 1000 K. The temperature must be in kelvins. 812 Appendix A Table A.5 Surface Tension, Vapor r,°c Y,N/m kPa a, m/s Table A.6 Properties of z, m 7’,K p. Pa p, kg/m^ a, m/s Pressure, and Sound 0 0.0756 0.611 1402 the Standard -500 291.41 107,508 1.2854 342.2 Speed of Water 10 0.0742 1.227 1447 Atmosphere 0 288.16 101,350 1.2255 340.3 20 0.0728 2.337 1482 500 284.91 95,480 1.1677 338.4 30 0.0712 4.242 1509 1000 281.66 89,889 1.1120 336.5 40 0.0696 7.375 1529 1500 278.41 84,565 1.0583 334.5 50 0.0679 12.34 1542 2000 275.16 79,500 1.0067 332.6 60 0.0662 19.92 1551 2500 271.91 74,684 0.9570 330.6 70 0.0644 31.16 1553 3000 268.66 70,107 0.9092 328.6 80 0.0626 47.35 1554 3500 265.41 65,759 0.8633 326.6 90 0.0608 70.11 1550 4000 262.16 61,633 0.8191 324.6 100 0.0589 101.3 1543 4500 258.91 57,718 0.7768 322.6 5000 255.66 54,008 0.7361 320.6 120 0.0550 198.5 1518 5500 252.41 50,493 0.6970 318.5 140 0.0509 361.3 1483 6000 249.16 47,166 0.6596 316.5 160 0.0466 617.8 1440 6500 245.91 44,018 0.6237 314.4 180 0.0422 1002 1389 7000 242.66 41,043 0.5893 312.3 200 0.0377 1554 1334 7500 239.41 38,233 0.5564 310.2 220 0.0331 2318 1268 8000 236.16 35,581 0.5250 308.1 240 0.0284 3344 1192 8500 232.91 33,080 0.4949 306.0 260 0.0237 4688 1110 9000 229.66 30,723 0.4661 303.8 280 0.0190 6412 1022 9500 226.41 28,504 0.4387 301.7 300 0.0144 8581 920 10,000 223.16 26,416 0.4125 299.5 320 0.0099 11,274 800 10,500 219.91 24,455 0.3875 297.3 340 0.0056 14,586 630 11,000 216.66 22,612 0.3637 295.1 360 0.0019 18,651 370 11,500 216.66 20,897 0.3361 295.1 374 0.0 22,090 0 12,000 216.66 19,312 0.3106 295.1 12,500 216.66 17,847 0.2870 295.1 Critical point. 13,000 216.66 16,494 0.2652 295.1 13,500 216.66 15,243 0.2451 295.1 14,000 216.66 14,087 0.2265 295.1 14,500 216.66 13,018 0.2094 295.1 15,000 216.66 12,031 0.1935 295.1 15,500 216.66 11,118 0.1788 295.1 16,000 216.66 10,275 0.1652 295.1 16,500 216.66 9496 0.1527 295.1 17,000 216.66 8775 0.1411 295.1 17,500 216.66 8110 0.1304 295.1 18,000 216.66 7495 0.1205 295.1 18,500 216.66 6926 0.1114 295.1 19,000 216.66 6401 0.1029 295.1 19,500 216.66 5915 0.0951 295.1 20,000 216.66 5467 0.0879 295.1 22,000 218.60 4048 0.0645 296.4 24,000 220.60 2972 0.0469 297.8 26,000 222.50 2189 0.0343 299.1 28,000 224.50 1616 0.0251 300.4 30,000 226.50 1197 0.0184 301.7 40,000 250.40 287 0.0040 317.2 50,000 270.70 80 0.0010 329.9 60,000 255.70 22 0.0003 320.6 70,000 219.70 6 0.0001 297.2 Table B.l Isentropic Flow of a Perfect Gas, k = 1.4 Appendix B Compressible Flow Tables Ma P/Po P/Po T/n A! A Ma P/Po P/Po T/T„ A! A 0.00 1.0000 1.0000 1.0000 00 2.10 0.1094 0.2058 0.5313 1.8369 0.10 0.9930 0.9950 0.9980 5.8218 2.20 0.0935 0.1841 0.5081 2.0050 0.20 0.9725 0.9803 0.9921 2.9635 2.30 0.0800 0.1646 0.4859 2.1931 0.30 0.9395 0.9564 0.9823 2.0351 2.40 0.0684 0.1472 0.4647 2.4031 0.40 0.8956 0.9243 0.9690 1.5901 2.50 0.0585 0.1317 0.4444 2.6361 0.50 0.8430 0.8852 0.9524 1.3398 2.60 0.0501 0.1179 0.4252 2.8960 0.60 0.7840 0.8405 0.9328 1.1882 2.70 0.0430 0.1056 0.4068 3.1830 0.70 0.7209 0.7916 0.9107 1.0944 2.80 0.0368 0.0946 0.3894 3.5001 0.80 0.6560 0.7400 0.8865 1.0382 2.90 0.0317 0.0849 0.3729 3.8498 0.90 0.5913 0.6870 0.8606 1.0089 3.00 0.0272 0.0762 0.3571 4.2346 1.00 0.5283 0.6339 0.8333 1.0000 3.10 0.0234 0.0685 0.3422 4.6573 1.10 0.4684 0.5817 0.8052 1.0079 3.20 0.0202 0.0617 0.3281 5.1210 1.20 0.4124 0.5311 0.7764 1.0304 3.30 0.0175 0.0555 0.3147 5.6286 1.30 0.3609 0.4829 0.7474 1.0663 3.40 0.0151 0.0501 0.3019 6.1837 1.40 0.3142 0.4374 0.7184 1.1149 3.50 0.0131 0.0452 0.2899 6.7896 1.50 0.2724 0.3950 0.6897 1.1762 3.60 0.0114 0.0409 0.2784 7.4501 1.60 0.2353 0.3557 0.6614 1.2502 3.70 0.0099 0.0370 0.2675 8.1691 1.70 0.2026 0.3197 0.6337 1.3376 3.80 0.0086 0.0335 0.2572 8.9506 1.80 0.1740 0.2868 0.6068 1.4390 3.90 0.0075 0.0304 0.2474 9.7990 1.90 0.1492 0.2570 0.5807 1.5553 4.00 0.0066 0.0277 0.2381 10.7188 2.00 0.1278 0.2300 0.5556 1.6875 813 814 Appendix B Table B.2 Normal Shock Relations for a Perfect Gas, k = lA Ma„, Ma„2 PilPi II TJTi Poi/Pdi A%IA% 1.00 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.10 0.9118 1.2450 1.1691 1.0649 0.9989 1.0011 1.20 0.8422 1.5133 1.3416 1.1280 0.9928 1.0073 1.30 0.7860 1.8050 1.5157 1.1909 0.9794 1.0211 1.40 0.7397 2.1200 1.6897 1.2547 0.9582 1.0436 1.50 0.7011 2.4583 1.8621 1.3202 0.9298 1.0755 1.60 0.6684 2.8200 2.0317 1.3880 0.8952 1.1171 1.70 0.6405 3.2050 2.1977 1.4583 0.8557 1.1686 1.80 0.6165 3.6133 2.3592 1.5316 0.8127 1.2305 1.90 0.5956 4.0450 2.5157 1.6079 0.7674 1.3032 2.00 0.5774 4.5000 2.6667 1.6875 0.7209 1.3872 2.10 0.5613 4.9783 2.8119 1.7705 0.6742 1.4832 2.20 0.5471 5.4800 2.9512 1.8569 0.6281 1.5920 2.30 0.5344 6.0050 3.0845 1.9468 0.5833 1.7144 2.40 0.5231 6.5533 3.2119 2.0403 0.5401 1.8514 2.50 0.5130 7.1250 3.3333 2.1375 0.4990 2.0039 2.60 0.5039 7.7200 3.4490 2.2383 0.4601 2.1733 2.70 0.4956 8.3383 3.5590 2.3429 0.4236 2.3608 2.80 0.4882 8.9800 3.6636 2.4512 0.3895 2.5676 2.90 0.4814 9.6450 3.7629 2.5632 0.3577 2.7954 3.00 0.4752 10.3333 3.8571 2.6790 0.3283 3.0456 3.10 0.4695 11.0450 3.9466 2.7986 0.3012 3.3199 3.20 0.4643 11.7800 4.0315 2.9220 0.2762 3.6202 3.30 0.4596 12.5383 4.1120 3.0492 0.2533 3.9483 3.40 0.4552 13.3200 4.1884 3.1802 0.2322 4.3062 3.50 0.4512 14.1250 4.2609 3.3151 0.2129 4.6960 3.60 0.4474 14.9533 4.3296 3.4537 0.1953 5.1200 3.70 0.4439 15.8050 4.3949 3.5962 0.1792 5.5806 3.80 0.4407 16.6800 4.4568 3.7426 0.1645 6.0801 3.90 0.4377 17.5783 4.5156 3.8928 0.1510 6.6213 4.00 0.4350 18.5000 4.5714 4.0469 0.1388 7.2069 4.10 0.4324 19.4450 4.6245 4.2048 0.1276 7.8397 4.20 0.4299 20.4133 4.6749 4.3666 0.1173 8.5227 4.30 0.4277 21.4050 4.7229 4.5322 0.1080 9.2591 4.40 0.4255 22.4200 4.7685 4.7017 0.0995 10.0522 4.50 0.4236 23.4583 4.8119 4.8751 0.0917 10.9054 4.60 0.4217 24.5200 4.8532 5.0523 0.0846 11.8222 4.70 0.4199 25.6050 4.8926 5.2334 0.0781 12.8065 4.80 0.4183 26.7133 4.9301 5.4184 0.0721 13.8620 4.90 0.4167 27.8450 4.9659 5.6073 0.0667 14.9928 5.00 0.4152 29.0000 5.0000 5.8000 0.0617 16.2032 Compressible Flow Tables 815 Table B.3 Adiabatic Frictional Flow in a Constant- Area Duct for ^ = 1 .4 Ma fUD p/p~ yyj’ p/p = V/V Po/p% 0.00 oo OO 1.2000 0.0000 OO 0.10 66.9216 10.9435 1.1976 0.1094 5.8218 0.20 14.5333 5.4554 1.1905 0.2182 2.9635 0.30 5.2993 3.6191 1.1788 0.3257 2.0351 0.40 2.3085 2.6958 1.1628 0.4313 1.5901 0.50 1.0691 2.1381 1.1429 0.5345 1.3398 0.60 0.4908 1.7634 1.1194 0.6348 1.1882 0.70 0.2081 1.4935 1.0929 0.7318 1.0944 0.80 0.0723 1.2893 1.0638 0.8251 1.0382 0.90 0.0145 1.1291 1.0327 0.9146 1.0089 1.00 0.0000 1.0000 1.0000 1.0000 1.0000 1.10 0.0099 0.8936 0.9662 1.0812 1.0079 1.20 0.0336 0.8044 0.9317 1.1583 1.0304 1.30 0.0648 0.7285 0.8969 1.2311 1.0663 1.40 0.0997 0.6632 0.8621 1.2999 1.1149 1.50 0.1361 0.6065 0.8276 1.3646 1.1762 1.60 0.1724 0.5568 0.7937 1.4254 1.2502 1.70 0.2078 0.5130 0.7605 1.4825 1.3376 1.80 0.2419 0.4741 0.7282 1.5360 1.4390 1.90 0.2743 0.4394 0.6969 1.5861 1.5553 2.00 0.3050 0.4082 0.6667 1.6330 1.6875 2.10 0.3339 0.3802 0.6376 1.6769 1.8369 2.20 0.3609 0.3549 0.6098 1.7179 2.0050 2.30 0.3862 0.3320 0.5831 1.7563 2.1931 2.40 0.4099 0.3111 0.5576 1.7922 2.4031 2.50 0.4320 0.2921 0.5333 1.8257 2.6367 2.60 0.4526 0.2747 0.5102 1.8571 2.8960 2.70 0.4718 0.2588 0.4882 1.8865 3.1830 2.80 0.4898 0.2441 0.4673 1.9140 3.5001 2.90 0.5065 0.2307 0.4474 1.9398 3.8498 3.00 0.5222 0.2182 0.4286 1.9640 4.2346 3.10 0.5368 0.2067 0.4107 1.9866 4.6573 3.20 0.5504 0.1961 0.3937 2.0079 5.1210 3.30 0.5632 0.1862 0.3776 2.0278 5.6286 3.40 0.5752 0.1770 0.3623 2.0466 6.1837 3.50 0.5864 0.1685 0.3478 2.0642 6.7896 3.60 0.5970 0.1606 0.3341 2.0808 7.4501 3.70 0.6068 0.1531 0.3210 2.0964 8.1691 3.80 0.6161 0.1462 0.3086 2.1111 8.9506 3.90 0.6248 0.1397 0.2969 2.1250 9.7990 4.00 0.6331 0.1336 0.2857 2.1381 10.7187 816 Appendix B Table B.4 Frictionless Duct Flow with Heat Transfer for k = lA Ma To/T% pip jyj’ p/p = VIV Po/Po 0.00 0.0000 2.4000 0.0000 0.0000 1.2679 0.10 0.0468 2.3669 0.0560 0.0237 1.2591 0.20 0.1736 2.2727 0.2066 0.0909 1.2346 0.30 0.3469 2.1314 0.4089 0.1918 1.1985 0.40 0.5290 1.9608 0.6151 0.3137 1.1566 0.50 0.6914 1.7778 0.7901 0.4444 1.1141 0.60 0.8189 1.5957 0.9167 0.5745 1.0753 0.70 0.9085 1.4235 0.9929 0.6975 1.0431 0.80 0.9639 1.2658 1.0255 0.8101 1.0193 0.90 0.9921 1.1246 1.0245 0.9110 1.0049 1.00 1.0000 1.0000 1.0000 1.0000 1.0000 1.10 0.9939 0.8909 0.9603 1.0780 1.0049 1.20 0.9787 0.7958 0.9118 1.1459 1.0194 1.30 0.9580 0.7130 0.8592 1.2050 1.0437 1.40 0.9343 0.6410 0.8054 1.2564 1.0777 1.50 0.9093 0.5783 0.7525 1.3012 1.1215 1.60 0.8842 0.5236 0.7017 1.3403 1.1756 1.70 0.8597 0.4756 0.6538 1.3746 1.2402 1.80 0.8363 0.4335 0.6089 1.4046 1.3159 1.90 0.8141 0.3964 0.5673 1.4311 1.4033 2.00 0.7934 0.3636 0.5289 1.4545 1.5031 2.10 0.7741 0.3345 0.4936 1.4753 1.6162 2.20 0.7561 0.3086 0.4611 1.4938 1.7434 2.30 0.7395 0.2855 0.4312 1.5103 1.8860 2.40 0.7242 0.2648 0.4038 1.5252 2.0451 2.50 0.7101 0.2462 0.3787 1.5385 2.2218 2.60 0.6970 0.2294 0.3556 1.5505 2.4177 2.70 0.6849 0.2142 0.3344 1.5613 2.6343 2.80 0.6738 0.2004 0.3149 1.5711 2.8731 2.90 0.6635 0.1879 0.2969 1.5801 3.1359 3.00 0.6540 0.1765 0.2803 1.5882 3.4245 3.10 0.6452 0.1660 0.2650 1.5957 3.7408 3.20 0.6370 0.1565 0.2508 1.6025 4.0871 3.30 0.6294 0.1477 0.2377 1.6088 4.4655 3.40 0.6224 0.1397 0.2255 1.6145 4.8783 3.50 0.6158 0.1322 0.2142 1.6198 5.3280 3.60 0.6097 0.1254 0.2037 1.6247 5.8173 3.70 0.6040 0.1190 0.1939 1.6293 6.3488 3.80 0.5987 0.1131 0.1848 1.6335 6.9256 3.90 0.5937 0.1077 0.1763 1.6374 7.5505 4.00 0.5891 0.1026 0.1683 1.6410 8.2268 Compressible Flow Tables 817 Table B.5 Prandtl-Meyer Supersonic Expansion Function for ^ = 1 .4 Ma u>, deg Ma a;, deg Ma uj, deg Ma ui, deg 1.00 0.00 3.10 51.65 5.10 77.84 7.10 91.49 1.10 1.34 3.20 53.47 5.20 78.73 7.20 92.00 1.20 3.56 3.30 55.22 5.30 79.60 7.30 92.49 1.30 6.17 3.40 56.91 5.40 80.43 7.40 92.97 1.40 8.99 3.50 58.53 5.50 81.24 7.50 93.44 1.50 11.91 3.60 60.09 5.60 82.03 7.60 93.90 1.60 14.86 3.70 61.60 5.70 82.80 7.70 94.34 1.70 17.81 3.80 63.04 5.80 83.54 7.80 94.78 1.80 20.73 3.90 64.44 5.90 84.26 7.90 95.21 1.90 23.59 4.00 65.78 6.00 84.96 8.00 95.62 2.00 26.38 4.10 67.08 6.10 85.63 8.10 96.03 2.10 29.10 4.20 68.33 6.20 86.29 8.20 96.43 2.20 31.73 4.30 69.54 6.30 86.94 8.30 96.82 2.30 34.28 4.40 70.71 6.40 87.56 8.40 97.20 2.40 36.75 4.50 71.83 6.50 88.17 8.50 97.57 2.50 39.12 4.60 72.92 6.60 88.76 8.60 97.94 2.60 41.41 4.70 73.97 6.70 89.33 8.70 98.29 2.70 43.62 4.80 74.99 6.80 89.89 8.80 98.64 2.80 45.75 4.90 75.97 6.90 90.44 8.90 98.98 2.90 47.79 5.00 76.92 7.00 90.97 9.00 99.32 3.00 49.76 818 Appendix B Fig. B.l Mach number downstream of an oblique shock for k = 1.4. Weak shock Strong shock 85 Compressible Flow Tables 819 40 35 30 25 20 Fig. B.2 Pressure ratio downstream of an oblique shock for k = 1.4. 3.0 Appendix C Conversion Factors During this period of transition there is a constant need for conversions between BG and SI units (see Table 1.2). Some additional conversions are given here. Conversion factors are given inside the front cover. Length Volume 1 ft = 12 in = 0.3048 m 1 mi = 5280 ft = 1609.344 m 1 nautical mile (nmi) = 6076 ft = 1852 m 1 yd = 3 ft = 0.9144 m 1 angstrom (A) = 1.0 E— 10 m 1 ft^ = 0.028317 m’ 1 U.S. gal = 231 in’ = 0.0037854 m’ 1 L = 0.001 m’ = 0.035315 ft’ 1 U.S. fluid ounce = 2.9574 E— 5 m’ 1 U.S. quail (qt) = 9.4635 E— 4 m’ 1 barrel = 42 U.S. gal = 0.15899 m’ Mass Area 1 slug = 32.174 Ibm = 14.594 kg 1 Ibm = 0.4536 kg 1 short ton = 2000 Ibm = 907.185 kg 1 tonne = 1000 kg 1 U.S. ounce = 0.02835 kg 1 ft’ = 0.092903 m’ 1 mi’ = 2.78784 E7 ft’ = 2.59 E6 m’ 1 acre = 43,560 ft’ = 4046.9 m’ 1 hectare (ha) = 10,000 m’ Velocity Acceleration 1 ft/s = 0.3048 m/s 1 mi/h = 1.466666 ft/s = 0.44704 m/s 1 kn = 1 nmi/h = 1.6878 ft/s = 0.5144 m/s 1 kn = 1.1508 mi/h 1 ft/s’ = 0.3048 m/s’ Mass flow Volume flow 1 slug/s = 14.594 kg/s 1 Ibm/s = 0.4536 kg/s 1 gal/min = 0.002228 ft’/s = 0.06309 L/s 1 X lO^ gal/day = 1.5472 ft’/s = 0.04381 m’/s 1 ft’/s = 0.028317 m’/s 820 Conversion Factors 821 Pressure Force 1 Ibf/fF = 47.88 Pa 1 Ibf/in^ = 144 Ibf/fF = 6895 Pa 1 atm = 2116.2 Ibf/fF = 14.696 Ibf/in^ = 101,325 Pa 1 inHg (at 20°C) = 3375 Pa 1 bar = 1.0E5Pa 1 ton- = (l/760)atm = 133.32 Pa = 2.7845 Ibf/fF 1 Ibf = 4.448222 N = 16 oz 1 kgf = 2.2046 Ibf = 9.80665 N 1 U.S. (short) ton = 2000 Ibf 1 dyne = 1.0E-5N 1 ounce (avoirdupois) (oz) = 0.27801 N 1 poundal = 0.13826 N Energy Power 1 ft • Ibf = 1.35582 J 1 Btu = 252 cal = 1055.056 J = 778.17 ft ■ Ibf 1 kilowatt hour (kWh) = 3.6 E6 J 1 calorie = 4.1868 J 1 therm = 1 E5 Btu = 1.0551 E8 J 1 hp = 550 ft • Ibf/s = 745.7 W 1 ft ■ Ibf/s = 1.3558 W Specific weight Density 1 Ibf/fF = 157.09 nW 1 slug/fF = 515.38 kg/m’ 1 Ibm/ft’ = 16.0185 kg/m’ 1 g/cm’ = 1000 kg/m’ Viscosity Kinematic viscosity 1 slug/(ft • s) = 47.88 kg/(m ■ s) 1 poise (P) = 1 g/(cm • s) = 0.1 kg/(m • s) 1 ft’/h = 0.000025806 m’/s 1 stokes (St) = 1 cm’/s = 0.0001 m’/s Temperature scale readings Fp = \|Fc + 32 Tc = ICFf - 32) Fr = Tp + 459.69 Fr = Fc + 273.16 where subscripts F, C, K, and R refer to readings on the Fahrenheit, Celsius, Kelvin, and Rankine scales, respectively. Specific heat or gas constant Thermal conductivity 1 ft • lbf/(slug ■ °R) = 0.16723 N • m/(kg ■ K) 1 Btu/(lbm • °R) = 4186.8 J/(kg ■ K) 1 Btu/(h • ft • °R) = 1.7307 W/(m ■ K) Although the absolute (Kelvin) and Celsius temperature scales have different starting points, the intervals are the same size: 1 kelvin = 1 Celsius degree. The same holds true for the nonmetric absolute (Rankine) and Fahrenheit scales: 1 Rankine degree = 1 Fahrenheit degree. It is customary to express temperature differences in absolute temperature units. Appendix D Equations of Motion in Cylindrical Coordinates The equations of motion of an incompressible newtonian fluid with constant /i, k, and Cp are given here in cylindrical coordinates (r, 9, z), which are related to cartesian coordinates (x, y, z) as in Fig. 4.2: x=rcos9 y = r sin 9 z = z (D-1) The velocity components are v^, Vg, and v,. Here are the equations: Continuity: 1 d r dr (rVr) + r d9 {Vg) + a dz {vd = 0 Convective time derivative: Laplacian operator: „ a 1 a a V = V,. - 1 - Vg - h V, — dr r ^d9 ^dz 1 a / a \ 1 a^ a^ rdrV dr) r^ d 9^ dzT (D.2) (D.3) (D.4) The r-momentum equation: dV,. dt + (V • - i?g r 1 dp p dr + g, + V, v^ 2 dVe\ r^ d9 ) (D.5) The 0-momentum equation: dVg dt + {y-V)Vg + 1 -V,Vg = j_ap prd9 + gg+ V^Vg Vg 2 dV,\ ? ? d9) (D.6) 822 Equations of Motion in Cylindrical Coordinates 823 The z-momentum equation: dv, I dp , + (V • W)v, = + g, + uW V, dt p dz The energy equation: ' dT PCp where dt + (V • V)7’ = kW^T + p[2{4r + 4e + 4z) + 4. + 4 + 4e] Sj-r e.. = dVr dr dv, dz _ dv, dv, — - h - dz dr Viscous stress components: t,t “ 2p£„ T,0 = p,£,g Angular velocity components: \(dVg 1 dv, dVg £ez - TT + r dO dz £re — dv. r\ de Vg + dVg dr Tee ~ 2/iff flfl Te, = /iff'fe Tzz = 2/iff^ rz f-^^rz 1 dv, dVg 2uj, = - r d0 dz 2ilJg — dv, dv, dz dr la IdV, 2uJ, = -—(rvg) - - — r dr r d6 (D.7) (D.8) (D.9) (D.IO) (D.ll) Appendix E Estimating Uncertainty in Experimental Data Uncertainty is a fact of life in engineering. We rarely know any engineering properties or variables to an extreme degree of accuracy. The uncertainty of data is normally defined as the band within which one is 95 percent confident that the true value lies. Recall from Fig. 1.7 that the uncertainty of the ratio was estimated as ±20 percent. There are whole monographs devoted to the subject of experimental uncertainty, so we give only a brief summary here. All experimental data have uncertainty, separated into two causes: (1) a systematic error due to the instrument or its environment and (2) a random error due to scatter in repeated readings. We minimize the systematic error by careful calibration and then estimate the random error statistically. The judgment of the experimenter is of crucial importance. Here is the accepted mathematical estimate. Suppose a desired result P depends upon a single experimental variable x. If x has an uncertainty dx, then the uncertainty dP is estimated from the calculus: If there are multiple variables, P = P(xi, X2, x^, ... Xf^), the overall uncertainty dP is calculated as a root-mean-square estimate : \2 / TO \2 / TO \2-\| 1/2 SP = dp — c dxi Sxi ± dp 9X2^ 5x' ■p dp dXN &A (E.l) This calculation is statistically much more probable than simply adding linearly the various uncertainties thereby making the unlikely assumption that all variables simultaneously attain maximum error. Note that it is the responsibility of the experi¬ menter to establish and report accurate estimates of all the relevant uncertainties rix,. If the quantity P is a simple power-law expression of the other variables, for example, P = Const ..., then each derivative in Eq. (E.l) is proportional to P and the relevant power-law exponent and is inversely proportional to that variable. 824 Estimating Uncertainty in Experimental Data 825 If P = Const then dP tiiP dP n2P dP n^P dXi Xi ’ dX2 X2 ’ dX2 Thus, from Eq. (E.l), P (E.2) Evaluation of dP is then a straightforward procedure, as in the following example. EXAMPLE The so-called dimensionless Moody pipe friction factor/ plotted in Fig. 6.13, is calculated in experiments from the following formula involving pipe diameter D, pressure drop Ap, density p, volume flow rate Q, and pipe length L: \p 8 PQ^L Measurement uncertainties are given for a certain experiment: D = 0.5 percent, Ap = 2.0 percent, p = 1.0 percent. Q = 3.5 percent, and L = 0.4 percent. Estimate the overall un¬ certainty of the friction factor/. Solution The coefficient 7T^/8 is assumed to be a pure theoretical number, with no uncertainty. The other variables may be collected using Eqs. (E.l) and (E.2): Sf U = — = f .60 ’ D + 1 AApV + 1 6p + 2 Q + 1 SL L 2t1/2 = [{5(0.5%)}" -f (2.0%)" -f (1.0%)" -f (2(3.5%)}" -f (0.4%)"]"" « 7.8% Ans. By far the dominant effect in this particular calculation is the 3.5 percent error in Q, which is amplified by doubling, due to the power of 2 on flow rate. The diameter uncertainty, which is quintupled, would have contributed more had 6D been larger than 0.5 percent. References 1 . I. Hughes and J. Hase, Measurements and their Uncertainties, Oxford University Press, New York, 2010. 2. H. W. Coleman and W. G. Steele, Experimentation and Uncertainty Analysis for Engineers, 3d ed., Wiley, New York, 2009. 3. S. E. Serrano, Engineering Uncertainty and Risk Analysis, Hydroscience Inc., Toms River, NJ, 2011. 4. S. J. Kline and F. A. McClintock, “Describing Uncertainties in Single-Sample Experiments,” Mechanical Engineering, January, 1953, pp. 3-9. Answers to Selected Problems Chapter 1 P1.70 h = 2Y cos 6l(pgW) P1.2 5.7 El 8 kg; 1.2 E44 molecules P1.72 z ~ 4800 m P1.6 (a) = {L-‘} P1.74 8.6 km PI. 8 cr« 1.00 My// P1.76 (b) /3steam ~ 262 kPa PI. 10 Yes, all terms are [ML/T^] P1.78 (a) 25°C; (b) 4°C PI. 12 (5) = IL-^] P1.80 Ma = 1.20 PI. 14 Q = Const B g'^//^® PI. 82 y = jc tan 6 + constant PI. 16 All terms are {ML~^T~~] P1.86 Approximately 5.0 percent PI. 18 V = Voe-"""' P1.20 (b) 2080 Chapter 2 P1.24 (a) 41 kPa; {b) 0.65 kg/m^ P2.2 = -289 Ib/ft^ Taa = -577 Ib/ft^ P1.26 = 0.71 Ibf P2.4 Approximately 100 degrees P1.28 p„a = 1-10 kg/m^ pdjy = 1.13 kg/m^ P2.6 (a) 26.9 ft; (b) 30.0 in; (c) 10.35 m; (d) 13,100 mm P1.30 lVi.2 = 21 ft ■ Ibf P2.8 Approximately 2.36 E6 Pa P1.32 (a) 76 kN; (b) 501 kN P2.10 10,500 Pa P1.34 (a) pi = 5.05 kg/m^; (b) p2 = 2.12 kg/m^ (ideal gas) P2.12 8.0 cm P1.36 (b) p « 628 kg/m^ P2.14 hi = 6.0 cm, h2 = 52 cm P1.38 T= 1380Pa, Re^ = 28 P2.16 (a) 1885 Ibf/tf ; (b) 2165 Ibf/ft^ PI. 40 Approximately 25 N ■ m per meter P2.18 1.56 PI. 42 T « 539°C P2.20 14 Ibf PI. 44 p ~ 0.040 kg/m • s P2.22 0.94 cm P1.46 (d) 3.0 m/s; (e) 0.79 m/s; (/) 22 m/s P2.24 Psealevel ~ 1 15 kPa, OTe^ct = 5.3 E18 kg P1.48 F ~ (pi/hi + p2^h2)AV P2.28 (a) 454 ft P1.50 (a) Yes; (b) p ~ 0.40 kg/(m ■ s) P2.30 (a) 29.6 kPa; (b) K = 0.98 P1.52 P«73W P2.32 22.6 cm P1.54 M ~ TrpClRlh P2.34 ^p = A/l[7„,,er(l + - Tcild - P1.56 p = 3M sin OKlivaR^) P2.36 25° P1.58 p = 0.040 kg/(m • s), last 2 points are turbulent flow P2.38 Pa = 219 kPa P1.60 39,500 Pa P2.40 Pb = 17.6 Ibf/in^ P1.62 28,500 Pa P2.42 Pa- Pb = (Pi “ P1.64 D> 5 mm P2.44 (a) 171 Ib/ft^; (b) 392 Ib/ft^; manometer reads friction ! P1.66 F = 0.014 N P2.46 1.45 P1.68 h = (Y/pg)‘® cot e P2.48 (a) 132 kPa; (b) 1.38 m 826 Answers to Selected Problems 827 P2.50 (a) 220 ft; (b) 1 10,000 Ibf P2.152 (a) 224 r/min; (b) 275 r/min P2.52 (a) 38,400 Ibf; (b) 5.42 ft from A P2.154 552 r/min P2.56 16.08 ft P2.156 420 r/min P2.58 0.40 m P2.158 10.57 r/min P2.60 (a) Approximately 62,000 Ibf P2.62 10.6 ft Chapter 3 P2.64 1.35 m P3.2 r = position vector from point O P2.66 F = 1.18 E9 N, Me = 3.13 E9 N ■ m counterclockwise. P3.4 V= 4.38 m/s no tipping P3.6 Q = {2b/3){2gf\{h + -{h- Lf P2.68 18,040 N P3.8 (a) 5.45 m/s; (b) 5.89 m/s; (c) 5.24 m/s P2.70 0.79 m P3.10 (a) 3 m/s; (b) 18 m/s; (c) 5 cm/s out P2.72 Ma = 32,700 N • m P3.12 Af = 46 s P2.74 H = F[7r/4 + {(7r/4)^ + 2/3 P3.14 dh/dt = (Qi + 03 - 22)/(7rrf"/4) P2.76 (a) 239 kN; (c) 388 kN ■ m P3.16 anp = 3Uobm P2.78 {b) Fab = 4390 N, Fcd = 4220 N P3.18 V3 = 14.7 m/s (out) P2.80 e > 77.4° P3.20 (a) 7.8 mL/s; (b) 1.24 cm/s P2.82 F[t = 97.9 MN, Fy = 153.8 MN P3.22 (a) 0.06 kg/s; (b) 1060 m/s; (c) 3.4 P2.84 (a) Fy = 2940 N; F„ = 6880 N P3.24 h = [3Kt^d^/{8 tan^ 6»)]''^ P2.86 F = 59 kN P3.26 Q = 2Uobh/3 P2.88 Fh = 176 kN, Fv = 31.9 kN, yes P3.28 (a) 0.131 kg/s; (b) 115 kPa P2.90 Fy = 22,600 N; F„ = 16,500 N P3.30 (a) Mai = 1-00; (fc) Fj = 216 K P2.92 F one bolt ~ 1 1 ,300 N P3.32 Ehoie = 6.1 m/s P2.94 Forces on each panel are equal. P3.34 V2 = 4660 ft/s P2.96 Fa = 1 10 kN , Fy = 279 kN P3.36 U} = 6.33 m/s P2.98 Fa = 245 kN, Fy = 51 kN P3.38 V = Vor/(2h) P2.100 Fa = 0,Fy= 297 kN P3.40 500 N to the left P2.102 (a) 238 kN; (b) 125 kN P3.42 F= (Pi- Pa)Ai - PiAiVf[(Di/D2f - 1 P2.104 5.0 N P3.44 F = pU'^Lb/3 P2.106 D « 2.0 m P3.46 a = (1 + cos 6»)/2 P2.108 (a) 0.0427 m; (b) 1592 kg/m^ P3.48 Vo « 2.27 m/s P2.110 (a) 14.95 N, SG = 0.50 P3.50 102 kN P2.112 (a) 39 N; (b) 0.64 P3.52 8.65 Ibf P2.114 0.636 P3.54 163 N P2.116 (a) Yes; (b) Yes; (c)3.51 in P3.56 (fl) 18.5 N to left; (fc)7.1Nup P2.118 6.14 ft P3.58 40 N P2.120 34.3° P3.60 2100 N P2.122 alb « 0.834 P3.62 3100 N P2.124 6850 m P3.64 980 N P2.126 3130 Pa (vacuum) P3.66 8800 N P2.128 Yes, stable if 5 >0.789 P3.70 91 Ibf P2.130 Slightly unstable, MG = —0.007 m P3.72 Drag « 4260 N P2.132 Stable if F//; > 3.31 P3.74 F, = 0,Fy= - 17 N, F, = 126 N P2.134 (a) unstable; (b) stable P3.76 (a) 1670 N/m; (b) 3.0 cm; (c) 9.4 cm P2.136 MG = LV(37rF) - 4F/(37r) > 0 if L > 2F P3.80 F = {pl2)gb(hi - hi) - ph,bVl(hA2 - P2.138 2.77 in deep; volume = 10.8 fluid ounces P3.82 25 m/s P2.140 = —4.9 mVs (decelerating) P3.84 23 N P2.142 (a) 16.3 cm; (b) 15.7 N P3.86 274 kPa P2.144 (a) = 319 m/s^; (b) no effect, Pa — Pb P3.88 V=C + [C + 2CV^]^'\C = pQ/2k P2.146 Leans to the right at P = 27° P3.90 dV/dt = g P2.148 (a) backward; (b) forward P3.92 dV/dt = gh/(L + h) P2.150 5.5 cm; linear scale OK P3.94 Vje, = 23.7 ft/s 828 Answers to Selected Problems P3.96 (fZIdf- + IgZIL = 0 P3.100 (a) 507 m/s and 1393 m; (b) 14.5 km P3.102 h2/hi = + ^[1 + 8Vi/(ghi)f^ P3.104 fi = (-VJR) In (1 - mt/Mo) P3.106 F = (ttK) p P3.108 (a) V = Vo/(l + CVot/M), C = pbh(l - cos 6») P3.110 Fboits = 13401bf P3.112 (fl) 10.3 kg/s; (b) 760 kPa P3.1 14 (a) V2 = 9.96 m/s; (b) /? = 40 cm P3.116 X=2[h(H-h)f^ P3.120 (a) 0.495 ftVs; (b) 12.3 ftVs P3.122 “ 25.4 cm P3.126 (a) 0.197 ft^/s; (b) 0.089 ftVs P3.130 104 kPa (gage) P3.132 g=127cmVs P3.134 (a) 15 m; (Zr) 25 mm P3.136 0.037 kg/s P3.140 5.93 m (subcritical); 0.432 m (supercritical) P3.142 fo = [2(a - l)ho/gf\ a = {D/df P3.144 Approximately 294 gal/min P3.150 To=-305 kN-m P3.154 rB = 40ft-lbf P3.156 2.44 kg/s P3.158 P = P R Pn(cOt Qi + cot 02) P3.160 TB=-2400kN-m P3.162 197 hp; max power at 179 r/min P3.164 To = p!2P„(F + L/2)k P3.166 eo = 2.5mVs;ro = 23.15°C P3.168 Ar«0.7°F P3.170 (a) 699 kJ/kg; (&) 7.0 MW P3.172 8.7 m P3.174 (a) 410 hp; (fc) 540 hp P3.176 97 hp P3.178 8.4 kW P3.180 112hp P3.182 3.2°C/s P3.184 76.5 mVs and 138 mVs Chapter 4 P4.2 (a) du/dt = (2V^/L)(1 + 2x/L) P4.4 {b) = (C/o/L)(l + x/L); a, = (C/^L)(y/L) P4.6 (b) a^= 16 x; Oy = 16 y P4.8 (a) 0.0196 (7^/L; (b) at t = 1 .05 LIU P4.10 {b)a = 8r P4.12 If Vg =v^= 0, Vr = r~^ fen (9, (f>) P4.14 Vg = fcn(r) only P4.16 (a) Yes, continuity is satisfied. P4.18 p = PoLoKLo - Vt) P4.20 n = -Uo = const, {AT} = [LIT], {«} = {L“'} P4.22 Vy= -Brl2+f(z)lr P4.24 (b) B = 3vJ(2h^) P4.28 (a) Yes; (b) Yes P4.30 (a, b) Yes, continuity and Navier-Stokes are satisfied. P4.32 /i = CrJi = Cz/r P4.36 C = pg sin 6»/(2/i) P4.40 T=Ty, + ipU - yW) P4.48 = Uorsin 9 — V^r cos 9 + const P4.50 (a) Yes, 'i/t exists. P4.52 t/t = -4Q9/{Trb) P4.54 Q = ULb P4.60 Irrotational, zc = H ~ <^^R~l(2.g) P4.62 xp = VyV(2h) + const P4.66 xp = —K sin 9lr P4.68 (a) Yes, a velocity potential exists. P4.70 V'«2.33m/s, -16.5° P4.72 (a) xp = -0.0008 0; (b) (p = -0.0008 ln(r) P4.74 xp = B r sin 9 + B Llnr + const P4.76 Yes, t/) exists. P4.78 y = A r" cos{n9) + const P4.80 (a) w = {pg/2p){2Sx - x^) P4.82 Obsessive result: Vg = ilR^/r P4.84 -u, = (pgb^l2p) In (r/a) - (pg/4p){r^ - a^) P4.86 e = 0.0031 mV(s • m) P4.88 v,= U In (r/b)/[ln (alb)] P4.90 (a) 54 kg/h; (b) 7 mm P4.92 h = ho exp[-TrD‘^pgtl(l28pLAo)] P4.94 Vg = LlR^lr P4.96 (a) 11 30 Pa Chapter 5 P5.2 Prototype V = 22.8 mi/h P5.4 V= 1.55 m/s, F= 1.3 N P5.6 (a) 1.39; (0)0.45 P5.8 Ar = gH^(Xpflp^ P5.10 (a) (b) [MLT^] P5.12 St = pUI(pgD^) P5.14 One possible group is /jF/fc. P5.16 Stanton number = /7/(pV'Cp) P5.18 Qpl[(XplL)b‘'] = const P5.20 One possible group is FlDIU. P5.22 LIDIV = fcn(/V, HIL) P5.24 FI(pVV) = fcn(Q;, pVUp, LID, Via) P5.26 (a) Indeterminate; (b) T = 2.75 s P5.28 6IL = fcn[L/D, pVDIp, EI(pV^)] P5.30 MRTy^KpJ)^) = fcn(Cp/F) P5.32 QI(bg''^H^'^) = const P5.34 Fbydrogen ~ 0. 1 82 lV/(m • K) P5.36 (a) 2ioss^^('A‘^7) = constant P5.38 dID = icnipUDIp, pU^DIY) P5.42 Halving m increases/by about 41 percent. P5.44 (a) {(T) = {L^} P5.48 F « 0.17 N; (doubling U quadruples F) P5.50 Approximately 2000 Ibf (on earth) P5.52 (a) 0.44 s; (b) 768,000 Answers to Selected Problems 829 P5.54 Power = 7 hp P5.56 « 25 N/m P5.58 l/«2.8m/s P5.60 ib) 4300 N P5.62 (a) Ll) = 14.4 r/min P5.64 t^aluminum = 0.77 Hz P5.66 (a) V=27 m/s; (b) z = 27 m P5.68 (fc) Approximately 1800 N P5.70 F = 87 Ibf (extrapolated) P5.72 About 44 kN (extrapolated) P5.74 Prototype moment = 88 kN ■ m P5.76 Drag = 107,000 Ibf P5.78 Weber no. « 100 ifLJLp = 0.0090 P5.80 (a) 1.86 m/s; (b) 42,900; (c) 254,000 P5.82 561 kN P5.84 = 39 cm/s; T„, = 3.1 s; H,„ = 0.20 m P5.88 At 340 W, D = 0.109 m Chapter 6 P6.2 (a) Yes P6.4 (a) 106 m^/h; (b) 3.6 m^/h P6.6 (a) hydrogen, x = 43 m P6.8 (a) -3600 Pa/m; (b) -13,400 Pa/m P6.10 (a)FromAtoS;(ft)/2^=7.8m P6.12 fi = 0.29 kg/m • s P6.14 Q = 0.0067 mVh if // = 50 cm P6.16 19 mm P6.18 4.3 mVh P6.20 4500 cc/h P6.22 F = 4.0N P6.24 (a) 0.019 m^/h, laminar; (b) d = 2.67 mm P6.26 (a) ^2 = 5.95 cm P6.28 ^p = 65 Pa P6.30 (fl) 19.3 m^/h; {b) flow is up P6.32 (a) flow is up\ (b) 1.86 m^/h P6.36 (a) 0.029 Ibf/ft^; (b) 70 ft/s P6.38 5.72 m/s P6.42 16.7 mm P6.44 hf = 10.4 m, ^p = 1.4 MPa P6.46 46 hp/mi P6.48 238,000 barrels/day P6.50 (a) —4000 Pa/m; (b) 50 Pa; (c) 46 percent P6.52 Pi = 2.38 MPa P6.54 fdrain = [4 WF/fTTO’)] [2/to( 1 + /avF/D)/g] P6.56 (a) 2680 Ibf/in^ (b) 5300 hp P6.58 80 m^/h P6.60 (a) Not identical to Haaland P6.62 204 hp P6.64 Q = 19.6 m^/h (laminar. Re = 1450) P6.66 (a) 56 kPa; (b) 85 m^/h; (c) u = 3.3 m/s at r = 1 cm P6.70 e = 31 m^/h P6.72 D « 9.2 cm P6.74 L = 205 m P6.76 Q = 15m^/hor9.0m^/h P6.78 2 = 25 m^/h (to the left) P6.80 Q = 0.905 mVs P6.82 (a) 10.9 mVh; (b) 100 m^/h P6.84 D « 0.104 m P6.86 (a) 3.0 m/s; (b) 0.325 m/m; (c) 2770 Pa/m P6.88 About 17 passages P6.90 // = 36 in P6.92 (a) 1530 m^/h; (b) 6.5 Pa (vacuum) P6.94 a = 18.3 cm P6.96 (b) 12,800 Pa P6.98 Approximately 128 squares P6.100 4.85 m^/h P6.102 (a) 5.55 hp; (b) 5.31 hp with 6° cone P6.104 Approximately 34 kPa P6.106 Q = 0.0296 ftVs P6.108 (a) K « 9.7; (b) Q « 0.48 ftVs P6.110 840 W P6.112 Q = 0.0151 ftVs P6.114 Short duct: Q = 6.92 ftVs P6.116 Q = 0.027 mVs P6.118 Ap = 131 Ibf/in^ P6.120 Qi = 0.0281 mVs, Q2 = 0.0111 mVs, = 0.0164 mVs P6.122 Increased e/d and L/d are the causes P6.124 Qi = -2.09 ftVs, 02 = 1-61 ftVs, 23 = 0.49 ftVs P6.126 a — oco ^opening P6.128 Qab — 3.47, 2bc — 2.90, 2bo — 0.58, Qqd ~ Qac = 2.38 ftVs (all) 5.28, P6.130 Qab — 0.95, 2bc — 0.24, 2bo — 0.19, 2cfl — 2ac = 1-05 ftVs (all) 0.31, P6.132 29 = 6°, D, = 2.0 m, p, = 224 kPa P6.134 20 = 10°, W, = 8.4 ft,p, = 2180 Ibf/ft^ P6.136 (a) 25.5 m/s, (b) 0.109 mVs, (c) 1.23 Pa P6.138 46.7 m/s P6.140 333 Pa P6.142 2 = 18.6 gal/min, i/reducer = 0.84 cm P6.144 (a) /i = 58 cm P6.146 (a) 0.00653 mVs; (b) 100 kPa P6.148 (a) 1.58 m; (b) 1.7 m P6.150 Ap = 27 kPa P6.152 D = 4.12 cm P6.154 106 gal/min P6.156 2 = 0.924 ftVs P6.158 (a) 49 m^/h; (b) 6200 Pa Chapter 7 P7.2 This is probably helium. P7.4 (a) 4 /im; (b) 1 m P7.6 H = 2.5 (versus 2.59 for Blasius) P7.8 Approximately 0.073 N per meter of width P7.12 Does not satisfy d^u/dy^ = 0 at y = 0 830 Answers to Selected Problems P7.14 C = pVf/fi = const < 0 (wall suction) P7.16 (a)F= 181N;(Z))256N P7.18 (a) 3.41 m/s; (b) 0.0223 Pa P7.20 X « 0.91 m P7.22 (a) y = 3.2 mm P7.24 hi = 9.2 mm; h2 = 5.5 mm P7.26 = 2.83 Fj, Fj, = 2.0 Fj P7.28 (fl) Fjrag = 2.66 N\ppLf''^U^'^a P7.30 Predicted thickness is about 10 percent higher P7.32 F = 0.0245 pi/''’ L®'’ <5 P7.34 45 percent P7.36 7.2 m/s = 14 kn P7.38 (a) 7.6 m/s; {b) 6.2 m/s P7.40 L = 3.51 m,fe = 1.14 m P7.42 (a) 5.2 N/m P7.44 Accurate to about ±6 percent P7.46 e«9mm,(/= 11.1 m/s = 22kn P7.48 Separation at xlL = 0.158 (1 percent error) P7.50 Separation at 0 ~ 2.3 degrees P7.52 (a)Rei = 0.84< l;(/?)2a = 30mm P7.54 z = TJ[B(n + \)ln = gURB)- \ P7.56 (a) 14 N; (b) crosswind creates a very large side force P7.60 Tow power = 140 hp P7.62 Square side length « 0.83 m P7.64 Afiooo - 2000m ~ 202 s P7.68 69 m/s P7.70 40 ft P7.72 (a) L = 6.3 m; (fe) 120 m P7.74 About 130 mi/h P7.76 (a) 343 hp P7.78 28,400 hp P7.80 0 = 72° P7.82 (a) 46 s P7.84 V = 9 m/s P7.86 Approximately 2.9 m by 5.8 m P7.88 (a) 62 hp; (b) 86 hp P7.90 Vovertum = 145 ft/s = 99 mi/h P7.94 (a) 100 mi/h; (b) 88 mi/h P7.96 flavg « 0.21 U/D P7.98 (/))/!« 0.18 m P7 . 1 00 (fl) 73 mi/h; (b) 79 mi/h P7.104 29.5 knots P7.106 1130 m’ P7.108 A-Xbaii ~ 13 m P7.110 Ay « 1.9 ft P7.1 14 Vdown ~ 25 m/min; V^p « 30 m/min P7.1 16 (a) 87 mi/h; (b) 680 hp P7.118 (a) 27 m/s; (fc) 360 m P7.120 (L/D)n,ax = 21; a = 4.8° P7.122 (a) 6.7 m/s; (b) 13.5 m/s = 26 kn P7 . 1 24 flc,ade theory ~ 340 r/s P7 . 1 26 Approximately 850 ft Chapter 8 P8.2 r = ttOCFI - Ri) P8.4 No, 1/r is not a proper two-dimensional potential P8.6 'lp = B/ sin(26l) P8.8 r = 4S P8.10 (a) 1.27 cm P8.12 r = o P8.14 Irrotational outer, rotational inner; minimum p = Poo ~ puj^R^ at r = 0 P8.16 (fl) 0.106 m to the left of A P8.18 From afar: a single source 4m P8.20 Vortex near a wall (see Fig. 8.17fe) P8.22 Stagnation flow toward a bump P8.24 Cp = -{2(x/fl)/[l -f (x/fl)’]}’, = -1.0 atx = fl P8.26 (fl) 8.75 m; {b) 27.5 m on each side P8.28 Creates a source in a square comer P8.30 r = 25 m P8.32 m2 = 40 m’/s P8.34 Two stagnation points, at x = ± P8.36 Foo = 12.9 m/s, 2L = 53 cm, = 22.5 m/s P8.40 1.47 m P8.42 1 1 1 kPa P8.44 K = 3.44 m’/s; (a) 218 kPa; (b) 205 kPa upper, 40 kPa tower P8.46 Fi.boit = 5060 N P8.50 h = 3fl/2, Fpiax = 5(7/4 P8.52 Vboat = 10.4 ft/s with wind at 58° P8.54 Fparaiiei = 6700 Ibf, Fnormai = 2700 Ibf, power = 560 hp (very approximate) P8.60 This is Fig. 8.18fl, flow in a 60° comer P8.62 Stagnation flow near a “bump” P8.66 X = 0.45m/(5m -f 1) if (7 = Cx"' P8.68 Flow past a Rankine oval P8.70 Applied to wind tunnel “blockage” P8.72 Adverse gradient for x> a P8.74 Vs..c.ai = (8Fi + 4Fj)/(15fl) P8.78 Need an infinite array of images P8.82 (fl) 4.5 m/s; (b) 1.13; (e) 1.26 hp P8.84 (fl) 0.21; (fc) 1.9° P8.86 (fl) 26 m; (b) 8.7; (e) 1600 N P8.88 (fl) 11.1; ((i) 0.56 P8.92 (fl) 0.77 m; (b) V = 4.5 m/s at (r, 6») = (1.81, 51°) and (1.11, 88°) P8.94 Yes, they are orthogonal P8.96 (fl) 0.61 (7i/fl P8.98 Yes, a closed teardrop shape appears P8.100 V= 14.1 m/s, = 115 kPa P8.102 (fl) 1250 ft; (b) 1570 ft (crudely) Chapter 9 P9.2 (fl) V2 = 450 m/s, Ai = 515 J/(kg ■ K); (b) V2 = 453 m/s, Ai = 512 J/(kg • K) P9.4 (fl) -f372 J/(kg ■ K) Answers to Selected Problems 831 P9.6 (fl)381K P9.8 410 K P9.10 (fl)0.80 P9.12 (a) 2. 13 E9 Pa and 1460 m/s; {b)2.9\ E9Paand 1670 m/s; (c) 2645 m/s P9.14 Approximately 1300 m/s P9.18 Ma«0.24 P9.20 (a) 41 kPa; {b) 421 m/s; (c) 1.27 P9.22 (a) 267 m/s; (b) 286 m/s P9.24 (b) at Ma « 0.576 P9.28 (b) 232 K P9.30 Deviation less than 1 percent at Ma = 0.3 P9.32 (fl) 141 kPa; (b) 101 kPa; (c) 0.706 P9.34 (a) 3.74 cm P9.36 (fl) 0.142 kg/s P9.40 (a) 0.192 kg/s P9.42 (fl)Ma = 0.90, 7= 260 K,V = 291 m/s P9.44 Vg = 5680 ft/s, = 15.7 psia, 7^ = 1587°R, thrust = 4000 Ibf P9.46 (a) 0.0020 m^ P9.48 (a) 313 m/s; (b) 0.124 m/s; (c) 0.00331 kg/s P9.50 (a) 0.0970 kg/s P9.52 (a) 5.9 cm^; (b) 773 kPa P9.54 (a)Ma2 = 0.513 P9.56 At about Ai « 24.7 cm^ P9.58 (a) 3.50 P9.60 Upstream: Ma = 1.92, V = 585 m/s P9.62 C = 19,100 ft/s, = 15,900 ft/s P9.64 (a) 4.0 cm^; (b) 325 kPa P9.66 /i= 1.09 m P9.68 Patm = 92.6 kPa; max flow = 0.140 kg/s P9.70 119kPa P9.72 D « 9.3 mm P9.74 0.191 kg/s P9.76 23 S, Ar,.]ao]gjag.stops 39 s P9.78 Case A: 0.071 kg/s; B: 0.068 kg/s P9.80 A = 2.4 E-6 ft^ or 0^0,^ = 0.021 in P9.82 P, = 1 10 m/s, Ma, = 0.67 (yes) P9.84 (a) 0.96 kg/s; (b) 0.27; (c) 435 kPa P9.86 V2 = 107 m/s, Pz = 371 kPa, T2 = 330 K, = 394 kPa P9.88 (a) 12.7 m P9.90 (a) 0.030; (b) 16.5 Ibf/in^ P9.92 (a) 14.46 m P9.96 (a) 128 m; (fc) 80 m; (c) 105 m P9.98 (a) 430; (b) 0.12; (c) 0.00243 kg/h P9.100 0.345 kg/s P9.102 Flow is choked at 0.56 kg/s P9.104 p,ank=190kPa P9.106 about 91s P9.108 Mass flow drops by about 32 percent P9.112 (/))129kPa P9.114 (fl) 2.21; (/)) 779 kPa;(c) 1146K P9.116 Vpiane < 2640 ft/s P9.118 P = 204 m/s, Ma = 0.6 P9. 120 P is 3 m ahead of the small circle, Ma = 2.0, 7g(ag = 5 18 K P9.122 P = 23.13°, Maz = 2.75, pz = 145 kPa P9.124 (fl) 1.87; (b) 293 kPa; (c) 404 K; (d) 415 m/s P9.126 (fl)2.11 P9.128 (Jwedge « 15.5° P9.132 (fl) pyi = 18.0 psia; (b) pg = 121 psia P9.134 Maj = 1.02, p3 = 727 kPa, 0 = 42.8° P9.136 (fl) h = 0.40 m; (b) Maj = 2.43 P9.138 p, = 21.7kPa P9.140 Maz = 2.75, Pz = 145 kPa P9.142 (fl) Maz = 2.641, pz = 60.3 kPa; (b) Maz = 2.299, Pz = 24.1 kPa P9.144 (fl) 10.34 degrees P9.146 (fl) 2.385; (i) 47 kPa P9.148 (fl) 4.44; (fo) 9.6 kPa P9.150 (fl) a = 4.10°; (b) drag = 2150 N/m P9.152 Approximately 53 Ibf P9.156 (fl)Cz, = 0.139; Co = 0.0146 Chapter 10 P10.2 (fl) Fr = 2.69 P10.4 These are piezometer tubes (no flow) P10.6 (fl) Fr = 3.8; (b) P,a,„nt = 7.7 m/s P10.8 Aftravei = 6.3 h PIO.IO Acrit = 27r(Y/pg)‘'" PIO. 14 Flow must be fully rough turbulent (high Re) for Chezy to be valid PIO. 16 (fl)2.27mVs P10.18 (fl) 12.4 mVs; (b) about 22 Pa P10.20 0.0174 or 1.0° P10.22 So = 0.00038 (or 0.38 m/km) P10.24 (fl) n « 0.027; (b) 2.28 ft P10.28 (fl) 0.00106 PIO. 30 At ~ 32 min PIO. 32 A = 4.39 m^, 10 percent larger P10.34 If fe = 4 ft, y = 9.31 ft, P = 22.62 ft; if fc = 8 ft, y = 4.07 ft, P = 16.14 ft P10.36 yz = 3.6 m PIO. 38 Maximum flow at 0 = 60° P10.42 The two are equally efficient. P10.44 Hexagon side length b = 2.12 ft P10.46 ho/b « 0.49 P10.48 (fl) 0.00634; (b) 0.00637 P10.50 (fl) 2.37; (b) 0.62 m; (c) 0.0023 P10.52 W = 2.06 m P10.54 (fl) 1.98 m; (b) 3.11 m/s; (c) 0.00405 P10.56 (fl) 7.11 ft/s; (fc) 0.70 P10.58 (fl) 0.0033; (b) 0.0016 P10.60 yz = 0.679 m; Pz = 3.53 m/s P10.64 A/t= 15.94 cm 832 Answers to Selected Problems P10.66 (b) 1.39 P11.28 BEP at about 6 ftVs; N, « 1430, 2max = 12 ftVs P10.70 2600 mVs P11.30 (a) 640 r/min; (b) 75 ft P10.72 (a) 0.046 m; {b) 4.33 m/s; (c) 6.43 P11.32 (a) D ~ 15.5 in; (c) n ~ 2230 r/min P10.76 (a) 379 ftVs P11.34 (a) 1 1.5 in; (b) 28 hp; (c) 100 ft; (d) 78 percent P10.78 1.01 ft P11.36 (a) No; (b) 24.5 in at 960 r/min P10.80 (a) 0.395 P11.38 (a) 18.5 hp; (b) 7.64 in; (c) 415 gal/min; (d) 81 percent P10.82 (a) 1.46 ft; {b) 15.5 ft/s; (c) 2.26; (d) 13 percent; (e) 2.52 ft PI 1.40 (a) D, = P10.84 y2 = 0.82 ft; 3)3 = 5.11 ft; 47 percent PI 1.42 NPSHp,„,„ « 23 ft P10.86 (a) 1.18 ft; (b) 4.58 ft/s PI 1.44 No cavitation, required depth is only 5 ft P10.88 (a) 2.22 mVs/m; (b) 0.79 m; (c) 5.17 m; (d) 60 percent; PI 1.46 D, « CIN„ C = 7800 ± 7 percent (e) 0.37 m PI 1.48 {b) Approximately 130 ft P10.90 (a) y2 = 1.83 ft; yj = 7.86 ft P11.52 (a) 6.56 mVs; (b) 12.0 kW; (c) 28.3° P10.92 yi = 1.71 mm; Vi = 0.310 m/s P11.54 (a) 21 in; (b) 490 bhp P10.94 (a) 5.32; (b) 0.385 m/s; (c) 18.7 cm P11.56 (a) D = 5.67 ft, n = 255 r/min, P = 700 hp; P10.96 R ~ 4.92 cm (b) D = 1.76 ft, n = 1770 r/min, P = 740 hp P10.98 (a) steep S-3; (b) S-2; (c) S-1 P11.58 (b) Approximately 2500 r/min P10.106 No entry depth leads to critical flow PI 1.60 (b) No. P10.108 Approximately 6.6 m PI 1.62 D= 18.7 ft, Ap= 1160 Pa PlO.llO (a) Merest ~ 0.782 m; (b) y{L) ~ 0.909 m PI 1.64 (a) 15.4 in; (b) 900 r/min P10.112 M-1 curve, with y = 2matL = 214m PI 1.66 720 ftVmin, non-BEP efficiency 78 percent P10.114 11.5 ft P11.68 (a) 4.8 in; (b) 6250 r/min P10.120 y = 0.64 m,a = 34° P11.70 (a) 212 ft; (b) 5.8 ftVs P10.122 5500 gal/min P11.72 (a) 10 gal/min; (b) 1.3 in P10.124 M-1 curve, y = 10 ft at x = —3040 ft P11.74 (a) 14.9; (b) 15.9; (c) 20.7 kgal/min (all) P10.126 At.r = -100m,y = 2.81 m P11.76 Dpipe « 1.70 ft P10.128 At 300 m upstream, y = 2.37 m P11.78 P11.80 Approximately 10 stages Both pumps work with three each in series, the largest Chapter 11 being more efficient. P11.6 This is a diaphragm pump. P11.84 Two turbines: (a) D ~ 9.6 ft; (b) D ~ 3.3 ft P11.8 (a) 86 percent P11.86 Asp = 70, hence Erancis turbines Pll.lO (a) 12 gal/min; (b) 12 gal/min; (c) 87 percent P11.88 (a) Erancis; (c) 16 in; (d) 900 r/min; (e) 87 hp P11.12 (a) 11.3 m; (b) 1520 W PI 1.90 P « 800 kW P11.14 1870W PI 1.94 (a) 71 percent; (b) Asp ~ 19 P11.16 Q ~ 7100 gal/min; // « 38 m PI 1.96 (a) 0.45 m; (b) 0.17 m P11.18 Kane = (l/3)Vjet for max power Pll.lOO About 5.7 MW PI 1.20 (a) 2 roots: Q = 7.5 and 38.3 ftVs; (b) 2 roots; N = 180 ft PI 1.102 2 « 29 gal/min and 35 ft P11.104 (a) 69 MW P11.22 (fl) BEP = 92 percent at 2 = 0.20 mVs PI 1.106 Approximately 15 mi/h PI 1.26 (a) Both are fine, the largest is more efficient. PI 1.108 (a) About 15 Darrieus turbines Index A absolute atmospheric pressure, 66 absolute temperature, 599 absolute temperature scales, 17 acceleration field, 15, 222-223, 578 Ackeret, Jacob, 659 Ackeret drag theory, 659 Active flow control, 488 actuator disk, 784 added mass. See hydrodynamic mass add energy (pumps), 741 adiabatic and isentropic steady flow about, 600-601 air, useful numbers for, 604 Bernoulli’s equation, relationship to, 603 critical values at sonic point, 603-604 isentropic pressure and density relations, 602-603 Mach number relations, 602 adiabatic flow, 628-632 adverse pressure gradient, 452, 468, 532 aerodynamic forces on road vehicles, 484^88 aerodynamics, 485 aeronautics, trends in, 663-664 air, useful numbers for, 604 aircraft trailing vortices, 561-562 airfoil theory aircraft trailing vortices, 561-562 Kutta condition, 554-555 potential theory for thick cambered airfoils, 555-557 wings of finite span, 558-561 alternate states, 697 Analytical Theory of Heat (Fourier), 288 Anderson’s formula, 759 angle of attack, 492 angles, 289, 381 angular momentum theorem, 172-178 annular strip of water, 488 Anselmet, Fabien, 346 apparent viscosity, 28 approximate solution for irregular channels, 713-714 arbitrarily moving control volume, 141 arbitrary fixed control volume, 138-140 arbitrary flow pattern, 138 Archimedes’ laws, 86 Archimedes number, 326 aspect ration (AR), 496 available head, 183 axial-flow pump, performance of, 764 axial-flow pump theory, 763-764 axisymmetric potential flow about, 562-563 hydrodynamic mass, concept of, 566-568 point doublet, 564 point source or sink, 564 spherical polar coordinates, 563 uniform stream in x direction, 563 uniform stream plus point doublet, 565-566 uniform stream plus point source, 565 B backwater curves, 721-723 basic equations boundary conditions for, 241-245 nondimensionalization of {See nondimensionalization of basic equations) bathtub vortex, 526 Bernoulli, Daniel, 134, 163 Bernoulli equation, 228 about, 163-164 dynamic pressure, 166 energy grade line (EGL), 167 as energy relation, 165 833 834 Index Bernoulli equation — Cont. hydraulic grade line (HGL), 167 for incompressible flow, 289 jet exit pressure equals atmospheric pressure, 166 relationship to, 603 restrictions on, 165-166 in rotating coordinates, 748 stagnation pressure, 166 static pressure, 166 steady incompressible flow, 165 surface velocity condition for large tank, 169 Bernoulli obstruction theory, 412—414 Bernoulli’s incompressible equation, 9 Bernoulli-type devices, 406 best efficiency point (BEP), 750, 751, 794 Betz number, 787 Bingham plastic, 29 biological drag reduction, 491 blade angle on pump head, effect of, 749-750 Blasius, H., 359 Blasius equation, 459 blockage factor, 398 blunt-body flow, 45 1 body drag at high Mach numbers, 489—491 body force, 61, 231 boundaries, 134 boundary conditions, 35, 39, 244 boundary conditions for basic equations about, 241-243 incompressible flow with constant properties, 244 inviscid flow approximations, 244—245 simplified free surface conditions, 243-244 boundary element method (BEM), 571-572 boundary layer (BE) analysis, 449 boundary layer equations about, 456-457 derivation for two-dimensional flow, 457^59 boundary layer separation on half-body, 532-534 boundary layers with pressure gradient about, 468-470 laminar integral theory, 470-474 Boundary layer theory, 474 bourdon tube, 100 bow shock wave, 490 brake horsepower, 745 Bridgman, P. W., 288 Brinell hardness, 294 Brinkman number, 326 British gravitational (BG) system, 7, 8 broad-nested weirs analysis, 717-718 Buckingham, E., 134, 288 Buckingham Pi Theorem, 288, 294 buoyancy center, 86 neutral, 88 and stability, 85-91 buoyant particles, 402 buoyant rising light spheres, 484 butterfly valve, 383 c cambered airfoil, 492 capacitive sensor, 101 capillary viscometer, 423 cartesian vector form of velocity field, 222 catsup, 29 cavitation, 33 cavitation bubbles, 33, 34 cavitation number, 33, 306, 309 center of buoyancy, 86 center of pressure (CP), 73, 74 centrifugal pump about, 744-745 basic output parameters, 745-746 blade angle on pump head, effect of, 749-750 elementary pump theory, 746-749 centrifugal pumps, 759 centroidal moments of inertia, 76 CGS system, 8 characteristic area, 476 Chezy formula, 689-695. See also uniform law chimney flow, 166 choking, 609 due to friction, 632—634 effects for simple heating, 641 circulation, 530-531 closed blades, 745 closed-duct flows, 683 Coanda effect, 488 coefficient of surface tension, 30 coefficient of viscosity, 16 Colebrook, C. E, 362 collinear forces, 87 combination car and airplane, 498-501 commercial CFD codes, 575-577 commercial pipe sizes, 370 composite-flow transitions, illustrative, 714-716 compressibility, 38 compressibility parameters, 306-307 compressible duct flow with friction about, 626-628 adiabatic flow, 628-632 choking due to friction, 632-634 isothermal flow with friction, 634-635 long pipelines, 634-635 mass flow for given pressure drop, 635-636 minor losses in compressible flow, 634 compressible flow adiabatic and isentropic steady flow, 600-606 Index 835 compressible duct flow with friction, 626-637 converging and diverging nozzles, operation of, 621-626 defined, 593 frictionless duct flow with heat transfer, 637-642 isentropic flow with area changes, 606-613 Mach waves and oblique shock waves, 642-652 normal shock wave, 613-621 Prandtl-Meyer expansion waves, 652-664 speed of sound, 598-600 thermodynamics, review of, 593-597 compressible flow stream function, 251 compressible gas flow correction factor, 418^21 compression stress, 16 compressors, 772-774 computational fluid dynamics (CFD), 397, 456, 746, 767 concentric annulus, 375-378 cone-plate viscometer, 50 conformal mapping, 547 conical diffuser, 396, 398, 400 conservation of mass, 224. See also differential equations of mass conservation about, 134, 144—145 incompressible flow, 146-149 consistent units, 12 contact angle, 31 continuity relation, 224. See conservation of mass continuum, defined, 7 continuum fluid density, limit definition of, 6 contraction, 387 control volume, integral relations for angular momentum theorem, 172-178 basic physical laws of fluid mechanics, 133-137 Bernoulli equation, 163-172 conservation of mass, 144-149 energy equation, 178-188 linear momentum equation, 149-162 Reynolds transport theorem, 137-144 control volume analysis, 158, 598 control volume moving at constant velocity, 140 control volume of constant shape but variable velocity, 140 convective acceleration, 223 converging-diverging nozzle, 623-626 converging nozzles, 621-623 Coriolis acceleration, 161 Coriolis mass flowmeter, 406, 410—41 1 congelations of fluctuating velocities, 353 Couette flow, 262, 268 Couette flow between fixed and moving plate, 261-262 counting numbers, 290 creeping flow, 310, 479—482 creeping motion, 507 critical depth about, 697 critical uniform flow, 700-701 flow under sluice gate, 703-704 frictionless flow over bump, 701-703 nonrectangular channels, 699—700 rectangular channels, 698 water channel compressible flow analogy, 698-699 critical-point measurements, 25 critical point of substance, 6 critical uniform flow, 700-701 Cross, Hardy, 395 cup anemometer, 401 cylindrical polar coordinates, 226 D d’Alembert’s paradox, 541 Darcy, Henry, 348 Darcy friction factor, 348, 358 Darrieus, G. J. M., 784 Dash, Sukanta, 346 deflection measurement, 100 deformable control volumes, 137, 138, 141 density of fluid, 6, 17 dependent variables, 296 depression, 729 derivation for two-dimensional flow, 457-459 design condition, 704 design flow rate, 750 design pressure ratio, 623 diaphragm, 100 differential equation, basic, 708-710 differential equation of angular momentum, 237 differential equation of energy, 238-241 differential equation of linear momentum, 230-235 Euler’s equation, 234 Navier-Stokes equations, 234-235 differential equation of mass conservation, 224-228 cylindrical polar coordinates, 226 incompressible flow, 227-228 steady compressible flow, 227 differential relations for fluid flow about, 221 acceleration field of fluid, 222-223 boundary conditions for basic equations, 241-245 differential equation of angular momentum, 237-238 differential equation of energy, 238-241 differential equation of linear momentum, 230-235 differential equation of mass conservation, 224-228 836 Index differential relations for fluid flow — Cont. frictionless irrotational flows, 255-261 incompressible viscous flow examples, 261-269 stream function, 246-253 vorticity and irrotationality, 253-255 diffuser, 386 diffuser performance, 395^00 diffuser stall, 470 digital differential manometer, 99, 100 dilatant, 29 dimensional analysis, 7, 288 dimensional analysis and similarity about, 285-288 modeling and similarity, 313-325 nondimensionalization of basic equations, 30^313 Pi theorem, 294-304 principle of dimensional homogeneity (PDH), 288-294 dimensional constants, 289 dimensional homogeneity principle, 9 dimensionally consistent units, 1 1 dimensionally inconsistent equations vs. homogeneous equations, 13 dimensional matrix, 327 dimensional variables, 289 dimensionless pai'ameters, 305-306, 307-310 dimensionless pump performance, 753-758 dimensions and units British Gravitational (BG) system, 8 defined, 7 International System (SI), 8 primary dimensions, 7, 8 secondary dimensions, 9 direct numerical simulation (DNS), 3 discharge coefficient, 13, 410, 413 discrepancies in water and air testing, 318-325 disk, 381 displacement, 291 displacement, dimensionless, 291 displacement thickness, 455 doppler shift, 404 doublet, 535-537 doughnut- shaped diffuser, 744 drag, 474 drag coefficient, 309, 460, 466, 476 drag force, 158, 311 drag of immersed bodies, 474—476 drag of surface ships, 488—489 drag reduction, 488 drive magnet, 406 dry adiabatic lapse rate (DALR), 128 ducted-propeller meters, 402 duct flow, types of, 637 duct shape effect, 348 dynamic pressure, 166 dynamic pumps, 742-744 dynamic similaiity, 317-318 E Eckert number, 307, 309 eddy viscosity, 356 Ekman number, 334 elastic deformation, 99 electric output, 99 electric-output sensors, 101 electromagnetic current meter, 401 electromagneticfields, 545 electromagnetic meter, 403 elementary plane flow solutions circulation, 530-531 line irrotational vortex, 525-526 line source or sink at origin, 525 Rankine half-body, 528—529 sink plus vortex at origin, 527 source plus equal sink, 526-527 superposition, 526-527 uniform stream at angle a, 530 uniform stream in x direction, 524—525 uniform stream plus source at origin, 528 elementary pump theory, 746-749 energy equation about, 178-180 friction and shaft work in low-speed flow, 182-183 kinetic energy correction factor, 185-186 one-dimensional energy-flux terms, 180 steady flow energy equation, 182 energy grade line (EGL), 167, 697 energy relation, Bernoulli equation as, 165 energy transfer, 741 engineering equations, peculiar, 293-294 Engineer-in-Training (E-I-T) Examination, 43 enthalpy, 16 entrance losses, 384 entrance region, 344 entropy, 16 equation of continuity, 225 equilibrium of fluid element, 61-62 gage pressure, 62 vacuum pressure, 62 Euler, Leonhard, 163 Eulerian frame of reference, 222 Euler number, 306, 309 Euler’s equation, 234, 244 Euler turbomachine formulas, 747, 776 exit loss, 385 Index 837 experimental external flows aerodynamic forces on road vehicles, 484^88 biological drag reduction, 491 body drag at high Mach numbers, 489^91 buoyant rising light spheres, 484 characteristic area, 476 combination car and airplane, 498-501 creeping flow, 479-482 drag of immersed bodies, 474—476 drag of surface ships, 488—489 drag reduction methods, 488 forces on lifting bodies, 492—498 friction drag, 476—479 Kline-Fogelman airfoil, 498 pressure drag, 476^79 three-dimensional bodies, 482-484 two-dimensional bodies, 479 wing inspired by humpback whale, 498 experimental weir discharge coefficients, 718-719 explicit model, 573 external flows, 258 extract energy (turbines), 741 F falling-body relation, 290, 292 Fanno line, 627 favorable pressure gradient, 452, 468, 532 finite aspect ratio, 558 finite difference method (FDM), 569-571 finite differences, 570 finite element method, 568-569 Finlayson, Bruce, 389 fixed control volume, 137 fixed jet-turning vane, 153 flat diffuser stability map, 396 flat-plate boundaiy layer laminar flow, 459-462 transition to turbulence, 462 turbulent flow, 463-467 flat-walled diffuser, 396, 398, 400 Flettner design, 542 floating body, 87, 88 floating offshore turbines, 789 floats, 402 flow along single streamline, 165 flow analysis techniques, 39 flow around corner of arbitrary angle, 548-549 flow between long concentric cylinders, 266-267 flow between parallel plates, 372 flow between plates, 26-27, 26 flow coefficient, 413-414 flow due to pressure gradient between two fixed plates, 262-263 flow in noncircular ducts flow between parallel plates, 372 flow through concentric annulus, 375-378 hydraulic diameter, 371-372 laminar flow solution, 373 noncircular cross sections, 378-380 turbulent flow solution, 373-375 flow measurement and control by weirs about, 716-717 backwater curves, 721-723 broad-nested weirs analysis, 717-718 experimental weir discharge coefficients, 718-719 sharp-crested weirs analysis, 717 thin-plate weir designs, 719-721 flow normal to flat plate, 549-55 1 flow nozzle, 416 flow past circular cylinder with circulation, 538-540 flow past immersed bodies boundary layer equations, 456^59 boundary layers with pressure gradient, 468—474 experimental external flows, 474-501 flat-plate boundary layer, 459—467 momentum integral estimates, 453-456 Reynolds number and geometry effects, 449-452 flow patterns flow visualization, 41—42 pathline, 39, 40 streakline, 39, 40 streamline, 39 timeline, 39, 40 flow property, 356 flow rate, 366-367 flow through concentric annulus, 375-378 flow under sluice gate, 703-704 flow visualization, 41, 42, 498 fluctuation, 353 fluid, defined, 4 fluid circulation, 530 fluid mechanics basic flow analysis techniques, 39 defined, 3 dimensionless groups in, 309 dimensions and units, 7-14 flow patterns, 39—42 fluid, as continuum, 6-7, 6 fluid, concept, 4-6, 5 Fundamentals of Engineering (FE) Examination, 43 history of, 43 overview, 3-4 thermodynamic properties, 15-22 velocity field, properties of, 15 viscosity and other secondary properties, 23-38, 23f fluid mechanics properties, properties of, 296 838 Index fluid meters local velocity measurements, 400^05 volume flow measurements, 405^21 fluid pressure, 6, 59 force, 293 force coefficient, 286 forces on lifting bodies, 492-498 Fourier’s law of conduction, 238-241 Francis turbines, 775, 776 free-body concept, 138 free overfall, 716 free propeller, 767 free-propeller meter, 401 free stream, 521 free-streamline solution, 550 free vortex, 525 frictional choking, 632 friction and shaft work in low-speed flow, 182-183 friction drag, 476^79 friction factor, 309 frictionless duct flow with heat transfer about, 637-639 choking effects for simple heating, 641 duct flow, types of, 637 Mach number relations, 639-640 normal shock wave, relationship to, 642 frictionless flow, 165 frictionless flow over bump, 701—703 frictionless irrotational flows about, 255-256 generation of rotationality, 258-259 orthogonality of streamlines and potential lines, 257 velocity potential, 256-257 friction loss, 753 friction loss devices, 406 friction velocity, 354 frontal area, 476 Froude, William, 306 Froude number, 286, 309, 489 Froude scaling, 316-317 fully developed flow, 345, 373, 376 fully developed laminar pipe flow, 264-266 fully rough flow, 360 fully turbulent flow, 340 Fundamentals of Engineering (FE) Examination, 43,52 fused quartz, force-balanced bourdon tube, 100, 101 G gage pressure, 62, 75 gas behavior, 99 gases slip flow in, 37-38 state relations for, 1 8-2 1 gas turbines, 774—775 gas viscosity, 28 gate valve, 381 generation of rotationality, 258-259 geometric interpretation of streamlines of flow, 247-251 geometric similarity, 314-315 global properties, 400 globe, 381 gradients, 231 gradual expansion or diffuser, 386-388 gradually varied flow (GVF) approximate solution for iiTegular channels, 713-714 classification of solutions, 710-711 composite-flow transitions, illustrative, 714—716 differential equation, basic, 708-710 numerical solution, 711-713 open-channel flow and, 686 graphical method of superposition, 531 Grashof number, 307, 309 gravity-based pressure measurement instruments, 99 H Haaland, S. E., 363 Hagen, G., 264, 343 Hagen-Poiseuille flow, 265 Hazen-Williams formula, 294 head, 183 head loss, 347—348 chart, 366 nonrecoverable, 418 heat conduction equation, 241 heat flux, 238 heat transfer, 17, 241, 243 heat transfer coefficient, 327 Hele-Shaw apparatus, 545 Helmholtz-Kirchoff theory, 551 Herschel, Clemens, 417 Herschel-type venturi, 416 high-moleculai-weight polymer additive, 488 high-pressure compressors (HPC), 775 high-pressure turbine (HPT), 775 hodograph, 647 Holloway, Gordon, 387 homogeneous equations, 13 homogeneous equations v^. dimensionally inconsistent equations, 13 horizontal-axis wind turbines (HAWT), 784, 789 horseshoe vortices, 577 Index 839 hot-film anemometer, 401 hot-wire anemometer, 403—404 hot wires and hot films, 401 Hunsaker, J., 134 hydraulically smooth walls, 360 hydraulic diameter, 371-372 hydraulic efficiency, 746 hydraulic grade line (HGL), 167, 348, 683 hydraulic jump about, 704 classification, 705-706 theory for, 706-708 hydraulic model, 321 hydrodynamic mass, concept of, 566-568 hydrogen, 319 hydrometer, 120 hydrostatic forces on curved surfaces, 80-81 in layered fluids, 83-85 hydrostatic forces on plane surfaces gage pressure formulas, 75-79 pressure distribution, 72-79 hydrostatic paradox, 112 hydrostatic pressure distribution, 62-69 in gases, 66-67 linear approximation, 68-69 in liquids, 64-65 mercury barometer, 65-66 standard atmosphere, 67 variable gravity effects, 63-64 hydrostatic stress, 4 hypersonic flow, 594, 595 I idealized radial turbine theory, 776 idealized wind turbine theory, 784-789 ideal pump theory, deviations from, 753 images, 551-554 impeller recirculation loss, 753 impulse turbines, 791 incompressible axisymmetric flow, 252 incompressible flow, 146-149, 165, 227-228, 594 incompressible flow with constant properties, 244 incompressible plane flow in polar coordinates, 251 incompressible viscous flow examples Couette flow between fixed and moving plate, 261-262 flow between long concentric cylinders, 266-267 flow due to pressure gradient between two fixed plates, 262-263 fully developed laminar pipe flow, 264—266 instability of rotating inner cylinder flow, 268-269 inertial coordinate system, 149, 173 infinite row of vortices, 534-535 initial conditions, 241 inlet Mach number, 398 inlet Reynolds number, 398 inlets, 179 instability of rotating inner cylinder flow, 268-269 integral mass conseiwation law, 144 integral relations for control volume, see control volume, integral relations for integrated properties, 400 intensive value, 139 internal energy, 16, 17, 18 internal flows, 258 internal vs. external viscous flows, 344-347 International System of Units (SI), 7, 8 inviscid flow analysis, 36 approximations, 244-245 Euler’s equation, 234 inviscid theory, 522 inward pressure force, 151 Ipsen alternate step-by-step method, 301-304 irregular channels, approximate solution for, 713-714 irrotational flow, 247, 255 irrotational function, 525 irrotationality, vorticity and, 253-255 isentropic bulk modulus, 38 isentropic flow, 594 isentropic flow with area changes about, 606-607 choking, 609 local mass flow function, 610-613 perfect-gas area change, 608-609 isentropic pressure and density relations, 602-603 isentropic process, 596-597 isothermal atmosphere, 67 isothermal flow with friction, 634-635 isovelocity contours, 685 J jet exit pressure conditions, 152-159 jet exit pressure equals atmospheric pressure, 166 Joukowski, N., 541 K Karman, T, von, 307 Karman’s analysis of flat plate, 453-455 Karman’s constant, 356 840 Index Karman vortex street, 307 Kelvin oval, 544 kinematic boundary condition, 243 kinematic similarity, 315 kinematic viscosity, 24, 26, 247 kinetic energy, 1 8 kinetic energy correction factor, 185-186 King’s law, 403 Kirchhoff, Gustav, R., 550 Kline, R. R, 398 Kline-Fogelman airfoil, 498 Knudsen number, 37 Korotkoff sounds, 110 Kutta, W. M., 541 Kutta condition, 554-555 Kutta-Joukowski lift theorem, 540-542, 554 L Lagrange, Joseph Louis, 248 Lagrangian coordinates, 222 laminar boundary method, 532 laminar flow, 26, 159, 186, 266, 340, 459—462, 477 laminai' flow element (LFE), 406, 411, 425 laminai flow minor losses, 388-389 laminai flow solution, 373 laminar flow theory, 344 laminar fully developed pipe flow, 349-351 laminar integral theory, 470-474 laminai minor loss coefficients, 389 laminai pipe flow, fully developed, 264-266 laminai shear, 354 Lanchester, Frederick W., 494 Laplace equation, 247, 257, 259, 523 large-eddy breakup devices (LEBU), 488 laser-Doppler anemometer (LDA), 401, 404 laser measurements of doppler anemometry (LDA), 767 laser measurements of particle tracking velocimetry (LPTV), 767 law of thermodynamics, 135 law of wall, 354 layered fluid (LF), 87 lift, 474 lift and drag of rotating cylinders, 542-544 lift coefficient, 309, 493, 498 lift-drag polar plot, 496 Lilienthal, Otto, 494 linear differential equations, 227 linearizer, 403 linear momentum equation about, 149 linear momentum tips, 160 momentum flux correction factor, 159-160 net pressure force on closed control surface, 150-151 noninertial reference frame, 160-162 one-dimensional momentum flux, 150 pressure condition at jet exit, 152-159 line irrotational vortex, 525-526 line source, 525 line source at point Zq, 548 line source or sink at origin, 525 line vortex at point Zq, 548 liquids, state relations for, 22 liquid viscosity, 28 local acceleration, 222 local-gravity vector, 63 local losses, 380 local mass flow function, 610-613 local properties, 400 local velocity measurements about, 400^01 electromagnetic meter, 403 floats or buoyant particles, 402 hot-wire anemometer, 403^04 laser-Doppler anemometer, 404 particle image velocimetry (PIV), 404—405 pilot-static tube, 402^03 rotating sensors, 402 logarithmic overlap law, 354-356 logarithmic overlap layer, 355 long concentric cylinders, flow between, 266-267 loss coefficients, 380, 382, 383 low-pressure compressor (LPC), 775 low-pressure turbine (LPT), 775 luminescent coatings, 99 M Mach, Ernst, 307 Mach cone, 643 Mach number, 38, 228, 258, 286, 306, 307, 309, 489, 594-595 Mach number relations, 602, 615-618, 639-640 Mach waves, 642-645, 643, 653 magnetic counter sensor, 406 Magnus, Gustav, 540 Magnus-Robins force, 540 Manning roughness correlation, 690-692 manometer, 99 manometry applications downward pressure, 69 pressure distribution, 69-72 simple manometer application, 70-72 mass flow, 136-137, 145, 150 mass flow for given pressure drop, 635-636 mass measurement, 404 Index 841 mass unit, 9 matching pumps to system characteristics about, 767-769 compressors, 112-11 A gas turbines, 774-775 multistage pumps, 772 pumps combined in parallel, 771-772 pumps combined in series, 772 material derivative, 223 Maxwell, James Clerk, 37 measured performance curves, 751-752 mechanical efficiency, 746 meniscus, 72 mercury barometer, 65-66 metacenter, 88, 89 metered- volume chamber, 406 methanol barometer. 104 Metric Convention, 7 Meyer, Theodor, 652 microbubbles, 488 Millikan, C. B., 355 minor losses in compressible flow, 634 minor or local losses in pipe systems about, 380-386 gradual expansion or diffuser, 386-388 laminar flow minor losses, 388-389 mixed- and axial-flow pumps about, 760-761 axial-flow pump, performance of, 764 axial-flow pump theory, 763-764 computational fluid dynamics, 767 free propeller, 767 pump performance vs. specific speed, 764-767 specific speed, 762 suction specific speed, 762 modeling and similarity. See also dimensional analysis and similarity about, 313-314 discrepancies in water and air testing, 318-325 dynamic similarity, 317—318 Froude scaling, 316-317 geometric similarity, 314-315 kinematic similarity, 315 Mohr’s circle, 4, 5 molecular dynamics model, 595 momentum flow term, 150 momentum flux correction factor, 159-160 momentum integral estimates displacement thickness, 455 Karman’s analysis of flat plate, 453—455 momentum integral theory, 454 momentum thickness, 453 Moody chart, 362-365 Moody-type pipe friction, 333 moving normal shocks, 618-621 multiphase flows, 5 multiple-fluid manometer, 71 multiple-pipe systems pipe in series, 389-391 pipe networks, 394-395 pipes in parallel, 391—392 three-reservoir pipe junction, 393-394 multistage pumps, 772 N NACA aorfoils, 556, 557 natural convection, 594 Navier, C. L. M. H., 234 Navier-Stokes equations, 234-235 net positive-suction head (NPSH), 752-753 net pressure force on closed control surface, 150-151 neutral buoyancy, 88 Newton, Isaac, 24 newtonian fluids, 24, 234 Newton’s law, 8, 9 in noninertial coordinates, 161 second, 134, 149, 222 Newton’s viscosity law, 238 Nikuradse, J., 360 noncirculai' duct, 371, 372 nondeformable inertial control volume, 173 nondimensionalization of basic equations. See also dimensional analysis and similarity about, 304-305 application, 310-313 compressibility pai'ameters, 306-307 dimensionless parameters, 305-306, 307—310 oscillating flows, 307 noninertial coordinates, 160 noninertial effects, 258 noninertial reference frame, 160-162 nonnewtonian fluids apparent viscosity, 28 Bingham plastic, 29 defined, 28 dilatant, 29 pseudoplastic, 29 rheology, 29 nonrectangular channels, 699-700 nonwetting liquid, 3 1 normal depth estimates, 692-694 normal power, 782 normal shock wave, 642 about, 613-615 Mach number relations, 615-618 moving normal shocks, 618-621 no-slip conditions, 24, 35-36 no-temperature-jump conditions, 35-36 842 Index nozzle geometry, 622, 624 numerical analysis about, 568 boundary element method (BEM), 571-572 commercial CFD codes, 575-577 finite difference method, 569-571 finite element method, 568-569 one-dimensional unsteady flow, 572-573 steady two-dimensional laminar flow, 573-575 viscous flow computer models, 572 nutating disc meter, 406—407 o oblique shock wave, 645-650 one-dimensional approximation, 160 one-dimensional energy-flux terms, 180 one-dimensional flux term approximations, 142 one-dimensional momentum flux, 150 one-dimensional unsteady flow, 572-573 onset point, 360 open blades, 745 open-channel flow about, 683-689 Chezy formula, 689-695 critical depth, 697-704 defined, 683 flow classification by depth variation, 686-687 flow classification by Froude number, 687 flow measurement and control by weirs, 716-723 gradually varied flow, 708-716 hydraulic jump, 704-708 one-dimensional approximations, 684-686 surface wave speed, 687-689 uniform flow, 689 uniform-flow channels, efficient, 695-697 orifice theory, 704 orthogonal flow net, 523 orthogonality of streamlines and potential lines, 257 oscillating flow pressure, 99 oscillating flows, 307 oscillating jump, 706 outer wall layer, 354 outlets, 179 outward normal unit vector, 137 outward unit vector, 151 overlap layer, 354 p Paddlewheel turbines, 741 paint, 29 parallel-disk rheometer, 29 parallel plates, flow between, 372 parameters, 290 Parshall flume, 734 particle image velocimetry (PIV), 42, 401, 404—405 Pascal’s law, 70 pathline, 39, 40 Pelton wheels, 778 perfect gas, 595-596 perfect-gas area change, 608-609 perfect gas law, 18 perfect-liquid law, 22 performance charts, 750 permeability, 329 Perspex, 767 Phillips, Horatio Frederick, 494 physical laws of fluid mechanics, basic about, 133-134 systems versus control volumes, 134-136 volume and mass rate of flow, 136-137 piezoelectric transducer, 101, 103 piezometric pressure, 305 pilot-static tube, 402^03 pipe diameter, 367-370 pipe flow problems, types of commercial pipe sizes, 370 flow rate, 366-367 pipe diameter, 367-370 pipe length, 370-371 pipe in series, 389-391 pipe length, 370-371 pipe networks, 394-395 pipes in parallel, 391-392 pipe velocity, 384 pitching moment, 474 Pi theorem. See also dimensional analysis and similarity about, 294-301 Ipsen alternate step-by-step method, 301-304 Pitot, Henri de., 403 Pitot formula, 403 pitot-static tube, 166, 401, 402 plane flow past closed-body shapes flow past circulai' cylinder with circulation, 538-540 Kelvin oval, 544 Kutta-Joukowski lift theorem, 540-542 lift and drag of rotating cylinders, 542-544 potential flow analogs, 544—546 Rankine oval, 537-538 plane polar coordinates, 524 plane potential flows about, 547-548 flow around comer of arbitrary angle, 548-549 flow normal to flat plate, 549-551 line source at point Zq, 548 Index 843 line vortex at point Zq, 548 uniform stream at angle of attack, 548 planform area, 476 plastic, 29 Pode’s angle, 512 point doublet, 564 point of inflection (PI), 464 point source or sink, 564 Poiseuille flow, 349 Poiseuille parabola, 263 portable barometer, 66 positive-displacement pumps (PDP), 741—742, 743 potential energy, 18 potential flow analogs, 544-546 potential flow and computational fluid dynamics airfoil theoiy, 554-562 axisymmetric potential flow, 562-568 elementary plane flow solutions, 524-531 images, 551-554 numerical analysis, 568-577 plane flow past closed-body shapes, 537-546 plane potential flows, 547-551 review, 521-524 superposition of plane flow solutions, 531—537 potential lines, 256, 525 potential theory, 523 potential theoiy for thick cambered airfoils, 555-557 potentiometer pointer, 545 pound of mass, 9 power coefficient, 787 power law, 28, 48 power specific speed, 776-778, 111 Prandtl, Ludwig, 3, 134, 344, 457, 652 Prandtl-Meyer expansion waves about, 652 aeronautics, trends in, 663-664 Prandtl-Meyer perfect-gas function, 652-656 supersonic airfoils, application to, 656-659 thin-airfoil theory, 659-661 three-dimensional supersonic flow, 662-663 Prandtl-Meyer perfect-gas function, 652-656 Prandtl-Meyer supersonic expansion function, 654, 655 Prandtl number, 307, 309 prefixes for engineering units, 13 pressure, 16 pressure coefficient, 309 pressure condition at jet exit, 152-159 pressure distribution, 622 buoyancy and stability, 85-91 equilibrium of fluid element, 61-62 hydrostatic forces in layered fluids, 83-85 hydrostatic forces on curved surfaces, 80-83 hydrostatic forces on plane surfaces, 72-79 hydrostatic pressure distribution, 62-69 manometry, application to, 69-72 pressure and pressure gradient, 59-60 pressure measurement, 99-103 in rigid-body motion, 91-99 pressure drag, 476—479 pressure force computation, 150 pressure force on fluid, 60-61 pressure gradient, 61, 233, 263 pressure head, 64, 182 pressure measurement, 99-103 pressure-recovery coefficient, 396 pressure sensors, 102 primary dimensions, 7, 8 principal chord line of body, 475 principle of corresponding states, 25 principle of dimensional homogeneity, 9 principle of dimensional homogeneity (PDH) about, 288-289 engineering equations, peculiar, 293-294 scaling (repeating) variables, 293 variables and constants, 289-290 variables and scaling parameters, choice of, 290-293 problem-solving techniques, fluid mechanics, 27 propeller meter, 401, 407 propeller-style pump, 766 propeller turbine, 775 prototype, 287 pseudoplastic fluids, 29 pump performance curves and similarity rules about, 750-751 deviations from ideal pump theory, 753 dimensionless pump performance, 753-758 measured performance curves, 751-752 net positive-suction head, 752-753 similarity rules, 758-759 viscosity effects, 759—760 pump performance vs. specific speed, 764-767 pumps, 741 classification of, 741-744 combined in parallel, 771-772 combined in series, 772 pump surge, 750 pure constants, 289 Q quartz gages, 101 R Rankine, W. J. M., 528 Rankine half-body, 528-529 844 Index Rankine-Hugoniot relations, 614 Rankine oval, 537-538 rapidly varying flow (RVF), 686 rarefied gas, 242 rated power, 782 Rayleigh, Lord, 134, 288, 637 Rayleigh flow, 637 Rayleigh line, 637 Rayleigh number, 309 reaction turbines, 775-776, 791 rectangular channels, 698 rectangular weir, 718 relaminarization region, 342 relative velocity, 141 repeating variable method of dimensional analysis, 301 repeating variables. See scaling parameters Resistance coefficients, 382, 384 revolutions, 289 Reynolds, Osborne, 306, 344 Reynolds number, 41, 266 dimensionless, 25-26, 309 extrapolation, 318 force coefficient and, 286 historical outline of, 343-344 local, 451 low-speed viscous flows with no free surface, 306 regimes, 339-344 transition, 462 vs. pipe friction, 349 Reynolds number and geometry effects, 449—452 Reynolds’ time-averaging concept, 352-354 Reynolds transport theorem about, 137-138 arbitrarily moving control volume, 141 arbitrary fixed control volume, 138-140 control volume moving at constant velocity, 140 control volume of constant shape, 140 deformable control volume, 141 one-dimensional flux term approximations, 142 rheology, 5, 29 rheometers, 29-30 rheopectic fluids, 29 Riabouchinsky, D., 288 Richardson number, 326 Rightmire, B., 134 rigid-body motion pressure distribution in, 91-99 rotation, 94 uniform linear acceleration, 92 road vehicles, aerodynamic forces on, 484—488 Robins, Benjamin, 540 Rockwell hardness, 294 rolling moment, 474 rolling resistance coefficient, 486 Rossby number, 309 rotameter, 406, 410 rotating cylinders, lift and drag of, 542-544 rotating inner cylinder flow, instability of, 268-269 rotating sensors, 402 rotational flow, 259 rotationality, generation of, 258-259 rotodynamic pumps, 742 rotor blades, 763 roughness, 311 roughness ratio, 309 turbulent flow, 286 rough walls effect, 360-362 S salinity, 22 savonius rotor, 401 Saybolt viscosity, 294 scaling laws, 287 scaling parameters, 290 scaling (repeating) variables selection, 293 seawater, salinity of, 22 secondary dimensions, 9 secondary turbulent flow, 379 separation bubble, 494 separation point, 464 shaft work, 179 shape factor, 46 1 sharp-crested weirs analysis, 717 shear strain rate, 255 shear stress, 4, 5, 23, 23, 237 shear-thickening fluid, 29 shear-thinning fluid, 29 shock expansion theory, 656 shock loss, 746, 753 shock polar, 647 side force, 474 silicon sensor, 101 similarity, 287, 313 similarity rules, 758-759 simple manometer application, 70-72 simplified free surface conditions, 243-244 sink plus vortex at origin, 527 skin friction coefficient, 309, 454 skin friction law, 464 slip flow in gases, 37-38 slip velocity, 37, 234 sluice gate, flow under, 703-704 smoke flow visualization, 494 solid, defined, 4, 5 solid surface, 179 sonic barrier, 490 sonic boom, 643 sonic point, critical values at, 603-604 Index 845 sound, speed of, 38 sound speed of materials, 600 source plus equal sink, 526-527 specific energy, 697. See also critical depth specific gravity, 17-18 specific-heat ratio, 306, 309, 595 specific speed, 762 specific speed pumps, 160-161 . See also mixed- and axial-flow pumps specific weight, 17, 64 speed of sound, 38, 51-52, 598-600 spherical polar coordinates, 226, 563 sphygmomanometer, 110 stability about, 88-89 and buoyancy, 85-91 and waterline area, 89-91 stability map of diffuser flow patterns, 398 stagnation enthalpy, 182, 601 stagnation point, 166 stagnation pressure, 166, 402 stall point, 497 stall speed, 497 standard atmosphere, 67 Stanton number, 327 starting vortex, 493 state relations for gases, 18-21 state relations for liquids, 22 static equilibrium of floating body, 87 static pressure, 166 stator blades, 763 steady compressible flow, 227 steady flow, 165 steady flow energy equation, 182 steady-flow-mass-conservation, 145 steady incompressible flow, 165 steady jump, 706 steady plane compressible flow, 25 1 steady two-dimensional laminar flow, 573-575 Stokes, George G., 234, 479 Stokes number, 326 Stokes-Oseen formula, 45 stopping vortex, 493 stratified flows, 306 streakline, 39, 40 stream function, 228 about, 246-247 geometric interpretation of streamlines of flow, 247-251 incompressible axisymmetric flow, 252 incompressible plane flow in polar coordinates, 251 steady plane compressible flow, 25 1 stream function, review of, 523 streamline for stagnation flow, 550 streamline in two-dimensional flow, 247 streamline of flow, 164 streamlines, 39, 40, 52, 525 streamline surface, 179 streamlining, 488 streamtube, 40 streamtube, defined, 40 strength of doublet, 536 strength of vortex, 526 stresses, 231 strong jump, 706 strong shock, 647 Strouhal number, 307, 309, 408 Strutt, John William, 637 submerged exits, 384 subsonic diffuser, 624 subsonic flow, 594 subsonic inlet, 632 substantial derivative, 223 suction specific speed, 762 sudden contraction (SC), 384 sudden expansion (SE), 384, 385 supercritical nappe, 716 superposition, 526-527 superposition of plane flow solutions boundary layer separation on half-body, 532-534 doublet, 535-537 graphical method of superposition, 53 1 infinite row of vortices, 534-535 vortex sheet, 535 supersonic airfoils, 656-659, 657 supersonic flow, 594 three-dimensional, 662-663 supersonic inlet, 632 supersonic jet, 152 surface force, 61, 231 surface of machine, 179 surface roughness, 360 surface tension, 30, 30, 31, 32, 50-51 surface velocity condition for large tank, 169 sun'oundings, 134 Sutherland law, 28 swallow float, 88 swing-check valve, 381 systems vs. control volumes, 134-136 T Taylor number, 268-269 temperature, 17 temperature-jump relation for gases, 242 temperature on viscosity, 28 temperature ratio, 309 tensors, 231 terminal fall velocity, 48, 50 846 Index tethered shrouded-balloon turbines, 789 Theory of Sound (Rayleigh), 288 thermal conductivity, 16, 238-241 thermodynamic properties of fluid, 15-22 density, 17 potential and kinetic energy, 1 8 pressure, 16 specific gravity, 17-18 specific weight, 17 state relations for gases, 18-21 state relations for liquids, 22 temperature, 17 transport properties, 16 thermodynamics about, 593-594 isentropic process, 596-597 Mach number, 594—595 perfect gas, 595-596 specific-heat ratio, 595 thin-airfoil theory, 659-661 thin-plate orifice, 415-416 thin-plate weir designs, 719-721 thixotropic fluids, 29 three-dimensional bodies, 482-484 three-dimensional supersonic flow, 662—663 three-reservoir pipe junction, 393-394 thrust, 766 Thwaites, B., 472 Thwaites’s method, 532 tidal bore, 731 timeline, 39, 40 total head, 183 Trans-Alaska Pipeline System (TAPS), 488 Transition® (car-plane), 499 transitional roughness, 360 transition to turbulence, 339, 462 transonic flow, 594 transport properties, 16 trapezoid angle, 696-697 troposphere, 67 turbine meter, 401, 406, 407^08 turbines. See also turbomachinery gas, 774-775 idealized radial turbine theory, 776 idealized wind turbine theory, 784-789 impulse, 778-783 power specific speed, 776—778 reaction turbines, 775-776 wind, 783-784 wind turbine technology, developments in, 789 turbomachinery. See also turbines centrifugal pump, 744-750 classification of pumps, 741-744 design, 768 matching pumps to system characteristics, 767-775 mixed- and axial-flow pumps, 760-767 pump performance curves and similarity rules, 750-760 turbomachines, 174 turbulence, 3, 340 intensity of, 353 transition to, 462 turbulence modeling about, 351-352 logai'ithmic overlap law, 354-356 modeling concepts, advanced, 356-357 Reynolds’ time-averaging concept, 352-354 turbulence models, 575 turbulent flow, 26, 159, 186, 352, 463^67 turbulent flow solution, 373-375 turbulent pipe flow about, 358-359 Moody chart, 362-365 rough walls, effect of, 360-362 turbulent separation, 478 turbulent stresses, 353 two-dimensional bodies, 479 u ultrasonic flowmeter, 406, 409—410 undular jump, 705 uniform-flow channels, efficient about, 695-696 trapezoid angle, 696-697 uniform law about, 689-690 manning roughness correlation, 690-692 normal depth estimates, 692-694 in partly full circular pipe, 694-695 uniform linear acceleration, 92 uniform stream, 525 at angle a, 530 at angle of attack, 548 plus point doublet, 565-566 plus point source, 565 plus source at origin, 528 in x direction, 524-525, 563 units, see also dimensions and units defined, 7 unsteady frictionless flow along streamline, 164 V vacuum pressure, 62 valve flow coefficient, 13 valve geometries, 381 Index 847 vaneless diffuser, 745 vapor pressure, 33-34 vainables, 290 vaiiables and constants, 289-290 variables and scaling parameters selection, 290-293 Vaschy, A., 288 vee-groove microriblets, 488 velocity drag force v.?., 293 falling displacement and, 292 mixing-length eddy, 356 pipe, 384 potential, 256-257 potential concepts, review of, 522-523 slip, 37 terminal fall, 48, 50 upstream, 729 volume-average, 146 zero, 169 velocity-defect law, 355 velocity field, properties of acceleration field, 15 velocity field, 15 velocity head, 182 velocity-of-approach factor, 413 vena contracta, 386 Venturi, Giovanni, 417 Venturi flume, 733 Venturi meter, 417^18 vertical-axis wind turbines (VAWT), 784, 789 very weak shock waves, 650-652 virtual mass. See hydrodynamic mass viscometer, 48 viscosity, 3, 23-25, 47-50, 293 viscosity and other secondary properties about, 23 flow between plates, 26-27 nonnewtonian fluids, 28-29 no-slip conditions, 35-36 no-temperature jump conditions, 35-36 problem-solving techniques, 27 Reynolds number, 25-26 rheometers, 29-30 slip flow in gases, 37-38 speed of sound, 38 surface tension, 30-32 temperature on, 28 vapor pressure, 33-34 variation with temperature, 28 viscosity coefficient, 234 viscosity effects on pump performance curves and similarity rules, 759-760 viscous-dissipation function, 240 viscous flow computer models, 572 viscous flow in ducts about, 339 experimental duct flows, 395^00 flow in noncircular ducts, 371-380 fluid meters, 400^21 head loss, 347-348 internal vs. external viscous flows, 344—347 laminar fully developed pipe flow, 349-351 minor or local losses in pipe systems, 380-389 multiple-pipe systems, 389-395 pipe flow problems, types of, 366-371 Reynolds number regimes, 339-344 turbulence modeling, 351-357 turbulent pipe flow, 358-365 viscous heat-conducting fluid flow analysis, 242 viscous-inviscid patching concept, 451 viscous stress tensor, 233 volume-average velocity, 146 volume flow, 136-137, 146, 251 volume flow measurements, 405-406, 410^1 1 about, 405^06 Bernoulli obstruction theory, 412^14 compressible gas flow correction factor, 418^21 Coriolis mass flowmeter, 410-411 flow nozzle, 416 laminar flow element, 411 nutating disc meter, 406—407 rotameter, 410 thin-plate orifice, 415—416 turbine meter, 407-408 ultrasonic flowmeters, 409^10 Venturi meter, 417^18 vortex flowmeters, 408—409 volume measurement, 405 volumetric efficiency, 746 vortex flowmeters, 330, 408^09 vortex meter, 406 vortex shedding, 308, 330 vortex sheet, 535 vorticity, 255 vorticity and irrotationality, 253-255 w wall layer, 354 wall oscillation, 488 wall temperature ratio, 307 water and air testing, discrepancies in, 318-325 water channel compressible flow analogy, 698-699 848 Index water horsepower, 745 waterline area, 89 wave-making drag, 489 wave trough, 489 weak shock, 647 Weber, Moritz, 306 Weber number, 306, 309, 320 weir, 716 Weisbach, Julius, 348 wetted area, 476 wind turbines, 783-784 wind turbine technology, developments in, 789 wing inspired by humpback whale, 498 wings of finite span, 558-561 Y yaw, 474 z zero-lift angle of NACA airfoils, 556 zero pressure gradient, 464 zone of action, 643 zone of silence, 643 EQUATION SHEET Ideal-gas law; p = pRT, = 287 J/kg • K Surface tension: Ap = Y{Ri ^ -1- R2 ') Hydrostatics, constant density: Pi - Pi = -lizi - Zi), J = pg Hydrostatic panel force: F = 'yhccA, }'cp = --^xtSin O/ihcoA), Xcp = -4ySin OlihcoA) Buoyant force; = 7£iui(j(displaced volume) CV mass: dldt{J^j)dv) -l-2(pAy)^^j -2(pAy)i„ = o CV momentum: d/dt{ J^^pYdv) + 2[(ME)V]oa,-2[(pAy)V]i„= SF CV angular momentum: dldt{J p(ro X\)dv) CV + 2pAy(roXV)ou,-2pAy(roXV)in=SMo Steady flow energy: (p/'y-\-aV^/2g-\-z)m ^ (pl'y CrV I2g 4" Z)out ^friction ^pump ^turbine Acceleration: dY/dt = dY/dt + u(dY/dx) + v{dY/dy) +w(5V/az) Incompressible continuity: V • V = 0 Navier-Stokes: p{dYldt) =pg—Vp+ pV^V Incompressible stream function ip{x,y)'. U = dtp/ dy, v = —dip! dx Velocity potential (p{x,y,z)'. u = d(p/dx', V = d(p/dy, w = d‘i'" Prandtl-Meyer expansion: K = (k + !)/(/: — 1), u; = ^^'itan“'[(Ma^ - \)IKt'^ - tan“'(Mai - 1)"^ Uniform flow, Manning’s n, SI units: yo(m/s) = {l.O/n)[R,(m)f'X'" Gradually varied channel flow: dy/dx = {So - S)/{1 - Fr^), Fr = V/V.^t Euler turbine formula: Power = pQ{u2V,2 — UiV,i), u = rui Image 11
1854	https://tobybartels.name/MATH-2080/2020FA/params/	Standard parametrizations The simplest curve to parametrize is a straight line. A line through the origin along a vector ⟨a, b⟩ can be parametrized with x = at, y = bt. If you want a ray (half-line) starting at the origin and travelling in the direction of this vector, then use the same formulas for x and y but add the restriction t ≥ 0 for a closed ray (including the endpoint at the origin) or t > 0 for an open ray (not including that endpoint). For a line segment running along the length of that vector, use the restriction 0 ≤ t ≤ 1 for a closed line segment (including both endpoints) or 0 < t < 1 for an open line segment (including neither endpoint). For both rays and line segments, the closed version is the usual standard, although there are times when the open version is needed instead. If the line (or ray or line segment) doesn't go through the origin, then you'll need some point (x1, y1) that it does go through. Then you can use x = x1 + at, y = y1 + bt. Again, without any restriction on t, this is a line; but you can restrict t as above to get a ray or a line segment. Or if you have two points on the line, then you can subtract them to get the relevant vector. Then the parametrization becomes x = x1 + (x2 − x1)t, y = y1 + (y2 − y1)t. All of this works in any number of dimensions; the line through P1 along the vector v has the parametrization P = P1 + tv, and the line through P1 and P2 is P = P1 + t(P2 − P1). The same restrictions on t as before will turn these into rays or line segments. Going back to 2 dimensions, the unit circle (whose radius is 1 and whose centre is at the origin) is usually parametrized like this: x = cos t, y = sin t. If there are no restrictions on t, then you are effectively going around and around the circle forever, counterclockwise (in a counterclockwise coordinate system). If you want the parametrization to be one-to-one, so that every point on the circle is covered exactly once, then you need a restriction on t; the usual one is 0 ≤ t < 2π. It's even more common to use 0 ≤ t ≤ 2π; this is almost one-to-one (since only the point (1, 0) is covered twice, once when t = 0 and once when t = 2π), and it has a compact domain (which is helpful for some things). So this restriction is the standard one for a circle. If the radius of the circle is r, then the parametrization becomes x = r cos t, y = r sin t. If the circle is centred at (h, k) instead of at the origin, then the parametrization becomes x = h + r cos t, y = k + r sin t. You use the same restrictions as before to make the parametrization one-to-one or almost one-to-one. Another useful parametrization is the graph of a function f. For this, you can use x itself as the parameter: x = t, y = f(t). Since you can always call the parameter something else instead of t, you can even call it x: x = x, y = f(x). If you only want the graph of the function restricted to an interval [a, b], then place this restriction on the parameter: a ≤ t ≤ b (or a ≤ x ≤ b if you are calling the parameter x instead of t). This works more generally any time you have an equation that you can solve for y; if you get a unique solution, then this equation defines a function, and the equation y = f(x) in the parametrization above is the equation that you get when you solve for y. If you solve for x instead of for y, then you can say that x is some function g of y. This isn't the graph of that function exactly, since the variables come in the wrong order, but you can still parametrize the curve using y as the parameter: x = g(t), y = t. Again, you can put a restriction on t if you only want certain values of the independent variable, which is now y. Go back to the course homepage. This web page was written by Toby Bartels, last edited on 2020 November 19. Toby reserves no legal rights to it. The permanent URI of this web page is
1855	https://physicspages.com/pdf/Electrodynamics/Gauss's%20law%20in%20electrostatics.pdf	GAUSS’S LAW IN ELECTROSTATICS Link to: physicspages home page. To leave a comment or report an error, please use the auxiliary blog. Post date: 28 Dec 2020. Gauss’s law in electrostatics is a relation between the charge contained by a closed surface and the electric ﬁeld that crosses that surface. The easiest way to see how it works is to begin with a point charge at the origin and a spherical surface centred at the origin. By symmetry the electric ﬁeld due to the point charge is E = 1 4πϵ0 q r2 ˆ r (1) The ﬂux of this ﬁeld through the sphere is deﬁned as the surface integral of the component of the ﬁeld that is normal to the surface. That is the ﬂux Φ is deﬁned as Φ = Z E·da (2) where the integral extends over the surface, and da is a differential vec-tor whose magnitude is a differential area element and whose direction is normal to the surface at each point. In the case of a sphere, it is not surprisingly easiest to use spherical coor-dinates, and in that case da = r2 sinθ dθ dφ ˆ r (3) That is, the area element points radially outwards at each point on the sphere. Combining these results, we see that for a point charge 1 GAUSS’S LAW IN ELECTROSTATICS 2 Φ = Z E·da (4) = Z 2π 0 Z π 0 1 4πϵ0 q r2 ˆ r · r2 sinθ dθ dφ ˆ r (5) = q 4πϵ0 Z 2π 0 Z π 0 sinθ dθ dφ (6) = q ϵ0 (7) That is, the ﬂux due to a point charge depends only on the magnitude of the charge and not on the radius of the sphere that contains it. This makes intu-itive sense, since if we imagine a point charge ’emitting’ the electric ﬁeld, then as long as we provide a surface that wraps up the charge completely, the ﬂux through that surface will be the same regardless of the size of the surface. In fact, it shouldn’t depend on the shape of the surface either, so long as that surface completely encloses the charge. To see how this works, suppose we have a charge q enclosed by a closed surface of some arbitrary shape (my diagramming skills aren’t up to producing a graphic of this, so try to use your imagination). Draw a vector E E E representing the electric ﬁeld from the charge to some element da a a on the surface of the shape. This vector makes an angle θ with the unit normal to da a a, so the component of E E E normal to the surface at this point is E E E · da a a = E da cosθ. If the distance from the charge to this point on the surface is r, then the electric ﬁeld has a strength of q/4πϵ0r2 at the element da a a, so that the normal component of the ﬁeld at this point is E⊥= E E E ·da a a = qcosθ 4πϵ0r2 (8) The solid angle subtended by the area element da a a at the location of the charge is the projection of da a a onto a unit sphere centred at the charge. That is, if we take r = 1, then the element of solid angle is dΩ= ˆ n n n·da a a = da cosθ (9) If r ̸= 1, then da a a is scaled by 1/r2 to give the solid angle, so in this case we have dΩ= da r2 cosθ (10) Thus from 8 we have GAUSS’S LAW IN ELECTROSTATICS 3 E E E ·da a a = q 4πϵ0 dΩ (11) Thus we see that E E E · da a a, when expressed in terms of the solid angle sub-tended by the area element da a a, does not depend on r. Thus we can integrate over all solid angle which gives, since R surface dΩ= 4π: Z surface E E E ·da a a = Z surface q 4πϵ0 dΩ= q ϵ0 (12) If the charge is outside the closed surface, then the integral R surface E E E ·da a a will give some value Ω+ for the solid angle subtended by the surface at the charge, for the portion of the surface where E E E · da a a > 0, that is, for the portion of the surface where the ﬁeld E E E has a component pointing towards the outside of the surface. However, since the surface is closed, there will be a portion of the surface where E E E ·da a a < 0, that is, where the ﬁeld E E E has a component opposite to the surface normal. This portion of the surface subtends a solid angle that is exactly equal to Ω+, but the contribution to R surface E E E·da a a has the opposite sign, so the integral R outward E E E·da a a exactly can-cels R inward E E E ·da a a, meaning that the integral R surface E E E ·da a a = 0 if the charge lies outside the surface. Notice that this result depends on the fact that the ﬁeld is an inverse-square force, since it is only in that case that the factor of 1 r2 in 8 for the ﬁeld matches the 1 r2 in the expression 10 for the solid angle. From here, we can generalize the idea to a collection of point charges using the principle of superposition, and get, for a collection of n point charges, all enclosed by the surface: Z E·da = 1 ϵ0 n ∑ i=1 qi (13) For a continuous charge distribution, where the charge density is ρ(r), we get Z surface E·da = 1 ϵ0 Z volume ρ(r)d3r (14) where it is important to note that the integral on the left is over the enclosing surface, while that on the right is over the volume enclosed by that surface. This is the integral form of Gauss’s law for electrostatics. Using the divergence theorem, we can equate the charge density with the divergence of the electric ﬁeld: GAUSS’S LAW IN ELECTROSTATICS 4 ∇·E = ρ ϵ0 (15) This is the differential form of Gauss’s law. Both these forms are very pow-erful in solving various types of problems since they allow electric ﬁelds to be calculated, often without requiring complicated integrals. REFERENCES (1) Grifﬁths, David J. (2007), Introduction to Electrodynamics, 3rd Edi-tion; Pearson Education, Chapter 2. (2) Jackson, John David (1999), Classical Electrodynamics, 3rd Edi-tion; Wiley, Chapter 1. PINGBACKS Pingback: Gauss’s law in electrostatics - examples Pingback: Conductors Pingback: Electrostatic boundary conditions Pingback: Laplace’s equation - average values of solutions Pingback: Laplace and Poisson equations - uniqueness of solutions Pingback: Green’s reciprocity theorem Pingback: Divergence of magnetic ﬁeld - magnetic monopoles Pingback: Maxwell’s equations Pingback: Potential of two charged wires Pingback: Electromagnetic ﬁeld tensor - justiﬁcation Pingback: Energy (not mass) is the source of gravity
1856	https://countingwell.com/playing_with_numbers.html	Introduction Do you enjoy solving number puzzles? Exploring numbers in detail helps us to understand their amazing properties. Did you know there are two numbers who are best friends? Watch the video to find out the relationship among numbers. The video helped us to know that numbers have interesting relationships. In this chapter, we will explore more about them. Concepts The chapter ‘Playing With Numbers’ covers the following concepts: Factors and Multiples Special Properties of Numbers Divisibility Rules Relationship between Numbers Prime Factorisation LCM and HCF Factors and Multiples A factor of a number is an exact divisor of that number. Eg: Factors of 6 are 1, 2, 3, and 6. Some properties of factors of a number are: 1 is a factor of every number Every number is a factor of itself Every factor is less than or equal to the given number The number of factors of a given number is finite Did you know? A number for which sum of all its factors is equal to twice the number is called a perfect number. 6 is the first perfect number. The first 5 perfect numbers are: 6 28 496 8,128 3,35,50,336 Now, a multiple of a number is the product of that number and another number. The numbers that appear in the times/multiplication table of a number are its multiples. For example, the multiples of 3 are 3, 6, 9, 12, 15,… etc. Remember, any number is a multiple of each of its factors, and any number is a multiple of itself since any number can be a product of 1 and itself. Let us conclude the properties of multiples of a number: Every number is a multiple of itself Every multiple is greater than or equal to the given number The number of multiples of a given number is infinite Do you know how factors and multiples are related? Any number is a multiple of each of its factors. For example, 3 × 5 = 15. 3 and 5 are the factors of 15 15 is a multiple of both 3 and 5 Special Properties of Numbers Let us learn some interesting terms about numbers: The numbers whose only factors are 1 and the number itself are called prime numbers. Numbers having more than two factors are called composite numbers. The numbers 0 and 1 are neither a prime number nor a composite number. Some pairs of prime numbers are formed using the same digits. Let us learn about the interesting properties of prime numbers: Two prime numbers whose difference is 2 are called twin primes. Let us find out how many twin primes exist from 1 to 100. Did you know, numbers can be expressed as a sum of two prime numbers? Let us look at some examples. When playing with numbers, the first thing you will be taught is to differentiate between even and odd numbers. The numbers which are multiples of 2 have a special name. They are called even numbers. E.g. 2, 4, 6, 8, … The numbers which are not multiples of 2 are called odd numbers. E.g. 1, 3, 5, 7, 9.. Is 9876433648 an even number? You need not divide the given number by 2 and check if it is a multiple of 2. Just look at the digit which is at the ones place of the number. If the digit at the ones place is 0, 2, 4, 6 or 8, then the number is even. If the digit at the ones place is 1, 3, 5, 7, or 9, then the number is odd. Remember, 0 is an even number, 2 is the only number which is both even and prime, and all other prime numbers are odd numbers. Let us list the properties of addition of even and odd numbers. For example, Sum of two odd numbers 3 and 5 is 8, which is an even number. Sum of two even numbers 4 and 6 is 10, which is an even number. Sum of an odd number 3 and an even number 6 is 9, which is an odd number. Let us list the properties of multiplication of even and odd numbers. For example, Product of two odd numbers 3 and 5 is 15, which is an ODD number. Product of two even numbers 2 and 4 is 8, which is an EVEN number. Product of an odd number 3 and an even number 4 is 12, which is an EVEN number. Divisibility Rules Divisibility refers to a number's quality of being evenly divided by another number, without a remainder left over. In other words, a number is said to be divisible by another number if the remainder is 0. Listed below are the divisibility rules for certain numbers: Divisibility Rule For 2: Small numbers can be divided by 2 and checked if they are divisible or not. But consider a very large number like 98,87,664. Is this number divisible by 2? EVEN numbers are divisible by 2 and even numbers have the digits 2, 4, 6, 8, or 0 in the one's place, i.e., the numbers are divisible by 2 if the digit at the ones place is 2, 4, 6, 8 or 0. Divisibility Rule For 3: Add the digits of the number. Check if the sum is divisible by 3. If the sum is divisible by 3, then the number is also divisible by 3. If the sum is not divisible by 3, then the number is also not divisible by 3. Divisibility Rule For 4: If the last two digits (i.e., ones and tens) in the number are divisible by 4 then the number is divisible by 4. Divisibility Rule For 5: If the digit at the ones place is 0 or 5 then the number is divisible by 5. Divisibility Rule For 6: If the number is divisible by both 2 and 3 then it is divisible by 6. Divisibility Rule For 8: If the last three digits in the number are divisible by 8 then the number is divisible by 8. Divisibility Rule For 9: Add the digits of the number Check if the sum is divisible by 9 If the sum of digits is divisible by 9, then the number is also divisible by 9. If the sum of digits is not divisible by 9, then the number is also not divisible by 9. Divisibility Rule For 10: If the digit at the ones place is 0 then the number is divisible by 10. Divisibility Rule For 11: Find sum of the digits at odd places (from the right) Find sum of the digits at even places (from the right) Find their difference If the difference is either 0 or divisible by 11, then the number is divisible by 11. We know the rules for checking the divisibility of given numbers by 2, 3, 4, 5, 6, 8, 9, 10 and 11. Sometimes, the given numbers are divisible by more than one number. For example, all even numbers are divisible by 2. Also, if in the even number, the last two digits are divisible by 4, then it is also divisible by 4. Let us take the number 2,34,126 Here, the given number is divisible by both 3 and 9. The divisibility rules are concisely given in the chart below: Now let us learn how to find the missing digit to make a number divisible by a given number. Follow the steps below: Take unknown digit as 'x' Identify the number which is divisible by the given number Apply the divisibility rule Find the value of 'x' Consider this example: 35_64 is divisible by 3. How can we find the missing digit? Given that, the number is divisible by 3. So, we must apply the divisibility rule for 3. As we know, if the sum of digits of the given number is divisible by 3 then the number is divisible by 3. Sum of the digits = 3 + 5 + x + 6 + 4 = 18 + x 18 + x is divisible by 3. Multiples of 3 are 3, 6, …., 18, 21, 24, 27, 30….. If 18 + x = 18, then x = 0. The number is 35,064 If 18 + x = 21, then x = 3. The number is 35,364 If 18 + x = 24, then x = 6. The number is 35,664 If 18 + x = 27, then x = 9. The number is 35,964 If 18 + x = 30, then x = 12, which is greater than 10 and hence, cannot be a digit of the number. The possible values x can take is 0, 3, 6 and 9. What is the smallest digit to replace the blank to make the number divisible by 3? It is 0. By replacing 0 as the value of x, the number becomes 35,064. Listed below are the properties of divisibility of numbers: Remember, if a number is divisible by another number, then it is divisible by each of the factors of that number. If a number is divisible by two numbers, then it is divisible by their product also. If two given numbers are divisible by a number, then their sum or difference is also divisible by that number. Relationship Between Numbers A factor of a number is an exact divisor of that number. When we find the factors of two or more numbers, and then find that some factors are the same ("common"), they are the "common factors". 1 is always a common factor of any number (two or more) because 1 is the factor of every number. Let us see how to find the common factors of 5, 15, and 25. For 5, 15, and 25 the factors 1 and 5 are common. Did you know? Two numbers having only 1 as a common factor are called co-prime numbers. A multiple of a number is the product of that number and another number. When we find the multiples of two or more numbers, and then find that some multiples are the same ("common"), then they are the "common multiples". Let us see how to find the common multiples of 2, 4, and 8. Hence, the first three common multiples of 2, 4, and 8 are 8, 16, 24. Prime Factorisation Do you know the prime factorisation method? Let us take an example of 12, which can be written as the product of 3 prime numbers: We can start with different pairs (2, 6) and (3, 4), but we will arrive at the same set of prime factors in the end, which is 2 × 2 × 3. This method of finding the prime factors of a number is called factor tree method. The numbers at the end of the factor tree are all the prime factors of the original number. Representing a number as a product of its prime factors is called prime factorisation. Every number has a unique set of prime factors. No two numbers can have the same prime factorisation. Now let us consider an incomplete factor tree. Start from the bottom of the tree: Find the next missing number: This is the complete factor tree. We saw how to find the prime factorisation of a number by finding the pairs of numbers which when multiplied gives the number. But for larger numbers, it may not be easy to find such pairs of numbers. To prime factorise such numbers easily, we use division method. The division method of prime factorisation involves the following steps: Start dividing the number by the least prime factor Continue the division until we get 1 as the result Collect all prime numbers and write them as a product Consider prime factorising 48 using the division method: Prime factorisation of 48 = 2 × 2 × 2 × 2 × 3 LCM and HCF The Highest Common Factor (HCF) of two or more given numbers is the highest (or greatest) of their common factors. It is also known as Greatest Common Divisor (GCD). Remember these points: HCF of co-prime numbers is 1 HCF of two consecutive numbers is 1 HCF of two consecutive even numbers is 2 HCF of two consecutive odd numbers is 1 HCF of two or more numbers cannot be greater than any of the numbers Let us learn how to find the HCF of a number using prime factorisation. Prime factorisation of 18, 54 and 81 is as follows: Prime factorise each number Identify common prime factors The product of common prime factors gives the HCF HCF = 3 × 3 = 9 Hence, HCF of 18, 54 and 81 is 9. We use HCF to solve different types of problems. Based on the type of the problem, the strategy to solve it will differ. To find the greatest number that divides exactly Strategy Find the HCF of the numbers. To find the greatest number that leaves the same remainder Strategy 1. Subtract the remainder from the numbers 2. Find the HCF of new numbers To find the greatest number that leaves a different remainder Strategy 1. Subtract the respective remainder from each number 2. Find the HCF of new numbers The Lowest Common Multiple (LCM) of two or more given numbers is the lowest (or smallest or least) of their common multiples. Remember these points: LCM of two or more numbers cannot be less than any of the numbers LCM of co-prime numbers is the product of the numbers When we have two numbers, a number, and its factor, then their LCM will be the number itself You know how to find the LCM of two numbers by listing their multiples and identifying the least common multiple. This may be a time-consuming process for large numbers. In such situations, we use prime factorisation method. Let us find the LCM of 18, 24 and 36. Thus, LCM of 18, 24 and 36 = 2 × 2 × 2 × 3 × 3 = 72 LCM can also be found using the division method. Consider the numbers 18, 25 and 30. LCM is the product of all these prime numbers. LCM of 18, 25 and 30 = 2 × 3 × 3 × 5 × 5 = 450 We use LCM to solve different types of problems. To find the smallest number which is exactly divisible by the given numbers To find the smallest number, upon decreasing by a quantity, is exactly divisible by the given numbers To find the smallest number, upon increasing by a quantity, is exactly divisible by the given numbers Now that we know HCF and LCM, let us look at the relationship between them. Consider the numbers 48 and 60. LCM of two numbers = product of two numbers / HCF of two numbers HCF of two numbers = product of two numbers / LCM of two numbers Common Errors The following are the topics in which students make common mistakes when dealing with numbers: 1. Do not omit 1 and the given number in the list of factors 2. Confusion between odd and prime numbers 3. Prime factor and factor 4. HCF is never ‘0’ 5. HCF is the product of common factors 6. LCM by prime factorisation method Do Not Omit 1 And The Given Number In The List Of Factors What are the factors of 6? Are they 2 and 3? No! 1 and the number itself are also factors of the given number. So, the factors of 6 are not 2 and 3 alone. The factors of 6 are 1, 2, 3 and 6. Confusion Between Odd And Prime Numbers Which of the following are prime numbers? 2, 3, 4, 5, 7, 9, 13, 16. The prime numbers are 2, 3, 5, 7 and 13. Many times, the terms ‘prime’ and ‘odd’ numbers may cause confusion. We know that 2 is the only even prime number and all other prime numbers are odd. That does not mean that all the odd numbers are prime numbers. 9 is not a prime number since it has more than 2 factors, 1, 3 and 9. For better understanding let us categorise the numbers as follows. Prime Factor And Factor 20 is divisible by 4. Can you say that 4 is a prime factor of 20? No! 4 is a factor, but it is not a prime number. The prime factors of 20 are 2 and 5. 20 = 2 × 2 × 5 Remember! Prime factor is the factor of a number that is itself, a prime number. HCF Is Never ‘0’ 4 = 2 × 2 9 = 3 × 3 What is the HCF of 4 and 9? 4 and 9 do not have any common factors. Can you say that the HCF is ‘0’? No! ‘0’ cannot be a factor of any number, hence, it cannot be the HCF of any two numbers. Also, you know that 1 is a factor of every number. We can write 4 and 9 as: 4 = 1 × 2 × 2 9 = 1 × 3 × 3 Hence, 1 is the common factor of 4 and 9. HCF (4, 9) = 1. HCF Is The Product Of Common Factors Care should be taken while finding the HCF. Consider the numbers 4 and 12. 4 = 2 × 2 12 = 2 × 2 × 3 What is HCF of 4 and 12? Is it 2? No! HCF is not the common factor, but it is the product of all the common factors. We can see that 2 is repeated twice. Hence, HCF of 4 and 12 = 2 × 2 = 4. LCM By Prime Factorisation Method Care should be taken while finding LCM by the prime factorisation method. Consider the numbers 70 and 84. 70 = 2 × 5 × 7 84 = 2 × 2 × 3 × 7 What is the LCM of 70 and 84? The LCM of two numbers is the product of the prime factors that occurs the maximum number of times in any of the numbers. 2 occurs a maximum number of 2 times in the prime factorisation of 84. Hence, LCM (70, 84) = 2 × 2 × 3 × 5 × 7 = 420. Conclusion Now you know about numbers, their properties, divisibility, and the relationship between numbers. Can you solve the puzzle below? Replace the question marks with the correct answer. #### Jyothi Author A math enthusiast and an educationist, Jyothi has a strong academic background and has demonstrated an impressive history of working in the education industry. Jyothi is skilled in developing and reviewing lesson plans, learning content, assessments, and video story boards and scripts. 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1857	https://www.reddit.com/r/chemhelp/comments/1i61v67/help_with_this_simple_lewis_diagram_problem_that/	Help with this simple Lewis diagram problem that I'm not understanding, please : r/chemhelp Skip to main contentHelp with this simple Lewis diagram problem that I'm not understanding, please : r/chemhelp Open menu Open navigationGo to Reddit Home r/chemhelp A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to chemhelp r/chemhelp r/chemhelp r/chemhelp is place where you can seek help with whatever chemistry problem you may encounter. Whether you are a high school student trying to understand the fundamentals, a college student struggling through upper division courses, or a professional seeking pointers on a real problem you are trying to solve, you have a place here. 99K Members Online •8 mo. ago AstrophysicsStudent Help with this simple Lewis diagram problem that I'm not understanding, please General/High School This is the problem. Here's my thinking. Neutral Magnesium has 2 valence electrons, so its Lewis diagram would have to 2 dots. Since this Magnesium ion has a 2+ charge, this ion has 2 less electrons than a neutral magnesium. Since removing 2 electrons would take away the only 2 electrons in the outer shell, the new outer shell would have a full 8 electrons. Therefore, the Lewis structure of Mg 2+ should have 8 dots. This is wrong, but I don't understand why. Read more Share Related Answers Section Related Answers Magnesium valence electrons count Magnesium electron configuration Mastering organic chemistry reactions Applications of thermodynamics in real life Role of catalysts in chemical reactions New to Reddit? Create your account and connect with a world of communities. Continue with Google Continue with Google. Opens in new tab Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of January 20, 2025 Reddit reReddit: Top posts of January 2025 Reddit reReddit: Top posts of 2025 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
1858	https://math.stackexchange.com/questions/3413977/example-of-a-calculus-optimization-problem-where-the-answer-occurs-at-an-endpoin	derivatives - Example of a calculus optimization problem where the answer occurs at an endpoint - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Example of a calculus optimization problem where the answer occurs at an endpoint Ask Question Asked 5 years, 11 months ago Modified5 years, 11 months ago Viewed 1k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I'm teaching optimization problems in calculus right now. An easy example would be something like: Find the dimensions of a rectangle with perimeter 100 100 m whose area is as large as possible. The goal would be to find the value of x x that maximizes A(x)=x(100−2 x 2)=x(50−x),A(x)=x(100−2 x 2)=x(50−x), where x∈[0,50].x∈[0,50]. When working over a closed interval, I teach them to find the critical points, and then evaluate A(x)A(x) at the critical points and also the endpoints of [0,50].[0,50]. They know that critical points aren't necessarily absolute max's or min's, so that's why they need to check the endpoints as well. My question is if there are any examples of some optimization problems where the absolute max or absolute min occurs at the endpoints of a closed interval. Every time we do these problems, the answer is always at the critical point, and it makes the last step of checking the endpoints seem pointless. calculus derivatives optimization maxima-minima extreme-value-theorem Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Oct 29, 2019 at 16:30 user525033 user525033 asked Oct 29, 2019 at 16:24 user525033 user525033 3 I assume you mean application-ish examples? Generally this will mean that somehow there is no tradeoff to making x x as large or small as possible, so this amounts to situations like a profit function where the revenue per unit always exceeds the cost per unit. Or else that there is a tradeoff but it only comes into play outside the domain that you're limited to.Ian –Ian 2019-10-29 16:34:15 +00:00 Commented Oct 29, 2019 at 16:34 Yes that's the idea user525033 –user525033 2019-10-29 16:35:16 +00:00 Commented Oct 29, 2019 at 16:35 Just keep your existing applications. The maximum of a convex function over a convex set occurs at the boundary, so as long as the objective function is nonlinear, you can always pick a sense (max or min) such that the optimal solution is at the boundary.LinAlg –LinAlg 2019-10-29 16:36:22 +00:00 Commented Oct 29, 2019 at 16:36 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. The height of an object above the surface of the water is modeled by the function h(t)=−t 3+5 t 2−8 t+4 h(t)=−t 3+5 t 2−8 t+4 On the interval t∈[0,3]t∈[0,3], when was the object furthest above the water? Furthest underwater? Critical numbers are t=4 3,2 t=4 3,2. Test and compare h(0),h(4 3),h(2),h(3)h(0),h(4 3),h(2),h(3) to get extreme values. The object is furthest above the water at t=0 t=0 and is furthest underwater at t=3 t=3. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Oct 29, 2019 at 16:36 Andrew ChinAndrew Chin 7,384 1 1 gold badge 14 14 silver badges 34 34 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Suppose your rectangle is limited to having width ≤20≤20 (i.e., same maximization problem but x∈[0,20]x∈[0,20]). Now the maximum will occur at the endpoint x=20 x=20. This example may be unsatisfactory since now the critical point doesn't even lie in the interval, but you can easily cook up an example where the critical points do occur inside the interval yet the extrema are on the boundary. For instance, finding the maximum and minimum of y=x 3−x y=x 3−x on [−2,2][−2,2]. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Oct 29, 2019 at 16:34 kccukccu 21.2k 1 1 gold badge 25 25 silver badges 43 43 bronze badges Add a comment\| You must log in to answer this question. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 5Good lecture optimization problem involving ln x ln⁡x or e x e x 0Optimization Calculus.. a box/shelter with sides missing.. 2Kepler's Wine Barrel Problem (Applied Optimization) 1Steps of finding an absolute extremum on an open interval 1Closed Interval Method 1Find the absolute extrema of the function f(x)=x 2−2 x−2 f(x)=x 2−2 x−2 on [0,1][0,1] 0Inconsistency in definitions of "critical points" and "differentiable"? 1An endpoint of a closed interval where the derivative is zero is considered a critical point? 0First and Second derivative test for local extrema to find dimensions of rectangle with maximum area Hot Network Questions Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term? 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1859	https://www.dummies.com/article/academics-the-arts/math/trigonometry/the-tangent-function-opposite-over-adjacent-149212/	Meet Earth’s mightiest heroes! Avengers For Dummies is here! Order your copy today. Book & Article Categories Collections Custom Solutions Dummies AI Main Menu Book & Article Categories Technology Academics & The Arts Home, Auto, & Hobbies Body, Mind, & Spirit Business, Careers, & Money Dummies AI Main Menu Book & Article Categories Technology Academics & The Arts Home, Auto, & Hobbies Body, Mind, & Spirit Business, Careers, & Money Dummies AI Main Menu Collections Explore all collections BYOB (Be Your Own Boss) Be a Rad Dad Career Shifting Contemplating the Cosmos For Those Seeking Peace of Mind For the Aspiring Aficionado For the Budding Cannabis Enthusiast For the College Bound For the Exam-Season Crammer For the Game Day Prepper Dummies AI Home Academics & The Arts Articles Math Articles Trigonometry Articles The Tangent Function: Opposite over Adjacent By Mary Jane Sterling Updated 2016-03-26 10:56:20 From the book Trigonometry For Dummies Share Download E-Book Trigonometry For Dummies Explore Book Trigonometry For Dummies Explore Book Buy Now Buy on AmazonBuy on Wiley Subscribe on Perlego Download E-Book Trigonometry For Dummies Explore Book Trigonometry For Dummies Explore Book Buy Now Buy on AmazonBuy on Wiley Subscribe on Perlego The third trig function, tangent, is abbreviated tan. This function uses just the measures of the two legs and doesn’t use the hypotenuse at all. The tangent is described with this ratio: opposite/adjacent. No restriction or rule on the respective sizes of these sides exists — the opposite side can be larger, or the adjacent side can be larger. So, the tangent ratio produces numbers that are very large, very small, and everything in between. You see that the tangents are And in case you’re wondering whether the two tangents of the acute angles are always reciprocals (flips) of one another, the answer is yes. The following example shows you how to find the values of the tangent for each of the acute angles in a right triangle where the hypotenuse is 25 inches and one leg is 7 inches. Find the measure of the missing leg. Using the Pythagorean theorem, a2 + b2 = c2, putting in 7 for a and 25 for c,and solving for the missing value, b, you find that the unknown length is 24 inches: 2. Select names for the acute angles in order to determine the opposite and adjacent designations. The easiest way to do this is to draw a picture and label it. The two acute angles are named with the Greek letters theta and lambda. The side opposite theta measures 7 inches, and the side adjacent to it measures 24 inches. For angle lambda, the opposite side measures 24 inches, and the adjacent side measures 7 inches. 3. Form the two tangent ratios by using the values 7, 24, and 25. About This Article This article is from the book: Trigonometry For Dummies About the book author: Mary Jane Sterling (Peoria, Illinois) is the author of Algebra I For Dummies, Algebra Workbook For Dummies, Algebra II For Dummies, Algebra II Workbook For Dummies, and many other For Dummies books. She taught at Bradley University in Peoria, Illinois for more than 30 years, teaching algebra, business calculus, geometry, and finite mathematics. This article can be found in the category: Trigonometry Get a Subscription
1860	https://www.youtube.com/watch?v=VNjUpnKwOlc	How to Prove That a Point Lies on a Line (or not) The Complete Guide to Everything 100000 subscribers 99 likes Description 14146 views Posted: 25 Feb 2020 In this video I will teach you how you can prove that a point lies on a line or not. This will allow you to check is a point fits a trend or not and is a really useful verification step in many fields. I will do this through two worked examples with a step by step method that should be nice and easy to follow. 13 comments Transcript: in this short tutorial I'm going to teach you how you can prove that a point is on a line so the process that you follow for every question about proving whether a point on the line is really just a three step process so we start with step one you put the point into an equation of the lines you're given an equation you put that point into the equation and that will help you on your way to find out if it's on the line the second step is simplify down so I have one number on one side one number on the other side so you've got a comparison the third step is to reach your conclusion so the rule is if both sides of the equation are equal you get something like five equals five then the point is on the line otherwise the point isn't on the line so when you simplify down you get five equals four let's tell you it's not on the line because if it was on the line both sides of the equation would be equal so we've either take a note of these or remember these we're going to look at an example of this in action so does the point five seven lie on the line y equals 6x plus five this is a typical question the first thing we need to do is identify our X&Y coordinates so the first number in a coordinate is your X the second number is your Y and I've marked those on for convenience so we write the equation of the line exactly the same as the equation before so we need to do the substitution so there's some arrows so the X is going to be replaced with the five and the Y is going to be replaced with seven so you can see how that matches up which is replacing the letters with the numbers that we know that match up with those letters so we then do the substitution so six brackets five just means six times five so we're saying seven equals six times five plus five well no it doesn't seven is not equal to 40 so obviously the point five seven does not lie on that line so the conclusion is the point is not on the line if they were equal it would be on the line but they're totally different so it's not on the line if we look at an example where it actually does work for example the point 217 so we know that we've got the same equation we substituted the first number for X set member for y so do this one a little bit quicker so a seventeen is six times two plus five is that true well yes 17 equals 17 so you simplify that old you get 17 equals 17 and that tells you that the point is on the line so that in a nutshell is how you prove that a point is or isn't on a line so hopefully this was helpful to you and thank you very much for watching
1861	https://artofproblemsolving.com/wiki/index.php/2025_AIME_II_Problems?srsltid=AfmBOopuSlo8lekyVJWPrffXmR7nRkR8VM25bwk0m6jCW3jxdxjHm8kK	Art of Problem Solving 2025 AIME II Problems - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2025 AIME II Problems Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2025 AIME II Problems 2025 AIME II (Answer Key) Printable version \| AoPS Contest Collections • PDF Instructions This is a 15-question, 3-hour examination. All answers are integers ranging from to , inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers. No aids other than scratch paper, rulers and compasses are permitted. In particular, graph paper, protractors, calculators and computers are not permitted. 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15 Contents [hide] 1 Problem 1 2 Problem 2 3 Problem 3 4 Problem 4 5 Problem 5 6 Problem 6 7 Problem 7 8 Problem 8 9 Problem 9 10 Problem 10 11 Problem 11 12 Problem 12 13 Problem 13 14 Problem 14 15 Problem 15 16 See also Problem 1 Six points and lie in a straight line in that order. Suppose that is a point not on the line and that , , , , , , and Find the area of . Solution Problem 2 Find the sum of all positive integers such that divides the product Solution Problem 3 Four unit squares form a grid. Each of the unit line segments forming the sides of the squares is colored either red or blue in such a way that each unit square has red sides and blue sides. One example is shown below (red is solid, blue is dashed). Find the number of such colorings. Solution Problem 4 The productis equal to where and are relatively prime positive integers. Find Solution Problem 5 Suppose has angles and Let and be the midpoints of sides and respectively. The circumcircle of intersects and at points and respectively. The points and divide the circumcircle of into six minor arcs, as shown. Find where the arcs are measured in degrees. Solution Problem 6 Circle with radius centered at point is internally tangent at point to circle with radius . Points and lie on such that is a diameter of and . The rectangle is inscribed in such that , is closer to than to , and is closer to than to , as shown. Triangles and have equal areas. The area of rectangle is , where and are relatively prime positive integers. Find . Solution Problem 7 Let be the set of positive integer divisors of . Let be a randomly selected subset of . The probability that is a nonempty set with the property that the least common multiple of its elements is is , where and are relatively prime positive integers. Find . Solution Problem 8 From an unlimited supply of -cent coins, -cent coins, and -cent coins, Silas wants to find a collection of coins that has a total value of cents, where is a positive integer. He uses the so-called , successively choosing the coin of greatest value that does not cause the value of his collection to exceed . For example, to get cents, Silas will choose a -cent coin, then a -cent coin, then -cent coins. However, this collection of coins uses more coins than necessary to get a total of cents; indeed, choosing -cent coins and -cent coins achieves the same total value with only coins. In general, the greedy algorithm succeeds for a given if no other collection of -cent, -cent, and -cent coins gives a total value of cents using strictly fewer coins than the collection given by the greedy algorithm. Find the number of values of between and inclusive for which the greedy algorithm succeeds. Solution Problem 9 There are values of in the interval where . For of these values of , the graph of is tangent to the -axis. Find . Solution Problem 10 Sixteen chairs are arranged in a row. Eight people each select a chair in which to sit so that no person sits next to two other people. Let be the number of subsets of chairs that could be selected. Find the remainder when is divided by . Solution Problem 11 Let be the set of vertices of a regular -gon. Find the number of ways to draw segments of equal lengths so that each vertex in is an endpoint of exactly one of the segments. Solution Problem 12 Let be an -sided non-convex simple polygon with the following properties: • For every integer , the area of is , • For every integer , , • The perimeter of the -gon is equal to . Then can be expressed as where are positive integers, is not divisible by any square, and no prime divides all of ,, and . Find . Solution Problem 13 Let the sequence of rationals be defined such that and for all . Then can be expressed as for relatively prime positive integers and . Find the remainder when is divided by . Solution Problem 14 Let be a right triangle with and There exist points and inside the triangle suchThe area of the quadrilateral can be expressed as for some positive integer Find Solution Problem 15 There are exactly three positive real numbers such that the function defined over the positive real numbers achieves its minimum value at exactly two positive real numbers . Find the sum of these three values of . Solution See also 2025 AIME II (Problems • Answer Key • Resources) Preceded by 2025 AIME IFollowed by 2026 AIME I 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15 All AIME Problems and Solutions American Invitational Mathematics Examination AIME Problems and Solutions Mathematics competition resources These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1862	https://www.youtube.com/watch?v=E1zK8NXcesU	Q45 (2016). Four-digit numbers are to be formed using the digits 1, 2, 3 and 4; and none of these.. Binge Solver 3190 subscribers 68 likes Description 4256 views Posted: 8 May 2023 Q45 (2016). Four-digit numbers are to be formed using the digits 1, 2, 3 and 4; and none of these four digits are repeated in any manner.Further, 1. 2 and 3 are not to immediately follow each other 2. 1 is not to be immediately followed by 3 3. 4 is not to appear at the last place 4. 1 is not to appear at the first place How many different numbers can be formed ? (a)6 (b)8 (c)9 (d)None of the above UPSC Civil Service (Preliminary) 2016 - GS2 - CSAT 2 comments Transcript: next question question 45 four digit numbers are to be formed using the digits one two three and four and none of these four digits are repeated in any manner for the four conditions are given the question is how many different numbers can be formed so there is four restrict four conditions which Narrows down our possible combinations and the solution options are six eight nine and none of the above from the option itself we can see we can manually count and find the answer um if the numbers were very high we should use the concept of permutation and use some logic to remove the duplicates and find the final answer here since the maximum number here is nine we can safely assume that the number will be less than 10 and find the number of cases first we will apply each conditions so first condition is 2 and 3 are not to immediately follow each other so we'll write down all such possible combinations so one possible combination is two three so two and three are separated by like this or this could be like interchanging 3 and 2 will also have a case like this then we will have here integing 2 and 3 will have two three two so another case will be like this three two so these are the all the cases were two and three are not to immediately follow each other so by that it means two three should not appear like this or 3 2 it should not appear like this also all such cases are removed from all these scenarios now we will see how we can fill one and four uh with to fill one and four we have such three conditions here one is not to immediately followed by three so one one is not to immediately followed by three means one three it should not one and three should not appear like this 3 and 1 is possible since uh here uh one is not immediately followed by three instead three is followed by one so uh this is possible but this is not possible so we should keep that in mind and one three should not appear so uh from so just using that we can see one shell one cannot appear here then one three will come so only possible combination is four three one so we will see whether other two conditions satisfied four is not to appear at the last place so in none of the cases four should not appear at the last place and one is known to appear at the first place so all the all the conditions are made for this case so this is a possible case now I will see two uh into uh one can appear here and four if you if it appears four E4 comes here then 4 appears at the last place so that condition is false so the only condition that is possible is three four two one one can appear at the last place and all the if you see all the four conditions are met so this is a possible case and if you take three one shall not appear here since one cannot appear at the first place so only possibility is four two one three but again if you see as I already said one three cannot come so one cannot come here so both the possible cases are false for this so this is not a possible so none of the we cannot fill one and four in any place here in this in our in this Arrangement so none of the valid cases come here so we will see next one so here we have one one cannot appear in the first place so four three one two so three one is Possible only one three is the problematic case so this is a valid case now we'll go to file here uh again one cannot appear here so it will become one three so only possibility is two one four three this is the valid case now here if you see one can one can appear here or this is a valid case also there is one more case so we'll write 6A which is 3 4 1 2 this is also a valid case four one one four four one both cases are valid case so all the cases are done so in total we have one two three four five six cases six stick one are the possible cases here we do not have any case so that is striked off so in total we have six cases so how many different numbers can be found the answer is a six solution to question 45 is a 6.
1863	https://www.voicetube.com/videos/48570	Photon Energy \| Physical Processes \| MCAT \| Khan Academy - VoiceTube: Learn English through videos! Download VoiceTube Learn English through FUN videos! Download VoiceTube Open in app Learning Package Channel Business & Finance Learning Resources Arts & Entertainment Music Animation News & Current Affairs Kids Leisure & Travel Science & Technology Health & Wellness Levels A1 A2 B1 B2 C1 C2 Level Description Pronunciation Challenge Saved Search Vocabulary Services Learning Package Log in Subtitles section Play video [x] Keywords ###### Exam Level [x] All - [x] TOEIC ###### Level [x] All - [x] A1 - [x] A2 - [x] B1 - [x] B2 - [x] C1 - [x] C2 - [x] Saved - [x] Searched Apply Print subtitles We've been treating light as a wave, and we've been drawing it with this continuous wave pattern of oscillating electric and magnetic fields that are traveling in some direction. And why shouldn't we treat it as a wave? If you sent it through a small opening, this electromagnetic radiation would spread out, There'd be diffraction, and that's what waves do. Or, if you let it overlap with itself, if you had some wave in some region, and it lined up perfectly with some other electromagnetic wave, you'd get constructive interference. If it was out of phase, you'd get destructive interference. That's what waves do. Why shouldn't we call electromagnetic radiation a wave? And that's what everyone thought. But, in the late 1800s and early 1900s, physicists discovered something shocking. They discovered that light, and all electromagnetic radiation, can display particle-like behavior, too. And I don't just mean localized in some region of space. Waves can get localized. If you sent in some wave here that was a wave pulse, well, that wave pulse is pretty much localized. When it's traveling through here, it's going to kind of look like a particle. That's not really what we mean. We mean something more dramatic. We mean that light, what physicists discovered, is that light and light particles can only deposit certain amount of energy, only discrete amounts of energy. There's a certain chunk of energy that light can deposit, no less than that. So this is why it's called quantum mechanics. You've heard of a quantum leap. Quantum mechanics means a discrete jump, no less than that. And so what do we call these particles of light? We call them photons. How do we draw them? That's a little trickier. We know now light can behave like a wave and a particle, so we kind of split the difference sometimes. Sometimes you'll see it like this, where it's kind of like a wavy particle. So there's a photon, here's another photon. Basically, this is the problem. This is the main problem with wave particle duality, it's called. The fact that light, and everything else, for that matter, can behave in a way that shows wavelike characteristics, it can show particle-like characteristics, there's no classical analog of this. We can't envision in our minds anything that we've ever seen that can do this, that can both behave like a wave and a particle. So it's impossible, basically, to draw some sort of visual representation, but, you know, it's always good to draw something. So we draw our photons like this. And so, what I'm really saying here is, if you had a detector sitting over here that could measure the light energy that it receives from some source of light, what I'm saying is, if that detector was sensitive enough, you'd either get no light energy or one jump, or no light energy or, whoop, you absorbed another photon. You couldn't get in between. If the quantum jump was three units of energy ... I don't want to give you a specific unit yet, but, say, three units of energy you could absorb, if that was the amount of energy for that photon, if these photons were carrying three units of energy, you could either absorb no energy whatsoever or you could absorb all three. You can't absorb half of it. You can't absorb one unit of energy or two units of energy. You could either absorb the whole thing or nothing. That's why it's quantum mechanics. You get this discrete behavior of light depositing all its energy in a particle-like way, or nothing at all. How much energy? Well, we've got a formula for that. The amount of energy in one photon is determined by this formula. And the first thing in it is Planck's constant. H is the letter we use for Planck's constant, and times f. This is it. It's a simple formula. F is the frequency. What is Planck's constant? Well, Planck was basically the father of quantum mechanics. Planck was the first one to figure out what this constant was and to propose that light can only deposit its energy in discrete amounts. So Planck's constant is extremely small; it's 6.626 times 10 to the negative 34 th joule times seconds. 10 to the negative 34 th? There aren't many other numbers in physics that small. Times the frequency -- this is regular frequency. So frequency, number of oscillations per second, measured in hertz. So now we can try to figure out, why did physicists never discover this before? And the reason is, Planck's constant is so small that the energy of these photons are extremely small. The graininess of this discrete amount of energy that's getting deposited is so small that it just looks smooth. You can't tell that there's a smallest amount, or at least it's very hard to tell. So instead of just saying 'three units,' let's get specific. For violet light, what's the energy of one violet photon? Well, the frequency of violet light is 7.5 times 10 to the 14 th hertz. So if you take that number times this Planck's constant, 6.626 times 10 to the negative 34 th, you'll get that the energy of one violet photon is about five times 10 to the negative 19 th joules. Five times ten to the negative 19 th, that's extremely small. That's hard to see. That's hard to notice, that energy's coming in this discrete amount. It's like water. I mean, water from your sink. Water flowing out of your sink looks continuous. We know there's really discrete water molecules in there, and that you can only get one water molecule, no water molecules, 10 water molecules, discrete amounts of these water molecules, but there's so many of them and they're so small, it's hard to tell that it's not just completely continuous. The same is happening with this light. This energy's extremely small. Each violet photon has an extremely small amount of energy that it contributes. In fact, if you wanted to know how small it is, a baseball, a professional baseball player, throwing a ball fast, you know, it's about 100 joules of energy. If you wanted to know how many of these photons, how many of these violet photons would it take to equal the energy of one baseball thrown at major league speed? It would take about two million trillion of these photons to equal the energy in a baseball that's thrown. That's why we don't see this on a macroscopic scale. For all intents and purposes, for all we care, at a macroscopic level, light's basically continuous. It can deposit any energy whatsoever, because the scale's so small here. But if you look at it up close, light can only deposit discrete amounts. Now, I don't mean that light can only deposit small amounts. Light can deposit an enormous amount of energy, but it does so in chunks. So think about it this way ... Let's get rid of all this. Think about it this way: let's say you had a detector that's going to register how much energy it's absorbing, and we'll graph it. We'll graph what this detector's going to measure, the amount of energy per time that it measures. So we'll get the amount of energy per time. Now, you can absorb huge amounts of energy. And on the detector, on a macroscopic scale, it just might look like this. You know, you're getting more and more light energy. You're absorbing more and more energy, collecting more and more energy. But what I'm saying is that, microscopically, if you look at this, what's happening is, you've absorbed one photon here. You absorbed another one, absorbed another one, absorbed a bunch of them. You keep absorbing a bunch of these photons. You can build up a bunch of energy. That's fine. It's just if you looked at it close enough, you have this step pattern that's absorbing photons at a time, certain numbers of them. Maybe it absorbs three at one moment, four at another moment. But you can't absorb anything in between. It can't be completely continuous. It has to be a discrete all-or-nothing moment of absorption of energy that, on a macroscopic scale, looks smooth but on a microscopic scale is highlighted by the fact that light energy is coming in discrete chunks, described by this equation that gives you the energy of individual photons of light. We've been treating light as a wave, x1.0 x 0.5 x 0.75 x 1.0 x 1.5 Repeat video - [x] Font size Shortcuts Return 175% 150% 125% 100% 90% Subtitles and vocabulary B1USenergydiscretephotonplanckwaveparticle Photon Energy \| Physical Processes \| MCAT \| Khan Academy 35 8 許藝菊 posted on 2017/02/18 More Share Save Report Video vocabulary Filter [x] Keywords Exam Level [x] All [x] TOEIC Level [x] All [x] A1 [x] A2 [x] B1 [x] B2 [x] C1 [x] C2 [x] Saved [x] Searched Apply Keywords No related vocabulary under this filter specific Share vocabulary US /spɪˈsɪfɪk/ ・ UK /spəˈsɪfɪk/ adjective Precise; particular; just about that thing Concerning one particular thing or kind of thing Clearly defined or identified. Relating to a particular species, structure, etc. A2 More bunch Share vocabulary US /bʌntʃ/ ・ UK /bʌntʃ/ other A group of things of the same kind A group of people. other To group people or things closely together other (Cloth) to gather/be gathered together in folds B1 More figure Share vocabulary US /ˈfɪɡjɚ/ ・ UK /ˈfiɡə/ other To appear in a game, play or event To calculate how much something will cost To understand after thinking; work out To consider, believe, or conclude. noun Your body shape Numbers in a calculation Doll-like thing meant to represent a person Picture or diagram giving information in a text Person who is very important or famous Shape of a person seen indistinctly or in outline Amount that is expressed in numbers A1TOEIC More extremely Share vocabulary US /ɪk'strimlɪ/ ・ UK /ɪkˈstri:mli/ adverb In a way that is much more than usual or expected Remarkably; unusually. From an extreme point of view. B1 More pattern Share vocabulary US /ˈpætən/ ・ UK /'pætn/ other Model to follow in making or doing something Colors or shapes which are repeated on objects Regular repeated behavior A regular or repeated way in which something happens or is done. A set of paper shapes used as a guide for cutting cloth when making clothes. other To copy the way something else is made To decorate with a pattern. A2TOEIC More enormous Share vocabulary US /ɪˈnɔrməs/ ・ UK /iˈnɔ:məs/ adjective Huge; very big; very important Very great in size, amount, or degree. Having a very great effect or influence. A2 More basically Share vocabulary US /ˈbesɪkəli,-kli/ ・ UK /ˈbeɪsɪkli/ adverb Used before you explain something simply, clearly In essence; when you consider the most important aspects of something. In a simple and straightforward manner; simply. A2 More constant Share vocabulary US /ˈkɑnstənt/ ・ UK /'kɒnstənt/ adjective Happening frequently or without pause Remaining the same over time or not changing. Faithful and dependable. noun Thing that happens always or at a regular rate A physical quantity that is believed to have a fixed value and is used in calculations. A2TOEIC More scale Share vocabulary US /skel/ ・ UK /skeɪl/ other Size, level, or amount when compared Small hard plates that cover the body of fish Device that is used to weigh a person or thing Range of numbers from the lowest to the highest Dimensions or size of something other To change the size of but keep the proportions To climb something large (e.g. a mountain) To remove the scales of a fish A2TOEIC More determine Share vocabulary US /dɪˈtɚmɪn/ ・ UK /dɪ'tɜ:mɪn/ other To control exactly how something will be or act To establish the facts about; discover To be the deciding factor in; to control or influence directly. To decide firmly on a course of action; to resolve. A2TOEIC More completely Share vocabulary US /kəmˈpliːtli/ ・ UK /kəmˈpli:tli/ adverb In every way or as much as possible To the greatest extent; thoroughly. Including all or everything; without anything lacking. A1 More negative Share vocabulary US /ˈnɛɡətɪv/ ・ UK /'neɡətɪv/ noun The opposite to a positive electrical charge In grammar, containing words such as 'no' or 'not' Reply to a question or statement that means 'no' Image on camera film used to make a photo adjective Being harmful, unwanted or unhelpful In mathematics, being less than zero Focusing on the bad aspects; pessimistic Expressing disagreement or refusal. Indicating the absence of something, such as a disease or condition. Carrying a negative electric charge. A2 More tricky Share vocabulary US /ˈtrɪki/ ・ UK /'trɪkɪ/ adjective Difficult, so needing skill to do or deal with Likely to use tricks; dishonest or deceptive A2 More treat Share vocabulary US /trit/ ・ UK /tri:t/ other To pay for the food or enjoyment of someone else To use medical methods to try to cure an illness To act in a certain way toward someone To subject to some process or action; to apply a substance to. other Something you buy for others as a surprise present something that tastes good and that is not eaten often Something special that gives pleasure. A1TOEIC More molecule Share vocabulary US /ˈmɑlɪˌkjul/ ・ UK /ˈmɒlɪkju:l/ noun Two or more atoms chemically combined B2 More amount Share vocabulary US /əˈmaʊnt/ ・ UK /ə'maʊnt/ other Quantity of something other To add up to a certain figure A1TOEIC More measure Share vocabulary US /ˈmɛʒɚ/ ・ UK /ˈmeʒə(r)/ other Plan to achieve a desired result Tool used to calculate the size of something A standard unit or system used for measuring. A certain amount or degree of something. A division of time in music, usually consisting of a fixed number of beats. other To determine the value or importance of something To calculate size, weight or temperature of To take actions to achieve a particular purpose. A1TOEIC More chunk Share vocabulary US /tʃʌŋk/ ・ UK /tʃʌŋk/ other A large bit of something; thick lump A significant amount or portion. A distinct segment or group. A contiguous block of data. other To divide something into chunks. To throw something forcefully (slang). B2 More contribute Share vocabulary US /kənˈtrɪbjut/ ・ UK /kən'trɪbju:t/ other To be a factor in causing something to happen To donate, give (money) or help to something To write articles for a magazine or newspaper A2 More absorb Share vocabulary US /əbˈsɔrb, -ˈzɔrb/ ・ UK /əb'sɔ:b/ other To take up all attention / energy of something To take in a liquid; soak up To take in and understand information or facts. To reduce the effect of a force, shock, or change. To bear (costs) without passing them on to others. B1TOEIC More region Share vocabulary US /ˈridʒən/ ・ UK /'ri:dʒən/ other Part of a country, of the world, area, etc. A part of the body An administrative district of a country A2TOEIC More dramatic Share vocabulary US /drəˈmætɪk/ ・ UK /drəˈmætɪk/ adjective Gripping the attention; causing an effect (Of an event) sudden and extreme Sudden and striking; sensational. B1 More sensitive Share vocabulary US /ˈsɛnsɪtɪv/ ・ UK /'sensətɪv/ adjective Taking offense easily; easily upset or hurt (Private information) needing careful treatment Concerning awareness of small differences Able to recognize the feelings of others Responsive to physical touch Becoming unstable A2TOEIC More draw Share vocabulary US /drɔ/ ・ UK /drɔ:/ other To attract attention to someone or something To influence a person's involvement in something To move an object by pulling To take one thing out of a container, etc. other Something that attracts people to visit a place A lottery or prize Result of a game, contest where the score is equal other To get closer to or approach something or someone (Of 2 teams) to finish a game with the same score other To create an image using pen or pencil and paper A1TOEIC More formula Share vocabulary US /ˈfɔrmjələ/ ・ UK /'fɔ:mjələ/ other A mathematical or scientific rule Plan, rule or method for doing or making something A symbolic representation of the composition of a chemical compound. A milk substitute for babies. The ingredients and their proportions in a mixture, especially a cosmetic or pharmaceutical product. A2TOEIC More leap Share vocabulary US /lip/ ・ UK /li:p/ other To enter or start something eagerly (Of prices) to increase quickly by a large amount To jump over something To move forwards often in big steps or jumps To move rapidly or suddenly noun (Of prices) a quick large increase A large forward step or jump B1 More behavior Share vocabulary US /bɪˈhevjɚ/ ・ UK /bɪ'heɪvjə/ other The way a person or thing acts; manner The way in which an animal or person acts in response to a particular situation or stimulus. The way in which a system or component functions or performs. A1TOEIC More register Share vocabulary US /ˈrɛdʒɪstɚ/ ・ UK /'redʒɪstə(r)/ other To record your name on an official list; sign up To show an amount on a measurement device (Of feelings) to show or make visible To send (a letter or parcel) by registered post. noun Till; machine used add up things you buy Range of notes of a person's (singing) voice An official list or record of names, items, or events. A mail that is registered. A variety of a language used for a particular purpose or in a particular social setting. A2 More equal Share vocabulary US /ˈikwəl/ ・ UK /'i:kwəl/ adjective Same in shape, size, or number other To add up or be the same as noun A person or thing that is the same as another in status or quality A1TOEIC More envision Share vocabulary US ・ UK other To imagine or picture in your mind C2TOEIC More sink Share vocabulary US /sɪŋk/ ・ UK /sɪŋk/ other To push something down into the ground To dig (a well) other Large bowl in a kitchen or washroom for washing other To move down toward a lower point To decrease in value or amount A2TOEIC More throw Share vocabulary US /θroʊ/ ・ UK /θrəʊ/ other To use your arm to make something fly in the air To move part of your body suddenly and forcefully To confuse or upset someone (E.g. judo) to forcibly put someone on the ground To put something somewhere suddenly and roughly other Arm movement to make a thing fly through the air Loose cloth or blanket (usually over a chair) Forcibly putting someone on the ground A1 More equation Share vocabulary US /ɪˈkweʒən, -ʃən/ ・ UK /ɪˈkweɪʒn/ other Mathematical statement showing things to be equal Complex set of different facts, ideas or issues A symbolic representation of a chemical reaction using chemical formulas. A complex of factors influencing a situation. A state of equilibrium or balance; a harmonious relationship. B1 More frequency Share vocabulary US /ˈfrikwənsi/ ・ UK /'fri:kwənsɪ/ other Number of repetitions of (radio or sound) wave How often something happens The number of times a particular value occurs in a set of data. B1TOEIC More moment Share vocabulary US /'moʊmənt/ ・ UK /'məʊmənt/ other Very short or brief period of time A particular period in time A1 More particle Share vocabulary US /ˈpɑrtɪkəl/ ・ UK /ˈpɑ:tɪkl/ noun Adverb or preposition that joins with a verb Tiny piece or quantity of something A minute portion of matter. B1 More receive Share vocabulary US /rɪˈsiv/ ・ UK /rɪ'si:v/ other To get something someone has given or sent to you To allow someone to become a member (of a club) To welcome someone as a guest into your home To respond to (e.g. news) in a particular way To get a signal To suffer an injury A1TOEIC More radiation Share vocabulary US /ˌrediˈeʃən/ ・ UK /ˌreɪdiˈeɪʃn/ other Energy transmitted as rays, waves or particles Harmful waves of energy from nuclear activity The emission of energy as electromagnetic waves or as moving subatomic particles, especially high-energy particles which cause ionization. The use of high-energy radiation to treat disease, especially cancer. The transfer of energy as electromagnetic waves without the involvement of a physical substance. B1 More destructive Share vocabulary US /dɪˈstrʌktɪv/ ・ UK /dɪ'strʌktɪv/ adjective Causing a very large amount of damage Tending to demolish or tear down; negative; unhelpful. Characterized by actions that destroy or harm. B2 More fact Share vocabulary US /fækt/ ・ UK /fækt/ other Something that is known or proved to be true A1 More kind Share vocabulary US /kaɪnd/ ・ UK /kaɪnd/ adjective In a caring and helpful manner other One type of thing A1TOEIC More constructive Share vocabulary US /kənˈstrʌktɪv/ ・ UK /kənˈstrʌktɪv/ adjective Having positive effective results B1 More magnetic Share vocabulary US /mæɡˈnɛtɪk/ ・ UK /mægˈnetɪk/ adjective Having the properties of a magnet; able to attract Having great power to attract attention, interest B2 More whatsoever Share vocabulary US /ˌhwɑtsoˈɛvɚ,ˌhwʌt-,ˌwɑt,ˌwʌt-/ ・ UK /ˌwɒtsəʊ'evə/ determiner Used to add emphasis to an idea being expressed B2 More overlap Share vocabulary US /ˌoʊvərˈlæp/ ・ UK /ˌəʊvəˈlæp/ other To lie over or cover a part of something else To happen at the same time as another event To share qualities or functions other Amount shared in the same space, form, idea B1TOEIC More electromagnetic Share vocabulary US /ɪˌlektroʊmægˈnetɪk/ ・ UK /ɪˌlektrəʊmægˈnetɪk/ adjective Having magnetic and electrical parts A1 More hard Share vocabulary US /hɑː(r)d/ ・ UK /hɑ:d/ adjective Difficult to do; difficult to understand (Of facts) not able to be misunderstood; clear (Of alcohol) strong Involving or requiring lots of work, effort, care Being full of pain, trouble, and problems Strong (Of edge) clear; sharp Difficult to bend, break or cut; solid adverb With lots of force, power, or impact A1TOEIC More quantum Share vocabulary US /ˈkwɑ:ntəm/ ・ UK /ˈkwɒntəm/ other The smallest unit or amount of energy B2 More pulse Share vocabulary US /pʌls/ ・ UK /pʌls/ other Regular beating of a heart A single vibration or short burst of sound, light, electricity, or other energy. A strong feeling or urge. The edible seeds of various leguminous plants, such as beans, lentils, and peas. other To beat in a rhythm To emit or produce in pulses. B1 More absorption Share vocabulary US /əbˈsɔ:rpʃn/ ・ UK /əbˈsɔ:pʃn/ other Process of being absorbed e.g. liquid, attention The process of understanding and assimilating information or ideas. The state of being completely engrossed or interested in something. The process by which energy, such as light or sound, is taken in and not transmitted. The process of one company merging with or being taken over by another. The process by which nutrients or drugs are taken into the body. B2 More microscopic Share vocabulary US /ˌmaɪkrəˈskɑ:pɪk/ ・ UK /ˌmaɪkrəˈskɒpɪk/ adjective Too small to be seen with the eyes Relating to or involving the use of a microscope. Extremely detailed or precise. B1 More unit Share vocabulary US /ˈjunɪt/ ・ UK /ˈju:nɪt/ other One apartment in a building Group of people, e.g. soldiers, that work together A chapter of a textbook A standard of measurement of something Single item made by a company B2TOEIC More interference Share vocabulary US /ˌɪntərˈfɪrəns/ ・ UK /ˌɪntəˈfɪərəns/ other Unwanted involvement in the concerns of others Signal interrupting a radio or television signal The act of hindering or obstructing an activity or process. Unwanted electrical signals that disrupt or degrade the quality of a signal. The combination of two or more waves to form a resultant wave of greater, lower, or the same amplitude. An act of illegally obstructing an opponent in sports. The influence of one language on another, especially in second language acquisition. B1 More discrete Share vocabulary US /dɪˈskrit/ ・ UK /dɪˈskri:t/ adjective Separate; composed of distinct parts Constituing a separate thing; distinct. B2TOEIC More violet Share vocabulary US /ˈvaɪəlɪt/ ・ UK /'vaɪələt/ noun Small fragrant plant with usually purplish flowers B2 More photon Share vocabulary US /ˈfoʊtɑ:n/ ・ UK /ˈfəʊtɒn/ noun Particle with energy but no mass, electric charge B2 More classical Share vocabulary US /ˈklæsɪkəl/ ・ UK /ˈklæsɪkl/ adjective Concerning the culture of ancient Greece or Rome (Of music) serious and traditional in form Concerning ideas considered to be traditional B1 More duality Share vocabulary US /duˈælɪti, dju-/ ・ UK /dju:ˈæləti/ other Tech having two aspects, e.g. mind and matter C2 More whoop Share vocabulary US /hup, hwup, wup/ ・ UK /wu:p/ other Loud cry expressing excitement or happiness other To shout loudly in a happy or excited manner B2 More oscillate Share vocabulary US ・ UK other To keep moving back and forth many times B2 More diffraction Share vocabulary US /dɪ'frækʃn/ ・ UK /dɪ'frækʃn/ other Bending light, etc. around something/through a hole C1 More hertz Share vocabulary US /hɜ:rts/ ・ UK /hɜ:ts/ noun Unit for measuring the frequency of sound waves B2 More More vocabulary Auto Play - [x] 03:27 ###### ScienceCasts: Draconid Meteor Outburst 52 B204:37 ###### The Gravity Gun 93 A204:20 ###### The Origin of Quantum Mechanics (feat. 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1865	https://www.youtube.com/watch?v=7G-MQBuwCRk	5.3 Monotonicity And Concavity Lumist 815 subscribers 5 likes Description 153 views Posted: 19 Mar 2021 Everything you need to know about Value Theorem, Monotonicity, and Concavity, Curve Sketching, Optimization Problems, etc. To ace the AP Calc AB&BC exam. Lumi will navigate you through the concepts, methods, and applications of differential and integral calculus, including topics such as Using Linear Partial Fractions, Slope Field, Euler’s Method, and Logistic Models. Now let's get you ready for the exam day. Subscribe for more info: Transcript: Introduction hi welcome to lumi in this video we will introduce another important application of derivative the monotecity and the concavity monotacity means increasing or decreasing of a function let f be a function defined on the interval i f is increasing on i when for all x1 and 2's belonging to the interval i x1 is lesser than x2 and f of x1 lesser than f x2 f would be considered non-decreasing on i when x1 is lesser than x2 and f of x one is less than or equal to f of x two f is considered decreasing on that interval i when x one is lesser than x two and f of x one is bigger than f of x two and f is considered not increasing on interval i when x one is less than x two and f of x one is bigger or equal to f of x 2. in some books including our book we use increasing and non-decreasing however in other books they strictly use increasing and increase it recall that a derivative at a point is equal to the slope of this point so if the derivative at a point is greater than zero then the slope at this point is greater than zero if the derivative is greater than zero on that interval it means that the slope of every point on that interval is greater than zero then we can say that the function is increasing on this interval similarly if the derivative is less than zero on that interval it means that the slope at every point on that interval is less than zero then we can say that the function is decreasing on that interval if you pay attention to the graph as you may see on the first graph x1 is less than x2 and f x 1 is less than f x 2 meaning that this function is increasing but in the next example x 1 is lesser than x 2 f of x 1 is bigger than f x 2 meaning that this function is decreasing well that was just an intuitive explanation now we introduce the theory of monotecity which allows us to use the first derivative to find the interval where a function is increasing or decreasing let a be lesser than b let f be a function defined at the interval a to b if for all x's belonging to a to b the derivative of f of x is bigger than zero then the function f is increasing on that interval a to b if for all x's that belong to a to b f prime of x is bigger or equal to zero then the function f is non-decreasing on the interval a to b if for all x's belonging to a to b the derivative of the function f of x is lesser than zero then the function f is decreasing on the interval a to b in that same concept if for all x is belonging to a to b the derivative of the function f is lesser or equal to zero then the function f is not increasing on the interval a to b this theorem can help us analyze the monotony of a function for example find where the function f x is increasing and where it is decreasing so we take the derivative of the function f of x and we end up with f prime of x to be equal to 12 x times x minus two times x plus one then we create a plus minus chart to find out in which intervals the function is increasing and when it is decreasing we find out that on the interval x being lesser than negative one f prime of x is less than zero meaning that this function is decreasing on the interval negative infinity to negative one on the interval between negative one to zero f prime of x is bigger than zero so we see that the function f of x is increasing on the interval negative one to zero then on the interval 0 to 2 f prime of x is lesser than 0 so that the function f of x is decreasing on the interval 0 to 2. and the interval x bigger than 2 f prime of x is bigger than 0 meaning that f of x is increasing on interval 2 to positive infinity here is the graph of the function and we can see that the monotony of this function is the same as we got in the solution set as you may see the function is decreasing between negative infinity to negative 1 increasing between negative 1 to 0 and is decreasing between 0 to positive 2 and is increasing again from 2 to positive infinity First derivative test the discussion of monotony can help us determine whether a function has a local maximum or a minimum or a saddle point at a critical point we call it the first derivative test suppose that c is a critical number on a continuous function f if f prime of x is bigger than 0 on x less than c and f prime of x is less than 0 on x is bigger than c then f has a local maximum at the point c as you may see their derivative of the function f of x is bigger than 0 here is lesser than zero right at the maximum point f prime of x equal to zero has a local maximum and if f prime of x is lesser than zero on x lesser than c and f prime of x is bigger than 0 on x bigger than c then f has a local minimum at the point c so as you see on the graph f prime of c here is less than 0 f prime of c here is bigger than 0 and at the point f prime of x equal to 0 we have a local minimum if f prime does not change signs for example if f prime is positive on both sides of c or negative on both sides then there is no local maximum or minimum at c so you can say that it has a saddle point so for example there's no maximum or minimum in the first example that we have you see that the f prime of x is bigger than zero and f prime of x is bigger than zero on both sides of c in this example meaning that c in this case is a saddle point and in the second example f prime of x is less than zero and f prime of x is less than zero on both left and right side of c meaning that c here is a saddle point for example if you want to find all the local maximum and minimum values of the function g of x where x is between 0 to 2 pi before this video what you will needed to do is you can approach this problem by using the local extreme value theorem however this theorem only helps you find where the maximum minimum or saddle point is and you can find the absolute extreme by comparing the function values of those to the critical points however you can't determine which critical point is a local maximum and which critical point is a local minimum now with the first derivative test you can definitely do that first we need to differentiate the function and find the g prime of x to be equal to 1 plus 2 cos of x then we analyze g by using the following table and find out that is increasing in the interval 0 to 2 pi over 3 decreasing on interval 2 pi over 3 to 4 pi over 3 and is increasing on the interval 4 pi over 3 to 2 pi since the derivative is greater than 0 on the 0 to 2 pi over 3 and less than 0 on 2 pi over 3 to 4 pi over 3 then the local maximum value is g of 2 pi over 3 which you get a solution of 3.38 since the derivative is lesser than 0 on the 2 pi over 3 to 4 pi over 3 and greater than 0 on 4 pi over 3 and 2 pi then the local minimum value is g of 4 pi over 3 which is in this case 2.46 and here we have a graph that describes this notation as you may see there is a local maximum at two pi over three and there is a local minimum at four pi over three next we will introduce the concept of concavity there are two types of concavities concave up and concave down in some books there may be the use of convex or convey it is really easy to mess up using convex and convey so in our video we will just use the vocabulary concave up and concave down here's the definition of the ladder if the graph of a function f lies above its tangent on the interval i then we say that the function f is concave up on i if the graph of the function f lies below its tangent line on the interval i then we say that the function f is concave down on the interval i to put it in the most simplest way possible here from the graph that you see concave down mean that the graph is open downwards and concave up means that the graph is open upwards with these definition in mind we can define the inflection point a point p on a curve y equals to f of x is called an inflection point if f is continuous there and the curve changes from concave upwards to concave downwards or from concave downwards to concave upwards so now let us consider the function f x as we have the graph in front of us as you see on the graph point b point c point d and point p are all inflection points since we are changing concavity from concave down to concave up to concave down to concave up and add e to p you change from concave up and to concave down however point e is not a point of inflection why because you're changing from concave up to concave up again meaning that it does not fall into the definition that we just talked about Second derivative test concavity is strongly related to the second derivative if a function f is defined and is twice differentiable on an open interval i then we can say if the second derivative of the function f is bigger than zero for all x's then the graph of f is concave upwards on the interval i and if the second derivative of the function f is lesser than zero for all x is in the interval i then the graph is concave downwards on the interval i also we have the second derivative test to check if a critical point is a local max or a local minimum or a saddle point suppose the second derivative of f is continuous near the point c if f prime of c is equal to zero and f double prime of c is bigger than zero then f has a local minimum at the point c if f prime of c is equal to zero and f double prime of c is lesser than zero then f has a local maximum at point c and if f prime of c is equal to zero and f double prime of c is equal to 0 then f has a saddle point at point c here is an example let us discuss the curve y equals to x to the power of 4 minus 4 x cubed with respect to concavity points of inflection and local maxima and minimum in order to use the second derivative test to discuss these we need to find the second derivative from the first derivative of this function so i take the first derivative of this function to be equal to 4x squared times x minus 3 and the second derivative is 12x times x minus 2. from the first derivative the critical points of this function is zero and three and we can say that the second derivative of zero is equal to zero and the second derivative of three is going to be bigger than zero so that the function has a saddle point at 0 and a local minimum at 3 and f double prime of 0 and f double prime of 2 is also equal to 0 so there is a point of inflection at x equals to zero and x equals two two here in this table of the second derivative of this function we can see that the function in the interval negative infinity to 0 is concave up the function on the interval 0 to 2 is concave down and the function on interval 2 to infinity is concave upwards here's the graph of this function to help you guys better visualize our results so as you may see there are two points of inflection at 0 0 and 2 and negative 16. and you can also see that there's going to be a local minimum or an absolute minimum at three and negative 27 and a saddle point on zero on zero and you can discuss the concavity which you see from negative infinity to zero is concave up from zero to three is concave down and from 3 to positive infinity is concave upwards Summary in summary of today's video we first described what an increasing and decreasing function means and what it looks like then we discussed first derivative test to find out the local maximum or a local minimum then we discussed the behavior of concavity where there would be concave up or concave down then we talked about the second derivative test to help us determine whether a critical point is a local maximum or a minimum or a saddle point that is all for our video today thank you guys for watching you
1866	https://pmc.ncbi.nlm.nih.gov/articles/PMC11134690/	Treatment outcomes of pulpotomy versus pulpectomy in vital primary molars diagnosed with symptomatic irreversible pulpitis: protocol for a non-inferiority randomised controlled trial - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. 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Learn more: PMC Disclaimer \| PMC Copyright Notice BMC Oral Health . 2024 May 28;24:626. doi: 10.1186/s12903-024-04411-6 Search in PMC Search in PubMed View in NLM Catalog Add to search Treatment outcomes of pulpotomy versus pulpectomy in vital primary molars diagnosed with symptomatic irreversible pulpitis: protocol for a non-inferiority randomised controlled trial Nebu Philip Nebu Philip 1 College of Dental Medicine, QU Health, Qatar University, Doha, Qatar Find articles by Nebu Philip 1,✉, Joe Mathew Cherian Joe Mathew Cherian 2 Department of Pedodontics and Preventive Dentistry, Christian Dental College, Ludhiana, India Find articles by Joe Mathew Cherian 2, Mebin George Mathew Mebin George Mathew 2 Department of Pedodontics and Preventive Dentistry, Christian Dental College, Ludhiana, India Find articles by Mebin George Mathew 2, Abi M Thomas Abi M Thomas 2 Department of Pedodontics and Preventive Dentistry, Christian Dental College, Ludhiana, India Find articles by Abi M Thomas 2, Sunaina Jodhka Sunaina Jodhka 3 The Dentist Multi-Speciality Clinic, Ludhiana, India Find articles by Sunaina Jodhka 3, Nino John Nino John 4 Primary Health Care Corporation, Doha, Qatar Find articles by Nino John 4, Bharat Suneja Bharat Suneja 5 The Dental Care Centre, Ludhiana, India Find articles by Bharat Suneja 5, Mandeep Duggal Mandeep Duggal 1 College of Dental Medicine, QU Health, Qatar University, Doha, Qatar Find articles by Mandeep Duggal 1 Author information Article notes Copyright and License information 1 College of Dental Medicine, QU Health, Qatar University, Doha, Qatar 2 Department of Pedodontics and Preventive Dentistry, Christian Dental College, Ludhiana, India 3 The Dentist Multi-Speciality Clinic, Ludhiana, India 4 Primary Health Care Corporation, Doha, Qatar 5 The Dental Care Centre, Ludhiana, India ✉ Corresponding author. Received 2024 Feb 4; Accepted 2024 May 24; Collection date 2024. © The Author(s) 2024 Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit The Creative Commons Public Domain Dedication waiver ( applies to the data made available in this article, unless otherwise stated in a credit line to the data. PMC Copyright notice PMCID: PMC11134690 PMID: 38807160 Abstract Background Pulpectomy continues to be the standard treatment recommendation for management of vital primary molars diagnosed with symptomatic irreversible pulpitis. The recent decade has seen a paradigm shift in the treatment concepts of how vital mature permanent molars diagnosed with irreversible pulpitis can be more conservatively managed using vital pulp therapy techniques like pulpotomy. However, despite emerging evidence indicating similarities between primary and permanent tooth pulp response to dental caries, there is limited research on whether pulpotomy can be similarly used as a definitive treatment modality for vital primary teeth with irreversible pulpitis. This randomised controlled trial (RCT) aims to compare the treatment effectiveness of pulpotomy versus pulpectomy in management of vital primary molars diagnosed with symptomatic irreversible pulpitis over a two-year period. Methods/design This clinical study is a parallel, two-armed, open label, non-inferiority RCT with a 1:1 allocation ratio between the experimental intervention arm (pulpotomy) and the active comparator arm (pulpectomy). Healthy cooperative children, between 4–9 years of age, who have painful primary molars with clinical symptoms typical of irreversible pulpitis will be recruited after obtaining informed consent from their parents/legal guardians. 50 vital primary molars clinically diagnosed with symptomatic irreversible pulpitis will be randomly distributed between the two treatment arms. The primary outcomes that will be assessed are clinical and radiographic success after six-months, one-year and two-years of the trial interventions. The influence of baseline pre-operative variables (age; gender; tooth type; site of caries; pre-operative furcal radiolucency; pre-operative pain intensity) and intra-operative factors (time taken to achieve haemostasis) on treatment outcomes will also be assessed. The secondary outcome evaluated will be the immediate (24 h and 7 d) post-operative pain relief afforded by the two treatment interventions. Discussion This trial seeks to provide evidence on whether pulpotomy treatment can be no worse than the standard pulpectomy treatment for the management of symptomatic irreversible pulpitis in vital primary molars. Trial registration ClinicalTrials.gov (NCT06183203). Registered on 30 January 2024. Keywords: Pulpectomy, Pulpotomy, Vital primary molars, Irreversible pulpitis Background/rationale Current guidelines from the American Academy of Pediatric Dentistry (AAPD) and the British Society of Paediatric Dentistry (BSPD) recommend pulpectomy as the standard treatment for vital primary molars diagnosed with irreversible pulpitis [1, 2]. Pulpectomy is a non-vital treatment procedure where the entire pulp tissue is extirpated, and the root canals debrided and shaped to receive a resorbable material to fill the canal space in the affected primary tooth. The main objectives of pulpectomy in painful primary molars are to keep them symptomless and functional, while also preserving the arch space until their exfoliation and replacement by their permanent successors . Treatment indications for the more conservative pulpotomy procedure (where only the coronal pulp tissue is removed), are restricted to vital primary teeth diagnosed with reversible pulpitis or in case of carious/mechanical pulp exposures in symptomless vital primary teeth . Traditionally, the clinical diagnosis of irreversible pulpitis in a deeply carious primary molar was based on patient reported symptoms of spontaneous lingering pain or pain that persists even after the removal of any provoking stimuli. Radiographic examination of such a tooth would reveal carious exposure of the pulp with or without furcal rarefaction. As conventional pulp sensibility tests are not very reliable in primary teeth, the vitality of a primary tooth diagnosed with irreversible pulpitis is usually ascertained intra-operatively after deroofing the pulp chamber, based on the colour and volume of the observed pulp tissue . However, even if the pulp of such a tooth is judged to be vital, radicular pulp is still extirpated based on the clinical diagnosis of irreversible pulpitis. This treatment approach is adopted due to the traditional notion of the poor healing capacity of primary dental pulp . Early histological studies seemed to indicate that primary teeth with a history of spontaneous pain are often associated with extensive degenerative changes extending into the radicular pulp and are poor candidates for vital pulpotomy techniques [6–8]. Another long-established rationale for performing pulpectomy in symptomatic vital primary teeth was the suggestion that primary tooth pulps exhibit a more pronounced and widespread inflammatory reaction to dental caries compared to permanent tooth pulps and hence would be poor candidates for pulpotomy . The 21st century has seen an improved understanding of pulp biology, pulpal inflammatory processes, and the potential for pulp healing and repair. Histopathologic and histobacteriologic studies in cariously exposed “irreversibly” inflamed permanent tooth pulps have found healthy pulpal architecture, free from inflammation and bacteria, few millimetres away from the bacterially colonised necrotic tissue in the pulp chamber [10–12]. If the infected coronal pulp is completely removed, a favourable environment can be created for radicular pulpal healing as the immunoinflammatory cells get eliminated by apoptosis and the odontoblast-like cells induce dentine bridge formation. This contemporary understanding of pulp pathophysiology, along with the introduction of bioactive calcium silicate cements, has revolutionised treatment modalities for the management of irreversible pulpitis in mature permanent teeth, even when they are associated with apical periodontitis . There is now abundant evidence from randomised controlled trials (RCTs) and systematic reviews favouring conservative vital pulp therapy in permanent teeth diagnosed with irreversible pulpitis [14–23]. A corresponding paradigm shift in the treatment approaches to “irreversibly” inflamed vital primary pulps has not been seen despite new evidence emerging regarding primary pulp biology. Immunocytochemical and vascular studies have now established that primary and permanent pulps have similar vascularity and showed a comparable degree of vasodilation and angiogenesis in response to the caries insult [24, 25]. The pulpal tissues in both carious primary and permanent teeth also showed similar neural changes when mounting a defense to deep caries . Furthermore, although primary pulp tissue contains more immune cells in both normal and carious states, these immune cells appear to localize in a manner identical to that seen in permanent teeth pulp tissue . These histological findings suggest the need to re-evaluate treatment approaches to primary dentition pulp therapy similar to what is currently underway within adult endodontics . This is especially important as literature reports also suggest that clinical symptoms, radiographic findings, and conventional pulp sensibility tests do not provide accurate information about pulp status in primary teeth [27, 28]. There are several advantages if pulpotomy can be offered as a substitute to the standard pulpectomy treatment for the management of irreversible pulpitis in vital primary molars. The ribbon-like torturous root canal anatomy in primary molars makes pulpectomy a difficult procedure for general dentists to perform, often requiring a child in pain to be referred to a specialist paediatric dentist for definitive treatment. In many situations, specialist care may not be easily accessible, or parents may not be able to afford its costs. Consequently, some parents may opt to extract the painful primary molar tooth of their child, resulting in complications such as the loss of function and arch integrity. The technically simpler pulpotomy procedure is not only less challenging for general dentists but is also potentially less time-consuming and easier for young patients to tolerate, both very important advantages in the dental treatment of children. Moreover, the pulpotomized primary tooth retains the regenerative/repair potential of the remnant radicular pulp and the proprioceptive sensation of the tooth. Finally, treatment complications sometimes associated with pulpectomy (e.g., extrusion of root filling material beyond the primary root apex) can be avoided if pulpotomy is the treatment of choice. There is limited research on whether pulpotomy can be offered as an alternate treatment to pulpectomy in vital primary molars diagnosed with irreversible pulpitis. A prospective cohort study of pulpotomy in 50 vital primary teeth with irreversible pulpitis reported 93% clinical success and 90% radiographic success after one year of treatment . A more recent retrospective study found higher clinical success rates for pulpotomy (99%) than pulpectomy (88%) in primary molars with carious pulp exposures or symptomatic irreversible pulpitis over an 18-month period . To the best of our knowledge, there have been no RCTs comparing treatment outcomes of pulpotomy vs. pulpectomy for management of irreversible pulpitis in vital primary molars. This study aims to fill this research gap and could be of great importance in initiating a paradigm shift in the treatment of vital primary molars diagnosed with irreversible pulpitis. The objectives and hypothesis of the study are detailed below. Objectives Compare the treatment effectiveness of full pulpotomy vs. single-visit pulpectomy in the management of vital primary molars diagnosed with symptomatic irreversible pulpitis based on the clinical and radiographic outcomes of the treated teeth over a two-year period. Assess influence of baseline pre-operative variables (age; gender; tooth type; site of caries; pre-operative furcal radiolucency; pre-operative pain intensity) and intra-operative factors (time taken to achieve haemostasis) on treatment outcomes. Hypothesis The null hypothesis of the study is that the treatment outcomes of the experimental pulpotomy intervention will not be inferior to the highly successful outcomes of the standard pulpectomy treatment for vital primary molars diagnosed with irreversible pulpitis. Methods Trial design This clinical study is a parallel, two-armed, open label, non-inferiority randomised controlled trial with a 1:1 allocation ratio between the experimental intervention arm (pulpotomy) and the active comparator arm (pulpectomy). Trial setting This clinical trial will be conducted in the Paediatric Dentistry Department of Christian Dental College Ludhiana, India. Trial registration and protocol As mandated by the International Committee of Medical Journal Editors (ICMJE) and jurisdictional legislation for conducting clinical trials in India, this trial has been prospectively registered on ClinicalTrials.gov and the Central Trial Registry—India (CTRI). Ethics approval has been obtained from the Institutional Ethics Committee of Christian Medical College Ludhiana (IEC Approval Number: 23-12-517/CDC) This clinical trial protocol follows the SPIRIT guidelines (Standard Protocol Items: Recommendations for Interventional Trials). Any protocol changes required in the future will be carried out on the ClinicalTrials.gov. Study participants: eligibility criteria Healthy (ASA I and II) co-operative children (Frankl Scale + and + +) between the ages of 4 to 9 years will be invited to participate in the study if they have vital primary molars that meet the following inclusion criteria: (i) symptoms typical of irreversible pulpitis i.e., spontaneous unprovoked pain lasting few seconds to several hours in the days before the dental visit or pain that is exacerbated by hot and/or cold stimuli with the pain lingering even after removal of stimuli; (ii) post-deroofing the pulp chamber, vitality of the primary molar tooth is confirmed by visual inspection of pulpal haemorrhage (uniformly reddish pink vascular tissue indicates healthy pulp); (iii) post-coronal pulp amputation, radicular pulp health is confirmed by attainment of radicular pulp haemostasis within 8-min of compression with a 5% sodium hypochlorite (NaOCl)-dampened cotton pellet; (iv) the primary molar is restorable with stainless steel crown; and iv) any physiologic root resorption, if present, is less than 1/3 the normal root length. The exclusion criteria include: (i) clinical examination reveals signs of pulpal infection i.e., pathologic tooth mobility, parulis/fistula, or soft tissue swelling; (ii) pre-operative periapical radiograph suggests presence of furcal radiolucency more than ½ the furcation to periapical area; (iii) pre-operative periapical radiograph suggests presence of periapical radiolucency; (iv) pre-operative periapical radiograph suggests presence of pathological root resorption (v) post-deroofing the pulp chamber, visual examination of pulp tissue reveals signs of necrosis i.e., avascular/minimally bleeding pulp tissue or yellowish necrotic areas/purulent exudate; (vi) post-coronal pulp amputation, there are signs of extensive radicular pulp inflammation i.e., bleeding continues even after 8-min compression with NaOCl-soaked cotton pellet; and (vii) parents not willing to place full coverage crowns post-treatment; and (vii) if both primary molars in the quadrant are painful and clinical diagnosis of irreversible pulpitis between the teeth is not sharply defined. Trial interventions The proposed trial interventions, pulpotomy and pulpectomy, will adhere to standard clinical procedural guidelines. Treatment group allocation of the consenting trial participants to one of the two intervention groups will be carried out only after intra-operative confirmation of pulp vitality and achievement of radicular pulp haemostasis. The choice of the treatment group (pulpotomy or pulpectomy) will be revealed to the operating clinician by an independent allocator only after coronal pulp amputation and confirmation of pulp vitality and haemostasis. Participants with pulp assessed to be non-vital or with uncontrolled bleeding from the pulp stumps will be excluded prior to allocation. Standard clinical protocol for the pulpotomy group: Detailed history of symptoms will be verified and a thorough clinical examination shall be performed. Pre-operative pain intensity will be recorded using the validated five-face visual analogue scale (VAS) commonly used for pain intensity measurement in children (Fig.1). After giving explanations of the VAS, children will be requested to select one of the faces that best reflect their pain intensity. A standard pre-operative periapical radiograph will be taken using a paralleling technique and a film holder device. Care will be taken to ensure that the radiographic image extends beyond the root tip of the primary molar tooth and there is no evidence of distortion, overlap, or processing errors. Local anaesthesia will be achieved with 2% lidocaine or 4% articaine containing 1:100,000 epinephrine and the tooth will be isolated under rubber dam. To minimize further bacterial contamination of the pulp, carious tissues will be progressively removed, starting at the periphery of the cavity and then over the pulp chamber roof. Once pulp chamber deroofing is complete, a fresh sterile bur will be used to remove all coronal pulp tissue to the level of the root canal orifices, under copious water irrigation. During this step, intra-operative assessment of pulp vitality will also be carried out. Healthy vital pulp will present as uniformly reddish pink vascular tissue, while non-vital necrotic pulp will present as dark avascular tissue with minimal bleeding or as yellowish liquefied areas. If pulp is judged to be necrotic, the tooth will be excluded from the study. Further treatment of such excluded teeth will be delivered following local management protocols. Haemostasis and disinfection of the radicular pulp tissue will be carried out by compressing a 5% NaOCl-soaked sterile cotton pellet over the pulp stumps for up to 8-min. Up to three attempts will be made to control pulp stump bleeding using moderate pressure applied with the NaOCl-dampened cotton pellet. Pulpal haemostasis will be assessed at three timepoints (3-, 6-, and 8-min) and the pulpotomy medicament placed over the pulp stumps as soon as haemostasis is achieved. If bleeding persists beyond 8-min, the radicular pulp shall be considered irreversibly inflamed, and the tooth will be excluded from the study. Further treatment of such excluded teeth will be delivered following local management protocols. Once haemostasis is achieved, 2 mm of a pre-mixed mineral trioxide aggregate (MTA) (NeoPutty, NuSmile, Houston, TX, U.S.A) medicament will be directly adapted over the pulp stumps ensuring that there is no porosity or any excess cement on the pulp chamber walls. The access cavity will be restored with a bulk-fill high-strength glass ionomer cement (3M ESPE Ketac Molar, Seefeld, Germany). Post-operative radiograph will be taken to confirm fidelity to the pulpotomy procedure. The pulpotomy-treated tooth will be prepared for receiving a full coverage stainless steel crown (SSC) one week after the pulpotomy procedure. Fig.1. Open in a new tab Visual analogue scale for pain assessment in children Standard clinical protocol for the pulpectomy group: The initial seven procedural steps of the pulpotomy treatment group shall be followed in the pulpectomy treatment group too. Once vitality is confirmed and haemostasis achieved, the entire radicular pulp will be extirpated, and the root canals prepared as described below: Radicular pulp will be extirpated from all the root canals using appropriate endodontic instruments. Maintaining a working length 1–1.5 mm short of the radiographic apex, chemo-mechanical preparation of the canals will be done using a series of 21 mm long Kerr-type endodontic files up to file no. 30 or 35 (depending on the initial size of the canal). Regular irrigation of the prepared canals will be carried out using 2.5% NaOCl alternating with saline. Sterile absorbent paper points will be used to dry the root canals prior to obturation with root filling material that contains zinc oxide-eugenol, Ca(OH)2, and iodoform (Endoflas; Sanlor Laboratories, Cali, Colombia). The Endoflas material that is available in a powder-liquid form will be mixed into a medium consistency such that it forms a 1 cm string when lifted from the glass slab with a cement spatula. A reamer coated with the mixed Endoflas will be used to insert the material into the prepared root canals and pressure applied with a sterile cotton pellet to ensure optimal canal obturation. Access cavity will be restored with a bulk-fill high-strength glass ionomer cement (3M ESPE Ketac Molar, Seefeld, Germany). Post-operative radiograph will be taken to confirm fidelity to the pulpectomy procedure. The pulpectomy-treated tooth will be prepared for receiving a full coverage SSC one week after the pulpectomy. Post-operative instructions for both pulpotomy and pulpectomy treatment groups will advise patients to take analgesics if necessary and to return if the pain is intolerable. Follow-up timeline Children enrolled in the study will be followed from the trial intervention until the end of the study (two years post-treatment). Trial participants and their parents/guardians will be initially contacted by phone 24 h after the treatment intervention to record their post-operative pain intensity score using the VAS. Post-operative pain intensity score will also be recorded 7 d after the treatment intervention before the placement of the SSC. Clinical and radiographic outcome data of the treated tooth will be recorded at 6 months, 12 months, and 24 months post-treatment. Children participating in the trial will be advised to continue their routine dental visits, with the recall period determined by their caries risk status. Any symptomatic visits to the dentist outside these follow-up timepoints will be identified and reason for the same recorded on the trial record form. Trial outcomes Primary outcomes Clinical outcomes will be assessed by a single experienced paediatric dentist who shall be blinded to the treatment allocation. Radiographic outcomes will be assessed independently by two experienced paediatric dentists under optimum viewing conditions. Clinical success of the treated tooth will be determined if all the following conditions are met: Absence of pain or discomfort Absence of tenderness on percussion and palpation Absence of any associated parulis/fistula Absence of any associated soft tissue swelling Absence of any associated pathological mobility Radiographic success will be determined as follows: Absence of pathosis on recall periapical radiograph i.e., no signs of pathological internal/external root resorption or new furcal/periapical lesions. Complete radiographic healing or reduction/no change in size of the any pre-operative furcal rarefaction. Teeth that show an increase in the size of any pre-operative furcal radiolucency will be considered as failed. Secondary outcomes Immediate post-treatment pain relief: post-operative pain felt by the child in the immediate aftermath of the trial interventions will recorded at two time-points (24 h and 7 d) using the VAS. The pain scores recorded will be used to evaluate pain reduction afforded by the two trial interventions. Target sample size Sample size determination for this binary outcome non-inferiority randomised trial was calculated using a computer-generated system available online ( Based on previous clinical studies [29–31], a 95% success rate was assumed for both treatment interventions along with a 17% non-inferiority margin. Even an overall success rate of 78% for the experimental pulpotomy intervention would be considered non-inferior to the standard pulpectomy intervention. To obtain a power of 80% (β = 0.20) and a 2-sided α equal to 0.05, approximately 42 patients for both groups will be needed. To compensate for loss during follow-up and other causes of attrition, the targeted sample size will be 50 vital primary molar teeth diagnosed with irreversible pulpitis, distributed equally between the two trial treatment arms. Recruitment All children in the 4 to 9-year age group who report to the Paediatric Dentistry Departments of Christian Dental College with symptoms typical of irreversible pulpitis in their primary molars will be considered for recruitment. Participant recruitment will be based on the defined inclusion and exclusion criteria. Recruitment is planned to start in February 2024 and expected to last for up to 8–10 months. No financial or non-financial incentives will be provided to trial investigators or participants for enrolment. A trial schematic outlining the screening, recruitment, randomisation, allocation, follow-up time-points, and planned data analysis is shown in Fig.2. Fig.2. Open in a new tab Trial schematic of the proposed study Randomisation and sequence generation The randomization list will be generated before participant enrolment. Permuted block randomisation will be used to randomly assign one of the two treatment interventions to consenting trial participants who meet all the inclusion/exclusion criteria. Block lengths will vary according to a sequence generated by an external researcher. The on-site trial investigators will be blinded to the block sizes and the randomization sequence. Allocation concealment The generated randomization sequence will be distributed in sequentially numbered opaque sealed envelopes (SNOSE). An independent allocator, not involved in the assessment and recruitment of the patient, will open the sequentially numbered envelope, and inform the clinical investigator which treatment group the patient is assigned to. Patient details will be written on the outside of the envelopes (that contain carbon paper overlying the treatment allocation paper), so that patient names are transferred to the allocation paper before the envelope is opened and can be audited at the end of the trial. Blinding Blinding of trial participants or clinical investigators will not be possible due to evident differences between the two trial interventions. However, post-operative clinical outcomes will be assessed by an experienced paediatric dentist who is blinded to the treatment allocation, while the radiographic outcomes will be independently assessed by two different paediatric dentists. An external researcher, not directly involved in the clinical part of the study, will generate the randomization sequence, provide the SNOSE for treatment group allocation, and perform the statistical analysis of the de-identified data. Data collection, management, and confidentiality Relevant demographic, clinical, radiographic, and pain intensity data will be collected at baseline and subsequently at the follow-up time points described earlier. All data collected will be recorded electronically at the trial site using unambiguous, standardized terminology and abbreviations to avoid misinterpretation. Checks will be applied at the time of data entry before data is committed to the electronic database to ensure accuracy of entered trial data. Any later modifications to data entered in the database will be documented. All participant data will be de-identified and stored on an institutional password protected computer. Data will only be accessible to authorized members of the research team. Data will be treated confidentially and stored for 3 years before being destroyed. Statistical methods All analyses will use the ‘intention to treat’ principle. Baseline demographic and clinical data will be summarised for each treatment group. Binary logistic regression models will be applied to analyse all outcome measures stated in the protocol adjusting for covariates thought to be of prognostic importance. Appropriate parametric or non-parametric tests will be used to determine any significant relationship between treatment outcomes and variables in the study (age, gender, tooth/caries type, and pre-operative furcal rarefaction/pain intensity). For all statistical tests, we will use 2-sided p-values with α ≤ 0.05 and the treatment effect estimates will be presented with 95% confidence intervals. Up-to-date versions of SPSS (Chicago, IL, USA) will be used to conduct the statistical analysis. Expected risks Whilst performing pulpotomy in vital primary molars diagnosed with irreversible pulpitis is a novel treatment, it is more conservative than the standard pulpectomy treatment. Pulpotomy can be considered as the first stage of pulpectomy and an effective emergency measure to manage pain associated with an infected pulp. Thus, we do not anticipate any significant safety concerns with the treatment interventions of this trial. The clinicians taking part in the trial are fully trained in the pulpotomy and pulpectomy techniques and all subjects enrolled in the trial will receive the usual standard of care treatment during and following the trial intervention. The trial site principal investigators (PIs) will record adverse events (AEs) and serious adverse events (SAEs) only if they have a reasonable casual relationship with the treatment interventions of the study. Rare but potential AEs and SAEs include perforation and hypochlorite injury. Patients who continue to experience pain following trial interventions will be able to access emergency care to address their symptoms. Trial site PIs will report AEs/SAEs to the trial co-ordinator who will escalate them as deemed necessary. AE reporting period for this trial begins upon enrollment and ends at the two-year follow-up period for participants. Ethical considerations Participants and their guardians will be invited to enrol in the trial voluntarily and will be assured that should they chose not to participate the standard treatment will be offered as normal with no prejudice. Study participants will be recruited to the trial only after obtaining the child’s verbal assent and their parents or guardians sign the Informed Consent Form containing detailed information about the study. Patients and their guardians will have the opportunity to ask questions and take any additional time they may need prior to their decision to participate. Participants and their guardians will be given the option of withdrawing from the study should they choose to do so at any time in the future. Dissemination Study results will be reported according to the Consolidated Standards Of Reporting Trials (CONSORT) guidelines through conference presentations and scientific articles in peer-reviewed journals. Authorship of the publications emerging from this study will be determined based on the ICMJE guidelines. Any significant amendments to this study protocol will be reported when the study results are disseminated. Discussion The proposed study is unique in that for the first time a non-inferiority RCT is evaluating whether the experimental pulpotomy treatment can prove to be no worse than the standard pulpectomy treatment for the management of symptomatic irreversible pulpitis in vital primary molars. Considering that pulpotomy has been shown to have highly successful treatment outcomes in vital mature permanent molars diagnosed with irreversible pulpitis, there exists the possibility of similar successful outcomes in primary molars too. Consequently, this trial has the potential to produce new knowledge and add information to an area that is currently under researched. It is important to highlight that the root filling material proposed to be used in the pulpectomy intervention group (Endoflas) has among the highest success rates reported for primary root canal filling materials [31–35]. Similarly, the new generation MTA product proposed to be used in the pulpotomy intervention group not only retains the biocompatibility, immunomodulatory, osteogenic, and sealing properties of traditional MTA, but being a resin-free pre-mixed bioceramic has superior handling characteristics, faster washout and setting times [36, 37]. The NaOCl proposed to be used as the haemostatic/disinfection agent in the pulpotomy intervention group demonstrated comparable clinical/radiographic success to traditional pulpotomy medicaments like formocresol and ferric sulphate without their drawbacks [38–40]. The two-year follow-up period proposed in this study should be adequate to assess treatment success of the trial interventions in primary molar teeth. Studies with longer follow-up periods can be planned based on the initial results of this study. If the hypothesis of the proposed study is confirmed, it will support the clinical application of more conservative treatment approaches to vital primary teeth with irreversible pulpitis. Such an approach has the potential to cause less pain and discomfort to paediatric patients, have a shorter chair-time, and lower the technical difficulty and cost of managing complicated pulp conditions in vital primary molars. Acknowledgements Not applicable. Abbreviations AAPD American Academy of Pediatric Dentistry AE Adverse Event ASA American Society of Anesthesiology BSPD British Society of Paediatric Dentistry CONSORT Consolidated Standards Of Reporting Trials CTRI Central Trial Registry India GIC Glass Ionomer Cement ICMJE International Committee of Medical Journal Editors MTA Mineral Trioxide Aggregate RCT Randomised Controlled Trials SAE Serious Adverse Event SNOSE Sequentially Numbered Opaque Sealed Envelopes SPIRIT Standard Protocol Items: Recommendations for Interventional Trials VAS Visual Analogue Scale Authors’ contributions NP conceived and designed the research study. JMC is the trial co-ordinator, MGM and AMT are the principal investigators. JMC, MGM, SJ, and BS undertook the ethics applications. JMC and MGM are responsible for patient recruitment and treatment. AMT, SJ, and BS are responsible for patient outcome evaluation over time. SJ, BS, NJ and MD provided intellectual input on the study design, statistical analysis, and trial outcomes. NP, NJ, and MD drafted the manuscript. All authors were involved in critically revising the manuscript and approved the final manuscript as submitted. Funding Open Access funding was provided by Qatar National Library. Availability of data and materials Deidentified data sets that will be generated and/or analysed in the current study shall be made available from the corresponding author upon reasonable request. A summary of final findings, in layman’s language, will be made available to the trial participants. Declarations Ethics approval and consent to participate Human research ethics approval was obtained for the study from the Biomedical and Health Institutional Ethics Committee of Christian Medical College Ludhiana, India (IEC Approval Number: 23-12-517/CDC). 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Data Availability Statement Deidentified data sets that will be generated and/or analysed in the current study shall be made available from the corresponding author upon reasonable request. A summary of final findings, in layman’s language, will be made available to the trial participants. Articles from BMC Oral Health are provided here courtesy of BMC ACTIONS View on publisher site PDF (1.6 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Background/rationale Methods Discussion Acknowledgements Abbreviations Authors’ contributions Funding Availability of data and materials Declarations Footnotes References Associated Data Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
1867	https://link.springer.com/chapter/10.1007/978-3-642-41498-5_6	Advertisement Uniform Polyhedrals Part of the book series: Bolyai Society Mathematical Studies ((BSMS,volume 24)) 1116 Accesses Abstract Half a century ago H. S. M. Coxeter, M. S. Longuet-Higgins and J. C. P. Miller published a very influential paper on “Uniform Polyhedra” . These are finite polyhedra with regular polygons as faces, and with vertices in a single orbit under symmetries. Uniform polyhedrals are defined by the same conditions, but with finite replaced by locally finite, and the additional explicit requirement that there are no coinciding elements (vertices, edges or faces); this was self-understood in . Coplanar faces, collinear edges, and partial overlaps are allowed for uniform polyhedrals, as they are for uniform polyhedra. It is somewhat surprising that no systematic study of infinite uniform polyhedrals has been undertaken so far. There are three distinct classes of such polyhedrals — rods, slabs, and sponges. The beginnings of their investigation form the core of this article, and many open problems become evident. Illustrations serve to shorten the explanations, but also to highlight the difficulty of presenting polyhedrals graphically. Applications of such polyhedrals and their relatives in fields such as architecture, biology, engineering, and others are discussed as well, as are the shortcomings of the mathematical reviewing journals in reporting these and related applications of geometry. This survey is meant to honor Laszlo Fejes Tóth, who for many years was one of few proponents of visual geometry This is a preview of subscription content, log in via an institution to check access. Access this chapter Subscribe and save Buy Now Tax calculation will be finalised at checkout Purchases are for personal use only Institutional subscriptions Preview Unable to display preview. Download preview PDF. Unable to display preview. Download preview PDF. Similar content being viewed by others An Exploration on the Form Design of Movable Structures Based on Uniform Convex Polyhedral Expansion On the Composite ( RR )-Polyhedra of the Second Type Rigidity of Nonconvex Polyhedra with Respect to Edge Lengths and Dihedral Angles Explore related subjects References Barker, R. J. P. and Guest, S. D., Inflatable triangulated cylinders, in: IUTAMIASS Symposium on Deployable Structures: Theory and Applications, S. Pellegrino and S. D. Guest, eds. Kluver, Dordrecht 2000, pp. 17–26. Chapter Google Scholar Badoureau, A., Mémoire sur les figures isoscèles, J. École Polytechn, 30 (1881), 47–172. Google Scholar Boerdijk, A. H., Some remarks concerning close-packing of equal spheres, Philips Research Reports, 7 (1952), 303–313. MATH MathSciNet Google Scholar Coxeter, H. S. M., Regular skew polyhedra in three and four dimensions, and their topological analogues, Proc. London Math. Soc. 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Meeting of Origami Science and Scientific Origami, Otsu, Japan, 1994 (ed.: K. Miura), Seian Univ. of Art and Design, 1997, 83–91. Google Scholar Thompson, D’A. W., On Growth and Form, Cambridge Univ. Press, 1948. Google Scholar Turing, A. M., The chemical basis of morphogenesis, Philos. Trans. Roy. Soc. London, Ser. B, 237 (1952), 37–72. Reprinted, together with previously unpublished manuscripts, in Morphogenesis, Collected Works of A. M. Turing, P. T. Saunders, ed. North-Holland, Amsterdam, 1992. Article Google Scholar Turing, S., Alan M. Turing, W. Heffer & Sons, Cambridge, 1959. Google Scholar van Iterson, G., Jun., Mathematische und mikroskopisch-anatomische Studien über Blattstellungen, nebst Betrachtungen über den Schalenbau der Miliolinen, Gustav Fischer, Jena, 1907. Chapter Google Scholar Wachman, A., Burt, M. and Kleinmann, M., Infinite Polyhedra, Technion, Haifa 1974; reprinted 2005. Google Scholar Weisstein, E. W., “Uniform Polyhedron.” From MathWorld — A Wolfram Web Resource. Google Scholar Wikipedia, Expanded list of uniform tilings. Google Scholar You, Z. and Pelegrino, S., Cable-stiffened pantographic deployable structures. I. Triangular mast, AIAA Journal, vol. 34, no. 4 (1996), 813–820. Article Google Scholar Download references Author information Authors and Affiliations Department of Mathematics, University of Washington, Box 354350, Seattle, WA, 98195-4350, USA Branko Grünbaum Search author on:PubMed Google Scholar Editor information Editors and Affiliations Alfréd Rényi Institute of Mathematics, Hungarian Academy of Sciences, Reáltanoda u. 13-15, Budapest, 1053, Hungary Imre Bárány , Károly J. Böröczky , Gábor Fejes Tóth & János Pach , , & Rights and permissions Reprints and permissions Copyright information © 2013 János Bolyai Mathematical Society and Springer-Verlag About this chapter Cite this chapter Grünbaum, B. (2013). Uniform Polyhedrals. 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1868	https://math.stackexchange.com/questions/1848398/sum-of-squares-using-gaussian-integers	Skip to main content Sum of squares using Gaussian integers Ask Question Asked Modified 9 years, 1 month ago Viewed 3k times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. Using Gaussian integers a+ib (a,b∈Z), one can prove that a prime p∈Z is sum of two squares in Z if and only if p≡1(mod4). Question: Using Gaussian integers, is there any result which says about what integers can be expressed as sum of two (non-zero) squares in Z? I do not know how factorization of m∈Z into product of primes will help for this, since 2 is sum of two squares, but the product 2.2 is not. 4 is not sum of two squares, 2 is sum of two squares, but 4.2 is sum of two squares. number-theory algebraic-number-theory Share CC BY-SA 3.0 Follow this question to receive notifications edited Jul 4, 2016 at 6:35 p Groups asked Jul 4, 2016 at 6:09 p Groupsp Groups 10.5k2323 silver badges5656 bronze badges 7 1 Perhaps, we have to use multiplicative property of norm: if N(a+ib):=a2+b2 then N((a+ib)(c+id))=N(a+ib)N(c+id). – p Groups Commented Jul 4, 2016 at 6:12 So zero is not a square for you? – Claudius Commented Jul 4, 2016 at 6:34 yes, thanks for informing it. I will edit question. – p Groups Commented Jul 4, 2016 at 6:35 1 An integer is a sum of two squares if and only if every prime of the form 4k+3 in its prime factorization occurs as an even power. This is the famous Fermat's theorem. – user348749 Commented Jul 4, 2016 at 6:50 2 If n=2a∏pi≡−1(mod4)pbii∏pj≡1(mod4)pcjj, then the first factor is always a sum of two squares, but a sum of two non-zero squares iff a is odd. The second factor is a sum of two squares iff it is a square, and never a sum of two non-zero squares. The last factor is always a sum of two non-zero squares. I think you can write n as a sum of two non-zero squares, iff all the three factors are sums of two squares, and at least one of them is a sum of two non-zero squares. Should follow from the UFD property of Z[i] (and multiplicativity of norm). – Jyrki Lahtonen Commented Jul 4, 2016 at 6:51 \| Show 2 more comments 2 Answers 2 Reset to default This answer is useful 3 Save this answer. Show activity on this post. Your definition of "0 is not a square" is confusing you. With this problem, it turns out to be much easier to allow 0 to be a square. You can then introduce your "0 is not a square" condition at the end. A number is a sum of two non-zero squares if (a) it is a sum of two squares and (b) it is not itself a square. Once you have allowed yourself to think like this, the answer will probably become clear to you. A sketch goes something like this. Let us deal with square-free n only, because every number N can be expressed as N=k2n, where k≥0 is an integer and n≥1 has no repeated prime factors. Your "two non-zero squares" condition amounts to insisting that n>1. If n is a product of primes which equal 1 modulo 4, it is a product of norms of Gaussian integers and therefore the norm of a product of Gaussian integers and therefore the norm of a Gaussian integer and therefore the sum of two squares. If n has prime factors which equal 3 modulo 4, it has prime factors which are not norms of Gaussian integers. Could the product of such prime factors be the norm of a Gaussian integer? No. For example, if 3×7 were the norm of a Gaussian integer, then it would be factorisable as (a+bi)(a−bi) and also as 3×7. But the Gaussian integers are a unique factorisation domain, so this is impossible. Hence a necessary and sufficient condition for square-free n to be expressible as a sum of squares is that it should have no prime factors which equal 3 modulo 4. When you are putting back the squares, translating from n language to N language, you will see that your "sum of non-zero squares" condition is equivalent to "n≥1". Share CC BY-SA 3.0 Follow this answer to receive notifications answered Jul 4, 2016 at 7:19 Martin KochanskiMartin Kochanski 3,8621111 silver badges1919 bronze badges 1 1 N=25=42+32 is a sum of two non-zero squares even though here k=5,n=1. You also need to allow n=1 when k has a prime factor ≡1(mod4). Otherwise I like the idea of factoring out a square. – Jyrki Lahtonen Commented Jul 4, 2016 at 7:40 Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. Assume that the prime factorization of a natural number n is n=2a∏i,pi≡−1(mod4)psii∏j,pj≡1(mod4)prjj.(∗) We need the multiplicativity of the norm N(u+vi)=(u+vi)(u−vi)=u2+v2 and the known splitting behavior of rational primes in Z[i]. Namely: 2 ramifies and is up to a unit factor equal to (1+i)2=2i, where 1+i is a prime of Z[i]. If p≡−1(mod4), then p is also a prime of Z[i]. If p≡1(mod4), then p factors as p=(a+bi)(a−bi) for some two primes a±bi of Z[i]. Here ab≠0. Let's look at the three factors on the r.h.s. of (∗), call them n1,n2,n3, individually. If n1=2a and n=u2+v2, then N(u+vi)=2a. Unique factorization in Z[i] implies that u+vi=ik(1+i)a. We easily see that if a is even, then either u=0 or v=0. But if a is odd, then \|u\|=\|v\|. The conclusion is that 2a is always a sum of two squares, but a sum of two non-zero squares iff a is odd. If n2=∏i,pi≡−1(mod4)psii it is well known that n2 is a sum of two squares iff it is a square (i.e. if all the exponents si are even). Furthermore, n2 cannot be a sum of two non-zero squares in this case. If n3=∏j,pj≡1(mod4)prjj. Then each of the primes pj is a sum of two squares: pj=(aj+ibj)(aj−ibj)=a2j+b2j, ajbj≠0. Consider the product u+iv=∏j(aj+ibj)rj. I claim that if n3> both u and v are non-zero. This follows from the UFD-property. For if uv=0 then u+iv and u−iv are associates in Z[i], i.e. their ratio is a power of i. But, u−iv=∏j(aj−ibj)rj, and for all j the primes aj+ibj are aj−ibj are non-associates, so uv=0,n3>1 violates uniqueness of factorization of the Gaussian integer u+iv. The conclusion is that if n3>1 it is a sum of two non-zero integers. We need to put these pieces together. Fermat's theorem (see the above link, but we actually got that as a by-product) states that n is a sum of two squares, iff n2 is a square. I claim that n is a sum of two non-zero squares, iff n2 is a square and either n1 or n3 is a sum of two non-zero squares. If n1 or n3 (or both) is a sum of two non-zero squares, then using the multiplicativity of the norm gives rise to a presentation of n as a sum of two non-zero squares. If n1=N(a1+ib1), n2=a22 and n3=N(a3+ib3), then then product u+iv=(a1+ib1)a2(a3+ib3) cannot be real or pure imaginary by the argument from item 3 above, and N(u+iv)=n. OTOH, if n=N(u+iv) with uv≠0, then the prime factorization of u+iv must either contain a factor of n3>1 or an odd power of (1+i). Summary: n is a sum of two non-zero squares, iff all the exponents si are even, and in addition either a is odd, or at least one of the exponents rj>0. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Jul 4, 2016 at 7:46 answered Jul 4, 2016 at 7:36 Jyrki LahtonenJyrki Lahtonen 142k3030 gold badges304304 silver badges738738 bronze badges Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions number-theory algebraic-number-theory See similar questions with these tags. Featured on Meta Community help needed to clean up goo.gl links (by August 25) Linked 5 p=4n+3 never has a Decomposition into 2 Squares, right? 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1869	https://math.stackexchange.com/questions/2681361/discriminant-of-a-polynomial-of-arbitrary-degree-and-positivity	Discriminant of a polynomial of arbitrary degree and positivity - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Discriminant of a polynomial of arbitrary degree and positivity Ask Question Asked 7 years, 6 months ago Modified7 years, 6 months ago Viewed 485 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Maybe a trivial question: Is there any relation between the discriminant of a polynomial p(x)=a n x n+…+a 0 p(x)=a n x n+…+a 0 with an arbitrary degree n n and the positivity of it for all x x, just like quadratic case? polynomials Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Mar 7, 2018 at 20:20 user344662 user344662 1 @Ihf So there is no general statement (for arbitrary n n) about this problem?user344662 –user344662 2018-03-07 21:06:55 +00:00 Commented Mar 7, 2018 at 21:06 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. This answer turned out somewhat long due to the derivation of the fact that the discriminant has sign (−1)s(−1)s. The point is that the sign of the discriminant and the congruence of n n modulo 4 4 may sometimes be sufficient to decide whether the polynomial has any real roots. For a monic polynomial of degree n n with roots α 1,…,α n α 1,…,α n (counting multiplicities), the discriminant is given by Δ=n∏i<j(α i−α j)2 Δ=∏i<j n(α i−α j)2 If the polynomial has coefficients in R then all non-real roots come in complex conjugate pairs. Write n=r+2 s where r is the number of real roots and s is the number of complex conjugate pairs. Denote the real roots by x 1,…,x r and the complex roots by y 1±i z 1,…,y s±i z s. Any term of the form (x i−x j)2 where i<j is a square of a real number and hence is positive. Whenever a term of the form (x i−y j−i z j)2 occurs, there will also be a term of the form (x i−y j+i z j)2 (its complex conjugate) and the combined contribution to the product is ((x i−y j)2+z 2 j)2 which is again a positive number. Terms of the form (y i±i z i−y j±i z j)2 where i≠j will also occur in complex conjugate pairs whose combined contribution to the product is again positive. So we are left to consider terms of the form (y i+i z i−y i+i z i)2 which are their own complex conjugates. These give a contribution of −4 z 2 i to the product. There are s such terms (for each pair of non-real roots). We conclude that sgn Δ=(−1)s. Therefore the sign of the discriminant is (−1)s where n=r+2 s and r is the number of real roots, s is the number of pairs of complex conjugate roots. To get back to the question, the positivity of p(x) means that r=0. The sign of the discriminant tells us whether s is even or odd. If Δ>0 then s is even. If Δ<0 then s is odd. If Δ=0 then the polynomial has repeated roots and we cannot decide using this method. Knowing the degree n and the parity of s, we do not get a lot of information about r. But we do know that r=n−2 s. If Δ>0 then s=2 k and r=n−4 k for some k=0,1,…. In order for r=0 to be possible we need n=4 k, i.e. the degree must be a multiple of 4. In such a case, both r=0 and r>0 are possible (i.e. not distinguishable by the sign of the discriminant). If Δ<0 then s=2 k+1 and r=n−4 k−2 for some k=0,1,…. In order for r=0 to be possible we need n=4 k+2, i.e. n≡2(mod 4). In such a case, if n=2 we are forced to conclude r=0 because r cannot be negative, otherwise if n>2 we cannot decide using this method. In conclusion we have the following decision procedure: If n is odd then the polynomial always has a real root (from Intermediate Value Theorem). If Δ>0 and n is not divisible by 4 then the polynomial always has a real root. If Δ<0 and n is divisible by 4 then the polynomial always has a real root. If Δ<0 and n=2 then the polynomial does not have any real roots. If Δ=0 and n=2 then the polynomial has a repeated root which must be real. In any other case, the test is inconclusive. Perhaps more information can be deduced using the value of Δ rather that just its sign. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Mar 7, 2018 at 22:25 answered Mar 7, 2018 at 21:59 Tob ErnackTob Ernack 5,389 1 1 gold badge 14 14 silver badges 36 36 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. I assume you mean polynomials with real coefficients. The discriminant tells us about the nature (real or complex) of the roots. Since complex roots occurs in pairs, when n=2, having no real roots means the polynomial is always positive or always negative. Already when n=3, something like this cannot be true because every cubic has at least one real zero. Nevertheless, the discriminant tells us about the nature of the roots, see Wikipedia: If Δ>0, then the equation has three distinct real roots. If Δ=0, then the equation has a multiple root and all of its roots are real. If Δ<0, then the equation has one real root and two non-real complex conjugate roots. When n=4, the sign of the discriminant is not enough to give complete information about the nature of the roots. See Wikipedia. When n≥5, things get even more complicated. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Mar 7, 2018 at 21:07 lhflhf 222k 20 20 gold badges 254 254 silver badges 585 585 bronze badges Add a comment\| You must log in to answer this question. 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1870	https://simons.berkeley.edu/workshops/discrete-optimization-continuous-relaxation	Discrete Optimization via Continuous Relaxation Skip to main content Search Utility navigation Calendar Contact Login MAKE A GIFT Main navigation Home Programs & Events Research Programs Workshops & Symposia Public Lectures Research Pods Internal Program Activities Algorithms, Society, and the Law People Scientific Leadership Staff Current Long-Term Visitors Research Fellows Postdoctoral Researchers Scientific Advisory Board Governance Board Industry Advisory Council Affiliated Faculty Science Communicators in Residence Law and Society Fellows Participate Apply to Participate Plan Your Visit Accessibility Location & Directions Postdoctoral Research Fellowships Law and Society Fellowships Science Communicator in Residence Program Circles Breakthroughs Workshops and Goldwasser Exploratory Workshops Support Annual Fund All Funders Institutional Partnerships News & Videos News Videos About Image Discrete Optimization via Continuous Relaxation Program Bridging Continuous and Discrete Optimization Location Calvin Lab auditorium Date Monday, Sept. 11 – Friday, Sept. 15, 2017 Back to calendar Breadcrumb Home Workshop & Symposia Discrete Optimization Via Continuous Relaxation Secondary tabs The Workshop Schedule Videos About Much of the progress in solving discrete optimization problems, especially in terms of approximation algorithms, has come from designing novel continuous relaxations. The primary tools in this area are linear programming and semidefinite programming. Other forms of relaxations have also been developed, such as multilinear relaxation for submodular optimization.In this workshop, we explore the state-of-the-art techniques for performing discrete optimization based on continuous relaxations of the underlying problem, as well as our current understanding of the limitations of this kind of approach. We focus on LP/SDP relaxations and techniques for rounding their solutions, as well as methods for submodular optimization, both in the offline and online setting. We also investigate the limits of such relaxations and hardness of approximation results. All presentations take place at the Calvin Lab auditorium. Chairs/Organizers Image Chandra Chekuri ((None); chair) Image Nikhil Bansal (University of Michigan) Image Seffi Naor (Technion Israel Institute of Technology) Image Mohit Singh (Georgia Institute of Technology) Invited Participants Anna Adamaszek (University of Copenhagen), Nima Anari Ahmadipouranari (Stanford University),Farid Alizadeh (Rutgers University),Jason Altschuler (MIT),Kyriakos Axiotis (MIT),Nikhil Bansal (Eindhoven University of Technology), Charles Carlson (University of Illinois Urbana-Champaign),Deeparnab Chakrabarty (Dartmouth College),Parinya Chalermsook (Aalto University),Moses Charikar (Stanford University), Chandra Chekuri (University of Illinois Urbana-Champaign), Ken Clarkson (IBM Almaden), Michael Cohen (MIT), Daniel Dadush (Centrium Wiskunde & Informatica), Damek Davis (Cornell University), Nikhil R. Devanur (Microsoft Research),Jelena Diakonikolas (Boston University),Reza Eghbali (University of Washington),Friedrich Eisenbrand (École Polytechnique Fédérale de Lausann),Alina Ene (Boston University),Maryam Fazel(University of Washington), Uri Feige(Weizmann Institute of Science), Sam Fiorini (Université Libre de Bruxelles),Shashwat Garg (TU Eindhoven),Michel Goemans (MIT), Fabrizio Grandoni(IDSIA, University of Lugano),Ben Grimmer (Cornell University),Krystal Guo (Université Libre de Bruxelles),Anupam Gupta (Carnegie Mellon University),Swati Gupta (MIT),Nick Harvey (University of British Columbia),Robert Hildebrand (IBM T.J. Watson Research Center),Rebecca Hoberg (University of Washington),Fatma Kılınç-Karzan (Carnegie Mellon University), Ravindran Kannan (Microsoft Research India),Sanjeev Khanna (University of Pennsylvania), Matthias Koppe (UC Davis), Rasmus Kyng (Yale University),Bundit Laekhanukit (Weizmann Institute of Science),Lap-Chi Lau (University of Waterloo),Euiwoong Lee (Carnegie Mellon University),Jon Lee (University of Michigan),Shi Li (SUNY Buffalo),Chris Liaw (University of British Columbia),Cong Han Lim (University of Wisconsin-Madison),Mike Luby (Qualcomm),Shiqian Ma (University of California, Davis),Konstantin Makarychev (Northwestern University),Alex Makelov (MIT),Aryan Mokhtari (University of Pennsylvania),Sarah Maria Morell (EPFL),Walaa Moursi (University of British Columbia),Seffi Naor (Technion – Israel Institute of Technology),Huy Nguyen (Northeastern University), Sasho Nikolov (University of Toronto), Debmalya Panigrahi (Duke University),Pablo Parrilo (MIT),Kostya Pashkovich (University of Waterloo),Sebastian Pokutta (Georgia Institute of Technology),Yuval Rabani (Hebrew University of Jerusalem), Prasad Raghavendra (UC Berkeley), R Ravi (Carnegie Mellon University),Jim Renegar (Cornell University),Alireza Rezaei (University of Washington),Thomas Rothvoß (University of Washington),Amin Saberi (Stanford University),Piotr Sankowski (University of Warsaw), Andreas Schmid (Max-Planck-Institut für Informatik), Ludwig Schmidt (MIT), Tselil Schramm (UC Berkeley),Roy Schwartz (Technion - Israel Institute of Technology), David Shmoys (Cornell University),Aaron Sidford (Stanford University), Mohit Singh (Georgia Institute of Technology),Sahil Singla (Carnegie Mellon University),Chaitanya Swamy (University of Waterloo), Prasad Tetali (Georgia Institue of Technology),Dimitris Tsipras (MIT),Levent Tunçel (University of Waterloo),Lieven Vandenberghe (UCLA),László Végh (London School of Economics),Soledad Villar (University of Texas at Austin),Cynthia Vinzant (North Carolina State University), Adrian Vladu (Boston University),Jan Vondrák (Stanford University),David Wajc (Carnegie Mellon University), Karol Węgrzycki (Unversity of Warsaw),David Williamson (Cornell University),Steve Wright (University of Wisconsin-Madison), Chenyang Yuan (MIT),Rico Zenklusen (ETH Zürich). Share this page Copy URL of this page The Simons Institute for the Theory of Computing is the world's leading venue for collaborative research in theoretical computer science. Footer Programs & Events About Participate Workshops & Symposia Contact Us Calendar Accessibility Footer social media Twitter Facebook Youtube © 2013–2025 Simons Institute for the Theory of Computing. All Rights Reserved. link to homepage Close Main navigation Home Programs & Events Research Programs Workshops & Symposia Public Lectures Research Pods Internal Program Activities Algorithms, Society, and the Law People Scientific Leadership Staff Current Long-Term Visitors Research Fellows Postdoctoral Researchers Scientific Advisory Board Governance Board Industry Advisory Council Affiliated Faculty Science Communicators in Residence Law and Society Fellows Participate Apply to Participate Plan Your Visit Accessibility Location & Directions Postdoctoral Research Fellowships Law and Society Fellowships Science Communicator in Residence Program Circles Breakthroughs Workshops and Goldwasser Exploratory Workshops Support Annual Fund All Funders Institutional Partnerships News & Videos News Videos About Utility navigation Calendar Contact Login MAKE A GIFT link to homepage Close Search
1871	https://ocw.mit.edu/courses/9-01-introduction-to-neuroscience-fall-2007/3a2fa42578499c13587be8fa06437abe_wk03_9_01_r02.pdf	MIT OpenCourseWare 9.01 Introduction to Neuroscience Fall 2007 For information about citing these materials or our Terms of Use, visit: 9.01 Recitation (R02) RECITATION #2: Tuesday, September 18th Review of Lectures: 3, 4 Reading: Chapters 3, 4 or Neuroscience: Exploring the Brain (3rd edition) Outline of Recitation: I. Previous Recitation: a. Questions on practice exam questions from last recitation? II. Review of Material: a. Exploiting Axoplasmic Transport b. Types of Glia c. THE RESTING MEMBRANE POTENTIAL d. THE ACTION POTENTIAL III. Practice Exam Questions IV. Questions on Pset? Exploiting Axoplasmic Transport: Maps connections of the brain Rates of transport: - slow: - fast: Examples: Uses anterograde transport: - Uses retrograde transport: - - - Types of Glia: - Microglia: - Astrocytes: - Myelinating Glia: 1 THE RESTING MEMBRANE POTENTIAL: The Cast of Chemicals: The Movement of Ions: Influences by two factors: (1) Diffusion: (2) Electricity: Ohm’s Law: I = gV Ionic Equilibrium Potentials (EION): Example: ENs + diffusional and electrical forces are equal 2 Nernst Equation: EION = 2.303 RT/zF log [ion]0/[ion]i Calculates equilibrium potential for a SINGLE ion. Inside Outside EION (at 37°C) [K+] [Na+] [Ca2+] [Cl ] Pumps maintain concentration gradients (ex. sodiumpotassium pump; calcium pump) Resting Membrane Potential (VM at rest): Measured resting membrane potential: 65 mV Goldman Equation: VM = 61.54 mV log (PK[K+]o + PNa[Na+]o)/ (PK[K+]I + PNa[Na+]i) Calculates membrane potential when permeable to both Na+ and K+ . Remember: at REST, gK >>> gNa therefore, VM is closer to EK THE ACTION POTENTIAL (Nerve Impulse): Phases of an Action Potential: Vm (mV) ENa 0 EK Time 3 Conductance of Ion Channels during AP: Remember: Changes in conductance, or permeability of the membrane to a specific ion, changes the membrane potential. Vm (mV) ENa 0 EK Time Voltagegated Sodium Channels: opens at 40 mV (a) (b) (c) (d) Initiation of AP: Mechanicallygated sodium channels Inject current: Injected current time Vm (mV) Time Conduction: A __ axon diameter and ______ increase conduction velocity. 4 Practice Exam Problems: From Brown Exam I 2000: 9. Ion X is negatively charged and more concentrated inside than outside of the cell. Therefore: (a) Ex is negative (b) EX is positive (c) EX is zero (d) The cell will fail its Olympic drug test 10. The resting membrane potential is closer to EK than ENa because at rest: (a) Vm – ENa = 0 (b) gNa = 0 (c) gK >> gNa (d) gK = 0 11. You discover an ion channel that, when open, is permeable to all ions. This channel lacks: (a) gating (b) quaternary structure (c) selectivity (d) an alpha subunit 14. Increasing extracellular K+ would lead to: (a) depolarization of the membrane (Vm becomes more negative) (b) depolarization of the membrane (Vm becomes less negative) (c) hyperpolarization of the membrane (Vm becomes more negative) (d) hyperpolarization of the membrane (Vm becomes less negative) 15. Increasing intracellular Na+ in a typical neuron would have what effect on ENa? (a) ENa would become more positive (b) ENa would become less positive (c) ENa would stay the same (d) That is not enough information to determine ENa 16. How would the mental pain caused by this exam be encoded as a neuronal signal? (a) By the amplitude of action potentials transmitted (b) By the duration of action potentials transmitted (c) By the frequency of action potentials transmitted (d) All of the above (e) None of the above 19. What would happen if the S4 domains of the voltagegated sodium channels important in action potential generation had their primary structure significantly altered? (a) Nothing important of interest (b) The action potentials generated would be of shorter duration (c) The gating properties of these channels would be disrupted (d) The channels will be phosphorylated 20. I. Axon diameter II. Axon myelination III. Axon length IV. Axonal membrane resistance Which of the above are important factors in the determination of action potential conduction velocity? (a) I only (b) I and II (c) I and III (d) I, II, IV (e) All of the above 5 Answers: 9. B 10. C 11. C 14. B 15. B 16. C 19. C 20. D 6
1872	https://brighterly.com/math/factorial/	Math tutors / Knowledge Base / Factorial – Meaning, Definition With Examples Factorial – Meaning, Definition With Examples Jo-ann Caballes Updated on January 2, 2024 Table of Contents Welcome to the enchanting world of mathematics at Brighterly! Here, we make learning exciting and engaging for young minds, empowering them to grasp complex mathematical concepts with ease and enjoyment. One of those fascinating concepts is the “Factorial,” a mathematical function that multiplies numbers in a sequence, weaving a tapestry of numerical possibilities. From its notations to its applications, the concept of factorial reveals endless avenues of exploration, creating opportunities for problem-solving, creativity, and logical thinking. Join us at Brighterly as we delve into the meaning, definition, and examples of factorials, and unravel a subject that is as fun as it is profound. Let’s get started on this mathematical adventure! What Is the Factorial? Factorial, often denoted by the symbol “!”, is a function applied to non-negative integers. When you see a number followed by an exclamation mark, it means that you multiply that number by every positive whole number less than itself. The factorial of 0 is specially defined as 1. This concept is essential in fields such as permutation, combinations, and other mathematical analysis. It’s like building a tower of numbers, each floor multiplying the one below! Definition of Factorial The factorial of a non-negative integer n is the product of all positive integers less than or equal to n. Mathematically, it is defined as: For example, the factorial of 5 (denoted as 5!) is 5 × 4 × 3 × 2 × 1 = 120. Understanding the Symbol and Notation The symbol “!” might seem strange, but it’s the hallmark of factorials. When you see 4!, you know to multiply 4 × 3 × 2 × 1. It’s like a secret code in the world of mathematics! This notation was introduced by the famous French mathematician Christian Kramp in 1808. It helps to simplify expressions and provides a concise way to represent complex products. Properties of Factorials Multiplicative Property: Factorials grow very quickly, and their values can be large even for small numbers. Recursive Property: You can find n! by multiplying (n−1)! by n. Predefined Values: 0! is defined as 1, and 1! is 1. Division: n!/(k!×(n−k)!) is a critical expression in combinations. Understanding these properties can be a big help when solving mathematical problems involving factorials. Examples of Factorials 3! = 3 × 2 × 1 = 6 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 0! = 1 (by definition) These examples demonstrate how the factorial function behaves and how quickly it can grow. Difference Between Factorials and Other Mathematical Concepts Factorials may seem similar to exponential growth, but they’re quite distinct. While exponential functions grow faster, factorial growth is unique due to the multiplication of descending integers. Understanding the difference between factorials and other mathematical concepts like powers and logarithms is essential in mastering higher-level math. Conclusion Factorials are not just numbers and symbols; they are a gateway to a world of mathematical wonder, exploration, and discovery. At Brighterly, we believe that mathematics is a language that tells fascinating stories, and the tale of factorials is one that resonates with curious minds of all ages. By understanding factorials, you unlock the door to permutations, combinations, series, and many other complex concepts. Whether you’re a parent, educator, or a young learner, we hope this comprehensive guide has ignited a passion for mathematics and illuminated the intriguing world of factorials. Remember, the journey with Brighterly doesn’t end here; it’s just the beginning of a lifetime of learning, questioning, and growing. Frequently Asked Questions on Factorials What is 0 factorial? Zero factorial, denoted as 0!, is uniquely defined as 1. This definition might seem surprising, but it’s essential for mathematical consistency, particularly in combinatorial contexts. Why are factorials used? Factorials are fundamental in various mathematical fields. They’re used to calculate permutations (arrangements of objects) and combinations (selections of objects), making them vital in probability and statistics. Factorials also appear in calculus, physics, and computer science algorithms. They’re a versatile tool that helps us understand and model different mathematical phenomena. How does factorial grow? Factorials grow very quickly. Even for small numbers, the values can become enormous. For example, 5! is 120, but 10! is 3,628,800. This rapid growth makes factorials an intriguing subject in mathematics, especially in analyzing algorithms and computational efficiency. Can you have a factorial of a negative number? Factorials are typically defined only for non-negative integers. The concept of factorial for negative numbers doesn’t hold in the usual sense. However, there’s an advanced mathematical function called the Gamma function that extends the concept of factorials to complex numbers, excluding negative integers. Is there a connection between factorials and real-life applications? Absolutely! Factorials are used in real-life scenarios like project planning, resource allocation, probability predictions, statistical modeling, and more. From planning seating arrangements at events to predicting outcomes in sports and finance, the applications of factorials extend far and wide, transcending the boundaries of the classroom and making them a valuable concept to explore with Brighterly. Information Sources Wikipedia – Factorial MathWorld – Factorial Government Educational Site – Factorial Functions Jo-ann Caballes 13 articles As a seasoned educator with a Bachelor’s in Secondary Education and over three years of experience, I specialize in making mathematics accessible to students of all backgrounds through Brighterly. My expertise extends beyond teaching; I blog about innovative educational strategies and have a keen interest in child psychology and curriculum development. My approach is shaped by a belief in practical, real-life application of math, making learning both impactful and enjoyable. Table of Contents What Is the Factorial? Definition of Factorial Understanding the Symbol and Notation Properties of Factorials Examples of Factorials Difference Between Factorials and Other Mathematical Concepts Conclusion Frequently Asked Questions on Factorials What is 0 factorial? Why are factorials used? How does factorial grow? Can you have a factorial of a negative number? Is there a connection between factorials and real-life applications? Math & reading from 1st to 9th grade Looking for homework support for your child? Choose kid's grade Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Grade 6 Grade 7 Grade 8 Math & reading from 1st to 9th grade Looking for homework support for your child? Book free lesson Related math Dividing Polynomials – Synthetic Division, Definition With Examples Welcome to another enlightening journey into the beautiful world of mathematics with Brighterly. Today, we dive deep into an essential concept in algebra and calculus: the division of polynomials through synthetic division. This often-overlooked tool in the mathematical arsenal offers a simplified alternative to the traditional long division method when dealing with polynomials. As we […] Read more Right Angle – Definition, Examples, Facts Welcome to Brighterly, the best place for children to learn and explore math concepts in a fun and engaging way! In this adventure, we’re going to dive deep into the world of right angles and unveil their secrets. So buckle up, and let’s start our exciting journey into the realm of right angles! What is […] Read more 400 in Words The number 400 is written in words as “four hundred”. It’s four times one hundred. For instance, if you collect 400 stamps, you start with four groups of one hundred stamps each. Hundreds Tens Ones 4 0 0 How to Write 400 in Words? The number 400 is expressed in words as ‘Four Hundred’. It […] Read more Close a child’s math gaps with a tutor! Book a free demo lesson with our math tutor and see your kid fill math gaps with interactive lessons Book demo lesson Get full test results See Your Child’s Test Results Enter your name and email to view your child’s test results now! 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1873	https://www.youtube.com/watch?v=1JhO0dBRUVs	Add Two-Digits by Two-Digits (Fastest & Easiest Method) (2.NBT.B.5) MyMathTA 74700 subscribers 11 likes Description 428 views Posted: 16 Jun 2024 In this video, we show the fastest and easiest way to Add any two-digit number by two-digits. Students can now learn this pivotal foundational skill within minutes. This is a part of the Common Core Math Standard 2.NBT.B.5 where the objective is to "Fluently add and subtract within 100 using strategies based on place value, properties of operations, and/or the relationship between addition and subtraction." You can see all 40+ examples and other lessons for foundational math skills at 1 comments Transcript: in this video we're going to take a look at how to add any two-digit number by another two-digit number and how to do so mentally so when you come across a problem such as this you can solve it in just a few seconds without even having to write it down so let's take a look at this first example 11 + 57 the key is we're first going to break down the second two-digit number into tens and ones 11 + 57 is the same thing as 11 and instead of 57 we're going to break down 57 into 50 + 7 11 + 50 is equal to 61 and now what we have to do is add the remaining ones digit 61 + 7 is 68 so 11 + 57 is 68 next example 29 + 32 we're going to think of this problem as 29 plus instead of 32 30 + 2 29 + 30 is equal to 59 now we add the two 59 + 2 is 61 so 29 + 32 is 61 how about 61 + 68 we're going to think of this one as 61 + 60 + 8 when we Add 61 and 60 we get 121 121 + 8 is 129 how about 42 + 28 we can think of that as 42 + 20 + 8 42 + 20 is 62 62 + 8 is 70 35 + 55 we look at that is 35 plus 50 + 5 first add the 35 and 50 and we get 85 we add the remaining five 85 + 5 is 90 83 + 62 we're going to think of this as 83+ 60 + 2 83 + 60 is 143 if we add the two 143 + 2 is 1 145 54 + 11 we're going to think of this as 54 plus 10 + 1 so we can easily add 54 and 10 and get 64 we add the remaining one 64 + 1 is 65 how about 75 + 36 we're going to look at this as 75+ 30 + 6 75 + 30 is going to give us 105 then we add 105 + 6 which gives us 111 95 + 73 95+ 70 + 3 now 95 + 70 is 165 then we add the three 165 + 3 is 168 72 + 13 72 + 10 + 3 72 + 10 is 82 82 + 3 is 855 21 + 29 21 + 20 + 9 21 + 20 is 41 and 41 + 9 is 50 52 + 74 we're going to look at that as 52 + 70 + 4 52 + 70 is 122 122 + 4 is 126 43 + 22 we look at this one as 43 + 20 + 2 43 + 20 is 63 63 + 2 is equal to 65 36 + 27 this can be thought of as 36 + 20 + 7 the 36 + 20 gives us 56 56 + 7 is 63 1 13 + 61 is the same thing as 13 + 60 + 1 13 + 60 is 73 we add the remaining one and we get 74 86 + 37 is the same thing as 86 + 30 + 7 86 + 30 is 116 116 + 7 is 123 922 + 35 that's the same thing as 92 + 30 + 5 92 + 30 is 122 and 122 + 5 is 127 62 + 45 we think of that one as 62 + 40 +5 62 + 40 is 102 we add the remaining five and we have 107 23 + 78 is the same thing as 23 Plus 70 + 8 the 23 and 70 give us 93 and 93 + 8 gives us the total of 101 37 + 81 we could think of that as 37 + 80 + 1 37 + 80 is 117 we add the one and we get 18 77 + 31 is the same thing as 77 + 30 + 1 77 + 30 is 107 we add the 1 and we have 108 45 + 19 45 + 10 + 9 45 + 10 is 55 we add the remaining nine and we have 64 98 + 21 98 + 20 + 1 98 + 20 is 118 add the remaining one and we have 119 59 + 71 that's the same thing as 5 9 + 70 + 1 add 59 and 70 and we get 129 when we add the one we have 130 and lastly 68 + 25 that's the same thing as 68 + 20 + 5 68 + 20 is 88 88 + 5 is 93 so as you can see we can easily add any two-digit number by another two-digit number by first breaking down the second two-digit number into tens and ones which makes it easier to add the first two-digit number by 10 and then we just add the ones digit to the remaining answer
1874	https://www.pw.live/neet/exams/difference-between-cell-wall-and-cell-membrane	Difference Between Cell Wall and Cell Membrane, Structure Share Difference Between Cell Wall and Cell Membrane: It is important to understand the difference between cell wall and the cell membrane because they are commonly confused. \| \| \| --- \| \| NEET Biology Syllabus \| NEET Biology Diagrams \| \| NEET Biology MCQ \| NEET Biology Chapter wise Weightage \| \| NEET Biology Notes \| NEET Previous Year Question papers \| Difference Between Cell Wall and Cell Membrane Overview Free Study Material for NEET Preparation Difference Between Cell Wall and Cell Membrane Cell wall and membrane stand out as vital organelles for all living organisms to function. Despite differences in appearance and function, the cell membrane and cell wall play important roles in the proper functioning of living entities. While a cell wall has a rigid structure and cannot change shape, a cell membrane has natural flexibility and can change shape and size. However, the inherent rigidity of a cell wall is essential in providing structural support to the cells. The table below outlines the key difference between cell wall and cell membrane. \| Difference Between Cell Wall and Cell Membrane \| \| \| --- \| Parameter \| Cell Wall \| Cell Membrane \| \| Meaning \| A sturdy, substantial structure is visible under a light microscope, present in plant, fungal, and bacterial cells, providing rigidity and shape. \| A delicate, thin structure visible under an electron microscope, present in all cell types, offering protection to the protoplasm and facilitating molecule passage. \| \| Composition \| Formed from chitin (fungi), cellulose (plant cells), and peptidoglycan (bacteria). \| Formed from proteins, lipids, and carbohydrates. \| \| Presence \| Found in plant cells, fungi, bacteria, and algae. \| Present in all cell types. \| \| Receptors \| Lacks receptors. \| Contains receptors facilitating cell-to-cell communication. \| \| Structure \| Thick layers visible under a light microscope. \| Thin layers visible under an electron microscope. \| \| Permeability \| Completely permeable. \| Semi-permeable. \| \| Thickness \| Thickness increases with cell life. \| Thickness remains constant throughout the cell's lifespan. \| \| Nutrition \| No nutritional requirements from the cell; serves for deposition. \| Requires proper nutrition for survival. \| Definition of Cell Wall A cell wall is the outermost layer of a plant cell. The cell wall is a flexible and complex structure outside the cell membrane. The cell wall is often composed of cellulose, hemicellulose, the long fiber of carbohydrates, lignin, and pectin. It regulates cell proliferation while also protecting it from physical injury. Endosmosis can cause a cell to perish. It allows smaller molecules to enter due to its rough and porous surface. In addition, if the pressure inside the cell rises, the cell wall stops it from expanding or rupturing. The cell wall also serves as a barrier for some biomolecules to enter and depart and a conduit for many other metabolic chemicals to enter and exit. In certain plants, the cell wall is composed of a single layer, but in others, the cell wall is composed of two layers. The cell is waterproofed thanks to these two layers. Cell Wall Structure Function of Cell Wall Definition of Cell Membrane Explore - NEET Online Courses by PW Function of Cell Membrane Similarities Between Cell Wall and Cell Membrane \| \| \| --- \| \| NEET Biology Difference Between Important Links \| \| \| Difference Between Biology and Biotechnology \| Difference Between Wildlife Sanctuary and National Park \| \| Difference Between Mitosis And Meiosis \| Difference Between Flora and Fauna \| \| Difference between Arteries and Veins \| Difference Between Sympathetic And Parasympathetic \| \| Difference between Osmosis and Diffusion \| Difference between Nucleotide and Nucleoside \| \| Difference Between Food Chain And Food Web \| Difference Between Blood and Lymph \| \| Difference Between Breathing and Respiration \| Difference Between Dicot And Monocot Root \| \| Difference Between Self Pollination and Cross Pollination \| Difference Between Active and Passive Immunity \| \| Difference between Bone and Cartilage \| Difference Between Globular and Fibrous Protein \| \| Difference between Sexual and Asexual Reproduction \| Difference Between Moth and Butterfly \| \| Difference between Endocrine and Exocrine Glands \| Difference between Herbs and Shrubs \| \| difference between Cell Wall and Cell Membrane \| difference between Biotic And Abiotic \| \| Difference Between Antibody and Antigen \| Difference Between Fragmentation and Regeneration \| \| Difference between Serum and Plasma \| Difference Between C3 And C4 Plants \| \| Difference between Active and Passive Transport \| Difference Between Diabetes Mellitus And Diabetes Insipidus \| \| Difference Between Pollination and Fertilization \| Difference Between Heart Rate and Pulse Rate \| \| Difference between Unicellular and Multicellular Organisms \| Difference Between RBC And WBC \| \| Difference Between Spermatogenesis and Oogenesis \| Difference between Leopard and Jaguar \| \| Difference Between Ligaments and Tendons \| Difference Between Cerebellum and Cerebrum \| \| Difference Between Leopard and Cheetah \| Difference Between Enzymes and Hormones \| \| Difference between Turtle and Tortoise \| Difference Between Disinfection and Sterilization \| Difference Between Cell Wall and Cell Membrane FAQs What are three differences between cell walls and cell membranes? Where is DNA located in the cell? Who discovered the cell wall? Does the cell wall and cell membrane have the same function? Physics Wallah We understand that every student has unique needs and abilities, that’s why our curriculum is designed to adapt to your needs and help you grow!
1875	https://www.doubtnut.com/qna/646661313	Maths Physics Chemistry Biology English Maths Physics Chemistry Biology English Maths Physics Chemistry Biology English Maths Physics Chemistry Biology English Maths Physics Chemistry Biology English Maths Physics Chemistry Biology English Class 11 Class 12 Class 12+ Class 11 Class 12 Class 12+ Class 9 Class 10 Class 11 Class 12 Class 9 Class 10 Class 11 Class 12 Class 9 Class 10 Class 11 Class 12 Class 9 Class 10 Class 11 Class 12 Class 9 Class 10 Class 11 Class 12 Class 9 Class 10 Class 11 Class 12 Class 9 Class 10 Class 11 Class 12 Class 9 Class 10 Class 11 Class 12 Class 9 Class 10 Class 11 Class 12 Class 9 Class 10 Class 11 Class 12 IIT-JEE NEET CBSE UP Board Bihar Board JEE NEET Class 6-10 Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 Online Class Search Doubtnut Blog Get App Select Theme Acetylene on heating in the presence of red hot Fe tube gives : A Benzene B Cyclooctatetraene C naphthalene D Anthracene Video Solution Video Player is loading. Current Time 0:00 / Duration -:- Loaded: 0% Stream Type LIVE Remaining Time 0:00 1x 2x 1.5x 1x , selected 0.5x 0.25x Chapters descriptions off , selected captions settings , opens captions settings dialog captions off , selected This is a modal window. Beginning of dialog window. Escape will cancel and close the window. End of dialog window. Text Solution Generated By DoubtnutGPT The correct Answer is: A To solve the question regarding what acetylene (ethyne) produces when heated in the presence of a red hot iron tube, we can follow these steps: Step 1: Identify the Structure of Acetylene Acetylene, also known as ethyne, has the chemical formula C₂H₂. Its structure can be represented as H-C≡C-H, where there is a triple bond between the two carbon atoms. Hint: Remember that acetylene is a simple alkyne with a triple bond between two carbon atoms. Step 2: Understand the Reaction Conditions The question specifies that acetylene is heated in the presence of a red hot iron tube. The temperature is typically around 873 Kelvin. Under these conditions, acetylene can undergo a reaction known as polymerization. Hint: Heating acetylene can lead to various reactions, including polymerization, especially in the presence of a catalyst like iron. Step 3: Describe the Polymerization Process When acetylene is heated, it can trimerize, meaning three molecules of acetylene can combine to form a larger molecule. In this case, the trimerization of acetylene leads to the formation of a cyclic structure. Hint: Trimerization involves the combination of three monomer units to form a polymer. Step 4: Determine the Product of the Reaction The trimerization of acetylene results in the formation of benzene (C₆H₆). The reaction can be summarized as three acetylene molecules combining to form one benzene molecule. Hint: Benzene is a cyclic compound with alternating double bonds, which can be formed from the trimerization of acetylene. Step 5: Write the Overall Reaction The overall reaction can be represented as: 3C2H2red hot Fe−−−−−−→C6H6 This indicates that three molecules of acetylene yield one molecule of benzene when heated in the presence of red hot iron. Hint: Always balance the chemical equation to ensure that the number of atoms on both sides is equal. Final Answer When acetylene is heated in the presence of a red hot iron tube, it gives benzene. --- 2 \| Class 12 CHEMISTRY CHEMISTRY AT A GLANCE Topper's Solved these Questions CHEMISTRY AT A GLANCE BOOK - ALLEN CHAPTER - CHEMISTRY AT A GLANCE EXERCISE - INORGANIC CHEMISTRY 300 Videos Chemical Equilibrium BOOK - ALLEN CHAPTER - Chemical Equilibrium EXERCISE - All Questions 30 Videos [ELECTROCHEMISTRY BOOK - ALLEN CHAPTER - ELECTROCHEMISTRY EXERCISE - EXERCISE -05 [B] 38 Videos](/qna/646655788) Similar Questions Which metals upon heating in presence of air followed by hydrolysis gives ammonia. View Solution A black mineral (A) on heating in presence of air gives a gas (B). (A) and (B) are respectively: View Solution (A): OX− on heating in the presence of OH− gives X− and XO−3 (R): Conversion of OX− to X− and XO−3, is called disproportionation. View Solution Propyne on passing through red hot copper tube forms View Solution Give balanced equations for the following: Acetone reacts with hydrogen in the presence of heated copper. View Solution A non-metal A which is the largest constituent of air, when heated with H2 in 1 : 3 ratio in the presence of catalyst (Fe) gives a gas B. 0n heating with O2 it gives an oxide C. If this oxide is passed into water in the presence of air, it gives an acid D which acts as a strong oxidising agent. (a) Identify A, B, C and D. (b) To which group of the periodic table does this non-metal belongs? View Solution Aniline is prepared in presence of Fe/HCl from View Solution Give balanced equations for the following reactions : Acetaldehyde is heated with hydroiodic acid in the presence of red phosphorous. View Solution Red algae is red due to the presence of :- View Solution Chlorination of toluene in the presence of light and heat followed by treatment with aqueous NaOH gives View Solution ALLEN - CHEMISTRY AT A GLANCE - ORGANIC CHEMISTRY Reagent x can be 05:39 2. Which one of the following compounds will give white precipitate wit... 02:27 3. Acetylene on heating in the presence of red hot Fe tube gives : 02:27 4. Which among the following alkenes will be most stable : 05:35 5. Which among the following compounds will give Wurtz reaction ? 02:52 6. In the following reaction : C(2)H(2)underset(HgSO(4)//H(2)SO(4))ove... 03:16 7. One of the following which does not observe the anti-Markovnikoff's ad... 02:12 8. Propyne and propene can be distinguished by conc. H2SO4 Br2 in "CCl"... 05:17 9. which of these will not react with acetylene ? 02:56 10. What happens when acetylene is treated with hypochlorous acid 02:30 Complete the following reaction 04:15 12. CH(3)-underset(CH(3))underset(\|)overset(CH(3))overset(\|)C-CH=CH-CH(3)o... 03:38 13. During debromination of meso-dibromobutane, the major compound formed ... 14. Which is maximum reactive towords acid catalysed dehydration : 15. Which gives ketonic group after hydroboration oxidation : 16. Complete the following reaction 17. on oxymercuration demercuaration produces the major product 18. How many chiral compounds are possible on monochlorination of 2-methyl... 19. Ozonolysis of an organic compound A produces acetone and propionaldehy... 20. Arrange the following in decreasing order of reactivity towards electr... 03:38
1876	https://math.stackexchange.com/questions/3470068/expanding-inner-product	linear algebra - Expanding inner product - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Expanding inner product Ask Question Asked 5 years, 9 months ago Modified5 years, 9 months ago Viewed 3k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. Let V V be an inner product space. Prove that: ∥x±y∥2=∥x∥2±2 R e⟨x,y⟩+∥y∥2.‖x±y‖2=‖x‖2±2 R e⟨x,y⟩+‖y‖2. After foiling ∥x±y∥2‖x±y‖2, I arrive at ∥x∥2±2(x⋅y)+∥y∥2‖x‖2±2(x⋅y)+‖y‖2. I'm unsure how 2(x⋅y)2(x⋅y) becomes 2 R e⟨x,y⟩2 R e⟨𝑥,𝑦⟩. Also, I'm new to StackExchange, so I apologize for the sloppy formatting. Any tips with the math and any tips with formatting is much appreciated! Thank you! linear-algebra inner-products Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Dec 9, 2019 at 20:05 Arturo Magidin 419k 60 60 gold badges 864 864 silver badges 1.2k 1.2k bronze badges asked Dec 9, 2019 at 20:00 thetimidsoulthetimidsoul 33 1 1 silver badge 4 4 bronze badges 4 1 Welcome to Mathematics Stack Exchange. Is V V an inner product space over the field C C? In that case, the inner product involves taking the complex conjugate of one of the vectors J. W. Tanner –J. W. Tanner 2019-12-09 20:05:23 +00:00 Commented Dec 9, 2019 at 20:05 1 For a comparison of real & complex spaces' identities, see here.J.G. –J.G. 2019-12-09 20:16:01 +00:00 Commented Dec 9, 2019 at 20:16 @J.W.Tanner It didn't specify whether V is over the complex or real numbers. I think we are to assume V can be over either.thetimidsoul –thetimidsoul 2019-12-09 20:19:19 +00:00 Commented Dec 9, 2019 at 20:19 1 here's a MathJax tutorial J. W. Tanner –J. W. Tanner 2019-12-09 20:20:01 +00:00 Commented Dec 9, 2019 at 20:20 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. You seem to be using the Euclidean definition of norm using the dot product in a real vector space. You should be using the general definition of norm induced by an inner product, namely ∥x∥=⟨x,x⟩−−−−−√‖x‖=⟨x,x⟩ where ⟨⋅,⋅⟩⟨⋅,⋅⟩ is an inner product on V V; that is, a function that assigns to a pair of vectors a scalar (in either R R or C C, depending on whether V V is a real or complex vector space), satisfying the following properties: Additive in the first coordinate: ⟨x+y,z⟩=⟨x,z⟩+⟨y,z⟩⟨x+y,z⟩=⟨x,z⟩+⟨y,z⟩ for all x,y,z∈V x,y,z∈V. Homogeneous in the first coordinate: ⟨α x,y⟩=α⟨x,y⟩⟨α x,y⟩=α⟨x,y⟩ for all x,y∈V x,y∈V, and all scalars α α. Conjugate symmetric: ⟨y,x⟩=⟨x,y⟩¯¯¯¯¯¯¯¯¯¯¯¯⟨y,x⟩=⟨x,y⟩¯ for all x,y∈V x,y∈V, where α¯¯¯α¯ is the complex conjugate of α α. Positive definite: ⟨x,x⟩≥0⟨x,x⟩≥0 for all x∈V x∈V, and ⟨x,x⟩=0⟨x,x⟩=0 if and only if x=0 x=0. So the correct expansion would be: ∥x±y∥2=⟨x±y,x±y⟩=⟨x,x⟩±⟨x,y⟩±⟨y,x⟩+⟨y,y⟩=∥x∥2+∥y∥2±(⟨x,y⟩+⟨y,x⟩).‖x±y‖2=⟨x±y,x±y⟩=⟨x,x⟩±⟨x,y⟩±⟨y,x⟩+⟨y,y⟩=‖x‖2+‖y‖2±(⟨x,y⟩+⟨y,x⟩). Now you should use the conjugate symmetry to get the desired result. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Dec 9, 2019 at 20:21 answered Dec 9, 2019 at 20:10 Arturo MagidinArturo Magidin 419k 60 60 gold badges 864 864 silver badges 1.2k 1.2k bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. You wrote I'm unsure how 2(x⋅y)2(x⋅y) becomes 2 R e⟨x,y⟩2 R e⟨𝑥,𝑦⟩. It's actually x⋅y+y⋅x,x⋅y+y⋅x, and by conjugate symmetry y⋅x=x⋅y¯¯¯¯¯¯¯¯¯y⋅x=x⋅y¯, so it's x⋅y+x⋅y¯¯¯¯¯¯¯¯¯=2 R e(x⋅y).x⋅y+x⋅y¯=2 R e(x⋅y). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Dec 9, 2019 at 20:13 J. W. TannerJ. W. Tanner 64.1k 4 4 gold badges 44 44 silver badges 89 89 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions linear-algebra inner-products See similar questions with these tags. 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1877	https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/27%3A_Biomolecules_-_Lipids	Skip to main content 27: Biomolecules - Lipids Last updated : Sep 30, 2024 Save as PDF 26.16: Additional Problems 27.0: Why This Chapter? Page ID : 448857 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives When you have completed Chapter 27, you should be able to fulfill all of the detailed objectives listed under each individual section. distinguish among fats and oils, phospholipids, prostaglandins, terpenes and steroids, and be familiar with the sources and biological roles of these substances. define, and use in context, the key terms introduced in this chapter. Lipids are organic compounds with limited water solubility and can be extracted using nonpolar solvents. They include fats, oils, waxes, vitamins, hormones, and nonprotein components of cell membranes. Unlike carbohydrates and proteins, lipids are characterized by their solubility rather than their structure. Key types of lipids discussed in this context include triacylglycerols, eicosanoids, terpenoids, and steroids. 27.0: Why This Chapter? : This chapter introduces lipids, essential biomolecules that play critical roles in biological systems, including energy storage, membrane structure, and signaling. It highlights the diversity of lipid structures, their classifications, and their functional significance in living organisms. Understanding lipids is crucial for comprehending their involvement in health, disease, and biochemistry. 27.1: Waxes, Fats, and Oils : This section discusses waxes, fats, and oils, which are types of lipids distinguished by their structures and properties. Waxes are long-chain fatty acids esterified to long-chain alcohols, providing waterproofing. Fats and oils are triglycerides, formed from glycerol and fatty acids. The key difference lies in saturation: fats (saturated) are solid at room temperature, while oils (unsaturated) are liquid. Their biological roles include energy storage and insulation. 27.2: Soap : This section explores soap, a product derived from the saponification process, where triglycerides react with a strong base, producing glycerol and soap. Soaps consist of long-chain fatty acids with a hydrophilic (water-attracting) head and a hydrophobic (water-repelling) tail, enabling them to emulsify fats and oils in water. This property makes soap effective for cleaning. Additionally, the section discusses different types of soaps and their roles in personal care and household cleaning. 27.3: Phospholipids : Phospholipids are essential components of cellular membranes, composed of two fatty acids, a glycerol backbone, and a phosphate group. Their structure creates a hydrophilic (water-attracting) "head" and two hydrophobic (water-repelling) "tails," allowing them to form bilayers in aqueous environments. This unique arrangement is crucial for membrane fluidity and permeability, facilitating cellular processes like signaling and transport. Phospholipids are involved in the formation of liposomes. 27.4: Prostaglandins and Other Eicosanoids : Prostaglandins were first discovered and isolated from human semen in the 1930s by Ulf von Euler of Sweden. Thinking they had come from the prostate gland, he named them prostaglandins. It has since been determined that they exist and are synthesized in virtually every cell of the body. Prostaglandins, are like hormones in that they act as chemical messengers, but do not move to other sites, but work right within the cells where they are synthesized. 27.5: Terpenoids : Prostaglandins and other eicosanoids are signaling molecules derived from fatty acids, particularly arachidonic acid. They play crucial roles in various physiological processes, including inflammation, pain, fever, and regulation of blood flow. Prostaglandins are synthesized through cyclooxygenase enzymes, leading to diverse biological effects. Eicosanoids can also influence immune responses and are implicated in numerous diseases. 27.6: Steroids : Steroids are a class of lipids characterized by a core structure of four fused carbon rings. They function as hormones and signaling molecules, influencing various biological processes, including metabolism, immune response, and reproductive functions. Examples include cholesterol, which is vital for cell membrane structure and as a precursor to other steroids, and sex hormones like estrogen and testosterone. Understanding steroids is essential for exploring their roles in health and disease. 27.7: Biosynthesis of Steroids : Steroid biosynthesis involves complex biochemical pathways that convert cholesterol into various steroid hormones. Key enzymes, such as cytochrome P450s, play crucial roles in the oxidation and rearrangement of steroid precursors. The process is regulated by factors like hormonal signals and tissue specificity, leading to the synthesis of hormones like cortisol, testosterone, and estrogen. Understanding steroid biosynthesis is vital for insights into metabolic disorders. 27.8: Chemistry Matters—Saturated Fats, Cholesterol, and Heart Disease : This section discusses the relationship between saturated fats, cholesterol, and heart disease. It highlights how high levels of saturated fats can increase low-density lipoprotein (LDL) cholesterol, contributing to atherosclerosis. The text emphasizes the importance of dietary choices, suggesting that replacing saturated fats with unsaturated fats may reduce heart disease risk. It also covers the complexities of cholesterol's role in the body. 27.9: Key Terms 27.10: Summary : The summary of the chapter on lipids highlights the diverse roles of lipids in biological systems, including energy storage, membrane structure, and signaling. It covers different types of lipids, such as fatty acids, triglycerides, phospholipids, and steroids, emphasizing their unique structures and functions. The chapter also discusses the importance of dietary fats and the implications of lipid metabolism for health. 27.11: Additional Problems 26.16: Additional Problems 27.0: Why This Chapter?
1878	https://math.libretexts.org/Courses/Los_Angeles_City_College/Math_230-Mathematics_for_Liberal_Arts_Students/05%3A_Logic/5.02%3A__Statements_and_Quantifiers	Skip to main content 5.2: Statements and Quantifiers Last updated : Feb 12, 2024 Save as PDF 5.1: Introduction 5.3: Compound Statements Page ID : 135413 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Figure Construction of a logical argument, like that of a house, requires you to begin with the right parts. (credit: modification of work “Barn Raising” by Robert Stinnett/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Identify logical statements. Represent statements in symbolic form. Negate statements in words. Negate statements symbolically. Translate negations between words and symbols. Express statements with quantifiers of all, some, and none. Negate statements containing quantifiers of all, some, and none. Have you ever built a club house, tree house, or fort with your friends? If so, you and your friends likely started by gathering some tools and supplies to work with, such as hammers, saws, screwdrivers, wood, nails, and screws. Hopefully, at least one member of your group had some knowledge of how to use the tools correctly and helped to direct the construction project. After all, if your house isn't built on a strong foundation, it will be weak and could possibly fall apart during the next big storm. This same foundation is important in logic. In this section, we will begin with the parts that make up all logical arguments. The building block of any logical argument is a logical statement, or simply a statement. A logical statement has the form of a complete sentence, and it must make a claim that can be identified as being true or false. When making arguments, sometimes people make false claims. When evaluating the strength or validity of a logical argument, you must also consider the truth values, or the identification of true or false, of all the statements used to support the argument. While a false statement is still considered a logical statement, a strong logical argument starts with true statements. Identifying Logical Statements Figure Not all roses are red. (credit: “assorted pink yellow white red roses macro” by ProFlowers/Flickr, CC BY 2.0) An example of logical statement with a false truth value is, “All roses are red.” It is a logical statement because it has the form of a complete sentence and makes a claim that can be determined to be either true or false. It is a false statement because not all roses are red: some roses are red, but there are also roses that are pink, yellow, and white. Requests, questions, or directives may be complete sentences, but they are not logical statements because they cannot be determined to be true or false. For example, suppose someone said to you, “Please, sit down over there.” This request does not make a claim and it cannot be identified as true or false; therefore, it is not a logical statement. Example Identifying Logical Statements Determine whether each of the following sentences represents a logical statement. If it is a logical statement, determine whether it is true or false. Tiger Woods won the Master’s golf championship at least five times. Please, sit down over there. All cats dislike dogs. Answer : 1. This is a logical statement because it is a complete sentence that makes a claim that can be identified as being true or false. As of 2021, this statement is true: Tiger Woods won the Master’s in 1997, 2001, 2002, 2005 and 2019. 2. This is not a logical statement because, although it is a complete sentence, this request does not make a claim that can be identified as being either true or false. 3. This is a logical statement because it is a complete sentence that makes a claim that can be identified as being true or false. This statement is false because some cats do like some dogs. Your Turn Determine whether each of the following sentences represents a logical statement. If it is a logical statement, determine whether it is true or false. The Buffalo Bills defeated the New York Giants in Super Bowl XXV. Michael Jackson’s album Thriller was released in 1982. Would you like some coffee or tea? Representing Statements in Symbolic Form When analyzing logical arguments that are made of multiple logical statements, symbolic form is used to reduce the amount of writing involved. Symbolic form also helps visualize the relationship between the statements in a more concise way in order to determine the strength or validity of an argument. Each logical statement is represented symbolically as a single lowercase letter, usually starting with the letter . To begin, you will practice how to write a single logical statement in symbolic form. This skill will become more useful as you work with compound statements in later sections. Example Representing Statements Using Symbolic Form Write each of the following logical statements in symbolic form. Barry Bonds holds the Major League Baseball record for total career home runs. Some mammals live in the ocean. Ruth Bader Ginsburg served on the U.S. Supreme Court from 1993 to 2020. Answer : 1. : Barry Bonds holds the Major League Baseball record for total career home runs. The statement is labeled with a Once the statement is labeled, use as a replacement for the full written statement to write and analyze the argument symbolically. 2. : Some mammals live in the ocean. The letter is used to distinguish this statement from the statement in question 1, but any lower-case letter may be used. 3. : Ruth Bader Ginsburg served on the U.S. Supreme Court from 1993 to 2020. When multiple statements are present in later sections, you will want to be sure to use a different letter for each separate logical statement. Your Turn Write each of the following logical statements in symbolic form. The movie Gandhi won the Academy Award for Best Picture in 1982. Soccer is the most popular sport in the world. All oranges are citrus fruits. Who Knew? Mathematics is not the only language to use symbols to represent thoughts or ideas. The Chinese and Japanese languages use symbols known as Hanzi and Kanji, respectively, to represent words and phrases. At one point, the American musician Prince famously changed his name to a symbol representing love. BBC Prince Symbol Article Negating Statements Consider the false statement introduced earlier, “All roses are red.” If someone said to you, “All roses are red,” you might respond with, “Some roses are not red.” You could then strengthen your argument by providing additional statements, such as, “There are also white roses, yellow roses, and pink roses, to name a few.” The negation of a logical statement has the opposite truth value of the original statement. If the original statement is false, its negation is true, and if the original statement is true, its negation is false. Most logical statements can be negated by simply adding or removing the word not. For example, consider the statement, “Emma Stone has green eyes.” The negation of this statement would be, “Emma Stone does not have green eyes.” The table below gives some other examples. Table 2.1 \| Logical Statement \| Negation \| \| Gordon Ramsey is a chef. \| Gordon Ramsey is not a chef. \| \| Tony the Tiger does not have spots. \| Tony the Tiger has spots. \| The way you phrase your argument can impact its success. If someone presents you with a false statement, the ability to rebut that statement with its negation will provide you with the tools necessary to emphasize the correctness of your position. Example Negating Logical Statements Write the negation of each logical statement in words. Michael Phelps was an Olympic swimmer. Tom is a cat. Jerry is not a mouse. Answer : 1. Add the word not to negate the affirmative statement: “Michael Phelps was not an Olympic swimmer.” 2. Add the word not to negate the affirmative statement: “Tom is not a cat.” 3. Remove the word not to negate the negative statement: “Jerry is a mouse.” Your Turn Write the negation of each logical statement in words. Ted Cruz was not born in Texas. Adele has a beautiful singing voice. Leaves convert carbon dioxide to oxygen during the process of photosynthesis. Negating Logical Statements Symbolically The symbol for negation, or not, in logic is the tilde, ~. So, not is represented as . To negate a statement symbolically, remove or add a tilde. The negation of not (not ) is . Symbolically, this equation is Example Negating Logical Statements Symbolically Write the negation of each logical statement symbolically. : Michael Phelps was an Olympic swimmer. : Tom is not a cat. : Jerry is not a mouse. Answer : 1. To negate an affirmative logical statement symbolically, add a tilde: . 2. Because the symbol for this statement is , its negation is . 3. The symbol for this statement is , so to negate it we simply remove the ~, because The answer is . Your Turn Write the negation of each logical statement symbolically. Ted Cruz was not born in Texas. Adele has a beautiful singing voice. Leaves convert carbon dioxide to oxygen during the process of photosynthesis. Translating Negations Between Words and Symbols In order to analyze logical arguments, it is important to be able to translate between the symbolic and written forms of logical statements. Example Translating Negations Between Words and Symbols Given the statements: : Elmo is a red Muppet. : Ketchup is not a vegetable. Write the symbolic form of the statement, “Elmo is not a red Muppet.” Translate the statement into words. Answer : 1. “Elmo is not a red muppet” is the negation of “Elmo is a red muppet,” which is represented as . Thus, we would write “Elmo is not a red muppet” symbolically as . 2. Because is the symbol representing the statement, “Ketchup is not a vegetable,” is equivalent to the statement, “Ketchup is a vegetable.” Your Turn Given the statements: r: Woody and Buzz Lightyear are best friends. ~p: Wonder Woman is not stronger than Captain Marvel. Write the symbolic form of the statement, "Wonder Woman is stronger than Captain Marvel." Translate the statement ~r into words. Expressing Statements with Quantifiers of All, Some, or None A quantifier is a term that expresses a numerical relationship between two sets or categories. For example, all squares are also rectangles, but only some rectangles are squares, and no squares are circles. In this example, all, some, and none are quantifiers. In a logical argument, the logical statements made to support the argument are called premises, and the judgment made based on the premises is called the conclusion. Logical arguments that begin with specific premises and attempt to draw more general conclusions are called inductive arguments. Consider, for example, a parent walking with their three-year-old child. The child sees a cardinal fly by and points it out. As they continue on their walk, the child notices a robin land on top of a tree and a duck flying across to land on a pond. The child recognizes that cardinals, robins, and ducks are all birds, then excitedly declares, "All birds fly!" The child has just made an inductive argument. They noticed that three different specific types of birds all fly, then synthesized this information to draw the more general conclusion that all birds can fly. In this case, the child's conclusion is false. The specific premises of the child's argument can be paraphrased by the following statements: Premise: Cardinals are birds that fly. Premise: Robins are birds that fly. Premise: Ducks are birds that fly. The general conclusion is: “All birds fly!” All inductive arguments should include at least three specific premises to establish a pattern that supports the general conclusion. To counter the conclusion of an inductive argument, it is necessary to provide a counter example. The parent can tell the child about penguins or emus to explain why that conclusion is false. On the other hand, it is usually impossible to prove that an inductive argument is true. So, inductive arguments are considered either strong or weak. Deciding whether an inductive argument is strong or weak is highly subjective and often determined by the background knowledge of the person making the judgment. Most hypotheses put forth by scientists using what is called the “scientific method” to conduct experiments are based on inductive reasoning. In the following example, we will use quantifiers to express the conclusion of a few inductive arguments. Example Drawing General Conclusions to Inductive Arguments Using Quantifiers For each series of premises, draw a logical conclusion to the argument that includes one of the following quantifiers: all, some, or none. Squares and rectangles have four sides. A square is a parallelogram, and a rectangle is a parallelogram. What conclusion can be drawn from these premises? and Of these, 1 and 2, 6 and 7, and 23 and 24 are consecutive integers; 3, 13, and 47 are odd numbers. What conclusion can be drawn from these premises? Sea urchins live in the ocean, and they do not breathe air. Sharks live in the ocean, and they do not breathe air. Eels live in the ocean, and they do not breath air. What conclusion can be drawn from these premises? Answer : 1. The conclusion you would likely come to here is “Some four-sided figures are parallelograms.” However, it would be incorrect to say that all four-sided figures are parallelograms because there are some four-sided figures, such as trapezoids, that are not parallelograms. This is a false conclusion. 2. From these premises, you may draw the conclusion “All sums of two consecutive counting numbers result in an odd number.” Most inductive arguments cannot be proven true, but several mathematical properties can be. If we let represent our first counting number, then would be the next counting number and . Because 2n2n is divisible by two, it is an even number, and if you add one to any even number the result is always an odd number. Thus, the conclusion is true! 3. Based on the premises provided, with no additional knowledge about whales or dolphins, you might conclude “No creatures that live in the ocean breathe air.” Even though this conclusion is false, it still follows from the known premises and thus is a logical conclusion based on the evidence presented. Alternatively, you could conclude “Some creatures that live in the ocean do not breathe air.” The quantifier you choose to write your conclusion with may vary from another person’s based on how persuasive the argument is. There may be multiple acceptable answers. Your Turn For each series of premises, draw a logical conclusion to the argument that includes one of the following quantifiers: all, some, or none. 1 + 2 = 3, 5 + 6 = 11, and 14 + 15 = 29. Of these, 1 and 2 are consecutive integers, 5 and 6 are consecutive integers, and 14 and 15 are consecutive integers. Also, their sums, 3, 11, and 29 are all prime numbers. Prime numbers are positive integers greater than one that are only divisible by one and the number itself. What conclusion can you draw from these premises? A robin is a bird that lays blue eggs. A chicken is a bird that typically lays white and brown eggs. An ostrich is a bird that lays exceptionally large eggs. If a bird lays eggs, then they do not give live birth to their young. What conclusion can you draw from these premises? All parallelograms have four sides. All rectangles are parallelograms. All squares are rectangles. What additional conclusion can you make about squares from these premises? Checkpoint It is not possible to prove definitively that an inductive argument is true or false in most cases. Video Logic Part 1A: Logic Statements and Quantifiers Check Your Understanding A _________ __________ is a complete sentence that makes a claim that may be either true or false. The _________________ of a logical statement has the opposite truth value of the original statement. If //p// represents the logical statement, “Marigolds are yellow flowers,” then ______ represents the statement, “Marigolds are not yellow flowers.” The statement //\text{~}(\text{~}p)// has the same truth value as the statement _______. The logical statements used to support the conclusion of an argument are called ____________. _______________________ arguments attempt to draw a general conclusion from specific premises. All, some, and none are examples of ______________________, words that assign a numerical relationship between two or more groups. The negation of the statement, “All giraffes are tall,” is _______________________________. Section 2.1 Exercises For the following exercises, determine whether the sentence represents a logical statement. If it is a logical statement, determine whether it is true or false. A loan used to finance a house is called a mortgage. All odd numbers are divisible by 2. Please, bring me that notebook. Robot, what’s your function? In English, a conjunction is a word that connects two phrases or parts of a sentence together. 8 - 3 = 5. 7 + 3 = 11. What is 7 plus 3? For the following exercises, write each statement in symbolic form. Grammy award winning singer, Lady Gaga, was not born in Houston, Texas. Bruno Mars performed during the Super Bowl halftime show twice. Coco Chanel said, “The most courageous act is still to think for yourself. Aloud.” Bruce Wayne is not Superman. For the following exercises, write the negation of each statement in words. Bozo is not a clown. Ash is Pikachu’s trainer and friend. Vanilla is the most popular flavor of ice cream. Smaug is a fire breathing dragon. Elephant and Piggy are not best friends. Some dogs like cats. Some donuts are not round. All cars have wheels. No circles are squares. Nature’s first green is not gold. The ancient Greek philosopher Plato said, “The greatest wealth is to live content with little.” All trees produce nuts. For the following exercises, write the negation of each statement symbolically and in words. p: Their hair is red. ~q: My favorite superhero does not wear a cape. Wolves howl at the moon. ~u: I do not love New York. ~v: Some cats are not tigers. t: Nobody messes with Texas. ~q: No squares are not parallelograms. ~p: The President does not like broccoli. For the following exercises, write each of the following symbolic statements in words. Given: p: Kermit is a green frog; translate ~p into words. Given: ~r: Mick Jagger is not the lead singer for The Rolling Stones; translate r into words. Given: q: All dogs go to heaven; translate ~q into words. Given: ~s: Some pizza does not come with pepperoni on it; translate s into words. Given: ~p: No pizza comes with pineapple on it; translate ~~p into words. Given: r: Not all roses are red; translate ~~r into words. Given: ~t: Thelonious Monk is not a famous jazz pianist; translate ~~t into words. Given: ~v: Not all violets are blue; translate ~~v into words. For the following exercises, draw a logical conclusion from the premises that includes one of the following quantifiers: all, some, or none. The Ford Motor Company builds cars in Michigan. General Motors builds cars in Michigan. Chrysler builds cars in Michigan. What conclusion can be drawn from these premises? Michelangelo Buonarroti was a great Renaissance artist from Italy. Raphael Sanzio was a great Renaissance artist from Italy. Sandro Botticelli was a great Renaissance artist from Italy. What conclusion can you draw from these premises? Four is an even number and it is divisible by 2. Six is an even number and it is divisible by 2. Eight is an even number and it is divisible by 2. What conclusion can you draw from these premises? Three is an odd number and it is not divisible by 2. Seven is an odd number and it is not divisible by 2. Twenty-one is an odd number and it is not divisible by 2. What conclusion can you draw from these premises? The odd number 5 is not divisible by 3. The odd number 7 is not divisible by 3. The odd number 29 is not divisible by 3. What conclusion can you draw from these premises? Penguins are flightless birds. Emus are flightless birds. Ostriches are flightless birds. What conclusion can you draw from these premises? Plants need water to survive. Animals need water to survive. Bacteria need water to survive. What conclusion can you draw from these premises? A chocolate chip cookie is not sour. An oatmeal cookie is not sour. An Oreo cookie is not sour. What conclusion can you draw from these premises? 5.1: Introduction 5.3: Compound Statements
1879	https://www.chegg.com/homework-help/questions-and-answers/statistical-physics-harmonic-oscillators-canonical-ensemble-want-understand-calculate-part-q34231932	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: Exercise 7.3: Harmonic oscillators in the canonical ensemble Calculate the thermodynamic properties of a set of N distinguishable harmonic oscillators of frequency The Hamiltonian for this system reads Solution 2m2 This must now be used to calculate the partition function without the Gibbs' factor) Z(T, V, N)- D9 Here all integrals factor, too, since H is Statistical Physics , Harmonic oscillators in the canonical ensemble I want to understand why we calculate the partition function without the Gibbs factor 1/N! in the Harmonic oscillators while we used the factor on the ideal gas problem Ps : the book is " Thermodynamics and Statistical Mechanics Walter Greiner, Ludwig Neise, Horst Stöcker. " Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
1880	https://www.studocu.com/row/document/debre-markos-university/development-economics/chapter-two-prod/39976255	Chapter TWO PROD - CHAPTER TWO THE THEORY OF PRODUCTION Introduction: Definition and basic concepts - Studocu Skip to document Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes New Home My Library AI Notes Ask AI AI Quiz Chats Recent You don't have any recent items yet. My Library Courses You don't have any courses yet. Add Courses Books You don't have any books yet. Studylists You don't have any Studylists yet. Create a Studylist Home My Library Discovery Discovery Universities High Schools Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI Chapter TWO PROD CHAPTER TWO PROD.pdf Course Development Economics (econ 2945) 32 documents University Debre Markos University Academic year:2021/2022 Uploaded by: gM gizaw Molla Debre Markos University 0 followers 16 Uploads3 upvotes Follow Recommended for you 9 Ch 4 - Lecture Notes on Development Economics: Growth and Lessons Development Economics Lecture notes 100% (2) 23 Ramsey - economic growth and economic development theory Development Economics Lecture notes 100% (1) 9 Monetary Policy Techniques: Overview of Quantitative & Qualitative Methods Development Economics Lecture notes 100% (1) 70 JavaScript Beginner's Handbook (JS101) - Essential Guide for New Coders Development Economics Tutorial work 100% (5) 4 Econometrics for Finance (AcFn3112) Course Outline and Objectives Development Economics Other 100% (3) Comments Please sign in or register to post comments. Report Document Students also viewed PGDMB15 Lecture 5 Notes:Basic Econometrics & Gauss-Markov Theorem Econ 509 Lecture Notes: Intro to Mathematical Economics I GUJARATI’S EBE SOLUTIONS MANUAL - CHAPTER 1 EXERCISES Learning Basic English Grammar: Parts of Speech Guide Monetary Policy Techniques: Overview of Quantitative & Qualitative Methods Ch 4 - Lecture Notes on Development Economics: Growth and Lessons Related documents Development Economics Moduleeu Development Economics for undergradute Chapter Seven Oligopoly Printed 2 Ramsey - economic growth and economic development theory The Formalist Revolution in Economics: An Analysis of the 1950s Modifying the title for clarity and relevance. New Title: Development Economics Insights: A Case Study (ECON 101) AGRICULTURAL ECONOMICS: DEFINITION, SCOPE, AND DEVELOPMENT MODELS Preview text CHAPTER TWO THE THEORY OF PRODUCTION Introduction: Definition and basic concepts In the production process/ activity, firms turn inputs into output. This transformation of inputs (factor of productions) into output at a particular time period and at a given technology (state of knowledge about the various methods that might be used to transform inputs into outputs) is described by a production function. The production function is a function that shows the highest output that a firm can produce for every specified combination of inputs. It is a purely technical relation which connects factor inputs to outputs. Fixed Vs variable inputs In economics, inputs can be classified as fixed & variable. Fixed inputs are those inputs whose quantity can not readily be changed when market conditions indicate that an immediate change in output is required. Buildings, machineries and managerial personnel are examples of fixed inputs because their quantity cannot be manipulated easily in short time periods. Variable inputs are those inputs whose quantity can be changed almost instantaneously in response to desired changes in output. The best example of variable input is unskilled labor. Short run Vs. long run Short run refers to that period of time in which the quantity of at least one input is fixed. For example, if it requires a firm one year to change the quantities of all the inputs, those time periods below one year are considered as short run. Thus, short run is that time period which is not sufficient to change the quantities of all inputs, so that at least one input remains fixed. One thing to be noted here is that short run periods of different firms have different duration. Some firms can change the quantity of all their inputs within a month while it takes more than a year to change the quantity of all inputs for another type of firms. For example, the time required to change the quantities of inputs in an automobile factory is not equal with that of flour factory. The later takes relatively shorter time. Long run is that time period (planning horizon) which is sufficient to change the quantities of all inputs. Thus there is no fixed input in the long -run. 2. Production in the short run: Production with one variable input Production with one variable input (while the others are fixed) is obviously a short run phenomenon because there is no fixed input in the long run. Total product (TPL), marginal product (MPL) and average product (APL) Total product: is the total amount of output that can be produced by efficiently utilizing a specific combination of labor and capital. The TP curve, thus, represents various levels of output that can be obtained from efficient utilization of various combinations of the variable input, and the fixed input. It shows the output produced for different amounts of the variable input, labor. Marginal Product (MPL) The MP of variable input is the addition to the TP attributable to the addition of one unit of the variable input to the production process, other inputs being constant (fixed). Before deciding whether to hire one more worker, a manager wants to determine how much this extra worker (∆L) will increase output, ∆Q. The change in total output resulting from using this additional worker (holding other inputs constant) is the MP of the worker. If output changes by ∆Q when the number of workers (variable input) changes by ∆L, the change in output per worker or MP of the variable input, denoted as MPL is found as MPL = dL orMPL dTP L Q = ∆ ∆ Thus, MPL measures the slope of the TP curve at a given point. In the short run, the MP of the variable input first increases reaches its maximum and then tends to decrease to the extent of being negative. That is, as we continue to combine more and more of the variable inputs with the fixed input, the MP of the variable input increases initially and then declines. Average Product (AP) The AP of an input is the ratio of total output to the number of variable inputs. L TP numberofL APlabour =totalproduct= The average product of labor (APL) first increases with the number of labor (i. TP increases faster than the increase in labor), and eventually it declines. Graphing the short run production curves It shows how the TPL, MPL and APL vary with the number of the variable input. Fig 3 relationship b/n TP, AP and MP Efficient Region of Production in the short-run We are now not in a position to determine the specific number of the variable input (labor) that the firm should employ because this depends on several other factors than the productivity of labor such as the price of labor, the structure of input and output markets, the demand for output, etc. However, it is possible to determine ranges over which the variable input (labor) be employed do best with this, let’s refer back to fig 3 and divide it into three ranges called stages of production. - Stage I – ranges from the origin to the point of equality of the APL and MPL. - Stage II – starts from the point of equality of MPL and APL and ends at a point where MP is equal to zero. - Stage III – covers the range of labor over which the MPL is negative Obviously, a firm should not operate in stage III because in this stage additional units of variable input are contributing negatively to the total product (MP of the variable input is negative) because of overcrowded working environment i., the fixed input is over utilized. Stage I is also not an efficient region of production though the MP of variable input is positive. The reason is that the variable input (the number of workers) is too small to efficiently run the fixed input; so that the fixed input is underutilized (not efficiently utilized). Thus, the efficient region of production is stage II. At this stage additional inputs are contributing positively to the total product and MP of successive units of variable input is declining (indicating that the fixed input is being optimally used). Hence, the efficient region of production is over that range of employment of variable input where the MP of the variable input is declining but positive. 2 Long run Production: Production with two variable inputs Long run is a period of time (planning horizon) which is sufficient for the firm to change the quantity of all inputs. For the sake of simplicity, assume that the firm uses two inputs (labor and capital) and both are variable. The firm can now produce its output in a variety of ways by combining different amounts of labor and capital. With both factors variable, a firm can usually produce a given level of output by using a great deal of labor and very little capital or a great deal of capital and very little labor or moderate amount of both. In this section, we will see how a firm can choose among combinations of labor and capital that generate the same output. To do so, we make the use of isoquant. So it is necessary to first see what is meant by isoquants and their properties. Isoquants An isoquant is a curve that shows all possible efficient combinations of inputs that can yield equal level of output. If both labor and capital are variable inputs, the production function will have the following form. Q = f (L, K) Given this production function, the equation of an isoquant, where output is held constant at q is Q = f (L, K) Isoquant maps: when a number of isoquants are combined in a single graph, we call the graph an isoquant map. An isoquant map is another way of describing a production function. Each isoquant represents a different level of output and the level of out puts increases as we move up and to the right. The following figure shows isoquants and isoquant map. Fig 3 Isoquant and isoquant map Properties of isoquants Isoquants have most of the same properties as indifference curves. The biggest difference between them is that output is constant along an isoquant where as indifference curves hold utility constant. Most of the properties of isoquants, results from the word ‘efficient’ in its definition. Isoquants slope down ward. Because isoquants denote efficient combination of inputs that yield the same output, isoquants always have negative slope. Thus, efficiently requires that isoquants must be negatively sloped. As employment of one factor increases, the employment of the other factor must decrease to produce the same quantity efficiently. The further an isoquant lies away from the origin, the greater the level of output it denotes. Higher isoquants (isoquants further from the origin) denote higher combination of inputs. The more inputs used, more outputs should be obtained if the firm is producing efficiently. Thus efficiency requires that higher isoquants must denote higher level of output. Isoquants do not cross each other. This is because such intersections are inconsistent with the definition of isoquants the following figure 3. When isoquants are L-shaped, there is only one efficient combination of labor and capital of producing a given level of output. To produce q 1 level of output there is only one efficient combination of labor and capital (L 1 and K 1 ). Output cannot be increased by keeping one factor (say labor) constant and increasing the other (capital). To increase output (say from q1 to q 2 ) both factor inputs should be increased by equal proportion. 3. Kinked isoquants This assumes limited substitution between inputs. Inputs can substitute each other only at some points. Thus, the isoquant is kinked and there are only a few alternative combinations of inputs to produce a given level of output. These isoquants are also called linear programming isoquants or activity analysis isoquants. See the figure below. Fig. 3 kinked isoquant In this case labor and capital can substitute each other only at some point at the kink (A, B, C, and D). Thus, there are only four alternative processes of producing q=100 output. 4. Smooth, convex isoquants This shape of isoquant assumes continuous substitution of capital and labor over a certain range, beyond which factors cannot substitute each other. Basically, kinked isoquants are more realistic: There is often limited (not infinite) method of producing a given level of output. However, traditional economic theory mostly adopted the continuous isoquants because they are mathematically simple to handle by the simple rule of calculus, and they are approximation of the more realistic isoquants (the kinked isoquants). From now on we use the smooth and convex isoquants to analyze the long run production. Fig: 3 the smooth and convex isoquant. This type of isoquant is the limiting case of the kinked isoquant when the number kink is infinite. The slope of the iso quant decrease as we move from the top (left) to the right (bottom) along the isoquant. This indicates that the amount by which the quantity of one input (capital)can be reduced when one extra unit of another inputs(labor)is used ( so that output remains constant) decreases as more of the latter input (labor)is used. The slope of an isoquant: marginal rate of technical substitution (MRTS) The slope of an isoquant (-∆K/∆L) indicates how the quantity of one input can be traded off against the quantity of the other, while output is held constant. The slope of the isoquant (dK/dL) defines the degree of substitutability of the factors of production. This slope decreases (in absolute terms) as we move downwards along the isoquant, showing the increasing difficulty in substituting L for K. The slope of the isoquant is called the rate of technical substitution, or the marginal rate of technical substitution (MRTS) of factors: MRTSL,K = -dK = slope of an isoquant. dL MRTSL,K is defined as the amount of K that the firm must sacrifice in order to obtain one more unit of L so that it produces the same level of output. It is the slope of an isoquant. It can be proved that the MRTS is equal to the ratio of the marginal products of the factors. That is, MRSL,K = -dK = ∂X/∂L = MPL dL ∂X/∂K MPK Proof: The production function can be written as Q = f(K,L)= C. It is equal to C because along an isoquant the TP is constant. The slope of a curve is the slope of a tangent line at that point. The slope of a tangent line is defined by the total differential. The total differential (dQ) is zero along an isoquant since the TP is constant. Thus, dQ = (∂X/∂K)∂K + (∂X/∂L)∂L = 0 (MPK)dK + (MPL)dL = 0 -(MPK)dK = (MPL)dL -dK/dL = MPL/MPK Elasticity of substitution MRTS as a measure of the degree of substitutability of factors has a serious defect. It depends on the units of measurement of factors. A better measure of the ease of factor substitution is provided by the elasticity of substitution, δ. The elasticity of substitution is defined as MRTS L K ∆ ∆ = % % δ = = MRTSLK L K L K d dMRTSLK , ( ) . ,       The elasticity of substitution σ = d(K/L)/(K/L) = 1 (Proof it) d(MRS)/(MRS) Factor intensity. In a Cobb-Douglas function factor intensity is measured by the ratio b 1 /b 2. The higher the ratio the more labor intensive the technique is and vice versa. The efficiency of production. The efficiency in the organization of factors of production is measured by the coefficient bo. It is clear that if two firms have the same K, L, b1, and b2 and still produce different quantities of output, the difference can be due to the superior organization and entrepreneurship of one firm which resulted in production difference. The more efficient firms will have a higher bo than the less efficient one. The returns to scale. In the Cobb-Douglas production function, the returns to scale are measured by the sum of the coefficients b 1 +b 2. It will be discussed the next section. Laws of Returns to Scale The law of variable proportion is a law for the case of short run where there is at least one fixed inputs. In our earlier discussion of the short run production function and stages of production, we have assumed labor as a variable input and capital as a fixed input. From that graph, what we can understand is that as the use of a variable input (labor) increases with other inputs (capital) fixed, the resulting addition to output will eventually decreases. This is shown by a downward sloping MPL curve after its maximum point. This principle is known as the law of variable proportion or the law of Diminishing returns. On the other hand, the law of returns to scale refers to the long run analysis of production. In the long run, where all inputs are variable output can be increased by changing all factors by the same proportion. The rate at which output increases as inputs are increased by the same proportion is called returns to scale. We have three cases of returns to scale: increasing, constant and decreasing returns to scale. I) Increase returns to scale: this is the case where increasing all factors by the same proportion, m, leads to an increase in output by more than m scale. II) Constant returns to scale: if we increase input by some factor, m and output is increased by the same proportion as inputs, m, then it is called constant returns to scale. In this case the size of the firm’s operation doesn’t affect the productivity of its factors. III) Decreasing returns to scale: if scaling up all inputs by m scales output up by less than m, it is called decreasing returns to scale. This is because, may be difficulties in organizing and running a large scale operation may lead to decreased production of both labor and capital. Returns to Scale and Homogeniety of the Production Function Suppose we increase both factors of production function X=f(L, K) by the same proportion m, and we observe the resulting new level of output X as X = f(mK, mL). If m can be factored out (that is, can be taken out of the bracket as a common factor), then the new level of output can be expressed as a function of m (to the power n) and the initial level of output as follows: X = mnf(L, K) or X = mnX. If so, the function is called homogeneous. If m cannot be factored out, the production function is called non- homogeneous. The above three examples are a homogeneous functions since m can be factored out. Thus, a homogeneous function a function such that if each of the inputs is multiplied by m, the m can be completely factored out of the function. The power n of m is called the degree of homogeneity and is a measure of the returns to scale. If n=1, we have a CRS. If n <1, “ DRS. If n >1, “ IRS. Given a Cobb-Douglas production function Q=boLb1Kb2, returns to scale is measured by the sum of the powers of the factors. That is, If b 1 + b 2 =1, then there is a CRS If b 1 + b 2 >1, “ “ IRS If b 1 + b 2 <1, “ “ DRS Proof Let L and K increases by m. The new level of output is X=bo(mL)b1(mK)b = bomb1Lb1mb2Kb X =mb1+b2(boLb1Kb2) X= mb1+b2X This implies the function is homogeneous of degree b 1 +b 2 and the returns to scale depend on the sum. Equilibrium of the firm: Choice of optimal combination of factors of production An isoquant denotes efficient combination of labor and capital required to produce a given level of output. But, this does not mean that the monetary cost of producing a given level of output is constant along an isoquant. That is, though different combinations of labor and capital on a given isoquant yield the same level of output, the cost of these different combinations of labor and capital could differ because the prices of the inputs can differ. Thus, isoquant shows only technically efficient combinations of inputs, not economically efficient combinations. Technical efficiency takes in to account the physical quantity of inputs where as economic efficiency goes beyond technical efficiency and seeks to find the least cost (in monetary terms) combination of inputs among the various technically efficient combinations. Hence, technical efficiency is a Case1: Maximization of output subject to cost constraint Suppose a firm having a fixed cost out lay (money budget) which is shown by its iso-cost line. Here, the firm is in equilibrium when it produces the maximum possible output, given the cost outlay and prices of input. The equilibrium point (economically efficient combination) is graphically defined by the tangency of the firm’s iso-cost line (showing the budget constraint) with the highest possible isoquant. At this point, the slope of the iso cost line ( r w ) is equal to the slope of the isoquant ( K L MP MP ). The condition of equilibrium under this case is, thus: r MP w MP or MP MP r w L K K = L = This is the first order (necessary) condition. The second order (sufficient) condition is that isoquant must be convex to the origin. See the following figure: The optimal combination of inputs ( L 1 andK 1 ) is defined by the tangency of the iso-cost line (AB) and the highest possible isoquant ( X 2 ), at point E. At this point the slope of iso-cost line ( r w ) is equal to the slope of isoquant X 2 ( K L MP MP ).The second order condition is also satisfied by the convexity of the isoquant. Mathematical derivation of the equilibrium condition A rational producer seeks the maximization of its output, given total cost outlay and the prices of factors. That means: maximize X = f (K, L) Subject to C = wL + rK This is a constrained optimization which can be solved by using the lagrangean method. The steps are: a. rewrite the constraint in the form : wL + rK – C = 0 b. multiply the constraint by a constant λwhich is the lagrangian multiplier: λ(wL + rK – C) = 0 c. form the composite function : Z = X - λ(wL + rK – C) d. partially derivate the function and then equate to zero = − λw= = λw λ = ................................................................ (1) Similarly, = − λr= = λr λ = ......................................................... (2) λ = wL + rK − C=0 .................................................................... (3) From equation (1) and (2) we understand that = or = This shows that the firm is in equilibrium when it equates the ratio of the marginal productivities of factors to the ratio of their prices. It can be shown that the second order conditions for the equilibrium of the firm require that the marginal product curves of the two factors have a negative slope. Slope of MPL = => < 0 where as slope of MPK = => < 0 And . > ! " Example Suppose the production function of a firm is given as X = 0. 5 L 1 / 2 K 1 / 2 prices of labor and capital are given as $ 5 and $ 10 respectively, and the firm has a constant cost out lay of $ 600 the combination of labor and capital that maximizes the firm’s output and the maximum output. Solution The condition of equilibrium is r w MP orMP r MP w MP K L = K L = 0. 25 L 1 / 2 K 1 / 2 L X MPL = − ∂ ∂ = and = 0. 251 / 2 − 1 / 2 ∂ ∂ = L K K X MPK Thus, the equilibrium exists when, $ 10 $ 5 0. 25 0. 25 1 / 2 1 / 2 1 / 2 1 / 2 − − = − − L K L K 2 ...................................( 1 ) 2 1 L K L K = ⇒ = The constraint equation is: 5 + 10 = 600 ......................................( 2 ) + = L K wL rK C Solving equation (1) and (2) would give us the optimal combination of L and K. 5 10 600 2 + = = L K L K ⇒ L=60 units and K=30 units, which is graphically represented as: => λ = . ............................................................................ (1) = r − λ ,% ,& =0 => r − λ - / =0 => r = λ => λ = . ............................................................................ (2) λ = Q − f%L, K&=0 ................................................................................. (3) From equation (1) and (2): . = 1 => . 1 = = 234, This is the same as the condition in case one. In a similar way, the second condition will be: Slope of MPL = => < 0 where as Slope of MPK = => < 0 And . > ! " Example: Suppose a certain contractor wants to maximize ∏ from building one bridge. The contractor uses both labor and capital, and efficient combinations of Labor and capital that are sufficient to make a bridge is by the function 0 L 2 1 K 2 1 .If the prices of labor (w) and capital (r) are $ 5 and $ 10 respectively. Find the least cost combination of L and K, and the minimum cost. Solution: The contractor wants to build one bridge. Thus, the constraint equation can be written as 0 L 2 1 2 1 k = MPL = 0 L 2 − 1 K 2 1 and MPK = 0 L 2 1 K 2 − 1 The equilibrium condition is r W MPK MPL = $ 10 $ 5 0. 125 0. 125 2 1 2 1 2 1 2 1 − = − L K L K => L K L K 2 2 1 = ⇒ = Substituting L = 2K in the constraint equation we obtain 0 (2k) 2 1 2 1 k =1 0 2. K= K= 2 4 0. 25 2 1 = K = and L = 2K ⇒ 2 8 Therefore, efficient combination (least cost combination) of L and K are 2 8 and 2 4 respectively. The least cost is C = 5       2 8 +       2 10 4 = $ 2 80 The Expansion Path In the above we explained how firms maximize their output with a given cost & minimize cost with a given output. Thus, it is called constrained maximization. Now we see how firms will change his factor combination as he expand his output & expenditure (cost) with a given factor price. We assume there are four isocost (figure below) AB, CD, UF, & GH, shows different level of total cost. All isocost parallel shows given price of factor(K and L). If firms wants to produce output level Q 1 (100 units), it will chose factor combination E 1 , which minimize cost & point of tangency between Q 1 , &isocost line AB. When he wants higher output Q 2 , the factor combination is E 2 which is the least–cost. For still higher output levels like Q 3 & Q 4 , firm will chose E 3 & E 4 respectively to minimize cost for the given output. The line joining the minimum cost combinations such as E 1 , E 2 , E 3 , E 4 is called the expansion path. Thus, the expansion path may be defined as the locus of the points of tangency between the isocost& the isoquant. Since the expansion path represents the minimum cost combination for various levels of output, it shows the cheapest way of producing each level of output, given the relative prices of the factors. a. The Production Function of Multi-product Firm For the purpose of analysis we assume that a firm produces two products X and Y. The Production Possibility Curve of the Firm (PPC) Each product assumed to be produced by two factors, L and K, for each product we have a production function. X=f 1 (L,K) Y=f 2 (L,K) Each production function is represent by a set of isoquants with the usual properties. We obtain PPC of the firm by using the device of Edgewoth box. we assume firm has total quantities of factors OL and OK(see fig below) measured along sides of Edgewood box. Any point of Edgewoth box shows a certain combination of quantities of X and Y produced by the available R=PxX +PyY, where Px and Py represent the price of commodities sale by the firm for X and Y respectively. Y=R/Py-Px/Py. Thus the slope of Isorevenue curve is dY/dX= -Px/Py Equilibrium of Multiproduct Firm The equilibrium of multiproduct firm is at point of tangency of PPC and the highest point of Isorevnenue curve. At equilibrium e: − 57 = MPL, Y MPL, X = MPk, Y MPk, X = Px Py Chapter TWO PROD Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 0 0 Save Chapter TWO PROD Course: Development Economics (econ 2945) 32 documents University: Debre Markos University Info More info Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 0 0 Save 1 \| P a g e CHAPTER TWO THE THEORY OF PRODUCTION Introduction: Definition and basic concepts In the production process/ activity, firms turn in puts int o output. This t ransf ormation of inputs (factor of productions) into output at a particular time period and at a given technology (state of knowledge about the various methods that might be used to transform i nputs into outputs) is described b y a product ion fun ction. The production function is a function that shows the highest output that a firm can produce f or e ver y specified comb ination of in puts. It is a purely technical relation which connects factor inputs to outputs. Fixed Vs variable in puts In economics, inputs can be classified as fixed &variable. Fixed inputs are those inputs whose quantity can not readily be changed when market co nditions indic ate that an immediate c hange in output is required. Building s, machineries an d manag erial personnel are examples of fix ed inputs because their quantit y cannot be manipulated easily in short time p eriods . Variable inputs are those in puts whose quantity can b e changed almost instantaneously in response to desired changes in output. The best example of variable input is unskilled labor. Short run Vs. long run Short run refers to that period of time in which the quantity of at least one input is fix ed. For example, if it requires a firm one year t o ch ange t he q uantities of all the inputs,those time periods below one year are considered as short run. Thus, sho rt run is that time period which is not sufficient to change the quantities of all inputs,so th at at least one input rema ins fixed. One thing to be noted here is that short run periods of different fir ms have differen t duration. Some firms can ch ange the q uantit y of all their inputs with in a month while it takes more than a y ear to change the quantit y of a ll i nputs for another t ype of firms. F or example, the time required to change the qu antities of inputs in an automobile factor y is not equal with that of fl our factory. The later ta kes relativel y shorter t ime. Long run is that time p eriod (plan nin g hori zon) which is sufficient to change the quantities of all inputs. Thus t here is no fixed input in the long -run. 2.1. Productio n in the shor t run: Production with one variable input Production with one variable input (while the others are fixed) i s obviousl y a sh ort run phenomenon because there is no fixed input in th e long run . Total product (TP L ), marginal product (MP L ) and average product (A P L ) Total product: is the tot al amount of output t hat can be p roduced b y effi ciently u tilizing a specific combination of l abor and capital. The TP curve, thus, represents various levels of output that can be obtained from e fficient utilization o f various co mbinations of the variable input, and the fixed input. It shows the output produced for different amounts of the variable input, labor. 2 \| P a g e Marginal Product (MP L ) The MP of variable input is the addition to the TP attributable to the addition of one unit of the variable i nput to the production process, o ther inputs bein g constant (fixed). Before d eciding whether t o hire one more worker, a manager w ants to d etermine how much this extra worker (∆L) wil l increase output, ∆Q. The change in tot al output resultin g from usin g this additional worker (holdin g other inputs constant) is the M P of the wo rker.If output changes by ∆Q whe n the number of wo rkers (variable input) changes b y ∆L, the change i n output per worker or MP of the variable input, denoted as MP L is found as MP L = dL dTP orMPL L Q= ∆ ∆ Thus, MP L measu res the s lope of th e TP curve at a given point. In the sh ort run, th e MP of the variable i nput first increases reaches its maximum and then tends to decrease to the extent of being negative. Th at is, a s we continue to combine more and more o f the variable inputs with t he fixed input, the MP of the variable input increases initi ally and then declines. Average Product (AP) The AP of an input is the ratio of total output to the number of variable inputs. L TP num berofL ct totalpro du APlabo ur== The average product of labor (AP L ) first i ncreases with the nu mber of l abor (i.e. TP incr eases faster than the increase in labor), and eventuall y it de clines. Graphing the short run p roduction curves It shows how the TP L , MP L and AP L var y with the number of the variable input. Fig 3.1The relationship b/n TP, AP and MP 3 \| P a g e As t he number of t he l abor h ired increases (capital b eing fixed), the TP curve first rises, reaches its ma ximum wh en L 3 a mount of labor i s emplo yed, be yond which it ten ds to decline Assuming that this sh ort run p roduction curve represents a certain car manuf acturing industr y, i t implies that L 3 numbers of workers are requi red to effi ciently run the machineries. If the nu mbers of workers fall b elow L 3 , the machine is not full y operating, resulting in a fall i n TP below L 3 . On the other hand,increasin g the n umber of workers above L 3 will do nothing fo r the p roduction process because onl y L 3 nu mber of workers c an efficiently run the machine. Increasing the number o f workers above L 3 , rather results in lower total product bec ause it results in overcrowded and unfavorable working environment. Marginal product curve increases until L 1 number of labo r reaches its maximum at L 1, and then it tends to fall. The MP L is zero at L 3 (when the TP is maximal); beyond which its value assumes zero indicatin g that eac h additional wo rker above L 3 tends to create o ver crowd ed wor king condition and reduces the to tal product. Thus, in the short run (where some inputs are fixed), the marginal product of successive units of labor hired increases initiall y, but n ot continuousl y, resulting in the limit to the total production. Geometricall y, th e MP cu rve measures the slope of the TP. The slope of the TP cur ve increases (MP increases) up to L 1 , it decreases from L 1 to L 3 and it becomes negative beyond L 3 . The average product curv e increases up to L 2 , beyond which it continuousl y declines. The AP curve can be measured b y the slope of rays originating fro m the origin to a point on t he TP curve. For example, the AP L at L 2 is the ratio of TP 2 to L 2 . This is identical to the slope of ray a. The relationship bet ween AP and MP of the variab le input The relationship between MP L and AP L can be stated as follows: •For all number of workers (Labor) below L 2 , MPL lies above AP L. •At L 2 , MP L and AP L are equal. • Beyond L 2 , MP L lies below the AP L Thus, the MP L curve passes throu gh the maxi mum of the AP L curve from above. The LDMR: short-r un law of production The LDMR (law of diminishing marginal returns) states that as the use of an in put increases in equal increments (with ot her inputs being f ixed), a p oint will eventually be reached at wh ich t he resulting additions t o output decreases. W hen t he l abor input is small (and capital is fix ed), extra labor adds considerabl y to o utput, often because workers get the c hance t o specialize in one or few tasks. Eventually, h owever, the LDMR operates: w hen the number of workers incr eases further, some workers wi ll inevitably become ineffective and the MPL falls(this happens when the number of workers ex ceeds L 1 in fig 3.1) Too long to read on your phone? Save to read later on your computer Save to a Studylist 4 \| P a g e Efficient Region o f Production in the sh ort-run We are now not in a position to determine the spe cific number of the variable input (labor) that the firm should emplo y because this depends on several other f actors th an the p roductivit y of labor such as the price of l abor, the structure o f input and outpu t markets, t he demand for output, etc. However, it is possible to determine ran ges over which the variable input (labor) be employed.To do best with this, let’s refer back to fig 3.1 and divide it into three ranges called stages of production. •Stage I – ranges from the origin to the point of equalit y of the APL and MPL. •Stage II – sta rts fro m the point of equalit y o f MPL and AP L and ends at a point wh ere MP is equal to zero. •Stage III – covers the range of labor over which the MPL is negative Obviously, a firm should not operate in stage III because in this stage addit ional units of variable input are contributing ne gatively to the total product (MP of the v ariable input is negative) because o f overcro wded wor king environment i.e.,the fix ed input i s over uti lized. Stage I is also not an efficient region o f production though the MP of variable input is positive. The rea son is that the va riable input (the number of wo rkers) is too sm all to efficiently run the fixed i nput; so that the fix ed inp ut is u nderutilized (not efficiently utilized). Thus, t he efficient r egion of production is stage II. At this stage additional inputs are contributin g positively to the total product and MP o f suc cessive units of var iable input is de clining (indicating that the fixed input is being optimally used). Hence, the efficient region of production is over that range of employment of variable input where the MP of the variable input is declining but positive. 2.2 Long run Produ ction: Production with t wo variable inputs Long run is a period of time (planning hor izon) which is sufficient for the firm to change the quantity of all inputs. For the s ake of si mplicity, ass ume that the firm u ses two inputs (labor and capital) and both are vari able. The firm can no w produce its output in a variet y of wa ys by combining differen t amounts of labor and capital. W ith both factors variable, a firm can usually produce a given level of output b y using a great deal of labor and ver y little capital or a great deal of capital and very littl e labor or moderate amount of both. In this secti on, we will see how a firm can choose among combinations of labor and capital that g enerate t he same output. To do so, we make the u se of i soquant. So i t is nece ssary to first see what is meant b y iso quants and their properties. Isoquants An isoquant is a curve that shows all possi ble efficient co mbinations o f inputs that can yield equal level of output. If both l abor and capital are variable inputs, the production function wi ll have the following form. Q = f (L, K) Given this production function, the equation of an isoquant, where output is held constant at q is 5 \| P a g e Q = f (L, K) Isoquant maps: when a number of isoquants are c ombined in a single graph, we call the graph an isoquant map. An i soquant map is another way of describing a p roduction function. E ach isoquant represen ts a d iff erent level of output and t he l evel of out puts increa ses as we move up and to the right. The followin g figure shows isoq uants and isoquant map. Fig 3.2 Isoquant and isoquant map Properties of isoqua nts Isoquants have most of the same p roperties as indifference curves. The bi ggest difference between them is that output is constant along an isoq uant where as ind ifference curves hold utilit y const ant. Most o f t he p roperties of isoquants, results from th e word ‘efficient’in its definition. Isoquants slope down ward. Because isoquants denote efficient combin ation of inputs that yield t he same output, isoquants a lways have negative slope. Thus, e fficiently r equires that isoquants must be n egatively sloped. As e mployment of one f actor increases, t he employment of the other factor must decrease to produce the same quantit y efficie ntl y. The furth er an isoquant l ies away from the origin, the g reater the level of output it denotes. Higher iso quants (isoquants further f rom t he o rigin) denot e hi gher combin ation of inputs. The more i nputs used, more outputs should be obtaine d if the firm is producing efficiently. Thus efficiency requires that hi g her isoquants must denote hi gh er level of output. Isoquants do not cross each oth er. This is becaus e such intersections are inconsistent with the definition of isoquants.Consider the following figure 3.4. 6 \| P a g e This figure sh ows that the firm can produ ce at either o utput l evel (20 o r 50) with the same combination o f labor an d capital (L and K). The firm must be producing in efficiently i f it produces q = 20, because it could produce q = 50 by the same combination of labor and capital (L and K). Thus, efficiency requires that isoquants do not cross each other. Shape of isoquants Isoquants can have diffe rent s hapes (curvatu re) d epending on the de gree t o which factor inputs can substitute each other. 1-Linear isoquants Isoquants would be linear when labor and capital are p erfect substitutes for ea ch other. In this case the slope of an iso quant is constant. As a result, the same o utput can be produced with o nly capital or only labor or an infinite combination of both. Graphicall y, Fig.3.5 linear isoquant. Capital and labor can perfectly substitute eac h other so that the same output (q=100) can be produced by using either 10k or 8K and 12L or 15L or an infinite combinations of both in puts. Input output isoq uant It i s also called Leontief isoquant. This assumes st rict complementarities or ze ro substitutabilit y of factors of production. In this case, it is impossi ble to make an y substitu tion among inputs. Each level of o utput requires a specific combination of labor and capital: Additio nal o utput cannot be ob tained unless mo re cap ital and labor are added in specifi c proportions. As a result, the isoquants are L-shaped. See following figure Fig:3.6 L-shaped Isoquant Document continues below Discover more from: Development Economicsecon 2945Debre Markos University 32 documents Go to course 70 JavaScript Beginner's Handbook (JS101) - Essential Guide for New Coders Development Economics Tutorial work 100% (5) 4 Econometrics for Finance (AcFn3112) Course Outline and Objectives Development Economics Other 100% (3) 9 Ch 4 - Lecture Notes on Development Economics: Growth and Lessons Development Economics Lecture notes 100% (2) 20 Chapter 2: Calculus Limits and Continuity for MATH101 Development Economics Practice materials 100% (1) 44 ውዳሴ ማርያም - አዋይ ዘገባ ዕለቱ ክብረ ገድሊ በሀይማኖት ዞር-ምንቱ ዕቅድ ያለው ሞክረ-ትብክህ ይወዳድር ይባል! Development Economics Tutorial work 100% (1) 9 Monetary Policy Techniques: Overview of Quantitative & Qualitative Methods Development Economics Lecture notes 100% (1) Discover more from: Development Economicsecon 2945Debre Markos University32 documents Go to course 70 JavaScript Beginner's Handbook (JS101) - Essential Guide for New Coders Development Economics 100% (5) 4 Econometrics for Finance (AcFn3112) Course Outline and Objectives Development Economics 100% (3) 9 Ch 4 - Lecture Notes on Development Economics: Growth and Lessons Development Economics 100% (2) 20 Chapter 2: Calculus Limits and Continuity for MATH101 Development Economics 100% (1) 44 ውዳሴ ማርያም - አዋይ ዘገባ ዕለቱ ክብረ ገድሊ በሀይማኖት ዞር-ምንቱ ዕቅድ ያለው ሞክረ-ትብክህ ይወዳድር ይባል! Development Economics 100% (1) 9 Monetary Policy Techniques: Overview of Quantitative & Qualitative Methods Development Economics 100% (1) 7 \| P a g e When isoquant s are L-shaped, there is onl y one efficien t combination o f labor and c apital of producing a given level of output. To produce q 1 level of output there is only one efficient combination of labor and cap ital (L 1 and K 1 ). Output cannot b e increased by keeping one f actor (say labor) constant and increasing the other (capit al). To increase ou tput (say from q1 to q 2 ) both factor inputs should be increased b y equal proportion. Kinked isoquants This assumes limited su bstitution between inputs. Inputs c an substitute eac h other onl y at some points. Thus, the isoquant is kinked and there are onl y a few alternative combinations of inputs to produce a given lev el of output. These isoquants are also called li near p ro gramming isoquants or activity analysis isoquants. See the figure below. Fig. 3.7 kinked isoquant In this case labor and capital can substitute each other only at some poi nt a t the kink (A, B, C, and D). Thus, there are only four alternative processes of producing q=10 0 output. Smooth, convex is oquants This shape of iso quant assumes continuous substitution of capital and l abor over a ce rtain rang e, beyond which factors can not substit ute each other. Basically, kinked isoquants are more realistic: There is often limited (not infinite) method of producin g a given level of o utput. However, traditional economic theory mostly adopted the c ontinuous isoquants because they are mathematically simple to handle by the simple rule of calculus, and they are approximation of the more realistic isoquants (the k inked isoquants). F rom n ow on we use t he smooth and convex isoquants to analyze the long run production. Fig: 3.9 the smooth and convex isoquant. 1 out of 19 Share Download Download More from:Development Economics(econ 2945) More from: Development Economicsecon 2945Debre Markos University 32 documents Go to course 70 JavaScript Beginner's Handbook (JS101) - Essential Guide for New Coders Development Economics Tutorial work 100% (5) 4 Econometrics for Finance (AcFn3112) Course Outline and Objectives Development Economics Other 100% (3) 9 Ch 4 - Lecture Notes on Development Economics: Growth and Lessons Development Economics Lecture notes 100% (2) 20 Chapter 2: Calculus Limits and Continuity for MATH101 Development Economics Practice materials 100% (1) More from: Development Economicsecon 2945Debre Markos University32 documents Go to course 70 JavaScript Beginner's Handbook (JS101) - Essential Guide for New Coders Development Economics 100% (5) 4 Econometrics for Finance (AcFn3112) Course Outline and Objectives Development Economics 100% (3) 9 Ch 4 - Lecture Notes on Development Economics: Growth and Lessons Development Economics 100% (2) 20 Chapter 2: Calculus Limits and Continuity for MATH101 Development Economics 100% (1) 44 ውዳሴ ማርያም - አዋይ ዘገባ ዕለቱ ክብረ ገድሊ በሀይማኖት ዞር-ምንቱ ዕቅድ ያለው ሞክረ-ትብክህ ይወዳድር ይባል! 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1881	https://math.stackexchange.com/questions/1755029/no-of-unit-cubes-in-a-nnn-cube	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams No of unit cubes in a nnn cube Ask Question Asked Modified 9 years, 5 months ago Viewed 7k times 2 $\begingroup$ Imagine a cubic array made up of an $n\times n\times n$ arrangement of unit cubes: the cubic array is n cubes wide, n cubes high and n cubes deep. A special case is a $3\times3\times3$ Rubiks cube, which you may be familiar with. How many unit cubes are there on the surface of the $n\times n\times n$ cubic array? As far as I can see there are 27 unit cubes in a $n\times n\times n$ rubik cube. But the answer says something different. There are total $6n^2$ squares are present in $n\times n\times n$ cube. But after that I cant proceed. Please help :) rubiks-cube Share edited Apr 23, 2016 at 4:39 Stella Biderman 31.5k66 gold badges4949 silver badges9494 bronze badges asked Apr 23, 2016 at 4:36 ViX28ViX28 64511 gold badge1212 silver badges2525 bronze badges $\endgroup$ 8 $\begingroup$ You mean in an $n \times n \times n$ cube composite, how many cubes are there which you can see from the outside? $\endgroup$ Sarvesh Ravichandran Iyer – Sarvesh Ravichandran Iyer 2016-04-23 04:39:19 +00:00 Commented Apr 23, 2016 at 4:39 $\begingroup$ @Ð°ÑÑÐ¾Ð½Ð²ÑÐ»Ð»Ð°Ð¾Ð»Ð¾ÑÐ¼ÑÐ»Ð»Ð±ÑÑÐ³ exactly $\endgroup$ ViX28 – ViX28 2016-04-23 04:41:11 +00:00 Commented Apr 23, 2016 at 4:41 $\begingroup$ In the $3\times 3\times 3$ there are $26$, all but the central cubelet. More generally there are $n^3-(n-2)^3$, which can be written in other less informative ways. $\endgroup$ André Nicolas – André Nicolas 2016-04-23 04:41:26 +00:00 Commented Apr 23, 2016 at 4:41 $\begingroup$ @AndréNicolas could you please explain a bit further .. the logic why $ (n-2)^3 $ is subtracted $\endgroup$ ViX28 – ViX28 2016-04-23 04:42:42 +00:00 Commented Apr 23, 2016 at 4:42 1 $\begingroup$ In the $n\times n\times n$ case, the invisible part is $(n-2)\times (n-2)\times (n-2)$, so the visible part is all $n^3$ of them except for the (n-2)^3$ invisibles. $\endgroup$ André Nicolas – André Nicolas 2016-04-23 04:45:01 +00:00 Commented Apr 23, 2016 at 4:45 \| Show 3 more comments 2 Answers 2 Reset to default 4 $\begingroup$ So, i think that thing you're missing is that you're counting the number of squares on the surface; not the number of cubes. For example, a corner piece of a Rubix cube is one cube but contributes three squares. In a $n\times n\times n$ cube, you have the outer layer of cubes counting, and the ones on the inside not counting. The easiest way to see this is recursive: for a $n\times n\times n$ cube, you have an $(n-2)\times (n-2)\times (n-2)$ cube on the inside that is coated by a number of cubes that form the cubes on the outside. Subtracting $$n^3-(n-2)^3=6n^2-12n+8$$ gives the final answer. This generalizes immediately to higher dimensions... In dimension $k$ the answer is $$n^k-(n-2)^k$$ Share edited Apr 23, 2016 at 4:51 answered Apr 23, 2016 at 4:44 Stella BidermanStella Biderman 31.5k66 gold badges4949 silver badges9494 bronze badges $\endgroup$ 2 $\begingroup$ that will be $12n$ instead of $6n$. $\endgroup$ ViX28 – ViX28 2016-04-23 04:50:54 +00:00 Commented Apr 23, 2016 at 4:50 $\begingroup$ @ViX28 thanks! Fixed that and added the general formula $\endgroup$ Stella Biderman – Stella Biderman 2016-04-23 04:51:36 +00:00 Commented Apr 23, 2016 at 4:51 Add a comment \| 1 $\begingroup$ There are $6$ faces with $n^2$ cubes on each face for a total of $6n^2$ cubes. The eight cubes on the vertices are counted $3$ times each so we must subtract $16$ to get $6n^2-16$ cubes. Likewise, there are $12$ edges, each with $n-2$ cubes that have been double counted so we must subtract $12(n-2)$ to get $6n^2-16-12n+24=6n^2-12n+8$ cubes. Share answered Apr 23, 2016 at 4:52 John DoumaJohn Douma 12.6k22 gold badges2626 silver badges2727 bronze badges $\endgroup$ 1 1 $\begingroup$ More simply: $6n^2$ cubes on each face. We have counted the edges too many times, so subtract $12n$. But now we have undercounted the vertices, so add $8$ again. This gives $6n^2-12n+8$. $\endgroup$ Théophile – Théophile 2016-04-23 05:03:18 +00:00 Commented Apr 23, 2016 at 5:03 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions rubiks-cube See similar questions with these tags. 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1882	https://www.ahajournals.org/doi/10.1161/strokeaha.110.586842	Stratified, Urgent Care for Transient Ischemic Attack Results in Low Stroke Rates \| Stroke Skip to main content Advertisement Become a member Volunteer Donate Journals BrowseCollectionsSubjectsAHA Journal PodcastsTrend Watch ResourcesCMEAHA Journals @ MeetingsJournal MetricsEarly Career Resources InformationFor AuthorsFor ReviewersFor SubscribersFor International Users Alerts 0 Cart Search Sign inREGISTER Quick Search in Journals Enter search term Quick Search anywhere Enter search term Quick search in Citations Journal Year Volume Issue Page Searching: This Journal This JournalAnywhereCitation Advanced SearchSearch navigate the sidebar menu Sign inREGISTER Quick Search anywhere Enter search term Publications Arteriosclerosis, Thrombosis, and Vascular Biology Circulation Circulation Research Hypertension Stroke Current Issue Archive Journal Information About Stroke Author Instructions Editorial Board Information for Advertisers Features Stroke Alert Podcast Blogging Stroke Early Career Program Journal Awards Journal of the American Heart Association Circulation: Arrhythmia and Electrophysiology Circulation: Cardiovascular Imaging Circulation: Cardiovascular Interventions Circulation: Cardiovascular Quality & Outcomes Circulation: Genomic and Precision Medicine Circulation: Heart Failure Stroke: Vascular and Interventional Neurology Annals of Internal Medicine: Clinical Cases Information For Authors For Reviewers For Subscribers For International Users Submit & Publish Submit to the AHA Editorial Policies Open Science Value of Many Voices Publishing with the AHA Open Access Information Resources AHA Journals CME AHA Journals @ Meetings Metrics AHA Journals Podcast Network Early Career Resources Trend Watch Professional Heart Daily AHA Newsroom Current Issue Archive Journal Information About Stroke Author Instructions Editorial Board Information for Advertisers Features Stroke Alert Podcast Blogging Stroke Early Career Program Journal Awards Submit Reference #1 Research Article Originally Published 14 October 2010 Free Access Stratified, Urgent Care for Transient Ischemic Attack Results in Low Stroke Rates Jason Wasserman, PhD, Jeff Perry, MD, MSc, Dar Dowlatshahi, MD, PhD, Grant Stotts, MD, Ian Stiell, MD, MSc, Jane Sutherland, RN, Cheryl Symington, RN, and Mukul Sharma, MD, MScAuthor Info & Affiliations Stroke Volume 41, Number 11 5,468 63 Metrics Total Downloads 5,472 Last 12 Months 159 Total Citations 63 Last 12 Months 2 View all metrics Track CitationsAdd to favorites PDF/EPUB Contents Abstract Methods Results Discussion Conclusion Acknowledgments Footnote Supplemental Material References eLetters Information & Authors Metrics & Citations View Options References Figures Tables Media Share Abstract Background and Purpose—Transient ischemic attack (TIA) is a marker for early risk of stroke. No previous studies have assessed the use of urgent stroke prevention clinics for emergency department (ED) patients with TIA. We hypothesized that an ABCD2-based ED triaging tool for TIA with outpatient management would be associated with lower 90-day stroke rate than that predicted by ABCD2. Methods—A cohort of prospectively identified patients presenting with symptoms suggestive of TIA seen in 2 tertiary-care EDs. These patients were divided into 3 strata based on their ACBD2 score, and triage targets were set for each stratum. All patients received the same standard of care in the Stroke Clinic regardless of their risk score. Primary outcome was stroke by 90 days of index TIA. Secondary outcomes were subsequent TIA, myocardial infarction, or death. Results—One-thousand ninety-three patients met the inclusion criteria; 982 patients completed 90-day follow-up and comprised the final cohort. After stratification, 32%, 49%, and 19% of patients were categorized as low-, moderate-, or high-risk, respectively. The overall 90-day risk of stroke in all patients was 3.2%, compared with the ABCD2-predicted risk of 9.2%. Only 1.6% of patients with TIA/minor stroke were admitted from the ED. The risk of subsequent TIA, myocardial infarction, or death by 90 days was 5.5%, 0.1%, and 1.7%, respectively. Conclusion—Outpatient care in a rapid-access stroke prevention clinic using the ABCD2 score for triage resulted in a low 90-day stroke rate for patients in the ED with TIA. Benefit occurred without requiring admission for most patients. Transient ischemic attack (TIA) is a marker for early risk of stroke. Stroke risk is 10% during the 90 days following TIA,1–4 and 25% of all strokes are preceded by a TIA.5 The occurrence of a TIA provides a unique opportunity for stroke prevention. Symptoms of TIA are nonspecific, and false-positive rates in the ED can be as high as 60%.6 Stratification techniques are required to effectively prioritize high-risk patients for urgent management. The ABCD2 score was devised to identify those patients with the highest risk of stroke following TIA; those classified as “high risk” carry a stroke risk of 17.8% at 90 days post TIA.7 Previous studies have suggested improved outcomes by referral to dedicated clinics 8,9 or by hospital admission.10,11 Application of these models is limited by resource constraints related to the random nature of events, which may overwhelm capacity and costs associated with inpatient care. No previous studies have assessed the use of urgent stroke prevention clinics for patients in the ED with TIA. In this study, we identified and followed a prospective cohort of patients presenting to 2 tertiary EDs (combined annual census of 120 000 visits); these patients were diagnosed with TIA and were referred to a comprehensive stroke prevention clinic. We hypothesized that an ABCD2-based ED triaging tool for TIA with outpatient management stratified by risk would be associated with lower 90-day stroke rate than that predicted by the individual patient ABCD2 scores. Methods Design and Setting This prospective cohort study was conducted at the 2 emergency departments of the Ottawa hospital. This is a 1172-bed, tertiary-care, regional stroke referral center with 126 850 emergency visits per year divided between 2 emergency departments. Local ethics board approval was obtained prior to the initiation of this study, and the data were collected by chart review, phone interview, and Stroke Clinic follow-up. Study Population We prospectively enrolled adult patients with a final ED diagnosis of TIA between January 2007 and April 2009. Patients were excluded if they fell into 1 of the following categories: diagnosis in emergency department with a confirmed stroke (ie, neurological deficit >24 hours); decreased level of consciousness (ie, Glasgow Coma Scale score of <15); cause for the deficit was other than TIA (e.g., hypoglycemia, seizure, electrolyte imbalance, or migraine); presentation to the enrolling ED more than 7 days following onset of most recent TIA. A standardized data collection form was completed by the emergency physician at the time of presentation and the following information was prospectively collected: history of neurological symptoms, physical findings, ECG and CT head results, past medical history, medication history and medications started or discontinued in the ED, and ABCD2 score for each patient. The data collection form also included an option to have the patient referred to the Stroke Clinic for follow-up. Only those patients referred to the Stroke Clinic for follow-up were included in the final analysis. Emergency Department Management All subjects had blood work taken, including complete blood count, electrolytes, renal and liver function, and international normalized ratio. An electrocardiogram and computed tommography head scan were performed while in the ED. Fasting blood glucose, lipids, carotid Dopplers, and echocardiogram 24-hour Holter monitor were scheduled as outpatient visits. Emergency physicians had the option of consulting the neurology service. They prescribed medications at their own discretion. Recommendations with respect to antiplatelet and antihypertensive agents were provided by the Stroke Clinic based on current guidelines. Follow-Up in the Stroke Clinic In January 2007, a dedicated and comprehensive Stroke Clinic was established at the Ottawa Hospital to provide rapid, standardized assessment and treatment for patients at risk for stroke. Only patients referred to the Stroke Clinic from the ED at the Ottawa Hospital were included in this study. These patients were divided into 3 strata based on their ACBD2 score, and triage targets were set for each stratum. Specifically, patients classified as high risk (ABCD2 ≥6), moderate risk (ABCD2 =4 to 5), or low risk (ABCD2 <4) were scheduled to see a stroke neurologist within 7 days, 7 to 14 days, or more than 14 days of the index TIA, respectively. In all instances, carotid Doppler, echocardiagram, fasting glucose, and lipids were obtained prior to the clinic visit. All patients received the same standard of care in the Stroke Clinic regardless of their risk score. Test results were reviewed with patients during their appointments, and any necessary medication changes were made. Primary and Secondary Outcomes The primary outcome was stroke within 90 days of index TIA. Secondary outcomes were recurrent TIA, myocardial infarction, or death within 90 days. Outcomes were assessed using a validated, standardized telephone questionnaire12 and chart review. If the occurrence of TIA or stroke was confirmed by a neurologist, it was deemed a subsequent event. Other outcomes were adjudicated by a 3-physician committee, which was blinded to the initial data collection form and ABCD2 score (and its components). Statistical Analysis Data were collected and analyzed in SPSS 17 (SPSS Inc, Chicago, Ill). The risk of stroke was calculated for the cohort as a whole and for the each stratum independently. Fisher exact test was used to compare the number of strokes at 90 days with that predicted by the ABCD2 score for each stratum, and for the cohort as a whole. Continuous data are presented as mean±SD where applicable. ABCD2 scores are presented as median with interquartile range. Results One-thousand, ninety-one patients were given a final diagnosis of TIA, and 1004 patients (92% of TIAs) were referred to the Stroke Clinic for assessment. After referral, follow-up data were not available for 22 patients, leaving a total of 982 patients for final analysis (Figure 1). Open in Viewer Figure 1.Study design. Only patients given a final ED diagnosis of TIA were included in this study. After referral to the Stroke Clinic, patients were triaged based on their ABCD2 score. Patients high risk, moderate risk, or at low risk were scheduled to see a stroke neurologist within 7 days, 7 to 14 days, or more than 14 days of the index TIA, respectively. All patients received the same standard of care regardless of their risk strata. Baseline characteristics of all patients referred to the Stroke Clinic are shown in Table 1. Sixty-eight percent of all patients arrived less than 24 hours after the onset of symptoms, and 63% of all patients reported that the symptoms lasted for at least 1 hour. Only 1.6% of all patients diagnosed with TIA were admitted from the ED. Eight percent (n=81) of cases were discussed with a neurologist while the patient was in the ED, whereas a neurologist was consulted in 5% (n=53) of all cases. Baseline characteristics between patients referred to the Clinic and those not referred were generally very similar; however, some differences were apparent. Specifically, there was no significant difference in the number of females (49.1% compared with 52.8%, P=0.50) or the number of patients with a prior history of stroke (11.7% compared with 16.0%, P=0.23). However, patients not referred were more likely to be admitted to the hospital (1.6% compared with 15.7%, P<0.0001) for their index event. Open in Viewer Table 1. Characteristics of Patients Presenting to the ED With TIA and Referred to the Stroke Clinic for Management N (%). Age, yr, mean (range)67 (19–97) Male 500 (50.9) Admitted to hospital 16 (1.6) Risk factors Hypertension 571 (58.1) Coronary artery disease 163 (16.6) Atrial fibrillation 84 (8.6) Diabetes 195 (19.9) Prior stroke 115 (11.7) Smoking 131 (13.3) Hyperlipidemia 319 (32.5) Clinical features Time from index TIA to presentation <24 hours 660 (67.2) <48 hours 184 (18.7) <1 week 132 (13.4) Duration of symptoms <1 minute 18 (1.8) 1–5 minutes 61 (6.2) 5–9 minutes 44 (4.5) 10–29 minutes 138 (14.1) 30–59 minutes 104 (10.6) ≥60 minutes 613 (62.4) Weakness 410 (41.8) Sensory 493 (50.2) Gait 225 (22.9) Speech 319 (32.5) Visual loss One eye 54 (5.5) Both eyes 61 (6.2) Pronator drift 65 (6.6) Atrial fibrillation on EKG 45 (4.6) Expand Table On arrival at the ED, 41% and 27% of all patients were already taking an antihypertensive medication or a statin, respectively (Table 2). Thirty-five percent (n=343) of patients were already taking an antiplatelet agent on arrival to the ED, while 90% (n=887) of patients left the ED on at least 1 antiplatelet agent. Of those patients not on an antiplatelet agent at discharge, 50% (n=48) were already taking or were started on Warfarin in the ED. Forty-five percent of patients (n=443) had evidence of carotid stenosis identified by carotid Doppler, while less than 1% of patients (n=5) underwent carotid endarterectomy. None of the patients who underwent carotid endarterectomy were admitted to the hospital during their initial visit. The time between presentation and outpatient carotid Doppler, echocardiogram, or Holter monitoring is shown in Table 3. Open in Viewer Table 2. Medications Already Being Taken, Started, or Discontinued in the ED Among Patients Diagnosed With TIA in Our Study \| Pharmaceutical \| Already Taking \| Started in ED \| Discontinued in ED \| --- --- \| \| Acetylsalicylic acid \| 284 (28.9) \| 429 (43.7) \| 57 (5.8) \| \| Clopidogrel \| 69 (7.0) \| 59 (6.0) \| 6 (0.6) \| \| Dipyridmole \| 26 (2.6) \| 172 (17.5) \| 5 (0.5) \| \| Statin \| 262 (26.7) \| 20 (2.0) \| 0 (0) \| \| Antihypertensive \| 403 (41.0) \| 16 (1.6) \| 0 (0) \| \| Ticlodipine \| 2 (0.2) \| 3 (0.3) \| 0 (0) \| \| Warfarin \| 66 (6.7) \| 7 (0.7) \| 0 (0) \| Expand Table Open in Viewer Table 3. Time Between Event and Outpatient Investigation \| \| Event to Investigation (Days) \| Performed Within 2 Weeks (%) \| --- \| \| Mean±SD. \| \| Carotid Doppler (n=831) \| 12.5±14.1 \| 71.5 \| \| Echocardiogram (n=792) \| 16.8±18.6 \| 57.1 \| \| Holter monitor (n=282) \| 30.7±18.2 \| 20.2 \| Expand Table The overall stroke risk at 90 days was 3.2% (Figure 2), with almost one third of the strokes occurring within 2 days of the index TIA (Table 4). The risk of subsequent TIA at 90 days was 5.5% (Figure 2), and in contrast to stroke, almost half of all the recurrent events occurred between 30 days and 90 days (Table 4). All-cause mortality was 1.7% at 90 days of index TIA, and in 3 of these patients, stroke was the cause of death. None of the patients not referred to the clinic had a stroke within 90 days of index TIA. Open in Viewer Table 4. Temporal Distribution of Recurrent Stroke or TIA or MI After Index TIA \| Event \| ≤2 Days \| >2 Days and≤7 Days \| >7 Days and≤30 Days \| >30 Days and≤90 Days \| All Events \| --- --- --- \| \| N (%). \| \| Stroke \| 10 (1.0) \| 9 (0.9) \| 7 (0.7) \| 5 (0.5) \| 31 (3.2) \| \| TIA \| 9 (0.9) \| 11 (1.1) \| 11 (1.1) \| 24 (2.4) \| 55 (5.5) \| \| MI \| 0 (0) \| 0 (0) \| 0 (0) \| 1 (0.1) \| 1 (0.1) \| Expand Table Open in Viewer Figure 2.Cumulative risk of recurrent stroke or TIA at 2, 7, 30, or 90 days after index TIA. The risk of stroke was significantly lower than that predicted by ABCD2 at 90 days for patients in each risk category and for all patients combined (Table 5). There was no significant difference in the stroke risk between patients in the moderate risk (ABCD2=4 to 5) and high-risk (ABCD2 ≥6) groups (P=0.40). The median ABCD2 score for patients referred to the Clinic was 4 (3, 5) compared with 5 (3, 6) for patients who were not referred. The median ABCD2 score for patients seen by a neurologist in the ED was 5 (4, 6). Open in Viewer Table 5. The 90-Day Stroke Rate in our Prospective Cohort Study of 982 TIA Patients Compared With the 90-Day Stroke Rate Predicted by the ABCD2 Score \| \| Strokes \| % (CI) \| Predicted (%) \| P Value \| --- --- \| \| ABCD ≤4 (n=321) \| 3 \| 0.9 (0–1.98) \| 3.1 \| 0.0364 \| \| ABCD 4–5 (n=469) \| 18 \| 3.8 (2.1–5.58) \| 9.8 \| <0.0001 \| \| ABCD ≥6 (n=192) \| 10 \| 5.2 (2.07–8.35) \| 17.8 \| <0.0001 \| \| Combined \| 31 \| 3.2 (2.07–4.25) \| 9.1 \| <0.0001 \| Expand Table Discussion In our study, patients with TIA presenting to a tertiary-care ED were risk stratified and managed urgently by an outpatient Stroke Clinic, yielding a 90-day stroke risk of 3.2%. This risk is one third of that which was predicted by the ABCD2 score for this cohort.7 We believe that the reduced stroke rate can be attributed to the system of care that combines the ED care and the clinic process. The ABCD2 score allows for the stratification of patients based on demographic and clinical features to expedite management for those at high risk of stroke.7 The present study is the first to use the ABCD2 score as a prospective triage tool for patients who were diagnosed with TIA in the ED and referred to an outpatient Stroke Clinic. Although all patients received similar medical care in the ED based on an institutional algorithm, the triage tool was used to determine how quickly a patient would be seen in the Stroke Clinic. Investigations and treatments were subsequently catered to each patient by the treating stroke neurologist. Most patients received carotid Doppler imaging within 2 weeks of their index event, a time frame previously identified as having the most significant benefit from endarterectomy.13 Study Strengths Our cohort was prospectively identified, consecutive, and followed for the occurrence of endpoints through a phone contact system. Medical records were routinely reviewed to ensure that all possible endpoints were collected. Standardized data forms were used along with adjudication of neurological events not identified by a neurologist. It is unlikely that endpoints were missed. Limitations Given the observational nature of this study, there are several limitations. First, this cohort was restricted to comparisons with predicted risks. Second, without baseline magnetic resonance imaging or computed tommography angiogram, we were unable to establish whether early strokes were due to progression of initial symptoms or a new event. This information would be relevant for developing interventions that differentiate between stroke prevention and infarct extension. Third, our data do not allow us to distinguish between the impact of ED interventions and the Stroke Clinic interventions. The low stroke rates suggest that outpatient care results in acceptable outcomes. Additional work is needed to identify the relative contribution of the initial management to these outcomes. Finally, we were unable to account for patient compliance with medications prescribed. The present study is unique in that after ED discharge, patients were followed by a stroke neurologist in a comprehensive and dedicated Stroke Clinic. In contrast, patients in the SOS-TIA study were referred back to their family physician with targets for modifiable risk factors, while in the EXPRESS study, no follow-up details were provided.8,9 In addition to ED treatment algorithms and time-to-stroke–clinic assessment, it is possible that ongoing management from the specialized clinic contributed to lower rates of 90-day stroke; the literature suggests risk factors are often suboptimally managed in the community and can benefit from Stroke Clinic involvement.14,15 Clinical Implications There is mounting evidence that specialized clinics for managing patients with TIA are associated with a reduced risk of stroke. The EXPRESS and SOS-TIA studies demonstrated that the implementation of rapid-access clinics to assess and initiate treatment following TIA was associated with a 2.1% and 1.2% risk of 90-day stroke, respectively.8,9 In these studies, patients with a suspected TIA were referred primarily from community clinics; in our study, patients originated from tertiary-care EDs, with one third presenting by ambulance, possibly reflecting a more acute population. Furthermore, the median length of symptoms was considerably longer in our study, suggesting a more severe presentation and a higher risk of stroke. Despite these differences, the community-based studies and our study all found significantly lower rates of stroke than would be expected based on ABCD2 scores. This lower event rate likely reflects more rapid, complete management than it did during the time of the ABCD2 study. Notably, our study found no difference in the rate of stroke between the medium- (ABCD2=4 to 5) and high-risk (ABCD2 >6) groups; thus, patients at highest risk may have benefitted most from improved management strategies. An alternate approach for TIA management is brief admission and observation. One study admitted nearly 70% of all patients diagnosed with a TIA and demonstrated an overall 2.4% risk of stroke at 90 days.11 More recently, a Canadian study proposed admitting high-risk TIA patients for rapid evaluation, resulting in a 4.7%, 90-day stroke rate with this approach.10 Conversely, only 3.6% of our high-risk TIA population was admitted, with a comparatively low 90-day stroke rate of 3.2%. Our study demonstrates that reductions in stroke risk can be achieved without admission to a hospital. Research Implications An intriguing finding in this study was the discrepancy between the timing of stroke as compared with recurrent TIA: almost one third of all strokes happened within 48 hours of the initial event, whereas almost half of recurrent TIAs occurred between 30 days and 90 days. Recent magnetic resonance imagaing studies suggest that early strokes are in fact a progression of the initial event, and it may be the case that a subset of our patients with early stroke represented either fluctuating lacunar syndromes, or infarction following partial large vessel occlusions.16 Additional studies with baseline computed tommography-angiography and/or magnetic resonance imaging are required to explore this. Conclusion Our study demonstrated that an ABCD2-based TIA triage rule in a tertiary-care ED with rapid-access Stroke Clinic follow-up resulted in a low 90-day stroke rate. This low rate was achieved without requiring hospital admission in high-risk patients. These results demonstrate that low rates of stroke are achievable without costly hospitalization through a well-organized regional outpatient stroke prevention clinic. Acknowledgments Sources of Funding The work was supported by a grant from CIHR. Disclosures None. Footnote Continuing medical education (CME) credit is available for this article. Go to to take the quiz. Supplemental Material File(wasserman_2601.pdf) Download 711.32 KB References 1. Gladstone DJ, Kapral MK, Fang J, Laupacis A, Tu JV. Management and outcomes of transient ischemic attacks in Ontario. CMAJ . 2004; 170: 1099–1104. Go to Citation Crossref PubMed Google Scholar 2. Hill MD, Yiannakoulias N, Jeerakathil T, Tu JV, Svenson LW, Schopflocher DP. The high risk of stroke immediately after transient ischemic attack: A population-based study. Neurology . 2004; 62: 2015–2020. Crossref PubMed Google Scholar 3. Johnston SC, Gress DR, Browner WS, Sidney S. Short-term prognosis after emergency department diagnosis of TIA. JAMA . 2000; 284: 2901–2906. Crossref PubMed Google Scholar 4. Lisabeth LD, Ireland JK, Risser JM, Brown DL, Smith MA, Garcia NM, Morgenstern LB. Stroke risk after transient ischemic attack in a population-based setting. Stroke . 2004; 35: 1842–1846. Crossref PubMed Google Scholar 5. Rothwell PM, Warlow CP. Timing of TIAs preceding stroke: Time window for prevention is very short. Neurology . 2005; 64: 817–820. Go to Citation Crossref PubMed Google Scholar 6. Prabhakaran S, Silver AJ, Warrior L, McClenathan B, Lee VH. Misdiagnosis of transient ischemic attacks in the emergency room. Cerebrovasc Dis . 2008; 26: 630–635. Go to Citation Crossref PubMed Google Scholar 7. Johnston SC, Rothwell PM, Nguyen-Huynh MN, Giles MF, Elkins JS, Bernstein AL, Sidney S. Validation and refinement of scores to predict very early stroke risk after transient ischaemic attack. Lancet . 2007; 369: 283–292. Crossref PubMed Google Scholar a [...] a stroke risk of 17.8% at 90 days post TIA. b [...] by the ABCD2 score for this cohort. c [...] for those at high risk of stroke. 8. Rothwell PM, Giles MF, Chandratheva A, Marquardt L, Geraghty O, Redgrave JN, Lovelock CE, Binney LE, Bull LM, Cuthbertson FC, Welch SJ, Bosch S, Alexander FC, Silver LE, Gutnikov SA, Mehta Z. Effect of urgent treatment of transient ischaemic attack and minor stroke on early recurrent stroke (EXPRESS study): A prospective population-based sequential comparison. Lancet . 2007; 370: 1432–1442. Crossref PubMed Google Scholar a [...] outcomes by referral to dedicated clinics b [...] study, no follow-up details were provided. c [...] 1.2% risk of 90-day stroke, respectively. 9. Lavallee PC, Meseguer E, Abboud H, Cabrejo L, Olivot JM, Simon O, Mazighi M, Nifle C, Niclot P, Lapergue B, Klein IF, Brochet E, Steg PG, Leseche G, Labreuche J, Touboul PJ, Amarenco P. A transient ischaemic attack clinic with round-the-clock access (SOS-TIA): Feasibility and effects. Lancet Neurol . 2007; 6: 953–960. Crossref PubMed Google Scholar 10. Wu CM, Manns BJ, Hill MD, Ghali WA, Donaldson C, Buchan AM. Rapid evaluation after high-risk TIA is associated with lower stroke risk. Can J Neurol Sci . 2009; 36: 450–455. Crossref PubMed Google Scholar a [...] or by hospital admission. b [...] 90-day stroke rate with this approach. 11. Stead LG, Bellolio MF, Suravaram S, Brown RD, Jr., Bhagra A, Gilmore RM, Boie ET, Decker WW. Evaluation of transient ischemic attack in an emergency department observation unit. Neurocrit Care . 2009; 10: 204–208. Go to Citation Crossref PubMed Google Scholar 12. Jones WJ, Williams LS, Meschia JF. Validating the Questionnaire for Verifying Stroke-Free Status (QVSFS) by neurological history and examination. Stroke . 2001; 32: 2232–2236. Go to Citation Crossref PubMed Google Scholar 13. Rothwell PM, Eliasziw M, Gutnikov SA, Warlow CP, Barnett HJ. Endarterectomy for symptomatic carotid stenosis in relation to clinical subgroups and timing of surgery. Lancet . 2004; 363: 915–924. Go to Citation Crossref PubMed Google Scholar 14. Mouradian MS, Majumdar SR, Senthilselvan A, Khan K, Shuaib A. How well are hypertension, hyperlipidemia, diabetes, and smoking managed after a stroke or transient ischemic attack? Stroke . 2002; 33: 1656–1659. Go to Citation Crossref PubMed Google Scholar 15. Mouradian MS, Hussain MS, Lari H, Salam A, Senthilselvan A, Dean N, Shuaib A. The impact of a stroke prevention clinic in diagnosing modifiable risk factors for stroke. Can J Neurol Sci . 2005; 32: 496–500. Crossref PubMed Google Scholar 16. Coutts SB, Hill MD, Campos CR, Choi YB, Subramaniam S, Kosior JC, Demchuk AM. Recurrent events in transient ischemic attack and minor stroke: What events are happening and to which patients? Stroke . 2008; 39: 2461–2466. Go to Citation Crossref PubMed Google Scholar Show all references eLetters eLetters should relate to an article recently published in the journal and are not a forum for providing unpublished data. Comments are reviewed for appropriate use of tone and language. Comments are not peer-reviewed. Acceptable comments are posted to the journal website only. Comments are not published in an issue and are not indexed in PubMed. Comments should be no longer than 500 words and will only be posted online. References are limited to 10. Authors of the article cited in the comment will be invited to reply, as appropriate. Comments and feedback on AHA/ASA Scientific Statements and Guidelines should be directed to the AHA/ASA Manuscript Oversight Committee via its Correspondence page. Sign In to Submit a Response to This Article Information & Authors Information Authors Information Published In Stroke Volume 41 • Number 11 • 1 November 2010 Pages: 2601 - 2605 PubMed: 20947856 Copyright © 2010. Versions You are viewing the most recent version of this article. 14 October 2010: Previous PDF (Version 1) History Received: 7 April 2010 Accepted: 29 July 2010 Published online: 14 October 2010 Published in print: 1 November 2010 Permissions Request permissions for this article. Request permissions Keywords ABCD2 ischemia recurrent event Authors Affiliations Expand All Jason Wasserman, PhD From the Divisions of Neurology (J.W., D.D., G.S., M.S.) and Emergency Medicine (J.P., I.S., J.S., C.S.), Ottawa Hospital, University of Ottawa, Ottawa, Ontario, Canada; Ottawa Hospital Research Institute (J.W., J.P., D.D., G.S., I.S., J.S., C.S., M.S.), University of Ottawa, Ottawa, Ontario, Canada. View all articles by this author Jeff Perry, MD, MSc From the Divisions of Neurology (J.W., D.D., G.S., M.S.) and Emergency Medicine (J.P., I.S., J.S., C.S.), Ottawa Hospital, University of Ottawa, Ottawa, Ontario, Canada; Ottawa Hospital Research Institute (J.W., J.P., D.D., G.S., I.S., J.S., C.S., M.S.), University of Ottawa, Ottawa, Ontario, Canada. View all articles by this author Dar Dowlatshahi, MD, PhD From the Divisions of Neurology (J.W., D.D., G.S., M.S.) and Emergency Medicine (J.P., I.S., J.S., C.S.), Ottawa Hospital, University of Ottawa, Ottawa, Ontario, Canada; Ottawa Hospital Research Institute (J.W., J.P., D.D., G.S., I.S., J.S., C.S., M.S.), University of Ottawa, Ottawa, Ontario, Canada. View all articles by this author Grant Stotts, MD From the Divisions of Neurology (J.W., D.D., G.S., M.S.) and Emergency Medicine (J.P., I.S., J.S., C.S.), Ottawa Hospital, University of Ottawa, Ottawa, Ontario, Canada; Ottawa Hospital Research Institute (J.W., J.P., D.D., G.S., I.S., J.S., C.S., M.S.), University of Ottawa, Ottawa, Ontario, Canada. View all articles by this author Ian Stiell, MD, MSc From the Divisions of Neurology (J.W., D.D., G.S., M.S.) and Emergency Medicine (J.P., I.S., J.S., C.S.), Ottawa Hospital, University of Ottawa, Ottawa, Ontario, Canada; Ottawa Hospital Research Institute (J.W., J.P., D.D., G.S., I.S., J.S., C.S., M.S.), University of Ottawa, Ottawa, Ontario, Canada. View all articles by this author Jane Sutherland, RN From the Divisions of Neurology (J.W., D.D., G.S., M.S.) and Emergency Medicine (J.P., I.S., J.S., C.S.), Ottawa Hospital, University of Ottawa, Ottawa, Ontario, Canada; Ottawa Hospital Research Institute (J.W., J.P., D.D., G.S., I.S., J.S., C.S., M.S.), University of Ottawa, Ottawa, Ontario, Canada. View all articles by this author Cheryl Symington, RN From the Divisions of Neurology (J.W., D.D., G.S., M.S.) and Emergency Medicine (J.P., I.S., J.S., C.S.), Ottawa Hospital, University of Ottawa, Ottawa, Ontario, Canada; Ottawa Hospital Research Institute (J.W., J.P., D.D., G.S., I.S., J.S., C.S., M.S.), University of Ottawa, Ottawa, Ontario, Canada. View all articles by this author Mukul Sharma, MD, MSc From the Divisions of Neurology (J.W., D.D., G.S., M.S.) and Emergency Medicine (J.P., I.S., J.S., C.S.), Ottawa Hospital, University of Ottawa, Ottawa, Ontario, Canada; Ottawa Hospital Research Institute (J.W., J.P., D.D., G.S., I.S., J.S., C.S., M.S.), University of Ottawa, Ottawa, Ontario, Canada. View all articles by this author Notes Correspondence to Mukul Sharma, MD, MSc, Regional Stroke Program, The Ottawa Hospital, Civic Campus, C2, Room 2182, 1053 Carling Ave, Ottawa, Ontario K1Y 4E9, Canada. 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Please select your download format: [x] Direct Import Raffaele Ornello, Matteo Foschi, Federico De Santis, Michele Romoli, Tiziana Tassinari, Valentina Saia, Silvia Cenciarelli, Chiara Bedetti, Chiara Padiglioni, Bruno Censori, Valentina Puglisi, Luisa Vinciguerra, Maria Guarino, Valentina Barone, Maria Luisa Zedde, Ilaria Grisendi, Marina Diomedi, Maria Rosaria Bagnato, Marco Petruzzellis, Domenico Maria Mezzapesa, Pietro Di Viesti, Vincenzo Inchingolo, Manuel Cappellari, Mara Zenorini, Paolo Candelaresi, Vincenzo Andreone, Giuseppe Rinaldi, Alessandra Bavaro, Anna Cavallini, Stefan Moraru, Pietro Querzani, Valeria Terruso, Marina Mannino, Umberto Scoditti, Alessandro Pezzini, Giovanni Frisullo, Francesco Muscia, Maurizio Paciaroni, Maria Giulia Mosconi, Andrea Zini, Ruggiero Leone, Carmela Palmieri, Letizia Maria Cupini, Michela Marcon, Rossana Tassi, Enzo Sanzaro, Cristina Paci, Giovanna Viticchi, Daniele Orsucci, Anne Falcou, Susanna Diamanti, Roberto Tarletti, Patrizia Nencini, Eugenia Rota, Federica Nicoletta Sepe, Delfina Ferrandi, Luigi Caputi, Gino Volpi, Salvatore La Spada, Mario Beccia, Claudia Rinaldi, Vincenzo Mastrangelo, Francesco Di Blasio, Paolo Invernizzi, Giuseppe Pelliccioni, Maria Vittoria De Angelis, Laura Bonanni, Giampietro Ruzza, Emanuele Alessandro Caggia, Monia Russo, Agnese Tonon, Maria Cristina Acciarri, Sabrina Anticoli, Cinzia Roberti, Giovanni Manobianca, Gaspare Scaglione, Francesca Pistoia, Alberto Fortini, Antonella De Boni, Alessandra Sanna, Alberto Chiti, Leonardo Barbarini, Maela Masato, Massimo Del Sette, Francesco Passarelli, Maria Roberta Bongioanni, Danilo Toni, Stefano Ricci, Simona Sacco, Eleonora De Matteis, Transient brain ischemic symptoms and the presence of ischemic lesions at neuroimaging: Results from the READAPT study, International Journal of Stroke, 20, 4, (426-437), (2024). Crossref Brit Long, Evie Marcolini, Michael Gottlieb, Emergency medicine updates: Transient ischemic attack, The American Journal of Emergency Medicine, 83, (82-90), (2024). Crossref Valentina Barone, Matteo Foschi, Lucia Pavolucci, Francesca Rondelli, Rita Rinaldi, Marianna Nicodemo, Roberto D’Angelo, Elisabetta Favaretto, Carlotta Brusi, Benilde Cosmi, Daniela Degli Esposti, Sergio D’Addato, Stefano Bacchelli, Fabrizio Giostra, Daniela Paola Pomata, Luca Spinardi, Luca Faccioli, Gianluca Faggioli, Andrea Donti, Claudio Borghi, Pietro Cortelli, Maria Guarino, Enhancing stroke risk prediction in patients with transient ischemic attack: insights from a prospective cohort study implementing fast-track care, Frontiers in Neurology, 15, (2024). Crossref F Cannizzaro, A Izquierdo, D Cocho, Rate of atrial fibrillation by Holter-Stroke Risk Analysis in undetermined TIA/rapidly improving stroke symptoms patients, Frontiers in Neurology, 15, (2024). Crossref Salih METİN, Hüseyin AYGÜN, Cemile HAKİ, The effect of COVID-19 pandemic on stroke admissions to a city, Journal of Health Sciences and Medicine, 6, 5, (893-897), (2023). Crossref Laura P. Rossi, Bradi B. Granger, Jeffrey T. Bruckel, Deborah L. Crabbe, Lucinda J. Graven, Kimberly S. Newlin, Megan M. Streur, Maya K. Vadiveloo, Benita Jeanne Walton-Moss, Bruce A. Warden, Annabelle Santos Volgman, Melissa Lydston, Person-Centered Models for Cardiovascular Care: A Review of the Evidence: A Scientific Statement From the American Heart Association, Circulation, 148, 6, (512-542), (2023)./doi/10.1161/CIR.0000000000001141 Abstract Shima Shahjouei, Homa Seyedmirzaei, Vida Abedi, Ramin Zand, Transient Ischemic Attack Outpatient Clinic: Past Journey and Future Adventure, Journal of Clinical Medicine, 12, 13, (4511), (2023). Crossref Thomas J. Jeerakathil, Amy Ying Xin Yu, Philip M.C. Choi, Shoufan Fang, Ashfaq Shuaib, Sumit R. Majumdar, Andrew M. Demchuk, Kenneth Butcher, Tim J. Watson, Naeem Dean, Deb Gordon, Michael D. Hill, Cathy Edmond, Shelagh B. Coutts, Effects of a Province-wide Triaging System for TIA, Neurology, 100, 20, (2023). Crossref M. Alonso de Leciñana, A. Morales, M. Martínez-Zabaleta, Ó. Ayo-Martín, L. Lizán, M. Castellanos, Characteristics of stroke units and stroke teams in Spain in 2018. Pre2Ictus project, Neurología (English Edition), 38, 3, (173-180), (2023). Crossref M. Alonso de Leciñana, A. Morales, M. Martínez-Zabaleta, Ó. Ayo-Martín, L. Lizán, M. Castellanos, Características de las unidades de ictus y equipos de ictus en España en el año 2018. Proyecto Pre2Ictus, Neurología, 38, 3, (173-180), (2023). Crossref See more Loading... View Options View options PDF and All Supplements Download PDF and All Supplements Download is in progress PDF/EPUB View PDF/EPUB Figures Open all in viewer Figure 1.Study design. Only patients given a final ED diagnosis of TIA were included in this study. After referral to the Stroke Clinic, patients were triaged based on their ABCD2 score. Patients high risk, moderate risk, or at low risk were scheduled to see a stroke neurologist within 7 days, 7 to 14 days, or more than 14 days of the index TIA, respectively. All patients received the same standard of care regardless of their risk strata. Go to FigureOpen in Viewer Figure 2.Cumulative risk of recurrent stroke or TIA at 2, 7, 30, or 90 days after index TIA. Go to FigureOpen in Viewer Tables Open all in viewer Table 1. Characteristics of Patients Presenting to the ED With TIA and Referred to the Stroke Clinic for Management Go to TableOpen in Viewer Table 2. Medications Already Being Taken, Started, or Discontinued in the ED Among Patients Diagnosed With TIA in Our Study Go to TableOpen in Viewer Table 3. Time Between Event and Outpatient Investigation Go to TableOpen in Viewer Table 4. Temporal Distribution of Recurrent Stroke or TIA or MI After Index TIA Go to TableOpen in Viewer Table 5. The 90-Day Stroke Rate in our Prospective Cohort Study of 982 TIA Patients Compared With the 90-Day Stroke Rate Predicted by the ABCD2 Score Go to TableOpen in Viewer Media Share Share Share article link Copy Link Copied! Copying failed. Share FacebookLinkedInX (formerly Twitter)emailWeChatBluesky References References 1. Gladstone DJ, Kapral MK, Fang J, Laupacis A, Tu JV. Management and outcomes of transient ischemic attacks in Ontario. CMAJ . 2004; 170: 1099–1104. Go to Citation Crossref PubMed Google Scholar 2. Hill MD, Yiannakoulias N, Jeerakathil T, Tu JV, Svenson LW, Schopflocher DP. The high risk of stroke immediately after transient ischemic attack: A population-based study. Neurology . 2004; 62: 2015–2020. Crossref PubMed Google Scholar 3. Johnston SC, Gress DR, Browner WS, Sidney S. Short-term prognosis after emergency department diagnosis of TIA. JAMA . 2000; 284: 2901–2906. Crossref PubMed Google Scholar 4. Lisabeth LD, Ireland JK, Risser JM, Brown DL, Smith MA, Garcia NM, Morgenstern LB. Stroke risk after transient ischemic attack in a population-based setting. Stroke . 2004; 35: 1842–1846. Crossref PubMed Google Scholar 5. Rothwell PM, Warlow CP. Timing of TIAs preceding stroke: Time window for prevention is very short. Neurology . 2005; 64: 817–820. Go to Citation Crossref PubMed Google Scholar 6. Prabhakaran S, Silver AJ, Warrior L, McClenathan B, Lee VH. Misdiagnosis of transient ischemic attacks in the emergency room. Cerebrovasc Dis . 2008; 26: 630–635. Go to Citation Crossref PubMed Google Scholar 7. Johnston SC, Rothwell PM, Nguyen-Huynh MN, Giles MF, Elkins JS, Bernstein AL, Sidney S. Validation and refinement of scores to predict very early stroke risk after transient ischaemic attack. Lancet . 2007; 369: 283–292. Crossref PubMed Google Scholar a [...] a stroke risk of 17.8% at 90 days post TIA. b [...] by the ABCD2 score for this cohort. c [...] for those at high risk of stroke. 8. Rothwell PM, Giles MF, Chandratheva A, Marquardt L, Geraghty O, Redgrave JN, Lovelock CE, Binney LE, Bull LM, Cuthbertson FC, Welch SJ, Bosch S, Alexander FC, Silver LE, Gutnikov SA, Mehta Z. Effect of urgent treatment of transient ischaemic attack and minor stroke on early recurrent stroke (EXPRESS study): A prospective population-based sequential comparison. Lancet . 2007; 370: 1432–1442. Crossref PubMed Google Scholar a [...] outcomes by referral to dedicated clinics b [...] study, no follow-up details were provided. c [...] 1.2% risk of 90-day stroke, respectively. 9. Lavallee PC, Meseguer E, Abboud H, Cabrejo L, Olivot JM, Simon O, Mazighi M, Nifle C, Niclot P, Lapergue B, Klein IF, Brochet E, Steg PG, Leseche G, Labreuche J, Touboul PJ, Amarenco P. A transient ischaemic attack clinic with round-the-clock access (SOS-TIA): Feasibility and effects. Lancet Neurol . 2007; 6: 953–960. Crossref PubMed Google Scholar 10. Wu CM, Manns BJ, Hill MD, Ghali WA, Donaldson C, Buchan AM. Rapid evaluation after high-risk TIA is associated with lower stroke risk. Can J Neurol Sci . 2009; 36: 450–455. Crossref PubMed Google Scholar a [...] or by hospital admission. b [...] 90-day stroke rate with this approach. 11. Stead LG, Bellolio MF, Suravaram S, Brown RD, Jr., Bhagra A, Gilmore RM, Boie ET, Decker WW. Evaluation of transient ischemic attack in an emergency department observation unit. Neurocrit Care . 2009; 10: 204–208. Go to Citation Crossref PubMed Google Scholar 12. Jones WJ, Williams LS, Meschia JF. Validating the Questionnaire for Verifying Stroke-Free Status (QVSFS) by neurological history and examination. Stroke . 2001; 32: 2232–2236. Go to Citation Crossref PubMed Google Scholar 13. Rothwell PM, Eliasziw M, Gutnikov SA, Warlow CP, Barnett HJ. Endarterectomy for symptomatic carotid stenosis in relation to clinical subgroups and timing of surgery. Lancet . 2004; 363: 915–924. Go to Citation Crossref PubMed Google Scholar 14. Mouradian MS, Majumdar SR, Senthilselvan A, Khan K, Shuaib A. How well are hypertension, hyperlipidemia, diabetes, and smoking managed after a stroke or transient ischemic attack? Stroke . 2002; 33: 1656–1659. Go to Citation Crossref PubMed Google Scholar 15. Mouradian MS, Hussain MS, Lari H, Salam A, Senthilselvan A, Dean N, Shuaib A. The impact of a stroke prevention clinic in diagnosing modifiable risk factors for stroke. Can J Neurol Sci . 2005; 32: 496–500. Crossref PubMed Google Scholar 16. Coutts SB, Hill MD, Campos CR, Choi YB, Subramaniam S, Kosior JC, Demchuk AM. Recurrent events in transient ischemic attack and minor stroke: What events are happening and to which patients? Stroke . 2008; 39: 2461–2466. Go to Citation Crossref PubMed Google Scholar Advertisement Recommended August 2008 Higher ABCD 2 Score Predicts Patients Most Likely to Have True Transient Ischemic Attack S. Andrew Josephson, Stephen Sidney, Trinh N. Pham, Allan L. Bernstein, and [...] S. Claiborne Johnston +1 authors April 2018 Abstract WP252: Reduction in Stroke Readmission Rates After Implementing an Emergency Department-based Transient Ischemic Attack Observation Unit Lori Fayas, Kathy Polum, and [...] Heather Stanko +0 authors December 2011 Patients With Transient Ischemic Attack With ABCD 2<4 Can Have Similar 90-Day Stroke Risk as Patients With Transient Ischemic Attack With ABCD 2 ≥4 Pierre Amarenco, Julien Labreuche, and [...] Philippa C. 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1883	https://www.dynamed.com/condition/malignant-otitis-externa/PlatformCE	Top of Page Condition Malignant Otitis Externa Editors: Bassem Hanna MBBCh, MSc, MD, PhD, FRCSC; Alexander Rae-Grant MD, FRCPC, FAAN; Esther Jolanda van Zuuren MD Overview and Recommendations Background Information Description Also called Epidemiology Etiology and Pathogenesis History and Physical Diagnosis Management Complications and Prognosis Prevention and Screening Guidelines and Resources Patient Information References < Previous Section Next Section > Background Information Description invasive infection of external auditory canal and skull base typically due to Pseudomonas aeruginosa1 , 2 rare fungal cases have also been reported secondary to aspergillosis1 Also called MOE malignant external otitis (MEO) malignant otitis necrotizing external otitis necrotizing otitis externa necrotizing otitis osteomyelitis of ear canal skull base osteitis invasive external otitis Epidemiology Who is most affected typically affects elderly patients1 , 2 most cases are reported in patients with diabetes1 , 2 can also affect immunocompromised patients, especially those with HIV-AIDS2 + tend to be younger patients + usually do not have diabetes rare in children1 , 3 + most cases are associated with immunocompromise (typically malignancy or malnutrition) + diabetes is less common symptom onset usually during spring and summer months (Eur Ann Otorhinolaryngol Head Neck Dis 2013 Jun;130(3):115) Incidence/Prevalence Laryngoscope 2017 Oct;127(10):2328 8,300 inpatients with malignant otitis externa reported in the United States from 2002 to 2013 based on data from the United States Nationwide Inpatient Sample (NIS) database from 2002 to 2013 8,300 inpatients with malignant otitis externa were identified 62.4% adults aged 18-64 years 32.7% elderly > 65 years old 4.9% children aged 1-17 years Reference - Laryngoscope 2017 Oct;127(10):2328 J Laryngol Otol 2015 Jun;129(6):600 incidence of malignant otitis externa reported to increase in England between 1999 and 2013 based on data from the United Kingdom's Hospital Episodes Statistics database cases rose steadily each year 67 cases reported in 1999-2000 421 cases reported in 2012-2013 Reference - J Laryngol Otol 2015 Jun;129(6):600 Likely risk factors untreated or inadequately treated acute otitis externa in patients with other risk factors for malignant otitis externa (such as diabetes or immunocompromise) (Am Fam Physician 2012 Dec 1;86(11):1055) older age1 diabetes mellitus type 1 or type 21 , 2 immunocompromised state due to1 , 2 , 3 + HIV infection/AIDS + neoplasia + acute monocytic leukemia + lymphoma + splenectomy + posttransplant immunosuppression, including bone marrow transplant + chemotherapy-induced aplasia + refractory anemia + immunoglobulin (Ig)G subclass deficiency + IgA deficiency + iatrogenic neutropenia due to induction chemotherapy for acute lymphoblastic leukemia + malnutrition water in the external auditory canal + CASE-CONTROL STUDYAnn Otol Rhinol Laryngol 1990 Feb;99(2 Pt 1):117 aural irrigation in patients with diabetes may increase risk for iatrogenic malignant otitis externa - based on case-control study - 13 elderly patients with diabetes and malignant otitis externa matched to 26 controls with diabetes and no malignant otitis externa - history of tap water aural irrigation in 61.5% of cases (8 of 13 patients) vs. 15% of controls (p < 0.05) - Reference - Ann Otol Rhinol Laryngol 1990 Feb;99(2 Pt 1):117 prolonged water exposure (spa therapy, swimming, or diving) reported in 8 of 22 patients with malignant otitis externa in retrospective cohort study (Eur Ann Otorhinolaryngol Head Neck Dis 2013 Jun;130(3):115) ear trauma (due to injury or surgery) (Eur Arch Otorhinolaryngol 2015 May;272(5):1269) malignant otitis externa following mastoidectomy in case report (BMJ Case Rep 2011 May 24;2011) Etiology and Pathogenesis Causes progressive infection of the external auditory canal and skull base1 the use of topical antibiotics before collection of culture often prevents identification of the pathogen1 of the known pathogens, Pseudomonas aeruginosa is reported to be the causative agent in most cases1 , 2 - reported to be isolated from exudate in about 75% of cases - pseudomonas is not a component of normal ear canal flora, so a positive culture is indicative of infection + less frequent bacterial causes include1 , 3 - Staphylococcus aureus - Staphylococcus epidermis - Klebsiella oxytoca - Proteus mirabilis - Klebsiella oxytoca - Proteus cepacia - Acinetobacter spp. and Enterococcus spp. (Otolaryngol Head Neck Surg 2014 Jul;151(1):112) fungal malignant otitis externa is rare, but has been reported in patients without diabetes who are immunosuppressed1 + Aspergillus subtypes (most common) - Aspergillus fumigatus - Aspergillus niger - Aspergillus flavus + some cases have been associated with candidal species, including - Candida glabrata - Candida parapsilosis (J Laryngol Otol 2000 May;114(5):366) - Candida albicans (Eur Ann Otorhinolaryngol Head Neck Dis 2013 Jun;130(3):115) - Candida ciferrii (Lancet Infect Dis 2004 Jan;4(1):34) + Scedosporium apiospermum, Pseudallescheria boydii, and Malassezia sympodialis have also been reported (Lancet Infect Dis 2004 Jan;4(1):34) Pathogenesis impaired immune function and microvascular abnormalities in the ear canal may predispose patients to malignant otitis externa (MOE) associated with noncommensal bacteria1 ear canal abnormalities are common in elderly patients and patients with diabetes bacteria invade blood vessels, resulting in vasculitis with thrombosis and coagulation necrosis of surrounding tissues other factors that may contribute to the development of an invasive pseudomonal infection include polymorphonuclear cells with impaired immune function (as found in patients with diabetes) increased pH of cerumen found in the external auditory canal of patients with diabetes increase in microbial growth following application of water (for example, aural irrigation) origination and spread of infection pseudomonal MOE usually originates at junction of cartilage and bone in the external auditory canal (EAC); involvement of the middle ear is a late finding1 , 3 + fungal MOE usually originates in the middle ear or mastoid1 + infection spreads from the EAC to the skull base via the fissures of Santorini (perforations in the cartilage portion of the EAC located along the floor of the canal)3 - once out of the canal, infection spreads medially to the tympanomastoid suture and along venous canals and fascial planes - normally compact skull bone is replaced with granulation tissues, resulting in bone destruction and skull base osteomyelitis - spread of infection to skull base foramina causes cranial neuropathies facial nerve is most commonly involved due to the proximity of the stylomastoid foramen to the EAC, followed by glossopharyngeal, vagus, and accessory nerves at the jugular foramen as infection spreads medially, the abducens, trigeminal, and optic nerves can be affected continued spread may result in septic thrombosis of the sigmoid sinus and internal jugular vein with complications of meningitis and cerebral abscess skull base osteomyelitis can also spread contralaterally and eventually invade the cervical spine extracranial spread can involve infratemporal fossa, parotid, and neck - neurotoxins released by Pseudomonas aeruginosa also contribute to the development of cranial nerve palsies2 < Previous Section Next Section > Related Topics Otitis ExternaAcute Otitis Media (AOM) in ChildrenOtitis Media with Effusion (OME) Published by EBSCO Information Services. 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1884	https://openstax.org/books/chemistry-2e/pages/14-2-ph-and-poh	Skip to ContentGo to accessibility pageKeyboard shortcuts menu Chemistry 2e 14.2 pH and pOH Chemistry 2e14.2 pH and pOH Search for key terms or text. Learning Objectives By the end of this section, you will be able to: Explain the characterization of aqueous solutions as acidic, basic, or neutral Express hydronium and hydroxide ion concentrations on the pH and pOH scales Perform calculations relating pH and pOH As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water (Kw). The concentrations of these ions in a solution are often critical determinants of the solution’s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is neutral if it contains equal concentrations of hydronium and hydroxide ions; acidic if it contains a greater concentration of hydronium ions than hydroxide ions; and basic if it contains a lesser concentration of hydronium ions than hydroxide ions. A common means of expressing quantities that may span many orders of magnitude is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where “X” is the quantity of interest and “log” is the base-10 logarithm: The pH of a solution is therefore defined as shown here, where [H3O+] is the molar concentration of hydronium ion in the solution: Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression: Likewise, the hydroxide ion molarity may be expressed as a p-function, or pOH: or Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the Kw expression: At 25 °C, the value of Kw is 1.0 10−14, and so: As was shown in Example 14.1, the hydronium ion molarity in pure water (or any neutral solution) is 1.0 10−7 M at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore: And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than 1.0 10−7 M and hydroxide ion molarities less than 1.0 10−7 M (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than 1.0 10−7 M and hydroxide ion molarities greater than 1.0 10−7 M (corresponding to pH values greater than 7.00 and pOH values less than 7.00). Since the autoionization constant Kw is temperature dependent, these correlations between pH values and the acidic/neutral/basic adjectives will be different at temperatures other than 25 °C. For example, the “Check Your Learning” exercise accompanying Example 14.1 showed the hydronium molarity of pure water at 80 °C is 4.9 10−7 M, which corresponds to pH and pOH values of: At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at other temperatures, such as enzyme reactions in warm-blooded organisms at a temperature around 36–40 °C. Unless otherwise noted, references to pH values are presumed to be those at 25 °C (Table 14.1). Summary of Relations for Acidic, Basic and Neutral Solutions \| Classification \| Relative Ion Concentrations \| pH at 25 °C \| --- \| acidic \| [H3O+] > [OH−] \| pH < 7 \| \| neutral \| [H3O+] = [OH−] \| pH = 7 \| \| basic \| [H3O+] < [OH−] \| pH > 7 \| Table 14.1 Figure 14.2 shows the relationships between [H3O+], [OH−], pH, and pOH for solutions classified as acidic, basic, and neutral. Figure 14.2 The pH and pOH scales represent concentrations of H3O+ and OH−, respectively. The pH and pOH values of some common substances at 25 °C are shown in this chart. Example 14.4 Calculation of pH from [H3O+] What is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of 1.2 10−3 M? Solution (The use of logarithms is explained in Appendix B. When taking the log of a value, keep as many decimal places in the result as there are significant figures in the value.) Check Your Learning Water exposed to air contains carbonic acid, H2CO3, due to the reaction between carbon dioxide and water: Air-saturated water has a hydronium ion concentration caused by the dissolved CO2 of 2.0 10−6 M, about 20-times larger than that of pure water. Calculate the pH of the solution at 25 °C. Answer: 5.70 Example 14.5 Calculation of Hydronium Ion Concentration from pH Calculate the hydronium ion concentration of blood, the pH of which is 7.3. Solution (On a calculator take the antilog, or the “inverse” log, of −7.3, or calculate 10−7.3.) Check Your Learning Calculate the hydronium ion concentration of a solution with a pH of −1.07. Answer: 12 M How Sciences Interconnect Environmental Science Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO2 which forms carbonic acid: Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including CO2, SO2, SO3, NO, and NO2 being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here: Carbon dioxide is naturally present in the atmosphere because most organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also originates from burning fossil fuels, which have traces of sulfur, and from the process of “roasting” ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine. Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure 14.3). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India. For further information on acid rain, visit this website hosted by the US Environmental Protection Agency. Figure 14.3 (a) Acid rain makes trees more susceptible to drought and insect infestation, and depletes nutrients in the soil. (b) It also is corrodes statues that are carved from marble or limestone. (credit a: modification of work by Chris M Morris; credit b: modification of work by “Eden, Janine and Jim”/Flickr) Example 14.6 Calculation of pOH What are the pOH and the pH of a 0.0125-M solution of potassium hydroxide, KOH? Solution Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding [OH−] = 0.0125 M: The pH can be found from the pOH: Check Your Learning The hydronium ion concentration of vinegar is approximately 4 10−3 M. What are the corresponding values of pOH and pH? Answer: pOH = 11.6, pH = 2.4 The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure 14.4). Figure 14.4 (a) A research-grade pH meter used in a laboratory can have a resolution of 0.001 pH units, an accuracy of ± 0.002 pH units, and may cost in excess of $1000. (b) A portable pH meter has lower resolution (0.01 pH units), lower accuracy (± 0.2 pH units), and a far lower price tag. (credit b: modification of work by Jacopo Werther) The pH of a solution may also be visually estimated using colored indicators (Figure 14.5). The acid-base equilibria that enable use of these indicator dyes for pH measurements are described in a later section of this chapter. Figure 14.5 (a) A solution containing a dye mixture, called universal indicator, takes on different colors depending upon its pH. (b) Convenient test strips, called pH paper, contain embedded indicator dyes that yield pH-dependent color changes on contact with aqueous solutions.(credit: modification of work by Sahar Atwa) PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. 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1885	https://www.ircmj.com/article_195129_b74a02fb7bae42f7a0772313350ca4f0.pdf	Iran Red Crescent Med J. 2017 April; 19(4):e44363. Published online 2017 February 13. doi: 10.5812/ircmj.44363. Research Article Prevalence of 2 UGT1A1 Gene Variations Related to Gilbert’s Syndrome in South of Iran: An Epidemiological, Clinical, and Genetic Study Mohammad Reza Heydari,1 Majid Fardaei,2, Mohammad Rahim Kadivar,3 Abbas Rezaianzadeh,4 Mohammad Reza Panjehshahin,1 Zeinab Gholami Bardeji,5 Mohammad Reza Miri,6 and Jamileh Saberzadeh6 1Department of Pharmacology, Medical School, Shiraz University of Medical Sciences, Shiraz, Iran 2Department of Medical Genetic, Medical School, Shiraz University of Medical Sciences, Shiraz, Iran 3Department of Pediatric, Namazi Hospital, Shiraz University of Medical Sciences, Shiraz, Iran 4Department of Epidemiology, Shiraz University of Medical Sciences, Shiraz, Iran 5Department of Radiology, Medical Imaging Research Center, Namazi Hospital, Shiraz University of Medical Sciences, Shiraz, Iran 6Medical Biotechnology Department, School of Advanced Medical Sciences and Technology, Shiraz University of Medical Sciences, Shiraz, Iran Corresponding author: Majid Fardaei, Department of Medical Genetic, Medical School, Shiraz University of Medical Sciences, Zand Street, Postal code: 71348-53185, Shiraz, Iran. Tel: +98-7132349610, Fax: +98-7132349610, E-mail: mfardaei@sums.ac.ir Received 2016 November 28; Revised 2017 January 13; Accepted 2017 February 01. Abstract Background: Gilbert’s syndrome can present as a chronic or benign asymptomatic condition, characterized by a slight increase in the serum bilirubin level without any hemolysis. In 1995, a genetic variation, located in the TATA box of UGT1A1 gene promoter, was identiﬁed in patients with Gilbert’s syndrome. Also, further analysis identiﬁed a new missense variation, Gly71Arg, within the codon region of UGT1A1 gene. Coincidence of TATA box and Gly71Arg variations and their relationship with clinical ﬁndings are mostly variable. Objectives: The aim of this study was to determine TATA box and Gly71Arg variations of UGT1A1 gene and assess their eﬀects on clinical ﬁndings in patients with Gilbert’s syndrome in southern provinces of Iran. Methods: In this cross sectional study, 213 unrelated infants and children, below 12 years, who were admitted to the pediatric ward of Namazi hospital, Shiraz, Iran, were enrolled from June 2015 to May 2016. Blood-extracted DNA was used for genotyping TATA box and Gly71Arg variations by sequencing. Further biochemical analyses were performed for each patient. Results: About 78.9% of the studied subjects had normal homozygous genotypes, and 21.1% were heterozygous for the Gly71Arg variation. In total, 34% of the cases were normal in the promoter region (TA6/6), and 55% were heterozygous with genotypes TA6/7, TA6/5, and TA 6/8. Three combinations of genotypes, ie, TA6/7-Gly/Gly, TA7/7-Gly/Gly, and TA7/7-Gly/Arg, showed signiﬁcant diﬀerences in the serum total bilirubin level. Also, creatinine phosphokinase in TA6/7-Gly/Arg, TA7/7-Gly/Gly, and TA7/7-Gly/Arg had a signiﬁcant increase. Conclusions: The present ﬁndings showed that the TA7/7 promoter of UGT1A1 gene accounted for a considerable number of Gilbert’s syndrome cases (11.3%). The studied variations had a signiﬁcant eﬀect on creatine phosphokinase and serum total bilirubin levels. Keywords: Bilirubin, Gilbert’s Disease, TATA Box, Genetic Variation, Glucuronosyltransferase 1. Background Gilbert’s syndrome (GS), ﬁrst described by Gilbert and Lereboullet, is an autosomal recessive disorder, caused by changes in uridine diphosphate (UDP) glucuronosyltrans-ferase 1A1 (UGT1A1) gene (1). GS is characterized as a mild, unconjugated hyperbilirubinemia in the absence of hep-atic disease and hemolysis (2). The main cause of increased serum bilirubin level in GS patients is the reduction of glu-curonosyltransferaseenzymeactivity(3). Inthesepatients, based on the genetic population makeup, UDP glucurono-syltransferase activity can be reduced by up to 30% of the normal level (2). So far, researchers have performed comprehensive studies on diﬀerent aspects of GS, thus leading to the bet-ter management and recognition of this condition (4). The main common genetic cause of GS is mutation in the TATA box of UGT1A1 gene promoter, which plays a regulatory role in gene transcription. It is well known that 6 TA repeats (UGT1A11) in this region are normal, while presence of 7 TA repeats (UGT1A12) reduces glucuronosyltransferase activ-ity; other variations are not well speciﬁed (5, 6). It is speculated that the enlarged TATA box sequence is less eﬀective in binding regulatory agents, which con-trol the transcription of UGT1A1 gene (7). Further studies re-vealed another cause of GS, ie, G > A variation at nucleotide 211 in exon 1, which results in the substitution of glycine Copyright © 2017, Iranian Red Crescent Medical Journal. This is an open-access article distributed under the terms of the Creative Commons Attribution-NonCommercial 4.0 International License ( which permits copy and redistribute the material just in noncommercial usages, provided the original work is properly cited. Heydari MR et al. to arginine at codon 71 (Gly71Arg) of UGT1A1 protein (8). In fact, the incidence of hyperbilirubinemia in neonates with Gly71Arg missense variation has been reported to be nearly 30% higher than the normal population in diﬀerent eth-nicities (9, 10). So far, 113 variants of UGT1A gene have been identiﬁed,whichcanleadtoawidespectrumof conditions ranging from jaundice and very mild conditions to severe, lethal diseases (11). GS may not require a special medical treatment as it has benign consequences (12, 13). Although the main feature of GS is an elevated unconjugated bilirubin level, metabolism of some drugs and xenobiotics may be also aﬀected (14). Acetaminophen, ethinylestradiol, indinavir, and irinotecan clearance from the serum is reduced in GS due to deﬁciency in drug glucuronidation. Accord-ingly, considering drug accumulation in the serum, some researchers have suggested a GS diagnostic test before irinotecan prescription to prevent severe side-eﬀects (15-17). Although all conducted studies have emphasized on the relationship between GS and increased bilirubin level, in the present study, in addition to conﬁrming previous ﬁndings in this area, the relationship between GS and el-evated creatine phosphokinase (CPK) level was analyzed. The present study aimed to determine 2 variations of UGT1A1 gene and assess their eﬀects on clinical ﬁndings in patients with GS. 2. Methods 2.1. Sample Collection The sample size for this cross sectional study was deter-mined to be 197, based on the following formula and previ-ous studies (18, 19) (P = 0.09 and d = 0.04): (1) N = (196)2 P (1 −P) d2 = 197 To achieve this sample size, the process of subject re-cruitment started from June 2015. Any unrelated infant or child (below 12 years), who was referred to the emer-gency ward of Namazi hospital, Shiraz, Iran and was trans-ferred to the pediatric ward, was included in this study. In May 2016, a total of 215 subjects were recruited, although 2 cases were eliminated, based on the exclusion criteria. Res-idence in Fars province or southern provinces of Iran and willingness to participate in the study were among the in-clusion criteria. Whole blood samples (1.5 mL) were collected from each patient in Greiner Bio-One Hematology K3-EDTA Lavender Evacuated Tubes and kept at -20°C until DNA extraction. 2.2. Ethical Considerations Sample collection and subsequent analysis were per-formed after obtaining a written informed consent from each patient according to the Declaration of Helsinki. The study was approved by the Medical Ethics Committee on Human Research of Shiraz University of Medical Sciences (code number, IR.SUMS.rec.1394.S95). 2.3. Inclusion and Exclusion Criteria The study population included children below 12years, who were living in Fars province or southern provinces of Iran. On the other hand, all cases who used liver enzyme inducers, such as barbiturates, aminopyrine, phenylbu-tazone, orphenadrine, or 3, 4-benzpyrene, were excluded from the study (20). According to these criteria, among 215 included cases, 2 cases were eliminated. 2.4. Demographic and Biochemical Information A demographic questionnaire (including name, gen-der, phone number, address, age, history of neonatal jaun-dice, parent’s history of jaundice, and history of speciﬁc diseases) was completed for each subject. Liver function test (LFT), as well as CPK, lactate dehydrogenase (LDH), blood urea nitrogen (BUN), and creatinine (Cr) levels, was measured by MAN Co. kits with a CS-400 auto-chemistry analyzer (DIRUI Industrial Co., Ltd, Eastern China). 2.5. Preparation of Genomic DNA Genomic DNA was extracted from the peripheral blood lymphocytes (100 - 200 µL) by the RIBO-prep nucleic acid extraction kit variant 100 (K2-9-Et-50-CE, Slovak Republic), according to the manufacturer’s instructions. The con-centration of the extracted genomic DNA was determined by measuring the ultraviolet absorbance at 260 nm with the NanoDrop Lite Spectrophotometer (Thermo Scientiﬁc, USA). About 70 - 110 ng of the genomic DNA was used for polymerase chain reaction (PCR). 2.5.1. PCR and DNA Sequencing The promoter and exon 1 regions of UGT1A1 gene were ampliﬁed, using in-house designed primers. Forward and reverse primers to amplify the TATA box promoter and exon 1 included 5’-TGAAATTCCAGCCAGTTCAA-3’ and 5’-TTGAAGACGTACCCTGTGC-3’, respectively. The reaction mixture (25 µL) contained 1 µL of genomic DNA, 1 µL of each forward and reverse primer (10 pmol/reaction), 13 µL of AMPLIQON TEMPase Hot start 2x Master Mix (CinnaGen Co., Iran), and 9 µL of ddH2O. Brieﬂy, PCR was performed by denaturation at 95°C for 15 minutes, followed by 35 cycles at 94°C for 40 seconds, 55°C for 40 seconds, and 72°C for 40 seconds, with a ﬁnal 2 Iran Red Crescent Med J. 2017; 19(4):e44363. Heydari MR et al. synthesisat72°Cfor5minutes. TheampliﬁedPCRproducts were sequenced by forward and reverse primers, using the Sangermethod(bi-directionalsequencing). Finally,thenu-cleotide sequences were analyzed by BioEdit software. 2.6. Statistical Analysis For data management and further analysis, SPSS ver-sion 21 was used. Descriptive analysis was performed for determining the frequency. Also, t test, Chi square, Fisher’s exact test, and one-way analysis of variance (ANOVA), fol-lowedbyDuncan’sandLSDmultiple-rangetests, wereused for pairwise comparisons. P value ≤0.05 was considered statistically signiﬁcant. For data which were not normally distributed, we used nonparametric Kruskal-Wallis test, as well as Bonferroni correction test. In these tests, P value ≤ 0.003 was considered statistically signiﬁcant. 3. Results 3.1. Demographic Characteristics A total of 213 unrelated infants and children, below 12 years, were included in this study. The mean age of the sub-jects, including 123 males (57.75%) and 90 females (42.25%), was 5.69 ± 3.4 months. Among all cases, 25.3 % (n = 54) had a history of diseases. Overall, 61% (n = 130) of the patients’ parents were consanguineously married. The mean body mass index (BMI) of the subjects was 16.36 ± 0.31 kg/m2. Based on the ﬁndings, no relationship was found between these characteristics and the laboratory data. It should be also noted that all cases were from southern provinces of Iran. 3.2. Analysis of UGT1A1 Variations In the evaluated population, 5 possible genotypes were identiﬁed, based on the ampliﬁcation and sequencing of UGT1A1 gene promoter. These genotypes included com-mon homozygous allele TA6/6, common heterozygous al-lele TA6/7, rare heterozygous alleles TA6/5 and TA6/8, and rare homozygous allele TA7/7. The minimum and max-imum frequencies of alleles carrying TA variations were 0.5% and 53.5%, respectively (Table 1). Thegenotypedistributionof Gly71Argvariationamong 213 Iranian children and neonates was determined as fol-lows: Gly/Gly (n = 168, 78.9%) and Gly/Arg (n = 45, 21.1%) (Table 1 and Figure 1). With respect to the coincidence of promoter TATA box and Gly71Arg variations, 8 genotypes were detected. TA6/7-Gly/Gly was recognized as the most frequent genotype (46.1%), while TA6/8-Gly/Gly and TA6/5-Gly/Gly were rare genotypes (0.5%) (Table 2). 3.3. Laboratory Parameters All the participants underwent LFT, CPK, LDH, BUN, and Cr tests. The results related to the association between laboratory tests and combined genotypes are presented in Tables 3 and 4. Among promoter and UGT1A16 variation groups, there was a statistically signiﬁcant diﬀerence in the serum total bilirubin (STB) level between TA7/7-Gly/Gly (P = 0.017), TA7/7-Gly/Arg (P < 0.001), and TA6/7-Gly/Gly (P = 0.009) genotypes. Also, a statistically signiﬁcant diﬀer-ence was found regarding CPK level between TA7/7-Gly/Gly (P = 0.034), TA7/7-Gly/Arg (P = 0.022), and TA6/7-Gly/Arg (P = 0.012) genotypes, compared to other groups. However, other liver laboratory tests showed no signiﬁcant diﬀer-ence between the genotype groups. Other than LDH test, gender had no eﬀects on the re-sults of laboratory tests in diﬀerent genotype groups. Sta-tistical analysis of UGT1A16 variations showed that ho-mozygosity of Gly71Arg alone had no impact on the in-crease in STB or CPK level. The variation in LDH was signif-icant in heterozygous Gly71Arg, while it was insigniﬁcant in combination with other variations in the promoter re-gion. Also, we found no signiﬁcant diﬀerence in STB level between children with and without a history of neonatal jaundice (0.53 ± 0.4; P = 0.9). Also, age and other demo-graphic factors had no eﬀects on liver laboratory test re-sults. 4. Discussion Although GS is known as a benign condition, it is of clinical importance due to its eﬀects on the patient’s liver enzyme proﬁle. Clinical and epidemiological ﬁnd-ings about GS in diﬀerent ethnical groups can help re-searchers, pharmacists, and physicians introduce more ef-fective medications. The prevalence of UGT1A1 gene poly-morphisms in the promoter and exon 1 (Gly71Arg) regions has been separately reported in only a limited number of cases (21, 22). The present study revealed the coincidence of these 2 polymorphisms and the clinical manifestations in the Iranian population for the ﬁrst time. In this regard, Hemmati et al. (23) determined the prevalence of GS, using rifampin test in Fars province, south of Iran. They reported an incidence rate of 19.1% in both genders for GS in the general population. The results of their study showed that the prevalence of GS in males (25.6%) was higher than females (12.8%), which is inconsis-tent with the present ﬁndings (14.3% in males and 9% in fe-males); also, the overall prevalence was much lower in the present study (11.3%). This discrepancy could be due to dif-ferences in the methods applied in these studies. Iran Red Crescent Med J. 2017; 19(4):e44363. 3 Heydari MR et al. Table 1. Distribution of Promoter TATA Box and Gly71Arg Variants Based on Age and Gender Genotype Frequency (%) Promoter TATA box Gly/Arg Gender TA6/6 TA6/7 TA6/8 TA6/5 TA7/7 Gly/Gly Gly/Arg Male 40 (32.8) 70 (56.6) 1 (0.8) 1 (0.8) 11 (9) 96 (78.7) 26 (21.3) Female 33 (36.3) 45 (49.4) 0.0 (0.0) 0.0 (0.0) 13 (14.3) 72 (79.1) 19 (20.9) Total (%) 73 (34.2) 114 (53.5) 1 (0.5) 1 (0.5) 24 (11.3) 168 (78.9) 45 (21.1) Agea 37.7 ± 42.9 51.3 ± 47.0 12.0 ± 0.0 144.0 ± 0.0 49.9 ± 44.88 47.5 ± 46.38 42.0 ± 44.6 aAge in months (mean ± SD). Figure 1. Sequencing chromatogram of UGT1A1 gene in the c.211G > A site. Arrows indicate the nucleotide substitution position. By using the restriction fragment length polymor-phism (RFLP) method, Dastgerdy et al. found that the fre-quency of Gly71Arg variation (homozygous and heterozy-gous) was 30.4%, which was higher than the rate reported in the present study (21.1%). The high frequency might be attributed to the low speciﬁcity and sensitivity of RLFP method (24). Also, similar to studies by Ergin and Kaveh, we found that homozygous TA7/7 genotype is associated with increased STB level in case of additional TA insertion in the promoter (21, 25). In the Malay population, Yossef et al. observed that the prevalence of homozygous TA7/7, heterozygous TA6/7, and heterozygous Gly71Arg (Gly/Arg) mutations was 7%, 18%, and 5.5%, respectively. In the present study, similar to the mentioned research, homozygous Gly71Arg (Arg/Arg) was not detected. However, we observed a higher frequency of heterozygous TA7/6 (55.4%) than homozygous TA7/7 (11.3%), compared to the study by Yusoﬀet al. (43.5% and 9.9%, re-4 Iran Red Crescent Med J. 2017; 19(4):e44363. Heydari MR et al. Table 2. Prevalence of Promoter TATA Box and Gly71Arg Variants Based on Gender in Southern Provinces of Iran Genotype Ggroups TATA Genotype Groups Gly71Arg Gender Total Female Male Frequency % Frequency % Frequency % TA6/6 Gly/Gly 29 13.62 30 14.08 59 27.70 TA6/6 Gly/Arg 4 1.88 10 4.69 14 6.57 TA6/7 Gly/Gly 38 17.84 60 28.17 98 46.01 TA6/7 Gly/Arg 6 2.82 10 4.69 16 7.51 TA7/7 Gly/Gly 4 1.88 5 2.35 9 4.23 TA7/7 Gly/Arg 9 4.23 6 2.82 15 7.04 TA6/5 Gly/Gly 0 0 1 0.47 1 0.47 TA6/8 Gly/Gly 0 0 1 0.47 1 0.47 Total 90 42.25 123 57.75 213 100 Table 3. The Results of Laboratory Tests (Mean & IQR) in the Genotype Groupsa Laboratory test (unit) Genotype Groups TA6/6 TA6/6 TA6/7 TA6/7 TA7/7 TA7/7 TA6/5 TA6/8 Gly/Gly Gly/Arg Gly/Gly Gly/Arg Gly/Gly Gly/Arg Gly/Gly Gly/Gly Median IQRb Median IQR Median IQR Median IQR Median IQR Median IQR Median IQR Median IQR ALP (U/L) 300.0 171.8 275.5 213.0 300.0 116.5 318.0 232.0 369.0 51.9 314.0 60.5 756.0 0.0 422.0 0.0 AST (U/L) 41.0 31.5 36.0 48.0 38.0 38.5 47.0 49.5 69.0 108.5 53.0 94.3 47.0 0.0 19.0 0.0 ALT (U/L) 24.0 30.0 21.5 24.3 19.5 47.1 24.0 55.0 20.0 114 19.0 50.0 23.0 0.0 8.0 0.0 LDH (U/L) 490.0 897.0 580.0 665.5 522.0 366.5 895.5 1571.3 1586 4049 478.0 238.4 48.0 0.0 427.0 0.0 BUN (mg/dL) 9 9.5 8.0 6.8 12.0 8.0 11.0 6.3 8.0 9.5 11.5 6.8 7.0 0.0 6.0 0.0 Cr (mg/dL) 0.5 0.30 0.5 0.35 0.5 0.2 0.6 .35 0.5 0.3 0.5 0.2 4.1 0.0 0.2 0.0 Abbreviations: ALP, alkaline phosphatase; ALT, alanine transaminase; AST, aspartate transaminase; BUN, blood urea nitrogen; Cr, creatinine; LDH, lactate dehydrogenase. aThe diﬀerence between the wild type and other genotypes is signiﬁcant (P ≤0.05). bInterquartile range (IQR) = Q3 - Q1. Table 4. The Results of Laboratory Tests (mean ± SD) in the Genotype Groups Laboratory test (unit) Genotype Groups TA6/6 TA6/6 TA6/7 TA6/7 TA7/7 TA7/7 TA6/5 TA6/8 Gly/Gly Gly/Arg Gly/Gly Gly/Arg Gly/Gly Gly/Arg Gly/Gly Gly/Gly STB (mg/dL) 0.40 ± 0.20 0.41 ± 0.18 0.54± 0.21a 0.46 ± 0.32 0.85 ± 0.12a 0.87 ± 0.21a 0.40 ± 0.00 0.20 ± 0.00 CPK (U/L) 358.20 ± 219.2 132.00 ± 80.0 204.58 ± 194.30 444.40 ± 134.14a 469.00 ± 83.20a 452.50 ± 9.7a 127.00 ± 0.00 114.00 ± 0.00 Abbreviations: BUN, blood urea nitrogen; CPK, creatine phosphokinase; Cr, creatinine; STB, serum total bilirubin. aThe diﬀerence between wild type and other genotypes is signiﬁcant (P≤0.05). spectively) (26, 27). Also, Nettles et al. showed that 11% of the Swiss population had GS, and the polymorphism had a signiﬁcant correlation with the total bilirubin level. The present ﬁndings were in line with the results reported by Nettles et al.; nevertheless, they did not study other liver enzymes (28). The results of the present study showed that Gly71Arg variation had no independent eﬀects on STB level; this in-dicates that this variation only aﬀects neonatal jaundice, not bilirubin level in the postneonatal period (9, 10). We found that none of the 2 variations (TATA and Gly71Arg) alone had a signiﬁcant eﬀect on CPK level, while in combi-Iran Red Crescent Med J. 2017; 19(4):e44363. 5 Heydari MR et al. nation with some other genotypes, an increase in CPK level was revealed. In some studies, the prevalence of TA7/7 and Gly71Arg variations has been reported to be 3 - 19% and 20 - 60%, respectively (29-33). One of the important ﬁndings of the present study was identifying 2 rare TA6/8 and TA6/5 geno-types with no signiﬁcant increase in STB level. These geno-types have been previously reported in African popula-tions (34). In the current study, we also showed a signif-icant increase in the CPK level in GS cases, which has not been previously reported; nevertheless, for further clariﬁ-cation, more studies are required. The strengths of the present study include the use of sequencing method, detection of 2 rare mutations (TA6/8 and TA6/5), and the relationship between GS and CPK level. On the other hand, the shortcoming of this study was the limited sample size due to restricted sources. In conclu-sion, the present study showed that the prevalence of GS in south of Iran was nearly similar to other populations. Acknowledgments This manuscript was extracted from a PhD thesis by Mohammad Reza Heydari and was funded by the Vice-Chancellor for Research Aﬀairs of Shiraz University of Med-ical Sciences, Shiraz, Iran (grant number, 94-7475). Footnote Conﬂicts of Interest: The authors declare no conﬂicts of interest. References 1. Song J, Sun M, Li J, Zhou D, Wu X. 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1886	https://chemistry.stackexchange.com/questions/32443/what-is-the-correct-value-of-the-avogadro-constant-and-how-was-it-derived	Skip to main content What is the correct value of the Avogadro constant? And how was it derived? Ask Question Asked Modified 5 years, 4 months ago Viewed 4k times This question shows research effort; it is useful and clear 11 Save this question. Show activity on this post. I found different values of Avogadro constant in different places. So what is the correct value? 6.0221367⋅1023 or 6.02214129⋅1023 or 6.0221415⋅1023 or anything else? mole Share CC BY-SA 4.0 Follow this question to receive notifications edited Jul 26, 2018 at 14:39 mhchem 3,35622 gold badges1919 silver badges3535 bronze badges asked Jun 4, 2015 at 13:06 SMYKSMYK 12111 silver badge55 bronze badges 4 2 physics.nist.gov/cgi-bin/cuu/Value?na user7951 – user7951 06/04/2015 13:15:13 Commented Jun 4, 2015 at 13:15 Also, I'm afraid the "how was it derived?" part might be too broad. There various methods and experiments to measure the Avogadro constant, most of which are usually explained well on the Net. M.A.R. – M.A.R. 06/04/2015 13:26:04 Commented Jun 4, 2015 at 13:26 3 en.wikipedia.org/wiki/Avogadro_constant#Measurement orthocresol – orthocresol 06/04/2015 13:28:17 Commented Jun 4, 2015 at 13:28 Please now delete such a "फालतू" question user156184 – user156184 09/05/2024 02:44:08 Commented Sep 5, 2024 at 2:44 Add a comment \| 3 Answers 3 Reset to default This answer is useful 24 Save this answer. Show activity on this post. Whenever you're looking for accurate fundamental physical constants, CODATA recommended values are the way to go. As of May 20, 2019, Avogadro's constant is now truly an exact value, with infinite significant digits. Behold! 6.02214076×1023 mol−1 It is unlikely this value will change within our lifetimes, which is exciting in its own way! For historical reasons, I'll leave my previous answer below: As of 2015, the latest data for the Avogadro constant is from 2014. According to CODATA, the most accurate value is: 6.022140857×1023 mol−1±0.000000074×1023 mol−1(CODATA 2014) The relative uncertainty in the measurement is thus only 12 parts per billion! Interestingly, the Avogadro constant may be redefined in the near future to be an exact value, that is, a constant with zero uncertainty by definition, much like the speed of light. This would come as a consequence of redefining the SI kilogram as a function of the number of atoms in an ultrapure monoisotopic X28X2282Si monocrystalline sphere engineered to extreme precision. A great video on this can be found in the Veritasium YouTube channel. All that said, I suspect you don't really have to care which constant should be used. All the suggested values differ by one part in a million, which makes essentially no difference for most chemistry. Edit: As pointed out by Loong in the comments, a few weeks after writing this answer, CODATA released updated values for the physical constants, so I updated this answer for accuracy. The next set of updated values will likely be announced in 2018-2019. For comparison, the previous value was: 6.02214129×1023 mol−1±0.00000027×1023 mol−1(CODATA 2010) This represents an uncertainty of 44 parts per billion. This means the uncertainty in the measurement has been cut to almost a fourth of its previous value in four years. Go science! Share CC BY-SA 4.0 Follow this answer to receive notifications edited Apr 24, 2020 at 11:11 answered Jun 4, 2015 at 13:32 Nicolau Saker NetoNicolau Saker Neto 29.3k44 gold badges9999 silver badges148148 bronze badges 7 7 Personally, I like memorizing Avogadro's constant as 6.022×1023 mol−1, because it nicely matches the fundamental charge 1.6022×10−19 C and I recall that 222=484, which reminds me of the Faraday constant 96484 C mol−1 (actually 96485 C mol−1 is a better approximation). Precision to spare for calculations, and pretty mathematical patterns to help my poor memory! Nicolau Saker Neto – Nicolau Saker Neto 06/04/2015 16:39:42 Commented Jun 4, 2015 at 16:39 +1 especially for link to Veritasium. If physicists really wanted to make things easier for everyone, they would get rid of both the Faraday constant and Avogadro's constant, defining a new "Faragadro constant" as being precisely equal to say 1021. That would be both the number of particles in a "new mole" (we need another name for that). Coloumbs and amperes would be made obsolete and current and charge would be counted in "new moles per second" or "new moles". Volts would be made obsolete, we could just say "kJ per new mole" instead. Curt F. – Curt F. 06/04/2015 19:03:30 Commented Jun 4, 2015 at 19:03 1 +1, very good answer. For what concerns the redefinition of the kilogram, this is part of the ongoing redefinition of the SI (don't hold your breath, it'll be in 2018 probably). A glimpse of the so-called new SI can be found on this BIPM webpage where there is a draft of the new SI brochure and other information. The new unit definitions are implicit, from the given, exact, values of the constants (the values given in the draft are not definitive). Massimo Ortolano – Massimo Ortolano 06/04/2015 20:53:50 Commented Jun 4, 2015 at 20:53 2 @Equinox If Avogadro's constant is changed to an exact value by definition (which seems likely this year or the next), I'll be sure to perform the last edit ever to this answer with the details! Nicolau Saker Neto – Nicolau Saker Neto 03/04/2018 21:38:03 Commented Mar 4, 2018 at 21:38 2 @NicolauSakerNeto "If Avogadro's constant is changed to an exact value by definition" seems to have happened prior to that comment of yours ;) See Felipe's answer below. Gaurang Tandon – Gaurang Tandon 07/25/2018 16:27:38 Commented Jul 25, 2018 at 16:27 \| Show 2 more comments This answer is useful 8 Save this answer. Show activity on this post. A new definition of Avogadro's constant will replace the current one soon (emphasis mine): The mole, symbol mol, is the SI unit of amount of substance. One mole contains exactly 6.02214076×1023 elementary entities. This number is the fixed numerical value of the Avogadro constant, NA, when expressed in mol−1, and is called the Avogadro number. This makes Avogadro number a fixed integer. The new definition was published on 8 January 2018 as an IUPAC Recommendation in Pure and Applied Chemistry, which is available online. It has yet to be approved by CGPM (expected November 2018), but "the revised definitions are expected to come into force on World Metrology Day, 20 May 2019" (thanks R.M.!). Share CC BY-SA 4.0 Follow this answer to receive notifications edited Jul 26, 2018 at 14:44 mhchem 3,35622 gold badges1919 silver badges3535 bronze badges answered Jul 25, 2018 at 16:10 schneiderfelipeschneiderfelipe 2,88322 gold badges2323 silver badges4242 bronze badges 2 2 It was published as an IUPAC Recommendation on January 8th 2018. It has yet to be approved by CGPM (expected Nov 2018), and as your article states: "the revised definitions are expected to come into force on World Metrology Day, 20 May 2019". R.M. – R.M. 07/25/2018 23:29:22 Commented Jul 25, 2018 at 23:29 Thanks for correcting it, I changed the text to reflect that. schneiderfelipe – schneiderfelipe 07/26/2018 00:35:08 Commented Jul 26, 2018 at 0:35 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. Avogadro's number has changed over time because it depends on how we standardize the atomic mass unit. Formerly, oxygen was used as a standard with a value of 16, and from a comment in another post I recall that physicists used 16 specifically for oxygen-16 (making natural oxygen slightly heavier than 16). When they went to the carbon 12 = 12 standard naturally occurring oxygen dropped below 16, so Avogadro's number represented less mass of oxygen (and everything else) and it dropped accordingly. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Jul 26, 2018 at 10:11 Oscar LanziOscar Lanzi 67.4k44 gold badges102102 silver badges208208 bronze badges Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions mole See similar questions with these tags. Featured on Meta Community Asks Sprint Announcement - September 2025 Linked 5 Avogadro's Number: 6.023 x 10^23 or 6.022 x 10^23? 5 Is Avogadro's number an integer? 23 Why was Avogadro's number chosen to be the value that it is? 7 Is it possible for me to derive Avogadro's number? 3 What is the dimension of Avogadro's constant? 4 Is the kilogram still consistent with the old definition? 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1887	https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_20?srsltid=AfmBOooG8vPIxMmQ3dMPaf7RUb81mG-b8AE0-qJR0yzD7GHoKrqqtd8c	Art of Problem Solving 2004 AMC 10A Problems/Problem 20 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 2004 AMC 10A Problems/Problem 20 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2004 AMC 10A Problems/Problem 20 Contents [hide] 1 Problem 2 Solution 1 3 Solution 2 (Non-trig) 4 Solution 3 (System of Equations) 5 Solution 4 6 Video Solution 7 Video Solution by TheBeautyofMath 8 See also Problem Points and are located on square so that is equilateral. What is the ratio of the area of to that of ? Solution 1 Since triangle is equilateral, , and and are congruent. Thus, triangle is an isosceles right triangle. So we let . Thus . If we go angle chasing, we find out that , thus . . Thus , or . Thus , and , and . Thus the ratio of the areas is z Solution 2 (Non-trig) WLOG, let the side length of be 1. Let . It suffices that . Then triangles and are congruent by HL, so and . We find that , and so, by the Pythagorean Theorem, we have This yields , so . Thus, the desired ratio of areas is Solution 3 (System of Equations) Assume . Then, is and is . We see that using , is congruent to EAB. Using Pythagoras of triangles and we get . Expanding, we get . Simplifying gives solving using completing the square (or other methods) gives 2 answers: and . Because , . Using the areas, the answer is Solution 4 First, since is equilateral and is a square, by the Hypothenuse Leg Theorem, is congruent to . Then, assume length and length , then . is equilateral, so and , it is given that is a square and and are right triangles. Then we use the Pythagorean theorem to prove that and since we know that and , which means . Now we plug in the variables and the equation becomes , expand and simplify and you get . We want the ratio of area of to . Expressed in our variables, the ratio of the area is and we know , so the ratio must be . So, the answer is Video Solution Education, the Study of Everything Video Solution by TheBeautyofMath ~IceMatrix See also 2004 AMC 10A (Problems • Answer Key • Resources) Preceded by Problem 19Followed by Problem 21 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25 All AMC 10 Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: Introductory Geometry Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1888	https://www.chegg.com/homework-help/questions-and-answers/use-binomial-theorem-prove-1-x-p-1-x-p-mod-p-every-prime-p-integer-x-q181885130	Solved Use the Binomial Theorem to prove that (1+ x)^p \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Math Advanced Math Advanced Math questions and answers Use the Binomial Theorem to prove that (1+ x)^p ≡1+ x^p (mod p) for every prime p and integer x. Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: Use the Binomial Theorem to prove that (1+ x)^p ≡1+ x^p (mod p) for every prime p and integer x. Use the Binomial Theorem to prove that (1+ x)^p ≡1+ x^p (mod p) for every prime p and integer x. There are 2 steps to solve this one.Solution Share Share Share done loading Copy link Step 1 To prove (1+x)p≡1+x p mod{p} for every prime p and integer x, we can use the Binomial Theorem and properties of modulo ... View the full answer Step 2 UnlockAnswer Unlock Previous questionNext question Not the question you’re looking for? Post any question and get expert help quickly. 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1889	https://www.math.stonybrook.edu/~ndang/mat126-fall20/sec_2.4.pdf	2.4 \| Arc Length of a Curve and Surface Area Learning Objectives 2.4.1 Determine the length of a curve, y = f(x), between two points. 2.4.2 Determine the length of a curve, x = g(y), between two points. 2.4.3 Find the surface area of a solid of revolution. In this section, we use definite integrals to find the arc length of a curve. We can think of arc length as the distance you would travel if you were walking along the path of the curve. Many real-world applications involve arc length. If a rocket is launched along a parabolic path, we might want to know how far the rocket travels. Or, if a curve on a map represents a road, we might want to know how far we have to drive to reach our destination. We begin by calculating the arc length of curves defined as functions of x, then we examine the same process for curves defined as functions of y. (The process is identical, with the roles of x and y reversed.) The techniques we use to find arc length can be extended to find the surface area of a surface of revolution, and we close the section with an examination of this concept. Arc Length of the Curve y = f(x) In previous applications of integration, we required the function f(x) to be integrable, or at most continuous. However, for calculating arc length we have a more stringent requirement for f(x). Here, we require f(x) to be differentiable, and furthermore we require its derivative, f′(x), to be continuous. Functions like this, which have continuous derivatives, are called smooth. (This property comes up again in later chapters.) Let f(x) be a smooth function defined over ⎡ ⎣a, b ⎤ ⎦. We want to calculate the length of the curve from the point ⎛ ⎝a, f(a) ⎞ ⎠ to the point ⎛ ⎝b, f(b) ⎞ ⎠. We start by using line segments to approximate the length of the curve. For i = 0, 1, 2,…, n, let P = {xi} be a regular partition of ⎡ ⎣a, b ⎤ ⎦. Then, for i = 1, 2,…, n, construct a line segment from the point ⎛ ⎝xi −1, f(xi −1)⎞ ⎠to the point ⎛ ⎝xi, f(xi)⎞ ⎠. Although it might seem logical to use either horizontal or vertical line segments, we want our line segments to approximate the curve as closely as possible. Figure 2.37 depicts this construct for n = 5. Figure 2.37 We can approximate the length of a curve by adding line segments. To help us find the length of each line segment, we look at the change in vertical distance as well as the change in horizontal distance over each interval. Because we have used a regular partition, the change in horizontal distance over each interval is given by Δx. The change in vertical distance varies from interval to interval, though, so we use Δyi = f(xi) −f(xi −1) to represent the change in vertical distance over the interval [xi −1, xi], as shown in Figure 2.38. Note that some (or all) Δyi may be negative. Chapter 2 \| Applications of Integration 169 Figure 2.38 A representative line segment approximates the curve over the interval [xi −1, xi]. By the Pythagorean theorem, the length of the line segment is (Δx)2 + ⎛ ⎝Δyi ⎞ ⎠2. We can also write this as Δx 1 + ⎛ ⎝ ⎛ ⎝Δyi ⎞ ⎠/(Δx)⎞ ⎠2. Now, by the Mean Value Theorem, there is a point xi ∈[xi −1, xi] such that f′(xi ) = ⎛ ⎝Δyi ⎞ ⎠/(Δx). Then the length of the line segment is given by Δx 1 + ⎡ ⎣f′(xi )⎤ ⎦2. Adding up the lengths of all the line segments, we get Arc Length ≈∑ i = 1 n 1 + ⎡ ⎣f′(xi )⎤ ⎦2 Δx. This is a Riemann sum. Taking the limit as n →∞, we have Arc Length = lim n →∞∑ i = 1 n 1 + ⎡ ⎣f′(xi )⎤ ⎦2 Δx = ∫ a b 1 + ⎡ ⎣f′(x) ⎤ ⎦2 dx. We summarize these findings in the following theorem. Theorem 2.4: Arc Length for y = f(x) Let f(x) be a smooth function over the interval ⎡ ⎣a, b ⎤ ⎦. Then the arc length of the portion of the graph of f(x) from the point ⎛ ⎝a, f(a) ⎞ ⎠to the point ⎛ ⎝b, f(b) ⎞ ⎠is given by (2.7) Arc Length = ∫ a b 1 + ⎡ ⎣f′(x) ⎤ ⎦2 dx. Note that we are integrating an expression involving f′(x), so we need to be sure f′(x) is integrable. This is why we require f(x) to be smooth. The following example shows how to apply the theorem. Example 2.18 Calculating the Arc Length of a Function of x Let f(x) = 2x3/2. Calculate the arc length of the graph of f(x) over the interval [0, 1]. Round the answer to three decimal places. 170 Chapter 2 \| Applications of Integration This OpenStax book is available for free at 2.18 2.19 Solution We have f′(x) = 3x1/2, so ⎡ ⎣f′(x) ⎤ ⎦2 = 9x. Then, the arc length is Arc Length = ∫ a b 1 + ⎡ ⎣f′(x) ⎤ ⎦2 dx = ∫ 0 1 1 + 9x dx. Substitute u = 1 + 9x. Then, du = 9 dx. When x = 0, then u = 1, and when x = 1, then u = 10. Thus, Arc Length = ∫ 0 1 1 + 9x dx = 1 9∫ 0 1 1 + 9x9dx = 1 9∫ 1 10 u du = 1 9 · 2 3u3/2\|1 10 = 2 27 ⎡ ⎣10 10 −1⎤ ⎦≈2.268 units. Let f(x) = (4/3)x3/2. Calculate the arc length of the graph of f(x) over the interval [0, 1]. Round the answer to three decimal places. Although it is nice to have a formula for calculating arc length, this particular theorem can generate expressions that are difficult to integrate. We study some techniques for integration in Introduction to Techniques of Integration. In some cases, we may have to use a computer or calculator to approximate the value of the integral. Example 2.19 Using a Computer or Calculator to Determine the Arc Length of a Function of x Let f(x) = x2. Calculate the arc length of the graph of f(x) over the interval [1, 3]. Solution We have f′(x) = 2x, so ⎡ ⎣f′(x) ⎤ ⎦2 = 4x2. Then the arc length is given by Arc Length = ∫ a b 1 + ⎡ ⎣f′(x) ⎤ ⎦2 dx = ∫ 1 3 1 + 4x2 dx. Using a computer to approximate the value of this integral, we get ∫ 1 3 1 + 4x2 dx ≈8.26815. Let f(x) = sin x. Calculate the arc length of the graph of f(x) over the interval [0, π]. Use a computer or calculator to approximate the value of the integral. Chapter 2 \| Applications of Integration 171 Arc Length of the Curve x = g(y) We have just seen how to approximate the length of a curve with line segments. If we want to find the arc length of the graph of a function of y, we can repeat the same process, except we partition the y-axis instead of the x-axis. Figure 2.39 shows a representative line segment. Figure 2.39 A representative line segment over the interval [yi −1, yi]. Then the length of the line segment is ⎛ ⎝Δy ⎞ ⎠2 + ⎛ ⎝Δxi ⎞ ⎠2, which can also be written as Δy 1 + ⎛ ⎝ ⎛ ⎝Δxi ⎞ ⎠/ ⎛ ⎝Δy ⎞ ⎠ ⎞ ⎠2. If we now follow the same development we did earlier, we get a formula for arc length of a function x = g(y). Theorem 2.5: Arc Length for x = g(y) Let g(y) be a smooth function over an interval ⎡ ⎣c, d ⎤ ⎦. Then, the arc length of the graph of g(y) from the point ⎛ ⎝c, g(c) ⎞ ⎠to the point ⎛ ⎝d, g(d) ⎞ ⎠is given by (2.8) Arc Length = ∫ c d 1 + ⎡ ⎣g′(y) ⎤ ⎦2 dy. Example 2.20 Calculating the Arc Length of a Function of y Let g(y) = 3y3. Calculate the arc length of the graph of g(y) over the interval [1, 2]. Solution We have g′(y) = 9y2, so ⎡ ⎣g′(y) ⎤ ⎦2 = 81y4. Then the arc length is Arc Length = ∫ c d 1 + ⎡ ⎣g′(y) ⎤ ⎦2 dy = ∫ 1 2 1 + 81y4 dy. Using a computer to approximate the value of this integral, we obtain ∫ 1 2 1 + 81y4 dy ≈21.0277. 172 Chapter 2 \| Applications of Integration This OpenStax book is available for free at 2.20 Let g(y) = 1/y. Calculate the arc length of the graph of g(y) over the interval [1, 4]. Use a computer or calculator to approximate the value of the integral. Area of a Surface of Revolution The concepts we used to find the arc length of a curve can be extended to find the surface area of a surface of revolution. Surface area is the total area of the outer layer of an object. For objects such as cubes or bricks, the surface area of the object is the sum of the areas of all of its faces. For curved surfaces, the situation is a little more complex. Let f(x) be a nonnegative smooth function over the interval ⎡ ⎣a, b ⎤ ⎦. We wish to find the surface area of the surface of revolution created by revolving the graph of y = f(x) around the x-axis as shown in the following figure. Figure 2.40 (a) A curve representing the function f(x). (b) The surface of revolution formed by revolving the graph of f(x) around the x-axis. As we have done many times before, we are going to partition the interval ⎡ ⎣a, b ⎤ ⎦and approximate the surface area by calculating the surface area of simpler shapes. We start by using line segments to approximate the curve, as we did earlier in this section. For i = 0, 1, 2,…, n, let P = {xi} be a regular partition of ⎡ ⎣a, b ⎤ ⎦. Then, for i = 1, 2,…, n, construct a line segment from the point ⎛ ⎝xi −1, f(xi −1)⎞ ⎠to the point ⎛ ⎝xi, f(xi)⎞ ⎠. Now, revolve these line segments around the x-axis to generate an approximation of the surface of revolution as shown in the following figure. Figure 2.41 (a) Approximating f(x) with line segments. (b) The surface of revolution formed by revolving the line segments around the x-axis. Notice that when each line segment is revolved around the axis, it produces a band. These bands are actually pieces of cones Chapter 2 \| Applications of Integration 173 (think of an ice cream cone with the pointy end cut off). A piece of a cone like this is called a frustum of a cone. To find the surface area of the band, we need to find the lateral surface area, S, of the frustum (the area of just the slanted outside surface of the frustum, not including the areas of the top or bottom faces). Let r1 and r2 be the radii of the wide end and the narrow end of the frustum, respectively, and let l be the slant height of the frustum as shown in the following figure. Figure 2.42 A frustum of a cone can approximate a small part of surface area. We know the lateral surface area of a cone is given by Lateral Surface Area = πrs, where r is the radius of the base of the cone and s is the slant height (see the following figure). Figure 2.43 The lateral surface area of the cone is given by πrs. Since a frustum can be thought of as a piece of a cone, the lateral surface area of the frustum is given by the lateral surface area of the whole cone less the lateral surface area of the smaller cone (the pointy tip) that was cut off (see the following figure). 174 Chapter 2 \| Applications of Integration This OpenStax book is available for free at Figure 2.44 Calculating the lateral surface area of a frustum of a cone. The cross-sections of the small cone and the large cone are similar triangles, so we see that r2 r1 = s −l s . Solving for s, we get r2 r1 = s −l s r2 s = r1(s −l) r2 s = r1s −r1 l r1l = r1s −r2 s r1 l = (r1 −r2)s r1 l r1 −r2 = s. Then the lateral surface area (SA) of the frustum is S = (Lateral SA of large cone) −(Lateral SA of small cone) = πr1 s −πr2 (s −l) = πr1⎛ ⎝ r1 l r1 −r2 ⎞ ⎠−πr2⎛ ⎝ r1l r1 −r2 −l⎞ ⎠ = πr1 2 l r1 −r2 −πr1 r2l r1 −r2 + πr2 l = πr1 2 l r1 −r2 −πr1 r2l r1 −r2 + πr2l(r1 −r2) r1 −r2 = πr1 2 l r1 −r2 −πr1 r2l r1 −r2 + πr1r2 l r1 −r2 −πr2 2l r1 −r2 = π⎛ ⎝r1 2 −r2 2⎞ ⎠l r1 −r2 = π(r1 −r2)(r1 + r2)l r1 −r2 = π(r1 + r2)l. Let’s now use this formula to calculate the surface area of each of the bands formed by revolving the line segments around the x-axis. A representative band is shown in the following figure. Chapter 2 \| Applications of Integration 175 Figure 2.45 A representative band used for determining surface area. Note that the slant height of this frustum is just the length of the line segment used to generate it. So, applying the surface area formula, we have S = π(r1 + r2)l = π⎛ ⎝f(xi −1) + f(xi)⎞ ⎠Δx2 + ⎛ ⎝Δyi ⎞ ⎠2 = π⎛ ⎝f(xi −1) + f(xi)⎞ ⎠Δx 1 + ⎛ ⎝ Δyi Δx ⎞ ⎠ 2 . Now, as we did in the development of the arc length formula, we apply the Mean Value Theorem to select xi ∈[xi −1, xi] such that f′(xi ) = ⎛ ⎝Δyi ⎞ ⎠/Δx. This gives us S = π⎛ ⎝f(xi −1) + f(xi)⎞ ⎠Δx 1 + ⎛ ⎝f′(xi )⎞ ⎠2. Furthermore, since f(x) is continuous, by the Intermediate Value Theorem, there is a point xi ∈[xi −1, xi] such that f(xi ) = (1/2)⎡ ⎣f(xi −1) + f(xi)⎤ ⎦, so we get S = 2π f(xi )Δx 1 + ⎛ ⎝f′(xi )⎞ ⎠2. Then the approximate surface area of the whole surface of revolution is given by Surface Area ≈∑ i = 1 n 2π f(xi )Δx 1 + ⎛ ⎝f′(xi )⎞ ⎠2. This almost looks like a Riemann sum, except we have functions evaluated at two different points, xi and xi , over the interval [xi −1, xi]. Although we do not examine the details here, it turns out that because f(x) is smooth, if we let n →∞, the limit works the same as a Riemann sum even with the two different evaluation points. This makes sense intuitively. Both xi and xi are in the interval [xi −1, xi], so it makes sense that as n →∞, both xi and xi approach x. Those of you who are interested in the details should consult an advanced calculus text. Taking the limit as n →∞, we get Surface Area = lim n →∞∑ i = 1 n 2π f(xi )Δx 1 + ⎛ ⎝f′(xi )⎞ ⎠2 = ∫ a b⎛ ⎝2π f(x) 1 + ⎛ ⎝f′(x) ⎞ ⎠2⎞ ⎠dx. As with arc length, we can conduct a similar development for functions of y to get a formula for the surface area of surfaces of revolution about the y-axis. These findings are summarized in the following theorem. 176 Chapter 2 \| Applications of Integration This OpenStax book is available for free at Theorem 2.6: Surface Area of a Surface of Revolution Let f(x) be a nonnegative smooth function over the interval ⎡ ⎣a, b ⎤ ⎦. Then, the surface area of the surface of revolution formed by revolving the graph of f(x) around the x-axis is given by (2.9) Surface Area = ∫ a b⎛ ⎝2π f(x) 1 + ⎛ ⎝f′(x) ⎞ ⎠2⎞ ⎠dx. Similarly, let g(y) be a nonnegative smooth function over the interval ⎡ ⎣c, d ⎤ ⎦. Then, the surface area of the surface of revolution formed by revolving the graph of g(y) around the y-axis is given by Surface Area = ∫ c d⎛ ⎝2πg(y) 1 + ⎛ ⎝g′(y) ⎞ ⎠2⎞ ⎠dy. Example 2.21 Calculating the Surface Area of a Surface of Revolution 1 Let f(x) = x over the interval [1, 4]. Find the surface area of the surface generated by revolving the graph of f(x) around the x-axis. Round the answer to three decimal places. Solution The graph of f(x) and the surface of rotation are shown in the following figure. Figure 2.46 (a) The graph of f(x). (b) The surface of revolution. We have f(x) = x. Then, f′(x) = 1/(2 x) and ⎛ ⎝f′(x) ⎞ ⎠2 = 1/(4x). Then, Chapter 2 \| Applications of Integration 177 2.21 Surface Area = ∫ a b⎛ ⎝2π f(x) 1 + ⎛ ⎝f′(x) ⎞ ⎠2⎞ ⎠dx = ∫ 1 4⎛ ⎝2π x 1 + 1 4x ⎞ ⎠dx = ∫ 1 4⎛ ⎝2π x + 1 4 ⎞ ⎠dx. Let u = x + 1/4. Then, du = dx. When x = 1, u = 5/4, and when x = 4, u = 17/4. This gives us ∫ 0 1⎛ ⎝2π x + 1 4 ⎞ ⎠dx = ∫ 5/4 17/4 2π u du = 2π⎡ ⎣2 3u3/2⎤ ⎦\|5/4 17/4 = π 6 ⎡ ⎣17 17 −5 5⎤ ⎦≈30.846. Let f(x) = 1 −x over the interval [0, 1/2]. Find the surface area of the surface generated by revolving the graph of f(x) around the x-axis. Round the answer to three decimal places. Example 2.22 Calculating the Surface Area of a Surface of Revolution 2 Let f(x) = y = 3x 3 . Consider the portion of the curve where 0 ≤y ≤2. Find the surface area of the surface generated by revolving the graph of f(x) around the y-axis. Solution Notice that we are revolving the curve around the y-axis, and the interval is in terms of y, so we want to rewrite the function as a function of y. We get x = g(y) = (1/3)y3. The graph of g(y) and the surface of rotation are shown in the following figure. 178 Chapter 2 \| Applications of Integration This OpenStax book is available for free at 2.22 Figure 2.47 (a) The graph of g(y). (b) The surface of revolution. We have g(y) = (1/3)y3, so g′(y) = y2 and ⎛ ⎝g′(y) ⎞ ⎠2 = y4. Then Surface Area = ∫ c d⎛ ⎝2πg(y) 1 + ⎛ ⎝g′(y) ⎞ ⎠2⎞ ⎠dy = ∫ 0 2⎛ ⎝2π⎛ ⎝1 3y3⎞ ⎠1 + y4⎞ ⎠dy = 2π 3 ∫ 0 2⎛ ⎝y3 1 + y4⎞ ⎠dy. Let u = y4 + 1. Then du = 4y3 dy. When y = 0, u = 1, and when y = 2, u = 17. Then 2π 3 ∫ 0 2⎛ ⎝y3 1 + y4⎞ ⎠dy = 2π 3 ∫ 1 171 4 udu = π 6 ⎡ ⎣2 3u3/2⎤ ⎦\|1 17 = π 9 ⎡ ⎣(17)3/2 −1⎤ ⎦≈24.118. Let g(y) = 9 −y2 over the interval y ∈[0, 2]. Find the surface area of the surface generated by revolving the graph of g(y) around the y-axis. Chapter 2 \| Applications of Integration 179 2.4 EXERCISES For the following exercises, find the length of the functions over the given interval. 165. y = 5x from x = 0 to x = 2 166. y = −1 2x + 25 from x = 1 to x = 4 167. x = 4y from y = −1 to y = 1 168. Pick an arbitrary linear function x = g(y) over any interval of your choice (y1, y2). Determine the length of the function and then prove the length is correct by using geometry. 169. Find the surface area of the volume generated when the curve y = x revolves around the x-axis from (1, 1) to (4, 2), as seen here. 170. Find the surface area of the volume generated when the curve y = x2 revolves around the y-axis from (1, 1) to (3, 9). For the following exercises, find the lengths of the functions of x over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it. 171. y = x3/2 from (0, 0) to (1, 1) 172. y = x2/3 from (1, 1) to (8, 4) 173. y = 1 3 ⎛ ⎝x2 + 2⎞ ⎠ 3/2 from x = 0 to x = 1 174. y = 1 3 ⎛ ⎝x2 −2⎞ ⎠ 3/2 from x = 2 to x = 4 175. [T] y = ex on x = 0 to x = 1 176. y = x3 3 + 1 4x from x = 1 to x = 3 177. y = x4 4 + 1 8x2 from x = 1 to x = 2 178. y = 2x3/2 3 −x1/2 2 from x = 1 to x = 4 179. y = 1 27 ⎛ ⎝9x2 + 6⎞ ⎠ 3/2 from x = 0 to x = 2 180. [T] y = sin x on x = 0 to x = π For the following exercises, find the lengths of the functions of y over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it. 181. y = 5 −3x 4 from y = 0 to y = 4 182. x = 1 2 ⎛ ⎝ey + e−y⎞ ⎠from y = −1 to y = 1 183. x = 5y3/2 from y = 0 to y = 1 184. [T] x = y2 from y = 0 to y = 1 185. x = y from y = 0 to y = 1 186. x = 2 3 ⎛ ⎝y2 + 1⎞ ⎠ 3/2 from y = 1 to y = 3 187. [T] x = tan y from y = 0 to y = 3 4 188. [T] x = cos2 y from y = −π 2 to y = π 2 180 Chapter 2 \| Applications of Integration This OpenStax book is available for free at 189. [T] x = 4y from y = 0 to y = 2 190. [T] x = ln(y) on y = 1 e to y = e For the following exercises, find the surface area of the volume generated when the following curves revolve around the x-axis. If you cannot evaluate the integral exactly, use your calculator to approximate it. 191. y = x from x = 2 to x = 6 192. y = x3 from x = 0 to x = 1 193. y = 7x from x = −1 to x = 1 194. [T] y = 1 x2 from x = 1 to x = 3 195. y = 4 −x2 from x = 0 to x = 2 196. y = 4 −x2 from x = −1 to x = 1 197. y = 5x from x = 1 to x = 5 198. [T] y = tan x from x = −π 4 to x = π 4 For the following exercises, find the surface area of the volume generated when the following curves revolve around the y-axis. If you cannot evaluate the integral exactly, use your calculator to approximate it. 199. y = x2 from x = 0 to x = 2 200. y = 1 2x2 + 1 2 from x = 0 to x = 1 201. y = x + 1 from x = 0 to x = 3 202. [T] y = 1 x from x = 1 2 to x = 1 203. y = x 3 from x = 1 to x = 27 204. [T] y = 3x4 from x = 0 to x = 1 205. [T] y = 1 x from x = 1 to x = 3 206. [T] y = cos x from x = 0 to x = π 2 207. The base of a lamp is constructed by revolving a quarter circle y = 2x −x2 around the y-axis from x = 1 to x = 2, as seen here. Create an integral for the surface area of this curve and compute it. 208. A light bulb is a sphere with radius 1/2 in. with the bottom sliced off to fit exactly onto a cylinder of radius 1/4 in. and length 1/3 in., as seen here. The sphere is cut off at the bottom to fit exactly onto the cylinder, so the radius of the cut is 1/4 in. Find the surface area (not including the top or bottom of the cylinder). 209. [T] A lampshade is constructed by rotating y = 1/x around the x-axis from y = 1 to y = 2, as seen here. Determine how much material you would need to construct this lampshade—that is, the surface area—accurate to four decimal places. 210. [T] An anchor drags behind a boat according to the function y = 24e−x/2 −24, where y represents the depth beneath the boat and x is the horizontal distance of the anchor from the back of the boat. If the anchor is 23 ft below the boat, how much rope do you have to pull to reach the anchor? Round your answer to three decimal places. Chapter 2 \| Applications of Integration 181 211. [T] You are building a bridge that will span 10 ft. You intend to add decorative rope in the shape of y = 5\|sin ⎛ ⎝(xπ)/5 ⎞ ⎠\|, where x is the distance in feet from one end of the bridge. Find out how much rope you need to buy, rounded to the nearest foot. For the following exercises, find the exact arc length for the following problems over the given interval. 212. y = ln(sin x) from x = π/4 to x = (3π)/4. (Hint: Recall trigonometric identities.) 213. Draw graphs of y = x2, y = x6, and y = x10. For y = xn, as n increases, formulate a prediction on the arc length from (0, 0) to (1, 1). Now, compute the lengths of these three functions and determine whether your prediction is correct. 214. Compare the lengths of the parabola x = y2 and the line x = by from (0, 0) to ⎛ ⎝b2, b⎞ ⎠as b increases. What do you notice? 215. Solve for the length of x = y2 from (0, 0) to (1, 1). Show that x = (1/2)y2 from (0, 0) to (2, 2) is twice as long. Graph both functions and explain why this is so. 216. [T] Which is longer between (1, 1) and (2, 1/2): the hyperbola y = 1/x or the graph of x + 2y = 3? 217. Explain why the surface area is infinite when y = 1/x is rotated around the x-axis for 1 ≤x < ∞, but the volume is finite. 182 Chapter 2 \| Applications of Integration This OpenStax book is available for free at
1890	https://www.jcproto.com/new/mechanical-properties-of-brass.html	The Mechanical Properties of Brass: Comprehensive Guide Views: 7 Author: Allen Xiao Publish Time: 2025-09-22 Origin: Site Brass, a versatile and widely used alloy, is primarily composed of copper and zinc. This metallic combination creates a material known for its excellent machinability, corrosion resistance, and attractive gold-like appearance. The specific properties of brass can be tailored by adjusting the proportions of copper and zinc and by adding other elements like lead, aluminum, or silicon. This adaptability makes it indispensable across numerous sectors, including plumbing, musical instruments, electrical components, and decorative hardware. Its fundamental characteristics stem from its copper base, which provides inherent antimicrobial properties and high ductility. content: An Overview of Brass's Mechanical Properties Understanding Brass Tensile Strength The Significance of Brass Yield Strength Examining Brass Ultimate Tensile Strength Measuring the Hardness of Brass The Density Question: Is Brass Heavy? The Stiffness Factor: Modulus of Brass An Overview of Brass's Mechanical Properties The mechanical properties of brass encompass its response to applied forces, including strength, ductility, hardness, and elasticity. These properties are not fixed but vary significantly with the alloy's chemical composition and its thermal history (e.g., whether it is annealed or cold-worked). Generally, brass offers a superb balance between strength and formability. It possesses good tensile and yield strength, especially in work-hardened conditions, and maintains high impact strength and fatigue resistance. Its low coefficient of friction and non-magnetic nature further expand its utility in mechanical and marine applications, making it a cornerstone material for engineers and designers. Understanding Brass Tensile Strength Tensile strength refers to the maximum stress a material can withstand while being stretched or pulled before necking, which is when the specimen's cross-section starts to significantly contract. For brass alloys, tensile strength values exhibit a broad range. Common yellow brass (C26000) may have a tensile strength of around 60,000 psi (414 MPa) in its annealed, soft state. However, this value can dramatically increase through cold working processes like rolling or drawing. In its hard temper, the same alloy can achieve a tensile strength exceeding 76,000 psi (524 MPa). This enhanced strength makes cold-worked brass suitable for springs, fasteners, and other components requiring high durability. The Significance of Brass Yield Strength Perhaps more critical than ultimate tensile strength for design engineers is the yield strength. This property denotes the stress level at which a material begins to deform plastically; beyond this point, the deformation is permanent and non-recoverable. Brass yield strength is highly sensitive to its temper. An annealed cartridge brass (C26000) has a very low yield strength, perhaps around 10,000 psi (69 MPa), making it easy to form and shape. Conversely, the same alloy in a full-hard temper can see its yield strength rise to approximately 63,000 psi (434 MPa). Designing around the yield strength ensures components will function within their elastic limit, avoiding permanent deformation under load. Examining Brass Ultimate Tensile Strength While often used interchangeably with tensile strength, the term "ultimate tensile strength" (UTS) specifically indicates the absolute maximum stress on the engineering stress-strain curve. It is the peak stress value reached during the test. The brass ultimate tensile strength is a key metric for comparing the load-bearing capacity of different alloys. For instance, a high-strength aluminum brass like C68700 can have a UTS of up to 79,000 psi (545 MPa) in the annealed condition, while a free-cutting leaded brass like C36000 might top out at approximately 47,000 psi (324 MPa) in its soft state. This value helps manufacturers select the appropriate grade for high-stress applications. Measuring the Hardness of Brass Hardness quantifies a material's resistance to localized plastic deformation, such as denting or scratching. Brass hardness is typically measured using scales like Rockwell B (HRB) or Vickers (HV). The hardness is directly influenced by the alloying elements and the amount of cold working. Annealed brass is relatively soft, with a Rockwell B hardness around 55 HRB. After severe cold working, its hardness can increase to 90 HRB or higher. Adding elements like aluminum or manganese also contributes to solid solution strengthening, thereby increasing the hardness. This property is crucial for applications involving wear, such as gears, bushings, and bearings. The Density Question: Is Brass Heavy? When pondering "is brass heavy," it's helpful to compare it to other common materials. Brass has a density ranging from approximately 0.303 lb/in³ (8,400 kg/m³) to 0.315 lb/in³ (8,730 kg/m³), depending on its zinc content. This makes it denser than aluminum (~0.098 lb/in³) and many types of plastic but notably less dense than steel (~0.284 lb/in³ for stainless, but carbon steel is around 0.282-0.292 lb/in³—so some brass can be slightly denser than some steels). While not the lightest metal, its favorable strength-to-weight ratio and other beneficial properties often justify its use in many applications where weight is a secondary concern. The Stiffness Factor: Modulus of Brass The modulus of brass, specifically its Young's Modulus of Elasticity, is a measure of its inherent stiffness. It defines the relationship between stress (force per unit area) and strain (proportional deformation) in the elastic region. For most brass alloys, this modulus falls within a relatively narrow range of 14-16 million psi (97-110 GPa). This is significantly lower than that of steel (around 29-30 million psi), meaning brass is more flexible and will deflect more under the same load. This property is vital for designing components where flexibility is desired, or where excessive stiffness could lead to failure. It is a fundamental property that is largely determined by the atomic bonds and is not significantly altered by cold working or heat treatment. In conclusion, the mechanical properties of brass form a complex and adaptable portfolio that engineers can leverage for a vast array of applications. From its fundamental tensile and yield strengths to its hardness and modulus of elasticity, understanding these characteristics is key to selecting the right brass alloy for any given task. Its unique combination of strength, workability, corrosion resistance, and aesthetic appeal ensures its continued prominence in modern manufacturing and design. Brass, a versatile and widely used alloy, is primarily composed of copper and zinc. This metallic combination creates a material known for its excellent machinability, corrosion resistance, and attractive gold-like appearance. The specific properties of brass can be tailored by adjusting the proportions of copper and zinc and by adding other elements like lead, aluminum, or silicon. This adaptability makes it indispensable across numerous sectors, including plumbing, musical instruments, electrical components, and decorative hardware. Its fundamental characteristics stem from its copper base, which provides inherent antimicrobial properties and high ductility. content: An Overview of Brass's Mechanical Properties Understanding Brass Tensile Strength The Significance of Brass Yield Strength Examining Brass Ultimate Tensile Strength Measuring the Hardness of Brass The Density Question: Is Brass Heavy? The Stiffness Factor: Modulus of Brass An Overview of Brass's Mechanical Properties The mechanical properties of brass encompass its response to applied forces, including strength, ductility, hardness, and elasticity. These properties are not fixed but vary significantly with the alloy's chemical composition and its thermal history (e.g., whether it is annealed or cold-worked). Generally, brass offers a superb balance between strength and formability. It possesses good tensile and yield strength, especially in work-hardened conditions, and maintains high impact strength and fatigue resistance. Its low coefficient of friction and non-magnetic nature further expand its utility in mechanical and marine applications, making it a cornerstone material for engineers and designers. Understanding Brass Tensile Strength Tensile strength refers to the maximum stress a material can withstand while being stretched or pulled before necking, which is when the specimen's cross-section starts to significantly contract. For brass alloys, tensile strength values exhibit a broad range. Common yellow brass (C26000) may have a tensile strength of around 60,000 psi (414 MPa) in its annealed, soft state. However, this value can dramatically increase through cold working processes like rolling or drawing. In its hard temper, the same alloy can achieve a tensile strength exceeding 76,000 psi (524 MPa). This enhanced strength makes cold-worked brass suitable for springs, fasteners, and other components requiring high durability. The Significance of Brass Yield Strength Perhaps more critical than ultimate tensile strength for design engineers is the yield strength. This property denotes the stress level at which a material begins to deform plastically; beyond this point, the deformation is permanent and non-recoverable. Brass yield strength is highly sensitive to its temper. An annealed cartridge brass (C26000) has a very low yield strength, perhaps around 10,000 psi (69 MPa), making it easy to form and shape. Conversely, the same alloy in a full-hard temper can see its yield strength rise to approximately 63,000 psi (434 MPa). Designing around the yield strength ensures components will function within their elastic limit, avoiding permanent deformation under load. Examining Brass Ultimate Tensile Strength While often used interchangeably with tensile strength, the term "ultimate tensile strength" (UTS) specifically indicates the absolute maximum stress on the engineering stress-strain curve. It is the peak stress value reached during the test. The brass ultimate tensile strength is a key metric for comparing the load-bearing capacity of different alloys. For instance, a high-strength aluminum brass like C68700 can have a UTS of up to 79,000 psi (545 MPa) in the annealed condition, while a free-cutting leaded brass like C36000 might top out at approximately 47,000 psi (324 MPa) in its soft state. This value helps manufacturers select the appropriate grade for high-stress applications. Measuring the Hardness of Brass Hardness quantifies a material's resistance to localized plastic deformation, such as denting or scratching. Brass hardness is typically measured using scales like Rockwell B (HRB) or Vickers (HV). The hardness is directly influenced by the alloying elements and the amount of cold working. Annealed brass is relatively soft, with a Rockwell B hardness around 55 HRB. After severe cold working, its hardness can increase to 90 HRB or higher. Adding elements like aluminum or manganese also contributes to solid solution strengthening, thereby increasing the hardness. This property is crucial for applications involving wear, such as gears, bushings, and bearings. The Density Question: Is Brass Heavy? When pondering "is brass heavy," it's helpful to compare it to other common materials. Brass has a density ranging from approximately 0.303 lb/in³ (8,400 kg/m³) to 0.315 lb/in³ (8,730 kg/m³), depending on its zinc content. This makes it denser than aluminum (~0.098 lb/in³) and many types of plastic but notably less dense than steel (~0.284 lb/in³ for stainless, but carbon steel is around 0.282-0.292 lb/in³—so some brass can be slightly denser than some steels). While not the lightest metal, its favorable strength-to-weight ratio and other beneficial properties often justify its use in many applications where weight is a secondary concern. The Stiffness Factor: Modulus of Brass The modulus of brass, specifically its Young's Modulus of Elasticity, is a measure of its inherent stiffness. It defines the relationship between stress (force per unit area) and strain (proportional deformation) in the elastic region. For most brass alloys, this modulus falls within a relatively narrow range of 14-16 million psi (97-110 GPa). This is significantly lower than that of steel (around 29-30 million psi), meaning brass is more flexible and will deflect more under the same load. This property is vital for designing components where flexibility is desired, or where excessive stiffness could lead to failure. It is a fundamental property that is largely determined by the atomic bonds and is not significantly altered by cold working or heat treatment. In conclusion, the mechanical properties of brass form a complex and adaptable portfolio that engineers can leverage for a vast array of applications. From its fundamental tensile and yield strengths to its hardness and modulus of elasticity, understanding these characteristics is key to selecting the right brass alloy for any given task. Its unique combination of strength, workability, corrosion resistance, and aesthetic appeal ensures its continued prominence in modern manufacturing and design. Related Articles content is empty! Categories Capabilities Resources Company
1891	https://www.khanacademy.org/math/algebra-home/alg-polynomials/alg-multiplying-monomials/v/multiplying-and-dividing-monomials-1	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
1892	https://mathematica.stackexchange.com/questions/311763/how-to-convert-a-cosine-or-sine-function-into-a-phasor-or-the-standard-form-of-a	complex - How to convert a cosine or sine function into a phasor or the standard form of a sinusoidal alternating quantity? - Mathematica Stack Exchange Join Mathematica By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematica helpchat Mathematica Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to convert a cosine or sine function into a phasor or the standard form of a sinusoidal alternating quantity? Ask Question Asked 6 months ago Modified6 months ago Viewed 193 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Edit 20250324 This is essentially a problem of trigonometric transformation. The key to the problem is how to convert a sine or cosine function into the standard form of a sinusoidal alternating quantity, that is: A sin(ω t+ϕ),where A>0,ω>0,ϕ∈[−180∘,180∘]A sin⁡(ω t+ϕ),where A>0,ω>0,ϕ∈[−180∘,180∘] . For example, mathematica Iin1 = 4SqrtSin[314t + 30 Degree]; Iin2 = -3SqrtCos[314t + 30 Degree]; Iin3 = 5SqrtSin[314t - 40 Degree]; Iin4 = 8SqrtCos[100t - 60 Degree]; Iin5 = -5SqrtSin[314t - 40 Degree]; the standard form of the sinusoidal alternating quantity is: mathematica 4SqrtSin[314t + 30 Degree]; 3SqrtSin[314t - 60 Degree]; 5SqrtSin[314t - 40 Degree]; 8SqrtSin[100t + 30 Degree]; 5SqrtSin[314t + 140 Degree]; Then, extract the effective value and initial phase of these standard sinusoidal quantities, and the complex number formed is the phasor. That is: mathematica "4 Exp[I 30 Degree]"; "3 Exp[I -60 Degree]"; "5 Exp[I -40 Degree]"; "8 Exp[I 30 Degree]"; "5 Exp[I 140 Degree]"; or mathematica {4, 30} {3, -60} {5, -40} {8, 30} {5, 140} Original question Extraction of RMS Phasor in Sinusoidal AC Circuits (Example: i(t)=5 2–√sin(100 t−40∘)i(t)=5 2 sin⁡(100 t−40∘)) Steps to Extract RMS Phasor: {A eff,ϕ A eff,ϕ}: 1.Standardize to Sine Form Ensure the expression is A sin(ω t+ϕ)A sin⁡(ω t+ϕ), where A>0,ω>0 A>0,ω>0, ϕ ϕ lies in [−180∘,180∘][−180∘,180∘]. Convert cosine terms (e.g., cos θ cos⁡θ) to ( sin(θ+90∘)sin⁡(θ+90∘)). 2.Extract Parameters Peak amplitude: A=5 2–√A=5 2 Angular frequency: ω=100 rad/s ω=100 rad/s Initial phase: ϕ=−40∘ϕ=−40∘ 3.Calculate RMS Value RMS amplitude: A eff=A 2–√=5 2–√2–√=5 A eff=A 2=5 2 2=5 4.Phase Normalization Ensure ϕ ϕ lies in [−180∘,180∘][−180∘,180∘]. Here, no adjustment is needed. 5.Construct Phasor The RMS phasor is: {5, -40} or 5 Exp[I -40 Degree] Key Points: Negative Amplitude Handling: A term like −sin(θ)−sin⁡(θ) is equivalent to sin(θ+180∘)sin⁡(θ+180∘) (phase reversal). The magnitude of a phasor is the effective value of a sinusoidal quantity, and this value is greater than zero. The angle of the phasor represents the initial phase of the sinusoidal quantity. ``mathematica ClearAll["Global"]; toPhasor[expr_, t_] := Module[{exprRad, exprSine, amp, arg, omega, phi, effectiveValue, adjustedPhase, adjustedPhaseDegree},(Convert to a unified sine form (including degree \ to radian conversion))exprRad = expr /. Degree -> Pi/180; exprSine = exprRad //. Cos[x_] :> Sin[x + Pi/2]; (Check if it is a single sine term) If[! MatchQ[exprSine, . Sin[]], Return["Expression cannot be converted to a single sine form."]]; (Extract the sine argument) arg = Cases[exprSine, Sin[a_] :> a, {1}]; (Extract the amplitude (handling both explicit and implicit coefficients)) amp = Coefficient[exprSine, Sin[arg]]; If[amp === 1, amp = exprSine /. Sin[_] -> 1]; (Adjust the angular frequency to be positive and synchronously correct \ the phase and amplitude)omega = Coefficient[arg, t]; If[omega < 0, arg = -arg; amp = -amp; omega = -omega;]; (Separate the static phase)phi = Expand[arg - omegat]; (Handle the negative sign of the amplitude and adjust the phase) If[amp < 0, amp = -amp; phi = phi + Pi;]; (Calculate the effective value)effectiveValue = amp/Sqrt; (Normalize the phase to[-[Pi],[Pi]]) adjustedPhase = Mod[phi + Pi, 2 Pi] - Pi; (Convert to degrees and ensure the correct sign) adjustedPhaseDegree = Round[adjustedPhase180/Pi, 10^-6]; adjustedPhaseDegree = If[adjustedPhaseDegree == 180, -180, adjustedPhaseDegree]; {effectiveValue, adjustedPhaseDegree}]; Iin1 = 4SqrtSin[314t + 30 Degree];({4,30}) Iin2 = -3SqrtCos[314t + 30 Degree];({3,-60}) Iin3 = 5SqrtSin[314t - 40 Degree]; ({5,-40}) Iin4 = 8SqrtCos[100t - 60 Degree]; ({8,30}) Iin5 = -5SqrtSin[314t - 40 Degree];({5,140}) IPhasor1 = toPhasor[Iin1, t] IPhasor2 = toPhasor[Iin2, t] IPhasor3 = toPhasor[Iin3, t] IPhasor4 = toPhasor[Iin4, t] IPhasor5 = toPhasor[Iin5, t] ``` Output: mathematica {4, 30} Expression cannot be converted to a single sine form. {5, -40} {8, 30} {5, 140} In the test function, only Iin2 = -3SqrtCos[314t + 30 Degree] fails to produce the correct phasor. Can someone help me modify the code? Thanks in advance! complex algebraic-manipulation trigonometry Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Mar 24 at 3:07 xzczd♦ 71.6k 9 9 gold badges 182 182 silver badges 520 520 bronze badges asked Mar 23 at 4:12 lotus2019lotus2019 2,763 6 6 silver badges 12 12 bronze badges 2 1 Have you seen the question Use "+" to add vectors in polar form (phasors) and the question Represent phasors in an easy way ?LouisB –LouisB 2025-03-23 08:29:09 +00:00 Commented Mar 23 at 8:29 @LouisB Thank you for the links. However, my problem is different. Those issues all start from the already standardized exponential function to provide phasors, whereas my code goes one step further. It can handle some non-standard forms of trigonometric functions, starting from more primitive and more common forms of AC voltage and current, such as -3SqrtCos[314t + 30 Degree]. The key step here is to first transform them into standard-form sinusoidal functions, and then convert them into phasors. So, the crucial part is how to transform them into standard-form sinusoidal functions.lotus2019 –lotus2019 2025-03-23 13:42:40 +00:00 Commented Mar 23 at 13:42 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. At least for the given examples, one rule is enough: ```mathematica lst = "Iin" <> ToString@# & /@ Range // Map@ToExpression mod = Mod[#, 2 Pi, -Pi] &; lst /. a_ (trig : Sin \| Cos)[b_ t + c_] :> {Abs[a]/Sqrt, mod[Sign[b] (b t + c) + Arg[a] + π/2 - Arg@trig[I] + Arg[b] /. t -> 0]/Degree} // FullSimplify ( {{4, 30}, {3, -60}, {5, -40}, {8, 30}, {5, 140}} ) ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Mar 24 at 3:01 xzczd♦xzczd 71.6k 9 9 gold badges 182 182 silver badges 520 520 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Mathematica Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. 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1893	https://www.expii.com/t/orders-of-magnitude-definition-examples-4448	Orders of Magnitude — Definition & Examples - Expii Expii Orders of Magnitude — Definition & Examples - Expii The order of magnitude of a number is the power of 10 used when the number is in scientific notation. HomeLog inSign up Search for math and science topics Search for topics Log InSign Up Pre-AlgebraScientific Notation Orders of Magnitude — Definition & Examples The order of magnitude of a number is the power of 10 used when the number is in scientific notation. Orders of Magnitude — Definition & Examples Practice Explanations (2) Alex Federspiel Text 9 The order of magnitude is the power of 10 a number is raised to when it's in scientific notation. For example, the order of magnitude for 19,400 would be 4. This is because when we convert it into scientific notation, 1.94×10 4, the 10 is being raised to the 4 th power. We can also compare orders for two different numbers in scientific notation. Consider for example A=6.7×10 8 B=6.7×10 9 C=6.7×10 10 One can quickly see that B is a bigger number than A, and C is bigger than B. However, we can say more. We can say C is two orders of magnitude larger than A. Also, C is ten times the smaller number B; B is ten times A. Now it's your turn to try a problem yourself. What is the order of magnitude for 6,032,095? 5 6 7 8 Show SolutionCheck ReportShare9 Like Related Lessons Scientific Notation — Definition & Examples Multiplying Scientific Notation — Examples & Practice Cube Root of Negative Numbers — Examples & Practice Fermi Problems — Definition & Examples View All Related Lessons Ben Ferrara Video 2 (Video) Orders of Magnitude by Truckee Truckee has a good video about orders of magnitude. Summary We use orders of magnitude for estimating. If you're looking for the volume of a lake or the area of a field, you can use estimates. Sometimes the numbers will be written in scientific notation. Other times you'll practice writing a number in scientific notation yourself. Regardless, these estimates are useful for small numbers and large numbers alike. The power of 10 could be positive or negative. If you're asked to find the order of magnitude for 2.3×10 4 you have two options. The first is rounding the scientific notation to the ones place. 2.3×10 4≈2×10 4 The other, is to just write out the power of ten. 2.3×10 4≈10 4 The numbers 2×10 4 and 10 4 are certainly different. The first number is 2 times larger than the second. However, the powers of 10 are the same. They are numbers of the same order. In addition, there are a total of four factors of ten. She then looks at an example. Estimate the volume of water in Gilmore Lake in the Desolation Wilderness. The lake is 2 km across and has an average depth of 50 m. We can think of the lake as a cylinder with a height of 50 m and a radius of 1 km. The volume of a cylinder is: V=π r 2 h We can write 1 km as 1×10 3 and 50 m as 5×10 1. Plugging these into the equation, we get: V=π r 2 h V=π(10 3)2(5×10 1)V=π×10 6×5×10 1 V=5 π×10 6+1 V=5 π×10 7 Be careful though. We see that 5 π is approximately 15.7. This is not between 1 and 10. In fact, 5 π is greater than10. V≈1.57×10 8 We can estimate this more as 2×10 8 or just 10 8. These are called examples of approximate comparisons. ReportShare2 Like You've reached the end TOP ABOUT US AboutJobsContact EXPLORE Daily ChallengeSolveModerate Zoom ChatLive Math Courses INITIATIVES SparkRamanujanShowcase INFO Terms of ServiceCode of ConductPrivacy NoticeLabor Condition Applications © 2019 Expii, Inc. How can we improve? close Feedback Type General Bug Feature Message Email (optional) Send Send Feedback
1894	https://www.quora.com/How-do-you-find-the-total-cost-from-the-marginal-cost-only	Something went wrong. Wait a moment and try again. Marginal Average Cost Introduction to Cost Acco... Full Costing Basic Microeconomics Cost Accountants Economics (subject) Total Cost Marginal Opportunity Cost 5 How do you find the total cost from the marginal cost only? · To find the total cost from the marginal cost, you can use the following approach: Understand Marginal Cost : Marginal cost (MC) is the cost of producing one additional unit of a good or service. It can vary with the level of production. Identify the Cost Function : If you have a function for marginal cost, M C ( q ) , where q is the quantity produced, you can find the total cost by integrating the marginal cost function. 3. Integrate the Marginal Cost : T C ( q ) = ∫ M C ( q ) d q + C Here, T C ( q ) is the total cost function, and C is the fixed cost (the cost incurred when production is zero). 4. Determine Fi To find the total cost from the marginal cost, you can use the following approach: Understand Marginal Cost: Marginal cost (MC) is the cost of producing one additional unit of a good or service. It can vary with the level of production. Identify the Cost Function: If you have a function for marginal cost, MC(q), where q is the quantity produced, you can find the total cost by integrating the marginal cost function. Integrate the Marginal Cost: TC(q)=∫MC(q)dq+C Here, TC(q) is the total cost function, and C is the fixed cost (the cost incurred when production is zero). Determine Fixed Costs: If you know the fixed costs, you can plug them into the equation. If not, you may need additional information to determine C. Calculate Total Cost: To calculate the total cost for a specific quantity q, evaluate the integral from 0 to q: TC(q)=∫q0MC(x)dx+C If the marginal cost is constant, say MC=5, then the total cost function becomes: TC(q)=5q+C If C (fixed cost) is $10, the total cost for producing 10 units would be: TC(10)=5(10)+10=50+10=60 To find total cost from marginal cost, integrate the marginal cost function and add any fixed costs. This will give you a total cost function depending on the quantity produced. Related questions How do we find the total cost from marginal cost in a short period? How do you find the marginal cost from a table? How do you find the marginal cost with a given total variable cost and quantity only? What is marginal cost, and how is it calculated? How do you calculate marginal cost from total cost? Tom Longwell A.B. from University of Chicago (Graduated 2003) · Author has 7.3K answers and 6.7M answer views · 8y You can’t. 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I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Jim Franz Studied Economics at San Diego State University (Graduated 1980) · Author has 57 answers and 81.3K answer views · 6y In the short run, you can’t find total cost from marginal cost, because as Tom Longwell posted in 2016 marginal cost doesn’t include fixed cost. Marginal cost will tell you the change in total cost. If you add the marginal cost of each unit produced you get total variable cost. Adding total fixed cost to total variable cost would give you total cost. In the long run, however, there are no fixed costs, so total cost could be derived from marginal cost. Isaac Agyei Nyamekye Works at REPLIB · 7y Marginal cost is the change in total cost as a results of producing an additional unit of output. By definition, mc¡=tc¡¡-tc¡. Where mc¡ is the current marginal cost Tc¡ is the previous Total cost Tc¡¡ is the current Total cost. From here, to obtain the Total cost, you need to sum up the the various Marginal cost. For example, to find the Total cost of production at the 5th Labourer, you have to sum up the various Marginal cost beginning from the 1st Labourer employed up to the 5th Labourer employed. Related questions Is it possible to derive variable cost from marginal cost? How does Marginal cost work in this? How do you calculate the total fixed costs, variable costs, and marginal costs for a company? What is the relationship between the average cost, marginal cost, and total cost? What is difference between total cost and marginal cost? Sonakshi Garg Former Learner on Quora · Author has 58 answers and 140.5K answer views · 7y TC is the submission of MC whereas MC is an addition made to TC by producing one more unit of output. MC=TCn-TCn-1 So we will conclude that TC=sigmaMC Now follow the given schedule Output 1 2 3 4 5 MC 10 8 4 6 8 TC 10 18 22 28 36 Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. 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Then multiplying quantity by marginal cost should give a good approximation of total cost. OR You could simulate fixed cost and take the best or worst case scenario based on your risk appetite. Scott Powell Worked at NTT Data · Author has 2.9K answers and 6M answer views · 3y Related How do I find the marginal cost when the total cost is not given? You don’t need to know the total cost to find the marginal cost. Let’s say your business is a bakery and you want to know the marginal cost of baking cupcakes. Add up the cost of the ingredients in a batch Calculate the cost of running the oven Calculate the labor cost to make the batch That’s roughly your marginal cost. What did we not include? The price of the oven The price of any of the other equipment you used to make the cupcakes: Cupcake tray Mixer Spatula Cooling racks Rent on the space Insurance on the business Cost of the counter staff Cost of cleanup staff Cost of staff that procures the ingredients You don’t need to know the total cost to find the marginal cost. Let’s say your business is a bakery and you want to know the marginal cost of baking cupcakes. Add up the cost of the ingredients in a batch Calculate the cost of running the oven Calculate the labor cost to make the batch That’s roughly your marginal cost. What did we not include? The price of the oven The price of any of the other equipment you used to make the cupcakes: Cupcake tray Mixer Spatula Cooling racks Rent on the space Insurance on the business Cost of the counter staff Cost of cleanup staff Cost of staff that procures the ingredients HVAC for the space Water, electrical, and trash pick up Numbers 1 and 2 are the fixed costs because you bought the gear but you could return/resell it. Numbers 3 and 4 are fixed and sunk costs because you will owe the rent even if you never open for business and unlike the equipment you can’t resell it. Numbers 5–9 you may not have to pay if you never open for business. These are the marginal costs of running the business but not the marginal cost of making cupcakes. Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Divyansh Ghuley Student of history economics and political science · 9y Actually marginal cost is the cost of a single good....if u multiply it with no of goods u will get total cost....but what if no of goods are not given and marginal cost is given as a function....then simply integrate the function nd the equation of total cost curve can be attained....as marginal cost is nothing but the slope of the total cost curve at a point... Leda Ramos Canellada-Celis Full Professor at Usls (1996–present) · 7y From the total cost you can derive your marginal cost by using the formula delta TC/delta Q (quantity). Reverse it to derive your marginal cost with TC in quantity one remaining constant. Sponsored by Betterbuck What are the stupidest things people overspend on in the U.S.? In my experience, the average American overpays for things constantly. Here are the 5 worst culprits. Nicholas Schubert Former Receipt · 8y Depends. If cost plus minus original bid from total cost. If straight bid then total cost (whatever was over budget plus bid) minus bid is your marginal cost. You need to know what the bid was. Ron Auerbach Degrees in: economics, finance, accounting, human resources · Author has 13.4K answers and 20.5M answer views · 3y Related How do I find marginal cost when the total cost is not given? You have to remember that total cost = total fixed cost + total variable cost. So while you might not know total cost, you can still figure out what marginal cost would be if you know your total variable cost. In other words, total fixed cost does not change with your level of output. But total variable cost does! So as you produce an additional unit, T... Sinja Fan B.A. in Medicine and Healthcare & Economics, Johns Hopkins University (Graduated 2020) · Author has 54 answers and 175.5K answer views · 7y Related How is marginal cost calculated? Marginal cost is the change in Total Cost after producing one more good. For example say you produce 4 goods and Total Cost is $450. You know that if you produce 5 goods, your total cost will be $475. That means your marginal cost of the 5th good is $25 since it took that amount of money to produce that one extra good. The brute force way of calculating marginal cost is to find the total cost of producing different numbers of goods and subtracting each total cost by the total cost of the preceding total cost of goods. Another way is through calculus. If you know the total cost function, you know Marginal cost is the change in Total Cost after producing one more good. For example say you produce 4 goods and Total Cost is $450. You know that if you produce 5 goods, your total cost will be $475. That means your marginal cost of the 5th good is $25 since it took that amount of money to produce that one extra good. The brute force way of calculating marginal cost is to find the total cost of producing different numbers of goods and subtracting each total cost by the total cost of the preceding total cost of goods. Another way is through calculus. If you know the total cost function, you know the marginal cost function. Why? Because marginal cost is just the change in cost or price over change in quantity which in this case is always 1. That means you need the instantaneous change of each point on the total cost curve! And how would one do that? Take the derivative of the Total Cost curve! And now you have the marginal cost function and can calculate the marginal cost of any amount of goods. Tom Longwell A.B. from University of Chicago (Graduated 2003) · Author has 7.3K answers and 6.7M answer views · 3y Related How do I find marginal cost when the total cost is not given? Impossible to say without more information. If you know variable cost, simply take the derivative, just like you would with total cost. If you know an average cost (that isn’t average fixed cost), multiply by the quantity to convert into total or variable cost and then take the derivative. Otherwise, I’d need to know exactly what you know. Related questions How do we find the total cost from marginal cost in a short period? How do you find the marginal cost from a table? How do you find the marginal cost with a given total variable cost and quantity only? What is marginal cost, and how is it calculated? How do you calculate marginal cost from total cost? Is it possible to derive variable cost from marginal cost? How does Marginal cost work in this? How do you calculate the total fixed costs, variable costs, and marginal costs for a company? What is the relationship between the average cost, marginal cost, and total cost? What is difference between total cost and marginal cost? How do I find marginal cost when the total cost is not given? How does marginal cost and average total cost relate? What is marginal cost and absorption costing? What are marginal costing techniques? Is absorption costing more relevant in decision making than marginal costing? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1895	https://www.omnicalculator.com/physics/kinematic-viscosity-of-air	Board Last updated: Kinematic Viscosity of Air Calculator In this kinematic viscosity of air calculator, you can easily examine the relationship between the dynamic and kinematic viscosity of air. Viscosity is an important property of fluids and gases, especially if you consider aerodynamic problems. Both viscosity terms are interconnected. For example, you can determine the kinematic viscosity of air if the dynamic viscosity and density are already known. In this calculator, we will learn more about viscosity, with a particular focus on the kinematic and dynamic viscosity of air. Keep reading to learn: What does viscosity mean? How to calculate the kinematic and dynamic viscosity of air? What does the viscosity of air depend on? What is the viscosity of air at 20 degrees Celsius? What is the viscosity of air at sea level? What does viscosity mean? Imagine an object moving in slow motion through the air. Aerodynamic forces are created between the air and the object because the air molecules near the object will become disturbed and move around the object. The magnitude of these forces depends on the object's shape and velocity, the gas's mass, and two other important properties of the air: viscosity and compressibility. When an object moves in a gas, the gas molecules stuck to its surface form moving surfaces between which friction is created. Viscosity tells us how much resistance air has to flow and how resistant it is to gradual deformation under stress. We can distinguish two types of viscosity: dynamic and kinematic viscosity of the air. 💡 We are using the same terms to describe viscosity both in air and liquids because air and all other gasses can be considered fluids. Their molecules are in constant motion, and they flow as in liquids. Dynamic viscosity of air A dynamic viscosity (or absolute viscosity) is generated if we have different surfaces of air molecules moving parallel to the object's surface. This results in a shear force. To overcome the resistance of the fluid, an external force F must be applied per unit of moving surface. The magnitude of F is proportional to the speed v and the area A of the moving surface and inversely proportional to their separation y: F=μAyv The dynamic viscosity μ (Greek letter mu) is the proportionality coefficient in this formula, and we can express it in SI units: μ = kg/m·s = N/m² = Pa·s. In practice, we can use the following empirical formula for engineering purposes: μ=T+110.41.458×10−6×T23 where: 1.458×10−6 — Constant; T — Temperature; and 110.4 — Another empirical constant. Kinematic viscosity of air Kinematic viscosity (marked with the Greek letter nu ν) describes a relationship between dynamic (absolute) viscosity and air density. For comparison, check water viscosity calculator. You can obtain the kinematic viscosity of air by dividing dynamic viscosity by the density ρ: ν=ρμ In the SI system, the unit of kinematic viscosity is m2/s, but other commonly used units are Stokes (St): 1 St = 10-4 m2/s = 1 cm2/s. Since 1 Stoke is a large unit, it is convenient to divide it by 100 to obtain smaller unit centiStokes (cSt): 1 St = 100 cSt; 1 cSt = 10-6 m2/s = 1 mm2/s; and 1 m2/s = 106 centiStokes You can estimate the density of air ρ at a given pressure and temperature directly from the ideal gas law equation: ρ=RTP where: P — Absolute pressure; R — Specific gas constant, which equals 287.05 J/(kg·K); and T — Absolute temperature. How to use kinematic viscosity of air calculator? To calculate the dynamic and kinematic viscosity of air at chosen conditions, you need to follow only a few steps: Determine the pressure P. If you want to convert different units, try our pressure conversion tool. Determine the temperature T. Calculate the air density ρ and dynamic and kinematic viscosity using the formulas above. Congratulations to you for working on our kinematic viscosity of air calculator. FAQs What does the viscosity of air depend on? In general, the viscosity of air is only slightly dependent on the pressure but is strongly related to temperature. The dynamic viscosity directly correlates with the square root of temperature μ ∝ √T, and density is inversely proportional to temperature: ρ ∝ 1/T. Therefore, we can relate these two proportionalities and derive the relationship between kinematic viscosity and temperature as ν = T3/2. As you can see now, when the temperature rises, the viscosity of the air increases. This distinguishes air and most gases from the viscosity of liquids. For liquids, viscosity generally decreases as temperature increases. What is the viscosity of air at 20 degrees Celsius? The kinematic viscosity of air at 20 degrees Celsius and 1 atm (101,325 Pa, sea level) pressure is approximately 1.51×10-5 m2/s or 15.20 cSt. The dynamic viscosity of air at sea level under the same temperature is 1.81×10-5 Pa·s. Here are the values of the kinematic and dynamic viscosity of air at a constant pressure of 1 atm and different temperatures: \| Temperature (°C) \| Density (kg/m3) \| Dynamic viscosity (Pa·s) \| Kinematic viscosity (m2/s) \| --- --- \| \| -10 \| 1.341 \| 1.666×10-5 \| 1.242×10-5 \| \| 0 \| 1.292 \| 1.716×10-5 \| 1.328×10-5 \| \| 10 \| 1.246 \| 1.765×10-5 \| 1.456×10-5 \| \| 20 \| 1.204 \| 1.813×10-5 \| 1.506×10-5 \| \| 30 \| 1.164 \| 1.861×10-5 \| 1.598×10-5 \| \| 40 \| 1.127 \| 1.908×10-5 \| 1.692×10-5 \| \| 50 \| 1.092 \| 1.954×10-5 \| 1.788×10-5 \| How do I calculate the kinematic viscosity at 1 bar and 50 °C? To calculate the kinematic viscosity of air at pressure 1 bar (105 Pa) and 50 °C, enter a few details into the kinematic viscosity of air calculator: Convert the temperature in Celsius to kelvin: T = 273.15 K + 50 = 323.15 K. Calculate the density of the air from the ideal gas law: ρ = P/(RT) = 105 Pa/(287.05 J/kg·K × 323.15 K) = 1.078 kg/m3 3. Calculate the dynamic viscosity at given conditions: μ = (1.458×10-6 × 323.15 K3/2)/(323.15 K + 110.4) = 1.95×10-5 Pa·s. 4. Calculate the kinematic viscosity: ν = 1.95×10-5Pa·s / 1.078 kg/m3 = 1.81×10-5 m2/s. How can I convert kinematic viscosity to dynamic viscosity? You can easily convert between the dynamic and kinematic viscosity of air as both quantities are related by the equation ν = μ/ρ. For example, if the density of air ρ is equal to 1.113 kg/m3 and ν = 1.697×10-5 m2/s, you can modify the equation above to obtain μ = ν×ρ = 1.697×10-5 m2/s × 1.113 kg/m3 = 1.889×10-5 Pa·s. Pa K J/(kg·K) kg/m³ Pa⋅s m²/s Did we solve your problem today? 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1896	https://zhuanlan.zhihu.com/p/367547952	组合数学--计数组合 (2) - 知乎关注推荐热榜专栏圈子 New付费咨询知学堂直答切换模式登录/注册组合数学--计数组合 (2) 首发于组合数学基础切换模式组合数学--计数组合 (2) Dave Zhou 做自己 2021 科学自立季收录于 · 组合数学基础 35 人赞同了该文章容斥原理及其应用主要概念：（1）错位排列：集合 {1,2,...,n}\left{ 1,2,...,n \right} 的一个排列 i 1 i 2...i n i_{1}i_{2}...i_{n} 若满足 i 1≠1,i 2≠2,...,i n≠n i_{1} \neq 1, i_{2} \neq 2, ... , i_{n} \neq n ，则称为错位排列（2）二重错排：设 a 1 a 2...a n a_{1}a_{2}...a_{n} 是 S={1,2,...,n}S=\left{ 1,2,...,n \right} 的一个错排， x 1 x 2...x n x_{1}x_{2}...x_{n} 是S的一个排列，如果 x i≠i,x i≠a i(1≤i≤n)x_{i} \neq i, \, x_{i} \neq a_{i} \, (1 \leq i \leq n) ，则称 [a 1 a 2...a n x 1 x 2...x n]\begin{bmatrix} a_{1} & a_{2} & ... & a_{n} \ x_{1} & x_{2} & ... & x_{n} \end{bmatrix} 是由错排 a 1 a 2...a n a_{1}a_{2}...a_{n} 限制的一个二重错排。取 a 1 a 2...a n=23...n 1 a_{1}a_{2}...a_{n}=23...n1 ，就是通常的二重错排即 x i≠i,x i≠(i+1)m o d n x_{i} \neq i, \, x_{i} \neq (i+1) \, mod \, n （3）置换：给定集合 S={1,2,...,n}S=\left{ 1,2,...,n \right} ，S到S的一个双射 σ:S→S\sigma: S \to S 称为S上的置换，其中 σ(i)\sigma(i) 是元素 i 在映射 σ\sigma 下的像。由双射可知 σ(1),σ(2),...,σ(n)\sigma(1), \sigma(2), ..., \sigma(n) 是S的一个排列。满足 σ(i)=i\sigma(i)=i 的数 i 称为 σ\sigma 的不动点（4）轮换：设 σ:S→S\sigma: S \to S 是 S={1,2,...,n}S=\left{ 1,2,...,n \right} 的一个置换，且 σ(i 1)=i 2,σ(i 2)=i 3,...,σ(i k−1)=i k,σ(i k)=i 1,\sigma(i_{1})=i_{2}, \sigma(i_{2})=i_{3}, ... , \sigma(i_{k-1})=i_{k}, \sigma(i_{k})=i_{1}, 对其余的元素 x≠i j(1≤j≤k)x \neq i_{j} \, (1 \le j \le k) 则不变 σ(x)=x\sigma(x)=x ，则称 σ\sigma 是S上的一个 k-轮换，或称为 k 阶轮换，轮换也称为循环，记作 (i 1,i 2,...,i k)(i_{1}, i_{2}, ... , i_{k}) 。不相交轮换：给定S的两个轮换 σ=(i 1,i 2,...,i k),τ=(j 1,j 2,...,j s)\sigma=(i_{1},i_{2},...,i_{k}), \, \tau=(j_{1}, j_{2}, ..., j_{s}) ，若 σ∩τ=∅\sigma \cap \tau = \varnothing ，则称它们是不相交的轮换。S中的任意一个置换均可表示成一组互不相交的轮换的乘积（5）置换的型：设 S={1,2,...,n}S=\left{ 1,2,...,n \right} 的一个置换为 σ:σ(i)=a i\sigma: \sigma(i)=a_{i} ，这样 a 1 a 2...a n a_{1}a_{2}...a_{n} 是S的一个排列，若 σ\sigma 的不相交轮换分解是由 λ 1\lambda_{1} 个1-轮换、 λ 2\lambda_{2} 个2-轮换、...、 λ n\lambda_{n} 个n-轮换的乘积组成，则称 1 λ 1 2 λ 2...n λ n 1^{\lambda_{1}}2^{\lambda_{2}} ... n^{\lambda_{n}} 为置换 σ\sigma 的型，也称为排列的 a 1 a 2...a n a_{1}a_{2}...a_{n} 的型，这里 λ 1+2 λ 2+...+n λ n=n\lambda_{1} + 2 \lambda_{2} + ... + n \lambda_{n}=n （6）带有限制位置的棋盘：对集合 {1,2,...,n}\left{ 1,2, ... , n \right} 的元素进行带限制位置的排列时，可用 n×n n \times n 方阵来表示，元素 i 不能出现在 j 位置上，则在 (i, j) 块位置涂上阴影。那些带阴影的块的全体称为棋盘，记作棋盘B，它表明了排列的所有限制位置。放 k 个元素的方法数：记 r k(B)r_{k}(B) 为从棋盘B中选出 k 个不同行、不同列的阴影块的选取方法数（即在不同行不同列放 k 个元素的方法数），并规定 r 0(B)=1 r_{0}(B)=1 。不相交子棋盘：把一个棋盘B的行和列做适当的调换，可以分解成 r 个互不相交的子棋盘 B 1,B 2,...,B r B_{1},B_{2},...,B_{r} ，即它们的行集合非交、列集合也非交。棋子多项式：定义数列 {r 0(B),r 1(B),...}\left{ r_{0}(B), r_{1}(B), ... \right} 的生成函数 R(x,B)=∑i=0∞r i(B)x i R(x,B)= \sum_{i=0}^{\infty}r_{i}(B)x^{i} 为棋盘B的棋子多项式， r i(B)r_{i}(B) 也称为棋子多项式的系数（7）偏序集上的函数卷积运算：设 (X,≤)(X, \leq) 是有限偏序集，X上所有满足只要 x⪇y x \nleqslant y 就有 f(x, y)=0 （即只有在 x≤y x \leq y 时 f(x, y) 可能不等于零）的所有二元实值函数 f:X×X→R f: X \times X \to R 的集合，记作 F(X) ，定义 F(X) 中的两个函数 f 和 g 的卷积 h=f∗g h=f \ast g 为 h(x,y)={∑x≤z≤y f(x,z)g(z,y),x≤y 0,o t h e r w i s e h(x,y) = \left{\begin{matrix} \sum_{x \leq z \leq y} f(x,z)g(z,y), & x \leq y \ 0, & otherwise \end{matrix}\right. 卷积满足结合律： f \ast (g \ast h) = (f \ast g) \ast h \,\,\,\, (f, g, h \in F(X)) （8）偏序集上的特殊函数：设 (X, \leq) 是有限偏序集， F(X) 是 X上所有满足只要 x \nleqslant y 就有 f(x, y)=0 （即只有在 x \leq y 时 f(x, y) 可能不等于零）的所有二元实值函数 f: X \times X \to R 的集合，F(X) 中的一些特殊函数： Kronecker delta函数： \delta(x,y) = \left{\begin{matrix} 1, & x = y \ 0, & otherwise \end{matrix}\right. ，有 \delta \ast f = f \ast \delta = f ，因此它是一个恒等函数； zeta函数： \zeta(x,y) = \left{\begin{matrix} 1, & x \leq y \ 0, & otherwise \end{matrix}\right. ，zeta函数是偏序集的一种表示（9）卷积的逆函数：设 (X, \leq) 是有限偏序集，若函数 f \in F(X) 对所有X中的 y 恒有 f(y,y) \neq 0 ，则存在函数 g \in F(X) 满足 g \ast f = f \ast g = \delta ，g 称为 f 的逆函数，g 的表达式为 g(x,y) = \left{\begin{matrix} \frac{1}{f(y,y)}, & x = y \ -\frac{1}{f(y,y)} \sum_{x \leq z \leq y} g(x,z) f(z, y), & x<y \end{matrix}\right. （10）偏序集上的Möbius函数：设 (X, \leq) 是有限偏序集，则zeta函数 \zeta(x, y) 关于卷积的逆函数就是莫比乌斯函数\mu(x, y) ，它的表达式为 \mu(x,y) = \left{\begin{matrix} 1, & x = y \ - \sum_{x \leq z \leq y} \mu(x,z), & x<y \end{matrix}\right. 例1：n 元集合 X_{n}=\left{ 1,2,...,n \right} 的幂集 (P(X_{n}), \subseteq) ，它的莫比乌斯函数为 \mu(A, B) = (-1)^{\left\| B \right\| - \left\| A \right\|} \,\, \, (A \subseteq B \subseteq X_{n}) ；例2：n 元集合 X_{n}=\left{ 1,2,...,n \right} 作为线性有序集 1<2< ... <n ，它的莫比乌斯函数为 \mu(k,l) = \left{\begin{matrix} 1, & k = l \ -1, & l = k+1 \ 0, & otherwise \end{matrix}\right. （11）偏序集的直积：设 (X, \leq_{1}), (Y, \leq_{2}) 是两个偏序集，在集合 X \times Y = \left{ (x,y): \, x \in X, y \in Y \right} 上定义关系为 (x,y) \leq ({x}', {y}') 当且仅当 x \leq_{1} {x}' 且 y \le_{2} {y}' ，则 (X \times Y, \le) 也是一个偏序集，称为 (X, \leq_{1}), (Y, \leq_{2}) 的直积。显然直积的定义可以扩展到任意个偏序集上（12）数论中的Möbius函数：对 n 元集合 X_{n}=\left{ 1,2,...,n \right} 的整除关系构成的偏序集 D_{n}=\left{ X_{n}, \| \right} ，即 a \, \| \, b 当且仅当 a 是 b 的因子，n 的唯一素数因子分解为 n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{k}^{\alpha_{k}} ，对 a \, \| \, b 有 \mu(a, b) = \mu(1, \frac{b}{a}) ，因此只要考虑 \mu(1, n) 即可，有 \mu(1,n) = \left{\begin{matrix} 1, & n=1 \ (-1)^k, & n=p_{1}p_{2}...p_{k} \ 0, & otherwise \end{matrix}\right. 这就是数论中的莫比乌斯函数，简记作 \mu(n) 。这里 p_{1}, p_{2}, ... , p_{k} 表示不同的素数，也就是当 n 有素数的幂作为因子时 \mu(n)=0 。它是积性函数，即 m,n 互素时，有 \mu(m)\mu(n)=\mu(mn) 。有恒等式： \sum_{d \, \| \, n} \mu(d)=0 \,\, (n \geq 2) 主要定理：（1）容斥原理：设S是有限集合， P_{1},P_{2},...,P_{m} 是与集合S有关的 m 个性质， A_{i} 是S中具有性质 P_{i} 的元素构成的集合， \overline{A_{i}} 是S中不具有性质 P_{i} 的则元素构成的集合，则S中至少具有一个性质 P_{i} 的元素个数为 \left\| A_{1} \cup A_{2} \cup ... \cup A_{m} \right\| = \sum_{1 \leq i \leq m} \left\| A_{i} \right\| - \sum_{1 \leq i<j \leq m} \left\| A_{i} \cap A_{j} \right\| + \sum_{1 \leq i<j<k \leq m} \left\| A_{i} \cap A_{j} \cap A_{k} \right\| - ... + (-1)^{m-1} \left\| A_{1} \cap A_{2} \cap ... \cap A_{m} \right\| 公式右边共有 C_m^1 + C_m^2 + ... + C_m^m = 2^m -1 项。也可以写成以下形式 \left\| \bigcup_{i=1}^{m} A_{i} \right\| = \sum_{k=1}^{m} (-1)^{k+1} \left( \sum_{1 \leq i_{1} < ... < i_{k} \leq m} \left\| A_{i_{1}} \cap A_{i_{2}} \cap ... \cap A_{i_{k}} \right\| \right) 另外由DeMorgan律\left\| \overline{A_{1}} \cap \overline{A_{2}} \cap ... \cap \overline{A_{m}} \right\| = \left\| S \right\| - \left\| A_{1} \cup A_{2} \cup ... \cup A_{m} \right\| ，可得S中不具有性质 P_{1},P_{2},...,P_{m} 的元素个数，称为筛法公式： \left\| \overline{A_{1}} \cap \overline{A_{2}} \cap ... \cap \overline{A_{m}} \right\| = \left\| S \right\| - \sum_{k=1}^{m} (-1)^{k+1} \left( \sum_{1 \leq i_{1} < ... < i_{k} \leq m} \left\| A_{i_{1}} \cap A_{i_{2}} \cap ... \cap A_{i_{k}} \right\| \right) 容斥原理也可以简洁地写成以下形式： \left\| \bigcup_{i=1}^{m} A_{i} \right\| = \sum_{\varnothing \neq J \subseteq \left{ 1,2,...,m \right}} (-1) ^{\left\| J \right\| + 1} \left\| \bigcap_{j \in J} A_{j} \right\| ； \left\| \bigcap_{i=1}^{m} \overline{A_{i}} \right\| =\left\| S \right\| - \sum_{\varnothing \ne J \subseteq \left{ 1,2,...,m \right}} (-1) ^{\left\| J \right\| + 1} \left\| \bigcap_{j \in J} A_{j} \right\| （2）容斥原理的推广：设S是有限集合， P=\left{ P_{1},P_{2},...,P_{m} \right} 是与集合S有关的 m 个性质，N(r) 表示S中恰好有P中 r 个性质的元素个数， N(P_{i_{1}}, P_{i_{2}}, ..., P_{i_{k}}) 表示S中至少有 k 个性质，即具有性质 P_{i_{1}},P_{i_{2}},...,P_{i_{k}} 的元素个数，设 w(k)=\sum_{1 \le i_{1} < ... < i_{k} \le m} N(P_{i_{1}}, P_{i_{2}}, ..., P_{i_{k}}) 规定 w(0)= \left\| S \right\| ，则有 N(r) = w(0) - \sum_{k=1}^{m-r} (-1)^{k+1} C_{r+k}^{r} w(r+k) （3）错位排列公式： n 元集合 X=\left{ 1,2,...,n \right} 的错位排列数目为 D_{n}=n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + ... + (-1)^{n} \frac{1}{n!} \right) = \left[ \frac{n!}{e} \right] 并且 D_{n} 是偶数当且仅当 n 是奇数。错排数也是X上没有任何不动点的置换 \varphi 的个数。证明思路：排列 i_{1}i_{2}...i_{n} 若满足 i_{j}=j 则称具有性质 P_{j} ，一个排列是错位排列当且仅当它不具有性质 P_{1}, ... , P_{n} 中的每一条性质，设 A_{j} 表示具有性质 P_{j} 的排列的集合，则错位排列的集合为 \overline{A_{1}} \cap \overline{A_{2}} \cap ... \cap \overline{A_{n}} ，因此 D_{n} = \left\| \overline{A_{1}} \cap \overline{A_{2}} \cap ... \cap \overline{A_{n}} \right\| ，再运用容斥原理。右边括号里的部分正是 e^{-1}=1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + ... 的部分和，因此有 \frac{1}{e} - \frac{D_{n}}{n!}=(-1)^{n+1} \frac{1}{(n+1)!} + (-1)^{n+2} \frac{1}{(n+2)!} + ... 由无穷交错级数的性质，可知 \left\| \frac{1}{e} - \frac{D_{n}}{n!} \right\| < \frac{1}{(n+1)!} ，得到 \left\| \frac{n!}{e} - D_{n} \right\| < \frac{1}{n+1} \leq 0.5 \,\,\, (n \ge 2) ， D_{n} 是最接近 \frac{n!}{e} 的整数，即 D_{n}=\left[ \frac{n!}{e} \right] 。错位排列的递推公式： D_{n}=nD_{n-1} + (-1)^n \,\,\, (n \geq 2) （4）图的顶点次数：设图G的顶点集X含有 n 个顶点，且不含有完全 k(k \ge 2) 子图，则它的顶点的次数 d(x) 满足不等式 \min_{x \in X} d(x) \le \left \lfloor \frac{(k-2)n}{k-1} \right \rfloor （5）带有禁止模式的排列：设 Q_{n} 是 S=\left{ 1,2, ... , n \right} 的排列中没有 12, 23, ..., (n-1)n 这些模式的排列数目，则有 Q_{n}= n! - \sum_{k=1}^{n-1} (-1)^{k+1} C_{n-1}^{k} (n-k)! 。有递推关系： Q_{n}=D_{n} + D_{n-1} \,\, (n \geq 2) 。证明思路：设 P_{i} 表示排列中出现 i(i+1) 模式， A_{i} 表示S中满足性质 P_{i} (1 \le i \le n-1) 的排列构成的集合，通过不断归纳求出 \left\| A_{i} \right\|=(n-1)! ，... ， \left\| A_{i_{1}} \cap A_{i_{2}} \cap ... \cap A_{i_{k}} \right\| =(n-k)! ，再运用容斥原理 Q_{n}= \left\| \overline{A_{1}} \cap \overline{A_{2}} \cap ... \cap \overline{A_{n}} \right\| （6）二重错排（ménage问题）：n 对夫妻参加宴会围着圆桌就座，要求男女相间且每对夫妻不相邻，有多少种不同的就座方案？这是一个二重错排的问题，记作 U_{n} 。它等于二重错排 \begin{bmatrix} 2 & 3 & ... & n & 1 \ x_{1} & x_{2} & ... & x_{n-1} & x_{n} \end{bmatrix} 的个数，公式为 U_{n}=w(0) - \sum_{k=1}^{n}(-1)^{k+1} w(k) = n! - \sum_{k=1}^{n}(-1)^{k+1} \frac{2n}{2n-k} C_{2n-k}^{k} (n-k)! 证明思路：不失一般性，我们先排女宾，方案数为 (n-1)! ，对于任意这样的方案数，顺时针给每个女宾以编号1, 2, ..., n 。设第 i 号与第 i+1 号女宾之间的位置为第i号位置（ 1 \le i \le n-1 ），第 n 号与第 1 号女宾之间的位置为第 n 号位置，设第 i 号女宾的丈夫的编号也在第 i 号（ 1 \le i \le n ），让 n 个男宾做到上述编号的第 n 个位置上。设 a_{i} 是坐在第 i 号位置上的男宾，则 a_{i} \neq i, \, a_{i} \neq (i+1) \, mod \, n \,\, (1 \le i \le n) ，因此 a_{1}a_{2}...a_{n} 是一个二重错排，该问题是求二重错排的个数。二重错排问题的详解和推广可参考：二重错排与拉丁矩的计数（7）对于带有限制位置的排列，如果棋盘B能分解成不相交的子棋盘 B_{1} , B_{2} ，则有 r_{k}(B)=\sum_{i=0}^{k}r_{i}(B_{1})r_{k-i}(B_{2}) 。若 B_{i} \, (1 \le i \le r) 为棋盘B的所有不相交子棋盘，则棋子多项式满足关系 R(x,B) = \prod_{i=1}^{r}R(x,B) （8）带有限制位置的排列：对 n 元集合 \left{ 1,2, ... , n \right} 进行带有限制位置的排列，设B是相应的棋盘， r_{k}(B) 为在棋盘B的不同行不同列放 k 个元素的方法数，则安排集合中的 n 个元素的方法数为 n! - \sum_{k=1}^{n} (-1)^{k+1} r_{k}(B)(n-k)! （9）Cauchy公式：给出了同类型的置换（或对应排列）的个数公式。在集合 S=\left{ 1,2,...,n \right} 的置换群 S_n 中，型为 1^{\lambda_1}2^{\lambda_{2}} 3^{\lambda_{3}} ... n^{\lambda_{n}} 的置换（或对应的S的排列）的个数为 \frac{n!}{\lambda_{1}! \lambda_{2}! \lambda_{3}! ... \lambda_{n}! 1^{\lambda_1} 2^{\lambda_{2}} 3^{\lambda_{3}} ... n^{\lambda_{n}}} （10）二重错排与型的关系：若 a_{1}a_{2}...a_{n} 和 b_{1}b_{2}...b_{n} 是集合 S=\left{ 1,2,...,n \right} 的两个错排，且有相同的型，则二重错排 \begin{bmatrix} a_{1} & a_{2} & ... & a_{n} \ x_{1} & x_{2} & ... & x_{n} \end{bmatrix} 与 \begin{bmatrix} b_{1} & b_{2} & ... & b_{n} \ x_{1} & x_{2} & ... & x_{n} \end{bmatrix} 的个数相同（11）Möbius反演公式：设 (X, \leq) 是偏序集且有最小元 0 ，它的莫比乌斯函数 \mu(x, y) : X \times X \to R 为 \mu(x,y) = \left{\begin{matrix} 1, & x = y \ - \sum_{x \leq z \leq y} \mu(x,z), & x<y \end{matrix}\right. F: X \to R 是X上的实值函数，函数 G: X \to R 定义为 G(x) = \sum_{z \leq x} F(z) \,\,\, (x \in X) ，则有反演公式 F(x) = \sum_{y \leq x} G(y) \mu(y,x) \,\,\, (x \in X) （12）一般容斥原理：设 n 元集合 S=\left{ 1,2,...,n \right} 的幂集关于包含关系的偏序集为 (P(S), \subseteq) ， A_{1},A_{2},...,A_{n} 是S的子集，函数 F: P(S) \to R 定义为对子集 K \subseteq S ，F(K) 为正好属于所有满足 i \notin K 的集合 A_{i} 的元素的个数，函数 G: P(S) \to R 定义为 G(K) = \sum_{L \subseteq K} F(L) \,\,\, (K \subseteq S) ，即有 G(K)=\left\| \bigcap_{i \notin K} A_{i} \right\| ，那么有 F(K)=\sum_{L \subseteq K} (-1)^{\left\| K \right\| - \left\| L \right\|} G(L) \,\,\, (K \subseteq S) 取 K=S ，用 L 的补集 J 代替 L ，则得到 \left\| \bigcap_{i=1}^{n} \overline{A_{i}} \right\| = \sum_{J \subseteq S} (-1) ^{\left\| J \right\|} \left\| \bigcap_{j \in J} A_{j} \right\| （13）直积的莫比乌斯函数：两个偏序集的直积的莫比乌斯函数，是它们的莫比乌斯函数的乘积。该结论对有限个有限偏序集也成立（14）经典的Möbius反演公式：设数论中的莫比乌斯函数为 \mu(n) = \left{\begin{matrix} 1, & n=1 \ (-1)^k, & n=p_{1}p_{2}...p_{k} \ 0, & otherwise \end{matrix}\right. f(n) 和 F(n) 是数论函数（即定义在正整数集上，值域为复数的函数），并且满足 F(n) = \sum_{d \, \| \, n}f(d) ，则有反演公式 f(n)=\sum_{d \, \| \, n} \mu(d) F(\frac{n}{d}) 。一般形式：设 F(x) 和 G(x) 是定义在 [1, +\infty) 上的复值函数，并且 G(x)=\sum_{1 \le n \le x}F(\frac{x}{n}) ，则有反演公式 F(x)=\sum_{1 \le n \le x} \mu(n) G(\frac{x}{n}) （15）欧拉函数：\varphi(n) 是欧拉函数，表示不超过 n 且与 n 互素的正整数个数，即 \varphi(n)=\left\| S_{n} \right\|, \,\,\, S_{n}=\left{ k: \, 1 \le k \le n, \, gcd(k,n)=1 \right} ，根据 n=\sum_{d \, \| \, n} \varphi(d) 作反演，可得 \varphi(n)=\sum_{d \, \| \, n}\mu(d) \frac{n}{d} = n \cdot \prod_{p \, \| \, n} \left(1- \frac{1}{p} \right) 其中乘积是对所有整除 n 的互不相同素数 p 求积。也可以写成 \varphi(n)=n \cdot \prod_{i=1}^{m}\left( 1 - \frac{1}{p_i} \right) ，其中 p_1, p_2,...,p_m 是 n 的所有质因数（16）多重集合的循环排列：k 个不同元素的 n 元循环排列（每个元素可以使用任意多次），也就是多重集合 \left{ n \cdot a_{1}, n \cdot a_{2}, ..., n \cdot a_{k} \right} 的 n 元循环排列，它的数目为 h(n, k)=\frac{1}{n} \sum_{d \, \| \, n}\varphi(\frac{n}{d}) k^{d} ，其中 \varphi(n) 为欧拉函数生成函数和递推关系主要概念：（1）形式幂级数：对实数域R上的数列 \left{ a_{0}, a_{1}, a_{2}, ... \right} ，x 是R上的未定元，表达式 A(x)=a_{0}+a_{1}x+a_{2}x^{2}+... 称为R上的形式幂级数。x 只是一个抽象符号，一般不需要对 x 赋值，因此不需要考虑收敛性。R 上的形式幂级数全体记作 R 。生成函数（母函数）：A(x) 也称为序列 \left{ a_{0}, a_{1}, a_{2}, ... \right} 的生成函数，记作 G\left{ a_{n} \right} （2）形式幂级数的导数：在整环 R 上可以定义形式导数，对任意 A(x)=\sum_{k=0}^{\infty}a_{k}x^{k} ，定义 DA(x)=\sum_{k=0}^{\infty}k a_{k}x^{k-1} ，称 DA(x) 为 A(x) 的形式导数，也记作 {A}'(x) 。n 次形式导数可递归地定义为 D^{0} A(x) \equiv A(x) ， D^{n} A(x) \equiv D\left[ D^{n-1} A(x) \right] \, (n \ge 1) 。形式导数像普通导数一样满足线性性质、Leibniz法则、链式求导法则（3）指数型生成函数：数列 \left{ a_{0}, a_{1}, a_{2}, ... \right} 的指数型生成函数定义为形式幂级数 \sum_{k=0}^{\infty} a_{k} \frac{x^{k}}{k!} （4）线性递推关系：设 k 是正整数，若数列 f(0), f(1), ... , f(n), ... 的相邻 k+1 项间满足关系 f(n)=c_{1}(n)f(n-1) + c_{2}(n)f(n-2) + ... + c_{k}(n)f(n-k) + g(n) 对 n \ge k 成立，其中 c_{k}(n) \ne 0 ，则称该关系为 \left{ f(n) \right} 的 k 阶线性递推关系。如果 c_{1}(n), c_{2}(n), ... , c_{k}(n) 都是常数，则称之为 k 阶常系数线性递推关系。如果 g(n)=0 ，则称之为齐次的。常系数线性齐次递推关系：f(n)=c_{1}f(n-1) + c_{2}f(n-2) + ... + c_{k}f(n-k) \,\, (n \ge k, c_{k} \ne 0) ，方程 x^{k}-c_{1}x^{k-1}-c_{2}x^{k-2}- ... - c_{k}=0 称为该递推关系的特征方程，它的 k 个复数根 q_{1}, q_{2},...,q_{k} （可能有重根）称为该递推关系的特征根（5）Dirichlet生成函数：数列 a_n 的Dirichlet生成函数为 f(s)=\sum_{n=1}^\infty \frac{a_n}{n^s} ，为区别于其它生成函数，Dirichlet生成函数的变量用 s 表示。对于 a_n=1 ，其Dirichlet生成函数 f(s)=\zeta (s)=\sum_{n=1}^\infty \frac{1}{n^s} 是黎曼Zeta函数主要定理：（1）形式幂级数整环：全体形式幂级数的集合为 R ，对任意两个形式幂级数 f(x)=\sum_{n=0}^\infty a_nx^n ， g(x)=\sum_{n=0}^\infty b_nx^n ，以自然的多项式方式定义加法、乘法 f(x)g(x)=\sum_{n=0}^n (\sum_{k=0}^na_kb_{n-k})x^n ，则 R 构成一个整环，即无零因子的交换幺环。加法零元是数列 \left{ 0, 0, 0, ... \right} 的形式幂级数 0 ，乘法单位元是数列 \left{ 1, 0, 0, ... \right} 的形式幂级数 1 （2）形式幂级数的乘法逆元：对整环 R 中的任意一个元素 A(x)=\sum_{k=0}^{\infty}a_{k}x^{k} 有乘法逆元当且仅当 a_{0} \neq 0 。若 \widetilde{A}(x) = \sum_{k=0}^{\infty} \widetilde{a}_{k} x^{k} 是 A(x) 的乘法逆元，则有 \widetilde{a}{0}=a{0}^{-1} ； \widetilde{a}{k}=(-1)^{k} a{0}^{-(k+1)} \begin{vmatrix} a_{1} & a_{2} & a_{3} & ... & a_{k-1} & a_{k} \ a_{0} & a_{1} & a_{2} & ... & a_{k-2} & a_{k-1} \ 0 & a_{0} & a_{1} & ... & a_{k-3} & a_{k-2} \ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \ 0 & 0 & 0 & ... & a_{1} & a_{2} \ 0 & 0 & 0 & ... & a_{0} & a_{1} \end{vmatrix} \,\,\, (k \ge 1) （3）生成函数的性质：设有两个数列的生成函数为 A(x)=\sum_{k=0}^{\infty}a_{k}x^{k} 和 B(x)=\sum_{k=0}^{\infty}b_{k}x^{k} 向前平移序列：若 b_{k} = \left{\begin{matrix} 0, & k<l \ a_{k-l}, & k \ge l \end{matrix}\right. ，则有 B(x) = x^{l} \cdot A(x) ；向后平移序列：若 b_{k}=a_{k+l} ，则有 B(x) = \frac{1}{x^{l}}\left[ A(x) - \sum_{k=0}^{l-1} a_{k}x^{k} \right] ；部分和序列：若 b_{k}=\sum_{i=0}^{k}a_{i} ，则有 B(x)=\frac{A(x)}{1-x} ；无穷和序列：若 b_{k}=\sum_{i=k}^{\infty}a_{i} 收敛，则有 B(x)=\frac{A(1)-xA(x)}{1-x} ；形式导数序列：若 b_{k}=ka_{k} ，则有 B(x)=x {A}'(x) ；积分序列：若 b_{k}=\frac{a_{k}}{k+1} ，则有 B(x)=\frac{1}{x} \int_{0}^{x} A(t)dt ；线性组合序列：若 c_{k}=\alpha a_{k} + \beta b_{k} ，则有 C(x) \equiv \sum_{k=0}^{\infty}c_{k}x^{k}=\alpha A(x) + \beta B(x) ；乘积序列：若 c_{k}=\sum_{i=0}^{k}a_{i}b_{k-i} ，则有 C(x) \equiv \sum_{k=0}^{\infty}c_{k}x^{k}=A(x) \cdot B(x) （4）指数型生成函数的性质：设有两个数列的指数型生成函数 f_e(x)=\sum_{n=0}^\infty a_n \frac{x^n}{n!} ， g_e(x)=\sum_{n=0}^\infty b_n \frac{x^n}{n!} 形式导数序列：xf^{'}e(x)=\sum{n=1}^\infty na_n \frac{x^n}{n!} ，因此若 b_n=na_n ，则 g_e(x)=x f_{e}^{'}(x) ；乘积序列：f_e(x)g_e(x)=\sum_{n=0}^\infty \left( \sum_{k=0}^n a_k \frac{x^k}{k!} b_{n-k} \frac{x^{n-k}}{(n-k)!} \right)=\sum_{n=0}^\infty \left( \sum_{k=0}^n \binom{n}{k} a_kb_{n-k} \right) \frac{x^n}{n!} ，因此 f_e(x)g_e(x) 为数列 c_n=\sum_{k=0}^n \binom{n}{k} a_kb_{n-k} 的指数型生成函数。这是指数型生成函数的乘积形式（5）排列逆序数的生成函数：集合 \left{ 1,2, ..., n \right} 的排列 \pi=i_{1}i_{2}...i_{n} 中的逆序是指满足 ki_{l} 的数对 (i_{k}, i_{l}) ，排列 \pi 中的逆序的数目记作 inv(\pi) ，易知 0 \le inv(\pi) \le n(n-1)/2 。设 h(n, t) 表示 \left{ 1,2, ..., n \right} 的排列中有 t 个逆序的排列的数目，于是 0 \le t \le n(n-1)/2 时有 h(n, t) \ge 1 ， t > n(n-1)/2 时有 h(n, t)=0 。它的生成函数 g(x)=\sum_{t=0}^{n(n-1)/2} h(n, t) x^t 为 g(x)=1(1+x)(1+x+x^2)(1+x+x^2+x^3)...(1+x+x^2+...+x^{n-1}) = \frac{\prod_{j=1}^{n}(1-x^j)}{(1-x)^n} （6）组合数的生成函数：设从 n 元集合 S=\left{ a_{1},a_{2},...,a_{n} \right} 中取 k 个元素的组合数为 b_{k} ，若限定元素 a_{i} 出现的次数集合为 M_{i} \, (1 \le i \le n) ，则该组合数序列的生成函数为 \sum_{k=1}^{\infty} b_k x^k =\prod_{i=1}^{n} \left( \sum_{m \in M_{i}}x^{m} \right) ，它的幂级数展开式中 x^{k} 的系数就是所求的组合数 b_{k} 。组合分配问题的生成函数：把 k 个相同的球放入 n 个不同的盒子 a_{1},a_{2},...,a_{n} ，限定盒子 a_{i} 的容量集合为 M_{i} \, (1 \le i \le n) ，则其分配方案数 b_{k} 的生成函数为 \sum_{k=1}^{\infty} b_k x^k =\prod_{i=1}^{n} \left( \sum_{m \in M_{i}}x^{m} \right) ，它的幂级数展开式中 x^{k} 的系数就是所求的分配方案数 b_{k} （7）排列数的生成函数：n 元集合的 k-排列数目 P(n, k) ，指数型生成函数为 \sum_{k=0}^{\infty} P(n,k) \frac{x^k}{k!} =\sum_{k=0}^{n} \frac{n!}{(n-k)!} \frac{x^k}{k!}=(1+x)^n （8）多重排列数的生成函数：多重集合 M=\left{ \infty \cdot a_{1}, \infty \cdot a_{2}, ... , \infty \cdot a_{n} \right} 的 k-排列中，若限定元素 a_{i} 出现的次数集合为 M_{i} \, (1 \le i \le n) ，则k-排列数序列 \left{ b_{k} \right} 的指数型生成函数为 \sum_{k=1}^{\infty} b_k \frac{x^k}{k!} =\prod_{i=1}^{n} \left( \sum_{m \in M_{i}} \frac{x^{m}}{m!} \right) ，它的幂级数展开式中 \frac{x^{k}}{k!} 的系数就是所求的可重排列数 b_{k} 。特别地，当每个元素可以出现任意次时，即通常的 k-排列，则 \sum_{k=1}^{\infty} b_k \frac{x^k}{k!} =\prod_{i=1}^{n} \left( \sum_{m=0}^{\infty} \frac{x^{m}}{m!} \right) =e^{nx}=\sum_{k=0}^{\infty}n^{k} \frac{x^{k}}{k!} ，从而 b_{k}=n^k 。有限多重集合的情形：M=\left{ n_{1} \cdot a_{1}, n_{2} \cdot a_{2}, ..., n_{t} \cdot a_{t} \right} ，k-排列数序列 \left{ b_{k} \right} 的指数型生成函数为 \sum_{k=1}^{\infty} b_k \frac{x^k}{k!} =\prod_{i=1}^{t} \left( \sum_{m=0}^{n_i} \frac{x^{m}}{m!} \right) ，展开式中 \frac{x^{k}}{k!} 的系数为 \sum_{i_1+...+i_t=k} \frac{k!}{i_1!i_2!...i_t!} ，此处对于 i_1+...+i_t=k, 0\leq i_1\leq n_1,...,0\leq i_t \leq n_t 的所有解求和，它就是M的 k-排列数。特别地，全排列数（即 k=t）为 \binom{n}{n_{1}...n_{t}} = \frac{(n_{1}+n_{2}+...+n_{t})!}{n_{1}! n_{2}! ... n_{t}!} ，其中 n=n_{1}+n_{2}+...+n_{t} 。排列分配问题的生成函数：把 k 个不同的球 1, 2, ..., k 放入 n 个不同的盒子 a_{1},a_{2},...,a_{n} ，限定盒子 a_{i} 的容量集合为 M_{i} \, (1 \le i \le n) ，则其分配方案数 b_{k} 的指数型生成函数为 \sum_{k=1}^{\infty} b_k \frac{x^k}{k!} =\prod_{i=1}^{n} \left( \sum_{m \in M_{i}} \frac{x^{m}}{m!} \right) ，它的幂级数展开式中 x^{k} 的系数就是所求的分配方案数 b_{k} （9）常系数线性齐次递推关系的解：若 q_{1}, q_{2},...,q_{k} 是 k 阶常系数线性齐次递推关系 f(n)=c_{1}f(n-1) + c_{2}f(n-2) + ... + c_{k}f(n-k) \,\, (n \ge k, c_{k} \ne 0) 的 k 个互不相等的特征根，则该递推关系的通解为 f(n)=b_{1}q_{1}^{n} + b_{2}q_{2}^{n} + ... + b_{k}q_{k}^{n} ，其中 b_{1},b_{2}, ... , b_{k} 为任意常数。特征根有重根的情形：设 q_{1}, q_{2},...,q_{t} 是该递推关系全部不同的特征根，其重数分别为 e_{1},e_{2},..., e_{t} \, (e_{1}+e_{2}+...+e_{t}=k) ，那么递推关系的通解为 f(n)=f_{1}(n) + f_{2}(n) + ... + f_{t}(n) ，其中 f_{i}(n)=(b_{1}+b_{2}n+ ... + b_{e_{i}} n^{e_{i}-1}) \cdot q_{i}^{n} \,\, (1 \le i \le t) （10）常系数线性齐次递推关系的生成函数：k 阶常系数线性齐次递推关系 f(n)=c_{1}f(n-1) + c_{2}f(n-2) + ... + c_{k}f(n-k) \,\, (n \ge k, c_{k} \ne 0) 的生成函数的形式为 g(x)=\frac{P(x)}{Q(x)} ，其中 Q(x) 是常数项不等于0的 k 次多项式，P(x) 是次数小于 k 多项式。反之，给定一个这样的多项式 P(x)/Q(x) ，则存在一个 k 阶常系数线性齐次递推关系以它为生成函数（11）常系数线性非齐次递推关系的解：f(n)=c_{1}f(n-1) + c_{2}f(n-2) + ... + c_{k}f(n-k) + g(n) ，它的通解形式是 f'(n) + f''(n) ，其中 f^{'}(n) 是它的一个非齐次特解， {f}''(n) 是相应齐次递推关系的 f(n)=c_{1}f(n-1) + c_{2}f(n-2) + ... + c_{k}f(n-k) 的通解。几种常见 g(n) 形式的特解：其中特征多项式为 P(x)=x^{k}-c_{1}x^{k-1}-c_{2}x^{k-2}- ... - c_{k} g(n)=\beta^{n}, \, P(\beta) \neq 0 ： f'(n)=a \beta^{n} ； g(n)=\beta^{n} ， \beta 是 P(x)=0 的 m 重根： f'(n)=a n ^{m} \beta^{n} ； g(n)=n^{s}, \, P(1) \neq 0 ： f'(n)=b_{s}n^{s}+b_{s-1}n^{s-1} + .. + b_{1}n + b_{0} ； g(n)=n^{s} ，1 是 P(x)=0 的 m 重根： f'(n)=n^{m} \left( b_{s}n^{s}+b_{s-1}n^{s-1} + .. + b_{1}n + b_{0} \right) ； g(n)=n^{s} \beta^{n}, \, P(\beta) \neq 0 ： f'(n)=\left( b_{s}n^{s}+b_{s-1}n^{s-1} + .. + b_{1}n + b_{0} \right) \beta^{n} ； g(n)=n^{s}\beta^{n} ， \beta 是 P(x)=0 的 m 重根： f'(n)=n^{m}\left( b_{s}n^{s}+b_{s-1}n^{s-1} + .. + b_{1}n + b_{0} \right) \beta^{n} （12）迭代归纳法求解递推关系：通过迭代找出解的规律，然后用数学归纳法证明（13）快速排序算法：通过一趟排序将要排序的数据分割成独立的两部分，其中一部分的所有数据都比另外一部分的所有数据都要小，然后再按此方法对这两部分数据分别进行快速排序，整个排序过程可以递归进行，以此达到整个数据变成有序序列。基本过程是首先设定一个分界值，通过该分界值将数组分成左右两部分，扫描一次元素并不断地交换，使右边部分的各个数都比左边部分大。然后用同样的方法处理这两个子序列（用递归调用）。快速排序是不稳定的，最理想情况算法时间复杂度O(nlgn)，最坏为O(n^2)，空间复杂度需要O(lgn) （14）用生成函数求解递推关系：给定 f(n) 的递推关系，求解 f(n) 的基本步骤是令 A(x)=\sum_{n=0}^{\infty} f(n) x^{n} ，将关于 A(x) 的递推关系式转化成关于 A(x) 的方程式，解出 A(x) ，将 A(x) 展开成 x 的幂级数， x^n 的系数即是 f(n) （15）一些递推关系问题： Hanoi塔问题：n 个圆盘，三根柱子，搬动次数为 f(n)=2f(n-1)+1, f(1)=1 ，解为 f(n)=2^n -1 。即最小移动次数为 2^n-1 。在 k \ge 4 根柱子的情况下，确定最小移动次数的问题仍然一个难题，尚未解决。地图着色问题：有公共交界点的 n 个连通区域用 k 种颜色着色，相邻区域要求不同色，着色方案数为 f(n)=(k-1)f(n-2) + (k-2)f(n-1) \,\, (n \ge 4) ， f(2)=k(k-1) ， f(3)=k(k-1)(k-2) （16）几何计数问题递推关系： n 条直线将平面分割的区域数（每两条直线交于一点但无三线共点）： h_n=h_{n-1}+n, \, h_0=1 ，解为 h_n=\frac{n(n+1)}{2}+1 ； n 个平面将空间分割的区域数（每两个平面交于一条直线但无三个平面交于一条直线，每三个平面交于一点但无四个平面交于一点）： h_n=h_{n-1}+\frac{n(n-1)}{2}+1, \, h_0=1 ，解为 h_n=\frac{n(n+1)(n-1)}{6}+n+1 ；凸 n 边形所有对角线分割的区域数（无三条对角线共点）： h_n=h_{n-1} + \frac{(n-3)(n-2)(n-1)}{6} + n-2 \, (n \ge 3) ， h_1=h_2=0, \, h_3=1 ，解为 h_n=\frac{(n-1)(n-2)(n^2-3n+12)}{24} ； n 个点的弦将圆分割的区域数（无三条弦在圆内共点）：与上面凸多边对角线分割问题类似地分析，解为 h_{n}=\binom{n}{4}+\binom{n}{2}+1 ； n 个圆中每一对圆交于两点且没有三圆共点，分割的区域数： h_n=h_{n-1} + 2(n-1) ， h_2=2 ，解为 h_n=n^2-n+2 （17）q二项式定理：(x+y)(x+qy)(x+q^2y)+...(x+q^{n-1}y) = \sum_{k=0}^{n}\binom{n}{k}{q} x^{n-k}y^{k} ，其中 n!{q}=\frac{\prod_{j=1}^{n}(1-q^j)}{(1-q)^n} 是 q 阶乘，而 \binom{n}{k}{q}=\frac{n!{q}}{k!{q}(n-k)!{q}} 是 q 二项式系数（18）Dirichlet生成函数的乘积形式：设 f(s)=\sum_{n=1}^\infty \frac{a_n}{n^s} ， g(s)=\sum_{n=1}^\infty \frac{b_n}{n^s} ，考虑 f(s)g(s)=\sum_{n=1}^\infty \frac{c_n}{n^s} ，其中每一项都具有形式 \frac{a_k}{k^s} \cdot \frac{b_{\frac{n}{k}}}{(\frac{n}{k})^s} 才能确保分母上具有 n^s 的形式。毫无疑问地，这蕴含了 k \mid n 的前提条件。故我们得到 f(s)g(s)=\sum_{n=1}^\infty \left( \sum_{k \mid n}a_kb_{\frac{n}{k}} \right) \frac{1}{n^s} ，它是 c_n=\sum_{k \mid n}a_kb_{\frac{n}{k}} 的Dirichlet生成函数。 c_n 可以看作是复合了 a_n,b_n 的一个Mobius变换。故Dirichlet生成函数的乘积形式对应了两个数列的复合Mobius变换（19）积性数论函数值列的Dirichlet生成函数：若 f 为积性数论函数即 f(nm)=f(n)f(m) ，则 \sum_{n=1}^\infty \frac{f(n)}{n^s}=\prod_p \left( \sum_{k=0}^\infty f(p^k)p^{-ks} \right) ，其中下标 p 表示对于所有素数 p 求乘积。特别地，当 f(n)=1 时，得到 \zeta(s)=\prod_p \left( \sum_{k=0}^\infty p^{-ks} \right)=\prod _p \frac{1}{1-p^{-s}} ，即Riemann-Zeta函数与Euler乘积等价（20）Mobius函数的Dirichlet生成函数： \sum_{n=1}^\infty \frac{\mu(n)}{n^s}=\prod_p \left( \sum_{k=0}^\infty \mu(p^k)p^{-ks} \right)=\prod_p \left( \sum_{k=0}^1\mu(p^k)p^{-ks} \right) （注意到当 k\geq 2 时， \mu(p^k)=0 ），故得到 \sum_{n=1}^\infty \frac{\mu(n)}{n^s}=\prod_p(1-p^{-s}) ，因此有 \sum_{n=1}^\infty \frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)} ，可见Mobius函数的Dirichlet生成函数是Riemann-Zeta函数的倒数（21）Mobius反演公式：若数列 a_n 为 b_n 的Mobius变换，即 a_n=\sum_{d \mid n}b_d ， \mu(n) 为Mobius函数，则有 b_n=\sum_{d \mid n}\mu(\frac{n}{d})a_d 。证明思路是利用Dirichlet生成函数特殊计数序列主要概念：（1）拟Catalan数： C_{n}^{\ast}=n! C_{n-1} \, (n \ge 1), \, C_{1}^{\ast}=1 ，其中 C_{n-1} 是Catalan数。递推关系为 C_{n}^{\ast}=(4n-6)C_{n-1}^{\ast} ，通项公式为 C_{n}^{\ast}=\frac{(2n-2)!}{(n-1)!} （2）差分序列：设 h_{0},h_{1},...,h_{2},...,h_{n},... 是一个序列，一阶差分序列为 \Delta h_{0}, \Delta h_{1}, ..., \Delta h_{n}, ... ，其中 \Delta h_{n}=h_{n+1}-h_{n} \, (n \geq 0) 。二阶差分序列为 \Delta^{2} h_{0}, \Delta^{2} h_{1}, ..., \Delta^{2} h_{n}, ... ，其中 \Delta^{2} h_{n}=\Delta(\Delta h_{n})=\Delta h_{n+1}-\Delta h_{n} 。递归地定义 p 阶差分序列 \Delta^{p} h_{n}=\Delta(\Delta^{p-1} h_{n}) ，规定 0 阶差分序列就是它自己即 \Delta^{0}h_{n}=h_{n} \, (n \geq 0) （3）阶乘幂：是基于自然数数列积的一种运算，分为递进阶乘（Rising factorial）和递降阶乘（Falling factorial）。递进阶乘（上升阶乘）：表示上升自然数数列的积，记作 x^{(n)} 或 x^{\overline{n}} ，定义为 x^{\overline{n}}=x(x+1)(x+2)...(x+n-1)=\frac{(x+n-1)!}{(x-1)!} ， x^{\overline{0}}=1 。递降阶乘（下降阶乘）：表示下降自然数数列的积，记作 (x)_{n} 或 x^{\underline{n}} ，定义为 x^{\underline{n}}=x(x-1)(x-2)...(x-n+1)=\frac{x!}{(x-n)!} ， x^{\underline{0}}=1 ，它与排列数 P(n, k) 相同。运用伽玛函数，阶乘幂的定义域可以扩展到实数： x^{\overline{n}}=\frac{\Gamma(x+n)}{\Gamma(x)}, \, x,x+n \neq 0,-1, -2, ... ； x^{\underline{n}}=\frac{\Gamma(x+1)}{\Gamma(x-n+1)}, \, x,x+n \neq -1, -2, -3, ... 阶乘幂也可以一般化至任意函数和公差： \left[ f(x) \right]^{k/h}=f(x)f(x+h)f(x+2h)...f(x+(k-1)h) ； \left[ f(x) \right]^{k/-h}=f(x)f(x-h)f(x-2h)...f(x-(k-1)h) 使用这个记号，原来的递进阶乘与递降阶乘便记作 \left[ x \right]^{k/1} 和 \left[ x \right]^{k/-1} （4）正整数分拆：正整数 n 的一个 k 分拆是把 n 表示成 k 个正整数的和 n=n_{1}+n_{2}+...+n_{k} \,\, (k \ge 1) 的一种表示法，其中 n_{i} >0 \, (1 \le i \le n) 称为该分拆的分部量。如果表示法是无序的，则称为无序分拆，简称为分拆。如果表示法是有序的，各项不同的次序被认为是不同的表示法，则称为有序 k 分拆。n 的所有分拆的个数称为 n 的分拆数。用 p(n, k) 表示 n 的无序 k 分拆数，p(n) 表示 n 的所有分拆数。显然有 p(n, k) >0 \, (k>n), \,\, p(n)=\sum_{k=1}^{n}p(n,k) 。无序分拆一般写成递降的顺序 n=n_{1}+n_{2}+...+n_{k}, \,\, n_{1} \ge n_{2} \ge ... \ge n_{k} ，若有 k_{i} 个分部量为 i （ 1 \le i \le n ），则分拆也可记为 n = k_{1} \cdot 1 + k_{2} \cdot 2 + ... + k_{n} \cdot n ，有时也简记为 n=1^{k_{1}}2^{k_{2}}...n^{k_{n}} （5）共轭分拆：把 n 的一个 k 分拆的Ferrers图的各行改成列，但其相对位置不变，这样得到的新Ferrers图称为原Ferrers图的共轭图。共轭Ferrers图所对应的分拆称为原分拆的共轭分拆。若原分拆与共轭分拆相同，即Ferrers图关于左上-右下的对角线是对称的，则称为自共轭分拆。更精确地说，设 n 的一个分拆为 \lambda: n=n_1+n_2+...+n_k ，它的共轭分拆是 \lambda^{\ast}: n=n_{1}^{\ast} + n_{2}^{\ast} + ... + n_{t}^{\ast} \, (t=n_1) ，其中 n_{i}^{\ast} 是分拆 \lambda 中大于等于 i 的分部量个数，即 n_{i}^{\ast}=\left\| \left{ j: \, n_{j} \geq i \right} \right\| \, (i=1,2,...,t) 。若 \lambda=\lambda^{\ast} ，则称为自共轭分拆（6）分拆集的偏序：设 P_n 是 n 的所有分拆的集合。可以对 P_n 的所有分拆建立偏序。允许分部量为0，这样任意两个分拆的分部量个数都可以看成是相等的。对任意两个 k-分拆 \lambda: n=n_1+n_2+...+n_k \,(n_1 \geq n_2 \geq ... \geq n_k \geq 0) ， \mu: n=m_1+m_2+...+m_k \,(m_1 \geq m_2 \geq ... \geq m_k \geq 0) ，如果前者的部分和小于等于后者的部分和，即 n_1+...+n_i \leq m_1+...+m_i \, (i=1,2,...,k) ，则称分拆 \mu 优超于 \lambda ，优超关系是自反、反对称且传递的，因此它是 P_n 上的偏序。字典序：字典序也可以在分折上产生 P_n 上的一个全序。如果存在一个整数 i ，使得 \lambda, \mu 在 i 处的分部量都有小于关系，而位于 i 之前的分部量都相等，即对于 j<i 有 n_j=m_j 且 n_i < m_i ，则称在字典序下 \lambda 先于 \mu （7）矩形格路径：从格点 (r, s) 到 (p, q) （其中 p \geq r, q \geq s ）的矩形格路径是指仅由从左到右的水平步 H=(1, 0) 和从下到上的垂直步 V=(0, 1) 组成的路径，也称为HV格路径。若限制格路径在对角线 y=x 的下方，则称为下对角线矩阵格路径。 HVD格路径：允许水平步 H=(1, 0) 、垂直步 V=(0, 1) 、和对角步 D=(1, 1) ，称这样的路径为HVD格路径。类似地有下对角线HVD格路径。 Schroder路径：从格 (0, 0) 到 (p, q) 的下对角线HVD格路径，称为Schroder路径。大Schroder数：从格 (0, 0) 到 (n, n) 的Schroder路径（即下对角线HVD格路径）的数目，记作 R_{n} 主要定理：（1）Fibonacci数列：递推关系f_n=f_{n-1}+f_{n-2} ， f_0=0, f_1=1 ，解为 f_n=\frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^{n} - \left( \frac{1-\sqrt{5}}{2} \right)^{n} \right] \,\,\, (n \ge 0) 生成函数： \sum_{n=0}^{\infty}f_nx^n=\frac{x}{1-x-x^2} 。 Fibonacci数列的性质：与黄金分割的关系： \lim_{n \to \infty} \frac{f_{n+1}}{f_{n}} = \frac{1+\sqrt{5}}{2} \approx 1.618... ；与平方数的关系：Fibonacci数列中只有三个平方数 0, 1, 144 ；与二项式系数的关系： f_n=\sum_{k=1}^{\left \lfloor \frac{n+1}{2} \right \rfloor} C_{n-k}^{k-1} ；各项求和： \sum_{k=0}^{n}f_k=f_{n+2}-1 ；奇数项求和： \sum_{k=1}^{n}f_{2k-1}=f_{2n} ；偶数项求和： \sum_{k=0}^{n}f_{2k}=f_{2n+1}-1 ；平方求和： \sum_{k=0}^{n}f_{k}^{2}=f_nf_{n+1} ；相邻项关系： f_{n-1}f_{n+1} - f_{n}^{2}=(-1)^n ；两倍项关系： \frac{f_{2n}}{f_{n}}=f_{n-1} + f_n ；分解定理： f_{n+m}=f_mf_{n+1}+f_{m-1}f_n ； f_n 整除 f_m ，当且仅当 n 整除 m （ n \ge 3 ）； gcd(f_m, f_n)=f_{gcd(m,n)} ；任意三个连续的Fibonacci数两两互素；与Lucas数列的关系：Lucas数列为 L_n=L_{n-1}+L_{n-2} ， L_0=2, L_1=1 ，通项公式 L_n= \left( \frac{1+\sqrt{5}}{2} \right)^{n} + \left( \frac{1-\sqrt{5}}{2} \right)^{n} ，有 L_n=f_{n-1}+f_{n+1} ； f_{2n}=L_nf_n ； f_n=\frac{L_{n-1} + L_{n+1}}{5} ； L_{n}^{2}=5f_{n}^{2} + 4(-1)^n ，因此可得 \lim_{n \to \infty} \frac{L_n}{f_n}=\sqrt{5} 。应用举例：用 2x1 方格的多米诺骨牌完全覆盖一个 nx2 的棋盘，覆盖的方案数为 h_n=f_{n+1} ；一个小孩上楼梯，每次可上一阶或两阶，上 n 阶楼梯的不同方法数为 h_n=f_{n+1} （2）Catalan数：递推关系 C_n=\sum_{k=0}^{n-1}C_kC_{n-k} ， C_0=C_1=1 ，可化简为 C_n=\frac{4n-2}{n+1}C_{n-1} 。通项公式：C_n=\frac{1}{n+1} \binom{2n}{n} ，经化简得 C_n=\binom{2n}{n}-\binom{2n}{n-1} 。生成函数： \sum_{n=0}^{\infty}C_n x^n=\frac{1-\sqrt{1-4x}}{2x} Catalan数的性质：渐近增长： C_n \sim \frac{4^n}{n^{3/2} \sqrt{\pi}} ，即左式除以右式的商当 n \to \infty 时趋于1，可用Stirling公式来证明；积分表示： C_n=\int_{0}^{4} \frac{x^n}{2\pi} \sqrt{\frac{4}{x}-1} \, dx 。 Catalan数的应用：矩阵乘法问题：n 个矩阵 A_{1}A_{2}...A_{n} 相乘通过加括号而形成的乘法方案数就是第 n-1 个Catalan数 C_{n-1}=\frac{1}{n} \binom{2n-2}{n-1} 。这里两个矩阵乘法是不可交换的，如果允许乘法可交换，例如 n 个数相乘，则总的乘法方案法为拟Catalan数 C_{n}^{\ast}=n!C_{n-1}=\frac{(2n-2)!}{(n-1)!} ；括号匹配问题：C_n 表示 n 对括号正确配对的合法运算式个数，例如这样配对 ((())) ()(()) ()()() (())() (()()) ；凸多边形三角划分：C_n 表示将一个凸 n+2 边形用不相交的对角线划分成三角形区域的方法数；不相交弦问题：在圆上选择 2n 个点，将这些点成对连接起来使得所得到的 n 条弦不相交的方法数，也是Catalan数 C_n ；二叉树计数：C_n 表示 n 个节点组成不同构二叉树的个数；满二叉树计数：C_n 表示有 2n+1 个节点组成不同构满二叉树的个数；格点的单调路径：C_n 表示所有在 nxn 格点中不越过对角线的单调路径的个数。一个单调路径从格点左下角出发，在格点右上角结束，每一步均为向上或向右；出栈次序：一个栈（无穷大）的进栈序列为 1, 2, ..., n ，有多少个不同的出栈序列？结论为Catalan数； 01序列：有 n 个0和 n 个1，问有多少个长度为 2n 的序列，使得序列的任意一个前缀中1的个数都大于等于0的个数。合法的序列个数为 C_n ；前缀序列之和大于0：由 n 个 +1 和 n 个 -1 构成的 2n 项序列 a_{1},a_{2},...,a_{2n} ，其前缀部分和总满足 a_{1}+a_{2}+...+a_{k} \ge 0 \, (k=1,2,...,2n) 序列的个数等于第 n 个Catalan数 C_n=\frac{1}{n+1} \binom{2n}{n} （3）差分的性质：通项为多项式的高阶差分为零：若序列 h_{n} 的通项是 n 的 p 次多项式，即 h_{n}=a_{p}n^{p}+a_{p-1}n^{p-1}+...+a_{1}n+a_{0} \,\, (n \geq 0) ，那么对所有的 n \geq 0 ， \Delta^{p+1}h_{n}=0 ；差分的线性性：若 g_n, f_n 是两个序列，c, d 是常数，则对每一个整数 p \geq 0 有 \Delta^{p}(cg_{n}+df_n)=c\Delta^{p}g_{n} + d\Delta^{p}f_{n} \,\, (n \geq 0) 。也就是说，序列的集合形成一个向量空间，而差分算子 \Delta 是这个向量空间上的线性变换（4）差分公式：序列 h_{n} 的 k 阶差分公式为 \Delta^{k}h_n=\sum_{j=0}^{k}(-1)^{k-j} \binom{k}{j} h_{n+j} 。证明思路是对k运用归纳法。用差分表计算差分：序列 h_{n} 的差分表的第 n(n \geq 0) 条对角线上的元素定义为 h_{n}=\Delta^{0}h_{n}, \Delta^{1}h_{n}, \Delta^{2}h_{n}, \Delta^{3}h_{n}, ... 。差分表（从左到右）中的元素由前一条对角线上的元素确定，即有 \Delta^{p}h_{n}= \Delta^{p} h_{n-1} + \Delta^{p+1}h_{n-1} （5）通过差分求序列的通项：若序列 h_{n} 的差分表的第 0 条对角线 h_{0}=\Delta^{0}h_{0}, \Delta^{1}h_{0}, \Delta^{2}h_{0}, \Delta^{3}h_{0}, ... 从某项开始为0，即为 c_{0}, c_{1},...,c_{p},0,0,... \,\, (c_{p} \neq 0) ，则 h_{n} 的通项是 n 的 p 次多项式 h_{n}=c_{0} \binom{n}{0} + c_{1}\binom{n}{1} + c_{2} \binom{n}{2} + ... + c_{p} \binom{n}{p} （6）通项为多项式的序列的部分和：若序列 h_{n} 的差分表的第 0 条对角线 h_{0}=\Delta^{0}h_{0}, \Delta^{1}h_{0}, \Delta^{2}h_{0}, \Delta^{3}h_{0}, ... 从某项开始为0，即为 c_{0}, c_{1},...,c_{p},0,0,... \,\, (c_{p} \neq 0) ，则 \sum_{k=0}^{n}h_{k}=c_{0} \binom{n+1}{1} + c_{1} \binom{n+1}{2} + ... + c_{p} \binom{n+1}{p+1} （7）阶乘幂的性质：递进阶乘与递降阶乘的关系： x^{\overline{n}}=(x+n-1)^{\underline{n}} ， 1^{\overline{n}}=n^{\underline{n}}=n! ；与二项式系数的关系： \frac{x^{\overline{n}}}{n!}=\binom{x+n-1}{n} ， \frac{x^{\underline{n}}}{n!}=\binom{x}{n} ；整除关系：n 个连续整数的积必定能被 n 整除，因此 n \mid x^{\overline{n}}, \, n \mid x^{\underline{n}} ；类二项式定理： (a+b)^{\overline{n}}=\sum_{r=0}^{n} \binom{n}{r} a^{\overline{n-r}} b^{\overline{r}} ， (a+b)^{\underline{n}}=\sum_{r=0}^{n} \binom{n}{r} a^{\underline{n-r}} b^{\underline{r}} ，其中系数为二项式系数；递降阶乘的线性展开： x^{\underline{m}} x^{\underline{n}}=\sum_{k=0}^{m} \binom{m}{k} \binom{n}{k} k! x^{\underline{m+n-k}} （8）第一类Stirling数：分为第一类有符号Stirling数和第一类无符号Stirling数，定义为对应递降阶乘和递进阶乘展开式的各项系数。 x^{\underline{n}}=x(x-1)(x-2)...(x-n+1)=\sum_{k=0}^{n}s(n,k) x^{k} ； x^{\overline{n}}=x(x+1)(x+2)...(x+n-1)=\sum_{k=0}^{n}c(n,k) x^{k} 。这里 s(n, k) 是第一类有符号Stirling数，c(n, k) 是第一类无符号Stirling数，也记作 \begin{bmatrix} n\ k \end{bmatrix} ，它们的关系为 s(n,k)=(-1)^{n+k} c(n,k) , \, c(n,k) = \left\| s(n,k) \right\| 。组合数学中的第一类Stirling数一般指第一类无符号Stirling数，它是 n 个不同元素排成 k 个圆排列的方案数。 Pascal型递推关系：s(n+1,k)=s(n,k-1)-n \cdot s(n,k) ；c(n+1,k)=c(n,k-1) + n \cdot c(n,k) ， c(0, 0)=c(n,n)=1, \, c(n,0)=c(0,n)=0 (n>0) ， c(n, 1)=(n-1)! ， c(n,k)=0 (k>n) 。通项公式：第一类Stirling数没有特别实用的通项公式。其中有一个是 s(n+r,n)=(-1)^r \sum_{1 \le k_1 < ... < k_r \le n+r} k_1k_2...k_r 指数型母函数：\sum_{n=0}^{\infty} s(n,k) \frac{x^n}{n!}=\frac{1}{k!} ln^{k}(1+x) ，用该式求 s(n, k) 的通项公式会比较繁琐。性质： c(n,n-1)=\binom{n}{2} ； c(n,n-2)=\frac{1}{4}(3n-1)\binom{n}{3} ； c(n,n-3)=\binom{n}{2} \binom{n}{4} ； \sum_{k=0}^{n} c(n,k)=n! ，证明可令递进阶乘中的x=1，比较两边系数；应用举例：第一类Stirling数 c(n, k) 表示置换群 S_n 中所有可分解为 k 个不相交轮换的置换个数。对 k 求和即得到所有的置换个数为 n! 。解锁仓库问题。有n个仓库，每个仓库有两把钥匙，共2n把钥匙。同时又有n位官员。问如何放置钥匙使得所有官员都能够打开所有仓库？（只考虑钥匙怎么放到仓库中，而不考虑官员拿哪把钥匙。）那如果官员分成m个不同的部，部中的官员数量和管理的仓库数量一致。那么有多少方案使得，同部的所有官员可以打开所有本部管理的仓库，而无法打开其他部管理的仓库？（同样只考虑钥匙的放置。）第一问很经典，就是打开将钥匙放入仓库构成一个环：1号仓库放2号钥匙，2号仓库放3号钥匙……n号仓库放1号钥匙。这种情况相当于钥匙和仓库编号构成一个圆排列方案数是 (n-1)! 种。而第二问就对应的将n个元素分成m个圆排列，方案数就是第一类无符号Stirling数 c(n, m) 。如要要考虑官员的情况，只需再乘上 n! 即可（9）第二类Stirling数：一个 n 元集合的全部 k 划分（即划分成 k 个两两不相交的非空子集的并）的个数，称为第二类Stirling数，记作 S(n, k) ，或者 \begin{Bmatrix} n\ k \end{Bmatrix} 。还可以用递降阶乘来定义 x^n=\sum_{k=0}^{n} S(n, k) x^{\underline{k}} 。它也可以理解成把 n 元集合划分到 k 个不可区分的盒子且无空盒子的划分个数。 Pascal型递推关系：S(n+1, k)=S(n, k-1)+kS(n,k) \, (1 \le k \le n) ， S(0, 0)=1, \, S(n,0)=S(0,n)=0 (n>0) ， S(n, 1)=S(n, n)=1 ， S(n,k)=0 (k>n) ；通项公式：S(n,k)=\frac{1}{k!} \sum_{i=0}^{k}(-1)^{i} \binom{k}{i}(k-i)^{n} 。这个通项公式有多种方法推导，常用的有容斥原理，生成函数和等式求解等；另一种通项公式： S(n+r,n)=\sum_{1 \leq k_1 \le ... \le k_r \le n} k_1k_2...k_r ；指数型母函数：\sum_{n=0}^{\infty} S(n,k) \frac{x^n}{n!}=\frac{1}{k!}(e^x-1)^k ，用它可以求 S(n, k) 的通项公式。性质： S(n,2)=2^{n-1}-1 ； S(n,3)=\frac{1}{2}(3^{n-1}+1)-2^{n-1} ； S(n,n-1)=\binom{n}{2}=\frac{1}{2}n(n-1) ； S(n,n-2)=\binom{n}{3}+3 \binom{n}{4} ； S(n,n-3)=\binom{n}{4}+10 \binom{n}{5}+15 \binom{n}{6} ；两类Stirling数的关系：可以看作是互为逆矩阵的关系，第二类Stirling数三角方阵 S_n^{(2)}=[S(i,j)] \, (i,j=0,1,...,n) 与第一类Stirling数三角方阵 S_n^{(1)}=[s(i,j)] \, (i,j=0,1,...,n) 互为逆矩阵，从而有 \sum_{j=0}^{n}s(n,j)S(j,m) = \sum_{j=0}^{n}S(n,j)s(j,m)=\delta_{nm} ，其中 \delta_{nm} 是Kronecker \delta 函数，n=m 时它的值为1，否则值为0 。应用举例： n个不同的球，放入m个无区别的盒子，不允许盒子为空。方案数为 S(n, m) 。这个跟第二类Stirling数的定义一致。 n个不同的球，放入m个有区别的盒子，不允许盒子为空。方案数为 m! S(n, m) 。因盒子有区别，乘上盒子的排列即可。这也 n 元集合到 m 元集合上满射的个数。 n个不同的球，放入m个无区别的盒子，允许盒子为空。方案数为 \sum_{k=0}^{n}S(n,k) ，枚举非空盒的数目便可。 n个不同的球，放入m个有区别的盒子，允许盒子为空。方案数为 \sum_{k=0}^{m} P(m,k) \cdot S(n,k)=m^n ，同样可以枚举非空盒的数目，注意到盒子有区别，乘上一个排列系数。既然允许盒子为空，且盒子间有区别，那么对于每个球有m种选择，每个球相互独立。因此有方案数 m^n 。该式子可以应用于第二类Stirling数通项的求解（10）Bell数：n 元集合的划分方法的数目，称为Bell数，记作 B_n 。集合 S 的一个划分是定义为S的两两不相交的非空子集的族，它们的并是S。与第二类Stirling数的区别是Bell数不限定一个划分中的子集个数。递推公式： B_{n+1}=\sum_{k=0}^{n} \binom{n}{k}B_{k} ， B_0=B_1=1 ； Dobinski公式： B_{n}=\frac{1}{e}\sum_{k=0}^{\infty} \frac{k^n}{k!} ，它是期望值为1的泊松分数的 n 次矩； Touchard同余： B_{n+p} \equiv B_{n}+B_{n+1} \, (mod \, p) ，其中 p 是任意素数；与第二类Stirling数的关系： B_{n}=\sum_{k=0}^{n}S(n,k) ；指数型母函数： \sum_{n=0}^{\infty}B_{n} \frac{x^n}{n!}=e^{e^x-1} （11）Lah数（第三类Stirling数）：拉赫数是用递进阶乘或递降阶乘来定义的，即 x^{\overline{n}}=\sum_{k=0}^{n} L(n,k) x^{\underline{k}} 或 x^{\underline{n}}=\sum_{k=0}^{n} (-1)^{n+k}L(n,k) x^{\overline{k}} ，其中 L(n, k) 即为无符号Lah数。它是 n 元集合的全序 k 划分（即划分为 k 个非空线性有序子集，也称全序子集）的个数。有符号Lah数的定义为 x^{\overline{n}}=(-1)^{n}\sum_{k=0}^{n} {L}'(n,k) x^{\underline{k}} 或 x^{\underline{n}}=\sum_{k=0}^{n} (-1)^{k}{L}'(n,k) x^{\overline{k}} ，其中 {L}'(n,k) 即为有符号Lah数。它们的关系为 {L}'(n,k)=(-1)^{n} L(n,k) 。递推公式：L(n+1,k)=L(n,k-1)+(n+k)L(n,k) ，或者 L(n,k+1)=\frac{n-k}{k(k+1)}L(n,k) ，其中 L(0, 0)=L(n,n)=1, L(n,0)=0 (n>0), L(n,k)=0(k>n) ；通项公式：L(n,k)=\binom{n-1}{k-1} \frac{n!}{k!} ；指数型母函数：\sum_{n=0}^{\infty} L(n,k) \frac{x^n}{n!}=\frac{1}{k!} \left( \frac{x}{1-x} \right)^{k} ；与两类Stirling数的关系： L(n,k)=\sum_{j=0}^{n}c(n,k) \cdot S(n,k) ；互逆关系：类似地，有 \sum_{j=0}^{n}L(n,j)L(j,m) =\delta_{nm} （12）正整数有序分拆数：正整数 n 的有序 k 分拆的数目为 \binom{n-1}{k-1} ，它等于方程 x_{1}+x_{2}+...+x_{k}=n 的正整数解的个数，也等于多重集合 M=\left{ \infty \cdot a_{1}, \infty \cdot a_{2}, ... , \infty \cdot a_{k} \right} 的 a_{1},a_{2},...,a_{k} 至少出现一次的 n 组合数（13）一些限定条件的有序分拆：正整数 n 的有序 k 分拆，要求各个分部量 n_{i} 大于或等于 p_{i} （ 1 \le i \le k ），其个数为 \binom{n-\sum_{i=1}^{k}p_{i} +k - 1}{k-1} ；正整数 2n 的有序 k 分拆，要求各分部量都是正偶数，其个数为 \binom{n-1}{k-1} ；正整数 n 的有序 k 分拆，若 n 与 k 同为奇数或同为偶数，则各分部量都是奇数的有序分拆数为 \binom{\frac{n+k}{2}-1}{k-1} （14）正整数的无序分拆数：n 的 k 分拆数 p(n, k) 满足递推关系 p(n+k, k) = \sum_{i=1}^{k}p(n, i) ，及 p(n, 1)=1，p(n, n)=1 ，k>n时 p(n, k)=0 。例如 p(n,2)=\left \lfloor \frac{n}{2} \right \rfloor 。另一个递推关系： p(n, k)=p(n-1, k-1)+p(n-k, k) ；通项公式： p(n,k)=\frac{n^{k-1}}{(k-1)! k!} + R_{k-2}(n) \, (k \ge 3) ，这里 R_{k-2}(n) 是 n 的次数 \le k-2 有有理系数多项式，其系数只与 n \, (mod \, k!) 有关； p(n, 3) 的公式： p(n,3)=\left[ \frac{(n+3)^2}{12} \right] ，这里 [a] 表示离实数 a 最近的整数（15）分拆数的性质：分拆数恒等式：n 的 k 分拆数，等于 n 的最大分部量为 k 的分拆数。证明思路是利用Ferrers图建立这两者之间的一一映射。自共轭分拆数恒等式：n 的自共轭分拆数，等于 n 的各分部量不相等且都是奇数的分拆数，也等于有奇数个不同分部量的分拆数。证明思路也是建立这两者之间的一一映射。欧拉恒等式：n 的各分部量两两不相等的分拆数，等于 n 的各分部量都是奇数的分拆数，也等于有奇数个分部量的分拆数（16）分拆数的生成函数：设 p(n) 表示正整数 n 所有分拆数，p_{r}(n) 表示各分部量都不超过 r 的分拆数， p_{H}(n) 表示 n 的各分部量都属于集合 H 的分拆数，p(n, k) 表示 k 分拆数，则它们相应的生成函数分别为 \sum_{n=0}^{\infty}p_{r}(n) x^{n}=\prod_{j=1}^{r} \frac{1}{1-x^{j}} ； \sum_{n=0}^{\infty}p_{H}(n)x^{n}=\prod_{j \in H} \frac{1}{1-x^{j}} ； \sum_{n=0}^{\infty}p(n,k) x^{n}= x^{k} \prod_{j=1}^{k} \frac{1}{1-x^{j}} ； P(x)=\sum_{n=0}^{\infty}p(n) x^{n}=\prod_{j=1}^{\infty} \frac{1}{1-x^{j}} ，即欧拉函数 \varphi(x)=\prod_{j=1}^{\infty}(1-x^j)=\sum_{n=0}^{\infty}q(n) x^{n} 的倒数是分拆数的生成函数， \varphi(x) 是 P(x) 在形式幂级数整环 R 中的乘法逆元，通过求系数 q(n) 来求出 p(n) 的递推式。有 q(n) = \left{\begin{matrix} (-1)^n, & n=k(3k \pm 1)/2 \ 0, & other \end{matrix}\right. ，可以得到欧拉五边形数定理（17）欧拉公式：分拆数 p(n) 的递推式为 p(n)=\sum_{k=1}^{\infty} (-1)^{k-1} \left[ p\left( n-\frac{k(3k-1)}{2} \right) + p\left( n-\frac{k(3k+1)}{2} \right) \right] \ =p(n-1)+p(n-2)-p(n-5)-p(n-7)+... 规定当 m<0 时 p(m)=0 （18）分拆数的渐近式：p(n) \sim \frac{e^{\pi \sqrt{2n/3}}}{4n \sqrt{3}} （19）分拆集偏序的线性扩展：对正整数 n 的分拆集 P_n ，其字典序是优超偏序的线性扩展（20）Jacobi三重积公式：\prod_{n=1}^{\infty} \left[ (1-q^{2n})(1+q^{2n-1}z)(1+q^{2n-1}z^{-1})\right] = \sum_{k=-\infty}^{\infty}q^{k^2}z^k ，其中 \left\| q \right\| <1 （21）欧拉五边形数定理：是一个由欧拉发现的数学定理，描述欧拉函数展开式的特性。欧拉函数的展开式为 \prod_{n=1}^{\infty}(1-x^n)=\sum_{k=-\infty}^{\infty}(-1)^{k} x^{\frac{k(3k-1)}{2}} = \sum_{k=0}^{\infty}(-1)^{k}x^{\frac{k(3k \pm 1)}{2}} ，即展开式中只留下次数为 \frac{k(3k \pm 1)}{2} 的项。推论：设 p_{n}^{'} 是把 n 分成偶数个不同分部量的分拆数，而 p_{n}^{'’} 是把 n 分成奇数个不同分部量的分拆数，则有 p_{n}^{'} = p_{n}^{''} + e_{n} ，其中 e_n 是误差项，如果 n 是形如 \frac{k(3k \pm 1)}{2} 的整数，则 e_n=(-1)^{k} ，否则 e_n=0 （22）几何区域计数问题：对 n 的前 k+1 个二项式系数之和 h_{n}^{(k)}=\sum_{i=0}^{k} \binom{n}{i} ，它表示 n 元集合的至多有 k 个元素的子集的个数。序列 h_{0}^{(k)}, h_{1}^{(k)}, ... , h_{n}^{(k)}, ... 的差分有性质 \Delta h_{n}^{(k)}= h_{n+1}^{(k)} - h_{n}^{(k)}= h_{n}^{(k-1)} 。 h_{n}^{(k)} 计数的是用 n 个一般位置上的 k-1 维超平面分割 k 维空间生成的最多区域数。 k=1： h_{n}^{(1)}=1+n ，它是 n 个0维超平面（即点）分割一条直线产生的最多区域数； k=2：h_{n}^{(2)}=1+n+\frac{n(n-1)}{2} ，它是 n 条一般位置上的直线（每两条直线交于一点但无三线共点）将平面分割的最多区域数； k=3：h_{n}^{(3)}=\sum_{k=0}^{3} \binom{n}{k}=\frac{n(n+1)(n-1)}{6}+n+1 ，它是 n 个一般位置上的平面（每两个平面交于一条直线但无三个平面交于一条直线，每三个平面交于一点但无四个平面交于一点）将3维空间分割的最多区域数。（23）矩形格路径数：从格 (r, s) 到 (p, q)（其中 p \geq r, q \geq s ）的矩形格路径的数目等于二项式系数 \binom{p-r+q-s}{p-r}=\binom{p-r+q-s}{q-s} 。它是 p-r 个水平步H和 q-s 个垂直步V的排列数，每个排列有 p-r 个H和 q-s 个V （24）下对角线矩阵格路径数：从 (0, 0) 到 (n, n) 的下对角线矩阵格路径数目等于第 n 个Catalan数 C_n=\frac{1}{n+1} \binom{2n}{n} 。更一般地，从格 (0, 0) 到 (p, q)（其中 p \geq q ）的下对角线矩形格路径的数目等于 \frac{p-q+1}{p+1} \binom{p+q}{q} （25）HVD格路径数：设 K(p, q) 是从格 (0, 0) 到 (p, q) 的HVD格路径数目，K(p, q: rD) 是正好使用了 r 个对角步D=(1, 1) 的HVD格路径数目，K(p, q: 0D) 就是矩阵格路径的数目，则当 r > \min \left{ p,q \right} 时 K(p, q: rD)=0 ，当 r \leq \min \left{ p,q \right} 时，有 K(p,q: rD)=\binom{p+q-r}{(p-r) \,\, (q-r) \,\, r}=\frac{(p+q-r)!}{(p-r)! (q-r)! r!} ； K(p,q)=\sum_{r=0}^{\min \left{ p,q \right}} \frac{(p+q-r)!}{(p-r)! (q-r)! r!} （26）下对角线HVD格路径数：设 R(p, q) 是从格 (0, 0) 到 (p, q)（其中 p \geq q ）的下对角线HVD格路径的数目，R(p, q: rD) 是正好使用了 r (r \leq q \leq p) 个对角步D=(1, 1) 的下对角线HVD格路径数目，R(p, q: 0D) 就是下对角线矩阵格路径的数目，则有 R(p,q: rD)=\frac{p-q+1}{p-r+1} \binom{p+q-r}{r \,\, (p-r) \,\, (q-r)} = \frac{p-q+1}{p-r+1} \cdot \frac{(p+q-r)!}{r! (p-r)! (q-r)!} ； R(p,q)=\sum_{r=0}^{q} \frac{p-q+1}{p-r+1} \cdot \frac{(p+q-r)!}{r! (p-r)! (q-r)!} （27）大Schroder数：定义为从格 (0, 0) 到 (n, n) 的Schroder路径（即下对角线HVD格路径）的数目，记作 R_{n} (n \geq 0) 。通项公式： R_n=R(n,n)=\sum_{r=0}^{n} \frac{1}{n-r+1} \cdot \frac{(2n-r)!}{r! ((n-r)!)^2} 生成函数： \sum_{n=0}^{\infty}R_{n}x^n=\frac{1}{2x} \left( 1-x- \sqrt{x^2-6x+1} \right) （28）小Schroder数：给 n 个符号 a_1,a_2,...,a_n 的序列添加括号的方法数，记作 s_n(n \geq 1) 。生成函数： \sum_{n=1}^{\infty} s_{n}x^{n}=\frac{1}{4} \left( 1+x - \sqrt{x^2-6x+1} \right) ；递推关系： (n+1)s_{n+2}-3(2n+1)s_{n+1}+(n-1)s_{n}=0 \, (n \geq 1) ；与大Schroder数的关系： R_n=2 s_{n+1} \, (n \geq 1) , R_{0}=1 。应用举例：凸多边形剖分：小Schroder数 s_n 表示将一个凸 n+1 边形用不相交对角线进行剖分的方法数。注意这里剖分得到的区域不一定是三角形。参考资料：（1）组合数学：第5版，Richard A. Brualdi （2）组合数学引论(第2版)：许胤龙、孙淑玲编辑于 2021-10-15 02:49 组合数学（Combinatorics）离散数学计数赞同 353 条评论分享喜欢收藏申请转载写下你的评论... 3 条评论默认最新 buming TOUCHARD问题是啥你知道不 2023-03-23 回复喜欢 fatzard 有用！！！ 2022-09-03 回复喜欢 23333 这学期上了一点对称函数代数方法的课搜了一下竟然也有计数组合的笔记 2021-12-05 回复喜欢关于作者 Dave Zhou 做自己回答 22文章 83关注者 870 关注他发私信推荐阅读组合数学--计数组合 (1) ============== 排列与组合主要概念：（1）组合数学：研究离散构造的存在、计数、分析和优化等问题的一门科学。常见的组合问题：排列的存在性、排列的列举或分类（计数问题）、排列的构造性问题、最优化… Dave ...发表于组合数学基...组合数学：偏序集与格 ========== 之前我谈到要介绍一下格理论，没曾想在组合数学中用的很多。相比于纯粹的代数性质定义，有应用案例自然更好。建议参考Enumerative Combinatorics, Volume 1第三章前面的内容。图论部分可以… 半数o阿白发表于数学想法组合数学学习笔记（九） =========== sola组合数学--组合设计 ========== 相异代表系主要概念：（1）相异代表系：设 A_1,A_2,...,A_n 是集合E的 n 个子集（这些集合可以是相交的或相等的），E中的 n 元素 e_1,e_2,...,e_n ，其中 e_1 \in A_1, e_2 \in A_2, ...,… Dave ...发表于组合数学基... 想来知乎工作？请发送邮件到 jobs@zhihu.com 打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码
1897	https://karger.com/sad/article/doi/10.1159/000545580/925790/Nail-Surgery-Complications-A-Review-of-the	Nail Surgery Complications: A Review of the Literature \| Skin Appendage Disorders \| Karger Publishers We Value Your Privacy! When you visit our website we and our partners collect personal information from you regarding your internet activity (e.g. online identifier, IP address, browsing history) and may set cookies or use similar technologies on your device to personalize the advertising that you see. This helps us to show you more relevant ads and improves your internet experience. We also use it to measure results or align our website content. Our partners may sell this information to third parties. You can opt out of our sharing of your information with third parties. Necessary Cookies Store and/or access information on a device Use precise geolocation data Functional Personalised advertising and content, advertising and content measurement, audience research and services development Accept AllSettingsReject AllSave Settings \| Cookies \| Privacy Policy \| Terms & Conditions \| Imprint Cookie Banner powered by consentmanager.net Skip to Main Content Open Menu Close Who We Serve Open Menu Overview Researchers Authors Reviewers Healthcare Professionals Patients & Their Supporters Librarians Health Sciences Industry Medical Societies Agents & Distributors What We Offer Open Menu Overview Journals Collections Subject Areas Karger Image Explorer Books & Book Series E-Learning Courses Open External Link Open Access Portfolio Impact Support Services Who We Are Open Menu Overview What We Solve Accessing Knowledge Presenting Knowledge Applying Knowledge Publish with Us Open Menu Overview Publish Your Paper Getting Started for Researchers Open External Link Calls for Papers Publication Services Partner Publications Submit & Publish Open Access Resources for You Open Menu Overview Exhibitions Webinars Case Studies Fast Facts Medical Information News & Media Releases AI Innovation Hub Our AI Principles Beyond the Cover Blog Open External Link The Waiting Room Blog Open External Link Research Viewpoints Blog Open External Link Karger Researcher Hub Open External Link Search Dropdown Menu header search search input Search input auto suggest filter your search Search Advanced Search User Tools Dropdown Register Login Open Menu Toggle Menu Menu Content Open Menu Articles Issues Online First Early View About Open Menu Details Contact Editorial Board Journal Guidelines Peer Review Affiliations Subscription Rates Rights & Permissions Advertising Submission Guide Open Menu Journal Guidelines Technical Instructions Editorial Policies What is Peer Review Submit now Skip Nav Destination Close navigation menu Article navigation Article Contents Abstract Plain Language Summary Introduction General Complications Complications Specific to the Nail Apparatus Complications Associated with Specific Nail Procedures Prevention of Surgical Complications Conflict of Interest Statement Funding Sources Author Contributions References Article Navigation Review Articles\|April 17 2025 Nail Surgery Complications: A Review of the Literature Free Subject Area:Dermatology Jessica J. Farzan; Jessica J. Farzan a University of Massachusetts Chan Medical School, Worcester, MA, USA Search for other works by this author on: This Site PubMed Google Scholar Bassel H. Mahmoud Corresponding Author Bassel H. Mahmoud b Department of Dermatology, University of Massachusetts Chan Medical School, Worcester, MA, USA Bassel.Mahmoud-Abdallah@umassmemorial.org Search for other works by this author on: This Site PubMed Google Scholar Bassel.Mahmoud-Abdallah@umassmemorial.org Skin Appendage Disord 432–440. Article history Received: November 16 2024 Accepted: March 21 2025 Published Online: April 17 2025 PubMed: 40385515 Content Tools Split-Screen Views Icon Views Open Menu Article contents Figures & tables Download Icon Download Open Menu Open the PDF for in another window Share Icon Share Facebook X LinkedIn Email Tools Icon Tools Open Menu Cite Icon Cite Get Permissions Search Site Citation Jessica J. Farzan, Bassel H. Mahmoud; Nail Surgery Complications: A Review of the Literature. _Skin Appendage Disord_ 2025; Download citation file: Ris (Zotero) Reference Manager EasyBib Bookends Mendeley Papers EndNote RefWorks BibTex toolbar search Search Dropdown Menu toolbar search search input Search input auto suggest filter your search Search Advanced Search Abstract Background: Nail surgery complications have not been addressed thoroughly. Summary: This review aimed to provide a robust literature review of nail surgery complications by identifying relevant data using a search of PubMed. It examines various complications, including general complications such as bleeding and infection; specific nail complications such as nail dystrophy; and common complications associated with specific nail procedures. Key Messages: This literature reviews better familiarizes dermatologists who perform nail procedures with relevant complications. Journal Section: Review Article Keywords: Nail surgery,Complications Plain Language Summary There is a dearth of literature reviewing nail surgery complications. The goal is to increase awareness of complications related to nail procedures among dermatologists that perform nail procedures. Journal Section: Review Article Keywords: Nail surgery,Complications Introduction There is a dearth of literature reviewing nail surgery complications . This literature review examines various complications, including general complications such as bleeding and infection; specific nail complications such as nail dystrophy, and common complications associated with specific nail procedures. We can improve our counseling by being more aware of complications associated with nail procedures. A comprehensive review of the literature summarizing the current knowledge of nail surgery complications was performed using PubMed. Publications say nail surgery complications dating from 1987 to 2023 are included. Search words to generate nail surgery complications included “nail surgery outcomes” and “nail surgery complications.” Once general complications and complications specific to nail procedures were identified using the aforementioned method, a more detailed search using the complication and “nail surgery” derived further publications. It is imperative to understand nail anatomy prior to reviewing nail surgery complications (Fig. 1). The nail plate comprises the hard exterior nail surface. It is anchored by the proximal and lateral nail folds. The nail folds, cuticle, and nail plate protect the underlying nail matrix and bed . View largeDownload slideGet Permissions Fig. 1. Anatomy of the fingernail: components of the nail. The anatomy of the fingernail is depicted. View largeDownload slideGet Permissions Fig. 1. Anatomy of the fingernail: components of the nail. The anatomy of the fingernail is depicted. Close modal The nail matrix harbors melanocytes and keratinocytes. The most proximal aspect of the matrix lies beneath the proximal nail fold. The proximal matrix produces most of the nail plate, namely, the dorsal part, and thus, damage to the proximal matrix causes more deformity than damage to the distal matrix. The distal matrix produces the ventral nail plate; though not seen in all nails, the distal aspect, lunula, is most visible in the thumbnail . The nail bed is the space beneath the nail plate between the matrix and hyponychium. An important feature of the nail bed is the onychodermal band, which traverses the distal nail bed. Material that penetrates beneath the hyponychium is blocked by the onychodermal band . Terminal branches of the palmar digital arteries supply the nails and digital tips. Periungual innervation comes from paired digital nerves, palmar and plantar . Specifically, the 1st and 5th fingers are innervated by the dorsal proper digital nerves, while the other fingers are innervated by the palmar proper digital nerves . General Complications Pain Pain is an important consideration as equally effective procedures can cause varying levels of postoperative pain. Several nail surgery studies include pain as an outcome (Table 1) [3‒8]. Possibly related to pain and other complications, postoperative work day loss has also been reported . Table 1. Summaries of data from studies of nail surgery complications \| . \| Author . \| Year . \| Nail condition . \| Treatment . \| Main finding . \| --- --- --- \| \| General complications \| \| Pain \| \| \| Vinay \| 2022 \| Onychocryptosis \| Phenol-based matricectomy \| Phenol matricectomies are associated with less pain (257 fewer cases of pain per 1,000 procedures) compared to other non-chemical matricectomies and sleeve procedures \| \| \| Ramesh \| 2020 \| Onychocryptosis \| NaOH vs. phenol vs. TCA matricectomy \| Phenol-based matricectomies resulted in a longer duration of pain (9 days) compared to NaOH (4 days) and TCA-based matricectomies (1 day) (p = 0.0092) \| \| \| Romero-Perez \| 2017 \| Onychocryptosis \| Surgical vs. phenol-based matricectomy \| Surgical matricectomy is associated with higher intensity of pain (5.7/10) compared to phenol-based chemical matricectomy (3.6/10) (p = 0.000) on scale 1–10 as well as longer pain duration (7.2 days in surgical vs. 4.2 days in phenol group, p = 0.002) \| \| \| Peyvandi \| 2011 \| Onychocryptosis \| Winograd vs. sleeve (gutter) surgical method \| The Winograd method resulted in a longer postoperative work day loss (2 weeks) compared to the sleeve method (1.1 weeks), which is statistically significant (p< 0.001) \| \| \| Ozan \| 2014 \| Onychocryptosis \| Partial matricectomy with curettage vs. with electrocautery \| Curettage resulted in a shorter duration of postoperative pain (2 days) compared to electrocautery (3 days), which is statistically significant (p< 0.05) \| \| \| Topin-Ruiz \| 2017 \| Squamous cell carcinoma of the nail unit \| Wide surgical excision with full-thickness skin graft \| 15% (8 out of 55) patients experienced severe postoperative pain \| \| \| Fritz \| 1997 \| Digital mucous cysts \| Surgical excision \| 9% (of 86 patients) had persistent pain and swelling postoperatively \| \| \| Kasdan \| 1994 \| Digital mucous cysts \| Surgical excision \| 14% (of 113 patients) experienced joint tenderness postoperatively \| \| \| Lee \| 2013 \| Nail matrix and bed subungual glomus tumors \| Surgical excision \| 14% (3 out of 22) patients with nail matrix or nail bed tumors experienced prolonged pain \| \| Complex regional pain syndrome \| \| \| Ingram \| 1987 \| Onychocryptosis \| Biopsy \| One case of complex regional pain syndrome was reported following a nail biopsy \| \| \| Guerrero-Gonzalez \| 2016 \| Nail matrix glomus tumor, melanonychia, myxoid cyst, nail unit tumor \| Excision or biopsy \| Four patients developed complex regional pain syndrome following surgical interventions for various nail conditions \| \| Bleeding \| \| \| Vinay \| 2022 \| Onychocryptosis \| Phenol-based matricectomy \| Phenol matricectomies are associated with less bleeding (177 fewer cases per 1,000 procedures) compared to other non-chemical matricectomies and sleeve procedures \| \| \| Ramesh \| 2020 \| Onychocryptosis \| NaOH vs. phenol vs. TCA matricectomy \| Phenol-based matricectomies resulted in a longer postoperative oozing (14 days) compared to NaOH (11 days) and TCA-based matricectomies (6 day) (p = 0.01) \| \| \| Romero-Perez \| 2017 \| Onychocryptosis \| Surgical vs. phenol-based matricectomy \| In a study, 63/202 (31.2%) surgical matricectomy patients had prolonged bleeding (>24 h) compared to 40/139 (28.8%) phenol-based chemical matricectomy patients (p = 0.634) \| \| Infection \| \| \| Grover \| 2005 \| Non-infectious and infectious nail disorders \| Various types of nail biopsies \| 11% of those who underwent nail biopsies developed secondary infections \| \| \| Romero-Perez \| 2017 \| Onychocryptosis \| Surgical matricectomy vs. phenol-based chemical matricectomy \| Surgical matricectomy showed higher infection risk (OR = 7.2, 95% CI 2.4–21.0) compared to phenol-based matricectomy \| \| \| Weinand \| 2014 \| Fingernail injury \| Splinting with native nail vs. silicone nail \| Splinting with the native nail was associated with half the infections compared to splinting with silicone nail \| \| \| Miranda \| 2012 \| Fingernail injury \| Nail replacement/foil placement vs. nail discarded \| Nail replacement group had higher infection rates vs. nail discarded (8% vs. 0%, p< 0.0001) \| \| \| Lee \| 2013 \| Nail matrix and bed subungual glomus tumors \| Surgical excision \| 1/13 (7.7%) nail matrix patients had mild wound dehiscence \| \| Inclusion cysts \| \| \| Topin-Ruiz \| 2017 \| Squamous cell carcinoma of the nail unit \| Wide surgical excision with full-thickness skin graft \| 9 out of 51 patients (18%) developed epidermal inclusion cysts \| \| Keloid \| \| \| Muzaffar \| 2004 \| Syndactyly \| Excision and reconstruction \| 8 of 681 patients developed keloids postoperatively \| \| Recurrence \| \| \| Fritz \| 1997 \| Digital mucous cysts \| Surgical excision \| 3% (of 86 excisions) developed recurrences \| \| \| Topin-Ruiz \| 2017 \| Squamous cell carcinoma of nail unit \| Wide local excision \| 2 of 55 patients (4%) developed recurrence \| \| \| Gou \| 2020 \| Squamous cell carcinoma of nail unit \| Mohs micrographic surgery \| Of 42 cases of nail unit SCC treated with Mohs surgery, 3 patients (7.1%) experienced recurrence \| \| \| Richert \| 2013 \| Longitudinal melanonychia of the proximal nail fold \| Tangential matrix excision/shave removal \| 16 out of 23 patients (70%) had pigmentation recurrence 8–12 months post-surgery \| \| \| Zhou \| 2019 \| Longitudinal melanonychia \| Modified shave surgery combined with the longitudinal-strip nail window technique \| 8 of 60 patients (13.3%) had pigmentation recurrence \| \| \| Lee \| 2013 \| Nail matrix and bed subungual glomus tumors \| Surgical excision \| No recurrence was observed in nail matrix (n = 13) or nail bed (n = 9) subungual glomus tumor excisions \| \| \| Moon \| 2004 \| Subungual glomus tumors \| Surgical excision \| None of the 16 patients experienced recurrence \| \| Complications specific to the nail apparatus \| \| Nail dystrophy \| \| \| Yang \| 2017 \| Large nail defects with exposed distal phalanxes \| Finger fascial flaps combined with thin split-thickness toenail bed grafts \| Two out of 6 patients (33.3%) developed slight nail deformities due to germinal layer injury \| \| \| Weinand \| 2014 \| Fingernail injury \| Splinting with native nail or silicone nail \| Splinting with the native nail resulted in fewer deformities compared to silicone nail splint group (p< 0.015) \| \| \| Zhou \| 2019 \| Longitudinal melanonychia \| Modified shave surgery combined with the longitudinal-strip nail window technique \| Nail dystrophy occurred in 15 out of 60 patients (25%) \| \| \| Richert \| 2013 \| Longitudinal melanonychia \| Tangential matrix excision/shave removal \| 8 of 23 patients (35%) developed dystrophy \| \| \| Grover \| 2005 \| Non-infectious and infectious nail disorders \| Various types of nail biopsies \| Among 65 patients, 19 (29%) developed dystrophy post-biopsy \| \| \| Lee \| 2013 \| Nail matrix and bed subungual glomus tumors \| Surgical excision \| 4 out of 13 patients (31%) with nail matrix tumors had persistent nail deformity \| \| \| Moon \| 2004 \| Subungual glomus tumors \| Transungual subungual glomus tumor excision \| 3 out of 16 patients (19%) experienced partial distal splitting of the nail \| \| Dysesthesia \| \| \| Topin-Ruiz \| 2017 \| Squamous cell carcinoma of the nail unit \| Wide surgical excision with full-thickness skin graft \| Among 51 patients, 39 (76.5%) reported hypersensitivity to mechanical shocks, and 38 (74.5%) reported increased cold sensitivity. 17 out of 35 patients (48.6%) noted loss of fine touch sensation \| \| \| Guerrero-Gonzalez \| 2016 \| Nail matrix glomus tumor, melanonychia, myxoid cyst, nail unit tumor \| Excision or biopsy \| 2 out of 4 patients (50%) with complex regional pain syndrome following nail surgery also reported hyperesthesia and cold sensitivity \| \| \| Lee \| 2013 \| Nail matrix and bed subungual glomus tumors \| Surgical excision \| 1 out of 13 patients (7.7%) in the nail matrix group experienced persistent decreased sensation at 9 months \| \| Spicules \| \| \| Topin-Ruiz \| 2017 \| Squamous cell carcinoma of the nail unit \| Wide surgical excision with full-thickness skin graft \| 7 (14%) patients experienced nail spicule regrowth \| \| . \| Author . \| Year . \| Nail condition . \| Treatment . \| Main finding . \| --- --- --- \| \| General complications \| \| Pain \| \| \| Vinay \| 2022 \| Onychocryptosis \| Phenol-based matricectomy \| Phenol matricectomies are associated with less pain (257 fewer cases of pain per 1,000 procedures) compared to other non-chemical matricectomies and sleeve procedures \| \| \| Ramesh \| 2020 \| Onychocryptosis \| NaOH vs. phenol vs. TCA matricectomy \| Phenol-based matricectomies resulted in a longer duration of pain (9 days) compared to NaOH (4 days) and TCA-based matricectomies (1 day) (p = 0.0092) \| \| \| Romero-Perez \| 2017 \| Onychocryptosis \| Surgical vs. phenol-based matricectomy \| Surgical matricectomy is associated with higher intensity of pain (5.7/10) compared to phenol-based chemical matricectomy (3.6/10) (p = 0.000) on scale 1–10 as well as longer pain duration (7.2 days in surgical vs. 4.2 days in phenol group, p = 0.002) \| \| \| Peyvandi \| 2011 \| Onychocryptosis \| Winograd vs. sleeve (gutter) surgical method \| The Winograd method resulted in a longer postoperative work day loss (2 weeks) compared to the sleeve method (1.1 weeks), which is statistically significant (p< 0.001) \| \| \| Ozan \| 2014 \| Onychocryptosis \| Partial matricectomy with curettage vs. with electrocautery \| Curettage resulted in a shorter duration of postoperative pain (2 days) compared to electrocautery (3 days), which is statistically significant (p< 0.05) \| \| \| Topin-Ruiz \| 2017 \| Squamous cell carcinoma of the nail unit \| Wide surgical excision with full-thickness skin graft \| 15% (8 out of 55) patients experienced severe postoperative pain \| \| \| Fritz \| 1997 \| Digital mucous cysts \| Surgical excision \| 9% (of 86 patients) had persistent pain and swelling postoperatively \| \| \| Kasdan \| 1994 \| Digital mucous cysts \| Surgical excision \| 14% (of 113 patients) experienced joint tenderness postoperatively \| \| \| Lee \| 2013 \| Nail matrix and bed subungual glomus tumors \| Surgical excision \| 14% (3 out of 22) patients with nail matrix or nail bed tumors experienced prolonged pain \| \| Complex regional pain syndrome \| \| \| Ingram \| 1987 \| Onychocryptosis \| Biopsy \| One case of complex regional pain syndrome was reported following a nail biopsy \| \| \| Guerrero-Gonzalez \| 2016 \| Nail matrix glomus tumor, melanonychia, myxoid cyst, nail unit tumor \| Excision or biopsy \| Four patients developed complex regional pain syndrome following surgical interventions for various nail conditions \| \| Bleeding \| \| \| Vinay \| 2022 \| Onychocryptosis \| Phenol-based matricectomy \| Phenol matricectomies are associated with less bleeding (177 fewer cases per 1,000 procedures) compared to other non-chemical matricectomies and sleeve procedures \| \| \| Ramesh \| 2020 \| Onychocryptosis \| NaOH vs. phenol vs. TCA matricectomy \| Phenol-based matricectomies resulted in a longer postoperative oozing (14 days) compared to NaOH (11 days) and TCA-based matricectomies (6 day) (p = 0.01) \| \| \| Romero-Perez \| 2017 \| Onychocryptosis \| Surgical vs. phenol-based matricectomy \| In a study, 63/202 (31.2%) surgical matricectomy patients had prolonged bleeding (>24 h) compared to 40/139 (28.8%) phenol-based chemical matricectomy patients (p = 0.634) \| \| Infection \| \| \| Grover \| 2005 \| Non-infectious and infectious nail disorders \| Various types of nail biopsies \| 11% of those who underwent nail biopsies developed secondary infections \| \| \| Romero-Perez \| 2017 \| Onychocryptosis \| Surgical matricectomy vs. phenol-based chemical matricectomy \| Surgical matricectomy showed higher infection risk (OR = 7.2, 95% CI 2.4–21.0) compared to phenol-based matricectomy \| \| \| Weinand \| 2014 \| Fingernail injury \| Splinting with native nail vs. silicone nail \| Splinting with the native nail was associated with half the infections compared to splinting with silicone nail \| \| \| Miranda \| 2012 \| Fingernail injury \| Nail replacement/foil placement vs. nail discarded \| Nail replacement group had higher infection rates vs. nail discarded (8% vs. 0%, p< 0.0001) \| \| \| Lee \| 2013 \| Nail matrix and bed subungual glomus tumors \| Surgical excision \| 1/13 (7.7%) nail matrix patients had mild wound dehiscence \| \| Inclusion cysts \| \| \| Topin-Ruiz \| 2017 \| Squamous cell carcinoma of the nail unit \| Wide surgical excision with full-thickness skin graft \| 9 out of 51 patients (18%) developed epidermal inclusion cysts \| \| Keloid \| \| \| Muzaffar \| 2004 \| Syndactyly \| Excision and reconstruction \| 8 of 681 patients developed keloids postoperatively \| \| Recurrence \| \| \| Fritz \| 1997 \| Digital mucous cysts \| Surgical excision \| 3% (of 86 excisions) developed recurrences \| \| \| Topin-Ruiz \| 2017 \| Squamous cell carcinoma of nail unit \| Wide local excision \| 2 of 55 patients (4%) developed recurrence \| \| \| Gou \| 2020 \| Squamous cell carcinoma of nail unit \| Mohs micrographic surgery \| Of 42 cases of nail unit SCC treated with Mohs surgery, 3 patients (7.1%) experienced recurrence \| \| \| Richert \| 2013 \| Longitudinal melanonychia of the proximal nail fold \| Tangential matrix excision/shave removal \| 16 out of 23 patients (70%) had pigmentation recurrence 8–12 months post-surgery \| \| \| Zhou \| 2019 \| Longitudinal melanonychia \| Modified shave surgery combined with the longitudinal-strip nail window technique \| 8 of 60 patients (13.3%) had pigmentation recurrence \| \| \| Lee \| 2013 \| Nail matrix and bed subungual glomus tumors \| Surgical excision \| No recurrence was observed in nail matrix (n = 13) or nail bed (n = 9) subungual glomus tumor excisions \| \| \| Moon \| 2004 \| Subungual glomus tumors \| Surgical excision \| None of the 16 patients experienced recurrence \| \| Complications specific to the nail apparatus \| \| Nail dystrophy \| \| \| Yang \| 2017 \| Large nail defects with exposed distal phalanxes \| Finger fascial flaps combined with thin split-thickness toenail bed grafts \| Two out of 6 patients (33.3%) developed slight nail deformities due to germinal layer injury \| \| \| Weinand \| 2014 \| Fingernail injury \| Splinting with native nail or silicone nail \| Splinting with the native nail resulted in fewer deformities compared to silicone nail splint group (p< 0.015) \| \| \| Zhou \| 2019 \| Longitudinal melanonychia \| Modified shave surgery combined with the longitudinal-strip nail window technique \| Nail dystrophy occurred in 15 out of 60 patients (25%) \| \| \| Richert \| 2013 \| Longitudinal melanonychia \| Tangential matrix excision/shave removal \| 8 of 23 patients (35%) developed dystrophy \| \| \| Grover \| 2005 \| Non-infectious and infectious nail disorders \| Various types of nail biopsies \| Among 65 patients, 19 (29%) developed dystrophy post-biopsy \| \| \| Lee \| 2013 \| Nail matrix and bed subungual glomus tumors \| Surgical excision \| 4 out of 13 patients (31%) with nail matrix tumors had persistent nail deformity \| \| \| Moon \| 2004 \| Subungual glomus tumors \| Transungual subungual glomus tumor excision \| 3 out of 16 patients (19%) experienced partial distal splitting of the nail \| \| Dysesthesia \| \| \| Topin-Ruiz \| 2017 \| Squamous cell carcinoma of the nail unit \| Wide surgical excision with full-thickness skin graft \| Among 51 patients, 39 (76.5%) reported hypersensitivity to mechanical shocks, and 38 (74.5%) reported increased cold sensitivity. 17 out of 35 patients (48.6%) noted loss of fine touch sensation \| \| \| Guerrero-Gonzalez \| 2016 \| Nail matrix glomus tumor, melanonychia, myxoid cyst, nail unit tumor \| Excision or biopsy \| 2 out of 4 patients (50%) with complex regional pain syndrome following nail surgery also reported hyperesthesia and cold sensitivity \| \| \| Lee \| 2013 \| Nail matrix and bed subungual glomus tumors \| Surgical excision \| 1 out of 13 patients (7.7%) in the nail matrix group experienced persistent decreased sensation at 9 months \| \| Spicules \| \| \| Topin-Ruiz \| 2017 \| Squamous cell carcinoma of the nail unit \| Wide surgical excision with full-thickness skin graft \| 7 (14%) patients experienced nail spicule regrowth \| View Large For surgical matricectomy compared to phenol-based chemical matricectomy, there is a higher intensity of pain and longer duration of pain in the surgical group . Partial matricectomy with curettage was found to have a shorter duration of postoperative pain compared to partial matricectomy with electrocautery (2 days vs. 3 days, p< 0.05) . Phenol-based matricectomies resulted in a longer duration of pain (9 days) compared to NaOH (4 days) and TCA-based matricectomies (1 day) (p = 0.0092) . In a study focused on the treatment of digital mucous cysts, 9% had persistent pain and swelling after excision [1, 10]. Another study demonstrated that 14% of patients experienced joint tenderness after mucous cyst excision [1, 11]. Among those who had nail bed and nail matrix subungual glomus tumor excised, 3 of the 22 patients (14%) had persistent pain which resolved spontaneously within a mean of 7 months . After wide surgical excision of squamous cell carcinoma of the nail unit with full-thickness grafting, 15% experienced postoperative pain . A rare but important complication of nail surgery is complex regional pain syndrome [13, 14]. Complex regional pain syndrome is classified as a neurovascular disorder after trauma to an extremity. This pain syndrome should be suspected if a patient, who underwent nail surgery, presents with sudomotor or vasomotor changes with pain that is out of proportion to the procedure . Bleeding Bleeding is a complication of surgical nail procedures [3, 5]. Phenol-based matricectomy resulted in less bleeding compared to non-chemical matricectomies and sleeve methods . Of note, bleeding was grouped with discharge, so it is unclear the exact number is due solely to bleeding. Phenol-based matricectomies resulted in a longer postoperative oozing (14 days) compared to NaOH (11 days) and TCA-based matricectomies (6 days) (p = 0.01) . Infection Nail biopsy can be associated with infection rates of up to 11% . For surgical matricectomy compared to phenol-based chemical matricectomy, there is a higher risk of infection in the surgical group (OR = 7.2 95% CI: 2.4–21.0) . There are a few common methods for treating fingertip injuries with variable infection rates. In 1 study with patients with fingertip injuries, splinting with the native nail was associated with half the infections compared to splinting with the silicone nail . Higher infection rates (8% vs 0% p< 0.0001) were also observed among pediatric patients who had their nail replaced or foil placed in repair of fingernail injury, compared to pediatric patients who did not have their nail replaced and nail was discarded . In one study, only 1 of 13 patients who underwent nail matrix subungual glomus tumor excision experienced mild wound dehiscence . Outcomes may vary depending on whether the excision site is in the nail bed or matrix. Inclusion Cysts The incidence of epidermal inclusion cyst development after excision of nail unit followed by full-thickness graft ranges from 0 to 38% of patients. After wide local excision of squamous cell carcinoma of the nail unit followed by full-thickness grafting, 18% developed epidermal inclusion cysts . Cysts may result from proximal and lateral matrix horn remnants [24, 25]. A surgeon, therefore, must thoroughly excise the nail matrix horn, while preserving delicate neighboring structures. It is important to attempt to reduce the risk of cyst development as these complications can require additional nail surgeries and cause further complications. Keloid Although keloids on hands and feet are uncommon, they remain a rare but possible complication of nail surgery. Eight of 681 patients developed keloids after syndactyly excision and reconstruction . Recurrence Nail surgery can be traumatic, and thus, recurrence rates of specific nail procedures should be closely examined to minimize harm. In a study focused on excision of digital mucous cysts, 3% developed recurrence [1, 10]. After wide local excision of squamous cell carcinoma of the nail unit, recurrence was observed in 4% of patients. There are widely reported recurrence rates for squamous cell carcinoma using Mohs micrographic surgery (0–25%) . Of note, the authors mention that the wide range of cure rates may be related to the smaller study sizes . Shave biopsy can be associated with high rates of recurrence. Richert et al. demonstrated that of 23 patients that underwent tangential matrix excision/shave removal of pigment from the proximal nail fold, 16 had pigmentation recurrence 8–12 months post-surgery. Another study demonstrated that 13% of patients had recurrence after modified shave surgery combined with the longitudinal-strip nail window technique for longitudinal melanonychia . Regardless of approach (including excision involving nail matrix or nail bed), recurrence was not observed with any surgical excision of subungual glomus tumors [12, 22]. The risk of recurrence can be minimized by complete removal of the subungual glomus tumor . Complications Specific to the Nail Apparatus Nail Dystrophy Scarring is the most common complication related to nail surgery . Scarring most often occurs when the nail matrix has been involved in the procedure . De Berker et al. found that 22% of nail surgery patients developed scarring, nail splitting, ridging, or pterygiums as postoperative complications. A comprehensive understanding of nail pathology is crucial for effective nail surgery and anticipating potential complications. The number of melanocytes is lower in the nail matrix compared to the skin epidermis, and furthermore, most proximal nail matrix melanocytes are dormant, while 50% of the distal nail matrix melanocytes are active and produce pigment and 50% of the distal nail matrix melanocytes are dormant. Longitudinal melanonychia may be the first sign of nail melanoma . Given the lesion contributing to the nail plate pigmentation, in the case of melanoma, is located in the matrix, matrix biopsy should be performed. We have reviewed potential complications associated with nail matrix biopsies, which vary depending on the type and location of the biopsy performed (Fig. 2). ![Image 8: Location of nail biopsies. The blue dotted lines indicate the sites of surgical incision. The procedures include longitudinal nail biopsy (a), nail plate biopsy (b), nail bed biopsy (c), and nail matrix biopsy (d) (adapted from Grover et al. 2005] ).]( View largeDownload slideGet Permissions Fig. 2. Location of nail biopsies. The blue dotted lines indicate the sites of surgical incision. The procedures include longitudinal nail biopsy (a), nail plate biopsy (b), nail bed biopsy (c), and nail matrix biopsy (d) (adapted from Grover et al. 2005] ). ![Image 9: Location of nail biopsies. The blue dotted lines indicate the sites of surgical incision. The procedures include longitudinal nail biopsy (a), nail plate biopsy (b), nail bed biopsy (c), and nail matrix biopsy (d) (adapted from Grover et al. 2005] ).]( View largeDownload slideGet Permissions Fig. 2. Location of nail biopsies. The blue dotted lines indicate the sites of surgical incision. The procedures include longitudinal nail biopsy (a), nail plate biopsy (b), nail bed biopsy (c), and nail matrix biopsy (d) (adapted from Grover et al. 2005] ). Close modal In one study, 6 patients underwent finger fascial flaps combined with thin split-thickness toenail bed grafts for the treatment of large nail bed defects with exposed distal phalanxes and 2 of the 6 patients had slight nail deformities due to germinal layer injury . In another study, nail dystrophy occurred in 25% of patients who underwent modified shave surgery combined with the longitudinal-strip nail window technique for management of longitudinal melanonychia . Nail biopsies can be associated with nail dystrophy. In one study, 29% of patients developed scarring, nail width reduction, and secondary dystrophy post-biopsy . Nail width reduction occurred in 18 of 38 patients (47%) that underwent longitudinal nail biopsies . Furthermore, midline nail unit biopsies are less recommended because they increase a patient’s risk of split nail deformity . Longitudinal biopsy on the center of the nail may result in chronic distal nail dystrophy or complete nail splitting. While lateral biopsy avoids these complications, de Berker and Baran report that excision of more than 3 mm of the lateral nail unit can lead to acquired nail malalignment . Thus, each approach carries its own set of potential risks and benefits. Shave removal/tangential matrix excision is associated with nail dystrophy. In Richert al. ’s study, 8 of 23 patients with longitudinal melanonychia developed dystrophy following removal of pigment from the nail matrix by shave removal/tangential matrix excision. In a study with patients with fingertip injuries, splinting with the native nail was associated with fewer deformities compared to splinting with the silicone nail (p< 0.015) . Another study demonstrated that 4 of the 13 patients who underwent nail matrix subungual glomus tumor excision developed persistent nail deformity . Partial distal splitting of the nail was a common complication reported after transungual subungual glomus tumor excision (19%) . Splitting may be due to the formation of a scar between the proximal nail fold and nail matrix or result from scar formation within the matrix . Dysesthesia Though not a serious complication, distorted sensation can be a nuisance to patients and has been reported as a complication after excision of different tumors [7, 12, 14]. One of 13 patients who underwent nail matrix subungual glomus tumor excision reported decreased sensation that persisted 9 months . Spicules Spicules may result from proximal and lateral matrix horn remnants . If the entire nail matrix is not excised when performing a lateral longitudinal excision, spicule formation may result [24, 25]. The incidence of nail spicule regrowth ranges from 0 to 11% of patients; furthermore, the authors highlighted a slightly higher rate of nail spicule regrowth (14%) after wide surgical excision of squamous cell carcinoma of the nail unit followed by full-thickness grafting . Complications Associated with Specific Nail Procedures Nail Plate Avulsion Nail plate avulsion, a surgical treatment required for late stage retronychia, can result in postoperative pain. Postoperative pain requiring analgesics was reported in a study with patients that underwent nail plate avulsion . Other less common complications of nail avulsion include infection, hematoma, nail deformity, malalignment, nail impaction (distal embedding), local spicule growth, and swelling . Full Unit Excision or en Bloc Excision for Nail Melanoma, Melanoma in situ, or Squamous Cell Carcinoma Topin-Ruiz et al. demonstrated that early complications of wide surgical excision of the nail unit followed by full-thickness skin graft include graft infection, delayed wound healing, and pain. Late postoperative complications include hypersensitivity to mechanical shocks, mildly increased sensitivity to cold, loss of fine touch sensation, and epidermal inclusion cysts. In addition, the formation of digital myxoid cysts after en bloc excision for nail melanoma in situ has been reported . Slow Mohs of Nail Melanoma in situ Pain has been reported to be associated with slow Mohs surgery of nail apparatus melanoma in situ . Excision of Myxoid Cysts Surgical removal of a myxoid cyst has been associated with postoperative nail deformities . In addition, excision of myxoid cysts is associated with bleeding, infection, and joint stiffness . Excision of Glomus Tumors Excision is one of the most common treatment methods for subungual glomus tumors. One of the main complications of excision of glomus tumors is nail dystrophy . The location of the glomus tumor may contribute to postoperative complications. One study showed that those with nail matrix involvement had complications such as nail deformity, decreased sensation, and prolonged pain sensation and those with nail bed lesions had prolonged pain sensation . Matricectomy Pain was greater in those who underwent surgical matricectomy compared to chemical matricectomy . In addition, infection rates were greater in those who underwent surgical matricectomy compared to chemical matricectomy . Subungual Hematoma Trephination When performing trephination for a subungual hematoma, the patient may have pain related to the pressure from the drilling or torquing motion . Despite local anesthesia, patients may have this brief discomfort due to the pressure. Prevention of Surgical Complications Most complications of nail surgery are related to damage to the nail matrix, especially in the proximal region . Damage can result in nail deformity and scarring, leading to nail splitting . Therefore, it is crucial that the surgeon accurately identifies the nail matrix and is deliberate to limit damage to the region. The surgeon must balance these considerations and the patient’s specific needs to determine a suitable treatment. A history of nail matrix damage can be associated with nail deformities . Patient medical records should be queried for previous nail injuries, and realistic patient expectations for surgical outcomes should be set accordingly. Awareness of medical conditions and medications that increase a patient’s risk of postoperative complications is also key. Nicotine use disorders, diabetes mellitus, and other vascular diseases can compromise blood flow to distal extremities, which then increases the risk of infection and delayed healing . Immunocompromised patients face a greater risk of infection. Additionally, patients using anticoagulants have an increased risk of prolonged bleeding. By screening for these risk factors, surgeons may possibly be able to mitigate foreseeable complications. It is especially important to sterilize the space beneath the hyponychium, a reservoir for microbes . There is no evidence that postoperative use of antibiotics decreases infection rates . One study found that patients who wore a gauze bandage developed significantly fewer infections than those who cleaned the surgical site postoperatively with alkaline or acidic soaps after partial nail ablation surgery . In conclusion, while nail surgery complications are not common, it is important for the surgeon to be aware of these complications specific to the nail apparatus as well as complications associated with specific nail procedures. Although this literature review may be limited by the bias of article selection, it remains valuable since there is a dearth of literature that summarizes complications related to nail procedures for dermatologists performing nail procedures. Given these studies did not discuss the experience of the surgeon performing nail procedures, further research is needed to determine the association of complications with the technique and experience of the nail surgeon. Conflict of Interest Statement The authors have no conflicts of interest to declare. Funding Sources This study was not supported by any sponsor or funder. Author Contributions B.H.M. conceived and designed analysis. J.J.F. collected data (literature) and wrote the paper. B.H.M. and J.J.F. reviewed the paper. References 1. Moossavi M , Scher RK . Complications of nail surgery: a review of the literature . Dermatol Surg . 2001 ; 27 ( 3 ): 225 – 8 . Google Scholar PubMed 2. de Berker D . Nail anatomy . Clin Dermatol . 2013 ; 31 ( 5 ): 509 – 15 . Google Scholar Crossref Search ADS PubMed 3. Vinay K , Narayan Ravivarma V , Thakur V , Choudhary R , Narang T , Dogra S , et al. Efficacy and safety of phenol-based partial matricectomy in treatment of onychocryptosis: a systematic review and meta-analysis . J Eur Acad Dermatol Venereol . 2022 ; 36 ( 4 ): 526 – 35 . Google Scholar Crossref Search ADS PubMed 4. Ramesh S , Shenoi SD , Nayak SUK . Comparative efficacy of 10% sodium hydroxide, 88% phenol, and 90% trichloroacetic acid as chemical cauterants for partial matricectomy in the management of great toe nail onychocryptosis . J Cutan Aesthet Surg . 2020 ; 13 ( 4 ): 314 – 8 . Google Scholar PubMed 5. Romero-Pérez D , Betlloch-Mas I , Encabo-Durán B . 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Indian J Dermatol Venereol Leprol . 2012 ; 78 ( 3 ): 299 – 308 . Google Scholar Crossref Search ADS PubMed 31. Cressey BDM , Jellinek NJ , Nathaniel JMD . Myxoid cyst as a probable complication of nail surgery . Dermatol Surg . 2018 ; 44 ( 12 ): 1647 – 9 . Google Scholar Crossref Search ADS PubMed 32. Zhang S , Wang Y , Fang K , Jia Q , Zhang H , Qu T . Slow Mohs micrographic surgery for nail apparatus melanoma in situ . Int J Dermatol . 2023 ; 62 ( 9 ): 1170 – 5 . Google Scholar Crossref Search ADS PubMed 33. Brown RE , Zook EG , Russell RC , Kucan JO , Smoot EC . Fingernail deformities secondary to ganglions of the distal interphalangeal joint (mucous cysts) . Plast Reconstr Surg . 1991 ; 87 ( 4 ): 718 – 25 . Google Scholar Crossref Search ADS PubMed 34. Reinders EFH , Klaassen KMG , Pasch MC . Transungual excision of glomus tumors: a treatment and quality of life study . Dermatol Surg . 2020 ; 46 ( 1 ): 103 – 12 . Google Scholar Crossref Search ADS PubMed 35. Akella A , Daniel AR , Gould MB , Mangal R , Ganti L . Subungual hematoma . Cureus . 2023 ; 15 ( 11 ): e48952 . Google Scholar PubMed 36. Queirós C , Garrido PM , Fraga A , Maia-Silva J , Filipe P . Nail surgery: general principles, fundamental techniques, and practical applications . J Cutan Aesthet Surg . 2022 ; 15 ( 4 ): 341 – 54 . Google Scholar Crossref Search ADS PubMed 37. Eekhof JA , Van Wijk B , Knuistingh Neven A , van der Wouden JC . Interventions for ingrowing toenails . Cochrane Database Syst Rev . 2012 ; 2012 ( 4 ): CD001541 . Google Scholar PubMed 38. Bernardshaw SV , Dolva Sagedal LH , Michelet KM , Brudvik C . Postoperative treatment after partial nail ablation of ingrown toenails - does it matter what we recommend? A blinded randomised study . Scand J Prim Health Care . 2019 ; 37 ( 2 ): 165 – 73 . Google Scholar Crossref Search ADS PubMed © 2025 S. Karger AG, Basel Copyright / Drug Dosage / Disclaimer Copyright: All rights reserved. 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1898	https://artofproblemsolving.com/wiki/index.php/Change_of_base_formula?srsltid=AfmBOorosB1hG2D4aNgT7Xz5sZsx7w7M0tvhaUAnMV_x8FH926CfVCD2	Art of Problem Solving Change of base formula - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Change of base formula Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Change of base formula The change of base formula is a formula for expressing a logarithm in one base in terms of logarithms in other bases. For any positivereal numbers such that neither nor are , we have This allows us to rewrite a logarithm in base in terms of logarithms in any base . This formula can also be written Proof Let . Then . And, taking the of both sides, we get By the properties of logarithms, Substituting for y, Use for computations The change of base formula is useful for simplifying certain computations involving logarithms. For example, we have by the change of base formula that The formula can also be useful when calculating logarithms on a calculator. Many calculators have only functions for calculating base-10 and base-e logarithms. But you can still calculate logs in other bases, you just need to use the change of base formula to put in in base 10. For example, if you wanted to calculate , you would first convert it to the form . Then you would evaluate it using the base-10 log function on the calculator. Special cases and consequences Many other logarithm rules can be written in terms of the change of base formula. For example, we have that . Using the second form of the change of base formula gives . One consequence of the change of base formula is that for positive constants , the functions and differ by a constant factor, for all . This article is a stub. Help us out by expanding it. Retrieved from " Category: Stubs Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1899	https://ximera.osu.edu/laode/linearAlgebra/linearMapsAndChangesOfCoordinates/linearMappingsAndBases	Linear Mappings and Bases - Ximera Statistics Get HelpContact my instructor Report error to authorsRequest help using XimeraReport bug to programmers Another Math Editor Failed Saved! Saving… Reconnecting…Save Update Erase Edit MeProfileSuperviseLogout Sign InSign In with GoogleSign In with TwitterSign In with GitHub Sign In Warning × You are about to erase your work on this activity. Are you sure you want to do this? No, keep my work.Yes, delete my work. Updated Version Available × There is an updated version of this activity. If you update to the most recent version of this activity, then your current progress on this activity will be erased. Regardless, your record of completion will remain. How would you like to proceed? Keep the old version.Delete my work and update to the new version. Mathematical Expression Editor × +–× ÷ x ⁿ √ⁿ√ π θ φ ρ ()\|\| sin cos tan arcsin arccos arctan e ˣ ln log [?]\blue{[?]}[?] Cancel OK Preface 1 Preliminaries 1.1 Vectors and Matrices 1.2 MATLAB 1.3 Special Kinds of Matrices 1.4 The Geometry of Vector Operations 2 Solving Linear Equations 2.1 2.2 The Geometry of Low-Dimensional Solutions 2.3 2.4 2.5 Linear Equations with Special Coefficients 2.6 Uniqueness of Reduced Echelon Form 3 Matrices and Linearity 3.1 3.2 Matrix Mappings 3.3 3.4 The Principle of Superposition 3.5 3.6 3.7 3.8 Determinants of 2 × 2 Matrices 4 Solving Ordinary Differential Equations 4.1 A Single Differential Equation 4.2 Graphing Solutions to Differential Equations 4.3 Phase Space Pictures and Equilibria 4.4 Separation of Variables 4.5 Uncoupled Linear Systems of Two Equations 4.6 Coupled Linear Systems 4.7 The Initial Value Problem and Eigenvectors 4.8 Eigenvalues of 2 × 2 Matrices 4.9 Initial Value Problems Revisited 4.10 5 Vector Spaces 5.1 Vector Spaces and Subspaces 5.2 Construction of Subspaces 5.3 5.4 5.5 Dimension and Bases 5.6 6 Closed Form Solutions for Planar ODEs 6.1 The Initial Value Problem 6.2 Closed Form Solutions by the Direct Method 6.3 Solutions Using Matrix Exponentials 6.4 Linear Normal Form Planar Systems 6.5 Similar Matrices 6.6 Formulas for Matrix Exponentials 6.7 Second Order Equations 7 Qualitative Theory of Planar ODEs 7.1 Sinks, Saddles, and Sources 7.2 Phase Portraits of Sinks 7.3 Phase Portraits of Nonhyperbolic Systems 8 Determinants and Eigenvalues 8.1 Determinants 8.2 Eigenvalues 8.3 Existence of Determinants 9 Linear Maps and Changes of Coordinates 9.1 Linear Mappings and Bases 9.2 9.3 9.4 10 Orthogonality 10.1 10.2 10.3 Least Squares Fitting of Data 10.4 Symmetric Matrices 10.5 11 Matrix Normal Forms 11.1 11.2 11.3 11.4 11.5 Markov Matrix Theory 11.6 Proof of Jordan Normal Form laode Linear Algebra Linear Maps and Changes of Coordinates Linear Mappings and Bases Martin Golubitsky and Michael Dellnitz The examples of linear mappings from that we introduced in Section?? were matrix mappings. More precisely, let be an matrix. Then defines the linear mapping . Recall that is the column of (see Chapter??, Lemma??); it follows that can be reconstructed from the vectors . This remark implies (Chapter??, Lemma??) that linear mappings of to are determined by their values on the standard basis . Next we show that this result is valid in greater generality. We begin by defining what we mean for a mapping between vector spaces to be linear. Let and be vector spaces and let be a mapping. The map is linear if for all and . Examples of Linear Mappings (a) Let be a fixed vector. Use the dot product to define the mapping by Then is linear. Just check that for every vector and in and for every scalar . (b) The map defined by is linear. Indeed, Similarly, . (c) The map defined by is linear. Indeed, Similarly, . It may be helpful to compute when . That is, Constructing Linear Mappings from Bases Let and be vector spaces. Let be a basis for and let be vectors in . Then there exists a unique linear map such that . Proof Let be a vector. Since , we may write as where in . Moreover, are linearly independent, these scalars are uniquely defined. More precisely, if then Linear independence implies that ; that is . We can now define We claim that is linear. Let be another vector and let It follows that and hence by (??) that Similarly Thus is linear. Let be another linear mapping such that . Then Thus and the linear mapping is uniquely defined. There are two assertions made in Theorem??. The first is that a linear map exists mapping to . The second is that there is only one linear mapping that accomplishes this task. If we drop the constraint that the map be linear, then many mappings may satisfy these conditions. For example, find a linear map from that maps to . There is only one: . However there are many nonlinear maps that send to . Examples are and . Finding the Matrix of a Linear Map from Given by Theorem?? Suppose that and . We know that every linear map can be defined as multiplication by an matrix. The question that we next address is: How can we find the matrix whose existence is guaranteed by Theorem??? More precisely, let be a basis for and let be vectors in . We suppose that all of these vectors are row vectors. Then we need to find an matrix such that for all . We find as follows. Let be a row vector. Since the form a basis, there exist scalars such that In coordinates where is an invertible matrix. By definition (see (??)) Thus the matrix must satisfy where is an matrix. Using (??) we see that and is the desired matrix. An Example of a Linear Map from to As an example we illustrate Theorem?? and (??) by defining a linear mapping from to by its action on a basis. Let We claim that is a basis of and that there is a unique linear map for which where We can verify that is a basis of by showing that the matrix is invertible. This can either be done in MATLAB using the inv command or by hand by row reducing the matrix to obtain Now apply (??) to obtain As a check, verify by matrix multiplication that , as claimed. Properties of Linear Mappings Let be vector spaces and and be linear maps. Then is linear. Proof The proof of Lemma?? is identical to that of Chapter??, Lemma??. A linear map is invertible if there exists a linear map such that is the identity map on and is the identity map on . Let and be finite dimensional vector spaces and let be a basis for . Let be a linear map. Then is invertible if and only if is a basis for where . Proof If is a basis for , then use Theorem?? to define a linear map by . Note that It follows by linearity (using the uniqueness part of Theorem??) that is the identity of . Similarly, is the identity map on , and is invertible. Conversely, suppose that and are identity maps and that . We must show that is a basis. We use Theorem?? and verify separately that are linearly independent and span . If there exist scalars such that then apply to both sides of this equation to obtain But the are linearly independent. Therefore, and the are linearly independent. To show that the span , let be a vector in . Since the are a basis for , there exist scalars such that Applying to both sides of this equation yields Therefore, the span . Let and be finite dimensional vector spaces. Then there exists an invertible linear map if and only if . Proof Suppose that is an invertible linear map. Let be a basis for where . Then Theorem?? implies that is a basis for and . Conversely, suppose that . Let be a basis for and let be a basis for . Using Theorem?? define the linear map by . Theorem?? states that is invertible. Exercises Use the method described above to construct a linear mapping from to with , , where and Let be the vector space of polynomials of degree less than or equal to . Show that is a basis for . Show that is a linear mapping. Show that is a linear mapping of . Use Exercises??, ?? and Theorem?? to show that is the identity map. Let denote the set of complex numbers. Verify that is a two-dimensional vector space. Show that defined by where is a linear mapping. Let denote the vector space of matrices and let be an matrix. Let be the mapping defined by where . Verify that is a linear mapping. Show that the null space of , , is a subspace consisting of all matrices that commute with . Let be defined by for . Verify that is a linear mapping. Let be the vector space of polynomials in one variable . Define by . 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